aa r X i v : . [ m a t h . C V ] A p r Growth Theory of Subharmonic Functions
Vladimir Azarin
Abstract.
In this course of lectures we give an account of the growth theory ofsubharmonic functions, which is directed towards its applications to entire functionsof one and several complex variables.
Contents R m
783 Asymptotic behavior of subharmonic functions of finite order 933.1 Limit sets 933.2 Indicators 1083.3 Densities 1244 Structure of the limit sets 1344.1 Dynamical systems 1344.2 Subharmonic functions with prescribed limit set 1524.3 Further properties of the limit sets 1684.4 Subharmonic curves. Curves with prescribed limit sets 1865 Applications to entire functions 1895.1 Growth characteristics of entire and meromorphic func-tions 1895.2 D ′ -topology and topology of exceptional sets 1915.3 Asymptotic approximation of subharmonic functions 1975.4 Lower indicator of A.A.Gol ′ dberg. Description of lowerindicator. Description of pair: indicator – lower indicator 2045.5 Asymptotic extremal problems. Semiadditive integral 2185.6 Entire functions of completely regular growth. Levin-Pfluger Theorem. Balashov’s theory 2235.7 General characteristics of growth of entire functions 227 Typeset by
AMS -TEX i i ρ -trigonometric convexity 2816.3 Completeness of exponential systems in a convex domain 288Notation 313List of terms 315References 321 This book aims to convert the noble art of constructing an entire function withprescribed asymptotic behavior to a handicraft. This is the aim of every Theory.For this you should only construct the limit set that describes the asymptoticbehavior of the entire function, i.e., you should consider the set U [ ρ, σ ] of subhar-monic functions (that is, { v is subharmonic : v ( re iφ ) ≤ σr ρ } ) and pick out thesubset U which characterizes its asymptotic properties.How to do it? The properties of limit sets are listed in §
3. All the standardgrowth characteristics are expressed in terms of limit sets in §§ §§ Acknowledgements.
I am indebted to many people. I start from Prof. I.V.Ostrovskii,who supported this idea for many years, and Prof. A.A.Goldberg, who stimulatedmy mathematical activity all my life by his letters and conversations.I am indebted to Prof’s A.Eremenko and M.Sodin, who, not being my “aspi-rants,” solved a lot of problems connected to limit sets, and also to Dr’s V.Giner,L.Podoshev and E.Fainberg that worked with me to develop the theory.I am indebted to Prof’s L.H¨ormander and R.Sigurdsson that have sent methe preprints of their papers that were not yet published. I am indebted toProf.I.F.Krasichkov-Ternovskii, who explained me many years ago the connectionbetween the multiplicator problem and completeness of the exponent system in aconvex domain.I am indebted to Prof’s. M.I.Kadec and V.P.Fonf for proving Theorem 4.1.5.2which is rather far off my speciality.I am indebted to my coauthors Prof’s D.Drasin and P.Poggi-Corradini; I haveexploited the results of our joint paper in § Let x ∈ R m be a point in an m -dimensional Euclidean space, E a Borel setand f ( x ) a function on E such that f ( x ) = ∞ .Set(2.1.1.1) M ( f, x, ε ) := sup { f ( x ′ ) : | x − x ′ | < ε, x ′ ∈ E } The function f ∗ ( x ) := lim ε → M ( f, x, ε )is called the upper semicontinuous regularization of the function f ( x ).In the case of a finite jump, the regularization ”raises” the values of the func-tion. However, there is no influence on f ∗ ( x ), if f ( x ) tends to −∞ ”continuously”. Proposition 2.1.1.1(Regularization Properties).
The following properties hold: (rg1) f ( x ) ≤ f ∗ ( x );(rg2) ( αf ) ∗ ( x ) = αf ∗ ( x );(rg3) ( f ∗ ) ∗ ( x ) = f ∗ ( x );( f + f ) ∗ ( x ) ≤ f ∗ ( x ) + f ∗ ( x );(max( f , f )) ∗ ( x ) ≤ max( f ∗ , f ∗ )( x );(rg 4) (min( f , f )) ∗ ( x ) ≤ min( f ∗ , f ∗ )( x );These properties are obvious corollaries of the definition of f ∗ ( x ). Exercise 2.1.1.1
Prove them.
The function f ( x ) is called upper semicontinuous at a point x if f ∗ ( x ) = f ( x ).We denote the class of upper semicontinuous functions on E by C + ( E ).The function f ( x ) is called lower semicontinuous if − f ( x ) is upper semicon-tinuous (notation f ∈ C − ( E ) ). Examples of semicontinuous functions are givenby Proposition 2.1.2.1(Semicontinuity of Characteristic Functions of Sets).
Let G ⊂ R m be an open set.Then its characteristic function χ G is lower semi-continuous in R m . Let F be a closed set, then χ F is upper semicontinuous . The proof is obvious.
Exercise 2.1.2.1
Prove this.
Proposition 2.1.2.2(Connection with Continuity). If f ∈ C + ∩ C − , then f is continuous. The assertion follows from the equalities f ∗ ( x ) = lim sup ε → { f ( x ′ ) : | x − x ′ | < ε } ; − ( − f ) ∗ ( x ) = lim inf ε → { f ( x ′ ) : | x − x ′ | < ε } . Proposition 2.1.2.3( C + -Properties). The following holds ( C + f ∈ C + ( E ) ⇒ αf ∈ C + ( E ) , f or α ≥ f , f ∈ C + ⇒ f + f , max( f , f ) , min( f , f ) ∈ C + . These properties follow from the properties of regularization (Prop.2.1.1.1).
Exercise 2.1.2.2
Prove them.Let G be an open set. Set G A := { x ∈ G : f ( x ) < A } . Theorem 2.1.2.4 (First Criterion of Semicontinuity).
One has f ∈ C + ifand only if G A is open for all A ∈ R .Proof. Let f ( x ) = f ∗ ( x ) , x ∈ G . Then { f ( x ) < A } = ⇒ { f ∗ ( x ) < A } = ⇒{ M ( f, x, ε ) < A } for all sufficiently small ε. Thus the neighborhood of xV ε,x := { x ′ : | x − x ′ | < ε } is contained in G A .Conversely, since the set G A is open for A = f ( x ) + δ , we have f ∗ ( x ) ≤ f ( x ) + δ for any δ >
0, hence for δ = 0. With property (rg1) ofProp.2.1.1.1 this gives f ∗ ( x ) = f ( x ) . (cid:3) Let F be a closed set. Set F A := { x ∈ F : f ( x ) ≥ A } . An obvious corollary ofthe previous theorem is Corollary 2.1.2.5.
One has f ∈ C + if and only if F A is closed for all A . Exercise 2.1.2.3
Prove the corollary.We denote compacts by K. Set M ( f, K ) = sup { f ( x ) : x ∈ K } . Theorem 2.1.2.6(Weierstrass).
Let K ⊂ R m be a compact set and f ∈ C + ( K ) . Then there exists x ∈ K such that f ( x ) = M ( f, K ) , i.e., f attains its supremum on any compact set. Proof.
Set K n := { x ∈ K : f ( x ) ≥ M ( f, K ) − /n } . The K n are closed by Cor.2.1.2.5, nonempty by definition of M ( f, K ) . Theirintersection is nonempty and is equal to the set K max := { x ∈ K : f ( x ) ≥ M ( f, K ) } . It means that there exists x in K such that f ( x ) ≥ M ( f, K ).The opposite inequality holds for any x in K . (cid:3) Exercise 2.1.2.4
Why?The following theorem shows that the functional M ( f, K ) is continuous withrespect to monotonic convergence of semicontinuous functions. Proposition 2.1.2.7(Continuity from the right of M ( f, K ) ). Let f n ∈ C + ( K ) , f n ↓ f, n = 1 , , ... . Then M ( f n , K ) ↓ M ( f, K ) . Proof.
It is clear that lim n →∞ M ( f n , K ) := M exists.Set K n := { x ∈ K : f n ( x ) ≥ M } . The intersection of the closed nonemptysets K n is nonempty and has the following form: T n K n = { x : f ( x ) ≥ M } . So M ( f, K ) ≥ M. The opposite inequality is obvious. (cid:3)
Exercise 2.1.2.5
Why?In the same way one prove
Proposition 2.1.2.8(Commutativity of inf and M ( · ) ). Let { f α ∈ C + ( K ) , α ∈ (0; ∞ ) } be an arbitrary decreasing family of semicontinuous functions.Then inf α M ( f α , K ) = M (inf α f α , K ) . Exercise 2.1.2.6
Prove this proposition.
Theorem 2.1.2.9(Second Criterion of Semicontinuity). f ∈ C + ( K ) iff thereexists a sequence f n of continuous functions such that f n ↓ f. Sufficiency.
Let f n ∈ C + ( K ) , f n ↓ f .Set K An := { x ∈ K : f n ( x ) ≥ A } . This is asequence of non-empty closed sets. If the set K A := { x : f ( x ) ≥ A } is nonempty,then K A is closed because T n K An = K A . Hence f ∈ C + ( K ) by Corollary 2.1.2.5. Necessity.
Set f n ( x, y ) := f ( y ) − n | x − y | . This sequence of functions has the following properties:a) it decreases monotonically in n andlim n →∞ f n ( x, y ) = (cid:26) f ( x ) , f or x = y ; −∞ , f or x = y ;b) for any fixed n the functions f n are continuous in x uniformly with respectto y , because | f n ( x, y ) − f n ( x ′ , y ) | ≤ n | x − x ′ | ;c) f n are upper semicontinuous in y .Prop.2.1.2.7 and c) imply that lim n →∞ M y ( f n ( x, y ) , K ) = M y ( lim n →∞ f n ( x, y ) , K )b) implies that the functions f n ( x ) := M y ( f n ( x, y ) , K ) are continuous, and a)implies that they decrease monotonically to f ( x ) . (cid:3) We will consider a family of upper semicontinuous functions: { f t : t ∈ T ⊂ (0 , ∞ ) } . It is easy to prove
Proposition 2.1.3.1. f t ∈ C + = ⇒ inf t ∈ T f t ( x ) ∈ C + . Exercise 2.1.3.1
Prove this Proposition.Set f T ( x ) := sup t ∈ T f t ( x ) . The function f T is not, generally speaking, upper semi-continuous even if T is countable and f t are continuous. It is not possible to replace sup T in the definition of f T by sup T , where T is a countable set .However, the fol-lowing theorem holds: Theorem 2.1.3.2(Choquet’s Lemma).
There exists such a countable set T ⊂ T , that (sup T f t ) ∗ ( x ) = (sup T f t ) ∗ ( x ) . Proof.
Let { x n } be a countable set that is dense in R m and ε j ↓ K n,j := { x : | x − x n | < ε j } cover every point x ∈ R m infinitely many times.Renumbering we obtain a sequence { K l : l ∈ N } . For any l there exists, bydefinition of sup K l , such a point x ∈ K l that(2.1.3.1) sup K l f T ( x ) ≤ f T ( x ) + 1 / l. By definition of sup T there exists t l such that f T ( x ) < f t l ( x ) + 1 / l. Thus(2.1.3.2) f T ( x ) < sup { f t l ( x ) : x ∈ K l } + 1 / l. The inequalities (2.1.3.1) and (2.1.3.2) imply that for any l there exists t l such that(2.1.3.3) sup { f T ( x ) : x ∈ K l } ≤ sup { f t l ( x ) : x ∈ K l } + 1 /l. Now set T = { t l } . Evidently, f T ( x ) ≤ f T ( x ) and thus(2.1.3.4) f ∗ T ( x ) ≤ f ∗ T ( x ) . Let us prove the opposite inequality.Let x ∈ R m . Choose a subsequence { K l j } that tends to x . From (2.1.3.3) weobtain f ∗ T ( x ) ≤ lim sup j →∞ sup x ′ ∈ K lj f T ( x ′ ) ≤ lim sup j →∞ sup x ′ ∈ K lj f t lj ( x ′ ) ≤ (2.1.3.5) lim sup j →∞ sup x ′ ∈ K lj f T ( x ′ ) = f ∗ T ( x ) . (2.1.3.4) and (2.1.3.5) imply the assertion of the theorem. (cid:3) Let G be an open set in R m and σ ( G ) a σ -algebra of Borel sets containingall the compact sets K ⊂ G .Let µ be a countably additive nonnegative function on σ ( G ), which is finite onall K ⊂ G . We will call it a measure or a mass distribution .Let G ( µ ) be the largest open set for which µ is zero. It is the union of all theopen sets G ′ that µ ( G ′ ) = 0 . The set supp µ := G \ G ( µ ) is called the support of µ . It is closed in G. We say that µ is concentrated on E ∈ σ ( G ) if µ ( G \ E ) = 0. Theorem 2.2.1.1(Support).
The support of a measure µ is the smallest closedset on which the measure µ is concentrated. Exercise 2.2.1.1
Prove this.A measure µ can be concentrated on a non-closed set E and then E ⋐ supp µ. Example 2.2.1.1.
Let E be a countable set dense in G . Then supp µ = G and ,of course, E = G .The set of all measures on G will be denoted a M ( G ) . The measure µ F ( E ) := µ ( E ∩ F ) is called the restriction of µ onto F ∈ σ ( G ).It is easy to see that µ F is concentrated on F and supp µ ⊂ F .
A countably additive function ν on σ ( G ) that is finite for all K ⊂ G is calleda charge . We consider only real charges. Example 2.2.1.2 ν := µ − µ , µ , µ ∈ M ( G ) . The set of all the charges will be denoted M d . Theorem 2.2.1.2(Jordan decomposition).
Let ν ∈ M d ( G ) . Then there existtwo sets G + , G − such thata) G = G + ∪ G − , G + ∩ G − = ∅ ; b) ν ( E ) ≥ f or E ⊂ G + ; ν ( E ) ≤ f or E ⊂ G − . One can find the proof in [Ha,Ch.VI Sec.29]
The measures ν + := ν G + and ν − := ν G − , where ν G + , ν G − are restrictions of ν to G + , G − , are called the positive and negative , respectively, variations of ν . Themeasure | ν | := ν + + ν − is called the full variation of ν or just a variation . Theorem 2.2.1.3(Variations).
The following holds: ν + ( E ) = sup E ′ ⊂ E ν ( E ′ ); ν − ( E ) = inf E ′ ⊂ E ν ( E ′ ); ν = ν + + ν − . The proof is easy enough.
Exercise 2.2.1.2
Prove this.
Example 2.2.1.3
Let ψ ( x ) be a locally summable function with respect tothe Lebesgue’s measure. Set ν ( E ) := R E ψ ( x ) dx .Then ν + ( E ) = Z E ψ + ( x ) dx, ν − ( E ) = Z E ψ − ( x ) dx ; | ν | ( E ) = Z E | ψ | ( x ) dx, where(2.2.1.1) ψ + ( x ) = max(0 , ψ ( x )); ψ − ( x ) = − min(0 , ψ ( x )) . The function f ( x ) , x ∈ G is called a Borel function if the set E A := { f ( x ) > A } belongs to σ ( G ) for any A in R . Let K ⋐ G be a compact set and f a Borel function. Then the Lebesgue-Stieltjes integrals of the form R K f + dµ, R K f − dµ with respect to a measure µ ∈M ( G ) are defined, and R K f dµ := R K f + dµ − R K f − dµ is defined if at least one ofthe terms is finite.We say that a property holds µ − almost everywhere on E if the set E of x for which it does not hold satisfies the condition µ ( E ) = 0 . We will denote all the compact sets in G as K (sometimes with indexes). Thefollowing theorems hold: Theorem 2.2.2.1(Lebesgue).
Let { f n , n ∈ N } be a sequence of Borel functionson K and g ( x ) ≥ a function on K ,that is summable with respect to µ (i.e., itsintegral is finite), | f n ( x ) | ≤ g ( x ) for x in K, and f n → f when n → ∞ .Then lim n →∞ R K f n dµ = R K f dµ. Theorem 2.2.2.2(B.Levy).
Let f n ↓ f when n → ∞ , and f be a summablefunction on K .Then lim n →∞ R K f n dµ = R K f dµ. Theorem 2.2.2.3(Fatou’s Lemma).
Let f n ( x ) ≤ const < ∞ for x in K .Then lim sup n →∞ R K f n dµ ≤ R K lim sup n →∞ f n dµ. The proofs can be found in [Ha,Ch.V, Sec.27].Let L ( µ ) be the space of functions that are summable with respect to µ . Wesay that f n → f in L ( µ ) if f n , f ∈ L ( µ ) and k f n − f k := Z | f n − f | ( x ) dµ → Theorem 2.2.2.4(Uniqueness in L ( µ ) ). Let f n → f in L ( µ ) and Z f n ψdµ → Z gψdµ for any ψ continuous on supp µ. Then k g − f k = 0 . For proof see, e.g, [H¨o,Th.1.2.5 ].
Let φ ( x ) be a Borel function on G . The set supp φ := { x : φ ( x ) = 0 } iscalled the support of φ ( x ). A function φ is called f inite in G if supp φ ⋐ G We say that a sequence µ n ∈ M converges weakly to µ ∈ M if the condition R φdµ n → R φdµ holds for any continuous function φ .We will not show the integration domain, because it is always supp φ. The weak (it is called also C ∗ -) convergence will be denoted as ∗ → . Theorem 2.2.3.1( C ∗ -limits). If µ n ∗ → µ , then for E ∈ σ ( G ) the following asser-tions hold lim sup n →∞ µ n ( E ) ≤ µ ( E );lim inf n →∞ µ n ( ◦ E ) ≥ µ ( ◦ E ); where ◦ E is the interior of E , E is the closure of E. Proof.
Let χ E be the characteristic function of the set E. It is upper semicontinuous.Thus there exists a decreasing sequence ϕ m of continuous functions finite in G thatconverges to χ E as m → ∞ . Then we have µ n ( E ) = Z χ E dµ n ≤ Z ϕ m dµ n . Passing to the limit as n → ∞ we obtainlim sup n →∞ µ n ( E ) ≤ Z ϕ m dµ. Passing to the limit as m → ∞ we obtain by Th.2.2.2.2lim sup n →∞ µ n ( E ) ≤ Z χ E dµ = µ ( E ) . The proof for ◦ E is analogous. (cid:3) Theorem 2.2.3.2.(Helly).
Let { µ α : α ∈ A } be a family of measures uniformlybounded on any compact set K ⊂ G , i.e., ∃ C = C ( K ) : µ α ( K ) ≤ C ( K ) , f or K ⋐ G. Then this family is weakly compact , i.e., there exists a sequence { α j : α j ∈ A } and a measure µ such that µ α j ∗ → µ. The proof can be found in [Ha] .A set E is called squarable with respect to measure µ ( µ -squarable ) if µ ( ∂E ) = 0 . Theorem 2.2.3.3.(Squarable Ring).
The following holdssqr1) if E , E are µ -squarable, the sets E ∩ E , E ∪ E , E \ E are µ -squarable;sqr2) for any couple: an open set G and a compact set K ⊂ G there exists a µ -squarable set E such that K ⊂ E ⊂ G .Proof. The assertion sqr1) follows from ∂ ( E ∪ E ) [ ∂ ( E ∩ E ) [ ∂ ( E \ E ) ⊂ ∂E ∪ ∂E . Let us prove sqr2). Let K t := { x : ∃ y ∈ K : | x − y | < t } be a t -neighborhoodof the K. It is clear that for all the small t we have K ⋐ K t ⋐ G. The function a ( t ) := µ ( K t ) is monotonic on t and thus has no more than a countable set ofjumps.Let t be a point of continuity of a ( t ) . Then µ ( ∂K t ) ≤ lim ǫ → [ µ ( K t + ǫ ) − µ ( K t − ǫ )] = 0 . Thus it is possible to set E := K t for this t . (cid:3) A family Φ of sets is called a dense ring if the following conditions hold:dr1) ∀ F , F ∈ Φ = ⇒ F ∪ F , F ∩ F ∈ Φ;dr2) ∀ K, G : K ⋐ G ∃ F ∈ Φ : K ⊂ F ⊂ G. The previous theorem shows that the class of µ -squarable sets is a dense ring.The following theorem shows how one can extend a measure from a dense ring theBorel’s algebra.Let Φ be a dense ring and ∆( F ) , F ∈ Φ a function of a set which satisfies theconditions∆1) monotonicity on Φ: F ⊂ F = ⇒ ∆( F ) ≤ ∆( F );∆2) additivity on Φ: ∆( F ∪ F ) ≤ ∆( F ) + ∆( F )and∆( F ∪ F ) = ∆( F ) + ∆( F ) if F ∩ F = ∅ ∆3) continuity on Φ: ∀ F ∈ Φ and ǫ > K and anopen set G ⊃ K such that ∀ F ′ ∈ Φ : K ⊂ F ′ ⊂ G the inequality | ∆( F ) − ∆( F ′ ) | < ǫ holds. Theorem 2.2.3.4.(N.Bourbaki).
There exists a measure µ such that µ ( F ) =∆( F ) , ∀ F ∈ Φ iff the conditions ∆
1) - ∆
3) hold.
Theorem 2.2.3.5.(Uniqueness of Measure).
Under the conditions ∆
1) - ∆ (2.2.3.1) µ ( K ) = inf { ∆( F ) : F ∈ Φ , F ⊃ K } ; (2.2.3.2) µ ( G ) = sup { ∆( F ) : F ∈ Φ , F ⊂ G } ;(2.2.3.3) µ ( E ) = sup { µ ( K ) : K ⊂ E } = inf { µ ( G ) : G ⊃ E } , and every F ∈ Φ is µ -squarable. For proof see [Bo,Ch.4,Sec 3,it.10]. The squarability follows from (2.2.3.3).The following theorem connects the convergence of measures on any dense ringand on the ring of sets squarable with respect to the limit measure.
Theorem 2.2.3.6.(Set-convergences). If µ n ( F ) → µ ( F ) for all F in a densering Φ , then µ n ( E ) → µ ( E ) for any µ -squarable set E. Proof.
Suppose ◦ E = ∅ . Let ǫ > K such that(2.2.3.4) µ ( K ) + ǫ ≥ µ ( ◦ E ) = µ ( E ) . One can also find an open set G such that(2.2.3.5) µ ( G ) − ǫ ≤ µ ( E ) = µ ( E ) . By property dr2) of a dense ring one can find
F, F ′ ∈ Φ such that K ⊂ F ⊂ ◦ E ⊂ E ⊂ E ⊂ F ′ ⊂ G. Thus µ n ( F ) ≤ µ n ( E ) ≤ µ n ( F ′ ) and hence(2.2.3.6) µ ( F ) ≤ lim n →∞ µ n ( E ) ≤ lim n →∞ µ n ( E ) ≤ µ ( F ′ ) . From (2.2.3.4) and (2.2.3.5) we obtain 0 ≤ µ ( F ′ ) − µ ( F ) ≤ µ ( G ) − µ ( K ) ≤ ǫ forarbitrary small ǫ. Thus from (2.2.3.6) we obtain(2.2.3.7) lim n →∞ µ n ( E ) = lim n →∞ µ n ( E ) = µ ( E ) . That is to say that µ n ( E ) → µ ( E ) . If ◦ E = ∅ , then µ ( E ) = 0 by the definition of a squarable set. One can show inthe same way that µ n ( E ) → . (cid:3) Now we connect the weak convergence to the convergence on squarable sets. Theorem 2.2.3.7.(Set- and C*-convergences).
The conditions (2.2.3.8) µ n ∗ → µ and µ n ( E ) → µ ( E ) on µ -squarable sets E are equivalent.Proof. Sufficiency of (2.2.3.8) follows from Th.2.2.3.1.
Exercise 2.2.3.1
Prove this.Let us prove necessity.For any compact set one can find a µ -squarable E such that K ⊂ E. Hence µ n ( K ) ≤ µ ( E ) + 1 := C ( K ) when n is big enough.By Helly’s theorem (Th.2.2.3.2) there exists a measure µ ′ and a subsequence µ n j ∗ → µ ′ .By the proved necessity µ ′ ( E ) = µ ( E ) on a dense ring of the squarablesets. Thus µ ′ = µ by Uniqueness theorem 2.2.3.5. And thus µ n ∗ → µ. (cid:3) Denote by µ E ( G ) := (cid:26) µ ( G ∩ E ) if G ∩ E = ∅ G ∩ E = ∅ the restriction of µ on the set E. Corollary 2.2.3.8.
Let µ n ∗ → µ and E be a squarable set for µ. Then ( µ n ) E ∗ → ( µ ) E . Indeed, if E is a squarable set for µ it is a squarable set for µ E . So Th.2.2.3.7implies the corollary. Let σ ( R m × R m ) be the σ -algebra of all the Borel sets, Φ i ⊂ σ ( R m i ) , i =1 , , be dense rings, Φ := Φ ⊗ Φ ⊂ σ ( R m × R m ) be a ring generated by all thesets of form F × F , F i ∈ Φ i . Theorem 2.2.4.1.(Product of Rings). If Φ i , i = 1 , are dense rings, then Φ ⊗ Φ is a dense ring; if they consist of squarable sets, then Φ consists of squarablesets.Proof. Let K ⊂ G ⊂ R m × R m . For every point x ∈ K one can (evidently) find F × F such that x ⊂ F × F ⊂ G. One can find a finite covering and obtain afinite union F of sets of such form. Thus F ∈ Φ ⊗ Φ and F ⊂ G. The second assertion follows from the formula: ∂ ( F × F ) = ( ∂F × F ) ∪ ( F × ∂F ) . (cid:3) Let µ i be a measure on σ ( R m i ) , i = 1 , , and µ := µ ⊗ µ the product ofmeasures , i.e., a measure on σ ( R m × R m ) such that µ ( E × E ) = µ ( E ) µ ( E )for all E i ∈ σ ( R m i ) , i = 1 , . Theorem 2.2.4.2.(Product of Measures).
A measure µ ⊗ µ is uniquely de-fined by its values on Φ ⊗ Φ . The assertion follows from Theorem 2.2.4.1 and Uniqueness Theorem 2.2.3.5.
Theorem 2.2.4.3.(Fubini).
Let f ( x , x ) be a Borel function on R m × R m . Then (2.2.4.1) Z R m × R m f ( x , x ) d ( µ ⊗ µ ) = Z R m dµ Z R m f ( x , x ) dµ = Z R m dµ Z R m f ( x , x ) dµ , if at least one of parts of (2.2.4.1) is well defined. The proof can be found in [Ha, Ch.VII, Sec.36]. Let us consider the set D ( G ) of all infinitely differentiable functions ϕ ( x ) , x ∈ G ⊂ R m . It is a linear space because for any constants c , c (D1) ϕ , ϕ ∈ D ( G ) = ⇒ c ϕ + c ϕ ∈ D ( G ) . It is a topological space with convergence defined by(D2) ϕ n D → ϕ : a ) supp ϕ n ⊂ K ⋐ R m f or some compact Kandb ) ϕ n → ϕ unif ormly on Kwith all their derivatives. We consider some examples of functions ϕ ∈ D . Denote(2.3.1.1) α ( t ) = ( Ce − − t , f or t ∈ ( −
1; 1)0 , f or t ∈ ( −
1; 1) . Evidently α ( | x | ) ∈ D ( R m ) and supp α ⊂ { x : | x | ≤ } . Exercise 2.3.1.1
Check this.Let us find C such that(2.3.1.2) Z α ( | x | ) dx = σ m Z α ( t ) t m − dt = 1where σ m is area of the unit sphere {| x | = 1 } . Set(2.3.1.3) α ε ( x ) := ε − m α (cid:18) | x | ε (cid:19) . For any ε we have α ε ∈ D and supp α ε ⊂ { x : | x | ≤ ε } . Let ψ ( y ) , y ∈ K ⊂ G be a Lebesgue summable function. Consider the function(2.3.1.4) ψ ε ( x ) := Z K ψ ( y ) α ε ( x − y ) dy. The function belongs to D ( G ) for ε small enough and its support is contained inthe ε -neighborhood of K . Let f ( x ) , x ∈ G ⊂ R m be a locally summable function in G. The formula(2.3.2.1) < f, ϕ > := Z f ( y ) ϕ ( y ) dy, ϕ ∈ D ( G )defines a linear continuous functional on D , i.e., one that satisfies the conditions(D’1) < f, c ϕ + c ϕ > = c < f, ϕ > + c < f, ϕ > ;(D’2) ( ϕ n D → ϕ ) = ⇒ < f, ϕ n > → < f, ϕ > . However, (2.3.2.1) does not exhaust all the linear continuous functionals asone can see further. An arbitrary linear continuous functional on D ( G ) is calledan L.Schwartz distribution and the linear topological space of the functionals isdenoted as D ′ ( G ).Following are some examples of functionals that do not have the form of(2.3.2.1):(2.3.2.2) < δ x , ϕ > := ϕ ( x )This distribution is called the Dirac delta-function . Further,(2.3.2.3) < δ ( n ) x , ϕ > := ϕ ( n ) ( x ) . This distribution is called the n-th derivative of the Dirac delta-function.
Exercise 2.3.2.1
Check that the functionals (2.3.2.2) and (2.3.2.3) are bothdistributions.A distribution of the form (2.3.2.1) is called regular . Theorem 2.3.2.1.(Du Bois Reymond ).
If two locally summable functions f and f define the same distribution, then they coincide almost everywhere. For proof see, e.g., [H/”o, Th.2.1.6].Remark that the converse assertion is obvious.A distribution µ is called positive if < µ, ϕ > ≥ ϕ ∈ D ( G ) such that ϕ ( x ) ≥ x ∈ R m . We shall write this as µ > D ′ . Example 2.3.2.1.
Let µ ( E ) be a measure in G. Then the distribution(2.3.2.4) < µ, ϕ > := Z ϕ ( x ) dµ is positive.This formula represents all the positive distributions as one can see from Theorem 2.3.2.2.(Positive Distributions).
Let µ > in D ( G ) .Then there ex-ists a unique measure µ ( E ) such that the distribution µ is given by (2.3.2.3). For proof see, e.g., [H/”o,Th.2.1.7].
Let us consider operations on distributions.A product of a distribution f by an infinitely differentiable function α ( x ) isdefined by(2.3.3.1) < αf, ϕ > := < f, αϕ > . It is well defined because αϕ ∈ D too.A sum of distributions f and f is defined by(2.3.3.2) < f + f , ϕ > := < f , ϕ > + < f , ϕ >, and the partial derivative ∂∂x k is defined by the equality(2.3.3.3) < ∂∂x k f, ϕ > := < f, − ∂∂x k ϕ > . These definitions look reasonable because of the following
Theorem 2.3.3.1.(Operations on Distributions).
The sum of regular distri-butions corresponds to the sum of the functions; the product of a regular distributionby an infinitely differentiable function corresponds to the product of the functions;the derivative of a regular distribution that generated by a differentiable functioncorresponds to the derivative of that function.Proof.
We have, for example, < α · ( f ) , ϕ > := Z f ( x )[ α ( x ) ϕ ( x )] dx = Z [ α ( x ) f ( x )] ϕ ( x ) dx := < ( αf ) , ϕ > For the sum we have < ( f ) + ( f ) , ϕ > := < f , ϕ > + < f , ϕ > = Z f ( x ) ϕ ( x ) dx + Z f ( x ) ϕ ( x ) dx == Z [ f ( x ) + f ( x )] ϕ ( x ) dx = < ( f + f ) , ϕ > . Let f ( x ) have the derivative ∂∂x k f. Then < ∂∂x k f, ϕ > := < f, − ∂∂x k ϕ > == Z f ( x , x , ...x m )[ − ∂∂x k ϕ ( x , x , ...x m )] dx dx , ...dx m == Z dx ...dx k − dx k +1 ...dx m Z f ( x , x , ...x m )[ − ∂∂x k ϕ ( x , x , ...x m )] dx k . Now we shall do integrating by parts and all the substitution will vanish, because ϕ is finite. So we obtain < ∂∂x k f, ϕ > = Z ∂∂x k f ( x ) ϕ ( x ) dx. That is to say the derivative of the distribution corresponds to the function deriv-ative. (cid:3)
We say that a sequence of distributions f n converges to a distribution f if(2.3.4.1) < f n , ϕ > → < f, ϕ > ∀ ϕ ∈ D ( G ) . Theorem 2.3.4.1.(Completeness of D ′ ). If the sequence of numbers < f n , ϕ > has a limit for every ϕ ∈ D ( G ) , then this functional is a linear continuous functionalon D ( G ) , i.e., a distribution. For proof see, e.g., [H¨o,Th.2.1.8].Differentiation is continuous with respect to convergence of distributions.
Theorem 2.3.4.2.(Continuity of Differential Operators). If f n → f in D ( G ) , then ∂∂x k f n → ∂∂x k f. Proof.
Set in (2.3.4.1) ϕ := − ∂∂x k ϕ. Then < ∂∂x k f n , ϕ > = < f n , − ∂∂x k ϕ > → < f, − ∂∂x k ϕ > = < ∂∂x k f, ϕ > . (cid:3) The following theorem shows that the D ′ - convergence is the weakest of theconvergences considered earlier. Theorem 2.3.4.3.(Connection between Convergences).
Let f n be a sequenceof Lebesgue summable functions on domain G such that at least one of the followingconditions holds:Cnvr1) f n → f uniformly on any compact set K ⋐ G and f is a locallysummable function;Cnvr2) f n → f on any K ⋐ G, satisfying the conditions of the Lebesguetheorem (Th.2.2.2.1);Cnvr3) f n ↓ f monotonically and f is a locally summable function.Then f n → f in D ′ ( G ) . Proof.
All the assertions are corollaries of the section 2.2.2 of passing to the limitunder an integral.Let us prove, for example, Cnvr3). Let f n ↓ f . Then(2.3.4.2) < f n , ϕ > = Z f n ( x ) ϕ ( x ) dx = Z f n ( x ) ϕ + ( x ) dx − Z f n ( x ) ϕ − ( x ) dx where ϕ + and ϕ − are defined in (2.2.1.1).Both last integrals in (2.3.4.2) have a limit by the B.Levy theorem( Th.2.2.2.2), and thus(2.3.4.3)lim n →∞ < f n , ϕ > = Z f ( x ) ϕ + ( x ) dx − Z f ( x ) ϕ − ( x ) dx = Z f ( x ) ϕ ( x ) dx = < f, ϕ > . (2.3.4.3) means that f n → f in D ′ . (cid:3) Exercise 2.3.4.1
Prove Cnvr 1) and 2).
Theorem 2.3.4.4.( D ′ and C ∗ convergences). Let µ n , µ be measures in G . Theconditions µ n → µ in D ′ ( G ) and µ n ∗ → µ are equivalent. It is clear that the first condition is necessary for the second one. The suffi-ciency holds, because every continuous function can be approximated with functionsthat belong to D . For more details see, e.g., [H¨o,Th.2.1.9].Let α ǫ ( x ) be defined as in (2.3.1.3).For any f ∈ D ′ ( D ) we can consider thefunction f ǫ ( x ) := < f, α ǫ ( x + • ) > . It is called a regularization of the distribution f. Theorem 2.3.4.5.(Properties of Regularizations).
The following holds:reg1) f ǫ ( x ) is an infinitely differentiable function in any K ⋐ D for sufficientlysmall ǫ ; reg2) f ǫ ( x ) → f in D ′ ( D ) as ǫ ↓ reg3) if f n → f in D ′ ( D ) , ( f n ) ǫ → f ǫ uniformly with all its derivatives on anycompact set in D. The property reg1) follows from the formula ∂∂x j f ǫ = < f, ∂∂x j α ǫ ( x + • ) > . The property reg2) follows from the assertion φ ǫ ( x ) := Z φ ( y ) α ǫ ( x + y ) dy → φ ( x ) in D ( D )as ǫ ↓ . For the proof of reg3) see [H¨o,Theorem’s 2.1.8, 4.1.5].Let us note the following assertion;
Theorem 2.3.4.6.(Continuity < • , • > ). The function < f, φ > : D ′ ( G ) × D ( G ) R is continuous in the according topology, i.e., f n → f in D ′ ( G ) and φ j → φ in D ( G ) imply < f n , φ j > → < f, φ > . For proof see [H¨o,Th.2.1.8].
Let G ⊂ G. Then D ′ ( G ) ⊂ D ′ ( G ) , because every functional on D ( G ) canbe considered as a functional on D ( G ) . A distribution f ∈ D ′ ( G ) considered as a distribution in D ′ ( G ) is called the restriction of f to G and is denoted f | G . Theorem 2.3.5.1.(Sewing Theorem).
Let G α ⊂ R m be a family of domainsand in every of them let there be a distribution f α ∈ D ( G α ) , such that If G α ∩ G α = ∅ , the equality (2.3.5.1) f α | G α ∩ G α = f α | G α ∩ G α holds.Then there exists one and only one distribution f ∈ D ′ ( G ) where G = S α G α such that f | G α = f α . In particular, it means that every distribution is defined uniquely by its re-striction to a neighborhood of every point.Let D ( S R ) be a space of infinitely differentiable functions on the sphere S R := { x : | x | = R } . The corresponding distribution space is denoted as D ′ ( S R ) . The sewing theorem holds for this space in the following form:
Theorem 2.3.5.2.( D ′ on Sphere). Let a family of domains Ω α cover S R and inevery of them let there be a distribution f α ∈ D (Ω α ) , such thatIf Ω α ∩ Ω α = ∅ ,the equality (2.3.5.1) f α | Ω α ∩ Ω α = f α | Ω α ∩ Ω α holds.Then there exists one and only one distribution f ∈ D ′ ( S R ) such that f | Ω α = f α . Let(2.3.6.1) L := X i,j ∂∂x i a i,j ( x ) ∂∂x j + q ( x )be a differential operator of second order with infinitely differentiable coefficients a i,j , q. We will consider only three types of differential operators: one dimensionaloperator with constant coefficients,the Laplace operator and the so called sphericaloperator (see Sec.2.4 ).For all these operators we have the following assertion which follows from thegeneral theory (see, e.g., [H¨o,Th.11.1.1] ): Theorem 2.3.6.1.(Regularity of Generalized Solution).
If the equation Lu = 0 has a solution u ∈ D ′ ( G ) , then u is a regular distribution and can berealized as an infinitely differentiable function . A distribution that satisfies the equation(2.3.6.2) Lu = δ y in D ′ ( G ) , where δ y is a Dirac delta function (see (2.3.2.2)), is called a fundamental solution of L at the point y. Every differential operator that we are going to consider has a fundamentalsolution (see, e.g., [H¨o,Th.10.2.1]).A restriction of the equation (2.3.6.2) to the domain G y := G \ y is a homoge-neous equation Lu = 0 in D ′ ( G y ). Thus we have Theorem 2.3.6.2.(Regularity of Fundamental Solution).
The fundamentalsolution is an infinitely differentiable function outside the point y. We will need further also the Fourier coefficients for the distribution on thecircle .Let D ( S ) be a set of all infinitely differentiable function on the unit circle S . The set of all linear continuous functionals over D ( S ) with the correspondingtopology (see 2.3.2) is the corresponding space of distributions D ′ ( S ) for which allthe previous properties of distributions holds.The functions { e ikφ , k = 0 , ± , ± , ... } belong to D ( S ) . The Fourier coeffi-cients of ν ∈ D ′ ( S ) are defined by(2.3.7.1) ˆ ν ( k ) := < ν, e − ikφ > . The inverse operator is defined by(2.3.7.2) < ν, g > = 12 π ∞ X k = −∞ ˆ ν ( k ) < g, e ikφ >, and the series converges, in any case, for those ν that are finite derivatives ofsummable functions, because Fourier coefficients of g decrease faster then everypower of x. The convolution of distribution ν ∈ D ′ ( S ) and g ∈ D ( S ) is defined by(2.3.7.3.) ν ∗ g ( φ ) = < ν, g ( φ − • ) >, This is a function from D ( S ) . The convolution of distributions ν , ν ∈ D ′ ( S ) is defined by(2.3.7.4) < ν ∗ ν , g > = ν ∗ ( ν ∗ g ) . In spite of the view it is commutative and \ ν ∗ ν ( k ) = ˆ ν ( k ) · ˆ ν ( k ) . Exercise 2.3.7.1
Count the Fourier coefficients of the functions(2.3.7.5) G ( re iφ ) = log | − re iφ | for r > , r = 1 , r <
1; the function defined by(2.3.7.6) ] cos ρ ( φ ) := cos ρφ, − π < φ < π, ρ ∈ (0 , ∞ )and 2 π -periodically extended; the function(2.3.7.7) ˜ φ sin pφ, p ∈ N where ˜ φ is the 2 π -periodical extension of the function f ( φ ) = φ, φ ∈ [0 , π ) . Exercise 2.3.7.2
Denote(2.3.7.8) P p − ( re iφ ) := ℜ ( p − X k =1 r k e ikφ k ) , p ∈ N . Prove that for every distribution ν :(2.3.7.9) ( P p − ( \ re i • ) ∗ ν )( p ) = 0The same for the function G p ( re iφ ) := G ( re iφ ) + P p ( re iφ )for r < . We will denote as ∆ the Laplace operator in R m :∆ := ∂ ∂x + ... + ∂ ∂x m . We introduce in R m the spherical coordinate system by the formulae: x = r sin φ sin φ ... sin φ m − ; x = r cos φ sin φ ... sin φ m − ; x = r cos φ sin φ ... sin φ m − ; .........................................................x k = r cos φ k − sin φ k − ... sin φ m − ; ...........................................................x m = r cos φ m − , where 0 < φ ≤ π ; 0 ≤ φ j < π, j = 1 , m −
2; 0 < r < ∞ . Passing to the coordinates ( r, φ , φ , ... φ m − ) in the Laplace operator weobtain ∆ = 1 r m − ∂∂r r m − ∂∂r + 1 r ∆ x . The operator ∆ x is called spherical , and has the form∆ x := m − X i =0 ∂∂φ i ΠΠ i ∂∂φ i , where Π := m − Y j =1 sin j φ j ; Π i := m − Y j = i +1 sin φ j ; Π m − := 1 . In particular, for m = 2, i.e., for the polar coordinates,∆ = 1 r ∂∂r r ∂∂r + 1 r ∂ ∂φ . A distribution H ∈ D ′ ( G ) is called harmonic if it satisfies the equation ∆ H = 0 . The next theorem follows from Theorem 2.3.6.1. Theorem 2.4.1.1.(Smoothness of harmonic functions).
Any harmonic dis-tribution is equivalent to an infinitely differentiable function.
This function, of course, satisfies the same equation and is a harmonic function in the ordinary sense. A direct proof can be found, e.g., in [Ro,Ch.1, § f ( z ) , z = x + ıy be a holomorphic function in a domain G ⊂ C . Then thefunctions u ( x, y ) := ℜ f ( z ) and v ( x, y ) := ℑ f ( z ) are harmonic in G. In particular,the functions r n cos nϕ and r n sin nϕ where r = | z | , ϕ = arg z are harmonic.Set(2.4.1.1) E m ( x ) := (cid:26) −| x | − m , for m ≥ | z | , for m = 2(We will often denote points of the plane as z ).It is easy to check that E m ( x ) is a harmonic function for | x | 6 = 0 . Set θ m := (cid:26) ( m − σ m , for m ≥ π, for m = 2;where σ m is the surface area of the unit sphere in R m . Theorem 2.4.1.2.(Fundamental Solution).
The function E m ( x − y ) satisfiesin D ′ ( R m ) the equation (2.4.1.2) ∆ x E m ( x − y ) = θ m δ ( x − y ) , where δ ( x ) is the Dirac δ -function (see 2.3.2).Proof. Let us prove the equality (2.4.1.2) for y = 0 . Suppose φ ∈ D ( R m ) andsupp φ ⊂ K ⋐ R m . We have < ∆ E m , φ > := Z E m ( x )∆ φ ( x ) dx = lim ǫ → Z | x |≥ ǫ E m ( x )∆ φ ( x ) dx. Transforming this integral by the Green formula and using the fact that φ is finitewe obtain Z | x |≥ ǫ E m ( x )∆ φ ( x ) dx = Z | x |≥ ǫ ∆ E m ( x ) φ ( x ) dx + Z | x | = ǫ E m ∂φ∂n ds − Z | x | = ǫ φ ∂ E m ∂n ds, E m is slightly a little different from the fundamental solution (see,(2.3.6.2)), but this istraditional in Potential Theory7 where ds is an element of surface area and ∂∂n is the differentiation in the directionof the external normal.Use the harmonicity of E m . Then the first integral is equal to zero. Furtherwe have Z | x | = ǫ E m ∂φ∂n ds = ǫ Z | x | =1 ∂∂r φ ( rx ) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = ǫ = O ( ǫ ) , for ǫ → Z | x | = ǫ φ ∂ E m ∂n ds = m − ǫ m − ǫ m − Z | x | =1 φ ( rx ) ds = [ φ (0) + o (1)]( m − σ m . Thus we obtain < ∆ E m , φ > = φ (0) θ m , and this proves (2.4.1.2) for y = 0 . It is clear that by changing φ ( x ) for φ ( x + y ) we obtain (2.4.1.2) in the generalcase. (cid:3) We will consider now a domain Ω with a
Lipschitz boundary (
Lipschitz do-main ). It means that every part of ∂ Ω can be represented in some local coordinates( x, x ′ ) , x ∈ R , x ′ ∈ R m − in the form x = f ( x ′ ) , where f is a Lipschitz function,i.e., | f ( x ′ ) − f ( x ′ ) | ≤ M ∂ Ω | x ′ − x ′ | where M depends only on the whole ∂ Ω and does not depend on this local part.Let G ( x, y, Ω) be the Green function of a Lipschitz domain Ω . It is known (see, e.g. [Vl,Ch.V, §
28] ) that the Green function has the followingproperties:(g1) G ( x, y, Ω) < , for ( x, y ) ∈ Ω × Ω; G ( x, y, Ω) = 0 for ( x, y ) ∈ Ω × ∂ Ω;(g2) G ( x, y, • ) = G ( y, x, • );(g3) G ( x, y, • ) − E m ( x − y ) = H ( x, y ) , where H is harmonic on x and on y within Ω . (g4) − G ( x, y, Ω ) ≤ − G ( x, y, Ω ) for Ω ⊂ Ω From (g3) follows Theorem 2.4.1.3.(Green Function).
The equality (2.4.1.3) ∆ x G ( x, y, Ω) = θ m δ ( x − y ) , holds in D ′ (Ω) . Let f ( x ) be a continuous function on ∂ Ω. It is known (see, e.g.[Vl,Ch.V, § H ( x, f ) := Z ∂ Ω f ( y ) ∂∂n G ( x, y, Ω) ds y is the only harmonic function that coincides with f on ∂ Ω . The unique solution of the Poisson equation∆ u = p, u | ∂ Ω = f for a continuous function p is given by the formula(2.4.1.5) u ( x, f, p ) := Z ∂ Ω f ( y ) ∂∂n y G ( x, y, Ω) ds y + θ − m Z Ω G ( x, y, Ω) p ( y ) dy. Let D be an arbitrary open domain.We can define a G ( x, y, D ) in the followingway. Consider a sequence Ω n of a Lipschitz domains such that Ω n ↑ D. The sequenceof the corresponding Green functions G ( x, y, Ω n ) monotonically decreases. If it isbounded from below in some point, it is bounded everywhere while x = y (as itfollows from Theorem 2.4.1.7). It can be shown that the limit exists for any domainthe boundary of which have positive capacity (see 2.5 and references there). Wewill mainly use the Green function for the Lipschitz domains.Let G ( x, y, K a,R ) be the Green function of the ball K a,R := {| x − a | < R } . Theorem 2.4.1.4.(Green Function for Ball). G ( x, y, K a,R ) = ( −| x − y | − m − ( | y − a || x − y ∗ a,R | R ) − m , for m ≥ | ζ − z | R | ζ − a || z − ζ ∗ a.R | for m = 2 where y ∗ a.R := a + ( y − a ) R | y − a | is the inversion of y relative to the sphere {| x − a | = R } . For the proof see, e.g., [Br],Ch.6, § Theorem 2.4.1.5.(Poisson Integral).
Let H be a harmonic function in K a,R and continuous in its closure. Then (2.4.1.6) H ( x ) = 1 σ m R Z | x − a | = R H ( y ) R − | x − a | | x − y | m ds y , x ∈ K ( a, R ) . In particular, for m = 2 H ( a + re ıφ ) = 12 π π Z H ( a + Re ıψ ) R − r R − Rr cos( φ − ψ ) + r dψ. This theorem follows from (2.4.1.4).
Theorem 2.4.1.6.(Mean Value).
Let H be harmonic in G ⊂ R m . Then (2.4.1.7) H ( x ) = 1 σ m R m − Z | x − a | = R H ( y ) ds y , where x ∈ G and R is taken such that K ( x, R ) ⋐ G. We must only set a := x in (2.4.1.6).We can rewrite (2.4.1.7) in the form H ( x ) = 1 σ m Z | y | =1 H ( x + Ry ) ds y . Theorem 2.4.1.7.(Harnack).
Suppose the family { H α ) , α ∈ A } of harmonicfunctions in G satisfies the conditions (Har1) H α ( x ) ≤ C ( K ) , for x ∈ K ; (Har2) H α ( x ) ≥ B > −∞ , for x ∈ K for every compact K ⋐ G and C ( K ) , B are constants not depending on α .Then the family is precompact in the uniform topology, i.e., there exists sucha sequence H α n , and a function H harmonic in the interior of K and continuousin K such that H α n → H uniformly in every K. One can prove by using (2.4.1.6) that | grad H α | are bounded on every compactset by a constant not depending on α. Thus the family is uniformly continuous andthus it is precompact by the Ascoli theorem .For details see, e.g., [Br,Supplement, § Theorem 2.4.1.8.(Uniform and D ′ -convergences). Suppose the sequence H n satisfies the conditions of the Harnack theorem and converges to a function H in D ′ ( G ) . Then H n converges to H uniformly on every compact K ⋐ G. Of course, H is harmonic in G. Proof.
By the Harnack theorem the family is precompact.Thus we must only provethe uniqueness of H. Suppose there exist two subsequences such that H k → H and H k → H uniformly on every compact K ⋐ G. By Connection between Convergences (Theorem 2.3.4.3) H k → H and H k → H in D ′ . Hence, H = H in D ′ ( G ) . By the De Bois Raimond theorem (Theorem2.3.2.1) H = H almost everywhere and hence everywhere because these functionsare continuous. (cid:3) Let D be a domain with a smooth boundary ∂D and let F ⊂ ∂D. Set ω ( x, F, D ) := Z F ∂G∂n y ( x, y ) ds y . It is called a harmonic measure of F with respect to D. A harmonic measure canbe defined for an arbitrary domain D by a limit process similar to the one we hadfor the Green function.In this case the formula (2.4.1.4) has the form H ( x, f ) := Z ∂D f ( y ) dω ( x, y, D ) . However we can not assert that H ( x, f ) coincides with f in any point x ∈ D. We canonly consider it as an operator that maps a function defined on ∂D to a harmonicfunction in D. By (2.4.1.3) we obtain
Theorem 2.4.1.9.(Two Constants Theorem).
Let H be harmonic in D andsatisfy the conditions H ( x ) ≤ A f or x ∈ F ; H ( x ) ≤ A f or x ∈ ∂D \ F where A and A are constants. Then H ( x ) ≤ A ω ( x, F, D ) + A ω ( x, ∂D \ F, D ) f or x ∈ D. Let y ∗ a,R be the inversion from Green Function for Ball (Theorem 2.4.1.4). Set y ∗ := y ∗ , , i.e., the inversion relative to a unit sphere with the center in the origin.Let G ∗ := { y ∗ : y ∈ G } be the inversion of a domain G. Theorem 2.4.1.10.(Kelvin’s Transformation). If H is harmonic in G, then (2.4.1.8) H ∗ ( y ) := | y | − m H ( y ∗ ) is harmonic in G ∗ . For the proof you must honestly compute Laplacian of H ∗ .”The computationis straightforward but tedious“ ([He,Theorem 2.24]). It is not so tedious if you usethe spherical coordinate system. Exercise 2.4.1.1
Do this.
Denote as S := { x : | x | = 1 } the unit sphere with center in the origin. Afunction Y ρ ( x ) , x ∈ Ω ⊂ S is called a spherical function of degree ρ if it satisfiesthe equation(2.4.2.1) ∆ x Y + ρ ( ρ + m − Y = 0 . For m = 2 (2.4.2.1) gets the form Y ′′ ( θ ) + ρ Y ( θ ) = 0 , i.e., Y ( θ ) = a cos ρθ + b sin ρθ, Spherical functions are obtained if we solve the equation ∆ H = 0 by the change H ( x ) = | x | ρ Y ( x ) . Theorem 2.4.2.1.(Sphericality and Harmonicity).
Let Y ρ ( x ) be spherical ina domain Ω ⊂ S if and only if the functions H ( x ) = | x | ρ Y ρ ( x ) and H ∗ ( x ) = | x | − ρ − m +2 Y ρ ( x ) are harmonic in the cone (2.4.2.2) Con (Ω) := { x = rx : x ∈ Ω , < r < ∞} . If ρ = k, k ≥ , k ∈ Z , and only in this case, Y k ( x ) is spherical on the whole S ,H ( x ) is a homogeneous harmonic polynomial of degree k and H ∗ is harmonic in R m \ . For the proof see, e.g.[Ax]The spherical functions of an integer degree k form a finite-dimension space ofdimension dim ( m, k ) = (2 k + m − k + m − m − k !In particular, d (2 , k ) = 2 for any k. For different k the spherical functions Y k ( x ) are orthogonal on S . In partic-ular, for m = 2, it means the orthogonality of the trigonometric functions system. Theorem 2.4.2.2.(Expansion of Harmonic Function ).
Let H ( x ) be a har-monic function in the ball K R := {| x | < R } . There exists an orthonormal systemof spherical functions Y k ( x ) , k = 0 , ∞ , depending on H such that (2.4.2.3) H ( x ) = ∞ X k =0 c k Y k ( x ) | x | k , f or | x | < R. For any such system we have (2.4.2.4) c k = 1 R k Z S H ( Rx ) Y k ( x ) ds x . For proof see, e.g., [Ax,Ch.10],[TT, Ch.4, §
10] .
Theorem 2.4.2.3.(Liouville).
Let H be harmonic in R m and suppose (2.4.2.5) lim inf R →∞ R − ρ max | x | = R H ( x ) < ∞ . holds.Then H is a polynomial of a degree q ≤ ρ. Proof.
We can suppose H (0) = 0 because H ( x ) − H (0) is harmonic and also satisfies(2.4.2.5). Let R n → ∞ be a sequence for which(2.4.2.6) R − ρn max | x | = R n H ( x ) ≤ const < ∞ From (2.4.2.4) we obtain(2.4.2.7) | c k | ≤ A k R − k Z S | H ( Rx ) | ds x , where A k = max S | Y k ( x ) | . From the mean value property (Theorem 2.4.1.6) Z S H ( Rx ) ds x = H (0) σ m = 0 . Thus(2.4.2.8) Z S | H ( Rx ) | ds x = 2 Z S H + ( Rx ) ds x ≤ σ m max | x | = R H ( x ) . From (2.4.2.8) and (2.4.2.7) we have(2.4.2.9) | c k | ≤ A k R − k σ m max | x | = R H ( x ) . Set R := R n and k > ρ. Passing to the limit when n → ∞ , we obtain c k = 0 for k > ρ. Then (2.4.2.3.) implies that H is a harmonic polynomial of degree q ≤ ρ. (cid:3) Let G ( x, y.D ) be the Green function of a Lipschitz domain D. We willsuppose it is extended as zero outside of D. Π( x, µ, D ) := − Z G ( x, y, D ) dµ y is called the Green potential of µ relative to D. The domain of integration will always be R m . Theorem 2.5.1.1.(Green Potential Properties).
The following holds:GPo1) Π( x, µ, D ) is lower semicontinuous;GPo2) it is summable over any ( m − - dimensional hyperplane or smoothhyper-surface;GPo3) ∆Π( • , µ, D ) = − θ m µ in D ′ ( D ); GPo4) the reciprocity law holds: Z Π( x, µ , D ) dµ x = Z Π( x, µ , D ) dµ x . GPo5) semicontinuity in µ : if µ n → µ in D ′ ( R m ) , then lim inf n →∞ Π( x, µ n , D ) ≥ Π( x, µ, D ) . GPo6) continuity in µ in D ′ : if µ n → µ , then Π( • , µ n , D ) → Π( • , µ, D ) in D ′ ( R m ) and in D ′ ( S R ) ,where S R is the sphere {| x | = R } . Proof.
Let us prove GPo1). Let
N > . Set G N ( x, y ) := max( G ( x, y ) , − N ) , atruncation of the function G ( x, y ) . The functions G N are continuous in R m × R m and G N ( x, y ) ↓ G ( x, y ) for every( x, y ) when N → ∞ . SetΠ N ( x, µ, D ) := − Z G N ( x, y, D ) dµ y . The functions Π N are continuous and Π N ( x, • ) ↑ Π( x, • ) by the B.Levy theorem(Theorem 2.2.2.2). Then Π N ( x, • ) is lower semicontinuous by the Second Criterionof semicontinuity (Theorem 2.1.2.9). Let us prove GPo5). Since Theorem 2.3.4.4.( D ′ and C ∗ convergences)lim n →∞ Π N ( x, µ n , D ) = Π N ( x, µ, D ) . Further Π( x, µ n , D ) ≥ Π N ( x, µ n , D ), hencelim inf n →∞ Π( x, µ n , D ) ≥ Π N ( x, µ, D ) Passing to limit while N → ∞ , we obtainGPo5)The assertion GPo2) follows from the local summability of the function | x | − m that can be checked directly.Let us prove GPo3). For φ ∈ D ( D ) we have < ∆Π , φ > : = < Π , ∆ φ > = − Z dµ y Z G ( x, y, D )∆ φ ( x ) dx = − Z < ∆ x G ( • , y, D ) , φ > dµ y = − θ m Z φ ( y ) dµ y = − θ m < µ, φ > . since < ∆ x G ( • , y, D ) , φ > = θ m φ ( y ) . by Theorem 2.4.1.3. The property GPo4) follows from the symmetry of G ( x, y, • ) (property (g2)).Let us prove GPo6). Note that integral R | x | m − dx converges locally in R m and in R m − . From this one can obtain by some simple estimates that functionsΨ( y ) := R G ( x, y, D ) ψ ( x ) dx while ψ ∈ D ( R m ) and Θ( y ) := R S R G ( x, y, D ) θ ( x ) ds x while θ ∈ D ( S R ) are continuous on y ∈ R m . Now we have < Π( • , µ n , D ) , ψ > = Z Ψ( y ) dµ n ( y ) → Z Ψ( y ) dµ ( y ) = < Π( • , µ, D ) , ψ > . Thus the first assertion in GPo6) is proved. The second one can be proved in thesame way. (cid:3)
Set ν := µ − µ , and let Π( x, ν, D ) := Π( x, µ , D ) − Π( x, µ , D ) be a potentialof this charge. Consider the boundary problem of the form∆ u = µ − µ , in D ′ ( D ) u | ∂D = f, (2.5.1.2)where f is a continuous function. Theorem 2.5.1.2.(Solution of Poisson Equation).
The solution of the bound-ary problem (2.5.1.2) is given by the formula u ( x ) = H ( x, f ) − θ − m Π( x, ν, D ) , where H ( x, f ) is the harmonic function from (2.4.1.4).Proof. Since Π( x, ν, D ) | ∂D = 0 the function u ( x ) satisfies the boundary condition.Using GPo3) we obtain ∆ u = ∆ H − [ θ m ] − ∆Π = µ − µ . (cid:3) A potential of the form Π( x, µ ) := Z dµ y | x − y | m − is called a Newton potential. It is the Green potential for D = R m . The potentialΠ( z, µ ) = − Z log | z − ζ | dµ ζ is called logarithmic . Let K ⋐ D .The quantity(2.5.2.1) cap G ( K, D ) := sup µ ( K )where the supremum is taken over all the mass distributions µ for which the fol-lowing conditions are satisfied:(2.5.2.2) Π( x, µ, D ) ≤ µ ⊂ K, is called the Green capacity of the compact set K relative to the domain D. Theorem 2.5.2.1.(Properties of cap G ). For cap G the following properties hold:capG1) monotonicity with respect to K : K ⊂ K implies cap G ( K , D ) ≤ cap G ( K , D ) . capG2) monotonicity with respect to D : K ⋐ D ⊂ D implies cap G ( K, D ) ≥ cap G ( K, D ) capG3) subadditivity with respect to K : cap G ( K ∪ K , D ) ≤ cap G ( K , D ) + cap G ( K , D ) Proof.
The set of all the mass distributions that satisfy (2.5.2.2) for K = K is notless than the analogous set for K = K . Thus capG1) holds.By the Green function property (g3) (see § − G ( x, y, D ) ≤ − G ( x, y, D ) . Thus the set of all µ that satisfy (2.5.2.2) for D = D is wider than for D = D . Hence capG2) holds.Let supp µ ⊂ K ∪ K and let µ , µ be the restrictions of µ to K , K respec-tively.If µ satisfies the (2.5.2.2) for K := K ∪ K then µ , µ satisfy (2.5.2.2) for K := K , K respectively.From the inequality µ ( K ∪ K ) ≤ µ ( K ) + µ ( K )we obtain that µ ( K ∪ K ) ≤ cap G ( K , D ) + cap G ( K , D )for any µ with supp µ ⊂ K ∪ K . Thus capG3) holds.The equivalent definition of the Green capacity is done by
Theorem 2.5.2.2.(Dual Property).
The following holds (2.5.2.4) cap G ( K, D ) = [inf µ sup x ∈ D Π( x, µ, D )] − where the infimum is taken over all the mass distributions µ such that µ ( K ) = 1 For proof see, e.g., [La,Ch.2, § D = R m , m ≥ , the Green capacityis called Wiener capacity ( cap m ( K )). It has the following properties in additionto those of the Green capacity:capW1) invariance with respect to translations and rotations, i.e. cap m ( V ( K + x )) = cap m ( K ) , where V K and K + x are the rotation and the translation of K respectively.The presence of the properties brings the notion of capacity closer to the notionof measure. Thus it is natural to extend the capacity to the Borel algebra of sets.The Wiener capacity of an open set is defined as cap m ( D ) := sup K cap m ( K ) , where the supremum is taken over all compact K ⋐ D. The outer and inner capacity of any set E can be defined by the equalities cap m ( E ) := inf D ⊃ E cap m ( D ); cap m ( E ) := sup K ⊂ E cap m ( D ) . A set E is called capacible if cap m ( E ) = cap m ( E ) Theorem 2.5.2.3.(Choquet).
Every set E belonging to the Borel ring is capaci-ble. For proof see, e.g., [La,Ch2, Th.2.8].Sets which have “small size” are sets of zero capacity. We emphasize thefollowing properties of these sets:capZ1)If cap m ( E j ) = 0 , j = 1 , , ... then cap m ( ∪ ∞ E j ) = 0;capZ2) The property to have the zero capacity does not depend of type of thecapacity: Green, Wiener or logarithmic capacity that we define below. Example 2.5.2.1.
Using Theorem 2.5.2.2 we obtain that any point has zero ca-pacity, because for every mass distribution concentrated in the point the potentialis equal to infinity. The same holds for any set of zero m − Example 2.5.2.2.
Any (m-1)- hyperplane or smooth hypersurface has positivecapacity, because the potential with masses uniformly distributed over the surfaceis bounded.The Wiener 2-capacity can be defined naturally only for sets with diameterless then one, because the logarithmic potential is positive only when this conditionholds.Instead, one can use the logarithmic capacity which is defined by the formulae(2.5.2.5) cap l ( K ) := exp[ − cap ( K )]for K ⊂ {| z | < } and cap l ( K ) := t − cap l ( tK )for any other bounded K, where t is chosen in such a way that tK ⊂ {| z | < } . One can check that this definition is correct, i.e. it does not depend on t. Let D be a domain such that ∂D ⋐ R m ,and supp µ ⋐ D . Then there exists a mass distribution µ b such that for m ≥ , or for m = 2 and for D which is a bounded domain, the following holds:bal1) Π( x, µ b ) < Π( x, µ ) for x ∈ D ; bal2) Π( x, µ b ) = Π( x, µ ) for x / ∈ D ; bal3) supp µ b ⊂ ∂D ; bal4) µ b ( ∂D ) = µ ( D ) . If m = 2 and the domain is unbounded, a potential of the form ˆΠ( z, µ ) := − Z log | − z/ζ | dµ ζ . satisfies all the properties.Proof. We will prove this theorem when ∂D is smooth enough. For y ∈ D, x ∈ R m \ D the function | x − y | − m is a harmonic function of y on D. Since | x − y | − m → y → ∞ we can apply the Poisson formula (2.4.1.4)even if D is unbounded. Thus(2.5.3.1.) | x − y | − m = Z ∂D | x − y ′ | − m ∂G∂n y ′ ( y, y ′ ) ds y ′ where G is the Green function of D. From this we have Z D | x − y | − m dµ y = Z ∂D | x − y ′ | − m ds y ′ (cid:18)Z D ∂G∂n y ′ ( y, y ′ ) dµ y (cid:19) . The inner integral is nonnegative, because ∂G∂n > y ′ ∈ ∂D . Let us denote dµ b,y ′ := (cid:18)Z D ∂G∂n ( y, y ′ ) dµ y (cid:19) ds y ′ . Then we obtain the properties bal2) and bal3).The potential Π( x, µ b ) is harmonic in D. Thus the function u ( x ) := Π( x, µ b ) − Π( x, µ )is a subharmonic function (see Theorem 2.6.4.1). Every subharmonic function sat-isfies the maximum principle (see Theorem 2.6.1.2), i.e. u ( x ) < sup y ∈ ∂D u ( y ) = 0 . Thus the property bal1) is fulfilled. To prove bal4) we can write the identity Z ∂G dµ b,y ′ = Z G dµ y Z ∂G ∂G∂n y ′ ( y, y ′ ) dy ′ . The inner integral is equal to one identically, because the function ≡ , y ∈ G isharmonic. Thus bal4) is true.Consider now the special case when m = 2, and D is an unbounded domain.Since log | − z/ζ | → ζ → ∞ , we obtain an equality like (2.5.3.1). Repeatingthe previous reasoning we obtain the last assertion for D with a smooth boundary. Exercise 2.5.3.1.
Check this in details. (cid:3)
For the general case see [La Ch.4, § x, µ b ) is also a solution of the Dirichletproblem in the domain D and the boundary function f ( x ) = Π( x, µ ) in the followingsense: Theorem 2.5.3.2.(Wiener).
The equality bal2) holds in the points x ∈ ∂D whichcan be reached by the top of a cone placed outside D. For m = 2 it can fail only forisolated points. For proof see [He],[La,Ch.4, § ∂D where the equality bal2) does not hold are called irregular . Theorem 2.5.3.3.(Kellogg’s Lemma).
The set of all the irregular points of ∂D has zero capacity. For proof see, e.g., [He], [La,Ch.4, § Theorem 2.5.3.4.(Equilibrium distribution).
For any compact K with cap m ( K ) > there exists a mass distribution λ K such that the following holds:eq1) Π( x, λ ) = 1 , x ∈ D \ E, cap m ( E ) = 0 ;eq2) supp λ K ⊂ ∂K ; eq3) λ K ( ∂K ) = cap m ( K ) . For proof see [He], [La,Ch.2, § E in the previous theorem is a set of irregular points.The mass distribution λ K is called equilibrium distribution , and the corre-sponding potential is called equilibrium potential . Let h ( x ) , x ≥ h (0) = 0 . Let { K ǫj } be a family of balls such that theirdiameters d j := d ( K ǫj ) are no bigger then ǫ. Let us denote m h ( E, ǫ ) := inf X h ( 12 d ( K ǫj )) , where the infimum is taken over all the coverings of the set E by the families { K ǫj } . The quantity m h ( E ) := lim ǫ → m h ( E, ǫ )is called h -Hausdorff measure [Ca,Ch.II] . Theorem 2.5.4.1.(Properties of m h ). The following properties hold:h1) monotonicity: E ⊂ E = ⇒ m h ( E ) ≤ m h ( E ); h2) countable additivity: m h ( ∪ E j ) = X m h ( E j ); E j ∩ E i = ∅ , for i = j ; E j ∈ σ ( R m ) . We will quote two conditions (necessary and sufficient) that connect the h -measure to the capacity (see,[La,Ch.3, § Theorem 2.5.4.2.
Let cap E = 0 . Then m h ( E ) = 0 for all h such that Z h ( r ) r m − dr < ∞ . Theorem 2.5.4.3.
Let h ( r ) = r m − for m ≥ and h ( r ) = (log 1 /r ) − for m = 2 . If the h -measure of a set E is finite, then cap m ( E ) = 0 . Side by side with the Hausdorff measure the
Carleson measure (see, [Ca,Ch.II],is often considered. It is defined by m Ch ( E ) := inf X h (0 . d j ) , where infimum is taken over all the coverings of the set E with balls of radii 0 . d j . The inequality m Ch ( E ) ≤ m h ( E ) obviously holds. Let β − mes C E be the Carlesonmeasure for h = r β . The following assertion connects the β − mes C to capacity. Theorem 2.5.4.4.
The following inequalities hold β − mes C E ≤ N ( m )( cap m ( E )) β/m − , f or m ≥ , β > m − β − mes C ( E ) ≤ cap l ( E ) , f or m = 2 , β > , where N depends only on the dimension of the space. For proof see [La Ch.III, § Now we will formulate an analog of the Luzin theorem for potentials.
Theorem 2.5.5.1.
Let supp µ = K and let the potential Π( x, µ ) be bounded on K. Then for any δ > there exists a compact K ′ ⊂ K such that µ ( K \ K ′ ) < δ and the potential Π( x, µ ′ ) of the measure µ ′ := µ | K (the restriction of µ to K ) iscontinuous. For proof see, e.g., [La,Ch.3, § Theorem 2.5.5.2.
Let cap
K > . Then for arbitrary small ǫ > there ex-ists a measure µ such that supp µ ⊂ K , the potential Π( x, µ ) is continuous and µ ( K ) > cap ( K ) − ǫ. Proof.
Consider the equilibrium distribution λ K on K. Its potential is bounded byTheorem 2.5.3.4. By Theorem 2.5.5.1 we can find a mass distribution µ such thatΠ( x, µ ) is continuous, supp µ ⊂ K and µ ( K ) > λ K ( K ) − ǫ = cap ( K ) − ǫ. (cid:3) Let u ( x ) , x ∈ D ⊂ R m be a measurable function bounded from above whichcan be −∞ on a set of no more than zero measure.Let us denote as(2.6.1.1) M ( x, r, u ) := 1 σ m r m − Z S x,r u ( y ) ds y the mean value of u ( x ) on the sphere S x,r := { y : | y − x | = r } . The function M ( x, r, u ) is defined if S x,r ⊂ D , but it can be −∞ a priori.A function u ( x ) is called subharmonic if it is upper semicontinuous ,
6≡ −∞ ,and for any x ∈ D there exists ǫ = ǫ ( x ) such that the inequality(2.6.1.2) u ( x ) ≤ M ( x, r, u )holds for all the r < ǫ. The class of functions subharmonic in D will be denoted as SH ( D ) . Example 2.6.1.1.
The function u ( x ) := −| x | − m , x ∈ R m belongs to SH ( R m ) for m ≥
3, and the function u ( z ) := log | z | , z ∈ R is subharmonic in R . Example 2.6.1.2.
Let f ( z ) be a holomorphic function in a plane domain D. Thenlog | f ( z ) | ∈ SH ( D ) . Example 2.6.1.3.
Let f = f ( z , z , ..., z n ) be a holomorphic function of z =( z , ...z n ) . Then u ( x , y , ...x n , y n ) := log | f ( x + iy , ..., x n + iy n ) | is subharmonicin every pair ( x j , y j ), and, as one can see in future, in all the variables. Example 2.6.1.4.
Every harmonic function is subharmonic as it follows fromTheorem 2.4.1.6.(Mean Value). Theorem 2.6.1.1.(Elementary Properties).
The following holds:sh1) if u ∈ SH ( D ) then Cu ∈ SH ( D ) for any constant C ≥ sh2) if u , u ∈ SH ( D ) , then u + u , max [ u , u ] ∈ SH ( D ); sh3) suppose u n ∈ SH ( D ) , n = 1 , , .. , and the sequence converges to u monotonically decreasing or uniformly on every compact set in D. Then u ∈ SH ( D ); sh4) suppose u ( x, y ) ∈ SH ( D ) for all y ∈ D , and be upper semicontinuousin D × D . Let µ be a measure in D such that µ ( D ) < ∞ . Then the function u ( x ) := R u ( x, y ) dµ y is subharmonic in D . sh5) let V ∈ SO ( m ) be an orthogonal transformation of the space R m and u ∈ SH ( R m ) . Then u ( V • ) ∈ SH ( R m ) . All the assertions follow directly from the definition of subharmonic functions,properties of semicontinuous functions and properties of the Lebesgue integral. Fordetailed proof see, e.g., [HK,Ch.2].
Theorem 2.6.1.2.(Maximum Principle).
Let u ∈ SH ( D ) , G ⊂ R m and u ( x ) const. Then the inequality u ( x ) < sup x ′ ∈ ∂D lim sup y → x ′ ,y ∈ D u ( y ) , x ∈ D holds, i.e. the maximum is not attended inside the domain.The assertion follows from (2.6.1.2) and the upper semicontinuity of u ( x ) . Fordetails see [HK,Ch.2].Let K ⋐ D be a compact set with nonempty interior ◦ K, and let f n be adecreasing sequence of functions continuous in K that tends to u ∈ SH ( D ) . Sucha sequence exists by Theorem 2.1.2.9.(The second criterion of semicontinuity).Consider a sequence { H ( x, u n ) } of functions which are harmonic in ◦ K and H | ∂K = f n . The sequence converges monotonically to a function H ( x ) harmonic in ◦ K by Theorem 2.3.4.3.(Connection between convergences), Theorem 2.4.1.8.(Uni-form and D ′ -convergences) and Theorem 2.6.1.2. The limit depends only on u as one can see, i.e. it does not depend on the sequence f n . This harmonic function H ( x ) := H ( x, u, K ) is called the least harmonic majorant of u in K. This name is justified because of the following
Theorem 2.6.1.3.(Least Harmonic Majorant).
Let u ∈ SH ( D ) .Then for any K ⋐ D u ( x ) ≤ H ( x, u, K ) , x ∈ K. If h ( x ) is harmonic in ◦ K and satisfies thecondition h ( x ) ≥ u ( x ) , x ∈ ◦ K, then H ( x, u, K ) ≤ h ( x ) , x ∈ ◦ K. For proof see [HK,Ch.3].
Let us study properties of the mean values of subharmonic functions. Let M ( x, r.u ) be defined by (2.6.1.1) and N ( x, r, u ) by N ( x, r, u ) := 1 ω m r m Z K x,r u ( y ) dy, where ω m is the volume of the ball K , . Theorem 2.6.2.1.(Properties of Mean Values).
The following holds:me1) M ( x, r, u ) and N ( x, r, u ) non-decreases in r monotonically;me2) u ( x ) ≤ N ( x, • ) ≤ M ( x, • ) ;me3) lim r → M ( x, r, u ) = lim r → N ( x, r, u ) = u ( x ) . Proof.
For simplicity let us prove me1) for m = 2 . We have M ( z , | z | , u ) = 12 π Z π u ( z + ze iφ ) dφ Since u ( z, φ ) := u ( z + ze iφ ) is a family of subharmonic functions that satisfies thecondition sh4) of Theorem 2.6.1.1, M ( z , | z | , u ) is subharmonic in z on any K ,r .By Maximum Principle (Theorem 2.6.1.2) we have M ( z , r , u ) = max S ,r M ( z , | z | , u ) ≤ max S ,r M ( z , | z | , u ) = M ( z , r , u )for r < r . Monotonicity of N ( x, r, u ) follows from the equality(2.6.2.1) N ( x, r, u ) = m Z s m − M ( x, rs.u ) ds and monotonicity of M ( x, r, u ) . The property me2) follows now from the definition of a subharmonic functionand (2.6.2.1).Let us prove me3). Let M ( u, x, r ) is defined by (2.1.1.1). We have M ( x, r, u ) ≤ M ( u, x, r ) and M ( u, x, r ) → u ( x ) because of upper semicontinuity of u ( x ) . Thusme2) implies me3). (cid:3)
It is clear from me2) that a subharmonic function is locally summable. Fromme3) we have the corollary
Theorem 2.6.2.2.(Uniqueness of subharmonic function). If u, v ∈ SH ( D ) and u = v almost everywhere, then u ≡ v. Let α ( t ) be defined by the equality (2.3.1.1), α ǫ ( x ) by (2.3.1.3).For a Borel set E let E ǫ := { x : ∃ y ∈ D : | x − y | < ǫ } . This is the ǫ -extension of E ; this is ,of course, an open set.For an open set D we set D − ǫ := [ K ǫ ⊂ D K ǫ This is the maximal set such that its ǫ -extension is a subset of D. One can see that D − ǫ is not empty for small ǫ and D − ǫ ↑ D when ǫ ↓ . Therefore for any D ⋐ D there exists ǫ such that D ⋐ D − ǫ . For u ∈ SH ( D ) set(2.6.2.2) u ǫ ( x ) := Z u ( x + y ) α ǫ ( y ) dy which is defined in D − ǫ . Theorem 2.6.2.3.(Smooth Approximation).
The following holds:ap1) u ǫ is an infinitely differentiable subharmonic function in any open set D ⊂ D − ǫ . ap2) u ǫ ↓ u ( x ) while ǫ ↓ for all x ∈ D. Proof.
The property ap1) follows from sh4) (Theorem 2.6.1.1) and the followingequality that one can obtain from (2.6.2.2):(2.6.2.3) u ǫ ( x ) = Z u ( y ) α ǫ ( x − y ) dy. Exercise 2.6.2.1
Prove this.Let us prove ap2). From (2.6.2.2) we obtain(2.6.2.4) u ǫ ( x ) = Z α ( s ) s m − M ( x, ǫs, u ) ds It follows from the property me1) (Theorem 2.6.2.1) that u ǫ ≤ u ǫ while ǫ < ǫ . Now we pass to limit in (2.6.2.4). Using me3) we have M ( x, ǫs, u ) ↓ u ( x ). We canpass to the limit under the integral because of Theorem 2.2.2.2.Thuslim ǫ ↓ u ǫ ( x ) = Z α ( s ) s m − u ( x ) ds = u ( x ) (cid:3) Theorem 2.6.2.4.(Symmetry of u ǫ ). If u ( x ) depends only on | x | then u ǫ dependsonly on | x | . Proof.
Let V ∈ SO ( m ) be a rotation of R m .Then u ǫ ( V x ) = Z u ( y ) α ǫ ( V x − y ) dy. Set y = V y ′ and change the variables. We obtain u ǫ ( V x ) = Z u ( V y ′ ) α ǫ ( V ( x − y ′ )) dy. Since α ǫ = α ǫ ( | x | ) and u = u ( | x | ), α ǫ ( V y ) = α ǫ ( y ) and u ( V y ) = u ( y ). Thus u ǫ ( V x ) = u ǫ ( x ) for any V and thus u ǫ ( x ) = u ǫ ( | x | ) . (cid:3) Since a subharmonic function is locally summable and defined uniquely byits values almost everywhere,every u ∈ SH ( D ) corresponds to a (unique) distribu-tion < u, φ > := Z u ( x ) φ ( x ) dx, φ ∈ D ′ . Theorem 2.6.3.1.(Necessary Differential Condition for Subharmonicity). If u ∈ SH ( D ) , then ∆ u is a positive distribution in D ′ ( D ) . Proof.
Suppose for beginning that u ( x ) has second continuous derivatives. By using(2.4.1.5) and (2.4.1.6) we can represent u ( x ) in the form(2.6.3.1) u ( x ) = M ( x, r, u ) + Z K x,r G ( x, y, K x,r )∆ u ( y ) dy, where G is negative for all r. Suppose ∆ u ( x ) <
0. Then it is negative in K x,r for some r. Thus the integral in (2.6.3.1) is positive and we obtain that u ( x ) −M ( x, r, u ) > . This contradicts the subharmonicity of u ( x ) . Now suppose u ( x ) is an arbitrary subharmonic function. Then ∆ u ǫ ( x ) ≥ x ∈ D when ǫ is small enough.For each x there is a neighborhood D x suchthat every u ǫ defines a distribution from D ′ ( D x ) . Hence ∆ u ǫ ( x ) defines a positivedistribution from D ′ ( D x ) . Passing to the limit in u ǫ when ǫ ↓ D ′ ( D x ) a distribution that is defined by function u ( x ) . Since the Laplace operatoris continuous in any D ′ (Theorem 2.3.4.2), ∆ u > D ′ ( D x ) . From Theorem2.3.5.1 we obtain that ∆ u is a positive distribution in D ′ ( D ) . (cid:3) The distribution ∆ u can be realized as a measure by Theorem 2.3.2.2.Themeasure ( θ m ) − ∆ u is called the Riesz measure of the subharmonic function u. Theorem 2.6.3.2.(Subharmonicity and Convexity).
Let u ( | x | ) be subhar-monic in x on K ,R . Then u ( r ) is convex with respect to − r − m for m ≥ andwith respect to log r for m = 2 . Proof.
By Theorem 2.6.2.4 u ǫ ( x ) depends on | x | only, i.e., u ǫ ( x ) = u ǫ ( | x | ), and thefunction u ǫ ( r ) is smooth. Passing to the spherical coordinates we obtain∆ u ǫ = 1 r m − ∂∂r r m − ∂∂r u ǫ ( r ) ≥ . By changing variables r = e v for m = 2 or r = ( − v ) − m for m ≥ u ǫ ( r ( v )] ′′ ≥ , i.e., u ǫ ( r ( v )) is convex in v. Passing to the limit on ǫ ↓ u ( r ( v )) is convex too, as a mono-tonic limit of convex functions. (cid:3) Now we will consider the connection between subharmonicity and potentials.
Theorem 2.6.4.1.(Subharmonicity of - Π ). − Π( x, µ, D ) ∈ SH ( D )It is because of GPo1) and GPo3) (Th.2.5.1.1).The following theorem is inverse to Theorem 2.6.3.1. Theorem 2.6.4.2.(Sufficient Differential Condition of Subharmonicity).
Let ∆ u ∈ D ′ ( D ) be a positive distribution.Then there exists u ∈ SH ( D ) thatrealizes u. Proof.
Set µ := θ − m ∆ u. Let Ω ⋐ Ω ⋐ D and Π( x, µ Ω ) be the Newtonian (orlogarithmic) potential of µ | Ω . By GPo5) (Th.2.5.1.1) the difference H := u + Π isa harmonic distribution in D ′ (Ω ) . Hence there exists a “natural” harmonic function H that realizes H (Theorem 2.4.1.1). Thus the function u := H − Π ∈ SH (Ω )and realizes u in D ′ (Ω) . Since Ω and Ω can be chosen such that a neighborhoodof any x ∈ D belongs to Ω , the assertion holds for D. (cid:3) By the way, we showed in this theorem that every subharmonic function canbe represented inside its domain of subharmonicity as a difference of a harmonicfunction and a Newton potential. Thus all the smooth properties of a subhar-monic function depend on the smooth properties of the potential only because anyharmonic function is infinitely differentiable.The following representation determines the harmonic function completely.
Theorem 2.6.4.3.(F.Riesz representation).
Let u ∈ SH ( D ) and let K be acompact Lipschitz subdomain of D. Then u ( x ) = H ( x, u, K ) − Π( x, µ u , K ) where µ u is the Riesz measure of u and H ( x, u, K ) the least subharmonic majorant.Proof. We can prove as above that the function H ( x ) := u ( x ) + Π( x, µ u , K ) isharmonic in ◦ K. Since H ( x ) ≥ u ( x ) we have H ( x ) ≥ H ( x, u, K ) . So we need thereverse inequality. Let us write the same equality for u ǫ that is smooth. u ǫ := H ( x, u ǫ ) − Π( x, µ u ǫ , K ) . Passing to the limit as ǫ ↓ u ( x ) = H ( x, u, K ) − lim ǫ ↓ Π( x, µ u ǫ , K ) , and the potentials converge because other summands converge. By Gpo5) lim ǫ ↓ Π( x, µ u ǫ , K ) ≥ Π( x, µ u , K ) . Hence H ( x ) ≤ H ( x, u, K ) (cid:3) In this item we will consider subharmonic functions in the ball K R := K ,R which are h a r m o n i c in some neighborhood of the origin and write u ∈ SH ( R ) . Set M ( r, u ) := max { u ( x ) : | x | = r } µ ( r, u ) := µ u ( K r ) M ( r, u ) := M (0 , r, u ) N ( r, u ) := A ( m ) Z r µ ( t, u ) t m − dt, where A ( m ) = max(1 , m − . (2.6.5.1) Theorem 2.6.5.1.(Jensen-Privalov).
For u ∈ SH ( R )(2.6.5.2) M ( r, u ) − u (0) = N ( r, u ) , f or < r < R. Proof.
By Theorem 2.6.4.3 we have u ( x ) = 1 σ m r Z | y | = r u ( y ) r − | x | | x − y | m ds y + Z K r G ( x, y, K r ) dµ y . For x = 0 we obtain u (0) = (cid:26) − R r (cid:0) t m − − r m − (cid:1) dµ ( t, u ) + M ( r, u ) , for m ≥ − R r log rt dµ ( t, u ) + M ( r, u ) , for m = 2 . Integrating by parts gives(2.6.5.3) u (0) − M ( r, u ) = ( − µ ( t, u ) (cid:0) t m − − r m − (cid:1) | r +( m − R r µ ( t,u ) t m − dt, for m ≥ − µ ( t, u ) log rt | r + R r µ ( t,u ) t dt, for m = 2 . We have µ ( t, u ) = 0 for small t because of harmonicity of u ( x ) . Thus (2.6.5.3)implies (2.6.5.2). (cid:3) Theorem 2.6.5.2.(Convexity of M ( r, u ) and M ( r, u ) ). These functions in-crease monotonically and are convex with respect to log r for m = 2 and − r − m for m ≥ . Proof.
Consider the case m = 2 . Set M ( z ) := max φ u ( ze iφ ) . One can see that M ( r ) = M ( r, u ) . Let u be a continuous subharmonic function.Then M ( z ) is subharmonic (Theo-rem 2.6.1.1, sh5) and continuous because the family { u φ ( z ) := u ( ze iφ ) } is uniformlycontinuous. The function M ( z ) depends only on | z | . Thus it is convex with respectto log r by Theorem 2.6.3.2.Let u ( z ) be an arbitrary subharmonic function and u ǫ ↓ u while ǫ ↓ . Then M ( r, u ǫ ) ↓ M ( r, u ) by Prop. 2.1.2.7 and is convex with respect to log r by sh3,Theorem 2.6.1.1.If m ≥ M ( x ) := max | y | = | x | u ( V y x ) where V y is a rotation of R m transferring x into y. The convexity of M ( r, u ) is proved analogously. Exercise 2.6.5.1
Prove it.The monotonicity of M ( r, u ) follows from the Maximum Principle (Th. 2.6.1.2).The monotonicity of M ( r, u ) was proved in Theorem 2.6.2.1. (cid:3) The following classical assertion is a direct corollary of Theorem 2.6.5.2.
Theorem 2.6.5.3.(Three Circles Theorem of Hadamard).
Let f ( z ) be aholomorphic function in the disc K R and let M f ( r ) be its maximum on the circle {| z | = r } . Then M f ( r ) ≤ ([ M f ( r )] log r r [ M f ( r )] log rr ) r r . for < r ≤ r ≤ r < R. For proof you should write down the condition of convexity with respect tolog r of the function log M f ( r ) which is the maximum of the subharmonic functionlog | f ( z ) | . Exercise 2.6.5.2.
Do this.For details see [PS, Part I, Sec.III,Ch.6,Problem 304]. We will formulate the following analogue for the Montel theorem of normalfamilies of holomorphic functions.The family(2.7.1.1) { u α , α ∈ A } ⊂ SH ( D )is called precompact in D ′ ( D ) if, for any sequence { α n , n = 1 , , ... } ⊂ A, there existsa subsequence α n j , j = 1 , , ... and a function u ∈ SH ( D ) such that u α nj → u in D ′ ( D ) . Example 2.7.1.1. u α := log | z − α | , | α | < Example 2.7.1.2. u α := log | f α | where { f α } is a family of holomorphic functionsbounded in a domain D form a precompact family.A criterion of precompactness is given by Theorem 2.7.1.1.(Precompactness in D ′ ). A family (2.7.1.1) is precompact iffthe conditions hold:comp1) for any compact K ⊂ D a constant C ( K ) exists such that (2.7.1.2) u α ( x ) ≤ C ( K ) for all α ∈ A and x ∈ K ; comp2) there exists a compact K ⋐ D such that (2.7.1.3) inf α ∈ A max { u α ( x ) : x ∈ K } > −∞ . For proof see [H¨o,Th.4.1.9].
Theorem 2.7.1.2.
Let u n → u in D ′ ( K R ) . Then u n → u in D ′ ( S r ) for any r < R. Proof.
We have µ n → µ. Let us choose R such that r < R < R. Then u n ( x ) = H ( x, u n , K R ) − Π( x, µ n , K R )by F. Riesz theorem (Theorem 2.6.4.3). Now, we have Π( x, µ n , K R ) → Π( x, µ, K R ) in D ′ ( R ) by GPo6), Theorem2.5.1.1. Thus H ( x, u n , K R ) → H ( x, u, K R ) in D ′ ( R ) . By Theorem 2.4.1.8 H ( x, u n , K R ) → H ( x, u, K R ) uniformly on any compactset in K R , in particular, on S r . Hence H ( x, u n , K R ) → H ( x, u, K R ) in D ′ ( S r ) . Also Π( x, µ n , K R ) → Π( x, µ, K R ) in D ′ ( S r ) by GPo6), Theorem 2.5.1.1. Hence, u n → u in D ′ ( S r ) . (cid:3) We say that a sequence f n of locally summable functions converges in L loc toa locally summable function f if for any x ∈ D there exists a neighborhood V ∋ x such that R V | f n − f | dx → . Theorem 2.7.1.3 (Compactness in L loc ). Under conditions of Theorem 2.7.1.1the family (2.7.1.1) is precompact in L loc . For the proof see [H¨o, Theorem 4.1.9].
Theorem 2.7.1.4.
Let u n → u in D ′ ( K R ) . Then u + n → u + in D ′ ( K R ) . This is because u + n ( x ) ≤ M, x ∈ K, for all compacts K ⋐ K R . The following theorem shows that a subharmonic function is much more“flexible” that harmonic or analytic functions.
Theorem 2.7.2.1.
Let D ⋐ R m be a Lipschitz domain and let u ∈ SH ( D ) satisfythe condition u ( x ) < C for x ∈ D. Then for any closed domain D ⋐ D there existsa function ˜ u ( x ) := ˜ u ( x, D ) such thatext1) u ( x ) = ˜ u ( x ) f or x ∈ D ; ext2) ˜ u ( x ) = C for x ∈ ∂D ; ext3) ˜ u ∈ SH ( D ) and is harmonic in D \ D ; ext4) u ( x ) ≤ ˜ u ( x ) f or x ∈ D. The function ˜ u is defined uniquely.Proof. We can suppose without loss of generality that C = 0 , because we canconsider the function u − C. Let u ( x ) be continuous in D . Consider a harmonic function H ( x ) which iszero on ∂D and u ( x ) on ∂D . We have H ( x ) ≥ u ( x ) for x ∈ D \ D because of Theorem 2.6.1.3. Set ˜ u ( x ) = (cid:26) H ( x ) , x ∈ D \ D ; u ( x ) , x ∈ D . The function ˜ u ( x ) is subharmonic in D . For x / ∈ ∂D it is obvious, and for x ∈ ∂D it follows from u ( x ) = ˜ u ( x ) ≤ M ( x, r, u ) ≤ M ( x, r, ˜ u )for r small enough.It is easy to check that all the assertions of the theorem are fulfilled for thefunction ˜ u. Exercise 2.7.1.1
Check this.Let u ( x ) be an arbitrary subharmonic function. Consider the family u ǫ ofsmooth subharmonic functions that converges to u ( x ) decreasing monotonically ina neighborhood of D . The sequence g ( u ǫ ) converges monotonically to a subharmonicfunction that has all the properties ext1) - ext4). (cid:3) Theorem 2.7.2.2.(Continuity of ˜ • ). Let u n → u in D ′ ( D ) and u n ( x ) < in D. Then for any K ⋐ D with a smooth boundary ∂K f u n ( • , K ) → ˜ u ( • , K ) in D ′ ( D ) . For proving, we need the following auxiliary statement:
Theorem 2.7.2.3.
Let u n → u in D ′ ( D ) . Then for any smooth surface S ⋐ D and any function g ( x ) continuous in a neighborhood of S the assertion (2.7.2.1) Z S u n ( x ) g ( x ) ds x → Z S u ( x ) g ( x ) ds x holds.Proof. Since u n → u in D ′ ( D ) also the Riesz measures of the functions converge.Hence µ n ( K ) ≤ C ( K ) for some K ⋑ S. Thus, for the sequence of potentialsΠ( x, µ n ) , we have Z S Π( x, µ n ) g ( x ) ds x = Z d ( µ n ) y Z S g ( x ) ds x | x − y | m − . The inner integral is a continuous function of y as can be seen by simpleestimates.Thus the assertion (2.7.2.1) holds for potentials. Now, one can represent u n in the form u n ( x ) = H n ( x ) − Π( x, µ n )in K. The sequence H n convergences in D ′ and, hence, uniformly on S. Thus(2.7.2.1) holds for every u n . (cid:3) Proof of Theorem 2.7.2.2.
Let φ ∈ D ( D ) and supp φ ⊂ ◦ K. Then < f u n , φ > = < u n , φ > → < u, φ > = < ˜ u, φ > . Let x ∈ D \ K. Then e u n ( x ) = Z ∂K ∂G∂n y ( x, y ) u n ( y ) ds y . By Theorem 2.7.2.3. f u n ( x ) → ˜ u ( x ) for x ∈ D \ K. The sequence e u n is precompactin D ′ ( D ) . Thus every limit u of the e u n coincides with ˜ u ( x ) in ◦ K and in D \ K. Hence, u ≡ ˜ u in D ′ ( D ) . (cid:3) The property sh2), Theorem 2.6.1.1, shows that maximum of any finitenumber of subharmonic functions is a subharmonic function too. However, it is notso if the number is not finite.
Example 2.7.3.1.
Set u n ( z ) = n log | z | , n = 1 , .... The functions u n ∈ SH ( K ) . Taking supremum in n we obtain u ( z ) =: sup n u n ( z ) = (cid:26) , f or z = 0; −∞ f or z = 0 . The function is not semicontinuous, thus it is not subharmonic. However, it differsfrom a subharmonic function on a set of zero capacity. The following theorem showsthat this holds in general.
Theorem 2.7.3.1.(H.Cartan).
Let a family { u α ∈ SH ( D ) , α ∈ A } be boundedfrom above and u ( x ) := sup α ∈ A u α ( x ) . Then u ∗ ∈ SH ( D ) and the set E := { x : u ∗ ( x ) > u ( x ) } is a zero capacity set. For proving this theorem we need an auxiliary assertion Theorem 2.7.3.2.
Let Π( x, µ n , D ) be a monotonically decreasing sequence of Greenpotentials and supp µ n ⊂ K ⋐ D. Then there exists a measure µ such that the in-equality lim n →∞ Π( x, µ n , D ) ≥ Π( x, µ, D ) holds for all x ∈ D with equality outside some set of zero capacity.Proof. The sequence Π( x, µ n , D ) converges monotonically and thus in D ′ (Theorem2.3.4.3). Then µ n → µ in D ′ (Theorem 2.2.4.2.) and thus in C ∗ - topology (Theorem2.3.4.4). By GPo5)(Theorem 2.5.1.1) we havelim n →∞ Π( x, µ n , D ) ≥ Π( x, µ, D ) . Suppose that the strict inequality holds on some set E of a positive capacity. ByTheorem 2.5.2.3 one can find a compact set K ⊂ E such that cap ( K ) > . Thenthere exists a measure ν concentrated on E such that its potential Π( x, ν, D ) iscontinuous (Theorem 2.5.5.2). Thus we have Z Π( x, µ, D ) dν < Z lim n →∞ Π( x, µ n , D ) dν = lim n →∞ Z Π( x, µ n , D ) dν lim n →∞ Z Π( x, ν, D ) dµ n = Z Π( x, ν, D ) dµ = Z Π( x, µ, D ) dν. The equalities use Theorem 2.2.2.2.(B. Levy), reciprocity law ( GPo4), Theorem2.5.1.1, C ∗ - convergence of µ n and once more reciprocity law respectively. So wehave a contradiction. (cid:3) Proof Theorem 2.7.3.1.
Suppose for beginning u n ( x ) ↑ u ( x ) . We can suppose alsothat u n < . For any domain G ⋐ D the sequence ˜ u n ( x ) → u ( x ) for x ∈ G (seeTheorem 2.7.2.1), because u n ( x ) = ˜ u n ( x ) for x ∈ G. Since ˜ u n = Π( x, ˜ µ n , D ) for x ∈ D , ˜ u ( x ) = Π( x, ˜ µ, D ) = u ( x ) for x ∈ G and coincides with lim n →∞ u n ( x )outside some set E G of zero capacity. Consider a sequence of domains G n thatexhaust D. Then u ( x ) = lim n →∞ u n ( x ) outside the set E := ∪ ∞ n =1 E G n which haszero capacity by capZ1) (see item 2.5.2).Now let { u n , n = 1 , ... } be a countable set that satisfies the conditions of thetheorem. Then the sequence v n := max { u k : k = 1 , ...n } ∈ SH ( D ) and v n ↑ u. Applying the previous reasoning we obtain the assertion of the theorem also in thiscase.Let { u α , α ∈ A } be an arbitrary set satisfying the condition of the theorem.By Theorem 2.1.3.2.(Choquet’s Lemma) one can find a countable set A ⊂ A suchthat (sup A u α ) ∗ = (sup A u α ) ∗ . Since sup A u α ≤ sup A u α , we have E := { x : (sup A u α ) ∗ > sup A u α } ⊂ E := { x : (sup A u α ) ∗ > sup A u α } . Thus cap ( E ) ≤ cap ( E ) = 0 . (cid:3) Corollary of Theorem 2.7.3.1 is
Theorem 2.7.3.3.(H.Cartan +).
Let { u t , t ∈ (0; ∞ ) } ⊂ SH ( D ) be a boundedfrom above family, and v := lim sup t →∞ u t . Then v ∗ ∈ SH ( D ) and the set E := { x : v ∗ ( x ) > v ( x ) } has zero capacity.Proof. Set u n := sup t ≥ n u t , E n := { x : ( u n ) ∗ > u n } , E := ∪ E n . Since cap ( E n ) =0 , cap E = 0 too.Let x / ∈ E. Then v ( x ) = lim n →∞ sup t ≥ n u t ( x ) = lim n →∞ ( u n ) ∗ ( x ) . The function v ∗ := lim n →∞ ( u n ) ∗ ( x )is the upper semicontinuous regularization of v ( x ) for all x ∈ D. (cid:3) In spite of Example 2.7.3.1 we have
Theorem 2.7.3.4.(Sigurdsson’s Lemma). [Si]
Let S ⊂ SH ( D ) be compact in D ′ . Then v ( x ) := sup { u ( x ) : u ∈ S } is upper semicontinuous and, hence, subharmonic. Proof.
Note that u ǫ ( x ) = < u, α ( x − • ) > (see (2.6.2.3),(2.3.2.1)); and it is continuous in ( u, x ) with respect to the producttopology on ( SH ( D ) ∩ D ′ ) × R m (Theorem 2.3.4.6).Let x ∈ D, a ∈ R and assume that v ( x ) < a. We have to prove that thereexists a neighborhood X of x such that(2.7.3.1) v ( x ) < a, x ∈ X. We choose δ > v ( x ) < a − δ . If u ∈ SH ( D ) and ǫ is chosen sufficientlysmall, then u ( x ) ≤ u ǫ ( x ) < a − δ by Theorem 2.6.2.3.(Smooth Approximation).Since u ǫ ( x ) is continuous, there exists an open neighborhood U of u in SH ( D )and an open neighborhood X of x such that u ǫ ( x ) < a − δ, u ∈ U , x ∈ X . The property ap2) (Theorem 2.6.2.3) implies(2.7.3.2) u ( x ) < a − δ, u ∈ U , x ∈ X . Since u is arbitrary and S is compact, there exists a finite covering U , U , ..., U n of S and open neighborhoods X , X , ..., X n of x such that (2.7.3.2) holds for all( u, x ) : u ∈ U j , x ∈ X j , j = 1 , ..., n. Set X := ∩ j X j . Then (2.7.3.1) holds. (cid:3)
Now we are going to connect D ′ -convergence to convergence outside a zerocapacity set, the so called quasi-everywhere convergence. Theorem 2.7.4.1.( D ′ and Quasi-everywhere Convergence). Let u n , u ∈ SH ( D ) and u n → u in D ′ ( D ) . Then u ( x ) = lim sup n →∞ u n ( x ) quasi-everywhereand u ( x ) = (lim sup n →∞ u n ( x )) ∗ everywhere in D. For proof we need the following assertion in the spirit Theorem 2.7.3.2. Theorem 2.7.4.2.
Let µ n → µ in D ′ ( D ) and supp µ n ⊂ K ⋐ D. Then lim inf n →∞ Π( x, µ n , D ) ≥ Π( µ, D ) with equality quasi-everywhere.Proof. The inequality was in GPo5), Theorem 2.5.1.1.Suppose the set E := { x : lim inf n →∞ Π( x, µ n , D ) > Π( x, µ, D )has a positive capacity. By Theorem 2.5.2.3 one can find a compact set K ⊂ E such that cap ( K ) > . By Theorem 2.5.5.2 one can find a measure ν concentratedon K with continuous potential. As in proof of Theorem 2.7.3.2 we have Z Π( x, µ, D ) dν < Z lim inf n →∞ Π( x, µ n , D ) dν ≤ lim inf n →∞ Z Π( x, µ n , D ) dν =lim inf n →∞ Z Π( x, ν, D ) dµ n = Z Π( x, ν, D ) dµ = Z Π( x, µ, D ) dν. The second inequality uses Theorem 2.2.2.3.(Fatou’s Lemma). The equalities usethe reciprocity law ( GPo4), Theorem 2.5.1.1), C ∗ - convergence of µ n and oncemore reciprocity law respectively. So we have a contradiction. (cid:3) Proof of Theorem 2.7.4.1.
Let D ⋐ D. Then the sequence u n is bounded in D by Theorem 2.7.1.1. We can assume that u n ( x ) < x ∈ D . For any domain G ⋐ D the sequence ˜ u n ( x, G ) → u ( x ) in D ′ ( D ) by Theorem2.7.2.2. We also have the equality ˜ u n = − Π( x, ˜ µ n , D ) . Thus ˜ µ n → ˜ µ in D ′ ( D ) . By Theorem 2.7.4.2 lim inf n →∞ Π( x, ˜ µ n , D ) = Π( x, ˜ µ, D ) quasi-everywhere in D .Hence(2.7.4.1) lim sup n →∞ u n = u quasi-everywhere in G because u n ( x ) = ˜ u n ( x ) for x ∈ G. Consider a sequence of domains G n that exhaust D. Then (2.7.4.1) holds out-side a set E n of zero capacity and (2.7.4.1) holds in D outside the set E := ∪ ∞ n =1 E n which has zero capacity by capZ1) (see item 2.5.2), i.e., quasi-everywhere. (cid:3) Now we connect the convergence of subharmonic functions in D ′ to theconvergence relative to the Carleson measure (see 2.5.4).We say that a sequence of functions u n converges to a function u relative to the α - Carleson measure if the sets E n := { x : | u n ( x ) − u ( x ) | > ǫ } possess the property(2.7.5.1) α − mes C E n → . Theorem 2.7.5.1.( D ′ and α − mes C Convergences).
Let u n , u ∈ SH ( D ) and u n → u in D ′ ( D ) . Then for an every α > and every domain G ⋐ D u n → u relative to the ( α + m − -Carleson measure. For proving this theorem we need some auxiliary definitions and assertions.Let µ be a measure in R m . We will call a point x ∈ R m ( α, α ′ , ǫ )- normal withrespect to the measure µ , ( α < α ′ ) if the inequality µ x ( t ) := µ ( K x,t ) < ǫ − α ′ t α + m − holds for all t < ǫ. Theorem 2.7.5.2.
In any ( α, α ′ , ǫ ) -normal point the following inequality holds − Z K z,ǫ [log | z − ζ | − log ǫ ] dµ ζ ≤ Cǫ α − α ′ , f or m = 2; Z K x,t [ | x − y | − m − ǫ − m ] dµ y ≤ Cǫ α − α ′ , f or m ≥ while C = C ( α, m ) depends on α and m only.Proof. Let us consider the case m = 2 . We have Z K z,ǫ log ǫ | z − ζ | dµ ζ = Z ǫ log ǫt dµ z ( t ) . Integrating by parts we obtain Z K z,ǫ log ǫ | z − ζ | dµ ζ = log ǫt µ z ( t ) | ǫ + Z ǫ µ z ( t ) t dt ≤≤ ǫ − α ′ Z ǫ t α − dt = 1 α ǫ α − α ′ . Let us consider the case m ≥ . We have Z K x,t [ | x − y | − m − ǫ − m ] dµ y = Z ǫ ( t − m − ǫ − m ) dµ x ( t ) == ( t − m − ǫ − m ) µ x ( t ) | ǫ +( m − Z ǫ µ x ( t ) t m − dt ≤≤ m − ǫ α ′ Z ǫ t α − dt = m − α ǫ α − α ′ (cid:3) Theorem 2.7.5.3.(Ahlfors-Landkof Lemma).
Let a set E ⊂ R m be coveredby balls with bounded radii such that every point is a center of a ball. Then thereexists an at most countable subcovering of the same set with maximal multiplicity cr = cr ( m ) , i.e., every point of E is covered no more than cr times.For proof see [La, Ch.III, §
4, Lemma 3.2].
Theorem 2.7.5.4.
Let K ⋐ D. The set E := E ( α, α ′ , ǫ, µ ) of points that belongto K and are not ( α, α ′ , ǫ ) -normal with respect to µ satisfies the condition (2.7.5.2) ( α + m − − mes C E ≤ cr ( m ) ǫ α ′ µ ( K ǫ ) where K ǫ is the ǫ -extension of K. Proof.
Let x ∈ E. Then there exists t x such that µ x ( t x ) ≥ t α + m − x ǫ − α ′ . Thus every point of E is covered by a ball K x,t x . By the Ahlfors-Landkof lemma(Theorem 2.7.5.3) one can find a no more than cr -multiple subcovering { K x j ,t xj } . Then we have X j t α + m − x j ≤ cr ( m ) ǫ α ′ µ ( K ǫ ) . By definition of the Carleson measure we obtain (2.7.5.2). (cid:3) Theorem 2.7.5.5.
Let µ n → µ in D ′ ( R m ) and supp µ n ⊂ K ⋐ R m . Then forevery α > and G ⋐ R m Π( x, µ n ) → Π( x, µ ) relative to the ( α + m − -Carlesonmeasure.Proof. Let m = 2 . Setlog ǫ | z − ζ | = (cid:26) log | z − ζ | , f or | z − ζ | > ǫ log ǫ, f or | z − ζ | ≤ ǫ . This function is continuous for ( z, ζ ) ∈ K × K. Set ν n := µ n − µ. Then we have − Z log | z − ζ | d ( µ n ) ζ + Z log | z − ζ | dµ ζ = − Z log | z − ζ | d ( ν n ) ζ == − Z log ǫ | z − ζ | dν n − Z K z,ǫ [log | z − ζ | − log ǫ ] dν n . The function log ǫ | z − ζ | is continuous in ζ uniformly over z ∈ K. Thus the sequenceΠ ǫ ( z ) := Z log ǫ | z − ζ | dν n converges uniformly to zero on K. Suppose now that z / ∈ E ( α, α ′ , ǫ, µ ) ∪ E ( α, α ′ , ǫ, µ n ) , i.e., it is an ( α, α ′ , ǫ )- normal point for µ and µ n . By Theorem 2.7.5.2 we have Z K z,ǫ [log | z − ζ | − log ǫ ] dν n < Cǫ α − α ′ . Thus for sufficiently large n > n ( ǫ ) | Π( z, µ n ) − Π( z, µ ) | == | Z log | z − ζ | d ( µ n ) ζ − Z log | z − ζ | dµ ζ | < δ = δ ( ǫ )while z / ∈ E ( α, α ′ , ǫ, µ ) ∪ E ( α, α ′ , ǫ, µ n ) := E n ( ǫ ) . By Theorem 2.7.5.3 the Carleson measure of E n ( ǫ ) satisfies the inequality α − mes C E n ( ǫ ) ≤ cr ( m ) ǫ α ′ [ µ ( K ) + µ n ( K )] ≤ Cǫ α ′ := γ ( ǫ )where C = C ( K ) does not depend on n because µ n ( K ) are bounded uniformly. Hence, for any ǫ > E ′ n ( ǫ ) := { z : | Π( z, µ n ) − Π( z, µ ) | > δ ( ǫ ) } satisfies the condition(2.7.5.2) α − mes C E ′ n ( ǫ ) ≤ γ ( ǫ ) . while n > n = n ( ǫ ) . Let us show that Π( z, µ n ) → Π( z, µ ) relative to α − mes C on K. Let γ , δ be arbitrary small. One can find ǫ such that δ ( ǫ ) < δ , γ ( ǫ ) < γ . One can find n = n ( ǫ ) such that (2.7.5.2) is fulfilled. Now the set E n,δ := { z : | Π( z, µ n ) − Π( z, µ ) | > δ } is contained in E ′ n ( ǫ ) . Thus α − mes C E n,δ < γ and this implies the convergencerelative to α − mes C . An analogous reasoning works for m ≥ . (cid:3) Proof of Theorem 2.7.5.1.
Let u n → u in D ′ . One can assume that u n , u arepotentials on any compact set(Theorem 2.7.2.2). Hence, by Theorem 2.7.5.5 itconverges relative ( α + m − − mes C . (cid:3) Let A be a class of nondecreasing functions a ( r ) , r ∈ (0 , ∞ ) such that a ( r ) ≥ a ( r ) → ∞ when r → ∞ . The quantity(2.8.1.1) ρ [ a ] := lim sup r →∞ log a ( r )log r is called the order of a ( r ) . Suppose ρ := ρ [ a ] < ∞ . The number(2.8.1.2) σ [ a ] := lim sup r →∞ a ( r ) r ρ is called the type number .If σ [ a ] = 0, we say a ( r ) has minimal type . If 0 < σ [ a ] < ∞ , a ( r ) has normaltype . If σ [ a ] = ∞ , it has maximal type . Example 2.8.1.1.
Set a ( r ) := σ r ρ . Then ρ [ a ] = ρ , σ [ a ] = σ . Example 2.8.1.2.
Set a ( r ) := (log r ) − r ρ . Then ρ [ a ] = ρ , σ [ a ] = 0 . Example 2.8.1.3.
Set a ( r ) := (log r ) r ρ . Then ρ [ a ] = ρ , σ [ a ] = ∞ . Theorem 2.8.1.1(Convergence Exponent).
The following equality holds: (2.8.1.3) ρ [ a ] = inf { λ : Z ∞ a ( r ) drr λ +1 < ∞} . If the integral converges for λ = ρ [ a ] , a ( r ) has minimal type. Exercise 2.8.1.1.
Prove this.For proof see, e.g., [HK, § Example 2.8.1.4.
Let r j , j = 1 , , ... be a nondecreasing sequence of positivenumbers. Let us concentrate the unit mass in every point r j and define a massdistribution n ( E ) := { the number points of the sequence { r j } in E } , E ⊂ R . Then(2.8.1.4) Z ∞ dnr λ = ∞ X r λj . The infimum of λ for which the series in (2.8.1.4) converges is usually called the con-vergence exponent for the sequence { r j } [ PS P.I,Sec.1,Ch.III, § a ( r ) = n (( −∞ , r )) . Theorem 2.8.1.1 shows that the convergenceexponent coincides with the order of this a ( r ) . A function ρ ( r ) is called a proximate order with respect to order ρ ifpo1) ρ ( r ) ≥ r →∞ ρ ( r ) = ρ po3) ρ ( r ) has a continuous derivative on (0 , ∞ )po4) lim r →∞ r log rρ ′ ( r ) = 0 . Two proximate orders ρ ( r ) and ρ ( r ) are called equivalent , if(2.8.1.5) ρ ( r ) − ρ ( r ) = o (cid:18) r (cid:19) . For a ∈ A set(2.8.1.6) σ [ a, ρ ( r )] := lim sup r →∞ a ( r ) r ρ ( r ) . It is called a type number with respect to a proximate order ρ ( r ) . It is clear thatthis type number is the same for equivalent proximate orders.
Theorem 2.8.1.2.(Proper Proximate Order).
Let a ∈ A and ρ [ a ] = ρ < ∞ . Then there exists a proximate order ρ ( r ) such that (2.8.1.7) 0 < σ [ a, ρ ( r )] < ∞ . For proof see [L(1980), Ch.1, Sec.12, Th.16].If a proximate order satisfies the condition (2.8.1.7), we will call it the proper proximate order of a ( r ) (p.p.o.). The function r ρ ( r ) inherits a lot of useful propertiesof the power function r ρ . Theorem 2.8.1.3.(Properties of P.O).
The following holds:ppo1) the function V ( r ) := r ρ ( r ) increases monotonically for sufficiently largevalues of r. ppo2) for q < ρ + 1 , Z r t ρ ( t ) − q dt ∼ r ρ ( r )+1 − q ρ + 1 − q and for q > ρ + 1 , Z ∞ r t ρ ( t ) − q dt ∼ r ρ ( r )+1 − q q − ρ − as r → ∞ . ppo3) the function L ( r ) := r ρ ( r ) − ρ satisfies the condition ∀ δ > , L ( kr ) /L ( r ) → when r → ∞ uniformly for k ∈ [ δ , δ ] . Exercise 2.8.1.2.
Prove these properties.For proof see, e.g., [L(1980), Ch.2, Sec12]. The following assertion allows toreplace any p.o. for a smooth one.
Theorem 2.8.1.4.(Smooth P.O).
Let ρ ( r ) be an arbitrary p.o. There exists aninfinitely differentiable equivalent p.o. ρ ( r ) such that (2.8.1.8) r k log rρ ( k )1 ( r ) → , k = 1 , , ... when r → ∞ . Proof.
Let α ǫ be defined by (2.3.1.3). Set ǫ := 0 . , po ( x ) := ρ ( e x ) and po ( x ) := po ( n ) + [ po ( n + 1) − po ( n )] Z xn α . ( t + 0 . dt for x ∈ [ n, n + 1) . The function po ( x ) is continuous and infinitely differentiable dueto properties of α ǫ and po ( n ) = po ( n ) for n = 1 , , ... . By property po3) of p.o.we have ( n + 1) | po ( n + 1) − po ( n ) | ≤ n + 1 n max y ∈ [ n,n +1] | y · po ′ ( y ) | → n → ∞ . Thusmax y ∈ [ n,n +1] | y · po ( k )1 ( y ) | ≤ const · ( n + 1) | po ( n + 1) − po ( n ) | → n → ∞ . So ρ ( r ) := po (log r ) is a p.o. that satisfies (2.8.1.8). Let us show that it isequivalent to ρ ( r ) . Indeed | po ( x ) − po ( x ) | = | Z xn [ po ( y ) − po ( y )] ′ y dyy | ≤≤ max y ∈ [ n,n +1] [ | y · po ′ ( y ) | + | y · po ′ ( y ) | ] log n + 1 n = o (cid:18) x (cid:19) , when x ∈ [ n, n + 1] and n → ∞ . (cid:3) We will further need (in 2.9.3) the following assertion
Theorem 2.8.1.5.(A.A.Goldberg).
Let ρ ( r ) → ρ be a p.o., and let f ( t ) be afunction that is locally summable on (0 , ∞ ) and such that (2.8.1.9) lim t → t ρ + δ f ( t ) = lim t →∞ t ρ +1+ γ f ( t ) = 0 for some < δ, γ < . Then lim r →∞ r − ρ ( r ) Z xcr − ( rt ) ρ ( rt ) f ( t ) dt = Z x t ρ f ( t ) dt lim r →∞ r − ρ ( r ) Z ∞ x ( rt ) ρ ( rt ) f ( t ) dt = Z ∞ x t ρ f ( t ) dt (2.8.1.10) for any c > and any x ∈ (0 , ∞ ) . Proof.
Set I ( r ) := Z ∞ cr − ( rt ) ρ ( rt ) r ρ ( r ) f ( t ) dt. It will be enough to prove that(2.8.1.11) lim r →∞ I ( r ) = Z ∞ t ρ f ( t ) dt because both functions f ( t, x ) := (cid:26) f ( t ) , f or t ∈ (0 , x )0 f or t ∈ [ x, ∞ )and f ∞ ( t, x ) := f ( t ) − f ( t, x ) also satisfy the condition of the theorem.Let us represent the integral as the following sum: (2.8.1.12) I ( r ) := Z ∞ cr − ( rt ) ρ ( rt ) r ρ ( r ) f ( t ) dt = I ( r, ǫ ) + I ( r, ǫ ) + I ( r, ǫ ) , where I ( r, ǫ ) := Z ǫcr − ( rt ) ρ ( rt ) r ρ ( r ) f ( t ) dtI ( r, ǫ ) := Z ǫ − ǫ ( rt ) ρ ( rt ) r ρ ( r ) f ( t ) dtI ( r, ǫ ) := Z ∞ ǫ − ( rt ) ρ ( rt ) r ρ ( r ) f ( t ) dt. We can represent I ( r, ǫ ) in the form I ( r, ǫ ) = Z ǫ − ǫ L ( rt ) L ( r ) t ρ f ( t ) dt. By ppo3) (Theorem 2.8.1.3),(2.8.1.13) lim r →∞ I ( r, ǫ ) = Z ǫ − ǫ t ρ f ( t ) dt. Let us estimate the “tails”. From (2.8.1.9) we have | f ( t ) | ≤ Ct − ρ − δ for 0 < t ≤ ǫ where C does not depend on ǫ and | f ( t ) | ≤ Ct − ρ − − γ for t ≥ ǫ − . We have | I ( r, ǫ ) | ≤ C Z ǫcr − ( rt ) ρ ( rt ) r ρ ( r ) t − ρ − δ dt := CJ ( r, ǫ )and(2.8.1.14) lim sup r →∞ | I ( r, ǫ ) | ≤ C lim r →∞ J ( r, ǫ )Let us calculate the last limit.We perform the change x = tr : J ( r, ǫ ) = r − ρ ( r )+ ρ + δ − Z ǫrc t − ρ ( x ) − ( ρ + δ ) dx. Now we use ppo2) for q = ρ + δ and ppo3):lim r →∞ J ( r, ǫ ) = 11 − δ lim r →∞ ( ǫr ) ρ ( ǫr ) − ( ρ + δ )+1 r ρ ( r ) − ( ρ + δ )+1 == ǫ − δ − δ lim r →∞ L ( ǫr ) L ( r ) = ǫ − δ − δ Substituting in (2.8.1.14) we obtain(2.8.1.15) lim sup r →∞ | I ( r, ǫ ) | ≤ C ǫ − δ − δ . Analogously one can obtain(2.8.1.16) lim sup r →∞ | I ( r, ǫ ) | ≤ C ǫ γ γ . Using (2.8.1.13), (2.8.1.15) and (2.8.1.16) one can pass to the limit in (2.8.1.12) as r → ∞ then let ǫ →
0, and obtain (2.8.1.11). (cid:3)
Let(2.8.2.1) u ( x ) := u ( x ) − u ( x )where u , u ∈ SH ( R m ) , u (0) > −∞ , u (0) = 0 and µ := µ u , µ := µ u areconcentrated on disjoined sets.Let m = 2 , u j ( z ) := log | f j ( z ) | , j = 1 , f j ( z ) , j = 1 , u ( z ) = log | f ( z ) | , where f ( z ) := f ( z ) /f ( z ) , is mero-morphic. The condition for masses means that f and f have no common zeros, u (0) = 0 corresponds to f (0) = 1 and u (0) > −∞ means f (0) = 0.The class of such functions is denoted as δSH ( R m ) . In spite of the standard-ization conditions the representation (2.8.2.1) is not unique. However for any pairof representations u − u and u ′ − u ′ (2.8.2.2) u j ( x ) − u ′ j ( x ) = H j ( x ) , j = 1 , H j are harmonic and H j (0) = 0 . Really, from the equality u − u = u ′ − u ′ we obtain µ − µ = µ ′ − µ ′ . Usingthe Theorem 2.2.1.2.(Jordan decomposition) we obtain µ = µ ′ , µ = µ ′ . Thus(2.8.2.2) holds. Obviously H (0) = 0 . Set(2.8.2.3) T ( r, u ) := 1 σ m Z | y | =1 max( u , u )( ry ) dy where σ m is the surface square of the unit sphere. It is called the Nevanlinnacharacteristic of u ∈ δSH ( R m ) . The Nevanlinna characteristic does not depend on the representation (2.8.2.1).Indeed, Z | y | =1 max( u , u )( ry ) dy = Z | y | =1 [( u − u ) + ( ry ) − u ( ry )] dy == Z | y | =1 [( u ′ − u ′ ) + ( ry ) − u ′ ( ry ) + H ( rx )] dy = Z | y | =1 [max( u ′ , u ′ )( ry ) + H ( rx )] dy == Z | y | =1 max( u ′ , u ′ )( ry ) dy + H (0) = Z | y | =1 max( u ′ , u ′ )( ry ) dy. Note also that the class δSH ( R m ) is linear.Actually, let u ∈ δSH ( R m ) . Then λu ∈ δSH ( R m ) for λ > . − u ∈ δSH ( R m ) , since − u ( x ) = [ u ( x ) − u (0)] − [ u ( x ) − u (0)] . Let us show that u + u ∈ δSH ( R m ) if u, v ∈ δSH ( R m ) . Set ν := ν u + ν v , where ν u , ν v are the corresponding charges. By Theorem2.2.1.2.(Jordan decomposition) ν = ν + − ν − , where ν u , ν v are measures concentratedon disjoint sets.Let u be a subharmonic function in R m the mass distribution of which coin-cides with ν + . Then u := u − ( u + v ) is a subharmonic function with the massdistribution ν − . Hence u ( x ) + v ( x ) = [ u ( x ) − u (0)] − [ u ( x ) − u (0)] . Theorem 2.8.2.1.(Properties T ( r, u ) ). The following holdst1) T ( r, u ) increases monotonically and is convex with respect to − r m − for m = 2 and with respect to log r for m = 2 We will give a construction of such function for the case of finite order (item 2.9.2), but itis possible actually always, see ,for example, [HK,Th.4.1]3 t2) For u ∈ SH ( R m ) , (i.e. u ≡ ) T ( r, u ) = 1 σ m Z | y | =1 u + ( ry ) dy t3) T ( r, u ) = T ( r, − u ) − u (0) t4) T ( r, u + u ′ ) ≤ T ( r, u ) + T ( r, u ′ ) , T ( r, λu ) = λT ( r, u ) for λ > . Proof.
Since v ( x ) := max( u , u )( x ) is subharmonic, t1) follows from Theorem2.6.5.2.(Convexity of M ( r, u ) and M ( r, u )).The property t2) is obvious, t3) follows from the equality − u ( x ) = u ( x ) − [ u ( x ) − u (0)] − u (0) . The properties t4) follow from the properties of maximum and t3). (cid:3)
Set ρ T [ u ] := ρ [ a ] (see, (2.8.1.1)) where a ( r ) := T ( r, u ) . It is called the order of u ( x ) with respect to T ( r ) . Theorem 2.8.2.2.( ρ T -property). For u , u ∈ δSH ( R m ) the following inequalityholds: (2.8.2.4) ρ T [ u + u ] ≤ max( ρ T [ u ] , ρ T [ u ]) , Equality in (2.8.2.4) is attained if ρ T [ u ] = ρ T [ u ] . Proof.
Set u := u + u . From t3) and t4) T ( r, u ) ≤ T ( r, u ) + T ( r, u ) + O (1) ≤ T ( r, u ) , T ( r, u )] + O (1) . From the definition of ρ T we obtain (2.8.2.4).Suppose, for example, ρ T [ u ] > ρ T [ u ] . Let us show that ρ T [ u ] = ρ T [ u ] . Fromthe equality u = u + ( − u ) we obtain ρ T [ u ] ≤ max( ρ T [ u ] , ρ T [ u ] If ρ T [ u ] <ρ T [ u ] , then from the previous inequality we would have the contradiction ρ T [ u ] <ρ T [ u ] . (cid:3) Let us define σ T [ u ] by (2.8.1.2) while ρ := ρ T [ u ] . Set also σ T [ u, ρ ( r )] := σ [ a, ρ ( r )] (see (2.8.1.6)), where a ( r ) := T ( r, u ) . The characteristics ρ T [ u ] , σ T [ u ] , σ T [ u, ρ ( r )] are defined for u ∈ δSH ( R m ) . For the class of subharmonic function we have the inclusion SH ( R m ) ⊂ δSH ( R m ) and, of course, all these characteristics can be applied to a subharmonic function.However, for the class SH ( R m ) the standard characteristic of growth is M ( r, u )that we can not apply to a δ − subharmonic function u ∈ δSH ( R m ) . Thus for u ∈ SH ( R m ) we define new characteristics ρ M [ u ] , σ M [ u ] , σ M [ u, ρ ( r )] in the sameway by replacing T ( r, u ) for M ( r, u ) . The following theorem shows that there is nota big difference between characteristics with respect T and M for u ∈ SH ( R m ) . Theorem 2.8.2.3.(T and M -characteristics).
Let u ∈ SH ( R m ) and ρ ( r )( → ρ ) any p.o.Then ρ MT1) ρ T [ u ] and ρ M [ u ] are finite simultaneously and ρ T [ u ] = ρ M [ u ] := ρ [ u ] ρ MT2) there exists A := A ( ρ, m ) such that Aσ M [ u, ρ ( r )] ≤ σ T [ u, ρ ( r )] ≤ σ M [ u, ρ ( r )]In particular, the last property means that the types with respect to T ( r ) and M ( r ) for the same p.o. are minimal, normal or maximal at the same time. Proof.
From t2), Theorem 2.8.2.1 we have T ( r, u ) ≤ M ( r, u ) for u ∈ SH ( R m ) . Thus ρ T [ u ] ≤ ρ M [ u ], proving the second part of ρ MT2).Let H ( x ) be the least harmonic majorant of u ( x ) in the ball K R . By thePoisson formula (Theorem 2.4.1.5) and Theorem 2.6.1.3(2.8.2.5) M ( R, u ) ≤ M ( R, H ) = max | x | = R σ m R Z | y | =2 R u ( y ) (4 R − | x | ) | x − y | m ds y ≤ (2.8.2.5) ≤ m − σ m Z | y | =1 | u (2 Ry ) | ds y = 2 m − [ T (2 R, u )+ T (2 R, − u )] = 2 m − [2 T (2 R, u ) − u (0)] . From here one can obtain ρ T [ u ] ≥ ρ M [ u ] . The left side of ρ MT2) with A ( ρ, m ) :=2 − ρ − m +2 follows from the properties of p.o. (cid:3) Exercise 2.8.2.1
Prove the first inequality from ρ MT2).
Let µ be a mass distribution (measure) in R m ( µ ∈ M ( R m )).The charac-teristic ρ [ µ ] := ρ [ a ] − m + 2 for a ( r ) := µ ( K r ) (see (2.8.1.1)) is called the convergence exponent of µ , and¯∆[ µ ] := σ [ a ]for the same a (see (2.8.1.2)) is called the upper density of µ. The least integer number p for which the integral(2.8.3.1) Z ∞ µ ( t ) t p + m dt converges is called the genus of µ and is denoted p [ µ ] . Theorem 2.8.3.1.(Convergence Exponent and Genus).
The following holds:ceg1) p [ µ ] ≤ ρ [ µ ] ≤ p [ µ ] + 1 ceg2) for ρ [ µ ] = p [ µ ] + 1 , ¯∆[ µ ] = 0 . Proof.
From Theorem 2.8.1.1 (Convergence Exponent) we have ρ [ µ ] + 1 + m − ≤ p [ µ ] + m. Thus ρ [ µ ] ≤ p [ µ ] + 1 . The same theorem implies ρ [ µ ] + m − ≥ p [ µ ] + m − . Thus p [ µ ] ≤ ρ [ µ ] , and ceg1) is proved.Let ρ ( µ ) = p [ µ ] + 1 . Then the integral (2.8.3.1) converges for p [ µ ] = ρ [ µ ] − . We use the inequality Z ∞ r µ ( t ) t ρ [ µ ]+ m − dt ≥ µ ( r ) Z ∞ r dtt ρ [ µ ]+ m − dt = µ ( r ) r ρ [ µ ]+ m − ( ρ [ µ ] + m − − . Since the left side of the inequality tends to zero we obtain¯∆[ µ ] = lim r →∞ µ ( r ) r ρ [ µ ]+ m − = 0 . (cid:3) Set(2.8.3.2) ¯∆[ µ, ρ ( r )] := σ [ a, ρ ( r ) + m − , where a ( r ) := µ ( r ) (see (2.8.1.6)). It is clear that ρ ( r ) + m − N ( r, µ ) := A ( m ) Z r µ ( t ) t m − dt, where A ( m ) = max(1 , m − . Set also ρ N [ µ ] := ρ [ a ] , ¯∆ N [ µ, ρ ( r )] := σ [ a, ρ ( r )] , where a ( r ) := N ( r, µ ) . This is the N-order of µ and the N-type of µ with respectto p.o. ρ ( r ) . Theorem 2.8.3.2.(N-order and Converges Exponent).
The following holds:Nce1) ρ N [ µ ] and ρ [ µ ] are finite simultaneously and ρ N [ µ ] = ρ [ µ ] Nce2) for ρ > there exists such A j := A j ( ρ, m ) , j = 1 , , that A ¯∆[ µ, ρ ( r )] ≤ ¯∆ N [ µ, ρ ( r )] ≤ A ¯∆[ µ, ρ ( r )] . Proof.
We have the inequality N (2 r, µ ) ≥ A ( m ) Z rr µ ( t ) t m − dt ≥ A ( m ) µ ( r ) Z rr dtt m − ≥≥ A ( m ) B ( m ) µ ( r )(2 r ) m − , where B ( m ) := 1 − − m for m ≥ B (2) := log 2 . From here one can obtain the inequality ρ [ µ ] ≥ ρ N [ µ ] and the left side of Nce2)for A ( ρ, m ) := A ( m ) B ( m )2 − ρ . For proving the opposite inequalities we use thel’Hˆopital rule (slightly improved):lim sup r →∞ N ( r, µ ) r ρ ( r ) ≤ lim sup r →∞ N ′ ( r, µ )( r ρ ( r ) ) ′ == lim sup r →∞ µ ( r ) r − m r ρ ( r ) [ ρ ( r ) + r log rρ ′ ( r )] = 1 ρ ¯∆[ µ ] . Thus ρ N [ µ ] ≤ ρ [ µ ] and the right side of Nce2) holds. (cid:3) We shall denote as δ M ( R m ) the set of charges (signed measures) of the form ν := µ − µ where µ , µ ∈ M ( R m ) . Let us remember that | ν | ∈ M ( R m ) is thefull variation of ν (see 2.2.1). Theorem 2.8.3.3.(Jensen).
Let u := u − u ∈ δSH ( R m ) and ν := µ − µ be acorresponding charge.ThenJ1) ρ [ | ν | ] ≤ max( ρ [ µ ] , ρ [ µ ]) ≤ ρ [ u ] J2) ¯∆[ | ν | , ρ ( r )] ≤ ¯∆[ µ , ρ ( r )] + ¯∆[ µ , ρ ( r )] ≤ Aσ T [ u, ρ ( r )] for some A := A ( ρ, m ) . Proof.
We can suppose without loss of generality that u (0) = 0 because the function u ( x ) − u (0) has the same order and the same number type if ρ > . We apply theJensen-Privalov formula (Theorem 2.6.5.1) to the functions u , u and obtain N ( r, µ j ) ≤ M ( r, u j ) ≤ T ( r, u ) . Thus N ( r, | ν | ) ≤ N ( r, µ ) + N ( r, µ ) ≤ T ( r, u ) . From here one can obtain J1) andJ2) for ρ N [ | ν | ] and ¯∆ N [ | ν | , ρ ( r )] . However, we can delete the subscript N becauseof Theorem 2.8.3.2. (cid:3) R m . Set(2.9.1.1) H ( z, cos γ, m ) := ( − log( z − z cos γ + 1) , f or m = 2( z − z cos γ + 1) − m − , f or m ≥ H ( z, cos γ, m ) is holomorphic on z in the disk {| z | < } . It can berepresented there in the form(2.9.1.2) H ( z, cos γ, m ) = ∞ X k =0 C m − k (cos γ ) z k where every coefficient C βk ( • ) , k = 0 , , ... is a polynomial of degree k. Such polynomials are called the
Gegenbauer polynomials . Note that C k ( • )are the Legendre polynomials and C k ( λ ) = (cid:26) , f or k = 0 k cos( k arccos λ ) , f or k ≥ , i.e. they are proportional to the Chebyshev polynomials.Thus for m = 2 we have the equality −
12 log( z − z cos γ + 1) = ∞ X k =1 cos kγk z k that can be checked directly.Let x ∈ R m . Set x := x/ | x | . Then the scalar product ( x , y ) is equal to cos γ where γ is the angle between x and y. Let E m ( x ) be defined by (2.4.1.1). For m ≥ E m ( x − y ) is theGreen function for R m . One can see that it is represented in the form G ( x, y, R m ) := E m ( x − y ) = −| y | − m H ( | x | / | y | , cos γ, m )where cos γ = ( x , y ) . For m = 2 the function − H ( | x | / | y | , cos γ,
2) plays the same role. Thus we willdenote it as G ( x, y, R ) . Theorem 2.9.1.1.(Expansion of G ( x, y, R m ) ). The following holds: (2.9.1.3) G ( x, y, R m ) = − ∞ X k =0 C m − k (cos γ ) | x | k | y | k + m − , for | x | < | y | , and the functions (2.9.1.4) D k ( x, y ) := C m − k (cos γ ) | x | k | y | k + m − are homogeneous harmonic functions in x and harmonic in y for y = 0 . Proof.
The expansion (2.9.1.3) follows from (2.9.1.2). The function G ( zx, y, R m ) isharmonic for | x | < | y | and, hence, for any real 0 ≤ z < . Hence, for any ψ ∈ D ( K r )while r := 0 . | y | the function g ( z ) := < G ( z • , y, R m ) , ∆ ψ > = 0 for z ∈ (0 , . Thefunction g is holomorphic for all the complex z ∈ {| z | < } because G ( zx, y, R m ) isholomorphic. Thus g ( z ) ≡ , i.e. all its coefficients are zero.From the expansion (2.9.1.3) we can see that the coefficients of G ( zx, y, R m )are D k ( x, y ) . Hence, < D k ( • , y ) , ∆ ψ > = 0 for every ψ ∈ D ( K r ) . Thus D k ( • , y ) isa harmonic distribution . By Theorem 2.4.1.1 it is an ordinary harmonic functionfor | x | < . | y | .C m − k (cos γ ) is a polynomial of degree k with respect to ( x , y ) . Thus D k ( x, y )is a homogeneous polynomial of x and is harmonic for all x. Let us prove the harmonicity in y. By Theorem 2.4.1.10 the function D k ( y ∗ , x ) | y | − m ( ∗ stands for inversion) is harmonic in y. We have D k ( y ∗ , x ) | y | − m = | y | − m D k ( y/ | y | , x ) = D k ( x , y ) . (cid:3) Set(2.9.1.5) H ( z, cos γ, m, p ) = H ( z, cos γ, m ) − p X k =0 C m − k (cos γ ) z k Theorem 2.9.1.2.
The following holds: (2.9.1.6) | H ( z, cos γ, m, p ) | ≤ A ( m, p ) | z | p +1 for | z | ≤ / , and (2.9.1.7) | H ( z, cos γ, m, p ) | ≤ A ( m, p ) | z | p for | z | ≥ , − π < arg z ≤ π. The factor | z | p should be replaced by log | z | if m = 2 , p = 0 . Proof.
Consider the function φ ( z ) := H ( z, cos γ, m, p ) z − p − . It is holomorphic inthe disk {| z | ≤ / } . We apply the maximum principle and obtain (2.9.1.6) where A ( m, p ) = 2 p +1 max | z | =1 / | φ ( z ) | . For proving (2.9.1.7) we consider the function ψ ( z ) := H ( z, cos γ, m, p ) z − p that isholomorphic in the domain D := { z : | z | ≥ , − π < arg z ≤ π } and continuous inits closure. Applying the maximum principle we obtain (2.9.1.7) where A ( m, p ) = 2 p max z ∈ ∂D | ψ ( z ) | . (cid:3) Set G p ( x, y, m ) := −| y | − m H ( | x | / | y | , cos γ, m, p )where cos γ = ( x , y ) . Note the equality G p ( x, y, m ) = G ( x, y, R m ) + p X k =0 D k ( x, y ) . Exercise 2.9.1.1.
Check this using (2.9.1.3),(2.9.1.4) and (2.9.1.5).It looks like a Green function for R m but it tends more quickly to zero atinfinity and generally speaking it is not negative.For m = 2 it can be represented in the form G p ( z, ζ,
2) = log | E ( z/ζ, p ) | where E ( z/ζ, p ) is the primary Weierstrass factor: E ( z/ζ, p ) := (cid:18) − zζ (cid:19) exp "(cid:18) zζ (cid:19) + 12 (cid:18) zζ (cid:19) + · · · + 1 p (cid:18) zζ (cid:19) p . We will call it the primary kernel analogously to the primary factor. Theorem 2.9.1.3.(Estimate of Primary Kernel).
The following holds: (2.9.1.8) | G p ( x, y, m ) | ≤ A ( m, p ) | x | p +1 | y | p + m − for | x | < | y | , (2.9.1.9) | G p ( x, y, m ) | ≤ A ( m, p ) | x | p | y | p + m − for | y | < | x | , and (2.9.1.10) G p ( x, y, m ) ≤ A ( m, p ) min (cid:18) | x | p +1 | y | p + m − , | x | p | y | p + m − (cid:19) for all x, y ∈ R m , where A ( m, p ) does not depend on x, y. For m = 2 , p = 0 we have G p ( z, ζ, ≤ A (0 ,
2) log(1 + | z || ζ | ) . Proof.
The inequality (2.9.1.8) follows directly from (2.9.1.6) and (2.9.1.9) followsfrom (2.9.1.7). By the condition 2 ≤ | x | / | y | (2.9.1.10) follows from (2.9.1.9).Suppose 1 / ≤ | x | / | y | ≤ . Since all the summands in (2.9.1.5) are boundedfrom below, for 1 / ≤ z ≤ G p ( x, y, m ) ≤ A ( m, p ) | y | − m ≤ A ( m, p ) min (cid:18) | x | p +1 | y | p + m − , | x | p | y | p + m − (cid:19) also under these conditions.The case m = 2 , p = 0 is obvious. (cid:3) Let µ ∈ M ( R m ) . We suppose below that its support does not contain theorigin.We will say that the integral R R m f ( x, y ) dµ y converges uniformly on x ∈ D ifsup x ∈ D (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z | y | >R f ( x, y ) dµ y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) → R → ∞ . Hence, the integral is permitted to be equal to infinity for some finite x. Let µ have genus p (see, 2.8.3). Set(2.9.2.1) Π( x, µ, p ) := Z R m G p ( x, y, m ) dµ y . It is called the canonical potential .In particular, let m=2 and µ := n be a zero distribution , i.e. it has unit massesconcentrated on a discrete point set { z j : j = 1 , , ... } . ThenΠ( z, n, p ) = log (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ Y j =1 E (cid:18) zz j , p (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) where ∞ Y j =1 E (cid:18) zz j , p (cid:19) is the canonical Weierstrass product . Theorem 2.9.2.1.(Brelot-Weierstrass).
The canonical potential (2.9.2.1) con-verges uniformly on any bounded domain. It is a subharmonic function with µ asits Riesz measure.Proof. Let | x | < R and | y | > R . From the estimate of the primary kernel(Theorem 2.9.1.3) we have | Z | y | >R G p ( x, y, m ) dµ y | ≤ A ( m, p ) | x | p +1 Z | y | >R | y | − p − m +1 dµ y == A ( m, p ) | x | p +1 Z ∞ R t − p − m +1 dµ ( t ) . Integrating by part we obtain Z ∞ R t − p − m +1 dµ ( t ) = µ ( R ) R p + m − + ( p + m − Z ∞ R µ ( t ) t p + m dt. The last integral converges since the genus of µ is p. Hence, both summands tendto zero when R → ∞ . Thussup | x |
Theorem 2.9.2.2.(Estimation of Canonical Potential).
The following in-equality holds: (2.9.2.2) M ( r, Π( • , µ, p )) ≤ A (cid:20)Z ∞ µ ( rτ ) r m − min(1 , τ − ) τ p + m − dτ + µ ( r ) r m − (cid:21) where A := A ( m, p ) does not depend on r and µ. Proof.
From (2.9.1.10)Π( x, µ, p ) ≤ A ( m, p ) Z R m min (cid:18) | x | p +1 | y | p + m − , | x | p | y | p + m − (cid:19) dµ y Set r := | x | , t := | y | . Then we have(2.9.2.3) M ( r, Π( • , µ, p )) ≤ A Z ∞ min (cid:18) r p +1 t p + m − , r p t p + m − (cid:19) dµ ( t )The integral on the right side of (2.9.2.3) can be represented in the form Z r r p t p + m − dµ ( t ) + Z ∞ r r p +1 t p + m − dµ ( t )Integrating every integral by parts we obtain( p + m − Z r r p t p + m − µ ( t ) dt + ( p + m − Z ∞ r r p +1 t p + m µ ( t ) dt + µ ( r ) r m − ≤≤ ( p + m − Z ∞ min (cid:16) , rt (cid:17) r p t p + m − µ ( t ) dt + µ ( r ) r m − . After the change t = rτ we obtain (2.9.2.2) where the new A ( m, p ) is equal to A ( m, p )( p + m − . (cid:3) Theorem 2.9.2.3.(Brelot - Borel).
The order of the canonical potential is equalto the convergence exponent of its mass distribution, i.e. ρ [Π( • , µ, p )] = ρ [ µ ] , if the genus of µ is equal to p. Proof.
First assume ρ [ µ ] < p + 1 . Let us choose λ such that ρ [ µ ] < λ < p + 1 . For some constant C that does not depend on t we have µ ( t ) ≤ Ct λ + m − . Actually, µ ( t ) /t λ + m − → , because λ > ρ [ µ ] . Since µ ( t ) = 0 for small t, thisfunction is bounded and we can take its lower bound as C. Now we have(2.9.2.4) f ( r, τ ) := µ ( rτ ) r λ + m − min(1 , τ − ) τ p + m − ≤ Cτ λ − p − min(1 , /τ ) . for all τ ∈ (0 , ∞ ) . We also have(2.9.2.5) lim r →∞ f ( r, τ ) = 0because of λ > ρ [ µ ] . Let us divide (2.9.2.2) by r λ and pass to the upper limit. By Fatou’s lemma(Theorem 2.2.2.3)(2.9.2.6)lim sup r →∞ M ( r, Π( • , µ, p )) r λ ≤ A ( m, p ) (cid:20)Z ∞ lim sup r →∞ f ( r, τ ) dτ + lim sup r →∞ µ ( r ) r λ + m − (cid:21) = 0Hence,(2.9.2.7) λ ≥ ρ [Π( • , µ, p )] . Since this holds for any λ > ρ [ µ ] , we have ρ [ µ ] ≥ ρ [Π( • , µ, p )] under theassumption λ < p [ µ ] + 1 . Let ρ [ µ ] = p [ µ ] + 1. By Theorem 2.8.3.1 ¯∆[ µ ] = 0 . Hence, µ ( t ) t − p − m +1 ≤ C and f ( r, τ ) := µ ( rτ )( rτ ) p + m − min(1 , τ − ) ≤ C min(1 , /τ ) . The function min(1 , /τ ) is not summable on (0 , ∞ ) . Therefore we will act in aslightly different way. From Theorem 2.9.2.2 we havelim sup r →∞ M ( r, Π( • , µ, p )) r p +1 ≤ A ( m, p ) (cid:20)Z lim sup r →∞ f ( r, τ ) dτ (cid:21) + + A ( m, p ) (cid:20) lim sup r →∞ Z ∞ r µ ( t ) t p + m dt + + lim sup r →∞ µ ( r ) r p + m (cid:21) The first integral is equal to zero because ¯∆[ µ ] = 0 . The second addend vanishessince the integral converges.Thus we have p + 1 = ρ [ µ ] ≥ ρ [Π( • , µ, p )] . The reverse inequality holds for any subharmonic function in R m by the Jensentheorem (Theorem 2.8.3.3). (cid:3) Let us denote as δSH ( ρ ) the class of functions u ∈ δSH ( R m ) for which ρ T [ u ] ≤ ρ. Theorem 2.9.3.1.(Brelot - Hadamard).
Let u = u − u ∈ δSH ( ρ ) , and let p , p be the genuses of the mass distributions µ j := µ u j , j = 1 , . Suppose supp [ µ − µ ] ∩ { } = ∅ . Then the following equality holds: u ( x ) = Π( x, µ , p ) − Π( x, µ , p ) + Φ q ( x ) where Φ q ( x ) is a harmonic polynomial of degree q ≤ ρ. Proof.
The function v ( x ) := u ( x ) − Π( x, µ , p ) + Π( x, µ , p ) is harmonic by theBrelot - Weierstrass theorem (Theorem 2.9.2.1). We also have the inequality(2.9.3.1) ρ T [ v ] ≤ max( ρ T [ u ] , ρ T [Π( • , µ , p )] , ρ T [Π( • , µ , p )])by Theorem 2.8.2.2 ( ρ T - properties). The property ρ MT1) (Theorem 2.8.2.3 )implies ρ T [Π( • , µ j , p j )] = ρ M [Π( • , µ j , p j )] := ρ [Π( • , µ j , p j )] , j = 1 , . The Brelot-Borel theorem (Theorem 2.9.2.3) implies ρ [Π( • , µ j , p j )] = ρ [ µ j ] , j = 1 , . The Jensen theorem (Theorem 2.8.3.3) impliesmax( ρ [ µ ] , ρ [ µ ]) ≤ ρ T [ u ] . From (2.9.3.1) we have ρ T [ v ] ≤ ρ T [ u ] ≤ ρ. Since v is subharmonic, ρ T [ v ] = ρ M [ v ] := ρ [ v ] by Theorem 2.8.2.3, and ρ [ v ] ≤ ρ. Therefore lim r →∞ M ( r, v ) r ρ + ǫ = 0for arbitrary small ǫ > . By the Liouville theorem (Theorem 2.4.2.3) v ( x ) is a harmonic polynomial ofdegree q ≤ ρ + ǫ, and thus v ( x ) = Φ q ( x ) for q ≤ ρ. (cid:3) For a non-integer ρ the Brelot-Hadamard theorem allows to connect the growthof functions and masses more tightly than in the Jensen theorem. Theorem 2.9.3.2.(Sharpening of Jensen).
Let ρ > and be non-integer, u = u − u ∈ δSH ( R m ) with ρ T [ u ] = ρ, and let ν u = µ − µ the correspondingcharge.ThenpJ1) ρ [ ν u ] = max( ρ [ µ ] , ρ [ µ ]) = ρ pJ2) A σ T [ u, ρ ( r )] ≤ ¯∆[ ν u , ρ ( r )] ≤ ¯∆[ µ , ρ ( r )] + ¯∆[ µ , ρ ( r )] ≤ A σ T [ u, ρ ( r )] , where A j = A j ( m, ρ ) and ρ ( r ) is an arbitrary proximate order such that ρ ( r ) → ρ when r → ∞ . For proving this theorem we need
Theorem 2.9.3.3.
Let Π( x, µ, p ) be a canonical potential with non-integer ρ [ µ ] :=[ ρ ] , and let ρ ( r )( → ρ ) be a proximate order.Then (2.9.3.2) σ [Π( • , µ, p ) , ρ ( r )] ≤ A ( m, ρ, p ) ¯∆[ µ, ρ ( r )] Proof.
We can suppose without loss of generality that ¯∆[ µ, ρ ( r )] < ∞ . By thiscondition and since µ ( t ) = 0 , < t < c for some c > , we have the inequality µ ( t ) t − ρ ( t ) − m +2 ≤ C for all t ∈ (0 , ∞ ) and some C > t. Set I ( r ) := Z ∞ c/r µ ( rt ) r ρ ( r )+ m − min(1 , /t ) t p + m − dt. By Theorem 2.9.2.2 we have(2.9.3.3) σ [Π( • , µ, p ) , ρ ( r )] = lim sup r →∞ M ( r, Π( • , µ, p ) r ρ ( r ) ≤ A ( m, p ) lim sup r →∞ I ( r ) . Let us choose r ǫ such thatsup r>r ǫ µ ( rǫ )( rǫ ) ρ ( rǫ )+ m − ≤ ¯∆[ µ, ρ ( r )] + ǫ. For such r we have I ( r ) = Z ∞ c/r µ ( rt )( rt ) ρ ( rt )+ m − ( rt ) ρ ( rt ) r ρ ( r ) min(1 , /t ) t p +1 dt ≤≤ sup c/r ≤ t ≤ ǫ µ ( rt )( rt ) ρ ( rt )+ m − Z ǫc/r ( rt ) ρ ( rt ) r ρ ( r ) min(1 , /t ) t p +1 dt +sup ǫ ≤ t ≤ /ǫ ... Z /ǫǫ ...dt + sup /ǫ ≤ t ≤∞ ... Z ∞ /ǫ ...dt ≤≤ C Z ǫc/r ( rt ) ρ ( rt ) r ρ ( r ) min(1 , /t ) t p +1 dt + ( ¯∆[ µ, ρ ( r )] + ǫ ) Z /ǫǫ ...dt + C Z ∞ /ǫ ...dt. The function f ( t ) := min(1 , /t ) t p +1 satisfies the conditions of Goldberg’s theorem (Theorem 2.8.1.5) with p + 1 − ρ <δ < < γ < p + 1 − ρ. Passing to the limit we havelim sup r →∞ I ( r ) ≤ C Z ǫ t ρ − p dt + ( ¯∆[ µ, ρ ( r )] + ǫ ) Z /ǫǫ t ρ − p − min(1 , /t ) dt + C Z ∞ /ǫ t ρ − p − dt. Passing to the limit as ǫ → σ [Π( • , µ, p ) , ρ ( r )] ≤ A ( m, p ) ¯∆[ µ, ρ ( r )] Z ∞ t ρ − p − min(1 , /t ) dt. (cid:3) Proof of Theorem 2.9.3.2.
The inequality ρ [ ν u ] ≤ ρ and the last inequality in pJ2)follow from the Jensen theorem (Theorem 2.8.3.3). Let us prove the reverse in-equality and the left side. Since ρ is non-integer, q < ρ in the Brelot-Hadamard theorem (Theorem2.9.3.1). Hence M ( r, Φ q ) = o ( r ρ ) and T ( r, u ) ≤ T ( r, Π( • , µ , p )) + T ( r, Π( • , µ , p )) + o ( r ρ ) . Thus ρ T [ u ] ≤ max( ρ [Π( • , µ , p )] , ρ [Π( • , µ , p ]) ,σ T [ u, ρ ( r )] ≤ max( σ T [Π( • , µ , p ) , ρ ( r )] , σ T [Π( • , µ , p ] , ρ ( r )]) . From Theorem 2.9.3.3 we obtain ρ T [ u ] ≤ max( ρ [ µ ] , ρ [ µ ]); σ T [ u, ρ ( r )] ≤ A ( m, ρ, p ) max( ¯∆[ µ , ρ ( r )] , ¯∆[ µ , ρ ( r )] == A ( m, ρ, p ) ¯∆[ | ν | , ρ ( r )] . We can set A := A − ( m, ρ, p ) and obtain the left side of pJ2). (cid:3) Let u ∈ δSH ( R m ) and ρ := ρ T [ u ] be an integer number. We can alwaysrepresent the function u in the form(2.9.4.1) u ( x ) = Π( x, ν, ρ ) + Φ ρ ( x )where Φ ρ ( x ) is a harmonic polynomial of degree at most ρ. Actually, such a rep-resentation can be obtained from Theorem 2.9.3.1 by addition and subtraction ofterms of the form Φ k j ( x ) := Z R m D k j ( x, y ) d ( µ j ) y , j = 1 , p j < k j ≤ ρ. All Φ k j ( x ) of such a kind are harmonic polynomials of degree at most ρ. Set(2.9.4.2) Π R< ( x, ν, ρ −
1) := Z | y |
The following holds A Ω[ u, ρ ( r )] ≤ σ T [ u, ρ ( r )] ≤ A Ω[ u, ρ ( r )] , where A j := A j ( m, ρ ) . For proving this theorem we will first study the function Π R< and Π R> . Set T ( r, λ, > ) := T ( r, Π λr> ( • , ν, ρ )) ,T ( r, λ, < ) := T ( r, Π λr< ( • , ν, ρ − . Theorem 2.9.4.3.(Estimate of T ( • , >, T ( • , < ) ). The following holds: (2.9.4.6) T ( r, λ, > ) ≤ A (cid:18)Z ∞ λ | ν | ( rt ) r m − min(1 , t − ) t ρ + m − dt + | ν | ( r ) r m − (cid:19) , (2.9.4.7) T ( r, λ, < ) ≤ A Z λ | ν | ( rt ) r m − min(1 , t − ) t ρ + m − dt ! ++ A (cid:18) | ν | ( rλ ) r m − Z ∞ λ min(1 , t − ) t ρ + m − dt + | ν | ( r ) r m − (cid:19) , where A := A ( m, ρ ) . Proof.
Let ν = µ − µ . Then | ν | = µ + µ . We have(2.9.4.8) Π R< ( x, ν, ρ −
1) = Π R< ( x, µ , ρ − − Π R< ( x, µ , ρ − R< (0 , µ , ρ −
1) = 0 , we have (see t3),t4), Theorem 2.8.2.1)(2.9.4.9) T ( r, Π R< ( • , ν, ρ − ≤ T ( r, Π R< ( • , µ , ρ − T ( r, Π R< ( • , µ , ρ − := Π R< ( • , µ , ρ − , Π := Π R< ( • , µ , ρ − . Let us estimate, for example, T ( r, Π ) . The masses of the canonical potential Π areconcentrated in K R . Applying Theorem 2.9.2.2 (Estimation of Canonical Potential)for p = ρ − T ( r, Π ) ≤ M ( r, Π ) ≤ A Z Rr µ ( rt ) r m − min(1 , t − ) t ρ + m − dt ++ A µ ( R ) r m − Z ∞ Rr min(1 , t − ) t ρ + m − dt + µ ( r ) r m − . Set R := rλ. Then we obtain the inequality (2.9.4.7) for ν := µ . Analogously onecan do for ν := µ . The inequality (2.9.4.9) allows to pass to limit in (2.9.4.7) inthe general case.Set Π := Π R> ( • , µ , ρ ) . Applying (2.9.2.2) for p = ρ we obtain T ( r, Π ) ≤ M ( r, Π ) ≤ Z ∞ Rr µ ( rt ) r m − min(1 , t − ) t ρ + m − dt + µ ( r ) r m − . In the same way we obtain (2.9.4.6). (cid:3)
Set σ [Π > , ρ ( r )] := lim sup r →∞ T ( r, Π r> ( • , ν, ρ )) r ρ ( r ) ,σ [Π < , ρ ( r )] := lim sup r →∞ T ( r, Π r< ( • , ν, ρ )) r ρ ( r ) . Theorem 2.9.4.4.
Let ν := µ − µ ∈ δ M ( ρ ) and ρ integer number. Then forany p.o. ρ ( r ) → ρ max( σ [Π > , ρ ( r )] , σ [Π < , ρ ( r )]) ≤ A ¯∆[ | ν | , ρ ( r )] where A := A ( m, ρ ) . Proof.
From (2.9.4.6) we have T ( r, Π r> ( • , ν, ρ )) = T ( r, , > ) ≤ A Z ∞ | ν | ( rt ) r m − t ρ + m dt + | ν | ( r ) r m − . Now we repeat the reasoning of Theorem 2.9.3.3 for µ := | ν | and p := ρ. We willobtain σ [Π > , ρ ( r )] ≤ A ¯∆[ | ν | , ρ ( r )] Z ∞ t − dt. For the other case we have from (2.9.4.7) T ( r, Π r< ( • , ν, ρ − T ( r, , < ) ≤ A Z | ν | ( rt ) r m − t ρ + m − dt ++ A | ν | ( r ) r m − (cid:18)Z ∞ t − ρ − m +1 dt + r − (cid:19) . We divide this inequality by r ρ ( r ) and pass to the upper limit while r → ∞ . The first summand of the right side gives A ¯∆[ | ν | , ρ ( r )] Z dt by the reasoning of Theorem 2.9.3.3.The second one can be computed directly, yielding A ¯∆[ | ν | , ρ ( r )] Z ∞ t − ρ − m +1 dt. Combining all these inequalities we obtain the assertion of the theorem. (cid:3)
Proof of Theorem 2.9.4.2.
Let us represent u ( x ) in the form(2.9.4.10) u ( ry ) = Π r< ( ry, ν u , ρ −
1) + Π r> ( ry, ν u , ρ ) + δ r ( ry, u, ρ ) + o ( r ρ − )where | y | = 1 . Then we have T ( r, u ) ≤ T ( r, Π r< ( • , ν u , ρ − T ( r, Π r> ( • , ν u , ρ )) + M ( r, δ ) + o ( r ρ − ) . Let us divide this by r ρ ( r ) and pass to the upper limit. By Theorem 2.9.4.4 weobtain σ T [ u, ρ ( r )] ≤ A max( ¯∆[ | ν | , ρ ( r )] , ¯∆ δ [ u, ρ ( r )]) = A Ω[ u, ρ ( r )]where A = A ( m, ρ ) . Let us write (2.9.4.11) in the form δ r ( ry, u, ρ ) = u ( ry ) − P i r< ( ry, ν u , ρ − − Π r> ( ry, ν u , ρ ) + o ( r ρ − ) . We obtain T ( r, δ r ( • , u, ρ )) ≤ T ( r, u ) + T ( r, Π r< ( • , ν u , ρ − T ( r, Π r> ( • , ν u , ρ ) + o ( r ρ − ) . Since δ R ( • , u, ρ ) is harmonic and homogeneous, we have by (2.8.2.5) M ( r, δ R ) ≤ m − T (2 r, δ R ) = 2 m − ρ T ( r, δ R ) . Therefore we obtain the inequality¯∆ δ [ u, ρ ( r )] ≤ σ T [ u, ρ ( r )] + 2 A ¯∆[ | ν | , ρ ( r )] . By the Jensen theorem (Theorem 2.8.3.3) we haveΩ[ u, ρ ( r )] ≤ A − σ T [ u, ρ ( r )]for some A = A ( m, ρ ). (cid:3) Let { V t : t ∈ (0 , ∞ ) } be a family of rotations of R m that form a one-parametric group, i.e.,(3.1.1.0) V t V t = V t t , V = I, where I is the identity map.The family of linear transformations(3.1.1.1) P t := tV t is also a one-parametric group.In particular, for m = 2 the general form of the rotations is V t z = z exp( iγ log t ) , where γ is real.The orbit { P t z : t ∈ (0 , ∞ ) } of every point z = 0 is a logarithmic spiral if γ = 0and a ray if γ = 0 . For m ≥ V t ≡ I, t ∈ (0 , ∞ ) the orbit of every point x = 0 is a ray fromthe origin. For other V • it is a spiral connecting the origin to infinity.It is clear that only one orbit { P t x : t ∈ (0 , ∞ ) } passes through every x = 0 . The behavior of every point y ( t ) := P t x is completely determined by a system ofdifferential equations with constant coefficients: ddt y = ( I + V ′ ) y, V ′ := ddt V t | t =1 with the initial condition of y (1) = x. Let u ∈ SH ( ρ ) and σ M [ u, ρ ( r )] < ∞ for some p.o. ρ ( r ) → ρ. We will write u ∈ SH ( R m , ρ, ρ ( r )) or shorter u ∈ SH ( ρ ( r )) . For u ∈ SH ( ρ ( r )) set(3.1.2.1) u t ( x ) := u ( P t x ) t − ρ ( t ) . We will denote this transformation as ( • ) t . Theorem 3.1.2.1.(Existence of Limit Set).
The following holds:els1) u t ∈ SH ( ρ ( r )) for any t ∈ (0 , ∞ ) els2) the family { u t } is precompact at infinity, i.e., for any sequence t k → ∞ there exists a subsequence t k j → ∞ and afunction v ∈ SH ( R m ) such that u t kj → v in D ′ ( R m ) (see, 2.7.1). Proof.
The functions u t are subharmonic by sh1) and sh5), Theorem 2.6.1.1.(Ele-mentary Properties), and M ( r, u t ) = M ( rt, u ) t − ρ ( t ) . Now we have σ M [ u t , ρ ( r )] = t − ρ ( t ) lim sup r →∞ M ( rt, u )( rt ) ρ ( rt ) · lim r →∞ ( rt ) ρ ( rt ) r ρ ( r ) = σ M [ u, ρ ( r )] t ρ − ρ ( t ) , because(3.1.2.2) lim r →∞ ( rt ) ρ ( rt ) r ρ ( r ) = t ρ lim r →∞ L ( rt ) L ( r ) = t ρ (see, ppo3), Theorem 2.8.1.3.(Properties of P.O)). Therefore els1) is proved.Let us check the conditions of Theorem 2.7.1.1.(Compactness in D ′ ). We have(3.1.2.3) lim sup t →∞ M ( r, u t ) = lim sup t →∞ M ( rt, u )( rt ) ρ ( rt ) · lim t →∞ ( rt ) ρ ( rt ) t ρ ( t ) = σ M [ u, ρ ( r )] r ρ . Thus, the family is bounded from above on every compact set andlim t →∞ u t (0) = lim t →∞ u (0) t − ρ ( t ) = 0 . Therefore u t (0) are bounded from below for large t . (cid:3) We will call the set of all functions v from Theorem 3.1.2.1 the limit set of thefunction u ( x ) with respect to V • and denote it Fr [ u, ρ ( r ) , V • , R m ] or shortly Fr [ u ] . The limit set does not depend on values of the subharmonic function on abounded set, hence, it is a characteristic of asymptotic behavior.Set U [ ρ, σ ] := { v ∈ SH ( R m ) : M ( r, v ) ≤ σr ρ , r ∈ [0 , ∞ ); v (0) = 0 } , (3.1.2.4) U [ ρ ] := [ σ> U [ ρ, σ ]and(3.1.2.4a) v [ t ] ( x ) := t − ρ v ( P t x ) , t ∈ (0 , ∞ ) . Let us emphasize that the transformation ( • ) [ t ] coincides with ( • ) t from (3.1.2.1)for ρ ( r ) ≡ ρ and satisfies the condition(3.1.2.4b) ( • ) [ tτ ] = (( • ) [ t ] ) [ τ ] Theorem 3.1.2.2.(Properties of
F r ). The following holds:fr1) Fr [ u ] is a connected compact set;fr2) Fr [ u ] ⊂ U [ ρ, σ ] , for σ ≥ σ M [ u ]; fr3) ( Fr [ u ]) [ t ] = Fr [ u ] , t ∈ (0 , ∞ ) , i.e., v ∈ Fr [ u ] implies v [ t ] ∈ Fr [ u ] . f4) if ρ ( r ) and ρ ( r ) are equivalent (see (2.8.1.5)), then Fr [ u, ρ ( r ) , • ] = Fr [ u, ρ ( r ) , • ] . We need the following assertion
Theorem 3.1.2.3.(Continuity u t ). The functions u t , v [ t ] : (0 , ∞ ) × D ′ ( R m )
7→ D ′ ( R m ) are continuous in the natural topology.Proof. For any ψ ∈ D ( R m ) consider < u t , ψ > := Z u t ( x ) ψ ( x ) dx = Z u ( y ) ψ ( y/t ) t m − ρ ( t ) dy := < u, ψ ( • , t ) >, where ψ ( y, t ) := ψ ( y/t ) t m − ρ ( t ) . The function ψ ( • , t ) is continuous in t in D ( R m ) . By Theorem 2.3.4.6 (Conti-nuity < • , • > ) < u, ψ ( • , t ) > is continuous in ( u, t ) . (cid:3) Proof of Theorem 3.1.2.2.
Let us denote as clos {•} the closure in D ′ -topology.The set F N := clos { u t : t ≤ N } ⊃ Fr [ u ] is compact in D ′ -topology. Indeed,let t j → t and t < ∞ ; then u t j → u t because of Theorem 3.1.2.3. If t j → ∞ and u t j → v, then v ∈ Fr [ u ] by its definition, hence, v ∈ F N . Since Fr [ u ] = ∩ ∞ N =1 F N , itis compact.Let us prove the connectedness. Suppose Fr [ u ] is not connected. Then it canbe written as a union of two disjoint non-empty closed sets F and F . Let V , V be disjoint open neighborhoods of F , F respectively in D ′ ( R m ) . Since F , F arenonempty there exist sequences { s j } , { t j } such that s j < t j , s j → ∞ , u s j ∈ V , u t j ∈ V . Since the mapping u t : (0 , ∞ )
7→ D ′ ( R m ) is continuous by Theorem3.1.2.3 its image is connected. Thus there exists a sequence { p j } with s j < p j < t j such that u p j / ∈ V ∪ V . This sequence has a subsequence that converges to afunction v ∈ Fr [ u ] and v / ∈ F ∪ F . This is a contradiction. Hence, Fr [ u ] isconnected and fr1) is proved.Set ψ ( r ) := lim sup r →∞ M ( r, u t ) . This function is convex with respect to − r − m for m ≥ r for m = 2 and hence continuous.Indeed, M ( | x | , u t ) are subharmonic (see Theorem 2.6.5.2.(Convexity M ( • , u )and M ( r, u )). By Theorem 2.7.3.3.(H.Cartan +) the function ψ ∗ ( | x | ) is subhar-monic and ψ ( | x | ) = ψ ∗ ( | x | ) quasi-everywhere. However, if ψ ( | x | ) < ψ ∗ ( | x | ) at somepoint, the same inequality holds on a sphere which has a positive capacity (seeExample 2.5.2.2). Hence, ψ ( | x | ) = ψ ∗ ( | x | ) everywhere, and ψ ( | x | ) is subharmonic.Thus ψ ( r ) is convex with respect to − r − m for m ≥ r for m = 2 by Theorem 2.6.3.2.(Subharmonicity and Convexity).One can also see that for u ∈ SH ( R m ) M ( r, u ǫ ) ≤ M ( r + ǫ, u ) , where ( • ) ǫ is defined by (2.6.2.3).Let v ∈ Fr [ u ] and u t j → v in D ( R m ) . By property reg3), Theorem 2.3.4.5( u t j ) ǫ → v ǫ uniformly on any compact set. Thus(3.1.2.5) v ǫ ( x ) = lim j →∞ ( u t j ) ǫ ≤ lim sup t →∞ ) M ( | x | , ( u t ) ǫ ) ≤ ≤ lim sup t →∞ ) M ( | x | + ǫ, u t ) = ψ ( | x | + ǫ ) . If ǫ ↓
0, then v ǫ ↓ v by Theorem 2.6.2.3 and ψ ( r + ǫ ) → ψ ( r ) because of continuity.Passing to the limit in (3.1.2.5) and using (3.1.2.3) we obtain(3.1.2.6) v ( x ) ≤ σ M [ u, ρ ( r )] | x | ρ . Since u (0) ≤ u ǫ (0) we have u (0) t − ρ ( t ) ≤ ( u t ) ǫ (0) . Let us pass to the limit as t := t j → ∞ . We obtain v ǫ (0) ≥ . Passing to the limit as ǫ ↓ v (0) ≥ . The inequalities (3.1.2.6) and (3.1.2.7) imply fr2).One can check the equality(3.1.2.8) ( u t ) [ τ ] = u tτ · ( tτ ) ρ ( tτ ) t ρ ( t ) τ ρ . By using properties of p.o. we havelim t →∞ ( tτ ) ρ ( tτ ) t ρ ( t ) τ ρ = 1(compare (3.1.2.2)).Let v ∈ Fr [ u ] and u t j → v. Set t := t j , τ := t in (3.1.2.8) and pass to the limit.Then v [ t ] = D ′ − lim j →∞ u t j t . Thus v [ t ] ∈ Fr [ u ] . The property f3) is proved.Let us prove f4).We have u ( P t x ) t ρ ( t ) = u ( P t x ) t ρ ( t ) × e ( ρ ( t ) − ρ ( t )) log t = u ( P t x ) t ρ ( t ) × (1 + o (1))as t → ∞ because of (2.8.1.5).This implies f4). Exercise 3.1.2.1
Check this in details. (cid:3) We can consider the limit sets as a mapping u Fr [ u ]. The following theoremdescribes some properties of this mapping.Set(3.1.2.9) U [ ρ ] := [ σ> U [ ρ, σ ]where U [ ρ, σ ] is defined by (3.1.2.4).Let X, Y be subsets of a cone (i.e. a subset of a linear space that is closed withrespect to sum and multiplication by a positive number). The set U [ ρ ] is such acone. Set(3.1.2.10) X + Y := { z = x + y : x ∈ X, y ∈ Y } ; λX := { z = λx : x ∈ X } . Theorem 3.1.2.4(Properties of u Fr [ u ] ). The following holds:fru1) Fr [ u + u ] ⊂ Fr [ u ] + Fr [ u ] fru2) Fr [ λu ] = λ Fr [ u ] .Proof. Let v ∈ Fr [ u + u ] . Then there exists t j → ∞ such that ( u + u ) t j → v in D ′ . We can find a subsequence t jk such that ( u ) t jk → v and ( u ) t jk → v . Then v = v + v . The property fru1) has been proved.The property fru2) is proved analogically. (cid:3)
We will write µ ∈ M ( R m , ρ ( r )) or shortly µ ∈ M ( ρ ( r )) if µ ∈ M ( R m ) (see2.8.3) and ¯∆[ µ, ρ ( r )] < ∞ (see 2.8.3.2).Let us define a distribution µ t for µ ∈ M ( ρ ( r )) by(3.1.3.1) < µ t , φ > := t − ρ ( t ) − m +2 Z φ ( P − t x ) dµ for φ ∈ D ( R m ) . It is positive. Hence, it defines uniquely a measure µ t . Theorem 3.1.3.1.(Explicit form of µ t ). For any E ∈ σ ( R m ) the following holds: (3.1.3.2) µ t ( E ) = t − ρ ( t ) − m +2 µ ( P t E ) Proof.
It is enough to proof the assertion for some dense ring (see Theorem 2.2.3.5),for example, for all compact sets.Let χ K be a characteristic function of a compact set K and let φ ǫ ↓ χ K be a monotonically converging sequence of functions that belong to D ( R m ) (seeTheorems 2.1.2.1, 2.1.2.9 and 2.3.4.4). Then Z φ ǫ ( x ) dµ t = t − ρ ( t ) − m +2 Z φ ǫ ( P − t x ) dµ. Since φ ǫ ( P − t x ) ↓ χ P t K ( x ) ,µ t ( K ) = Z χ K ( x ) dµ t = t − ρ ( t ) − m +2 Z χ P t K ( x ) dµ = t − ρ ( t ) − m +2 µ ( P t K ) . (cid:3) Theorem 3.1.3.2.(Existence of µ -Limit Set). The following holds:mels1) µ t ∈ M ( ρ ( r )) for any t ∈ (0 , ∞ ) ;mels2) the family { µ t } is precompact in infinity, i.e., for any sequence t k → ∞ there exists a subsequence t k j → ∞ and ameasure ν ∈ M ( R m ) such that µ t kj → ν in D ′ ( R m ) (see, 2.7.1). Proof.
We have µ t ( r ) = µ ( rt ) t − ρ ( t ) − m +2 . Thus(3.1.3.3)lim sup r →∞ µ t ( r ) r ρ ( r )+ m − = lim sup r →∞ µ ( rt )( rt ) ρ ( rt )+ m − ( rt ) ρ ( rt ) t ρ ( t )+ m − r ρ ( r ) = t ρ − ρ ( t ) − ( m − ¯∆[ µ, ρ ( r )] . Therefore mels1) holds.We also have lim sup t →∞ µ t ( r ) = ¯∆[ µ, ρ ( r )] r ρ + m − . Thus µ t satisfies the assumption of the Helly theorem (Theorem 2.2.3.2). Usingalso Theorem 2.3.4.4 we obtain mels2). (cid:3) We will call the set of all measures ν from Theorem 3.1.2.1 the limit set of themass distribution µ with respect to V • and denote it Fr [ µ, ρ ( r ) , V • , R m ] or shortly Fr [ µ ] . Set(3.1.3.4) M [ ρ, ∆] := { ν : ν ( r ) ≤ ∆ r ρ + m − , ∀ r > } . M [ ρ ] := [ ∆ > M [ ρ, ∆] , and(3.1.3.5) ν [ t ] ( E ) := t − ρ − m +2 ν ( P t E )for E ∈ σ ( R m ) . Theorem 3.1.3.3.(Properties of
F r [ µ ] ). The following holds:frm1) Fr [ µ ] is connected and compact;frm2) Fr [ µ ] ⊂ M [∆ , ρ ] , for ∆ ≥ ¯∆[ µ, ρ ( r )]; frm3) ( Fr [ µ ]) [ t ] = Fr [ µ ] , t ∈ (0 , ∞ ) , Proof.
We will only prove frm2) because frm1) and frm3) are proved word by wordas in Theorem 3.1.2.2.Suppose t n → ∞ and µ t n → ν ∈ Fr [ µ ] . Let us choose r ′ > r such that theopen ball K r ′ is squarable with respect to ν. It is possible because of Theorem2.2.3.3, sqr2). By Theorems 2.2.3.7 and 2.3.4.4 µ t n ( r ′ ) → ν ( r ′ ) . Thus (comparewith (3.1.2.3)) ν ( r ′ ) = lim t n →∞ µ t n ( r ′ ) ≤ lim sup t →∞ µ t ( r ′ ) = ¯∆[ µ, ρ ( r )]( r ′ ) ρ + m − . Choosing r ′ ↓ r we obtain lim r ′ → r ν ( r ′ ) = ν ( r )because (2.2.3.3). Thus frm2) holds. (cid:3) The following assertion is a “copy” of Theorem 3.1.2.4. Theorem 3.1.3.4.(Properties of µ Fr [ µ ] ). The following holds:frmu1) Fr [ µ + µ ] ⊂ Fr [ µ ] + Fr [ µ ] frmu2) Fr [ λµ ] = λ Fr [ µ ] . The proof is also a “copy” and we omit it.
We are going to study the class U [ ρ ] and obtain for it “non-asymptotic”analogies of Theorem 2.8.3.3 (Jensen), 2.9.2.3 (Brelot-Borel), 2.9.3.1 (Brelot - Hadamard) Theorem 3.1.4.1.(*Jensen).
Let v ∈ U [ ρ ] . Then its Riesz measure ν v ∈ M [ ρ ] . Proof.
As in Theorem 2.8.3.2 we have an inequality(3.1.4.1) ν v ( r ) r m − ≤ A ( m ) N (2 r, ν v )Since v (0) = 0 we have (Theorem 2.6.5.1.(Jensen-Privalov))(3.1.4.2) N (2 r, ν v ) = M (2 r, v ) ≤ M (2 r, v ) ≤ ρ σr ρ . Substituting (3.1.4.2) in (3.1.4.1) we obtain ν v ∈ M [ ρ, ∆] for some ∆. Thus ν v ∈M [ ρ ] . (cid:3) Let ρ be non-integer and ν ∈ M [ ρ ] . Consider the canonical potential Π( x, ν, p )where p := [ ρ ] (see (2.9.2.1)). Let us emphasize that the support of ν may containthe origin but ν (0) = 0 , i.e., there is no concentrated mass in the origin. Thus wemust also check its convergence in the origin. Theorem 3.1.4.2.(*Brelot-Borel).
Let ρ be non-integer and let ν ∈ M [ ρ ] . Then Π( x, ν, p ) converges and belongs to U [ ρ ] . Proof.
Using (2.9.1.9) we have(3.1.4.2) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z | y | < | x | G p ( x, y, m ) dν y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ A ( m, p ) | x | p Z | x | dν ( t ) t p + m − . Let us estimate the integral in (3.1.4.2). Integrating by parts we obtain I < ( x ) := Z | x | dν ( t ) t p + m − = ν ( t ) t p + m − | | x | + ( p + m − Z | x | ν ( t ) dtt p + m − . Since ν ∈ M [ ρ, ∆] for some ∆ ,I < ( x ) ≤ A ( m, ρ, p )∆ | x | ρ − p . Substituting this in (3.1.4.2) we obtain(3.1.4.3) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z | y | < | x | G p ( x, y, m ) dν y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ A ( m, ρ, p )∆ | x | ρ . Analogously, using (2.9.1.8) we obtain(3.1.4.4) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z | x | < | y | G p ( x, y, m ) dν y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ A ( m, ρ, p )∆ | x | ρ . In particular, these estimates show that Π( x, ν, p ) exists. Now using (2.9.1.10) wehave also Z | x | ≤| y |≤ | x | G p ( x, y, m ) dν y ≤ A ( m, p ) Z | x | | x | dν ( t ) min (cid:18) | x | p +1 t p + m − , | x | p t p + m − (cid:19) . The latter integral can also be easily estimated by ∆ A ( m, p, ρ ) | x | ρ . Thus we have Z | x | ≤| y |≤ | x | G p ( x, y, m ) dν y ≤ A ( m, p, ρ ) | x | ρ . Therefore by (3.1.4.4) and (3.1.4.3) we obtain M ( r, Π) ≤ σr ρ for some σ. Since G p (0 , y, m ) = 0 for all y = 0 and the integral converges,Π(0 , ν, p ) = 0 . (cid:3) We will need an assertion that looks like the Liouville theorem (Theorem2.4.2.3).
Theorem 3.1.4.3.(*Liouville).
Let H be a harmonic function in R m and H ∈ U [ ρ ] . Then H ≡ if ρ is non-integer and H is a homogeneous polynomial of degree p if ρ = p is integer. In particular, for m = 2 we have H ( re iφ ) = r p ℜ ( ce ipφ ) . Proof.
Like in the proof of the Liouville theorem we obtain the inequality (2.4.2.9)and | c k | ≤ AR − k max | x | = R H ( x ) ≤ AσR ρ − k for some σ > . If k > ρ, we will pass to the limit when R → ∞ and obtain c k = 0 . If k < ρ ,we will do that when R → c k = 0 . (cid:3) The following theorem can be considered as an analogy of the Brelot-Hadamardtheorem (Theorem 2.9.3.1):
Theorem 3.1.4.4.(*Hadamard).
Let ρ be non-integer and v ∈ U [ ρ ] . Then (3.1.4.5) v ( x ) = Π( x, ν v , p ) for p = [ ρ ] . Proof. . Consider the function H ( x ) := v ( x ) − Π( x, ν v , p ) . It is harmonic.We alsohave by(2.8.2.5) M ( r, H ) ≤ A ( m ) T ( r, H ) ≤ A ( m )[ T ( r, v ) + T ( r, Π)] ≤ σr ρ for some σ. Hence, H ( x ) ≡ (cid:3) Let us consider the case of integer ρ. Let ν ∈ M [ ρ ] for an integer ρ = p. Set(3.1.4.6) Π < ( x, ν, ρ ) := Z | y | < G p − ( x, y, m ) dν (3.1.4.7) Π > ( x, ν, ρ ) := Z | y |≥ G p ( x, y, m ) dν. Both potentials converge and belong to U [ ρ ]. Theorem 3.1.4.5 (**Hadamard).
Let ρ be integer and let v ∈ U [ ρ ] . Then (3.1.4.8) v = H ρ ( x ) + Π < ( x, ν, ρ ) + Π > ( x, ν, ρ ) , where H ρ is a homogeneous harmonic polynomial of degree ρ. The proof is exactly the same as in the *Hadamard theorem, but we use thesecond case of Theorem 3.1.4.3. We also note that the polynomial may be equal tozero identically. Exercise 3.1.4.1
Check this in details.Let as check that ν from (3.1.4.8) has the following property that is analogousto Theorem 4.9.4.2. Theorem 3.1.4.6 (*Lindel¨of ).
Let ρ be integer and let v ∈ U [ ρ ] . Then (3.1.4.9) lim ǫ → Z ǫ ≤| y | < D ρ ( x, y ) µ ( dy ) = H ρ ( x ) Proof.
Consider the function(3.1.4.10) v ∗ ǫ ( x ) := v ( x ) + R | y | <ǫ ν ( dy ) | x − y | m − f or m > − R | y | <ǫ log | x − y | ν ( dy ) , f or m = 2 . It is subharmonic with supp ν ∩{ } = ∅ . We represent this function like in (2.9.4.10)in the form v ∗ ǫ ( x ) = Π < ( x, ν ∗ ǫ , ρ ) + Π > ( x, ν ∗ ǫ , ρ ) + P ∗ ρ − ( x, v ∗ ǫ ) + δ ( x, v ∗ ǫ , ρ )In this representation we can pass to limit as ǫ → Exercise 3.1.4.2
Check this, using that all the integrals converge for ν ∈M ( ρ, ∆) and showing that the integral in (3.1.4.10) tends to zero.The last two summands forms a harmonic polynomial, the limit of which isalso a harmonic polynomial. Comparing the limit with the representation (3.1.4.8),we obtain that P ∗ ρ − ( • , v ∗ ǫ ) tends to zero and δ ( x, v ∗ ǫ , ρ ) tends to H ρ ( x ) . (cid:3) Theorem 3.1.4.7. (**Liouville). If v ∈ U [ ρ ] satisfies inequality v ( x ) ≤ for z ∈ R m then v ( x ) ≡ . Otherwise it contradicts to subharmonicity in 0.
Let us study the connection between Fr [ u ] and Fr [ µ u ].Note the following properties of the transformations ( • ) t and ( • ) [ t ] . Theorem 3.1.5.0 (Connection between u t and µ t ). One has (3.1.5.0) ( µ u ) t = µ u t ; ( µ v ) [ t ] = µ v [ t ] . Proof.
By the F.Riesz theorem (Theorem 2.6.4.3) and Theorem 2.5.1.1 ,GPo3) wehave for any ψ ∈ D ′ ( R m ) < µ u , ψ > = θ m < ∆ u, ψ > = θ m < u, ∆ ψ > . Using the definition (3.1.3.1), we obtain < ( µ u ) t , ψ > = < ( µ u ) t , ψ (( P t ) − • ) > t − ρ ( t ) − m +2 . Thus < ( µ u ) t , ψ > = θ m < u, ∆[ ψ (( P t ) − • )] > t − ρ ( t ) − m +2 . Since the Laplace operator is invariant with respect to V t for any t we have∆[ ψ (( P t ) − • )] = t − [∆ ψ ](( P t ) − • ) . Thus we obtain < ( µ u ) t , ψ > = θ m t − ρ ( t ) − m < u, [∆ ψ ](( P t ) − • ) > == θ m < u ( P t • ) t − ρ ( t ) , ∆ ψ > = θ m < u t , ∆ ψ > = < µ u t , ψ > . (cid:3) Exercise 3.1.5.1.
Do this for ( • ) [ t ] . We begin from the case of a non-integer ρ . Theorem 3.1.5.1 (Connection between Fr’s for non-integer ρ ). Let u ∈ U ( ρ ( r )) and µ u be its Riesz measure. Then (3.1.5.1) Fr [ µ u ] = { ν v : v ∈ Fr [ u ] } , (3.1.5.2) Fr [ u ] = { Π( • , ν, p ) : ν ∈ Fr [ µ u ] } . Proof.
Let ν ∈ Fr [ µ u ] . There exists t n → ∞ such that ( µ u ) t n → ν in D ′ . We canfind a subsequence t ′ n such that u t ′ n → v ∈ F r [ u ] . Thus ( µ u ) t ′ n → ν v and therefore ν = ν v . Hence,
F r [ µ u ] ⊂ { ν v : v ∈ Fr [ u ] } . Analogously we can prove that every ν v ∈ F r [ µ u ] and hence (3.1.5.1) holds.Let ν ∈ Fr [ µ u ] . We find a sequence t n → ∞ such that ( µ u ) t n → ν in D ′ . Wefind a subsequence t ′ n such that u t ′ n → v ∈ F r [ u ] and ν v = ν. By the *Hadamardtheorem (Theorem 3.1.4.4) v = Π( • , ν, p ) . Hence, { Π( • , ν, p ) : ν ∈ Fr [ µ u ] } ⊂ Fr [ u ] . And vice versa, since Fr [ u ] ⊂ U [ ρ ] (Theorem 3.1.2.2, fr2)), every v ∈ Fr [ u ] isrepresented as Π( • , ν v , p ) and ν v ∈ Fr [ µ ] by (3.1.5.1). (cid:3) Let ρ be integer and u ∈ U ( ρ ( r )). Let us consider the precompact family ofhomogeneous polynomials δ t ( x, u, ρ ) t ρ − ρ ( t ) from Theorem 2.9.4.1. For every t n →∞ we can find a subsequence t ′ n such that the pair ( δ t ′ n ( • , u, ρ ) t ′ nρ − ρ ( t ′ n ) , ( µ u ) t ′ n )tends to a pair ( H ν , ν ) where H ν is a homogeneous harmonic polynomial of degree p. We denote the set of all such pairs as ( H , F r )[ u ] . Every v ∈ U [ ρ ] can be representedin the form (3.1.4.7). Thus for every v the polynomial H v := H p is determined. Theorem 3.1.5.2 (Connection between Fr’s for integer ρ ). Let u ∈ U ( ρ ( r )) . Then (3.1.5.3) ( H , F r )[ u ] = { ( H v , ν v ) : v ∈ Fr [ u ] } , (3.1.5.4) Fr [ u ] = { v := H ν + Π < ( • , ν, ρ ) + Π > ( • , ν, ρ ) : ( H ν , ν ) ∈ ( H , F r )[ u ] } . The proof is clear.
Up to now we supposed that the family of rotations V • was fixed. Now wetake in consideration that it can be vary and use the notation Fr [ u, V • ] . Theorem 3.1.6.1.(Dependence of
F r on V • ). Let Fr [ u, V • ] and Fr [ u, W • ] belimit sets of u with respect to rotation families V • and W • accordingly. Then forany v ∈ Fr [ u, V • ] there exist a rotation V v and w v ∈ Fr [ u, W • ] such that v ( x ) = w v ( V v x ) for all x ∈ R m . Proof.
Let v ∈ Fr [ u, V • ] and let t n → ∞ be a sequence such that t − ρ ( t n ) n u ( t n V t n • ) → v. Since the family V t is obviously precompact there exists a subsequence (forwhich we keep the same notation),and a rotation V v such that W − t n V t n → V v and w ∈ Fr [ u, W • ] such that t − ρ ( t n ) n u ( t n W t n • ) → w. Now we have v ( • ) = D ′ − lim t − ρ ( t n ) n u ( t n V t n • ) = D ′ − lim t − ρ ( t n ) n u ( t n W t n W − t n V t n • ) = w ( V v • ) . (cid:3) Let u ∈ SH ( ρ ( r )) and let Fr [ u ] be the limit set. Set(3.2.1.1) h ( x, u ) := sup { v ( x ) : v ∈ Fr [ u ] } (3.2.1.2) h ( x, u ) := inf { v ( x ) : v ∈ Fr [ u ] } . These functionals reflect the asymptotic behavior of u along rays of the form(3.2.1.3) l x := { x = t x : t ∈ (0 , ∞ ) } and are called indicator of growth of u and lower indicator respectively.Of course, the indicators depend on ρ ( r ) and V t , but we will only note that ifnecessary. Theorem 3.2.1.1(Properties of Indicators).
The following holdsh1) h is upper semicontinuous, h is subharmonic;h2) they are semiadditive and positively homogeneous, i.e., (3.2.1.4) h ( x, u + u ) ≤ h ( x, u ) + h ( x, u );(3.2.1.5) h ( x, u + u ) ≥ h ( x, u ) + h ( x, u );(3.2.1.6) h, h ( x, Cu ) = Ch, h ( x, u ) , C ≥ h3) invariance: (3.2.1.7) h, h [ t ] ( x, u ) = h, h ( x, u ) . Proof.
Semicontinuity of h follows from Theorem 2.1.2.8.(Commutativity of inf andM(.). Semicontinuity and subharmonicity of h follow from Theorem 2.7.3.4.(Sig-urdsson’s Lemma). The properties h2) follow from properties of infimum and supre-mum. The invariance follows from invariance of Fr [ u ] (Theorem 3.1.2.2, fr3)). (cid:3) Set(3.2.1.8) x ( x ) := P − | x | ( x )where P t is defined by (3.1.1.1).This is an intersection of the orbit of P t that passes through a point x withthe unit sphere.If V t ≡ I (3.2.1.9) x ( x ) = x/ | x | := x Theorem 3.2.1.2(Homogeneity h, h ). One has (3.2.1.10) h, h ( x, • ) = | x | ρ h, h ( x ( x ) , • )Thus the indicators are determined uniquely by their values on the unit sphere,i.e.,they are “functions of direction.In particular, they are homogeneous for V t ≡ I :(3.2.1.11) h, h ( x, • ) = | x | ρ h, h ( x , • )The proof of (3.2.1.10) follows from h4), Theorem 3.2.1.1 if we set t := | x | ; x := P − t x. In this item we will suppose that V t ≡ I and study the indicator.Let ∆ x be defined in 2.4.1. Its coefficients depend on a choice of the sphericalcoordinate system. However, one has Theorem 3.2.2.1.
Let ψ ( y ) have continuous second derivatives on the unit sphere S . Then the differential form ∆ x ψ ( y ) dy is invariant with respect to the choice ofspherical coordinate system.Proof. Let φ ( x ) be a smooth function in R m . Then ∆ φ ( x ) dx is invariant withrespect to the choice of an orthogonal system because ∆ (the Laplace operator)and an element of volume are invariant. Set φ ( x ) = ψ ( y ) , where y := x / | x | . Then∆ φdx = ∆ x ψ ( y ) dy mr m − dr. Since r is invariant with respect to rotations of the coordinate system, ∆ x ψ ( y ) dy is invariant with respect to the choice of a spherical coordinate system. (cid:3) Note that for m = 2 this theorem is obvious because(3.2.2.1) ∆ x = d dθ and it does not depend on translations with respect to θ. We define the operator ∆ x on f ∈ D ′ ( S ) by < ∆ x f, ψ > := < f, ∆ x ψ >, ψ ∈ D ( S )in a fixed spherical coordinate system.The definition is correct. Indeed, suppose in a fixed system(3.2.2.2) supp ψ ⊂ S \{ θ j = 0; π : j = 1 , , ..., m − } . Then all the coefficients of ∆ x are infinitely differentiable and ∆ x ψ ( y ) ∈ D ( S ) . By Theorem 3.2.2.1 we obtain that the condition of Theorem 2.3.5.2.( D ′ on Sphere)are fulfilled.Note that for m = 2 the operator ∆ x is realized by the formula (3.2.2.1) onfunctions of the form f = f ( e iθ ) , i.e., on 2 π -periodic functions. Theorem 3.2.2.2(Subsphericality of Indicator).
One has (3.2.2.3) [∆ x + ρ ( ρ + m − h ( y, u ) := s > in D ′ ( S ) , i.e., s is a measure on S . Proof.
It is sufficient to prove this locally, in any spherical system. Let R ( r ) befinite, infinitely differentiable and non-negative in (0; ∞ ) and let ψ ∈ D ( S ) be non-negative and satisfy (3.2.2.2). Set φ ( x ) := R ( | x | ) ψ ( x ) . Using the subharmonicityof h ( x, u ) (h1), Theorem 3.2.1.1 and (3.2.2.2), we have0 ≤ Z h ( x, u )∆ φ ( x ) dx = = Z ( y,r ) ∈ S × (0; ∞ ) r ρ h ( y, u ) (cid:20) r m − ∂∂r r m − ∂∂r + 1 r ∆ x (cid:21) ψ ( y ) r m − dydr. Transforming the last integral we obtain Z h ( x, u )∆ φ ( x ) dx =(3.2.2.4) = Z ∞ r ρ (cid:20) r m − ∂∂r r m − ∂∂r R ( r ) (cid:21) r m − dr Z S h ( y, u ) ψ ( y ) dy ++ Z ∞ r ρ − r m − R ( r ) dr Z S h ( y, u )∆ x ψ ( y ) dy. Integrating by parts in the first summand we obtain(3.2.2.5) Z ∞ r ρ (cid:20) r m − ∂∂r r m − ∂∂r R ( r ) (cid:21) r m − dr = Z ∞ R ( r ) ρ ( ρ + m − r ρ + m − dr. Substituting (3.2.2.5) into (3.2.2.4), we have0 ≤ Z ∞ r ρ + m − R ( r ) dr Z S h ( y, u )[∆ x + ρ ( ρ + m − ψ ( y ) dy. Since R ( r ) is an arbitrary non-negative function, Z S h ( y, u )[∆ x + ρ ( ρ + m − ψ ( y ) dy ≥ ψ. (cid:3) We will call an upper semicontinuous function which satisfies (3.2.2.3) a ρ - subspherical one. Now we are going to study properties of these functions. We consider the case m = 2 . A ρ -subspherical function for m = 2 is called ρ - trigonometrically convex ( ρ -t.c.). We will obtain for such a function a representationlike in Theorems 3.1.4.4, 3.1.4.5.(*,** Hadamard). Set T ρ := d dφ + ρ . Let us find a fundamental solution of this operator.Let ρ be non-integer. Let us denote as ] cos ρ ( φ ) the periodic continuation ofcos ρφ from the interval ( − π, π ) . Theorem 3.2.3.1.(Fundamental Solution of T ρ ). One has ρ sin πρ T ρ g cos ρ ( φ − π ) = δ ( φ ) in D ′ ( S ) Proof.
Let f ∈ D ( S ) . We have(3.2.3.1) Z π ] cos ρ ( φ − π )[ f ′′ + ρ f ] dφ = lim ǫ → Z π − ǫǫ cos ρ ( φ − π )[ f ′′ + ρ f ] dφ Integrating by parts we obtain Z π − ǫǫ cos ρ ( φ − π )[ f ′′ + ρ f ] dφ = cos ρ ( φ − π ) f ′ ( φ ) | π − ǫǫ + ρ sin ρ ( φ − π ) f ( φ ) | π − ǫǫ ++ Z π − ǫǫ f ( φ ) T ρ cos ρ ( φ − π ) dφ. However, T ρ cos ρ ( φ − π ) = 0 for φ ∈ ( ǫ, π − ǫ ) . Thus the limit in (3.2.3.1) is equalto f (0)2 ρ sin πρ. (cid:3) Let s be a measure on the circle S . SetΠ( φ, s ) := Z π ] cos ρ ( φ − ψ − π ) s ( dψ ) . Theorem 3.2.3.2..
One has T ρ Π( φ, s ) = (2 ρ sin πρ ) ds in D ′ ( S ) . The proof is the same as GPo3) in Theorem 2.5.1.1.
Theorem 3.2.3.3(Representation of ρ -t.c.f for a non-integer ρ ). Let h be ρ -t.c. on S for non-integer ρ and let s := T ρ h. Then h ( φ ) = 12 ρ sin πρ Π( φ, s ) . The proof is like in Theorem 3.1.4.4(*Hadamard).
We will suppose in this item that V t = I, m = 2 , ρ is integer. Theorem 3.2.4.1(Condition on s ). Let ρ be integer, h be ρ -t.c. and T ρ h = s .Then (3.2.4.1) Z π e iρφ ds = 0 Proof.
We have for f ∈ D ( S ) : < s, f > = < T ρ h, f > = < h, T ρ f > . Since e iρφ ∈ D ( S ) for integer ρ and T ρ e iρφ = 0 , we have for f := e iρφ < s, e iρ • > = < h, T ρ e iρ • > = 0 . (cid:3) Let us denote the periodic continuation of the function f ( φ ) := φ from theinterval [0 , π ) to ( −∞ , ∞ ) as ˜ φ. Theorem 3.2.4.2.(Generalized Fundamental Solution for T ρ ). One has T ρ [ − πρ ˜ φ sin ρφ ] = δ ( φ ) − π cos ρφ in D ′ ( S ) . Proof.
Let φ ∈ ( ǫ, π − ǫ ) . Then T ρ ˜ φ sin ρφ = 2 ρ cos ρφ because ˜ φ = φ when φ ∈ ( ǫ, π − ǫ ) . We have also( φ sin ρφ ) ′ = sin ρφ + φρ cos ρφ. Thus < T ρ ˜ • sin ρ • , f > = Z π ˜ φ sin ρφ T ρ f dφ = lim ǫ → Z π − ǫǫ φ sin ρφ T ρ f dφ. Integrating by parts we obtain Z π − ǫǫ φ sin ρφ T ρ f dφ = φ sin ρφf ′ ( φ ) | π − ǫǫ − − f ( φ )[sin ρφ + φρ cos ρφ | π − ǫǫ + Z π − ǫǫ T ρ [ φ sin ρφ ] f ( φ ) dφ. Passing to the limit as ǫ → f is periodic and continuouswe obtain < T ρ [˜ • sin ρ • ] , f > = − πρf (0)+2 ρ Z π cos ρφf ( φ ) dφ = − πρf (0)+ < cos ρ • , f > . (cid:3) Set ˆΠ( φ, ds ) := Z π ^ ( φ − ψ ) sin ρ ( φ − ψ ) s ( dψ ) . Theorem 3.2.4.3..
One has T ρ ˆΠ( • , ds ) = − πρds in D ′ ( S ) for s that satisfies (3.2.4.1).Proof. Using Theorem 3.2.4.2 we obtain < T ρ ˆΠ( • , ds ) , f > = < s, f > − π < Z π cos ρ ( • − ψ ) ds ψ , f > . The last integral is zero because of Theorem 3.2.4.1. (cid:3)
Theorem 3.2.4.4.(Representation of ρ -t.c.f. for an integer ρ ). Let h be a ρ -t.c.f.for an integer ρ and T ρ h := s. Then h ( φ ) = ℜ ce iφ + ˆΠ( φ, ds ) . for some complex constant c. Proof.
The function H ( φ ) := h ( φ ) − ˆΠ( φ, ds ) satisfies the equation T ρ H = 0 in D ′ ( S ) because of Theorem 3.2.4.3 and it is real.Thus H ( φ ) = ℜ ce iφ . (cid:3) The class
T C ρ of ρ -t.c.functions has a number of properties of subharmonicfunctions.The function ] cos ρφ is continuous and e φ sin ρφ is continuous for integer ρ. Therefore any ρ -t.c.f is continuous as follows from Theorem 3.2.3.3 and 3.2.4.4. Set E ( φ ) := 12 ρ sin ρ | φ | . For any interval I := ( α, β ) ⋐ ( − π, π ) this function satisfies the equality T ρ E = δ in D ′ ( α, β ), where δ is the Dirac function in zero.Let G I ( ψ, phi ) be the Green function of T ρ for the interval I. By definition itmust be symmetric with respect to φ, ψ and have the form(3.2.5.1) G I ( φ, ψ ) := 12 ρ sin ρ | φ − ψ | + A I cos ρφ cos ρψ + B I sin ρφ sin ρψ, where A I , B I are chosen such that G I ( φ, ψ ) be equal to zero on ∂ { I × I } . An explicitform of G I is given by G I ( φ, ψ ) = ( sin ρ ( β − φ ) sin ρ ( ψ − α ) ρ sin ρ ( β − α ) , for ψ < φ ; sin ρ ( β − ψ ) sin ρ ( φ − α ) ρ sin ρ ( β − α ) , for φ < ψ. The following assertion analogous to the Riesz theorem (Theorem 2.6.4.3):
Theorem 3.2.5.1.(Representation on I ). Let h ∈ T C ρ and the let I be aninterval of length mesI < π/ρ. Then h ( φ ) = Y ρ ( φ, h ) − β Z α G I ( φ, ψ ) s ( dψ ) , where Y ρ ( φ, h ) is the only solution of the boundary problem: (3.2.5.2) T ρ Y = 0 , Y ( α ) = h ( α ) , Y ( β ) = h ( β ) and ds := T ρ h. Proof.
Set Π I ( φ, ds ) := β Z α G I ( φ, ψ ) s ( dψ )One can check like in Theorem 3.2.3.2 that T ρ Π I = − ds in D ′ ( I ) . Then the function Y ρ ( φ ) := h ( φ ) + Π I ( φ, ds ) satisfies the conditions (3.2.5.2). (cid:3) The explicit form of Y ρ ( φ ) is(3.2.5.3) Y ρ ( φ ) = h ( α ) sin ρ ( β − φ ) + h ( β ) sin ρ ( φ − α )sin ρ ( β − φ ) . Since Π I ( φ ) ≥ Theorem 3.2.5.2. ( ρ -Trigonometric Majorant). Suppose h ∈ T C ρ and Y ρ ( φ ) is the solution of (3.2.5.2). Then h ( φ ) ≤ Y ρ ( φ ) , φ ∈ I if β − α < π/ρ. This inequality can be written in the symmetric form(3.2.5.4) h ( α ) sin ρ ( β − φ ) + h ( φ ) sin ρ ( α − β ) + h ( β ) sin ρ ( φ − α ) ≥ α, φ, β ) − min( α, φ, β ) < π/ρ. It is called the fundamental relation of indi-cator.
Theorem 3.2.5.3.(Subharmonicity and ρ -t.c.). A function h ( φ ) ∈ T C ρ iff thefunction u ( re iφ ) := h ( φ ) r ρ is subharmonic in R . Proof.
Sufficiency follows from Theorem 3.2.2.2. Let us prove necessity. The func-tion u ( z ) := r ρ sin ρ | φ | is subharmonic. Actually, it is harmonic for φ = 0 , r = 0and can be represented in the form u ( z ) = max( r ρ sin φ, − r ρ sin φ )in a neighborhood of the line φ = 0 . Hence, it is subharmonic because of sh2),Theorem 2.6.1.1 (Elementary Properties).The function u ( z ) := r ρ Π I ( φ ) is subharmonic because of sh5) and sh4), The-orem 2.6.1.1. The function r ρ Y ρ ( φ ) is harmonic for r >
0. This can be checkeddirectly. Hence, u ( z ) is subharmonic for r > u ( z ) is also subharmonic for r = 0 , because it is, obviously, continuousat z = 0 . (cid:3) Theorem 3.2.5.4.(Elementary Properties of ρ -t.c.Functions). One hastc1) If h ∈ T C ρ , then Ah ∈ T C ρ for A > tc2) If h , h ∈ T C ρ , then h + h , max( h , h ) ∈ T C ρ . These properties follow from Theorem 3.2.5.3 and properties of subharmonicfunctions.
Exercise 3.2.5.1
Prove Th.3.2.5.4.Similarly to (usual) convexity, ρ -t.convexity of functions implies several ana-lytic properties. Theorem 3.2.5.5.
Let h ∈ T C ρ . then there exist right ( h ′ + ) and left ( h ′− ) deriva-tives and they coincide everywhere except, maybe,for countable set of points.Proof. It is enough to proof these properties for the potentialΠ( φ ) := β Z α sin ρ | φ − ψ | ds ψ , because of (3.2.5.1) and Theorem 3.2.5.1.We will prove the following(3.2.5.5) Π ′ + ( φ ) = ρ φ − Z α cos ρ ( φ − ψ ) ds ψ + ρµ ( φ ) − ρ β Z φ +0 cos ρ ( φ − ψ ) ds ψ ;(3.2.5.6) Π ′− ( φ ) = ρ φ − Z α cos ρ ( φ − ψ ) ds ψ − ρµ ( φ ) − ρ β Z φ +0 cos ρ ( φ − ψ ) ds ψ , where µ ( φ ) is the measure, concentrated in the point φ. We have for ∆ > φ + ∆∆ = φ − Z α sin ρ | φ + ∆ − ψ | − sin ρ | φ − ψ | ∆ ds ψ ++ sin ρ ∆∆ µ ( φ ) + φ +∆ Z φ +0 ... + β Z φ +∆ .... Let us estimate the second integral.We have φ +∆ Z φ +0 (cid:12)(cid:12)(cid:12)(cid:12) sin ρ | φ + ∆ − ψ | − sin ρ | φ − ψ | ∆ (cid:12)(cid:12)(cid:12)(cid:12) ds ψ ≤≤ ρ ∆∆ [ s ( φ + ∆) − s ( φ + 0)] = o (1)when ∆ → +0 . Passing to the limit, we obtain (3.2.5.5). The equality (3.2.5.6) is obtained bythe same way when ∆ < . Since µ ( φ ) = 0 at most in a countable set for all the other points Π ′ + ( φ ) =Π ′− ( φ ) . (cid:3) Now we consider the case m ≥ . We will obtain for the ρ -subsphericalfunction a representation like for the ρ -trigonometrically convex functions. Theorem 3.2.6.1 (Subharmonicity and Subsphericality).
Let h be subspher-ical in a neighborhood of y ∈ S . Then the function u ( x ) := h ( y ) r ρ , x = ry issubharmonic in the corresponding neighborhood of the ray x = ry : 0 < r < ∞ . Proof.
Let f ∈ D ′ ( R m \ . We can represent it in the form f := f ( rx ) , x ∈ S where f ( • , x ) ∈ D ′ (0 , ∞ ) for any x. Then < u, ∆ f > = ∞ Z Z S u ( rx )∆ f ( rx ) r m − drds x == ∞ Z Z S u ( rx ) 1 r m − ∂∂r r m − ∂∂r f ( rx ) r m − drds x + ∞ Z Z S u ( rx ) 1 r ∆ x f ( rx ) r m − drds x . Integrating by parts in the first integral, we obtain ∞ Z ρ ( ρ + m − r ρ + m − Z S h ( x ) f ( rx ) drds x . Set(3.2.6.1) S ρ := ∆ x + ρ ( ρ + m − Together with the second summand we obtain < u, f > = ∞ Z Z S h ( x ) S ρ f ( rx ) ds x r ρ + m − dr > f ( rx ) ≥ . (cid:3) Note that the Riesz measure for such u has the form µ ( r m − drds x ) = r ρ + m − drν h ( ds x ) , where ν h is a positive measure on S , that is equal to S ρ h in D ′ ( S ) . For a non-integer ρ set E ρ ( x, y ) := ∞ Z G p ( x, ry, m ) r ρ + m − dr, where G p is the primary kernel and x, y ∈ S . Theorem 3.2.6.2.
For non-integer ρ and any ρ -subspherical function h one has h ( x ) = Z S E ρ ( x, y ) dν h . Proof.
Set in (3.1.4.5) v := r ρ h ( x ) . It is clear that v ∈ U [ ρ ] . We have r ρ h ( x ) = Z S ∞ Z G p ( rx, ty, m ) t ρ + m − dtν h ( ds x ) . Now we make the change t ′ := t/r and use the homogeneity of G p ( rx, ty, m ) . (cid:3) Exercise 3.2.6.1.
Show that E ρ ( x, y ) is a fundamental solution of the operator S ρ . For an integer ρ = p set E ′ ρ ( x, y ) := Z G p − ( x, ry ) r ρ + m − dr + ∞ Z G p ( x, ry ) r ρ + m − dr. Exercise 3.2.6.2.
Prove. Theorem 3.2.6.3.
For any integer ρ = p and any ρ -subspherical function h onehas h ( x ) = Y p ( x ) + Z S E ′ ρ ( x, y ) dν h . where Y p is some p -spherical function.For any p -spherical function Y Z S Y ( x ) dν h = 0 . We return to the general case when x ∈ R m , V t is a one parametric group, ρ ( r ) is a proximate order and u ∈ SH ( ρ ( r )) . The following theorem representsindicators in a form of limits in usual topology.
Theorem 3.2.7.1(Classic Indicators).
One has (3.2.7.1) h ( x, u ) = sup T [lim sup t j →∞ u t j ] ∗ ( x ) = [lim sup t →∞ u t ( x )] ∗ where * can be deleted outside a set of zero capacity, and (3.2.7.2) h ( x, u ) = inf T [lim sup t j →∞ u t j ] ∗ ( x ) , where T is the set of all the sequences that tend to infinity.Proof. Let us prove (3.2.7.1). Set(3.2.7.3) h ( x, u, { t j } ) := lim sup t j →∞ u t j ( x ) . Let v ∈ Fr [ u ] and u t j → v in D ′ . Then(3.2.7.4) h ∗ ( x, u, { t j } ) = v ( x )by Theorem 2.7.3.3.(H.Cartan+). Thus(3.2.7.5) sup T h ∗ ( x, u, { t j } ) ≥ h ( x, u ) . Let ǫ > t j := t j ( x ) be a sequence such that h ∗ ( x, u, { t j } ) ≥ sup T h ∗ ( x, u, { t j } ) − ǫ We can find a subsequence { t j } (we keep the same notation for it) and v ∈ Fr [ u ]such that u t j → v in D ′ . From (3.2.7.4) we obtain h ( x, u ) ≥ v ( x ) ≥ sup T h ∗ ( x, u, { t j } ) − ǫ. Thus the reverse inequality to (3.2.7.5) holds. Therefore h ( x, u ) = sup T h ∗ ( x, u, { t j } )Let us prove the second equality in (3.2.7.1). Sincesup T h ( x, u, { t j } ) = lim sup t →∞ u t ( x )we have(3.2.7.6) h ( x, u ) ≥ [lim sup t →∞ u t ] ∗ ( x )Let us prove the opposite inequality. Let v ∈ Fr [ u ] . There exists a sequence t j → ∞ such that u t j → v in D ′ ( R m ) . By (3.2.7.4)[lim sup t →∞ u t ] ∗ ≥ h ∗ ( x, u, { t j } ) = v ( x )Since it holds for every v ∈ Fr [ u ] we have the reverse inequality to (3.2.7.6).Hence,(3.2.7.1) is proved completely.Let us prove (3.2.7.2). From (3.2.7.4) we haveinf T h ∗ ( x, u, { t j } ) ≤ v ( x )for all v ∈ Fr [ u ] . Therefore(3.2.7.7) inf T h ∗ ( x, u, { t j } ) ≤ h ( x, u ) . Let us prove the opposite inequality. Let { t j } be any sequence that tends to ∞ . Let us find a subsequence { t j ′ } such that u t j ′ → v in D ′ ( R m ) . Then h ( x, u, { t j } ) ≥ lim sup j ′ →∞ u t j ′ ( x ) . Taking * from the two sides of this inequality and using Theorem 2.7.3.3, we obtain h ∗ ( x, u, { t j } ) ≥ [lim sup j ′ →∞ u t j ′ ] ∗ ( x ) = v ( x ) ≥ h ( x, u ) . This implies the reverse inequality to (3.2.7.7). Hence (3.2.7.2) holds. (cid:3) Corollary 3.2.7.2.
If all the functions (3.2.7.3) are upper semicontinuous, then h ( x, u ) = lim sup t →∞ u t ( x ) h ( x, u ) = lim inf t →∞ u t ( x ) Proof.
We have h ∗ ( x, u, { t j } ) = h ( x, u, { t j } ) and thus h ( x, u ) = sup T [lim sup t j →∞ u t j ]( x ) = lim inf t →∞ u t ( x ) .h ( x, u ) = inf T [lim sup t j →∞ u t j ]( x ) = lim inf t →∞ u t ( x ) . (cid:3) Theorem 3.2.7.3(Indicators of Harmonic Function).
Let u ∈ SH ( ρ ( r )) beharmonic for all the large | y | in a “cone” of the form Co Ω := { y = P t x : x ∈ Ω , t ∈ (0; ∞ ) } where Ω ⊂ S . Then (3.2.7.7) h ( x, u ) = lim sup t →∞ u t ( x ) and (3.2.7.8) h ( x, u ) = lim inf t →∞ u t ( x ) for x ∈ Co Ω . Proof.
The harmonicity of u in Co Ω implies [ u t ] ǫ ( x ) = u t ( x ) for large t and suffi-ciently small ǫ when x ∈ Co Ω . The family [ u t ] ǫ is uniformly continuous by reg3), Theorem 2.3.4.5 (Propertiesof Regularizations).Thus the function (3.2.7.5) is continuous. Therefore we can useCorollary 3.2.7.2. (cid:3) Theorem 3.2.7.4.(Indicator for m = 2 ). Let u ∈ SH ( R ) .Then (3.2.7.9) h ( z, u ) = lim sup t →∞ u t ( x ) , i.e., the star in (3.2.7.1) can be deleted. Proof.
Let as denote as h ( z, u ) the right part of (3.2.7.9).The “homogeneity” ofthe indicator (3.2.1.10) and also of h ( z, u ) implies the following property: if theinequality h ( z, u ) < h ( z.u ) holds for some z , it holds on the whole orbit z = { P t z : 0 < t < ∞} that has a positive capacity in R . This contradicts Theorem 3.2.7.1. (cid:3) G is an open set, K is a compact and E - a bounded Borelset. Let µ ∈ M ( ρ ( r )) and Fr [ µ ] := Fr [ µ, ρ ( r ) , V t , R m ] be the limit set of µ . Set∆( G, µ ) := sup { ν ( G ) : ν ∈ Fr [ µ ] } ;∆( E, µ ) := inf { ∆( G, µ ) : G ⊃ E } ;∆( K, µ ) := inf { ν ( K ) : ν ∈ Fr [ µ ] } ;∆( E, µ ) := sup { ∆( K ) : K ⊂ E } . The quality ∆(
E, µ ) , (∆( E, µ )) is called the upper ( lower) density of µ relative tothe proximate order ρ ( r ) and the family V t . Theorem 3.3.1.1.(Properties of Densities).
The following properties holddens1) if E = ∅ , then ∆( E, • ) = ∆( E, • ) = 0 dens2) ∀ E, ∆( E, • ) ≤ ∆( E, • ); dens3) monotonicity: ∆ , ∆( E , • ) ≤ ∆ , ∆( E , • ) for E ⊂ E ;dens4) generalized semi-additivity with respect to a set: (3.3.1.1) ∆( E ∪ E , • ) + ∆( E ∩ E , • ) ≤ ∆( E , • ) + ∆( E , • )(3.3.1.2) ∆( E ∪ E , • ) + ∆( E ∩ E , • ) ≥ ∆( E , • ) + ∆( E , • ) . dens5) continuity from the right and from the left. (3.3.1.3) E n ↑ E = ⇒ ∆( E n , • ) ↑ ∆( E, • ); K n ↓ K = ⇒ ∆( K n , • ) ↓ ∆( K, • );(3.3.1.4) E n ↓ E = ⇒ ∆( E n , • ) ↓ ∆( E, • ); G n ↑ G = ⇒ ∆( G n , • ) ↑ ∆( G, • ) . dens6) semi-additivity and positive homogeneity with respect to µ, i.e., (3.3.1.5) ∆( E, µ + µ ) ≤ ∆( E, µ ) + ∆( E, µ ); see Exercise 3.3.1.125 (3.3.1.6) ∆( E, µ + µ ) ≥ ∆( E, µ ) + ∆( E, µ );(3.3.1.7) ∆ , ∆( E, λµ ) = λ ∆ , λ ∆( E, µ ) for λ ≥ dens7) invariance with respect to the map ( • ) [ t ] (see, 3.1.2.4a), i.e., t − ρ − m +2 ∆ , ∆( P t E, • ) = ∆ , ∆( E, • ) . Proof of Theorem 3.3.1.1.
The property dens1) holds because the empty set is openby definition. The properties dens2) and dens3) hold because of the monotonicityof ν. Let us prove dens4). Since ν is a measure we have ν ( G ∪ G , µ ) + ν ( G ∩ G , µ ) = ν ( G , µ ) + ν ( G , µ )for any G ⊃ E and G ⊃ E . From this we obtain(3.3.1.8) ν ( G ∪ G , µ ) + ν ( K ∩ K , µ ) ≤ ν ( G , µ ) + ν ( G , µ )for K ⊂ E and K ⊂ E . The right side of (3.3.1.8) is no larger than ∆( G , • ) + ∆( G , • ) . Now we cantake supremum over ν ∈ Fr µ in the first summand of the left side and infimum inthe second summand. Thus we obtain(3.3.1.9) ∆( G ∪ G , • ) + ∆( K ∩ K , • ) ≤ ∆( G , • ) + ∆( G , • )Since ∆( E, • ) and ∆( E, • ) are monotonic with respect to E, inf { ∆( G ∪ G , • ) : G ⊃ E , G ⊃ E } = ∆( E ∪ E , • )and sup { ∆( K ∩ K , • ) : K ⊂ E , K ⊂ E } . Thus we obtain the first inequality in dens4) from (3.3.1.9). The second one canbe obtained analogously .Let us prove dens5). For arbitrary G ⊃ K there exists n such that K n ⊂ G for n > n . According to dens3)∆( K, • ) ≤ ∆( K n , • ) ≤ ∆( G, • ) . Hence, ∆( K, • ) ≤ lim n →∞ ∆( K n , • ) ≤ ∆( G, • ) . Taking infimum over all G ⊃ K, we obtain the second assertion in (3.3.1.3).For G n ↑ G we have the equality(3.3.1.10) lim n →∞ ∆( G n , • ) = sup n ∆( G n , • ) = ∆( G, • )because one can change the order of taking the supremum on n and on ν ∈ Fr [ µ ] . Let E n ↑ E and let ǫ be arbitrarily small. One can find G n ⊃ E n such that∆( G n , • ) ≤ ∆( E n , • ) + ǫ. Since G := S ∞ G n ⊃ E we have∆( G n , • ) − ǫ ≤ ∆( E n , • ) ≤ ∆( E, • ) ≤ ∆( G, • ) . Using (3.3.1.10), we obtain∆( E, • ) − lim n →∞ ∆( E n , • ) ≤ ǫ. Since ǫ is arbitrary small ∆( E, • ) ≤ lim n →∞ ∆( E n , • )and hence the first assertion in (3.3.1.8) holds.The assertion (3.3.1.9) can be proved analogously. Exercise 3.3.1.2 see Exercise 3.3.1.327 Let us prove dens6).One has∆(
G, µ + µ ) = sup { ν ( G ) : ν ∈ Fr [ µ + µ ] } . Since Fr [ µ + µ ] ⊂ Fr [ µ ] + Fr [ µ ](see frmu1), Theorem 3.1.3.4 (Properties of µ Fr [ µ ])) one can continue theprevious inequality as ≤ sup { ν ( G ) : ν ∈ Fr [ µ ] + Fr [ µ ] } == sup { ν ( G ) : ν ∈ Fr [ µ ] } + sup { ν ( G ) : ν ∈ Fr [ µ ] } = ∆( G, µ ) + ∆( G, µ ) . Passing to the infimum over G ⊃ E, we obtain (3.3.1.5). The assertions (3.3.1.6)and (3.3.1.7) can be proved analogously. The properties dens7) follow from the invariance of Fr [ µ ] (see frm3), Theorem3.1.3.3. (Properties of F r [ µ ])). (cid:3) Exercise 3.3.1.1.
Prove the subadditivity of ∆( E, • ) :∆( E ∪ E , • ) ≤ ∆( E , • ) + ∆( E , • )and the superadditivity of ∆( E, • ) :∆( E ∪ E , • ) ≥ ∆( E , • ) + ∆( E , • ) . from the Theorem 3.3.1.1. Exercise 3.3.1.2.
Prove (3.3.1.2).
Exercise 3.3.1.3.
Prove (3.3.1.4).
Exercise 3.3.1.4.
Prove (3.3.1.6) and (3.3.1.7).Set for I ⊂ (0 , ∞ ) and Ω ⊂ S Co Ω ( I ) := { x = P t y : y ∈ Ω , t ∈ I } . Also set I t := (0 , t ) . see Exercise 3.3.1.428 Theorem 3.3.1.2.(Cone’s Densities).
One has ∆ , ∆( Co Ω ( I t )) = t ρ + m − ∆ , ∆( Co Ω ( I )) . We obtain this from dens7), Theorem 3.3.1.1, taking E := Co Ω ( I ) . Exercise 3.3.1.5.
Show that for m = 2 , S = {| z | = 1 } , Ω = { z = e iφ : φ ∈ ( α, β ) } Co Ω ( I t ) is a sector of radius t corresponding to the arc ( α, β ) on the unit circle. Let δ ( E ) be a monotonic function of E ∈ R m . A set E is called δ -squarableif(3.3.2.1) sup K ⊂ E δ ( K ) = inf G ⊃ E δ ( G ) . Example 3.3.2.1.
Let δ ( E ) be a measure. Then (3.3.2.1) implies δ ( ∂E ) = 0 , i.e., E is δ -squarable in sense of item 2.2.3. Exercise 3.3.2.2.
Prove.
Theorem 3.3.2.1. If ∆( ∂E ) = 0 then E is ∆ -squarable. If E is ∆ -squarable, then ∆( ∂E ) = 0 . Set E t := { x : ∃ y ∈ E : | x − y | < t } . This is a t -extension of E. A family of sets A is said to be dense in a family A if for each set E ∈ A and an arbitrary small ǫ > E ∈ A such that(3.3.2.2) E ∆ E := ( E \ E ) ∪ ( E \ E ) ⊂ ( ∂E ) ǫ . Exercise 3.3.2.3.
Prove
Theorem 3.3.2.2.
The relation “to be dense in” is reflexive and transitive, i.e., A is dense in A , and(3.3.2.3) {A is dense in A } ∧ {A is dense in A } = ⇒ {A is dense in A } . There are lots of squarable sets. Theorem 3.3.2.3.
For any monotonic δ ( E ) the class of δ -squarable sets is densein the class of all the subsets of R m . Proof.
For any E ⊂ R m set(3.3.2.4) E ( t ) := E ∪ ( ∂E ) t . One can check that(3.3.2.5) E ∆ E ( t ) ⊂ ( ∂E ) t and(3.3.2.6) E ( t ) ⊂ ◦ E ( t )for t < t . The function f ( t ) := δ ( ◦ E ( t )) is monotonic. Hence, its set of continuity pointshas a concentration point at t = 0.Suppose ǫ > t < ǫ is a continuity point for f ( t ) . From (3.3.2.6) we havelim t → t − ǫ δ ( E ( t )) ≤ sup K ⊂ E t δ ( K ) ≤ inf G ⊃ E t δ ( G ) ≤ lim t → t + ǫ δ ( ◦ E ( t ))Hence, E t is δ -squarable. From (3.3.2.5) we have E ∆ E ( t ) ⊂ ( ∂E ) ǫ . (cid:3) Set ∆ cl ( E ) = lim sup t →∞ µ t ( E ); ∆ cl ( E ) = lim inf t →∞ µ t ( E ) . These are classic densities determined without D ′ -topology. They are monotonic.The following assertion connects these densities to ∆ and ∆ . Theorem 3.3.2.4.(Classic Densities).
For any ∆ cl -squarable set E (3.3.2.7) ∆ cl ( E ) = sup { ν ( E ) : ν ∈ Fr [ µ ] } = ∆( E, µ ) . For any ∆ cl -squarable set E (3.3.2.7’) ∆ cl ( E ) = inf { ν ( E ) : ν ∈ Fr [ µ ] } = ∆( E, µ ) . The theorem follows obviously from the following assertion
Theorem 3.3.2.5.
One has (3.3.2.8) sup K ⊂ E ∆ cl ( K ) ≤ sup ν ∈ Fr ν ( E ) ≤ ∆( E ) ≤ inf G ⊃ E ∆ cl ( G );(3.3.2.9) sup K ⊂ E ∆ cl ( K ) ≤ inf ν ∈ Fr ν ( E ) ≤ ∆( E ) ≤ inf G ⊃ E ∆ cl ( G ); Proof.
Let us prove, for example, (3.3.2.9). Let us choose any G and K such that K ⊂ E ⊂ G. We can find a sequence t j → ∞ such thatlim j →∞ µ t j ( G ) = ∆ cl ( G ) . Choose a subsequence t j n such that µ t jn → ν in D ′ for some ν ∈ Fr . Using Theorems 2.3.4.4.(D’and C*) and 2.2.3.1.(C*-limits), we obtain(3.3.2.10) ν ( G ) ≤ lim inf n →∞ µ t jn ( G ) = ∆ cl ( G ) . By the same theorems(3.3.2.11) ∆ cl ( K ) ≤ lim sup n →∞ µ t jn ( K ) ≤ ν ( K ) . From (3.3.2.10) and (3.3.2.11) we obtain(3.3.2.12) ∆ cl ( K ) ≤ ν ( E ) ≤ ν ( G ) ≤ ∆ cl ( G )because of monotonicity of ν ( E ) . Taking supremum over all K ⊂ E and infimumover all G ⊃ E, we obtain (3.3.2.9). (cid:3) Exercise 3.3.2.4.
Prove (3.3.2.8). Corollary 3.3.2.6.
The following holds (3.3.2.13) ∆ cl ( K t ) = ∆( K t , µ ) = t ρ + m − ∆( K , µ ) , t ≥ . (3.3.2.13’) ∆ cl ( K t ) = ∆( K t , µ ) = t ρ + m − ∆( K , µ ) , t ≥ . where K t = { x : | x | < t } is the ball.Proof. The right equalities follow from Theorem 3.3.1.2 with Ω := S . The leftequalities hold at least for one t because of Theorem 3.3.2.4 and hence for all t. (cid:3) Let us note generally speaking that values of ∆ and ∆ on the sets Co Ω ( I t )do not determine their values even on the sets Co Ω ( I ) for I = ( t , t ) . However thefollowing assertion holds.
Theorem 3.3.3.1.(Existence of Density).
Let Φ be a dense ring (see, 2.2.3)on S . Then the conditions (3.3.3.1) ∆( Co Ω ( I t )) = ∆( Co Ω ( I t )) for Ω ∈ Φ and some t determine uniquely a measure ∆(Ω) on S . Fr [ µ ] consists ofone single measure ν and (3.3.3.2) ν ( Co Ω ( I t )) = t ρ + m − ∆(Ω) for all the t ∈ (0 , ∞ ) . To prove this we need an assertion that is valuable by itself. Set(3.3.3.3) ∆(Ω) := ∆( Co Ω ( I )); ∆(Ω) := ∆( Co Ω ( I ))forΩ ∈ S . We will call them angular densities because for m = 2 and V t ≡ I Ω determinesan angle in the plane.Let Ω G denote an open set in S and Ω K a closed one. Theorem 3.3.3.2.(Angular Densities).
One has (3.3.3.4) ∆(Ω) = sup Ω G ⊃ Ω ∆(Ω G ); ∆(Ω) = inf Ω K ⊂ Ω ∆(Ω K ) . Proof.
We need to prove two assertions(3.3.3.5) ∀ ǫ > ∃ Ω G : ∆(Ω G ) < ∆(Ω) + ǫ ;(3.3.3.6) ∀ ǫ > ∃ Ω K : ∆(Ω K ) > ∆(Ω) − ǫ ;Let us prove (3.3.3.5). SetΩ G ( ǫ ) := Co Ω ( I ǫ ) ∪ {| x | < ǫ } . This is an open set that contains Co Ω ( I ) . One can show the following:
Exercise 3.3.3.1.
For every open set G ⊃ Co Ω ( I ) there exists ǫ > G ⊂ S such that Ω G ( ǫ ) ⊂ G. We will show(3.3.3.7) ∆(Ω G ( ǫ )) < ∆(Ω G ) + o (1)uniformly with respect to Ω G ⊂ S while ǫ → . We have from Exercise 3.3.1.1(3.3.3.8) ∆(Ω G ( ǫ )) ≤ ∆( Co Ω ( I ǫ )) + ∆( {| x | < ǫ } ) . The property dens7), Theorem 3.3.1.1, gives∆( Co Ω G ( I ǫ )) = ∆( Co Ω G ( I ))(1 + ǫ ) ρ + m − . Since ∆( Co Ω G ( I )) ≤ ∆( {| x | < } ) we have(3.3.3.9) ∆( Co Ω G ( I ǫ )) = ∆( Co Ω G ( I )) + o (1)uniformly with respect to Ω G ⊂ S as ǫ → . By dens7) we also have(3.3.3.10) ∆( {| x | < ǫ } ) = ∆( {| x | < } ) ǫ ρ + m − = o (1) . From (3.3.3.10),(3.3.3.9) and (3.3.3.8) we obtain (3.3.3.7). Hence (3.3.3.5) is proved.Let us prove (3.3.3.6). SetΩ K ( ǫ ) := Co Ω K ( ¯ I − ǫ ) \ {| x | < ǫ } where ¯ I is the closure of I. One can show the following:
Exercise 3.3.3.2.
For any compact K ⊂ Co Ω ( I ) there exist Ω K ⊂ Ω and ǫ > K ⊂ Ω K ( ǫ ) ⊂ Co Ω ( I ) . From the definition of ∆(Ω) and the monotonicity we obtain (3.3.3.6). (cid:3)
Proof of Theorem 3.3.3.1.
Suppose (3.3.3.1) holds .The property dens7), Theorem3.3.1.1, implies (3.3.3.1) for all the t ∈ (0 , ∞ ) . Set ∆(Ω) := ∆(Ω) = ∆(Ω) for Ω ∈ Φ . Let us prove that ∆ satisfies the conditions ∆1) − ∆3) from 2.2.3. The con-ditions ∆1) and ∆2) follow from dens3) and dens4), Theorem 3.3.1.1, Exercise3.3.1.1.Let us prove ∆3) . By Th.3.3.3.2 for arbitrary Ω ∈ Φ and ǫ > G ⊃ Ω such that ∆(Ω) > ∆(Ω G ) − ǫ and Ω K ⊂ Ω such that ∆(Ω) < ∆(Ω K )+ ǫ. Suppose Ω ′ ∈ Φ satisfies the condition Ω K ⊂ Ω ′ ⊂ Ω G . Then∆(Ω ′ ) = ∆(Ω ′ ) ≤ ∆(Ω G ) ≤ ∆(Ω) + ǫ = ∆(Ω) + ǫ and ∆(Ω) − ǫ = ∆(Ω) − ǫ ≤ ∆(Ω K ) ≤ ∆(Ω ′ ) = ∆(Ω ′ ) , implying ∆3) . (cid:3) The most complete and effective description of an arbitrary limit set can bedone in terms of dynamical systems (see, [An]).A family of the form T t : M M, t ∈ R , on a compact metric space ( M, d ) with a metric d ( • , • ) is a dynamical system ( T • , M ) if it satisfies the condition T t + τ = T t ◦ T τ , t, τ ∈ R and the map ( t, m ) T t m is continuous with respect to ( t, m ) , for all t ∈ R , m ∈ M. ****Let m, m ′ ∈ M, and ǫ, s > . An ( ǫ, s ) -chain from m to m ′ is a finite sequence m = m, m , ..., m n = m ′ , satisfying the conditions d ( T t j m j , m j +1 ) < ǫ, j =0 , , ..., n − , for some t j ≥ s. A dynamical system ( T t , M ) is called chain recurrent (see, [HS]), an arbitrarilysmall ǫ > s > ǫ, s )-chain in M from m to m. Theorem 4.1.1.1 (Properties of Chain Recurrence).
Let ( T t , M ) be a dy-namical system on a compact set.Then the following conditions are equivalent:cr1) M is connected and ( T t , M ) is chain recurrent;cr2) for every open proper U ⊂ M satisfying (4.1.1.1) T t U ⊂ U, −∞ < t < , the boundary ∂U contains a non-empty T -invariant subset of M ; cr3) for every closed proper K ⊂ M satisfying (4.1.1.2) T t K ⊂ K, t ≥ , the boundary ∂K contains a non-empty T -invariant subset of M ; cr4) there does not exist any open proper V ⊂ M satisfying T τ clos V ⊂ V forsome τ > cr5) for any small ǫ > , large s > , and every pair of points m, m ′ thereexists an ( ǫ, s ) -chain from m to m ′ . Proof.
The conditions cr2) and cr3) are equivalent. Let us prove, for example,cr2)= ⇒ cr3). Set U := M \ K. It is open. Applying to (4.1.1.2) T − t and, usingthe invariance of M, we obtain (4.1.1.1) for U . Hence ∂U contains a non-emptyinvariant subset of M. Since ∂K = ∂U we obtain cr2).Let us prove the implication cr1)= ⇒ cr3). Let K ⊂ M be closed, proper andsatisfy (4.1.1.2). Since M is proper ∂K is non-empty.Let W denote the interior of K in M .The continuity of T and (4.1.1.2) imply(4.1.1.3) T t W ⊂ W for t ≥ . Indeed, T t W ⊂ K. It must be open.Thus it can not contain any point of ∂K , since else it would contain some neighborhood of this point, contradicting thedefinition of ∂K. Suppose that ∂K does not contain any non-empty T -invariant set.Let us show that there exists s > T s K ⊂ W. For any m ∈ ∂K there exists t = t ( m ) such that T t m ∈ W. There exists a neigh-borhood V m of m in ∂K that passes to W under T t ( m ) -action because of continuityof T t m on m .We also have T t V m ⊂ W for t > t ( m ) because of (4.1.1.3). Since ∂K is compactwe can cover it by a finite number of neighborhoods and obtain s such that(4.1.1.5) T s ∂K ⊂ W. (4.1.1.5) and (4.1.1.3) give (4.1.1.4). Set ǫ := 0 . d ( ∂K, T s K ) . From (4.1.1.2) we see that T t K ⊂ T s K for t > s. Therefore there does not exist any ( ǫ, s )-chain from a small neighborhood of a point m ∈ ∂K to itself. This contradicts the chain recurrence of M. Let us prove cr3)= ⇒ cr4).Assume that there exists an open proper V ⊂ M satisfying T τ clos V ⊂ V forsome τ > . We will construct K that does not satisfy cr3). Set W := S ≤ t ≤ τ T t V and K := clos W. Then(4.1.1.6) T s W ⊂ W, ∀ s ≥ . Indeed, let s = kτ + s ′ , s ′ ∈ [0 , τ ) , k ∈ Z . Then(4.1.1.7) T s W = [ t ∈ [0 ,τ ] T t + s V. Since T τ V ⊂ V we have T t + kτ V ⊂ T t V for t > . From (4.1.1.7) we obtain T s W = [ t ∈ [0 ,τ ] T t + s ′ + kτ V ⊂ [ t ∈ [0 ,τ ] T t + s ′ V = [ t ′ ∈ [ s ′ ,τ + s ′ ] T t ′ V = [ t ∈ [ s ′ ,τ ] T t V ∪ [ t ∈ [ τ,τ + s ′ ] T t V := W ∪ W . Further we have W ⊂ W by definition. W can be represented in the form W = [ t ∈ [0 ,s ′ ] T t + τ V. Since T t + τ V = T t T τ V and T τ V ⊂ V by the assumption we get: W ⊂ W ⊂ W. This implies (4.1.1.6).The same holds for K because of continuity of T t , i.e. K satisfies (4.1.1.2).Let us prove the equality(4.1.1.8) K = [ ≤ t ≤ τ T t clos V. Denote as K ′ the right side of (4.1.1.8).The set K ′ is closed because of compactness of [0 , τ ] . Indeed, let the sequence { T t j v j : j = 1 , , ... } ∈ T t j (clos V ) converge to w. Choose a subsequence t j k → s ∈ [0 , τ ] . Then v := lim k →∞ v j k = lim k →∞ T − t jk w = T − s w. Since clos V is closed, v ∈ clos V. Thus w = T s v for some s ∈ [0 , τ ] and some v ∈ clos V, i.e. w ∈ K ′ . Now, W ⊂ K ′ because T t V ⊂ clos T t V = T t clos V. Hence, K := clos W ⊂ clos K ′ = K ′ . We also have( T t V ⊂ W ∀ t ∈ [0 , τ ]) = ⇒ (clos T t V = T t clos V ⊂ clos W = K, ∀ t ∈ [0 , τ ]) . Hence, K ′ ⊂ K. Therefore K = K ′ , i.e. (4.1.1.8) holds.From (4.1.1.8) and T τ clos V ⊂ V we obtain T τ clos W ⊂ W. Hence T τ ∂K ⊂ W. This and ∂K ∩ W = ∅ imply(4.1.1.9) T τ ∂K ∩ ∂K = ∅ . To obtain a contradiction and complete the proof of cr3)= ⇒ cr4) we have toshow that K is a proper subset, because both cases: ∂K = ∅ and ∂K = ∅ willcontradict cr3). Since V is proper T t V is proper for any t ∈ ( −∞ , ∞ ) . Otherwise T t V = M implies V = T − t M = M that is a contradiction.Since V is a neighborhood of the compact set T τ clos V we can find α > T t ◦ T τ clos V ⊂ V for t ∈ [0 , α ] . Then T t clos V ⊂ T − τ V for t ∈ [0 , α ] . By iteration of this inclusion we obtain T jt clos V ⊂ T − jτ V for any integer j. When jα > τ it follows that K ⊂ T − jτ V. The last set is proper because wementioned already that T t V is proper for any t ∈ ( −∞ , ∞ ) . Hence K is proper.So K satisfies the conditions of cr3) but ∂K does not contain a non-empty T -invariant set. This contradiction proves the implication cr3)= ⇒ cr4).Let us prove cr4)= ⇒ cr5). Let ǫ > s > V denote the set of all m ′ ∈ M such that there exists an ( ǫ, s )-chain from m to m ′ . This set is open and closed. Indeed, let m ′ ∈ V. There exists an ( ǫ, s )-chain m = m , ..., m n − , m n = m ′ from m to m ′ . Choose ǫ < ǫ − d ( m n , m n − ) andconsider the closed neighborhood W := { m ′′ : d ( m ′ , m ′′ ) ≤ ǫ } . Then for any m ′′ ∈ W the chain m = m , ..., m n − , m n = m ′′ is an ( ǫ, s )-chain from m to m ′′ . Hence, with every point V contains its closed neighborhood. Therefore it is openand closed.Therefore it is a connected component of M. We also have T s m ∈ V because for that case n = 1 , m = m, m = T s m. Hence T s clos V ⊂ V. If V does not coincide with the whole M the latter contradicts tocr4). Hence V = M. Finally, let us prove cr5)= ⇒ cr1). If M is a union of two non-empty disjointsets A and B, then both of them are open end closed. Since M is compact, thedistance ǫ between A and B is positive . Hence every ( ǫ/ , s )-chain starting at apoint of A remains in A, contradicting cr5).Since for every point m ∈ M the set V from the proof of cr4)= ⇒ cr5) coincideswith M , cr1) holds. (cid:3) Theorem 4.1.1.2.
Let T t be chain recurrent on M α , α ∈ A. Then T t is chainrecurrent on M = S α ∈ A M α . This is because every ( ǫ, s )- chain from m to m ′ in M α is also ( ǫ, s )-chain in M. Here we prove two auxiliary assertions that will be used further.
Theorem 4.1.2.1.
Let T be chain recurrent on a connected compact M and let { q j } be a sequence in M. Then there exist sequences { α ν } and { ω ν } of real numbersand a sequence { p ν } in M having { q j } as a subsequence, such that (4.1.2.1) α ν → −∞ ; ω ν → ∞ and (4.1.2.2) d ( T ω ν p ν , T α ν +1 p ν +1 ) → as ν → ∞ . Proof.
In addition to { α ν } , { ω ν } and { p ν } we define, by induction, a sequence { ǫ ν } of positive real numbers, tending to zero, and an increasing sequence { ν j } ofpositive integers, such that { p ν j } = { q j } and(4.1.2.3) d ( T ω ν p ν , T α ν +1 p ν +1 ) < ǫ ν , ν = 1 , , .... We start by setting α = − , ǫ = 1 , ν = 1 , ω = 5 and p = q . Assume nowthat α ν , ǫ ν , ω ν and p ν have been chosen for ν = 1 , , ..., ν j . Set(4.1.2.4) α = α ν j − , ǫ = ǫ ν j / , ω = ω ν j . By Theorem 4.1.1.1, cr5) there exists a sequence r := T ω q j , r , ..., r m := T α q j +1 such that d ( T t k r k , r k +1 ) < ǫ for k = 0 , , ..., m −
1, where t k ≥ ω . Now we set ν j +1 = ν j + m + 1 . For ν = ν j + k + 1 , k = 0 , , ..., m − , we set α ν = − t k / , ω ν = t k / , p ν = T t k / r k , and finally, for ν = ν j +1 we set α ν = α, ǫ ν = ǫ, ω ν = ω + 1 , p ν = q j +1 . Let us check that with this setting the properties (4.1.2.1) hold . Since ω ν j +1 = ω ν j + 1 we have ω ν j → ∞ as j → ∞ . From t k ≥ ω = ω ν j we obtain α ν → −∞ and ω ν → ∞ . Hence (4.1.2.1) holds. One can see from (4.1.2.4) that ǫ ν = ǫ ν j / → . To prove (4.1.2.2) it is enoughto check (4.1.2.3). For k = 0 we have p ν = T t / r = T t / ω q j = T t / ω p ν j . Hence, T α ν p ν = T ω p ν j = T ω νj p ν j . Thus(4.1.2.5) d ( T ω νj p ν j , T α ν p ν ) = 0for this case.For k = 1 , ..., m − ν we have T ω ν p ν = T t k / ◦ T t k / r k = T t k r k and T α ν +1 p ν +1 = T − t k +1 / ◦ T t k +1 / r k +1 = r k +1 . Hence,(4.1.2.6) d ( T ω ν p ν , T α ν +1 p ν +1 ) = d ( T t k r k , r k +1 ) < ǫ = ǫ ν Finally, for the last link of the chain we obtain k = m − , ν = ν j + m, ν + 1 = ν j +1 , α ν j +1 = α,T α ν +1 p ν +1 = T α νj +1 p ν j +1 = T α q j +1 = r m Thus (4.1.2.6) holds for k = m − . Hence, (4.1.2.3) also holds. Therefore (4.1.2.2)holds. (cid:3)
Lemma 4.1.2.2.
Let p k , q k ∈ M and d ( p k , q k ) → as k → ∞ . Then there existsa sequence { γ k ↑ ∞} such that (4.1.2.7) d ( T τ p k , T τ q k ) → uniformly with respect to τ ∈ [ − γ k +1 , γ k ] . Proof.
Let [ − γ, γ ] be a fixed segment.Then d ( T τ p k , T τ q k ) → Indeed, suppose there exist sequences τ j , k j such that d ( T τ j p k j , T τ j q k j ) ≥ ǫ > . Choosing a subsequence we can assume that τ j → τ ∈ [ − γ, γ ] , p k j → p ∈ M and q k j → q = p. Using continuity of T τ m on ( τ, m ) and continuity of d ( • , • ) in botharguments we obtain 0 = d ( p, p ) ≥ ǫ > . Denote ǫ ( γ, k ) := max τ ∈ [ − γ,γ ] d ( T τ p k , T τ q k ) . This function increases monotonically in γ and tends to zero for any γ as k → ∞ . Choose l n such that ǫ ( n, k ) ≤ /n for k ≥ l n . Set γ k +1 := n for l n < k ≤ l n +1 . One can see that ǫ ( γ k +1 , k ) → k → ∞ . Sincemax τ ∈ [ − γ k +1 ,γ k ] d ( T τ p k , T τ q k ) ≤ ǫ ( γ k +1 , k ) , { γ k } satisfies (4.1.2.7). (cid:3) Let us consider some corollaries of the previous results.
Theorem 4.1.3.1. ( T t , M ) is chain recurrent iff M is connected and for any m ∈ M, small ǫ > and large s > there exist an ( ǫ, s ) -chain from m to m. i.e. we can omit V from the definition of the chain recurrence. The assertionfollows from cr5). Exercise 4.1.3.1.
Prove Theorem 4.1.3.1.We connect the property of being chain recurrent with other well known char-acteristics of dynamical system .A point m ∈ M is called non-wandering (see [An]) if for any neighborhood O of m and arbitrarily large number s ∈ R there exists m ∈ O and t ≥ s such that T t m ∈ O . This means that the “returns” take place to an arbitrary small neighborhoodof the point m . We shall denote as Ω( T • ) the set of non-wandering points. It is aclosed invariant subset of M. The set A ⊂ M is called an attractor if it satisfies the following conditions:attr1) for any neighborhood O ⊃ A there exists a neighborhood O ′ , A ⊂ O ′ ⊂O such that T t O ′ ⊂ O t ∈ R , where T t O ′ is the image of O ′ ; attr2) there exists a neighborhood O ⊃ A such that T t m → A when t → ∞ for m ∈ O . Theorem 4.1.3.2. If Ω( T t ) = M, then ( T t , M ) is chain recurrent; if ( T t , M ) hasan attractor A = M, it is not chain recurrent.Proof. The property Ω( T t ) = M obviously implies the chain recurrence for m =1.Suppose there exists an attractor A = M. Take a point m that does not belongto A and choose a neighborhood O ⊃ A such that d ( m , clos O ) = 2 ǫ > . This ispossible because an attractor is closed. Let O ′ be chosen by attr1) and s be such that T s m ∈ O ′ . Then there does not exist any ( ǫ, s )-chain from a small neighborhood of m itself. By Theorem 4.1.3.1 ( T t , M ) is not chain recurrent. (cid:3) Let us give examples of dynamical systems on connected compacts that arechain recurrent.
Theorem 4.1.3.3.
Let M be a connected compact and let T t be the identity map.Then ( T t , M ) is chain recurrent. This theorem, of course, is trivial. However, if M consists of a single pointthis dynamical system determines an important class of subharmonic and entirefunctions of completely regular growth (see [L(1980),Ch.III]).Let m ∈ M. Set(4.1.3.1) C ( m ) := clos { T t m : −∞ < t < ∞} It is closed, connected and invariant.
Exercise 4.1.3.2.
Prove this.Let us denote as Ω( m ) the set of all limits of the form(4.1.3.2.) Ω( m ) := { m ′ ∈ M : ( ∃ t k → ∞ )( m ′ = lim k →∞ T t k m } This is a limit set as t → ∞ . It is the “tangle” at the end of the curve. Denote by A ( m ) the analogous set for t → −∞ . Exercise 4.1.3.3.
Prove that A ( m ) and Ω( m ) are invariant. Theorem 4.1.3.4. ( T t , C ( m ) ) is chain recurrent iff (4.1.3.3) A ( m ) ∩ Ω( m ) = ∅ . Proof.
Suppose B := A ( m ) ∩ Ω( m ) = ∅ . Then Ω( m ) is an attractor and ( T t , C ( m ))is not chain recurrent by Theorem 4.1.3.2.Suppose B = ∅ . We will use cr2) from Theorem 4.1.1.1.Let U be an open proper subset of C ( m ) satisfying (4.1.1.1). Consider twocases:i) B contains a point of U. Thus U contains a sequence of form T t k m, t k → ∞ . From (4.1.1.1) we obtain that U contains T t m for all t ∈ ( −∞ , ∞ ) . Thus U ⊃ C ( m )and clos U = C ( m ) . Set K = C ( m ) \ U. One can show that K satisfies (4.1.1.2)(seethe beginning of proof of Theorem 4.1.1.1). Hence K contains the set(4.1.3.4) K ∗ := \ t ≥ T t K that is invariant ( Exercise 4.1.3.4).Therefore K ∗ ⊂ K ⊂ clos U \ U = ∂U. By cr2) ( T t , C ( m )) is chain recurrent.ii) B contains no point of U. Then B ⊂ A ( m ) ⊂ ∂U. By cr2) ( T t , C ( m )) ischain recurrent. (cid:3) Exercise 4.1.3.4.
Let U satisfy (4.1.1.1) and K := M \ U. Prove that K ∗ from(4.1.3.4) is invariant. The connectedness of M is a necessary condition for a dynamical system tobe chain recurrent.Let M be a subset of a linear space. The set M is called polygonally connected if every pair of points m , m can be connected by polygonal path.Of course, polygonal connectedness implies connectedness and even arcwiseconnectedness. Theorem 4.1.4.1.
Let ( T t , M ) be a dynamical system such that M is a polygonallyconnected set. Then ( T t , M ) is chain recurrent. Proof.
Let U be an open proper subset of M, satisfying (4.1.1.1). We choose m ∈ U and m in an invariant subset K ∗ of K := M \ U. Then there exists a polygonalpath from m to m : m θ := ( j + 1 − θ ) m ′ j + ( θ − j ) m ′ j +1 , for θ ∈ [ j, j + 1] j = 0 , , ..., l − m ′ := m , m ′ l := m . Now M is invariant, so for each t the continuous path θ T t m θ lies in M. If t ∈ ( −∞ ,
0) its initial point T t m belongs to U and its endpoint T t m belongs to K ∗ ⊂ K. For each t ∈ ( −∞ ,
0) we set θ ( t ) := min[ θ ∈ [0; l ] : T t m θ ∈ K ] . Then θ ( t ) > , T t m θ ( t ) ∈ ∂U and (4.1.1.1) implies that t θ ( t ) is a decreasingfunction. Hence the limit θ ( −∞ ) := lim t →−∞ θ ( t )exists and is positive.Set m := m θ ( −∞ ) . We claim that A ( m ) ⊂ ∂U ( A ( · ) is a set defined beforeTheorem 4.1.3.4). If θ ( −∞ ) ∈ ( j, j + 1] for some j ∈ [0 , l ] then θ ( t ) ∈ ( j, j + 1] for t that is near to −∞ , and T t m = T t m θ ( t ) + ( θ ( t ) − θ ( −∞ )) T t m ′ j + ( θ ( −∞ ) − θ ( t )) T t m ′ j +1 . The first term in the right hand side lies in ∂U.
The set M is compact andinvariant so the other terms tend to zero as t → −∞ . Hence A ( m ) ⊂ ∂U. Thus ∂U contains this invariant subset and ( T t , M ) is chain recurrent by cr2),Theorem 4.1.1.1. (cid:3) We have the obvious
Corollary 4.1.4.2.
Let ( T t , M ) be a dynamical system such that M is a convexset. Then ( T t , M ) is chain recurrent. This is because the polygonal path can be taken as a line segment connectingevery pair of points. Let U ( ρ, σ ) be a set of subharmonic functions defined in (3.1.2.4). It isinvariant with respect to the transformation ( • ) [ t ] defined in (3.1.2.4a).Set (subindex!)(4.1.5.1) T t v := v [ e t ] . Since ( • ) [ t ] has the property (3.1.2.4b*)(4.1.5.2) T t + τ v = ( T t ◦ T τ ) v, ∀ t, τ ∈ R . By Theorem 3.1.2.3 T • is continuous in the appropriate topology and hence ( T • , U [ ρ, σ ])is a dynamical system. Theorem 4.1.5.1 (Universality of U [ ρ, σ ] ). Let ( T • , M ) be a chain recurrentdynamical system on a compact set M. Then for any ρ, σ there exists U ⊂ U [ ρ, σ ] and a homeomorphism imb : M U such that imb ◦ T t = T t ◦ imb, t ∈ ( −∞ , ∞ ) . i.e., any dynamical system can be imbedded in ( T • , U [ ρ, σ ]) . It is sufficient to prove the theorem by supposition P t x = tx because ( T Pt , U [ ρ, σ ])is a dynamical system for any P t and imb : ( T t , U [ ρ, σ ]) ( T Pt , U [ ρ, σ ]) where imb : u ( x ) T − t T Pt u ( x ) is also a homeomorphism of dynamical systems. Exercise 4.1.5.1.
Consider Theorem 3.1.6.1 from this point of view.We need some auxiliary definitions and results. Let us denote as M ( S m − ) theset of measures ν with bounded full variation on the unit sphere S m − . Introducethe metric d ( ν,
0) := Var ν and consider the set K := { ν : ν > , d ( ν, ≤ } , i.e., the intersection of the cone of positive measures with the unit ball.The following assertion is a corollary of the Keller’s theorem (see,e.g. [BP,Th.3.1, p.100]). Theorem 4.1.5.2 (Imbedding).
Every metric compact set can be homeomorphi-cally imbedded to K. Thus we can assume below that for any m ∈ M there exists a positive measure Y ( • , m ) = Y ( dx , m ) ∈ K such that(4.1.5.3) ( Y ( • , m ) = Y ( • , m )) = ⇒ ( m = m )and Y ( • , m ) is continuous with respect to the metrics.We also introduce a new coordinate system. For x := e y x ∈ R m \ P ol ( x ) = ( y, x ) . This formula gives a one-to-one map from R m \ Cyl := ( −∞ , ∞ ) × S m − . Thus, for any ( y, x ) ∈ Cyl, P ol − ( y, x ) = e y x . For m = 2 this is a common cylinder. Proof of Theorem 4.1.5.1.
We consider separately the cases of integer and non-integer ρ. Let ρ be non-integer and σ > . For any v ∈ U [ ρ, σ ] , one has the representationof Theorem 3.1.4.4.(4.1.6.1) v ( x ) = Π( x, µ, ρ )where µ ∈ M [ ρ, ∆] and ∆ depends only on σ (Theorem 2.8.3.3).Vice versa, every µ ∈ M [ ρ, ∆] generates v by (4.1.6.1) and v [ t ] ( x ) = Π( x, µ [ t ] , ρ ) . Let us “transplant” µ in Cyl. For µ that has a dense f µ ( rx ) , we set ν ( dy ⊗ dx ) := f µ ( e y x ) e ( − ρ − y ( dy ⊗ dx ) . i.e., the density f ν of ν is defined by f ν ( x , y ) := f µ ( e y x ) e ( − ρ − y . Respectively f µ ( x , r ) = f ν ( x , log r ) r ρ +2 We can extend this equality for all µ ∈ M [ ρ, ∆] by using limit process in D ′ topology. Exercise 4.1.6.3.
Do that using, for example, Theorem 2.3.4.5.We can also define ν as a distribution in D ′ ( Cyl ) . Namely, for ψ ∈ D ( Cyl ) weset ψ ∗ ( x , r ) := ψ ( P ol − ( x , log r )) r − ρ − m +2 and < ν, ψ > := Z ψ ∗ ( x , r ) µ ( dx ⊗ r m − dr ) Exercise 4.1.6.4.
Check that this definition gives the same ν. The transformation P t x = ( x , tr ) , rx ∈ R m \ P ol ◦ P t ◦ P ol − ( x , y ) = ( x , y + log t )Thus T e τ µ gives a transformation S τ ν defined by S t f ν ( x , y ) := f ν ( x , y + t )for densities or by(4.1.6.2) < S t ν, ψ > := Z ψ ( x , y − t ) ν ( dx ⊗ dy )for distributions ( ψ ∈ D ( Cyl ) . ) Exercise 4.1.6.5.
Check the equivalence.From µ ∈ M [ ρ, ∆] we obtain(4.1.6.3) Z y ≤ e ( ρ + m − y S t ν ( dy ⊗ dx ) ≤ ∆ , t ∈ R , Exercise 4.1.6.6.
Check this.Let X ( t ) be a positive function satisfying the condition ∞ Z −∞ X ( t ) dt = 1 and such that the linear hull of its translations are dense in L ( −∞ , ∞ ). We canchose, for example, the function X ( t ) := 1 √ π e − t because its Fourier transformation does not vanish in R (it is e − s ). Exercise 4.1.6.7.
Check these properties.Let us define ν ( • , m ) by(4.1.6.4) < ν ( • , m ) , ψ ) := Z ( x ,y ) ∈ Cyl ψ ( x , y ) ρ ∞ Z −∞ Y ( dx , T y − t m ) X ( t ) dt dy. Now we check the property S τ ν ( • , m ) = ν ( • , T τ m )Using (4.1.6.2), we obtain < S τ ν ( • , m ) , ψ > = Z ψ ( x , y ) ρ ∞ Z −∞ Y ( dx , T y + τ − t m ) X ( t ) dt dy = Z ψ ( x , y ) ρ ∞ Z −∞ Y ( dx , T y − t ( T τ m )) X ( t ) dt dy = < ν ( • , m ) , T τ m > . We also check the condition (4.1.6.3). Z y ≤ e ρy S t ν ( dy ⊗ dx ) = ∞ Z −∞ X ( t ) dt Z y ≤ e ρy ) ρdy Z S m − Y ( dx , T y + τ − t m ≤≤ sup τ ∈ R Y ( S m − , T τ m ) ∞ Z −∞ X ( t ) dt ≤ , since Y ( • , • ) ∈ K. Now we should “transplant” ν back to R m \ S τ passes to ( • ) [ e τ ] . Define µ ( • , m ) by(4.1.6.5) < µ ( • , m ) , ψ ∗ > := < ν ( • , m ) , ψ >, where ψ ∗ ( rx ) ∈ D ( R m \
0) and ψ ( x , y ) := ψ ∗ ( e y x ) e − ( ρ − m +2) y ∈ D ( Cyl ) . Then < ( µ ) [ e τ ] , ψ ∗ > = < ( µ ) , T − τ ψ ∗ > = < ν, S − τ ψ > = < S τ ν, ψ > . The condition µ ( • , m ) ∈ M [ ρ, σ ] is also satisfied. Exercise 4.1.6.8.
Check these properties.Now we use (4.1.6.1) to transplant the dynamical system to U [ ρ, σ ] . This com-pletes a construction of an homomorphism ( T t , M ) ( T t , U [ ρ, σ ]) . Let us check that it is an imbedding, i.e., we must check the one-to-one corre-spondence. One-to-one correspondence of v ( • , m ) and µ ( • , m ) is known (Theorem3.1.4.4). One-to-one correspondence of µ ( • , m ) and ν ( • , m ) can be also checkedeasily. Exercise 4.1.6.9.
Check this in details.So we should check the one-to-one correspondence of ν ( • , m ) and Y ( • , m ) . Suppose ν ( • , m ) = ν ( • , m ) . Then < ν ( • , m ) , ψ > = < ν ( • , m ) , ψ > ∀ ψ ∈ D ( Cyl ) . In particular, set ψ ( x , y ) = φ ( x ) R ( y ) , φ ∈ D ( S m − ) , R ∈ D ( −∞ , ∞ ) . Then(4.1.6.6) < ν ( • , m ) , ψ > = Z R ( y ) dy ∞ Z −∞ < Y ( • , T y − t m ) , φ > S m − X ( t ) dt == < ν ( • , m ) , ψ > = Z R ( y ) dy ∞ Z −∞ < Y ( • , T y − t m ) , φ > S m − . where < Y ( • ) , φ > S m − := Z S m − φ ( x ) Y ( dx ) . Set F j ( y ) := < Y ( • , T y m j ) , φ > S m − , j = 1 , . From (4.1.6.6) we obtain for the convolutions( F ∗ X )( y ) ≡ ( F ∗ X )( y ) , y ∈ ( −∞ , ∞ ) . Thus F ( y ) ≡ F ( y ) , y ∈ ( −∞ , ∞ )because of the property of X. Hence Y ( • , T y m ) ≡ Y ( • , T y m ) , y ∈ ( −∞ , ∞ ) . In particular, for y = 0 we have Y ( • , m ) = Y ( • , m ) . Hence m = m because of (4.1.5.3), and this completes the proof of one-to-onecorrespondence.Consider the case of an integral ρ. For this case we can use v ∈ U [ ρ, σ ] of theform v ( x ) = Π < ( x, µ, ρ ) + Π > ( x, µ, ρ )instead of (4.1.6.1). (cid:3) Exercise 4.1.6.10.
Check this.
The most simple set satisfying the conditions of Theorem 4.1.3.4 is the setthat is generated by a function v ∈ U [ ρ ] that has the property v [ te P ] = v [ t ] , t ∈ (0 , ∞ ) for some P. Then T t + P v = T t v, t ∈ (0 , ∞ )i.e., the dynamical system T • is periodic with the period P on the set C ( v ) := { T t v : 0 ≤ t ≤ P } . Theorem 4.1.7.1(Periodic Limit Set).
For all
P > , ρ > , σ > , there exists v ∈ U [ ρ, σ ] such that the dynamical system ( T • , C ( v )) is periodic.Proof. Suppose ρ is non-integer. Let us take µ ∈ M [ ρ, ∆] such that the canonicalpotential Π( x, µ, [ ρ ]) belongs to U [ ρ, σ ] . This is possible because of Theorem 3.1.4.2.Denote as µ ∗ P the restriction of µ on the spherical ring { x : 1 < | x | < e P } andset µ P := ∞ X k = −∞ T kP µ ∗ P . We have µ P ∈ M [ ρ, ∆] and T t + P µ P = T t ( ∞ X k = −∞ T ( k +1) P µ ∗ P ) = T t µ P . Then v := Π( x, µ P , [ ρ ]) ∈ U [ ρ, σ ] and T t + P v = T t v because of (3.1.5.0).For an integer ρ we use the function v ( x ) := Π < ( x, µ P , ρ ) + Π > ( x, µ P , ρ ) . (cid:3) The following two theorems describe structure of limit sets in terms of dy-namical systems.
Theorem 4.2.1.1 (Necessity).
Let u ∈ SH ( R m , ρ, ρ ( r )) . Then the dynamicalsystem ( T • , Fr [ u, • ]) is chain recurrent. The chain recurrence is also sufficient.
Theorem 4.2.1.2 (Sufficiency).
Let U be a compact connected and T • – invariantsubset of U [ ρ, σ ] for some σ > , such that the dynamical system ( T • , U ) is chainrecurrent.Then for any proximate order ρ ( r ) → ρ there exists u ∈ SH ( R m , ρ, ρ ( r )) suchthat Fr [ u, ρ ( r ) , V t , R m ] = U. Proof of Theorem 4.2.1.1.
We need the curve u t , t ≥ , and Fr [ u, • ] to be containedin a common metric space X. Thus we set X := { v ∈ SH ( R m ) : sup r ≥ M ( r, v ) r − ρ − ≤ sup r ≥ M ( r, u ) r − ρ − } . We want to use Theorem 4.1.1.1 cr 2). Let U be an open proper subset of Fr [ u, • ]satisfying (4.1.1.1) and let F be a T • –invariant subset of K := Fr [ u, • ] \ U. Such F exists. Indeed, K is closed and T t K ⊂ K for t > ⇐⇒ cr3)).Thus Ω( K ) ⊂ K where Ω( • ) was defined in (4.1.3.2).The set Ω( K ) is invariant with respect to T t (see Exercise 4.1.3.2). So the set ofsuch sets F is not empty.If F intersects ∂U at a point v , then A ( v ) ⊂ F ∩ ∂U. Since A ( v ) is invariant(Exercise 4.1.3.3) ∂U contains a nonempty T • –invariant set. So we obtain theassertion of the theorem using Theorem 4.1.1.1, cr2).Suppose F does not intersect ∂U. Let U be an open set in X such that(4.2.1.1) U ∩ Fr [ u, • ] = U, clos U ∩ Fr [ u, • ] = clos U. (see Exercise 4.2.1.1). Since clos U ∩ F = ∅ we can take a sequence of openneighborhoods U , U , ... of F in X such that the all sets clos U j , j = 1 , , ... donot intersect clos U and U j ↓ F. By definition of Fr [ u, • ] we can find intervals a j ≤ t ≤ b j with a j → ∞ suchthat u e aj ∈ ∂U j , u e bj ∈ ∂U , and u e t clos U ∪ clos U j for a j < t < b j . We canpass to a subsequence and assume that(4.2.1.2) u e aj → w ∈ F. Let us use the following identity :(4.2.1.3) u e t + aj = ( u e aj ) e t ρ ( e t ) ρ ( e a j ) ρ ( e t + a j ) . By (4.2.1.2),(4.2.1.3) and the property (3.1.2.2) of a proximate order we obtain u e t + aj → T t w ∈ F uniformly for any bounded set of t. Thus b j − a j → ∞ . Passing to a subsequence we may assume that u e bj → v ∈ Fr [ u, • ] ∩ ∂U = ∂U. Since u e t + bj → T t v and u e t + bj / ∈ U when a j − b j < t < T t v / ∈ U when t < . Hence the whole backward orbit { T t v : t < } lies in ∂U, which must thereforecontain the T • –invariant set A ( v ) . (cid:3) Exercise 4.2.1.1.
Prove that the set U := [ v ∈ U { w ∈ X : dist ( v, w ) < dist ( v, K ) / } satisfies the conditions (4.2.1.1). Proof.
We have U ⊃ U = ⇒ U ∩ Fr [ u, • ] ⊃ U ∩ Fr [ u, • ] = U Thus(4.2.1.4) U ∩ Fr [ u, • ] ⊃ U From (4.2.1.4) we have(4.2.1.5) clos U ∩ Fr [ u, • ] = clos U ∩ clos Fr [ u, • ] = clos ( U ∩ Fr [ u, • ]) ⊃ clos U Finally (4 . . . ∧ (4 . . .
5) = ⇒ (4 . . . . (cid:3) To prove Theorem 4.2.1.2 we need some preparation. Theorems of the nextitems form the basis of the construction that we will use in the proof.Let β be an infinitely differentiable function on R such that 0 ≤ β ( x ) ≤ , β ( x ) = 0 for x ≤ β ( x ) = 1 for x ≥ . We can set, for example, β ( x ) := A x Z −∞ α ( y + 1) dy where α is taken from (2.3.1.1) and A = ∞ Z −∞ α ( y + 1) dy. Suppose that the sequences { r k , σ k , k = 0 , , .. } satisfy the following conditions:(4.2.2.1) r = 1; r k < r k σ k < r k +1 /σ k +1 < r k +1 , k = 1 , , ... (4.2.2.2) σ k ↑ ∞ ; r k +1 σ k +1 r k σ k ↑ ∞ . Set ψ k ( r ) := β (cid:18) log r − log( r k /σ k )log( σ k r k ) − log( r k /σ k ) (cid:19) − β (cid:18) log r − log( r k +1 /σ k +1 )log( σ k +1 r k +1 ) − log( r k +1 /σ k +1 ) (cid:19) ψ ( r ) := 1 − β (cid:18) log r − log( r /σ )log( σ r ) − log( r /σ ) (cid:19) The sequence { ψ k } , k = 0 , , ... forms a partition of unity with the following prop-erties Theorem 4.2.2.1(Partition of Unity).
One has (prtu1) ∞ X k =0 ψ k = 1;(prtu2) supp ψ k ⊂ ( r k /σ k , r k +1 σ k +1 );(prtu3) ψ k ( r ) = 1 , for r ∈ ( r k σ k , r k +1 /σ k +1 );(prtu4) supp ψ k ∩ supp ψ l = ∅ for | k − l | > k →∞ max r ψ ′ k ( r ) r = lim k →∞ max r ψ ′′ k ( r ) r = 0 . Moreover (prtu6) max r | ψ ′ k ( r ) r | , max r | ψ ′′ k ( r ) r | ≤ γ k where γ k can be made to tend to zero arbitrarily fast by choosing the sequences { σ k } and { r k } . Proof.
Set β k ( r ) := β (cid:18) log r − log( r k /σ k )log( σ k r k ) − log( r k /σ k ) (cid:19) . The functions β k ( r ) and β k +1 ( r ) vanish for r < r k /σ k because β ( x ) = 0 for x ≤ r ≥ σ k r k because β ( x ) = 1 for x ≥ . Hence,(prtu2) holds.One has for any r ∈ (0 , ∞ ) n X k =0 ψ k = 1 − β n +1 ( r ) . As mentioned β n +1 ( r ) = 0 for n such that r n +1 /σ n +1 > r. Thus (prtu1) holds.Counting derivatives of ψ k , we have:max r | rψ ′ k ( r ) | ≤ (cid:2) (log( σ k r k ) − log( r k /σ k )) − + (log( σ k +1 r k +1 ) − log( r k +1 /σ k +1 )) − (cid:3) max x | β ′ | ( x ) . Thus we can take the right side of the inequality as γ k and regulate its vanishingby choice of the ratio in (4.2.2.2). The same holds for r ψ ′′ ( r ) . Hence (prtu5) and(prtu6) are proved.
Exercise 4.2.2.1.
Check (prtu4). (cid:3) Now we construct a function which is of zero type but has a “ maximalpossible” mass density.
Theorem 4.2.3.1 (Maximal Mass Density Function).
Let ρ ( r ) → ρ, ρ > be a smooth proximate order (i.e., having properties (2.8.1.8)), and let γ ( r ) , r ∈ [0 , ∞ ) , satisfy the conditions: γ ( r ) > and γ ( r ) → , as r → ∞ . Then there exists an infinitely differentiable subharmonic function Φ( x ) suchthat (4.2.3.1) ∆Φ( x ) ≥ γ ( x ) | x | ρ ( r ) − and (4.2.3.2) (Φ) t → in D ′ as t → ∞ . To prove Theorem 4.2.3.1 we need an elementary lemma.
Theorem 4.2.3.2 (Convex Majorization).
Let a ( s ) , s ∈ [ s , ∞ ) be a functionsuch that a ( s ) → −∞ as s → ∞ . Then there exists an infinitely differentiable, convex function k ( s ) such thatk1) k ( s ) ≥ a ( s ); k2) k ( s ) ↓ −∞ as s → ∞ ;k3) k ( n ) ( s ) → for all n = 1 , , ... . Proof.
Set a ∗ ( s ) := sup { a ( t ) : t ≥ s } . Then a ∗ ( s ) ↓ −∞ as s → ∞ . Set b := − a ∗ ( s ) and denote as s ( b ) , b ∈ [ b , + ∞ ) the function inverse to thefunction − a ∗ ( s ) . Let us construct a convex function that majorates s ( b ) and tendsto infinity monotonically with all its derivatives. It can be done in the followingway. First we construct a piece-wise linear convex function. Set s ( b ) := s + 1 + α ( b − b ) , b ∈ [ b , b + 1] , and chose α such that the inequality s ( b ) > s ( b ) holds for b ∈ [ b , b + 1] . For this we choose α ≥ sup b ∈ [ b ,b +1] s ( b ) − s − b − b Since s ( b ) − s − < s ( b ) := s ( b + j ) + α j ( b − b − j ) , b ∈ [ b + j, b + j + 1] , where α j ≥ α j − and satisfies the condition α j ≥ sup b ∈ [ b + j,b + j +1] s ( b ) − s ( b + j ) b − b − j To obtain a smooth function set(4.2.3.3) s ( b ) := Z α ( b − x ) s ( x ) dx, where α ( x ) is defined by (2.3.1.1). Then s ( b ) is infinitely differentiable, monotonicand convex . Exercise 4.2.3.1.
Check this.Set(4.2.3.4) k ( s ) := − s − ( s ) , where s − ( s ) is the inverse function to s . One can check that k ( s ) satisfies theproperties k1), k2), k3). (cid:3) Exercise 4.2.3.2.
Check that k ( s ) satisfies k1),k2),k3). Proof of Theorem 4.2.3.1.
We are going to show that Φ can be taken in the form(4.2.3.5) Φ( x ) := ce k (log | x | ) | x | ρ ( | x | ) . where c and k ( s ) will be chosen later.Note that Φ( x ) = Φ( | x | ) depends only on r = | x | and pass to the variable s := log r . Then for φ ( s ) := Φ( e s/ ) we have∆Φ( x ) = r − m ∂∂r r m − ∂∂r ce k (log r ) r ρ ( r ) =(4.2.3.6) ce − s (cid:18) ∂ ∂s + m − ∂∂s (cid:19) φ ( s ) ≥ cme − s min[ φ ′′ ( s ) , φ ′ ( s )] . Let us chose k as in Theorem 4.2.3.2 with a ( s ) := log γ ( r ) = log γ ( e s ) . Now weestimate the derivatives from below. φ ′ ( s ) = φ ( s )[ k ′ ( s ) + 14 sρ ′ ( e s ) + 12 ρ ( e s )] . By k3) and k1) k ′ ( s ) → k ( s ) ≥ a ( s ) . Also sρ ′ ( e s ) → ρ ( e s ) → ρ byproperties of proximate order (Theorem 2.8.1.4). Thus we can chose c such that(4.2.3.7) φ ′ ( s ) > m e log γ ( e s )+ s ρ ( e s ) . Differentiating once again, we obtain φ ′′ ( s ) = φ ( s )[ k ′ ( s ) + 14 sρ ′ ( e s ) + 12 ρ ( e s )] + k ′′ ( s ) + 12 ρ ′ ( e s ) + 18 sρ ′′ ( e s )] . From here we obtain by choosing c :(4.2.3.8) φ ′′ ( s ) > m e log γ ( e s )+ s ρ ( e s ) . Using (4.2.3.6) ,(4.2.3.7) and (4.2.3.8) we obtain:∆Φ( s ) > e log γ ( e s )+ s ρ ( e s ) . Returning to the variable r we obtain (4.2.3.1). Correctness of (4.2.3.2) can bechecked directly using k2) and properties of the proximate order (Theorem 2.8.1.3). Exercise 4.2.3.3.
Check this. (cid:3) We have already approximated distributions and subharmonic functions byinfinitely differentiable functions (Theorems 2.3.4.5 and 2.6.2.3). Now we need tomake more precise this approximation.Namely, we are going to make it uniformwith respect to v ∈ U [ ρ, σ ] . We will denote(4.2.4.1) ∂ l := ∂ | l | ( ∂x ) l ( ∂x ) l ... ( ∂x m ) l m where l = ( l , l , ...l m ) , | l | = l + l + ... + l m . Set for v ∈ U [ ρ, σ ](4.2.4.2) R ǫ v ( x ) := Z α ǫ ( x − y ) v ( y ) dy where α ǫ is taken from (2.3.1.3).We have changed the notation from 2.3.1 and 2.6.2 because a subindex of v was already engaged for t. For a fixed 0 < δ ≤ . Str ( δ ) := { x : δ ≤ | x | ≤ δ − } Theorem 4.2.4.1 (Estimation of R ǫ ). Let v ∈ U [ ρ, σ ] . ThenR1. for a fixed g ∈ D ( R m \ with supp g ⊂ Str ( δ )(4.2.4.4) | < R ǫ v − v, g > | ≤ o (1 , g )2 σδ − ρ where o (1 , g ) → as ǫ → R2. the inequality (4.2.4.5) | ∂ l R ǫ v ( x ) | ≤ A ( m ) σǫ −| l |− m +1 | x | −| l | + ρ , with A ( m ) depending only on m, holds for ǫ < | x | / .Proof. One has(4.2.4.6) < R ǫ v, g > = < v, R ǫ g > . Thus(4.2.4.7) < R ǫ v − v, g > = < v, R ǫ g − g > . Exercise 4.2.4.1.
Check (4.2.4.6) and (4.2.4.7).Now(4.2.4.8) | < v, R ǫ g − g > | ≤ max Str ( δ ) | R ǫ g − g | ( x ) Z Str ( δ ) | v | ( x ) dx The first factor is o (1) because g is smooth. For the second one we have(4.2.4.9) Z Str ( δ ) | v | ( x ) dx ≤ Z Str ( δ ) v + ( x ) dx ≤ σδ − ρ . This and (4.2.4.8) imply R1).Differentiating the equality R ǫ v ( x ) := C m Z ǫ − m α ( | x − y | /ǫ ) v ( y ) dy, we have | ∂ l R ǫ v ( x ) | ≤ C m ǫ −| l |− m max {| y | <ǫ } | ∂ l α ( | y | ) | Z {| y | <ǫ } | v ( x − y ) | dy. Suppose | x | = 1 . Then for 0 < ǫ ≤ . , we have Z | y | <ǫ | v | ( x − y ) dy ≤ Z − ǫ< | x | < ǫ | v | ( x ) ≤ Z − ǫ< | x | < ǫ v + ( x ) dx ≤ σ m · ǫσ (1 + ǫ ) ρ ≤ σ m σǫ where σ m is the square of the unit sphere. Hence for | x | = 1(4.2.4.10) | ∂ l R ǫ v ( x ) | ≤ A ( m ) σǫ −| l |− m +1 . with A ( m ) = 6 σ m max y ∈ R m | ∂ l α ( | y | ) | . Set t = | x | . Apply the inequality (4.2.4.10) to v := v [ t ] ( y ) with y := x/ | x | . Then | ∂ l R ǫ v [ t ] ( y ) | ≤ A ( m ) σǫ −| l |− m +1 . Computing the derivatives, we obtain ∂ l R ǫ v [ t ] ( x ) = t − ρ t | l | ∂ l R ǫ v ( x ) | x = ty Thus one has R2. (cid:3) In this item we describe the main part of a construction that will be usedin the proof of the Theorem 4.2.1.2.Let { v j ∈ U [ ρ, σ ] , j = 1 , , ... } and { ψ j , j = 1 , ... } be the partition of unityfrom Theorem 4.2.2.1. Let us chose ǫ j ↓ γ j ǫ − mj → ∞ holds for γ j taken from Theorem 4.2.2.1, (prtu 6).Set(4.2.5.2) v ( x | t ) := ∞ X j =0 ψ j ( t )( v j ) [ t ] ( x ) , where ( · ) [ t ] defined by (3.1.2.4a).One can see that v ( x | t ) ∈ U [ ρ, σ ] for all t. Exercise 4.2.5.1.
Show this, using properties of { ψ j } and invariance of U [ ρ, σ ]with respect to ( · ) [ t ] .We can consider v ( x | t ) as a curve (a pseudo-trajectory ) in U [ ρ, σ ] . Set(4.2.5.3) u ( x ) := ∞ X j =0 ψ j ( | x | ) R ǫ j ( v j )( x ) | x | ρ ( | x | ) − ρ . where R ǫ is defined by (4.2.4.2).It is infinitely differentiable function in R m . Theorem 4.2.5.1(Construction ).
One has (4.2.5.4) u t − v ( •| t ) → in D ′ ( R m ) , and (4.2.5.5) ∆ u ( x ) = f ( x ) + γ ( x ) | x | ρ ( | x | ) − with f ( x ) ≥ and γ ( x ) = o (1) as | x | → ∞ . Let us note that the function u ( x ) is “almost-subharmonic” and can be madesubharmonic by summing with the function Φ from Theorem 4.2.3.1. Exercise 4.2.5.2.
Prove this.So we have Theorem 4.2.5.2(Pseudo-Trajectory Asymptotics).
For any v ( x | t ) of theform (4.2.5.2) there exists an infinitely differentiable function u ∈ SH ( ρ ( r )) thatsatisfies (4.2.5.4).Proof of Theorem 4.2.5.1. One has u t ( x ) := ∞ X j =0 ψ j ( t | x | )( R ǫ j ( v j )) [ t ] ( x ) a ( x, t ) , where a ( x, t ) := | tx | ρ ( t | x | ) − ρ t ρ ( t ) − ρ . For any 0 < δ < . x ∈ Str ( δ ) a ( x, t ) → | x | as t → ∞ . Thisfollows from Theorem 2.8.1.3 , ppo3).
Exercise 4.2.5.3.
Check this in details.We have(4.2.5.6) u t ( x ) − v ( x | t ) = ∞ X j =0 [ ψ j ( t | x | )( R ǫ j ( v j )) [ t ] ( x ) a ( x, t ) − ψ j ( t )( v j ) [ t ] ( x )] , and there are no more than three summands in the sum for sufficiently large t = t ( δ )because of Theorem 4.2.2.1, prtu4. Let us estimate every summand. One has b j ( x, t ) := [ ψ j ( t | x | )( R ǫ j ( v j )) [ t ] ( x ) a ( x, t ) − ψ j ( t )( v j ) [ t ] ( x )] =[ ψ j ( t | x | ) − ψ j ( t )]( R ǫ j ( v j ))) [ t ] a ( x, t )+ ψ j ( t )( R ǫ j ( v j )) [ t ] ( x )[ a ( x, t ) −
1] + ψ j ( t )[( R ǫ j ( v j )) [ t ] ( x ) − ( v j ) [ t ] ( x )] :=:= ( a + a + a )( x, t ) . Let us estimate < b j ( t, • ) , g > for every g ∈ D ( R m \ . We can assume that supp g ⊂ Str ( δ ) . Set M ( g ) := max x ∈ Str ( δ ) | g | ( x ) . We have | < a ( • , t ) , g > | ≤ M ( g ) max r ∈ (0 , ∞ ) | rψ ′ j ( r ) | δ − Z Str ( δ ) | ( R ǫ j ( v j )) [ t ] | ( x ) dx. One can check that Z Str ( δ ) | ( R ǫ j ( v j )) [ t ] | ( x ) dx ≤ σδ − ρ . Exercise 4.2.5.4.
Check this using (4.2.4.9) and the invariance of U [ ρ, σ ] withrespect to ( • ) [ t ] (see (3.1.2.4)).Hence(4.2.5.7) | < a ( • , t ) , g > | ≤ C ( g ) γ j . Let us estimate a ( x, t ). We have(4.2.5.8) < a ( • , t ) , g > ≤ max Str ( δ ) | a ( x, t ) − | ψ j ( t ) M ( g )3 σδ − ρ = C ( g ) o (1)where o (1) → t → ∞ .For estimating a ( x, t ) , we use Theorem 4.2.4.1(Estimation of R ǫ ), (4.2.4.4):(4.2.5.9) | < a ( • , t ) , g > | ≤ o ( ǫ j , g )2 σδ − ρ . where o ( ǫ j , g ) → j → ∞ . Hence (4.2.5.7), (4.2.5.8) and (4.5.5.9) imply(4.2.5.10) < b j ( • , t ) , g > → t → ∞ and j → ∞ . Suppose, for a large fixed t, the sum (4.2.5.6) contains b j ( x, t ) for j = j ( t ) , j = j ( t ) + 1 and j = j ( t ) + 2 . This implies that j ( t ) → ∞ as t → ∞ . Since < u t ( • ) − v ( •| t ) , g > = < b j ( t ) ( • , t ) , g > + < b j ( t )+1 ( • , t ) , g > + < b j ( t )+2 ( • , t ) , g > we obtain from (4.2.5.10) that < u t ( • ) − v ( •| t ) , g > → t → ∞ for any g ∈D ( R m \ . This is (4.2.5.4).Let us prove (4.2.5.5). We have(4.2.5.11)∆ u = ∞ X j =0 [∆( R ǫ j v j )( x ) ψ j ( x ) | x | ρ ( | x | ) − ρ + X l,m.k ∂ l ( R ǫ j v j )( x ) ∂ n ψ j ( x ) ∂ k | x | ( ρ | x | ) − ρ ] , where l, m, k are multi-indexes that satisfy the condition: in any summand thereare derivatives in the same variable, the derivatives of ψ j and | x | ρ ( | x | ) − ρ have nomore than second order and the derivatives of R ǫ j v j ( x ) have no more than firstorder. Exercise 4.2.5.5.
Check this.As usual, the derivative of zero order is the function itself.For any x ∈ Str ( δ ) , the outside sum contains no more then three summands.First we consider only the terms in the square brackets.The first term is non-negative because of subharmonicity of R ǫ j v j and non-negativity of all the other factors. Set(4.5.2.12) f ( x ) := X j =0 ∞ [∆( R ǫ j v j )( x ) ψ j ( x ) | x | ρ ( | x | ) − ρ ≥ . Using Theorem 4.2.4.1, R2) we obtain | ∂ l ( R ǫ j v j )( x ) | ≤ (4.2.5.13) A ( m ) σǫ −| l |− m +1 j | x | −| l | + ρ for | l | = 0 or | l | = 1 . From Theorem 4.2.2.1, prtu6), and inequality | ∂ x i | x || ≤ | ∂ n ψ j ( | x | ) ≤ | ψ ( n ) ( r ) || r = | x | ≤ γ j | x | −| n | for | n | = 1 , . Using properties of the smooth proximate order (Theorem 2.8.1.4), one canobtain(4.2.5.15) | ∂ | k | | x | ρ ( | x | ) − ρ | = ( | x | ρ ( | x | ) − ρ −| k | ) | r = | x | (1 + o (1)) , as | x | → ∞ . Exercise 4.2.5.6.
Check in details (4.2.5.13), (4.2.5.14) and (4.2.5.15).Thus, for every term of the inner sum, we have | ∂ l ( R ǫ j v j )( x ) ∂ n ψ j ( x ) ∂ k | x | ρ ( | x | ) − ρ | ≤ (4.2.5.16) A ( m ) σγ j ǫ −| l |− m +1 j | x | − ρ | x | ρ ( | x | ) − ρ ≤ β j | x | ρ ( | x | ) − , where β j → x the outside sum contains no more then threesummands, say, j = j ( x ) , j = j ( x ) + 1 and j = j ( x ) + 2 . Thus j ( x ) → ∞ as | x | → ∞ . Hence (4.2.5.12) and (4.2.5.16) imply (4.2.5.5). (cid:3)
Proof of Theorem 4.2.1.2.
Let v ( •| t ) have the form (4.2.5.2). We denote as Ω( v ) aset of the D ′ –limits of the form w := lim t k →∞ v ( •| t k ) . We are going to construct some v ( •| t ) for which(4.2.6.1) Ω( v ) = U, and at the next step to use Theorem 4.2.5.2 to obtain a subharmonic function withthe same limit set.First we describe the construction of the function v ( •| t ) . Let { r k , t k , k =1 , , ... } be an alternating sequences r = 1 , r k < t k < r k +1 such thatlim k →∞ t k r k = lim k →∞ r k +1 t k = ∞ . Let us chose in U a countable, dense set { g j } and form from it a sequence { w k } such that every element g j is repeated infinitely often. For example, w := g , w := g , w := g , w := g , w := g , w := g , ... . Set q k := ( w k ) [1 /t k ] = T − log t k w k in the notation (4.2.1.1).Now we use that ( T • , U ) is chain recurrent. Set α k := log r k t k ; ω k := log r k +1 t k and find, by Theorem 4.1.2.1, a sequence { v j } ⊃ { q k } such that the condition(4.1.2.1) holds, i.e.,(4.2.6.2) T ω k v k − T α k +1 v k +1 → k → ∞ . Set in Theorem 4.1.2.3 p k := T α k +1 v k +1 , q k := T ω k v k and find γ k such that the condition(4.2.5.3) T τ ◦ T ω k v k − T τ ◦ T α k +1 v k +1 → τ ∈ [ − γ k +1 , γ k ] . Set σ k := min (cid:20) e γ k , r t k r k (cid:21) . These σ k satisfy the conditions (4.2.2.1) and (4.2.2.2). Exercise 4.2.6.1.
Check this.We define v ( •| t ) by (4.2.5.2) with described v j and with ψ j from Theorem4.2.2.1, corresponding to the chosen r j and σ j . Let us prove (4.2.6.1).Consider for fixed k the following three cases.1. t ∈ [ r k σ k , r k +1 /σ k +1 );2. t ∈ [ r k +1 /σ k +1 , r k +1 );3. t ∈ [ r k , r k σ k ) . For the first case we have v ( •| t ) = ( v k ) [ t/t k ] = T log t/t k v k . For the second one v ( •| t ) = ψ k ( t )( v k ) [ t/t k ] + ψ k +1 ( t )( v k +1 ) [ t/t k +1 ] =( v k ) [ t/t k ] + ψ k +1 ( t )[( v k +1 ) [ t/t k +1 ] − ( v k ) [ t/t k ] We transform the expression in the square brackets( v k +1 ) [ t/t k +1 ] = T log( t/t k +1 ) v k +1 = T log( t/r k +1 ) ◦ T log( r k +1 /t k +1 ) v k +1 = T log( t/r k +1 ) ◦ T α k +1 v k +1 . For the second term, we obtain( v k ) [ t/t k ] = T log( t/r k +1 ) ◦ T ω k v k . Exercise 4.2.6.2.
Check this.Setting τ := log( t/r k +1 ) , we have(4.2.6.4) v ( •| t ) = ( v k ) [ t/t k ] + ψ k +1 ( t )[ T τ ◦ T α k +1 v k +1 − T τ ◦ T ω k v k ] , where τ ∈ [ − log σ k +1 , ⊂ [ − γ k +1 , γ k ]. For the third case, set τ := log( t/r k ) . Then(4.2.6.5) v ( •| t ) = ( v k ) [ t/t k ] + ψ k ( t )[ T τ ◦ T ω k − v k − − T τ ◦ T α k v k ] , where τ ∈ [0 , log σ k ) ⊂ [ − γ k , γ k − ].Let t N → ∞ be an arbitrary sequence. Choosing a subsequence, we maysuppose that there exist the limits ( v k ( t N ) ) [ t N /t k ( tN ) ] → v ∗ ∈ U and v ( •| t N ) → v ∞ . Choosing a subsequence, we may suppose that t N satisfies either 1. or 2. or 3.For the case 1., we obtain at once v ∞ = v ∗ ∈ U. For the case 2., from (4.2.6.4), (4.2.6.2) and Theorem 4.1.1.3 we obtain thatthe superfluous addends tend to zero, and hence v ∞ ∈ U. The same holds for the case 3. Hence Ω( v ) ⊂ U. Further, for t = t k , we have v ( •| t ) = w k . The sequence { w k } contains the set { g j } that is dense in U. Thus Ω( v ) ⊃ U. Thus equality (4.2.6.1) has been proved.As already said, the application of Theorem 4.2.5.2 concludes the proof. (cid:3) Let as mark the following property of the pseudo-trajectory v ( •| t ) definedin (4.2.5.2): Theorem 4.3.1.1.
One has (4.3.1.1) T τ v ( •| e t ) − v ( •| e t + τ ) → as t → ∞ uniformly with respect to τ ∈ [ a, b ] for any [ a, b ] ⊂ ( −∞ , ∞ ) . Proof.
Using definition of ( • ) t (see (3.1.2.1)) and (4.2.5.4) the remainder in (4.3.1.1)can be represented in the form b ( t, τ, • ) := T τ v ( •| e t ) − v ( •| e t + τ ) = T τ ( u e t ) − u e t + τ + o (1)where o (1) → τ ∈ [ a, b ] for any [ a, b ] ⊂ ( −∞ , ∞ ) . Exercise 4.3.1.1.
Check this in details.Then we obtain b ( t, τ, • ) = u e t + τ [ e ρ ( e t ) − ρ ( e t + τ ) −
1] + o (1) → { u e t } and prop-erties of the proximate order. Exercise 4.3.1.2.
Check this in details. (cid:3)
The property (4.3.1.1) shows that the pseudo-trajectory v ( •| t ) behaves asymp-totically like the dynamical system T • . Thus a pseudo-trajectory with this propertyis called an asymptotically dynamical pseudo-trajectory with dynamical asymptotics T • (a.d.p.t.).Theorem 4.2.5.1 shows that for any a.d.p.t. of the form (4.2.5.2) there exists u ∈ SH ( ρ ( r )) that satisfies the condition(4.3.1.2) u e t − v ( •| e t ) → . as t → ∞ . The following assertion shows that we can suppose v ( •|• ) to be an arbitrary,in some sense, a.d.p.t.We call a pseudo-trajectory w ( •|• ) piecewise continuous if the property(4.3.1.3) w ( •| t + h ) − w ( •| t ) → h → t except may be a countable set without points of condensa-tion.Let U ⊂ U [ ρ, σ ] for some σ > . A pseudo-trajectory w ( •|• ) is called ω –dense in U if Ω( w ) = U (see (4.1.3.2)),i.e.,(4.3.1.4) { v ∈ U [ ρ ] : ( ∃ t j → ∞ ) v = D ′ − lim w ( •| e t j ) } = U We have proved already that v ( •|• ) defined by (4.2.5.2) has this property (see(4.2.6.1)).Now we consider again the dynamical system ( T • , U ) where U ⊂ U [ ρ, σ ] forsome σ > T t is defined by (4.2.1.1). Theorem 4.3.1.2 (A.D.P.T. and Chain Recurrence). ( T • , U ) is chain recur-rent iff there exists an a.d.p.t. that is piecewise continuous and ω -dense in U. Necessity has been proved already, because the pseudo-trajectory (4.2.6.2) pos-sesses this property. Sufficiency will be proved later.The claim of piecewise continuity can be justified by
Theorem 4.3.1.3.
For any u ∈ SH ( ρ ( r )) there exists a piecewise continuouspseudo-trajectory w ( •|• ) such that (4.3.1.5) u t − w ( •| t ) → . as t → ∞ . Of course, w ( •|• ) is a.d.p.t. Exercise 4.3.1.2.
Check this. Proof of Theorem 4.3.1.3.
Let { t n } be any sequence such that(4.3.2.1) t n → ∞ , t n +1 /t n → , for example, t n = n. There exists a sequence { v n } ⊂ Fr [ u ] such that(4.3.2.2) u t n − v n → . Set(4.3.2.3) w ( •| t ) := v n , f or t n < t ≤ t n +1 . This is a piecewise continuous function.Let us prove that(4.3.2.4) u t − w ( •| t ) → . Assume the opposite; i.e., there exists a sequence { t ′ k } such that it is not true. Wecan suppose that(4.3.2.5) u t ′ k → w ∈ Fr [ u ] , w ( •| t ′ k ) → w ∈ Fr [ u ] , w = w . Let us find a sequence { n k } such that t n k < t ′ k < t n k +1 . Then(4.3.2.6) t n k /t ′ k → . From (4.3.2.5), (4.3.2.3) and (4.3.2.2) we have(4.3.2.7) u t nk → w . Then we have, using properties of ( • ) t and the proximate order,(4.3.2.8) u t ′ k = ( u t nk ) [ t ′ k /t nk ] (1 + o (log( t ′ k /t n k )) → w because of (4.3.2.6) and the continuity of u [ t ] on ( u, t ) . However (4.3.2.8) contradicts (4.3.2.5). Thus (4.3.2.4) holds. (cid:3) Now we will prepare the proof of Theorem 4.3.1.2.Let { v k , k = 1 , , ... } ⊂ U [ ρ, σ ] for some σ be a sequence of functions and { r k , k = 1 , , ... } , { t k k = 1 , , ... } be two sequences such that(4.3.3.1) 0 < r k < t k < r k +1 , k = 1 , , ... and(4.3.3.2) lim k →∞ t k /r k = lim k →∞ r k +1 /t k = ∞ . Set(4.3.3.3) w ∗ ( •| t ) := ( v k ) t/t k , for t ∈ [ r k , r k +1 )where k = 1 , , ... Theorem 4.3.3.1.
Let w ( •|• ) ⊂ U be an arbitrary ω -dense a.d.p.t. and { p j , j =1 , , ... } ⊂ U an arbitrary sequence. Then there exists a sequence { v k , k = 1 , , ... }⊃{ p j , j = 1 , , ... } and sequences { r k , k = 1 , , ... } and { t k , k = 1 , , ... } satisfying(4.3.3.1) and (4.3.3.2) such that for w ∗ ( •|• ) determined by (4.3.3.3) the condition (4.3.3.4) w ∗ ( •| t ) − w ( •| t ) → as t → ∞ is fulfilled. This proposition shows that any ω -dense a.d.p.t. is equivalent to one con-structed of long pieces of trajectories of the dynamical system T τ . Proof of Theorem 4.3.3.1.
We can take sequences { ǫ j ↓ , j = 1 , , ... } and { b j ↑ ∞ , j = 1 , , ... } and choose a sequence { τ j , j = 1 , , .. } such that the in-equalities(4.3.3.5) d ( T τ p j − T τ w ( •| τ j )) < ǫ j / d ( T τ w ( •| t ) − w ( •| e τ t )) < ǫ j / , t > τ j are fulfilled uniformly with respect to τ ∈ [ b − j +1 , b j ] . Indeed, w ( •|• ) is ω -dense in U, and hence we can find τ n → ∞ such that p n − w ( •| τ n ) → . Set in Lemma 4.1.2.2 p n := p n , q n := w ( •| τ n ) , γ n := 2 log b j Then for any ǫ j we can find τ j := τ n j such that (1.2.5) holds uniformly with respectto τ ∈ [ b − j +1 , b j ] . The inequality (4.3.3.6) holds, because w ( •|• ) is asymptotically dynamical (see(4.3.1.1)).We can also suppose without loss of generality that τ j > τ j − b j − , i.e., thatthe sequence { τ j } is rather thin.The inequality (4.3.3.5) shows that for intervals of t that are determined bythe inequality b − j +1 ≤ t/τ j ≤ b j our pseudo-trajectory is already close to sometrajectories.Now we divide the spaces between such intervals into equal parts in the log-arithmic scale such that their logarithmic lengths would be between log b j andlog b j +1 ,so that they tend to infinity.To this and set n j := (cid:20) log τ j +1 − log τ j b j (cid:21) where [ · ] means the entire part, and γ j := ( τ j +1 /τ j ) nj . It is clear that b j < γ j < b j . As centers of new intervals we take the points τ j,l := τ j γ lj , l = 0 , , ..., n j . Thus τ j, = τ j and τ j,n j = τ j +1 . The ends of the intervals are τ j,l /γ j and τ j,l γ j . Now we complete the sequence { p j } by the values of the pseudo-trajectory w ( •| t )in the centers of the intervals, i.e., we set p j,l := w ( •| τ j,l ) , l = 1 , ..., n j − For t ∈ ( τ j,l /γ j , τ j,l γ j ) , l = 1 , ..., n j − d (( p j,l ) t/τ j,l − w ( •| t )) < ǫ j / l = 0 and l = n j we set accordingly p j, := p j ; p j,n j := p j +1 . Using (4.3.3.5) and (4.3.3.6) we have the inequality like (4.3.3.7) for l = 0 , l = n j but with ǫ j instead of ǫ j / . We complete the proof re-denoting all the centers τ j,l as t k , all the ends as r k and all the p j,l as v k . (cid:3) Proof of sufficiency in Theorem 4.3.1.2.
A direct corollary of the previous Theorem4.3.3.1 is(4.3.4.1) w ∗ ( •| r k − − w ∗ ( •| r k ) → k → ∞ . Really, w ∗ ( •|• ) is an a.d.p.t. Exercise 4.3.4.1.
Check this as in Theorem 4.3.1.1 using that w ( •|• ) is asymp-totically dynamical.For τ ∈ [ − ǫ,
0] and t = r k we have uniformly on τT τ w ∗ ( •| t ) − w ∗ ( •| t ) = T τ ( v k ) r k /t k − ( v k +1 ) r k /t k → . Setting τ = 0 we obtain 4.3.4.1.Let V ⊂ U be an arbitrary open set, ǫ > s > ǫ, s )-chain from V to V. Choose s such thati. for r k > s , d ( w ∗ ( •| r k − , w ∗ ( •| r k )) < ǫ. It is possible by virtue of (4.3.4.1). ii. w ( •| s ) ∈ V . This is possible because of w ( •|• ) is ω -dense.iii. d ( w ∗ ( •| t ) , w ( •| t )) < d ( w ( •| s ) , ∂V ) for t > s . This is possible because ofTheorem 4.3.3.1.Choose s > s such that w ( •| s ) ∈ V. This is possible because w ( •|• ) is ω -dense. Then the pseudo-trajectory w ∗ ( •| e t ) for s ≤ e t ≤ s is an ( ǫ, s )-chainconnecting w ∗ ( •| s ) and w ∗ ( •| s that belong to V. Exercise 4.3.4.1.
Check this in details.Hence ( T • , U ) is chain recurrent. We will prove one more existence theorem that is a corollary of Theorem4.2.1.2.
Theorem 4.3.5.1.
Let Λ ⊂ U [ ρ ] be a compact connected and T • – invariant subsetof U [ ρ, ] Then for any proximate order ρ ( r ) → ρ there exists u ∈ SH ( R m , ρ, ρ ( r )) such that (4.3.5.1) h ( x, u ) = sup { v ( x ) : v ∈ Λ } (4.3.5.2) h ( x, u ) = inf { v ( x ) : v ∈ Λ } . Proof.
Let U := Conv
Λ be the convex hull of Λ . It is linearly connected and hencepolygonally connected (see 4.1.4). By Theorem 4.1.4.1 it is chain recurrent and byTheorem 4.2.1.2 for any proximate order ρ ( r ) → ρ there exists u ∈ SH ( R m , ρ, ρ ( r ))such that Fr [ u, ρ ( r ) , V t , R m ] = U. Since every v ∈ U can be represented in the form v = av + (1 − a ) v for 0 ≤ a ≤ , v , v ∈ Λ we obtain (4.3.5.1) and (4.3.5.2) from Theorem 3.2.1.1 (Properties ofIndicators),h2).
Exercise 4.3.5.1.
Check this.
In applications we need the following Theorem 4.3.6.1.
Let p ∈ P ⊂ R m and let P be a connected closed set. Let U P := { v ( z, p ) : p ∈ P ⊂ R m } be a family of functions with parameter p such thatfor every p ∈ P v ( • , p ) ∈ U [ ρ ] and satisfy the condition (4.1.3.3). Then there exists u ∈ SH ( ρ ( r ) such that Fr [ u ] = U P . This is a direct corollary of Theorems 4.1.1.2, 4.1.3.4 and 4.2.1.2.
Exercise 4.3.6.1
Explain this in details.
In the next three §§ we return to the periodic limit sets (see Th.4.1.7.1).We show that the limit set Fr [ u, ρ ( r ) , V • , R m ] of every subharmonic function u ∈ SH ( ρ ( r ) , R m ) , ρ ( r ) → ρ for noninteger ρ can be approximated in some sense byperiodic limit sets ([Gi(1987)],[GLO,Ch.3, § X n ⊂ U [ ρ ] , n = 1 , , ... be a sequence ofcompact sets. We say that X n converges to a compact set Y ⊂ U [ ρ ] , i.e.,(4.3.7.1) D ′ − lim n →∞ X n = Y if the following two conditions hold:converg1) ∀ x n ∈ X n , n = 1 , , ... ∃ x n j ∈ X n j , j = 1 , , ... and y ∈ Y such that D ′ − lim j →∞ x n j = y ;converg2) ∀ y ∈ Y ∃ x n ∈ X n , n = 1 , , .., such that x n → y. On every compact set K in D ′ - topology one can introduce a metric d ( • , • )such that the topology generated by this metric is equivalent to D ′ - topology(see,e.g.[AG(1982)]).Denote by X ǫ := { y ∈ K : ∃ x ∈ X such that d ( y, x ) < ǫ } the ǫ − neighborhood of X. Let
X, Y be two compact sets. Set d ( X, Y ) := inf { ǫ : X ⊂ Y ǫ , Y ⊂ X ǫ } . Exercise 4.3.7.1
Prove the assertion(4.3.7.2) (4 . . . ⇐⇒ { d ( X n , Y ) → } . We prove the following Theorem 4.3.7.1.(Approximation by Periodic Limit Sets).
Let u ∈ SH ( ρ ( r ) , R m ) , ρ ( r ) → ρ for noninteger ρ. Then for every V • there exists asequence u n ∈ SH ( ρ ( r ) , R m ) with periodic limit sets Fr [ u n ,ρ ( r ) , V • , R m ] such that Fr [ u n , • ] → Fr [ u, • ] . This theorem is a corollary of the following
Theorem 4.3.7.2.
Let µ ∈ M ( ρ ( r ) , R m ) , ρ ( r ) → ρ for noninteger ρ. Thenfor every V • there exists a sequence µ n ∈ SH ( ρ ( r ) , R m ) with periodic limit sets Fr [ µ n ,ρ ( r ) , V • , R m ] such that Fr [ µ n , • ] → Fr [ µ, • ] . Proof of Theorem 4.3.7.1.
The canonical potential u ( x ) := Π( x, µ, p ) (see (2.9.2.1))of a measure µ ∈ M ( ρ ( r ) , R m ) belongs to SH ( ρ ( r ) , R m ) by Th.2.9.3.3 and has alimit set Fr [ u, • ] = { Π( • , ν, p ) : ν ∈ Fr [ µ u ] } by Th.3.1.5.2. The potentials u n ( x ) := Π( x, µ n , p ) have periodic limit sets Fr [ u n , • ] = { Π( • , ν, p ) : ν ∈ Fr [ µ u n ] } by Th.3.1.5.0. Let us prove that Fr [ u n , • ] =: X n → Y := Fr [ u, • ] . If v n ∈ Fr [ u n , • ] then from the corresponding sequence of ν n := ν v n ∈ Fr [ µ n , • ]we can find a subsequence ν n j and ν ∈ Fr [ µ, • ] such that ν n → ν (by Th.2.2.3.2(Helly)). It is easy to check, using Th.3.1.4.3 (*Liouville), that v ∗ = D ′ − lim j →∞ Π( • , ν n j )exists and coincides with v = Π( • , ν, p ) ∈ Fr [ u, • ] . So the condition converg1) is verified. In the same way one can check con-verg2). (cid:3)
Exercise 4.3.7.2
Prove this in details.
Now we are going to prove Theorem 4.3.7.2. We begin from Proposition 4.3.8.1.
For any µ ∈ M ( ρ ( r ) , • ) there exists ˆ µ ∈ M ( ρ, • ) such that (4.3.8.0) Fr [ˆ µ, ρ, • ] = Fr [ µ, ρ ( r ) , • ] . In other words we can suppose further that ρ ( r ) ≡ ρ. Proof.
Set L ( r ) = r ρ ( r ) − ρ and(4.3.8.1) ˆ µ ( dx ) := L − ( | x | ) µ ( dx ) . Using properties of proximate order (2.8.1., po1)-po4)), it is easy to check that(4.3.8.2) [ L − ( r )] ′ = L − ( r ) o (1) and [ L ( r )] ′ = L ( r ) o (1) , as → . Exercise 4.3.8.1
Prove this.Let us show that ˆ µ ∈ M [ ρ, ∆] for some ∆ . Indeedˆ µ ( R ) R ρ + m − = R − ρ − m +2 R Z µ ( dr ) L ( r ) = µ ( r ) R ρ + m − L ( r ) | R + R − ρ − m +2 R Z µ ( r )( L − ) ′ dr =We suppose that µ ( r ) = 0 in some neigborhood of zero. Using (4.3.8.2) we obtainfurther = µ ( R ) R − ρ ( r ) + R − ρ − m +2 R Z µ ( r )( L − ) o (1) dr. Using the l’Hospital rule, we obtainlim R →∞ R − ρ − m +2 R Z µ ( r )( L − ( r )) o (1 /r ) dr = ( − ρ − m + 2) lim R →∞ µ ( R ) R − ρ ( R ) o (1 /R ) . Thus lim sup R →∞ ˆ µ ( R ) R ρ + m − ≤ lim sup R →∞ µ ( R )( L − ( R ))[1 + o (1 /R )] = ∆[ µ, ρ ( r )] < ∞ Let us note that µ t = L ( t )ˆ µ [ t ] . This implies equality (4.3.8.0) because L ( t ) → t → ∞ . Exercise 4.3.8.2.
Prove this in details. (cid:3)
Proof of Th.4.3.7.2.
As we already said we can suppose that µ ∈ M [ ρ ] . Let ν ∈ Fr [ µ ] . We can suppose that(4.3.8.2a) ν ( {| x | = 1 } ) = 0 . Otherwise we can find τ such that ν [ τ ] ( {| x | = 1 } ) = 0 and if ν n → ν τ and areperiodic then ( ν n ) [1 /τ ] are also periodic and ( ν n ) [1 /τ ] → ν. Let r n → ∞ be such that µ [ r n ] → ν. By passing to subsequences we can make r n +1 /r n > r n . Denote K n := { x : r n ≤ | x | < r n +1 } . Set for every E ⊂ K n µ n | E := µ ( E )and continue it periodically with the period P n = r n +1 /r n by the equality(4.3.8.3) µ n ( P kn E ) = P kρn µ ( E ) , k = ± , ± , ... Since every X ∈ R m can be represented in the form X = ∞ [ k = −∞ { X ∩ P kn K n } , we can define µ n ( X ) := ∞ X k = −∞ µ n ( { X ∩ P nk K n } ) . It is easy to check that µ n is periodic with the period P n and µ n ∈ M [ ρ, ∆] with∆ independent of n. Exercise 4.3.8.3
Check this.Let us prove that(4.3.8.4) Fr [ µ ] = lim n →∞ Fr [ µ n ] . Check the condition converg1). Let ν n j ∈ Fr [ µ n j ] and suppose D ′ − lim j →∞ ν n j := ν. Let us prove that ν ∈ Fr [ µ ] . Since Fr [ µ n j ] is a periodic limit set ν n j = ( µ n j ) [ τ j ] . Take k j such that τ ′ j := τ j P k j n j ∈ [ r n j , r n j +1 ) . From periodicity µ n j we obtain ν n j = ( µ n j ) [ τ ′ j ] . Passing to a subsequence if necessary, we can consider three cases:i) lim j →∞ τ ′ j /r n j = ∞ , lim j →∞ τ ′ j /r n j +1 = 0;ii) lim j →∞ τ ′ j /r n j = τ ; 1 ≤ τ < ∞ ;In this case we have also lim j →∞ τ ′ j /r n j +1 = 0 . iii) lim j →∞ τ ′ j /r n j +1 = τ ; 0 < τ ≤ j →∞ τ ′ j /r n j = ∞ . Consider the case i). Let φ ∈ D ( R m \ O ) . Then supp φ ( x/τ ′ j ) ⊂ ( r n j , r n j +1 ) for j ≥ j . It is easy to see that for j ≥ j < ( µ n j ) [ τ ′ j ] , φ > = < µ [ τ ′ j ] , φ > . Exercise 4.3.8.4
Check this.Since µ [ τ ′ j ] → ν ∈ Fr [ µ ] by definition the condition converg1) holds for the casei). Consider the case ii).Recall that O / ∈ supp φ. Then there exists 1 ≤ c < ∞ such thatsupp φ ⊂ { x : | x | ∈ (1 /c, c ) } . Define φ t ( x ) := φ ( x/t )(1 /t ) ρ . Represent τ ′ j in the form τ ′ j := e j τ r n j where e j := τ ′ j r n j τ . The condition ii) means that(4.3.8.4a) e j → . Compute < ν j , φ > := < ( µ n j ) [ τ ′ j ] , φ > = < µ n j , (( φ τ ) e j ) r nj > . Note that supp φ τ ⊂ { x : | x | ∈ ( τ /c, τ c ) } . We can increase c so that 1 ∈ ( τ /c, τ c ) . Consider the following partition of unity. Choose the functions η k ∈ D ( R m ) , k =1 , , η ( t ) + η ( t ) + η ( t ) = 1for t ≥ η ⊂ { x : | x | < − ǫ } , supp η ⊂ { x : | x | ∈ (1 − ǫ, ǫ ) } , supp η ⊂ { x : | x | > ǫ } , where ǫ is an arbitrary number, satisfying τ /c < − ǫ < ǫ < τ c. Represent φ τ in the form φ τ = ψ + ψ + ψ , where ψ k = φ τ η k , k = 1 , , . In this notation(4.3.8.5) < ν j , φ > = X k =1 < µ n j , ( ψ k ) e j r nj > . Choose j ǫ such that for j ≥ j ǫ the following inclusions hold:supp( ψ ) e j r nj ⊂ { x : | x | ∈ ( r n j τ /c, r n j (1 − ǫ )) } ;supp( ψ ) e j r nj ⊂ { x : | x | ∈ ((1 − ǫ ) r n j , r n j (1 + ǫ )) } ;supp( ψ ) e j r nj ⊂ { x : | x | ∈ ((1 + ǫ ) r n j , τ r n j } . Thus for ψ we have < µ n j , ( ψ ) e j r nj > = Z ( e j r n j ) − ρ ψ ( | x | / ( e j r n j )) µ n j ( dx ) = Z ( e j r n j ) − ρ ψ ( | x | / ( e j r n j )) µ ( dx ) = < µ [ r nj ] , ( ψ ) e j > . Since µ [ r nj ] D ′ → ν and ( ψ ) e j D → ψ we have (see Th.2.3.4.6)(4.3.8.6) lim j →∞ < µ n j , ( ψ ) e j r nj > = < ν, ψ > . Consider the addend with ψ . Because of periodicity µ n j we have < µ n j , ( ψ ) e j r nj > = < ( µ n j ) [ P nj ] , ( ψ ) e j r nj > . Transforming the RHS we obtain < ( µ n j ) [ P nj ] , ( ψ ) e j r nj > = < µ n j , (( ψ ) P nj ) e j r nj > . Since P n j = r n j +1 /r n j the following inclusion holds for j ≥ j ǫ :supp( ψ ) P nj e j r nj ⊂ { x : | x | ∈ ( P n j r n j τc , P n j r n j (1 − ǫ )) } = { x : | x | ∈ ( r n j +1 τc , r n j +1 (1 − ǫ )) } ⊂ { x : | x | ∈ ( r n j , r n j +1 ) } . Thus < µ n j , (( ψ ) P nj ) e j r nj > = < µ, ( ψ ) P nj e j r nj > = < µ, ( ψ ) e j r nj > = < µ [ r nj ] , ( ψ ) e j > . Hence(4.3.8.7) lim j →∞ < µ n j , ( ψ ) e j r nj > = < ν, ψ > because e j → µ [ r nj ] D ′ → ν (see Th.2.3.4.6).From (4.3.8.5), (4.3.8.6) and (4.3.8.7) we obtainlim j →∞ < ν j , φ > = < ν, ψ + ψ > + lim j →∞ < µ n j , (( ψ ) e j r nj > . Let us estimate the last limit. We havelim j →∞ < µ n j , (( ψ ) e j r nj > = < ( µ n j ) [ e j r nj ] , ψ > Define E ( ǫ ) := { x : | x | ∈ (1 − ǫ, } ; E ( ǫ ) := { x : | x | ∈ [1 , ǫ ) } Suppose ǫ is chosen so that(4.3.8.7a) ν ( ∂E k ) = 0 , k = 1 , . Recall that ν satisfies the condition (4.3.8.2a), hence E , E are ν -squarable andhence (see Th.2.2.3.7) lim n →∞ µ [ r n ] ( E k ( ǫ )) = ν ( E k ( ǫ )) , k = 1 , . Define C φ := max { φ ( x ) : x ∈ R m } . Then for j ≥ j ǫ | < ( µ n j ) [ e j r nj ] , ψ > | ≤ C φ ( µ n j ) [ e j r nj ] ( E ( ǫ ) ∪ E ( ǫ )) = C φ (( µ n j ) [ e j r nj ] ( E ( ǫ )) + ( µ n j ) [ e j r nj ] ( E ( ǫ )) . By definition ( µ n j ) [ e j r nj ] ( E ( ǫ )) = µ [ e j r nj ] ( E ( ǫ )) . Because of (4.3.8.4a) we obtainlim j →∞ µ [ e j r nj ] ( E ( ǫ )) = ν ( E ) Exercise 4.3.8.5
Check in details.To compute the limit of the first addend we use periodicity of µ n j : µ n j ( E ( ǫ )) = P − ρn j µ n j ( P n j E ( ǫ )) = ( µ n j ) [ P nj ] ( E ( ǫ )) , where r n j P n j E ( ǫ ) = { x : | x | ∈ ( r n j +1 (1 − ǫ ) , r n j +1 ) } . Thus ( µ n j ) [ e j r nj ] ( E ( ǫ )) = ( µ n j ) [ e j r nj P nj ] ( E ( ǫ )) =( µ n j ) [ e j r nj +1 ] ( E ( ǫ )) = µ [ e j r nj +1 ] ( E ( ǫ )) . From this we obtain lim j →∞ µ n j ) [ e j r nj ] ( E ( ǫ )) = ν ( E ( ǫ )) . Therefore(4.3.8.8) lim j →∞ | < ( µ n j ) [ e j r nj ] , ψ > | ≤ C φ ν ( E ( ǫ ) ∪ E ( ǫ )) . Because of (4.3.8.2a) we have(4.3.8.9) ν ( { x : | x | ∈ (1 − ǫ, ǫ ) } ) → ǫ → ǫ satisfying (4.3.8.7a). From (4.3.8.8) and (4.3.8.9) weobtain(4.3.8.10) < lim j →∞ < ( µ n j ) [ e j r nj ] , ψ > → and(4.3.8.11) < ν, ψ > → ǫ → . Hence if ǫ → j →∞ < ν j , φ > − < ν [ τ ] , φ > =lim ǫ → [ < ν, ψ + ψ > + lim j →∞ < ( µ n j ) [ e j r nj ] , ψ > − < ν, ψ + ψ + ψ > ] =lim ǫ → [ lim j →∞ < ( µ n j ) [ e j r nj ] , ψ > − < ν, ψ > ] = 0The last equality holds because every addend tends to zero.The case iii) can be considered in an analogous way. Exercise 4.3.8.6.
Consider it.Thus the condition converg1) was checked. (cid:3)
Now we should check the condition converg2). We need
Lemma 4.3.9.1.
Let µ ∈ M [ ρ ] , ν ∈ Fr [ µ ] and r n → ∞ , n = 1 , , .. is a sequencesuch that (4.3.9.1) D ′ − lim n →∞ µ [ r n ] = ν . Then passing if necessary to a subsequence, we can find { r n } such that for arbitrarily ν ∈ Fr [ µ ] a sequence t j → ∞ exists such that (4.3.9.2) D ′ − lim j →∞ µ [ t j ] = ν. and for every n we can find t j ∈ [ r n , r n +1 ] . Proof.
Note that if the assertion of the Lemma is satisfied for the sequence { r n , n =1 , , ... } it is satisfied for every subsequence of { r n , n = 1 , , ... } . Let M be a countable set that is dense in Fr [ µ ] . Since reduction D ′ - topologyon M [ ρ ] is metrizable, it is sufficient to prove that we can choose a subsequence r n for which assertion of the lemma is satisfied for all ν ∈ M. We can do it using adiagonal process. Let r n → ∞ be an arbitrary sequence such that D ′ − lim n →∞ µ [ r n ] = ν and let a sequence t j satisfies the condition D ′ − lim j →∞ µ [ t j ] = ν . Omitting in the sequence { r n } the ends of the segments [ r n , r n +1 ] that do not con-tain elements { t j } we obtain a subsequence { r n } . Continuing in such way we ob-tain a subsequence { r mn } satisfying (4.3.9.1) and the sequences { t j } , { t j } ... { t mj } , j =1 , ... satisfying(4.3.9.3) D ′ − lim j →∞ µ [ t lj ] = ν l , l = 1 , , ...m. Taking a diagonal sequence { r nn } , n = 1 , , ... we observe that it is a subsequenceof every subsequence { r mn } and hence satisfies the assertion of the lemma. (cid:3) Proof of converg2).
We can suppose that { r n } from the construction of µ n withperiodic limit sets satisfies the assertion of Lemma 4.3.9.1. Let ν ∈ Fr [ µ ] and µ t j → ν under condition t j ∈ [ r n , r n +1 ] . We should consider as in the proof ofconverg1) three cases i),ii) and iii). But all these cases were already considered andhence it was proved that ν n := ( µ n ) r n → ν. Exercise 4.3.9.1.
Check this. (cid:3) In this paragraph we consider subharmonic functions u ∈ SH ( ρ ( r )) in theplane of finite type with respect to some proximity order ρ ( r ) → ρ. The pair u := ( u , u ) , u , u ∈ SH ( ρ ( r )) is called a subharmonic curve orshort by curve .The family ( u ) t := (( u ) t , ( u ) t )is precompact in the topology of convergence in D ′ -topology on every component.The set of all limits Fr [ u ] := { v = ( v , v ) : ∃ t j → ∞ , v = D ′ − lim j →∞ u t j } is called the limit set of the curve u . Actually this set describes coordinated asymptotic behavior of pairs of subhar-monic functions.
Theorem 4.4.1.1. Fr [ u ] is closed, connected, invariant with respect to ( • ) [ t ] (see3.1.2.4a) and is contained in the set U [ ρ, σ ] := { v = ( v , v ) : v n ( z ) ≤ σ n | z | ρ , v n (0) = 0 , n = 1 , . } where σ := ( σ , σ ) Exercise 4.4.1.1.
Prove this by using Th.3.1.2.2.Let us define σ > σ n > , n = 1 , . Set U [ ρ ] := [ σ> U [ ρ, σ ]We will write U ⊂ U [ ρ ] if U ⊂ U [ ρ, σ ] for some σ. Since ( T • , U [ ρ ]) is a dynamical system we have two theorems analogous toTheorems 4.2.1.1 and 4.2.1.2. Exercise 4.4.1.2.
Formulate and prove this theorems.All the other assertions and definitions of §§ U ⊂ U [ ρ ] . Set U ′ := { v ′ : ∃ v ′′ : ( v ′ , v ′′ ) ∈ U } . This is a projection of U . Set for v ′ ∈ U ′ U ′′ ( v ′ ) := { v ′′ : ( v ′ , v ′′ ) ∈ U } . This is the fibre over v ′ . Theorem 4.4.1.2.
Let U ⋐ U [ ρ ] be closed and invariant and assume that everyfiber U ′′ ( v ′ ) is convex.Let U ′ = Fr [ u ′ ] for some u ′ ∈ U ( ρ ( r )) . Then there exists u ′′ ∈ U ( ρ ( r )) such that Fr ( u ′ , u ′′ ) = U . We construct a pseudo -trajectory asymptotics in the form 4.2.5.2 replacing u with u and v with v . We can directly check that this curve satisfies the assertionof the Theorem.
Exercise 4.4.1.3
Check this.
Theorem 4.4.1.3.(Concordance Theorem).
Let u ∈ U ( ρ ( r )) and v ∈ Fr [ u ] , and suppose v ∈ U [ ρ ] has the property lim τ →−∞ T τ v = lim τ → + ∞ T τ v = ˜ v Then there exists a function w ∈ U ( ρ ( r )) such that the limit set of the curve u =( u, w ) Fr [ u ] = ( Fr [ u ] , C ( v )) and for every sequence t n → ∞ such that lim n →∞ w t n = v (4.4.1.3.) lim n →∞ u t n = ( v , v )For proving this theorem we should use a.d.p.t. (4.2.5.2). If v j = v we replace v j by v j := ( v , v ) . If v j = v we replace v j by v j := ( v j , ˜ v ) . Exercise 4.4.1.3.
Do that and exploit Th.4.3.1.2 and Th.4.2.1.2. Corollary 4.4.1.4.
Under conditions of Th.4.4.1.3, if lim n →∞ w t n = T τ v, then lim n →∞ u t n = T τ v . We should apply T τ to (4.4.1.3) and use its continuity in D ′ -topology. (cid:3)
5. Applications to Entire functions5.1.Growth characteristics of entire functions5.1.1.
Let f ( z ) be an entire function. The function u ( z ) := log | f ( z ) | is subhar-monic in R (= C ) . Hence the scale of growth subharmonic functions considered in § u for index f. For example, M ( r, f ) := M ( r, log | f | ) , T ( r, f ) := T ( r, log | f | ) . If u ( z ) := log | f ( z ) | has order ρ [ u ] = ρ, then f ( z ) has order ρ [ f ] := ρ and so on.We will write f ∈ A ( ρ, ρ ( r )) and say “ f is an entire function of order ρ andnormal type with respect to proximate order ρ ( r )” if log | f | is a subharmonic functionof order ρ and normal type with respect to the same proximate order. Shortly, iflog | f | ∈ SH ( ρ, ρ ( r ) , R ) then f ∈ A ( ρ, ρ ( r )) . Exercise 5.1.1.1.
Give definitions of T ( r, f ) , M ( r, f ) , ρ T [ f ] , ρ M [ f ] , σ T [ f, ρ ( r )] , σ M [ f, ρ ( r )]and reformulate all the assertions of § A divisor of zeros of an entire function can be represented as an integermass distribution n on a discrete set { z j } ⊂ C . The multiplicity of a zero z j is themass concentrated at the point z j . The notation for characteristics of the behavior of zeros will mimic that of hebehavior of the behavior of masses, replacing µ for n. For example, n ( K r ) , n ( r ) isthe number of zeros (with multiplicities) in the disk K r , ρ [ n ] is the convergenceexponent, ∆[ n ] is the upper density and so on. Exercise 5.1.1.2.
Give definitions of N ( r, n ) , ρ N [ n ] , ∆ N [ n ] , p [ n ] . The limit set Fr [ f ] of an entire function f ∈ A ( ρ, ρ ( r )) is defined as the limitset of the subharmonic function u ( z ) := log | f ( z ) | ∈ SH ( ρ, ρ ( r ) , R ) (see § Fr [ f ] := Fr [log | f | ] . It possesses, of course, all the properties described in Ch.’s 3,4 but it is not clearnow if there exists an entire function with prescribed limit set, i.e., whether thesubharmonic function in Theorem 4.2.1.2 can be chosen to be log | f ( z ) | where f ∈ A ( ρ, ρ ( r )) . It turns out that this is possible and we prove this in § V • for the case of the plane is V t z = ze iγ log t , where γ is real.The limit set Fr [ n ] of a divisor n is the limit set of the corresponding massdistribution n (see 3.1.3).Of course generally speaking n t (see (3.1.3.2)) is not an integer mass distribu-tion. Exercise 5.1.3.1.
Give a complete definition of Fr [ f ] and Fr [ n ] , and reformulateall the theorems of §§ Fr [ f ] and Fr [ n ] is preserved completely (see § Exercise 5.1.3.2.
Reformulate the theorems of § Let f = f /f be a meromorphic function, where f , f have no commonzeros. If f (0) = 1 , f (0) = 0 and f , f ∈ A ( ρ, ρ ( r )) , then u := log | f | − log | f | ∈ δSH ( ρ, ρ ( r )) , and we write f ∈ M er ( ρ, ρ ( r )) and say “ f is a meromorphic func-tion of order ρ and normal type with respect to the proximate order ρ ( r ) . For f ∈ M er ( ρ, ρ ( r )) we use the following characteristics: T ( r, f ) , ρ T [ f ] , σ T [ f, ρ ( r )] . The charge of log | f | consists of integer positive and negative masses. D ′ -topology and Topology of exceptional sets.5.2.1. Let α − mes be the Carleson measure defined in 2.5.4. Set for C ⊂ R (5.2.1.1) α − mes C := lim sup R →∞ [ α − mes( C ∩ K R )] R − α . It is called the relative
Carleson α -measure. Theorem 5.2.1.1 (Properties of the Relative Carleson Measure).
One hasrCm1) If C is bounded α − mes C = 0; rCm2) α − mes( C ∩ C ) ≤ α − mes C + α − mes C , i.e., the relative Carleson measure is sub-additive;rCm3) C ⊂ C ⇒ α − mes( C ) ≤ α − mes C , i.e., the relative Carleson measure is monotonic with respect to sets;rCm4) α > α ⇒ α − mes C ≤ α − mes C, i.e., the relative Carleson measure is monotonic with respect to α. Exercise 5.2.1.1.
Prove this.A set C ⊂ R for which α − mes C = 0 is called a C α − set. If α − mes C = 0for all α > , C is called a C − set. Let us recall that if u , u ∈ SH ( ρ, ρ ( r ) , R ) , then u = u − u ∈ δSH ( ρ, ρ ( r ) , R )(see 2.8.2). Theorem 5.2.1.2 ( D ′ topology and Exceptional sets). Let u ∈ δSH ( ρ, ρ ( r ) , R ) In order that (5.2.1.2) u t → in D ′ as t → ∞ it is sufficient that (5.2.1.3) u ( z ) | z | − ρ ( | z | ) → as z → ∞ outside some C − set. If (5.2.1.2) holds, then (5.2.1.3) holds outside some C − set. To prove Theorem 5.2.1.2 we need some auxiliary assertions. Recall that dz is an element of area following the notation of the previous chapters. Proposition 5.2.2.1.
Let u ∈ SH ( ρ, ρ ( r ) , R ) , and C ,R := C ∩ K R . Then (5.2.2.1) Z C ,R | u | ( z ) dz = o ( R ρ ( R )+2 ) as R → ∞ . Proof.
Suppose (5.2.2.1) does not hold. Then there exists a sequence R j → ∞ suchthat(5.2.2.2) lim R j →∞ R − ρ ( R j ) − j Z C ,Rj | u | ( z ) dz = A > δ -subharmonic functions:(5.2.2.2a) u j ( ζ ) := R − ρ ( R j ) j u ( ζR j )It can be represented as a difference u j = u ,j − u ,j of subharmonic functions ofthe same form.Thus it is precompact in L loc (Theorem 2.7.1.3). Let us choose a convergentsubsequence for which we keep the same notation. Its limit v is a locally summablefunction.Now let χ j be the characteristic functions of the sets E j := R − j C ,R j . Since mes E j → χ j → R j arethe same for χ j and u j . Thus Z | ζ |≤ | χ j ( ζ ) u j ( ζ ) − · v ( ζ ) | dζ = Z | ζ |≤ | χ j ( ζ ) u j ( ζ ) | dζ → . By change of variables z = R j ζ we obtain that R − ρ ( R j ) − j Z C ,Rj | u | ( z ) dz = Z | ζ |≤ | χ j ( ζ ) u j ( ζ ) | dζ → . Hence the limit in (5.2.2.2) is equal to zero. Contradiction. (cid:3) Proposition 5.2.2.2.
Under condition (5.2.1.2) the set C := { z : | u ( z ) || z | − ρ ( | z | ) > ǫ } is a C − set for arbitrary ǫ. Proof.
Assume the contrary; that is , ∃ α > α − mes C = 2 δ > . One can see that for some η > R →∞ ( α − mes K ηR ) R − α ≤ δ/ . Exercise 5.2.2.1.
Check this.(5.2.2.3) and (5.2.2.4) imply that there exists a sequence R j → ∞ such thatlim R j →∞ α − mes[ C ∩ ( K R j \ K ηR j )] R − αj ≥ δ. Set E j := R − j C ∩ ( K R j \ K ηR j ) . It is clear that E j ⊂ K \ K η and for sufficiently large j (5.2.2.5) α − mes E j ≥ δ. Set u j as in (5.2.2.2a). We claim that for large j and ζ ∈ E j (5.2.2.6) | u j | ( ζ ) ≥ ǫ | ζ | ρ . Indeed, | u j | ( ζ ) = | u | ( R j ζ ) R ρ ( R j ) j = | u | ( z ) | z | ρ ( | z | ) (1 + o (1)) | ζ | ρ ≥ ǫ | ζ | ρ . We used here properties of the proximate order and the equivalence z = R j ζ ∈ C ∩ ( K R j \ K ηR j ) ⇔ ζ ∈ E j . Exercise 5.2.2.2.
Check this in details. Now we will show that the condition (5.2.1.2) contradicts (5.2.2.6). Since u ∈ δSH ( ρ, ρ ( r ) , R ) it is a difference of u , u ∈ SH ( ρ, ρ ( r ) , R ) . The correspondingsequences u ,j and u ,j are precompact in D ′ and there exist subsequences (withthe same notation) that converge to v and v , respectively.By Theorem 2.7.5.1 these sequences converge to v and v with respect to α − mes on K \ K η . Since u t → D ′ , it follows that v = v . Thus u j → α − mes on K \ K η . However, this contradicts (5.2.2.5) and (5.2.2.6). (cid:3)
Proposition 5.2.2.3.
Let { C j } ∞ be a sequence of C -sets. There exists a sequence R j → ∞ such that the set (5.2.2.7) C = ∞ [ j =1 { C j ∩ ( K R j +1 \ K R j ) } is a C -set.Proof. Choose ǫ j ↓ α j ↓ . Set R := 1 . Suppose R j − was already chosen.Take R j such that(5.2.2.8) α j − mes[ C ∩ K R j − ] < ǫ j R α j for R > R j . It is possible because of property rC1) Theorem 5.2.1.1. We can also increase R j so that(5.2.2.9) α j − mes[ C j ∩ K R ] < ǫ j R α j and(5.2.2.10) α j − mes[ C j +1 ∩ K R ] < ǫ j R α j for R > R j . It is possible because C j and C j +1 are C - sets.Let us estimate α j − mes[ C ∩ K R ] for R j ≤ R < R j +1 . From (5.2.2.8),(5.2.2.9)and (5.2.2.10) we obtain(5.2.2.11) α j − mes[ C ∩ K R ] ≤ ǫ j R α j . Let α > α j < α. For R j +1 ≥ R > R j we have α − mes[ C ∩ K R ] R − α ≤ α j − mes[ C ∩ K R ] R − α j ≤ ǫ j . Hence α − mes C = 0 . (cid:3) Proof of Theorem 5.2.1.2.
Let φ ∈ D ( C ) and supp φ ⊂ K R . Then for any ǫ > J ( t ) := Z φ ( z ) u t ( z ) dz = Z K R \ K ǫ + Z K ǫ φ ( z ) u t ( z ) dz := J ( t ) + J ( t )We have for J (see 2.8.2.3):(5.2.3.2) | J | ( t ) ≤ max | z |≤ ǫ | φ ( z ) | × const. ǫ Z T ( r, | u t | ) rdr ≤ const.T ( ǫ, | u t | ) ǫ . Further (see Theorem 2.8.2.1) T ( r, | u t | ) ≤ T ( r, u t ) + O ( t − ρ ( t ) ) ≤ T ( r, u ,t ) + T ( r, u ,t )] + O ( t − ρ ( t ) ) ≤ (5.2.3.3) ≤ M ( r, u ,t ) + M ( r, u ,t )] + O ( t − ρ ( t ) Using (5.2.3.2), (5.2.3.3) and (3.1.2.3) we obtain(5.2.3.4) lim sup t →∞ | J ( t ) | ≤ const.ǫ ρ +2 To estimate J ( t ) write(5.2.3.5) | J ( t ) | ≤ const. Z ˜ K t \ C ,Rt | u ( z ) | dz + Z C ,Rt | u ( z ) | dz t − ρ ( t ) − := J , ( t ) + J , ( t ) , where ˜ K t := { z : ǫt ≤ | z | ≤ Rt } . Exercise 5.2.3.1.
Check this using the change of variable z = tζ. The summand J , is o (1) as t → ∞ by (5.2.1.3). Exercise 5.2.3.2.
Check this using the properties of the proximate order(Theorem 2.8.1.3, ppo3).The summand J , , = o (1) by Theorem 5.2.2.1. Thuslim sup t →∞ | J ( t ) | ≤ const.ǫ ρ +2 for any ǫ. Hence it is equal to zero and the sufficiency of (5.2.1.3) has been proved.Let us prove sufficiency of (5.2.1.2). Let ǫ j ↓ . By Theorem 5.2.2.2 we choosea C -set C j outside which | u ( z ) || z | − ρ ( | z | ) < ǫ j . We construct the set C by 5.2.2.7. Outside C we have (5.2.1.3). And byTheorem 5.2.2.3 it is a C -set. (cid:3) One of the widely applied methods of constructing entire functions witha prescribed asymptotic behavior is the following: First construct a subharmonicfunction behaving asymptotically as he logarithm of modulus of the entire func-tion,and then approximate it in some sense by the logarithm of modulus of entirefunction such that the asymptotic is preserved.Various requests a precision of the approximation and on the metric in whichsuch approximation was implemented generated a specter of theorems of such kindwe will demonstrate.Historically the first theorems of such kind were proved for concrete functionsthe masses of which were concentrated on sufficiently smooth curves (in particularly,on lines, see ,e.g. [BM],[Ev],[Ki],[Ar], ...)In such cases the approximation was very precise and exceptional sets wherethe approximation failed were small and determined.The first general case was proved in [Az(1969)]. Next it was developed in[Yu(1982)], and made excellent in [Yu(1985)]. It is the following
Theorem 5.3.1.1.(Yulmukhametov).
Let u ∈ SH ( ρ ) . Then there exists anentire function f such that for every α ≥ ρ | u ( z ) − log | f ( z ) || < C α log | z | for z / ∈ E α , where E α is an exceptional set that can be covered by discs D z j ,r j := { z : | z − z j | < r j } satisfying the condition X | z j | >R r j = o ( R ρ − α ) , R → ∞ . This theorem is precise in the following sense: If || z | − log | f ( z ) || = o (log | z | ) , z → ∞ , z / ∈ E, then for every covering of E by discs D z j ,r j and every ǫ > X | z j |
Let u be a subharmonic func-tion in C with µ u satisfying the conditions: µ u ( C ) = ∞ and there exists α > , q > , R > such that µ u ( D ,qR \ D ,R ) > α for all R > R . Then there exists an entire function f such that for every ǫ > | u ( z ) − log | f ( z ) || < C ǫ for z ∈ C \ E ǫ with ∆( E ǫ ) < ǫ So if µ u has no “Hadamard’s gaps” such approximation is possible.In this book we restrict ourself to weaker and simply proved theorem that issufficient for our aim Theorem 5.3.1.4. (Approximation Theorem).
For every u ∈ SH ( ρ, ρ ( r )) there exists an entire function f such that D ′ − lim t →∞ ( u − log | f | ) t = 0 . Nevertheless this theorem has an important
Corollary 5.3.1.5.
For every u ∈ SH ( ρ, ρ ( r )) there exists an entire function f such that Fr [ u ] = Fr [ f ] . Now we prove Theorem 5.3.1.4. We can suppose, because of Theorem 3.1.6.1(Dependence Fr on V • ), that in the definition of ( • ) t (see 3.1.2.1) V t ≡ I We prove this theorem for the case non-integer ρ. For proving this theorem weneed
Lemma 5.3.2.1.
Let u ∈ δSH ( ρ, ρ ( r )) , f or ρ non-integer, and ν is its charge.Then u t → iff ν t → in D ′ as t → ∞ . Proof.
Sufficiency. Suppose u t := ( u ) t − ( u ) t . There exists a subsequence t j → ∞ and subharmonic functions v and v such that(5.3.2.1) u t j = ( u ) t j − ( u ) t j → v − v := v = 0Applying to (5.3.2.1) the continuity of ∆ in D ′ and using the conditions of thetheorem, we obtain ν t j → π ∆ v = 0 . Hence v is harmonic. Since v , v ∈ U [ ρ, ] also v ∈ U [ ρ ] (see Th. 2.8.2.1, t3),t4)and Th.2.8.2.3). Exercise 5.3.2.1.
Prove this in details.By Th.3.1.4.3 we obtain v = 0 . Contradiction.Necessity. Since the Laplace operator is continuous in D ′ -topology, the asser-tion u t → ν t := π ∆ u t → . (cid:3) Now we describe a construction of the zero distribution of the future entirefunction.Let u ∈ SH ( ρ ) and µ be its mass distribution. Set(5.3.2.2) R j +1 := R j ( j + 1) /κ where κ := min( ρ − [ ρ ] , [ ρ ] + 1 − ρ ) . Let us divide all the plane by circles of the form S R j := {| z | = R j } such that R j +1 /R j → ∞ and µ ( S R j ) = 0 . Exercise 5.3.2.2.
Prove that it is possible.Chose a sequence δ j ↓ . Divide every annulus K j := { z : R j ≤ | z | < R j +1 } bycircles S R j,n for R j,n := (cid:18) δ j − δ j (cid:19) n R j , n = 0 , , , ..., n j , where n j := log R j +1 R j log δ j − δ j , and by rays L k := z : arg z = kδ j , k = 0 , , ..., [2 π/δ j ] . They divide all the plane into sectors K j,n,k . We can choose δ j in such way that µ ( ∂K j,n,k ) = 0 because µ ( K j,n,k ) is monotonic function of δ j and have only count-able set of jumps. Exercise 5.3.2.3.
Explain this in details.Chose a point z j,n,k in every sector K j,n,k and concentrate all the mass of thesector at this point. In other words we consider a new mass distribution ˆ µ that hasmasses concentrated in the points z j,n,k and ˆ µ ( z j,n,k ) = µ ( K j,n,k ) . The next lemma shows that ˆ µ is close to µ. Lemma 5.3.2.3.
One has ˆ µ t − µ t → in D ′ as t → ∞ . Proof.
Assume the contrary, i.e., ˆ µ t − µ t . Chose a sequence t l → ∞ such thatˆ µ t l → ˆ ν and µ t l → ν, ν, ˆ ν ∈ M [ ρ ] , ν = ˆ ν. Then there exists a disc K z ,r := { z : | z − z | < r } such that ν ( K z ,r ) = ˆ ν ( K z ,r ) . We can assume that this disc doesnot contain zero since for all the ν ∈ M [ ρ ] the condition ν ( K r ) ≤ ∆ r ρ , ∀ r > ν ( K z ,r ) > ˆ ν ( K z ,r ) . Set a := ν ( K z ,r ) − ˆ ν ( K z ,r ) > . Chose ǫ such that(5.3.2.4) ν ( K z ,r ) < ν ( K z ,r − ǫ ) + a/ . This is possible because the countable additivity of ˆ ν implies lim r ′ ↑ r ν ( K z ,r ′ ) = ν ( K z ,r ) . Consider now the sets t l K z ,r , t l K z ,r − ǫ . For sufficiently large t l they arecontained in the union of the annuluses K j l ∪ K j l +1 . As j l → ∞ the diameters of all the sectors K j l ,n,k are o ( R j l ) uniformly. Thusthey are o ( t l ) . Hence for such t l ’s we can find a union Γ l of sectors covering t l K z ,r − ǫ that does not intersect the circle of t l K z ,r . We have ˆ µ (Γ l ) = µ (Γ l ) by definition of ˆ µ. Using the monotonicity of measures,we obtain µ ( t l K z ,r − ǫ ) ≤ ˆ µ ( t l K z ,r ) whence µ t l ( K z ,r − ǫ ) ≤ ˆ µ t l ( K z ,r ) . Passing to limit as l → ∞ and using Theorems 2.2.3.1 and 2.3.4.4, we obtain ν ( K z ,r − ǫ ) ≤ ˆ ν ( K z ,r ) . Using (5.3.2.4), we obtain ν ( K z ,r ) − / ν ( K z ,r ) − ˆ ν ( K z ,r )] ≤ ˆ ν ( K z ,r ) and hence ν ( K z ,r ) ≤ ˆ ν ( K z ,r ) , that contradicts (5.3.2.3).Since ν and ˆ ν are symmetric in this reasoning the lemma is proved. (cid:3) Let us finish the proof of the Theorem 5.3.1.4 for non-integer ρ. We construct a distribution n with integer masses concentrated at points z j,k,n . Set n ( z j,k,n ) := [ˆ µ ( z j,k,n )]and estimate the growth of the difference δµ := ˆ µ − n that is also a mass distribution concentrated at the same points.Since δµ ( z j,k,n ) ≤ K R . The number of points in the annulus { R j ≤ | z | < R } is found from (5.3.2.2) δµ ( { R j ≤ | z | < R } ) ≤ (cid:20) log (cid:18) δ j − δ j (cid:19)(cid:21) − πδ j log RR j ≤ const × log( j + 1) δ j = const × ( j + 1) log( j + 1) . The mass of the disc K R is estimated by the inequality(5.3.2.5) δµ ( K R ) ≤ const × n − X k =0 ( k + 1) log( k + 1) = o ( n ) = o ( R ǫ ) for any ǫ > R > R n − = (( n − /κ . Exercise 5.3.2.4.
Check this in details.The estimate (5.3.2.5) shows that(5.3.2.6) δµ t → t → ∞ .Lemma 5.3.2.2 and (5.3.2.6) implies that(5.3.2.7) µ t − n t → u ( z ) := Π( z, n, p )(see (2.9.2.1)) where Π is a canonical potential. This is a subharmonic functionin the plane with integral masses. Thus it is the logarithm of the modulus of theentire function f ( z ) = Y E ( z/z j,k,n ) . (5.3.2.7) implies by Lemma 5.3.2.1 that u t − ( u ) t → ρ. (cid:3) ′ dberg.Description of lower indicator.Description of the pair:indicator-lower indicator5.4.1. Now we consider the lower indicator . For an entire function of finite order ρ and normal type it can be defined in one of the following ways:(5.4.1.1) h ( φ, f ) := sup C ∈C { lim inf re iφ →∞ ,re iφ / ∈ C log | f ( re iφ ) | r − ρ ( r ) } , where C is the set of C -sets (see [L(1980,Ch.II, § K δ j ( z j ) := { z : | z − z j | < δ j } such thatlim R →∞ R X | z j | Check this.We are going to prove Theorem 5.4.1.1. Let g ( φ ) be a π -periodic function that is either semicontin-uous from above or ≡ −∞ and ρ ( r ) → ρ be an arbitrary approximate order.Thenthere exists an entire function f ∈ A ( ρ, ρ ( r )) such that (5.4.1.4) h ( φ, f ) = g ( φ ) for all φ ∈ [0 , π ) . We will use the following assertion that is a corollary of Theorem 4.3.5.1and Corollary 5.3.1.5: Theorem 5.4.2.1. Let Λ ⊂ U [ ρ ] be a compact, connected and T • – invariant subset.Then for any proximate order ρ ( r ) → ρ there exists f ∈ A ( ρ, ρ ( r )) such that (5.4.2.1) h ( φ, f ) = sup { v ( e iφ ) : v ∈ Λ } (5.4.2.2) h ( φ, f ) = inf { v ( e iφ ) : v ∈ Λ } . Exercise 5.4.2.1. Prove Theorem 5.4.2.1.For the sake of clarity let us restrict ourselves to non-integer ρ . We will con-struct a set Λ such that inf { v ( e iφ ) : v ∈ Λ } = g ( φ )Denote H ( z, p ) := log | − z | + ℜ p X k =1 z k k ; p = [ ρ ] γ ( z, K, λ ) := − λ + K | z − | , λ, K ≥ . Note the following properties of these functions:a) min | z − |≥ δ δH ( z, p ) | z | − ρ → 0, as δ → δH ( z, p ) | z | − ρ ≤ Aδ, for all z ∈ C , where A depends only on p ; c)(5.4.2.3) max | z − |≤ . K γ ( z, K, λ ) ≤ − Exercise 5.4.2.2 Prove properties a),b),c).Let us note that H (1 , p ) = −∞ . Consider the family:Λ ∞ = { v θ,τ ( z ) := H ( ze − iθ τ, p ) τ − ρ : θ ∈ [0 , π ) , τ ∈ (0 , ∞ ) } ∪ U [ ρ ] because of b) and closed in D ’-topology. It isalso T • – invariant, hence, satisfies the conditions of Theorem 5.4.2.1. For every φ ∈ [0 , π ) there exists θ (= φ ), and τ (= 1) such that v θ ,τ ( e iφ ) = H (1 , p ) = −∞ Hence(5.4.2.4) inf { v ( e iφ ) : v ∈ Λ ∞ } = −∞ For general case this construction will be improved, cutting the “trunk” of thefunction H ( ze − iθ , p ).Take δ small enough so that the following conditions hold(5.4.2.5) δH ( z, p ) | z | − ρ ≥ − , for | z − | ≥ δ (5.4.2.6) δH ( z, p ) ≥ − , for | z − | = δ, (5.4.2.7) δ ≤ K Then(5.4.2.8) δH ( z, p ) > γ ( z, K, λ ) , for | z − | = δ. Denote(5.4.2.9) W ( z, K, δ, λ ) := (cid:26) max { δH ( z, p ) , γ ( z, K, λ ) } , for | z − | < δδH ( z, p ) , for | z − | ≥ δ Lemma 5.4.2.2. The following holdsaw) The function W ( z, K, δ, λ ) is subharmonic in C ; bw)supp µ W ⋐ {| z − | < δ } ; cw) (5.4.2.10) sup z ∈ C W ( z, • , δ, λ ) | z | − ρ ≤ Aδ, where A depends only on p. Proof of Lemma. For | z − | < δ W is subharmonic as the maximum of two sub-harmonic functions. For | z − | ≥ δ it is harmonic even in the neighborhood ofthe circle | z − | = δ, because of inequality (5.4.2.8). So aw) and bw) hold. Theassertion cw) follows from b) and c) (5.4.2.3) above.Now we get to the proof of (5.4.1.4). Let g n ↓ g a sequence of continuouslydifferentiable functions that converges to g monotonically. This is possible, because g is semicontinuous from above. Exercise 5.4.2.3 Prove that Theorem 2.1.2.9 and the Weierstrass theorem ofapproximation of every periodic function by trigonometrical polioses imply the lastassertion.Denote M n := max φ g + n ( φ )where as usual a + = max( a, 0) .Set v θ,n ( z ) := W ( ze − iθ , K n , δ n , M n + 1 − g n ( θ )) + ( M n + 1) | z | ρ , where δ n is chosen small and K n is chosen large. Set z = τ e iφ . It is clear that(5.4.2.11) v φ,n ( e iφ ) = g n ( φ )for all K n , δ n . We can choose K n so large and δ n so small that γ ( z, K n , M n + 1 − g ( θ )) | z | − ρ ≥ g n ( φ ) for | z − | ≤ δ n , because g n has bounded derivative.After that we can make δ n smaller so that for | z − | ≥ δ n the inequality(5.4.2.8) would hold. Exercise 5.4.2.2. Estimate exactly K n and δ n via the derivative of g n . Then v θ,n ( z ) | z | − ρ ≥ g n ( φ )for all z = re iφ . Thus min θ,τ v θ,n ( τ e iφ ) τ − ρ = g n ( e iφ ) , and the minimum is attained for τ = 1 , θ = φ. Let us note that from (5.4.2.10) we havesup θ sup z ∈ C v θ,n ( z ) | z | − ρ ≤ Aδ n + M n + 1 ≤ A + M + 1 . Consider now the family of functionsΛ := { v θ,n ( zτ ) | τ | − ρ : θ ∈ [0; 2 π ) , n = 1 , , ..., τ ∈ (0; ∞ ) } It is contained in U [ ρ, σ ] for σ = A + M + 1 and is T • -invariant. Let Λ be itsclosure in D ′ . Let us show that(5.4.2.12) g ( φ ) = inf { v ( e iφ ) : v ∈ Λ } . Indeed, for every sequence v j ∈ Λ v j ( e iφ ) ≥ inf n g n ( φ ) = g ( φ )Let v ∈ Λ . By Theorem 2.7.4.1 ( D ′ and Quasi-everywhere Convergence) v ( z ) := ( D ′ − lim j →∞ v j )( z ) = (lim sup j →∞ v j ) ∗ ( z )Hence v ( e iφ ) ≥ g ( φ ) . However, the infimum is attained for every φ on the sequence v φ,n ( z ) because of(5.4.2.11). Hence (5.4.2.12) holds and Theorem 5.4.1.4 is proved. (cid:3) Now we describe of the pair: indicator-lower indicator. Let h be a 2 π -periodic, ρ -trigonometrically convex function ( ρ -t.c.f) and let g be a 2 π -periodicupper semicontinuous function. Further they are indicator and lower indicator ofentire function, and hence must satisfy the condition(5.4.3.1) h ( φ ) ≥ g ( φ ) , φ ∈ [0 , π ) . An interval ( a, b ) ⊂ [0 , π ) is called a maximal interval of ρ -trigonometricity of thefunction h if(5.4.3.2) h ( φ ) = A cos ρφ + B sin ρφ, φ ∈ ( a, b )for some constants A, B , and h has no such representation on any larger interval( a ′ , b ′ ) ⊃ ( a, b ) . A function h is said to be strictly ρ -t.c.f. if it is a ρ -t.c.f. and is not ρ -trigonometrical on any interval.If the function h is a strictly ρ -t.c.f., then h and g (satisfying other previousbounds) could be an indicator and lower indicator of an entire function f ∈ A ( ρ ( r )) . However this is not so if the function h has an intervals of trigonometricity.Recall, for example, the famous M.Cartwright Theorem [L(1980), Ch.IV, § a, b ) with b − a > π/ρ then the function is a CRG -function on this interval, i.e.,(5.4.3.3) h ( φ ) = g ( φ ) , φ ∈ ( a, b ) . Let us formulate all the necessary conditions of such kind. Let ( a, b ) be a max-imal interval of ρ -trigonometricity of the function h. The M. Cartwright theoremcan be formulated as the implication:(5.4.3.4) ( b − a > π/ρ ) ⇒ (5 . . . . The following implications are also necessary:(5.4.3.5) ( ∃ φ ∈ ( a, b ) : h ( φ ) = g ( φ )) ⇒ (5 . . . (5.4.3.6a) ( h ( a ) = g ( a ) ∧ h ′ + ( a ) = h ′− ( a )) ⇒ (5 . . . h ( b ) = g ( b ) ∧ h ′ + ( b ) = h ′− ( b )) ⇒ (5 . . . h ′± ( a ) and h ′± ( b ) are the right and left derivatives of the function h at thepoints a and b. (5.4.3.7a) ( b − a = π/ρ ∧ h ′ + ( a ) = h ′− ( a )) ⇒ (5 . . . b − a = π/ρ ∧ h ′ + ( b ) = h ′− ( b ) ⇒ (5 . . . (cid:18) lim inf φ → a +0 h ( φ ) − g ( φ ) φ − a = 0 (cid:19) ⇒ (5 . . . (cid:18) lim inf φ → b − h ( φ ) − g ( φ ) b − φ = 0 (cid:19) ⇒ (5 . . . h and g are said to be concordant if at least one of the following conditions holds:1. h is strictly ρ -t.c.;2.for each ( a, b ) that is a maximal interval of ρ -trigonometricity of the function h the implications (5.4.3.4)-(5.4.3.8b) are satisfied. Theorem 5.4.3.1. Let < ρ < ∞ , h ( φ ) be a π -periodic, ρ -t.c.f., g ( φ ) be anupper semicontinuous, π -periodic function, h ( φ ) ≥ g ( φ ) for all φ, and h g. A function f ∈ A ( ρ ( r )) which simultaneously satisfies the identity h f ≡ h, h f ≡ g with an arbitrary proximate order ρ ( r ) → ρ exists if and only if the functions h and g are concordant. Proof of necessity. Note that implication (5.4.3.4) is a corollary of (5.4.3.6a) or(5.4.3.6b), because every ρ -trigonometrical function is continuous and has continu-ous derivative in ( a, b ) . Recall that ( • ) [ t ] was defined by (3.1.2.4a). From properties of the limit set Fr [ f ] (Theorem 3.1.2.2, fr2),fr3)) and thedefinition of indicators ( (3.1.2.1),(3.1.2.2)) we can obtain for every function v ∈ Fr [ f ] the inequality(5.4.4.1) v ( τ e iφ ) ≤ τ ρ h ( φ ) , φ ∈ [0 , π ) , τ > . Since h ( φ ) is ρ -trigonometrical for φ ∈ ( a, b ), the function H ( re iφ ) := r ρ h ( φ )is harmonic in the angleΓ( a, b ) := { re iφ : φ ∈ ( a, b ) , r ∈ (0 , ∞ ) } , whence the function v − H is subharmonic and nonpositive in Γ( a, b ) . By virtueof the maximum principle, either v < H in Γ( a, b ) or v ≡ H in Γ( a, b ) for each v ∈ Fr [ f ] . Note that the condition v ≡ H in Γ( a, b ) implies v ≡ H in Γ[ a, b ] for theclosed interval because of the upper semicontinuity of v. Let us prove (5.4.3.5).For every v ∈ Fr [ f ] we have v ( re iφ ) − H ( re iφ ) = 0whence by the maximum principle v = H in Γ( a, b ) . Hence (5.4.3.3) holds.Let us prove (5.4.3.6a). Assume the contrary: h ( a ) = g ( a ) ∧ h ′ + ( a ) = h ′− ( a )holds, but there exists φ ∈ ( a, b ) such that h ( φ ) > g ( φ ) . Then there exists v ∈ Fr [ f ] such that g ( φ ) ≤ v ( e iφ ) < h ( φ )whence(5.4.4.2) v ( τ e iφ ) < τ ρ h ( φ ) ∈ Γ( a, b ) . Without loss of generality ,we can assume that v ( z ) > −∞ otherwise we can replace v with max( v, − C ) for a large positive constant C > . We choose 0 < τ < τ and to the every function W j ( re iφ ) := v [ τ j ] ( re iφ + a ) − r ρ h ( φ + a ) , j = 1 , , γ = b − a, re iφ ∈ Γ(0 , γ )we apply the following lemma due to A.E.Eremenko and M.L.Sodin [So](see also[PW],[Ho]): Lemma 5.4.4.1. (E.S.). Let W be a subharmonic nonpositive function inside theangle Γ(0 , γ ) , γ > . Then the following implication is valid lim sup φ → W ( e iφ ) φ = 0 ! ⇒ W ≡ . If the condition of this theorem is not satisfied for W ∗ ( re iφ ) = max τ ∈ [ τ ,τ ] v [ τ ]( re iφ )it would be possible to insert a ρ -t.c.function between h ( φ ) − ǫ ( φ − a ) (for a small ǫ ) and v ( e iφ ) .However, such function does not exist, because of negative jump ofderivative. So it will be a contradiction. See further for details.From Lemma 5.4.4.1 we getlim inf φ → a +0 h ( φ ) − v [ τ ] ( e iφ ) φ − a := α > φ → a +0 h ( φ ) − v [ τ ] ( e iφ ) φ − a := α > . So a ∆ > a + ∆ < b and the inequalities(5.4.4.3) H ( τ j e iφ ) − v [ τ j ] ( e iφ ) > ατ ρj ( φ − a ) , j = 1 , , where α := 1 / α , α ), hold for all φ ∈ [ a, a + ∆].We denote β := min τ ∈ [ τ ,τ ] ( H ( τ e i ( a +∆) ) − v ( τ e i ( a +∆) ))which is positive because of (5.4.4.2).Let us choose ǫ > ǫ < min( α, β ( τ ) − ρ ∆ − )and let us consider the ρ -trigonometrical function h ǫ ( φ ) := ρ − ( h ′ ( a ) − ǫ ) sin ρ ( φ − a ) + h ( a ) cos ρ ( φ − a ) , φ ∈ ( a, b ) that coincides with h ( φ ) = ρ − h ′ ( a ) sin ρ ( φ − a ) + h ( a ) cos ρ ( φ − a ) , φ ∈ ( a, b )in the point φ = a but has a tangent that is lower then the tangent of h. Further(5.4.4.5) h ( φ ) − h ǫ ( φ ) = ρ − ǫ sin ρ ( φ − a ) ≤ ǫ ( φ − a ) , φ ∈ [ a, a + ∆]Combining (5.4.4.3)-(5.4.4.5) we obtain(5.4.4.6) v [ τ j ] ( e iφ ) < τ j ρ h ( φ ) − α ( φ − a ) < τ jρ h ( φ ) − ǫ ( φ − a ) ≤ τ jρ h ǫ ( φ ) , φ ∈ [ a, a + ∆] , j = 1 , v ( τ e iφ ) ≤ τ ρ h ǫ ( a + ∆) + τ ρ ǫ ∆ − β <τ ρ h ǫ ( a + ∆) , τ ∈ [ τ , τ ]We denote(5.4.4.8) G := { re iφ : φ ∈ [ a, a + ∆] , τ ∈ [ τ , τ ] } It follows from (5.4.4.6),(5.4.4.7) that v ( re iφ ) < r ρ h ǫ ( φ ) , re iφ ∈ ∂G, where ∂G is the boundary of the domain G . Since the functions v ( re iφ ) and r ρ h ǫ ( φ )are subharmonic in G , by virtue of the maximum principle we have(5.4.4.9) v ( re iφ ) < r ρ h ǫ ( φ ) , re iφ ∈ G. Let us consider the function H ( re iφ ) := r ρ h ( φ ) , re iφ ∈ Γ( a − ∆ , a + ∆)where h ( φ ) := (cid:26) h ( φ ) , φ ∈ ( a − ∆ , a ] ,h ǫ ( φ ) , φ ∈ [ a, a + ∆) . The function H is continuous in Γ( a − ∆ , a + ∆) and subharmonic in the anglesΓ( a − ∆ , a ) and Γ( a, a +∆) . Let us prove that it is subharmonic at the point z = e ia . Let M ( z, R, v ) be the mean value of v over the circle { ζ : | ζ − z | = R } (see (2.6.1.1)).Taking into consideration (5.4.4.9) and subharmonicity of v (see (2.6.1.1)), for allsmall R we have M ( e ia , R, H ) ≥ M ( e ia , R, v ) ≥ v ( e ia ) = H ( e ia ) . Hence H is subharmonic for z = e ia . Since H is homogeneous, i.e. H ( kz ) = k ρ H ( z ), M ( ke ia , kR, H ) = k ρ M ( e ia , R, H ) ≥ k ρ H ( e ia ) = H ( ke ia )So H is subharmonic on the ray { z = ke ia : k ∈ (0 , ∞ ) } and hence in the angleΓ( a − ∆ , a + ∆) . Thus h ( φ ) is a ρ -t.c.f. for φ ∈ ( a − ∆ , a + ∆) . However, byconstruction ( h ) ′− ( a ) = h ′− ( a ) = h ′ + ( a ) = ( h ǫ ) ′ + + ǫ = ( h ) ′ ( a ) + ǫ and this contradicts the fact that h is ρ -t.c.f.Concordance of the implication (5.4.3.6a) is proved. Here we continue proof of necessity. Pass to the proof of necessity of thecondition (5.4.3.7a). Assume the contrary. Then there exists v ∈ Fr [ f ] and φ ∈ [ a, b ] such that g ( φ ) ≤ v ( e iφ ) < h ( φ ), whence by virtue of the maximum principle, v ( τ e iφ ) < τ ρ h ( φ ) for τ e iφ ∈ Γ( a, b ) . Actually v ( τ e iφ ) ≤ τ ρ h ( φ ) everywhere andon the circle we have strict inequality. If v ( τ e ia ) = H ( τ e ia ) for a τ > 0, then v [ τ ] ( e ia ) = h ( a ), and it will suffice to repeat the arguments used in proving (5.4.3.6a)with v [ τ ] instead of v. Exercise 5.4.5.1 Do that.So it is sufficient to examine the case v ( τ e ia ) < H ( τ e ia ) , τ > . Denote T ( φ ) := h ′ ( a ) ρ − sin ρ ( φ − a ) + h ( a ) cos ρ ( φ − a ) . This is a ρ -trigonometrical function the graph of which is tangent to the graph of h ( φ ) at the point a. There are two possibilities for T ( φ ) on some small interval φ ∈ ( a − γ, a ) , γ > T ( φ ) < h ( φ ) or T ( φ ) = h ( φ ) . Inequality T ( φ ) > h ( φ ) contradicts to ρ -t.convexity at the point a. The equal-ity on the sequence of points φ j → a − ρ -t.c.functions. Exercise 5.4.5.2 Why is it?If T ( φ ) = h ( φ ) , φ ∈ ( a − γ, a ) , then h is ρ -trigonometrical on the interval( a − γ, b ) ⊃ ( a, b ) that was already considered in the case (5.4.3.4) (M.Cartwright’sTheorem).So we assume T ( φ ) < h ( φ ) , φ ∈ ( a − γ, a ) . We set h ( φ ) := h ( φ ) − T ( φ ) , φ ∈ ( a − γ, a ) v ( re iφ ) := v ( re iφ ) − r ρ T ( φ ) , re iφ ∈ Γ( a − γ, b ) . Then h ( φ ) = 0 for φ ∈ [ a, b ], h ( φ ) > φ ∈ ( a − γ, a ) and h ′ ( a ) = 0 . The function v ( e iφ ) < , φ ∈ [ a, b ) . Let us analyze the behavior of the function v ( e iφ ) at the point b. Either v ( e ib ) < v ( e ib ) = 0 butlim sup φ → b − v ( e iφ )( b − φ ) − ≤ − C for some C > v ( e iφ ) is strictly negative also in some left (say, ( a − ∆ , a ))neighborhood of a because of upper semicontinuity. In any case v ( e iφ ) can bemajorated on the interval ( a − ∆ , b ) by the function h ǫ := − A sin( ρ − ǫ )( b − φ )with sufficiently small A. A point of intersection of the graph of h ǫ with the axis 0 , φ can be regulatedby ǫ and can be chosen so close to the point a that the graph of h ǫ also intersectthe graph of h ( φ ), at some point θ < a because h ( a ) = h ′ ( a ) = 0 . Exercise 5.4.4.1. Make the precise proof with all the estimates.Let the parameters A, ǫ, θ be fixed as above. Denote S := { re iφ : φ ∈ ( θ , b ) , < r < } . Then H ǫ ( re iφ ) := r ρ − ǫ h ǫ ( φ ) is harmonic in the sector S and satisfies the inequality H ǫ ( re iφ ) ≥ v ( re iφ ) on ∂S. Hence H ǫ ( re iφ ) ≥ v ( re iφ ) on S. Thus(5.4.5.1) v ( re iφ ) ≤ H ( re iφ ) + H ǫ ( re iφ ) , re iφ ∈ S. Let M ( r, v ) be the mean value of the function on the circle { ζ : | ζ | = r } (see2.6.1.1). Using (5.4.5.1) we have M ( r, v ) ≤ b Z θ [ H ( re iφ )+ H ǫ ( re iφ )] dφ + Z [0 , π ) \ ( θ ,b ) H ( re iφ ) dφ ≤ d r ρ − d r ρ − ǫ , d , d > M ( r, v ) < v (0) for sufficiently small r > v at zero. Now we complete proof of necessity, proving (5.4.3.8a,b). Assume the con-trary:suppose(5.4.6.1) lim inf φ → a +0 h ( φ ) − g ( φ ) φ − a = 0but there exists a φ ∈ ( a, b ) such that h ( φ ) > g ( φ ) . Then there exists a function v ∈ Fr [ f ] such that(5.4.6.2) g ( φ ) ≤ v ( e iφ ) < h ( φ )Then the function v ( re iφ ) := v ( re iφ ) − H ( re iφ ) is subharmonic and nonpositive inΓ( a, b ) . By virtue of the maximum principle v ( re iφ ) < , re iφ ∈ Γ( a, b ) . From (5.4.6.1) we obtain0 = lim inf φ → a +0 h ( φ ) − g ( φ ) φ − a ≥ lim inf φ → a +0 h ( φ ) − v ( e iφ ) φ − a = − lim inf φ → a +0 v ( e iφ ) φ − a whence, recollecting that v ( e iφ ) < , we getlim sup φ → a +0 v ( e iφ ) φ − a = 0 . Applying Lemma 5.4.4.1 (E.S.) to the function W ( re iφ ) = v ( re iφ + a ) , re iφ ∈ Γ(0 , γ ) , γ = b − a we get v ≡ a, b ) which leads to a contradiction. The implication (5.4.3.8b)is proved in the same way. So the proof of necessity in Theorem 5.4.3.1 is com-pleted. (cid:3) We do not include here the proof of sufficiency and refer the readers to theoriginal paper [Po(1992)]. Suppose some class of entire functions is determined by asymptotic behaviorof their zeros, and we want to know what is the restriction on asymptotic behaviorof functions: for example, to estimate indicator of such function. The first exampleof such problem was considered by B.Ya. Levin in [L(1980),Ch.IV, § ′ dberg[Go(1962)] and his pupils [Kon(1967)] , [KF(1972)].We consider this theory fromthe point of view of limit sets.Let M ⋐ M ( ρ ) (see (3.1.3.4)) be a convex set of measures which is closed in D ′ and is invariant with respect to the transformation ( • ) t (see (3.1.3.1),(3.1.3.2))) andlet A ( M ) be a class of entire functions f for which Fr [ n f ] ⊂ M . We suppose ρ isnon-integer. Recall that canonical potential Π( z, ν, p ) is defined by: (see (2.9.2.1))Π( z, ν, p ) := Z C G p ( z/ζ ) ν ( dζ ) , where ν is a measure and G p ( z ) := log | − z | + ℜ p X k =1 z k k . Theorem 5.5.1.1. [AP(1984)] The relation (5.5.1.1) h ( φ, f ) = sup { Π( e iφ , ν, p ) : ν ∈ M} is valid. There exists f ∈ A ( M ) for which the equality holds in (5.5.1.1) for all φ. Proof. We should only prove that there exists an entire function with such indicator.Consider the set Λ := { Π( e iφ , ν, p ) : ν ∈ M} It is a convex set contained in U [ ρ ] . Thus there exists a subharmonic (see Corollary4.1.4.2) and hence entire (see Corollary 5.3.1.5) function f such that Fr [ f ] = Λ . ByTheorem 5.4.2.1 (5.5.1.1) holds . (cid:3) For some M it is possible to compute the supremum in (5.5.1.1) and thus toobtain explicit precise estimates of indicators in the respective class A ( M ) . As anexample, we shall present an estimate given by A.A.Gol ′ dberg.We recall that the upper density of zeros of an entire function f ∈ A ( ρ ) isdefined by the equality ∆[ n f ] := lim sup r →∞ n f ( r ) r ρ where n f is the distribution of zeros of the function f, and denote(5.5.1.2) K ( t, φ ) := − [ ddt G + p ( e iφ /t )] − where a + := max( a, , a − := min( a, . This function is piece-wise continuous. Corollary 5.5.1.2. [Go(1962)] Let the distribution of zeros n f of a function f beconcentrated on the positive ray, and let ∆[ n f ] ≤ ∆ < ∞ . Then (5.5.1.3) h ( φ, f ) ≤ ∆ ∞ Z t ρ K ( t, φ ) dt, φ ∈ [0 , π ) and there exists a function from the same class for which equality is attained for all φ. Proof of Corollary. We exploit Theorem 5.5.1.1. The class of the functions f sat-isfying the assumption of the Corollary coincides with the class of f for which(5.5.1.4) Fr [ n f ] ⊂ M = { ν ∈ M ( ρ ) : supp ν ⊂ [0 , ∞ ] ∧ ν ( r ) ≤ ∆ r ρ } . C Exercise 5.5.1.1 Show this by using Corollary 3.3.2.6.Thus Π( e iφ , ν, p ) = ∞ Z G p ( e iφ /t ) ν ( dt ) ≤ ∞ Z G + p ( e iφ /t ) ν ( dt )Integrating by parts we obtainΠ( e iφ , ν, p ) ≤ − ∞ Z ν ( t )[ ddt G + p ( e iφ /t )] − dt By (5.5.1.4) we get (5.5.1.3). (cid:3) Denote M p ( r ) := max { G p ( re iφ ) : φ ∈ [0 , π ) } In the same way one can prove Corollary 5.5.1.3. [Go(1962),Th.4.1] .Let distribution of zeros of the function f ∈ A ( ρ ) satisfy only the condition ∆[ n f ] ≤ ∆ < ∞ . Then (5.5.1.5) h ( φ, f ) ≤ ∆ ρ ∞ Z t ρ − M p (1 /t ) dt, φ ∈ [0 , π ) and there exists a function from the same class for which equality is attained for all φ. Exercise 5.5.1.2. Prove this Corollary exploiting M := { ν ∈ M ( ρ ) : ν ( r ) ≤ ∆ r ρ , ∀ r > } . To be able to obtain explicit estimates for more diverse classes of entirefunctions defined by a restriction on the density of zeros, Gol ′ dberg introduced anintegral with respect to a non-additive measure and obtained estimates for indica-tors in terms of one-dimensional integral (along a circumference) with respect tosuch a measure ([Go(1962)].Gol ′ dberg initially constructed integral sum of a specialform.The construction presented here is based on the Levin–Matsaev–Ostrovskiitheorem (see (see [Go(1962),Th.2.10]). Fainberg (1983) developed this approachusing a two-dimensional integral. This made it possible to extend significantly theset of classes of entire functions for which the estimate expressed by a nonadditiveintegral is precise. We shall present these results after the necessary definitions.Let δ ( X ) be a non-negative monotonic function of X ⊂ C , the function beingfinite on bounded sets and δ ( ∅ ) = 0 . For a given family of sets X := { X } we denoteby N ( δ, X ) the class of countable-additive measures µ defined by the relation N ( δ, X ) := { µ : µ ( X ) ≤ δ ( X ) , X ∈ X } . For a Borel function f ≥ X ) Z f dδ := sup (cid:26)Z f dµ : µ ∈ N ( δ, X ) (cid:27) , called an ( X ) − integral with respect to a nonnegative measure δ. For a Borel set E ⊂ C we set ( X ) Z E f dδ := ( X ) Z f I E dδ, where I E is an indicator of the set E, i.e., I E ( z ) := (cid:26) , if z ∈ E ;0 if z / ∈ E. This integral possesses a number of natural properties: it is monotonic with respectto f and δ and the family X ,positively homogeneous and semi-additive with respectto the function f and δ. If δ is a measure, if X is a Borel ring, and if f is a measurablefunction, then ( X )-integral coincides with the Lebesgue -Stieltjes integral. Exercise 5.5.2.1. Check these properties.Let δ (Θ) be a nonadditive measure on the unit circle T , defined initially onthe family of all open sets Θ ⊂ T . It can be naturally extended to all closed setsΘ F using the equality δ (Θ F ) := inf { δ (Θ) : Θ ⊃ Θ F } . Let χ Θ be a set of open sets containing the set T . Denote D r, Θ := { z = te iθ : 0 < t < r, e iθ ∈ Θ } , χ z := { D r, Θ : r > , Θ ∈ χ Θ } The subscripts Θ and z at χ indicate that the families under consideration arelocated either on T or on the plane, respectively.Let us define a non-additive measure δ z on χ z by the equalities δ z ( D r, Θ ) := r ρ δ (Θ) , D r, Θ ∈ χ z . Now the integral ( χ z ) R G + p ( e iθ /ζ ) dδ z is defined. Recall that the classical angular upper density of zeros of an entire function f ∈ A ( ρ ) is defined by the equality (compare (3.3.2.7))∆ cl [ n f , Θ] := lim sup r →∞ n f ( D r, Θ ) r − ρ . Consider the class of entire functions A cl ( δ, χ Θ ) defined by the equality(5.5.2.6a) A cl ( δ, χ Θ ) := { f : ∆ cl [ n f , Θ] ≤ δ ( θ ) , ∀ Θ ∈ χ Θ } for a given non-additive measure δ (Θ) and a family χ Θ . Theorem 5.5.2.2. [Fa] Let δ (Θ) satisfy the condition (5.5.2.6) δ (Θ) = δ (Θ) , ∀ Θ ∈ χ Θ (the dash means the closure of a set). Then (5.5.2.7) h ( φ, f ) ≤ ( χ z ) Z G + p ( e iθ /ζ ) dδ z There exists a function f ∈ A cl ( δ, χ Θ ) such that equality in (5.5.2.7) is attained forall φ ∈ [0 , π ) simultaneously.Proof. Let us note the following: If we replace in this theorem∆ cl [ n f , Θ] with its D ′ counterpart ∆( Co Θ ( I )) (see Theorem 3.3.1.2) and con-sider the corresponding class of entire function A ( δ, χ Θ ) the assertion of the theoremholds without conditions(5.1.5.6). You should only apply Theorem 5.5.1.1 with thecorresponding M . The condition (5.5.2.7) is exploited only for replacing “ D ′ ” quan-tities by the classic ones using results of § (cid:3) Exercise 5.5.2.2 Prove this theorem in details.It is also worth to note that every family χ Θ can be replaced for a family χ ′ Θ that is dense in χ Θ (see 3.2.2) and such that for χ ′ Θ (5.5.2.6) already holds (seeTheorem 3.3.2.3). The most famous definition of a function of completely regular growth (CRG-function) is the following:A function f ∈ A ( ρ ( r )) is a function of completely regular growth , if the limitlim z →∞ r − ρ ( r ) log | f ( z ) | , r := | z | exists when z → ∞ uniformly outside some C -set (see § ′ dberg ([Go(1967)]) this definition was reduced to the following :A function f ∈ A ( ρ ( r )) is a function of completely regular growth , if h f ( φ ) = h f ( φ ) , ∀ φ ∈ [0 , π ) . Because of the formulae (3.2.1.1), (3.2.1.2) (see also § Theorem 5.6.1.1. A function f ∈ A ( ρ ( r )) is a function of completely regulargrowth (CRG -function) iff Fr [ f ] consists of only one subharmonic function h ( z ) . Because of (3.2.1.11) the function h ( z ) has the form(5.6.1.1) h ( z ) = r ρ h ( e iφ )The function h ( φ ) := h ( e iφ ) is ρ -trigonometrically convex and it was studied in §§ The initial definition of regular zero distribution [L(1980),Ch.II, § 1] is thefollowing :Let n be a zero distribution (divisor,or mass distribution) of convergence expo-nent ρ := ρ [ n ] (see § ρ > [ ρ ] . Let ρ ( r ) → ρ be a proper proximateorder of n ( r ) (see Th.2.8.1.2). It means that n ∈ M ( ρ ( r )) , ρ ( r ) → ρ (see § The initial definition of regular zero distribution for ρ being non-integer is:A zero distribution n is regular if the limitlim r →∞ n ( Co ( α,β ) ( I t )) t ρ ( t ) := ∆(( α, β )) exists for all α > β except may be for a countable set on the circle.By using results of § Theorem 5.6.2.1. The zero distribution n is regular iff F r [ n ] consists only onemeasure ν reg . Exercise 5.6.2.1. Prove this exploiting Th’s.3.3.3.1 and 3.3.2.4.Recall that for f ∈ A ( ρ ( r )) , ρ ( r ) → ρ we have n f ∈ M ( ρ ( r )) , ρ ( r ) → ρ. (seeTh.2.9.3.2). Now we can formulate Theorem 5.6.2.2 (Levin-Pfluger). [L(1980),Ch.II,Ch.III] An entire function f ∈ A ( ρ ( r ) , ρ ( r ) → ρ of non-integer order ρ is of completely regular growth functioniff its zero distribution is regular. After Theorems 5.6.1.1, 5.6.2.1 this theorem is a direct corollary of Th.3.1.5.1. Consider now the case of integer ρ. In general, this case is differs from thecase of non-integer ρ. For example, Th. 2.9.4.2 (Brelot-Lindel¨of) implies that( f ∈ A ( ρ ( r )) , ρ ( r ) → ρ ) ⇐⇒ n f ∈ M ( ρ ( r )) , ρ ( r ) → ρ iff the family of polynomials (2.9.4.4a) is compact.To describe the regularity of zero distribution for the case of integer ρ weassume that the limit(5.6.3.1) lim R →∞ δ R ( z, ν, ρ ) := ℜ [ δ ∞ z ρ ]exists, where δ ∞ := lim R →∞ Z | ζ | Proposition 5.6.3.1. The measure ν reg has the following form ν reg ( drdφ ) = ρr ρ dr ⊗ ∆( dφ ) where ∆ is a measure of bounded variation on the unit circle. This assertion is a corollary of invariance of Fr [ n ], Th.3.1.3.3, frm3). In the papers [Bal(1973)][Bal(1976)] functions of completely regular growthalong curves of regular rotation were considered. A curve of regular rotation is acurve that is described by the equation z = te i ( γ ( t ) log t + φ ) , < t < ∞ for a fixed φ. If γ ( t ) ≡ γ then this curve is a logarithmic spiral. In general case γ ( t ) is adifferentiable function such that γ ( t ) → γ, tγ ′ ( t ) → , t → ∞ . To describe this theory in terms of limit sets we consider the transformation P t = te iγ ( t ) log t u t ( z ) = u ( P t z ) t − ρ ( t ) The following Theorem is similar to Th.3.1.2.1 Proposition 5.6.4.1(Existence of spiral Limit Set). The following holdsesls 1) u t ∈ SH ( ρ ( r )) for all t ∈ (0 , ∞ ); esls 2) the family { u t } is precompact at infinity. The set of all limits D ′ − lim j →∞ u t j does not depend on γ ( t ) but only on theconstant γ since lim t →∞ ( γ ( t ) − γ ) log t = lim t →∞ tγ ′ ( t ) = 0So it is the same that for P t = te iγ log t , i.e., the case that was already considered in the general theory.In particular (3.2.1.8) for this case has the form(5.6.5.1) z ( z ) = e i ( − γ log r + φ ) Hence , from Theorem 3.2.1.2 the indicator (see (3.2.1.1)) has the form h ( re iφ ) = r ρ h ( − γ log r + φ ) , z = re iφ , where h ( φ ) is a ρ -trigonometrically convex 2 π -periodic function (see § §§ .Theorem 3.1.6.1 connects limit sets for every γ. Exercise 5.6.1.1. Formulate and prove Balashov’s analogy of the Levin-Pluger Theorem 5.6.2.2 and Theorem 3.1.6.1 for m = 2 . For other generalization of the Levin-Pfluger theory see [AD] A functional F ( u ) acting in the unit circle and defined on subharmonicfunctions u ∈ SH ( ρ ( r )) is called a growth characteristic if the following conditionsare fulfilled:1. continuity :(5.7.1.1) F ( u j ) → F ( u ) , if u j → u uniformly on compacts (of course, for continuous functions u ) or if u j ↓ u ;2. positive homogeneity :(5.7.1.2) F ( cu ) = c F ( r, u );for every constant c > . Here we shall list some widely used functionals that satisfy these conditions:(5.7.1.4) H φ ( u ) := u ( e iφ );(5.7.1.5) T ( u ) = 12 π π Z u + ( e iφ ) dφ ;(5.7.1.6) M α ( u ) := max { u ( e iφ ) : | φ | ≤ α } (5.7.1.7) M ( u ) := M π ( u );(5.7.1.8) I αβ ( u ) := β Z α u ( e iφ ) dφ (5.7.1.9) I ( u, g ) := π Z u ( e iφ ) g ( φ ) dφ, g ∈ L [0 , π ] . Exercise 5.7.1.1. Check properties 1. and 2. for these functionals.Let α ( t ) and α ǫ ( ζ ) be the “hats” defined by the equalities (2.3.1.1)-(2.3.1.3)and let R ǫ u be defined by (2.3.1.4).This averaging has the following properties. Proposition 5.7.1.1. 1. if u is subharmonic, then R ǫ u is subharmonic;2. R ǫ u ↓ u as ǫ ↓ for every subharmonic function;3. if u j → u in D ′ and u j , u are locally summable functions, R ǫ u j → R ǫ u uniformly on every compact set. Exercise 5.7.1.2. Prove this using Th.2.3.4.5, 2.6.2.3.Now we can define the asymptotic characteristics of growth of entire function f ∈ A ( ρ ( r )) :(5.7.1.12) F [ f ] := lim ǫ → lim sup t →∞ F ( R ǫ u t ( • )) , (5.7.1.13) F [ f ] := lim ǫ → lim inf t →∞ F ( R ǫ u t ( • )) , where u = log | f | and ( • ) t is defined by (3.1.2.1). Proposition 5.7.1.2. For F ( u ) defined by (5.7.1.4) F [ f ] = h f ( φ ); F [ f ] = h f ( φ ) . For other functionals from the list (5.7.1.5)-(5.7.1.9) one may replace R ǫ u by u and omit lim ǫ → . Exercise 5.7.1.3. Prove this.The following assertion connects the asymptotic growth characteristics withlimit sets. Theorem 5.7.1.4. The relations F [ f ] = sup {F ( v ) : v ∈ Fr [ f ] } , F [ f ] = inf {F ( v ) : v ∈ Fr [ f ] } are true.Proof. Let v ∈ Fr [ f ] and u t j → v in D ′ . Then R ǫ u t j → R ǫ v uniformly on everycompact set.Hence lim t j →∞ F ( R ǫ u t j ) = F ( R ǫ v ) . Passing to the limit as ǫ → ǫ → lim t j →∞ F ( R ǫ u t j ) = F ( v ) . Choosing a sequence that corresponds to lim sup or lim inf we obtain the assertionof the Theorem. (cid:3) Applying this theorem to the functional (5.7.1.4) we obtain the RHS’s of(3.2.1.1), (3.2.1.2) and hence another definition for the indicator and lower indi-cator. A family of growth characteristics χ A := {F α ( r, • ) : α ∈ A } is called total ifthe equation(5.7.2.1) F α ( v ) = F α ( v ) , ∀ r > , α ∈ A implies v ≡ v for v , v ∈ U [ ρ ] (see 3.1.2.4).Here are some examples of the total families:(5.7.2.2) χ H := { H φ ( u ( e iφ ) : φ ∈ [0 , π ) } ;(5.7.2.3) χ I := { I α,β ( u ) : α, β ∈ [0 , π ) } ;(5.7.2.4) χ F o := { c k ( u ) = I ( u, g k ) : k ∈ Z } ;where(5.7.2.5) g := 1 , g k := cos kφ ; g − k = sin kφ, k ∈ N It is easy to deduce from Th.5.6.1.1 Theorem 5.7.2.1. Let a family {F α ( • ) : α ∈ A } be a total family of characteris-tics. An entire function f is a CRG -function iff (5.7.2.6) F α [ f ] = F α [ f ] Exercise 5.7.2.1. Check this. Let us consider a total family of characteristics of the form(5.7.2.7) χ Ψ := { I ( u, ψ ) : ψ ∈ Ψ } , where Ψ is a set which is complete in L [0 , π ] . For instance, such are the families χ I and χ F o . Theorem 5.7.3.1. [Po(1985] Let f ∈ A ( ρ ( r )) is a CRG-function if and only if atleast one of the following assertions are equivalent:a) F [ f g ] = F [ f ] + F [ g ] , ∀F ∈ χ Ψ ,b) F [ f g ] = F [ f ] + F [ g ] , ∀F ∈ χ Ψ ,for all entire functions g ∈ A ( ρ ( r )) . c) f is a GRG-function. Let us prove c ) = ⇒ a ) and c ) = ⇒ b ) . Using the Theorem 5.7.1.4 we obtain for every characteristics F (5.7.3.1) F [ f g ] = sup {F ( w ) : w ∈ Fr [ f g ] } . Because of Th.3.1.2.4 fru1) Fr [ f g ] ⊂ Fr [ f ] + Fr [ g ]Since f is CRG-function F r [ f ] consists of only one subharmonic function v reg (seeTh 5.6.1.1) and it is easy to check that in this case we have equality Fr [ f g ] = v reg + Fr [ g ] . Exercise 5.7.3.1. Check this.Since F ( v reg + v g ) = F ( v reg ) + F ( v g ) We We obtains F [ f g ] = F ( v reg ) + sup {F ( v g ) : v g ∈ Fr [ g ] } = F [ f ] + F [ g ]So c ) = ⇒ a ) was proved. In the same way one can prove c ) = ⇒ b ) . Exercise 5.7.3.2. Prove this. In the proof of sufficiency of the conditions of this theorem we can supposethat ψ belong to the space D ( T ) of infinitely differentiable functions on the unitcircle T because D ( T ) is complete in L [0 , π ] . We prove now sufficiency of b) inthe Theorem 5.7.3.1.We recall that (see (3.1.2.4a)) v [ t ] ( z ) = v ( tz ) t − ρ , v ∈ U [ ρ ]to distinguish it from ( • ) t that defined by u t ( z ) = u ( tz ) t − ρ ( t ) , u ∈ SH ( ρ ( r ))The main constructive element for the proof of Theorem 5.7.3.1 is Lemma 5.7.3.2. Let ψ ∈ D ( S ) . There exists v ∈ U [ ρ ] with the following proper-ties: (5.7.3.2) D ′ − lim t → v [ t ] = D ′ − lim t →∞ v [ t ] = ˜ v, (5.7.3.3) h v [ t ] ( e i • ) , ψ i > h v ( e i • ) , ψ i for t ∈ (0 , ∞ ) , t = 1 , (5.7.3.4) h ˜ v ( e i • ) , ψ i > h v ( e i • ) , ψ i Proof. Let ψ be represented by Fourier series ψ ( φ ) = a ∞ X n =1 ( a n cos nθ + b n sin nθ )Since ψ a k = 0 or b k = 0 . Suppose there exists a k = 0 . In theproof we will consider three cases: 1. k = 0; 2. k = 0 ∧ k ≤ p ; 3. k ≥ p + 1 . Thenumber ρ is supposed noninteger and p = [ ρ ] . Consider the case a = 0 , a > 0. Set ψ ( x ) := log( − e − α | x | + C ) , α > , C > (5.7.3.4a) v ( z ) := | z | ρ e ψ (log | z | ) = exp( ρ log r + ψ (log r ))Applying the Laplace operator, we obtain:∆ v = 1 r r ∂∂r r ∂∂r v ( r ) = e − x ∂ ∂x e ρx + ψ ( x ) =(5.7.3.5) = exp(( ρ − x + ψ ( x ))[( ρ + ψ ′ ( x )) + ψ ′′ ( x )] , x = log r Since ψ ′ ( x ) = α sgn x exp( − α | x | ) → , ψ ′′ ( x ) = − α sgn x exp( − α | x | ) → x → ±∞ , it is possible to chose α such that the expression (5.7.3.5) be positive.So v ( z ) is subharmonic.It is easy to check that all the assertions of the lemma are satisfied and ˜ v ( z ) = b | z | ρ where b ( > 0) is a constant. Exercise 5.7.3.3. Check this. Continuation of the proof. If a = −| a | < , consider the function v ( z ) := (cid:26) log | z | , | z | ≥ , , | z | < < v t ] ( e i • ) , ψ > = a t − ρ log + t Since the RHS of (5.7.3.6) is minimized for t = e ρ − , the function v ( z ) := v t − ] ( z )satisfies the assertions of the lemma with ˜ v = lim t → , ∞ v [ t ] = 0Now let a = 0 , a k = 0 , < k < p. We will search for a function v of the form(5.7.3.7) v ( re iφ ) := π Z G p ( re i ( φ − θ ) )(1 − sgn a k cos kθ ) dθ This is the convolution G p ( re i • ) ∗ g of the primary kernel (see § G p ( z ) = log | − z | + ℜ p X n =1 z n /n with a positive function g ( θ ) := (1 − sgn a k cos kθ ) on the circle. So it is subharmonic.Recall that the cos -Fourier coefficients of the function G p ( re iθ ) are (see Exercise2.3.7.2)(5.7.3.8) ˆ G p ( m, r ) = (cid:26) , m = 0 , , ..., p (1 /m ) r m , m = p + 1 , ... if r ≤ G p ( m, r ) = log r, m = 0 m ( r m − r − m ) , m = 1 , ..., p (1 /m ) r m , m = p + 1 , ... if r ≥ g are 1 and − sgn a k . Using well known properties of Fourier coefficients, we obtain for 0 < k ≤ p ˆ v [ t ] (0) = (cid:26) , t ≤ log tt ρ , t ≥ v [ t ] ( k ) = (cid:26) , t ≤ − k t k − t − k t ρ sgn a k t ≥ h v [ t ] ( e i • ) , ψ i = (cid:26) , t ≤ − /k ( t k − ρ − t − k − ρ ) | a k | t ≥ t 7→ h v t ( e i • ) , ψ i tends to zero when t → , ∞ and has its onlyminimum at the point t = (cid:18) ρ + kρ − k (cid:19) /k Thus v ( t ) − satisfies the conditions of the lemma with ˜ v = 0 . For k ≥ p + 1 we should take the same g and then h v t ( e i • ) , ψ i = (cid:26) − (1 /k ) t k − ρ | a k | , t ≤ − (1 /k ) t − k − ρ | a k | t ≥ t 7→ h v t ( e i • ) , ψ i obtains minimum at the point t = 1 and the function v satisfies the assertions of the lemma with ˜ v = 0 . (cid:3) Exercise 5.7.3.4. Prove the lemma for the case b k = 0 Lemma 5.7.3.3. Let v ∈ U [ ρ ] with the following condition fulfilled D ′ − lim t → v [ t ] = D ′ − lim t →∞ v [ t ] = ˜ v and let u ∈ SH ( ρ ( r )) with some v ∈ Fr [ u ] .Then there exists w ∈ SH ( ρ ( r )) such that (5.7.3.10) Fr [ w ] = { v [ t ] : t ∈ (0 , ∞ ) } ∪ ˜ v and the following condition holds:1.if the sequence lim t n →∞ w t n = v [ t ] for some t ∈ (0 , ∞ ) and the sequence u t n converges in D ′ as t n → ∞ then lim n →∞ u t n = v t ] . For proof see Corollary 4.4.1.3. Lemma 5.7.3.4. Let w ∈ SH ( ρ ( r )) , ψ ∈ D ( S ) . Then the following holds: lim inf t →∞ h w, ψ i = min v ∈ Fr w h v, ψ i lim sup t →∞ h w, ψ i = max v ∈ Fr w h v, ψ i Exercise 5.7.3.5. Prove this exploiting completeness of Fr . Proof of sufficiency in Theorem 5.7.3.1. In assumption b) we should prove that f is a CRG-function, i.e.,by Th.5.6.1.1 its Fr [ f ] consists of only one function. Sincelog | f | ∈ SH ( ρ ( r )) and because of Th.5.3.1.4 (Approximation) it is enough to provethe corresponding theorem for subharmonic functions. Suppose(5.7.3.11) F [ u + w ] = F [ u ] + F [ w ] , ∀F ∈ χ Ψ for all w ∈ SH ( ρ ( r ) . We exploit Lemma 5.7.3.5 and write (5.7.3.10) in the form:min v ∈ Fr [ u + v ] h v, ψ i = min v ∈ Fr u h v, ψ i + min v ∈ Fr w h v, ψ i , ∀ ψ ∈ Ψ . Suppose the contrary, i.e., u is not a CRG-function and Fr u does not consists ofonly one v min ∈ U [ ρ ] . Then there exists v = v min . The family χ Ψ is total; thereforethere exists ψ ∈ Ψ such that h v , ψ i 6 = h v min , ψ i and hence(5.7.3.12) h v , ψ i > h v min , ψ i . Using Lemma 5.7.3.2, construct for the function ψ a function v ∈ U [ ρ ] satisfyingthe conditions (5.7.3.2),(5.7.3.3) and (5.7.3.4). Apply Lemma 5.7.3.3 to constructa function w satisfying (5.7.3.10a) and the condition 1. Under conditions of theTheorem(5.7.3.13) min ω ∈ Fr ( u + w ) h ω, ψ i = min ω ∈ Fr ( u ) h ω, ψ i + min ω ∈ Fr ( w ) h ω, ψ i Let γ ∈ Fr ( u + w ) be the function on which the minimum of LRH in (5.7.3.11) isattained. Using (5.7.3.3), (5.7.3.4) and (5.7.3.10), we can rewrite (5.7.3.11) in theform(5.7.3.14) h γ, ψ i = min ω ∈ Fr u h ω, ψ i + h v, ψ i Since γ ∈ Fr ( u + w ) , γ = D ′ − lim n →∞ ( u + w ) t n . Passing to subsequences, we cansuppose that the sequences { u t n } and { w t n } have limits. Since Fr w has the form(5.7.3.10), there are two possible cases : w t n → v [ t ] , t ∈ (0 , ∞ ) and w t n → ˜ v. Consider the first case. Because of condition 1. from Lemma 5.7.3.3 u t n → v t ] and γ = v t ] + v [ t ] . Substituting this in (5.7.3.13), we obtain h v t ] , ψ i − min ω ∈ Fr u h ω, ψ i = h v, ψ i − h v [ t ] , ψ i This equality leads to contradiction because for t = 1 it contradicts (5.7.3.12) andfor t = 1 it contradicts (5.7.3.3).Consider the second case, when w t n → ˜ v. Denote v = lim n →∞ u t n and rewrite(5.7.3.11) in the form h v , ψ i − min ω ∈ Fr u h ω, ψ i = h v, ψ i − h ˜ v, ψ i The last equality contradicts (5.7.3.4). (cid:3) Sufficiency of condition a) of Theorem 5.7.3.1 can be proved using the Lemmas5.7.3.3 ,5.7.3.4 and the following lemma. Lemma 5.7.3.2’. Let ψ ∈ D ( S ) . There exists v ∈ U [ ρ ] with the following prop-erties: (5.7.3.2’) D ′ − lim t → v [ t ] = D ′ − lim t →∞ v [ t ] = ˜ v, (5.7.3.3’) h v [ t ] ( e i • ) , ψ i < h v ( e i • ) , ψ i for t ∈ (0 , ∞ ) , t = 1 , (5.7.3.4’) h ˜ v ( e i • ) , ψ i < h v ( e i • ) , ψ i Exercise 5.7.3.6. Prove this lemma and sufficiency of a) in Theorem 5.7.3.1. In this § we consider the question of summing the asymptotic characteristicsconnected with the functional (5.7.1.4), i .e. indicator and lower indicator. Recallthat f ∈ A ( ρ ( r )) is completely regular on the ray { arg z = φ } ( f ∈ A reg,φ ) if(5.7.4.0) h f ( φ ) = h f ( φ )We are going to prove the following assertions: Theorem 5.7.4.1. Let f ∈ A reg,φ . Then for every g ∈ A ( ρ ( r ))(5.7.4.1) h fg ( φ ) = h f ( φ ) + h g ( φ )(5.7.4.2) h fg ( φ ) = h f ( φ ) + h g ( φ ) Theorem 5.7.4.2. Suppose the equality (5.7.4.1) holds for every g ∈ A ( ρ ( r )) . Then f ∈ A reg,φ . Let us note that the assertion of the Theorem 5.7.4.2 holds also if the equality(5.7.4.1) fulfilled for some sequence φ n → φ, because indicator is continuous func-tion (see § φ that is densein [0 , π ) (or the set(5.7.4.3) e i Φ := { e iφ : φ ∈ Φ } is dense on the unit circle), then f ∈ A reg,φ for all φ, i.e. f is a CRG -function.On the other hand, the following assertion holds Theorem 5.7.4.3. If the set Θ of θ is not dense in [0 , π ) , there exists f ∈ A reg,θ , θ ∈ Θ that is not a CRG-function. The situation with lower indicator is analogous, but in another topology.A set E is called non -rarefied at a point z if for every function v subharmonicin a neighborhood of z the following holds: v ( z ) = lim sup z ∈ E,z → z ,z = z v ( z ) = lim sup z ∈ E,z → z v ( z ) . A set is rarefied if it is not non-rarefied.Note that if h f ( φ ) = −∞ then h fg ( φ ) = −∞ for every g ∈ A ( ρ ( r )) . It isobvious that f / ∈ A reg,φ . The next theorems was proved in [GPS]. Theorem 5.7.4.4. Let (5.7.4.2) be fulfilled for ψ ∈ E for all g ∈ A ( ρ ( r )) and e iE is non rarefied at the point e iφ . Then f ∈ A reg,φ . Theorem 5.7.4.5. Let E be a set such that e iE is rarefied at all the points ofunit circle. Then there exists f ∈ A ( ρ ( r )) for which (5.7.4.2) fulfilled for all φ ∈ E and all g ∈ A ( ρ ( r )) , but f / ∈ A reg,φ for all φ and h f ( φ ) > −∞ , ∀ φ. Let us note that E can be dense in [0 , π ) and E from Theorem 5.7.4.4 caneven be of zero measure.The proof of Theorems 5.7.4.4 and 5.7.4.5 is based on the following assertionthat gives a criterion for (5.7.4.2) in terms of limit sets Fr [ f ] . Theorem 5.7.4.6. Let f ∈ A ( ρ ( r )) and h f ( φ ) > −∞ . The condition (5.7.4.2)holdsfor every g ∈ A ( ρ ( r )) , such that h g ( φ ) > −∞ iff (5.7.4.4) lim inf t → v ( te iφ ) = h f ( φ ) for all v ∈ Fr [ f ] . An analogous criterion holds for (5.7.4.1). Theorem 5.7.4.7. Let f ∈ A ( ρ ( r )) . (5.7.4.2) holds for every g ∈ A ( ρ ( r )) , iff (5.7.4.5) lim sup t → v ( te iφ ) = h f ( φ ) , for all v ∈ Fr [ f ] . Corollary 5.7.4.8. The equality (5.7.4.5) implies f ∈ A reg,φ . Actually , for every v ∈ Fr [ f ] we have, using semicontinuity of subharmonicfunctions and the definition (3.2.1.1) of the indicator, h f ( φ ) = lim sup t → v ( te iφ ) ≤ v ( e iφ ) ≤ h f ( φ )for all v ∈ Fr [ f ] . So Fr [ f ] consists of functions v that coincide at the point e iφ andhence on the ray { re iφ : r ∈ (0 , ∞ ) } . Note also that the set e iE for which (5.7.4.1) holds is closed and Theorem5.7.4.4 means that the set where (5.7.4.2) holds is thinly closed , i.e.,closed in thintopology (see [Br § e iφ is a limit point of e iE in the euclidian (respectively, thin)topology, then (5.7.4.1)((5.7.4.2), respectively)is also a sufficient condition for com-pletely regular growth at φ . The main constructive element for proving Theorem 5.7.4.6 is Lemma 5.7.5.1. Let ǫ > , t > and φ ∈ [0 , π ) be fixed. Then there exists v ∈ U [ ρ ] with the following properties: (5.7.5.1) D ′ − lim t → v [ t ] = D ′ − lim t →∞ v [ t ] = 0(5.7.5.2) v ( e iφ ) < v [ t ] ( e iφ ) , t ∈ (0 , ∪ (1 , ∞ )(5.7.5.3) −∞ < v ( e iφ ) < − ǫ, and the inequality (5.7.5.4) v [ t ] ( e iφ ) − v ( e iφ ) ≤ ǫ/ implies (5.7.5.5) t ∈ [1 /t , t ] The last condition means that the function ψ ( t ) := v [ t ] ( e iφ ) can be less than ψ (1) + ǫ/ t = 1 . Proof. Set w ( z ) := max(log | − ze − iφ | , − N ) + ℜ p X n =1 n ( ze − iφ ) n , (5.7.5.6) N > , p = [ ρ ] . It is obvious that w is subharmonic, with masses ν w concentrated in a neighborhoodof the point e iφ . Thus ν w ∈ M [ ρ ] (see (3.1.3.4)) and D ′ − lim t → ( ν w ) [ t ] = D ′ − lim t →∞ ( ν w ) [ t ] = 0Hence (see Th.3.1.4.2) w ∈ U [ ρ ] , and (see (3.1.5.0))(5.7.5.7) D ′ − lim t → w [ t ] = D ′ − lim t →∞ w [ t ] = 0Let us capitalize on the behavior of w [ t ] on the ray { arg z = φ } . (5.7.5.8) w [ t ] ( e iφ ) := ψ ( t ) = (max(log | − t | , − N ) + ℜ p X n =1 n ) t − ρ t n , It is possible to prove directly the following properties of ψ ( t ) . i) outside interval (1 − e − N , e − N ) , ψ ( t ) = G p ( t ) t − ρ ; where G p is the PrimaryKernel (see § − N ;ii) ψ ( t ) > t > t where t is a zero of the equation G p ( t ) = 0 , ψ ( t )decreases monotonically on the interval (0 , − e − N ) and increases monotonicallyon the interval (1 − e − N , t ) . Exercise 5.7.5.1. Prove this.Now set t := 1 − e − N and v ( z ) := Dw t ( z ) , where D is a constant. Thisfunction satisfies the conditions (5.7.5.1) and (5.7.5.2) of the lemma and v [ t ] ( e iφ )has the only one negative minimum for t = 1. Thus it is possible to take D sufficiently large to satisfy the conditions (5.7.5.3) and (5.7.5.4) for fixed ǫ and t . (cid:3) Exercise 5.7.5.2. Prove this in details.In the proof of Theorem 5.7.4.6 we also use Lemma 5.7.3.3. We can prove allthe assertions for subharmonic functions from SH ( ρ ( r )) . Proof of Theorem 5.7.4.6. Necessity. We should prove that if the equality(5.7.5.9) h ( e iφ , u + w ) = h ( e iφ , u ) + h ( e iφ , w )holds for a fixed u ∈ SH ( ρ ( r )) , φ and every w ∈ SH ( ρ ( r )) , then(5.7.5.10) lim inf t → v ( te iφ ) = h ( e iφ , u )for all v ∈ Fr u. Assume that h ( e iφ , u ) > −∞ and h ( e iφ , w ) > −∞ . Suppose thecontrary, i.e. there exists v ∈ Fr u such that(5.7.5.11) lim inf t → v ( te iφ ) > h ( e iφ , u )The inequality (5.7.5.11) implies that there exists ǫ > t > t ∈ [1 /t , t ] the inequality(5.7.5.11a) v ( te iφ ) > h ( e iφ , u ) + ǫ holds. Let us construct by Lemma 5.7.5.1 for these ǫ, t , φ a function v and byLemma 5.7.3.3 for the functions u, v and the already found v a function w . Letus show that for w the equality (5.7.5.9) does not hold.Compute h ( e iφ , w ) . From (3.2.1.2) h ( e iφ , w ) = min { , inf { v [ t ] ( e iφ : t ∈ (0 , ∞ ) }} . The inequalities (5.7.5.3) imply that 0 can be omitted and (5.7.5.2) implies thatthe infimum is attained at t = 1 , i.e.,(5.7.5.12) h ( e iφ , w ) = v ( e iφ ) . Find v ǫ ∈ Fr ( u + w ) such that h ( e iφ , u + w ) > v ǫ ( e iφ ) − ǫ/ . Let t n → ∞ and( u + w ) t n → v ǫ in D ′ . Passing to subsequences we can assume that u t n and w t n also converge. Consider two cases. The first, when(5.7.5.13) D ′ − lim w t n = v [ t ] , t ∈ (0 , ∞ ) By Lemma 5.7.3.3 lim u t n = v t ] and hence v ǫ = lim( w + u ) t n = v [ t ] + v t ] . If t / ∈ [1 /t , t ] then by (5.7.5.4)(5.7.5.14) v [ t ] ( e iφ ) > v ( e iφ ) + ǫ/ h ( e iφ , w ) + ǫ/ h ( e iφ , u + w ) ≥ v ǫ ( e iφ ) − ǫ/ ≥ v [ t ] + v t ] − ǫ/ h ( e iφ , u + w ) ≥ h ( e iφ , w ) + h ( e iφ , u ) + ǫ/ t ∈ [1 /t , t ] then from (5.7.5.11) we have(5.7.5.16) h ( e iφ , u + w ) ≥ h ( e iφ , w ) + h ( e iφ , u ) + 2 ǫ/ D ′ − lim w t n = 0 . In this case we have h ( e iφ , u + w ) ≥ v ǫ ( e iφ ) − ǫ/ ≥ h ( e iφ , u ) − ǫ + ǫ − ǫ/ h ( e iφ , u ) − ǫ + 2 ǫ/ . Using (5.7.5.12) and (5.7.5.3) we obtain h ( e iφ , u + w ) ≥ h ( e iφ , u ) + h ( e iφ , w ) + 2 ǫ/ . So we proved in any case that (5.7.5.9) does not hold if (5.7.5.10) does not hold .Let us prove sufficiency in Theorem 5.7.4.6. We prove it for subharmonicfunctions. Let u ∈ SH ( ρ ( r )) and for every v ∈ Fr u (5.7.5.10) holds. Let us showthat for all w ∈ SH ( ρ ( r )) (5.7.5.9) holds. It is sufficient to prove that(5.7.5.17) h ( e iφ , u + w ) ≤ h ( e iφ , u ) + h ( e iφ , w )holds since the inverse inequality holds for every w ∈ SH ( ρ ( r )) (see (3.2.1.5)).Let us note for beginning that for every v ∈ Fr w there exist v ∈ Fr ( u + w ) and v ∈ Fr u such that(5.7.5.19) v = v + v Indeed, let t n → ∞ be a sequence such that w t n → v . We can suppose , choosingsubsequence, that u t n → v and ( u + w ) t n → v. Then (5.7.5.19) holds.Let ǫ be arbitrarily small.Chose v ∈ Fr w such that v ( e iφ ) < h ( e iφ , w ) + ǫ holds. From upper semicontinuity of v we have(5.7.5.20) lim sup t → v ( e iφ ) ≤ h ( e iφ , w ) + ǫ. Let v ∈ Fr u and v ∈ Fr ( u + w ) satisfy (5.7.5.19).Then we have h ( e iφ , u + w ) ≤ ( v + v ) [ t ] ( e iφ ) = v t ] ( e iφ ) + v t ] ( e iφ ) , ∀ t. Hence h ( e iφ , u + w ) ≤ lim inf t → v t ] ( e iφ ) + lim sup t → v t ] ( e iφ ) . Using (5.7.5.10) and (5.7.5.20) we obtain h ( e iφ , u + w ) ≤ h ( e iφ , u ) + h ( e iφ , w ) + ǫ. This proves the inverse inequality and hence the equality (5.7.5.9), because ǫ isarbitrarily small. (cid:3) Now we are going to prove Theorem 5.7.4.4.We need the following assertionfrom Potential Theory. Lemma 5.7.6.1. Let E be a set that is non-rarefied at the point e iφ . Let E ′ bea set in C , such that ∀ e iφ ∈ E and ∀ δ > there exists a point z ′ ∈ E ′ on the ray { arg z = φ } such that | z ′ − e iφ | < δ . Then E ′ is also non-rarefied at the point e iφ . Proof. We can suppose without loss of generality that E ′ have no intersection withsome neighborhood of zero. Denote by P ( z ) the map z e i arg z . It is easy to seethat for all pairs z ′ , z ′ ∈ E ′ the inequality | P ( z ′ ) − P ( z ′ ) | < A | z ′ − z ′ | hold for someconstant A. Thus the logarithmic capacity (2.5.2.5) satisfies ([La,Ch.II, § cap l ( M ) < A cap l ( M ′ )where M ′ ⊂ E ′ , M = P ( M ′ ) . Now we exploit the following properties of non-rarefied sets. First, if E is non-rarefied at a point z , then there exists a compact set that is non-rarefied at z ([La,Ch.V, § § K that is non-rarefied at z (5.7.6.3) ∞ X n =1 n log( cap l K n ) − = ∞ where K n := K ∩ { z : q n +1 ≤ | z − z | ≤ q n } , < q < . Using the inequality (5.7.6.2),we obtain that divergence of the series (5.7.6.3)for a compact K ⊂ E implies divergence for K ′ ⊂ E ′ where K = P ( K ′ ) , i.e., E ′ isnon-rarefied at the point P ( e iφ ) = e iφ . (cid:3) Proof Theorem 5.7.4.4. Let ǫ ( φ ) → φ → φ and let v ∈ Fr u. Suppose(5.7.5.9)holds for e iφ ∈ E. By Theorem 5.7.4.6 the equality (5.7.5.10) holds. Thus ∀ ∆ > , ∃ z ′ = z ′ ( e iφ , ∆) such that(5.7.6.4) | z ′ − e iφ | < ∆ , arg z ′ = φ, v ( z ′ ) < h ( e iφ ) + ǫ ( φ )Set E ′ := [ φ ∈ E ∞ [ n =1 z ′ ( e iφ , /n )By (5.7.6.4) and upper semicontinuity of h ( e iφ ) we obtain(5.7.6.5) lim sup z ′ → e iφ , z ′ ∈ E ′ v ( z ′ ) ≤ h ( e iφ )Since E ′ is non-rarefied by Lemma 5.7.6.1 the upper limit of v coincides with v ( e iφ ) and hence v ( e iφ ) ≤ h ( e iφ ) . The inverse inequality holds always. Thus v ( e iφ ) = h ( e iφ ) , ∀ v ∈ Fr u. Hence h ( e iφ ) = h ( e iφ ) . (cid:3) Now we are going to prove Theorem 5.7.4.5. Before this we need to describea construction and prove some auxiliary assertions.Let B j := { z : T j < | z | < T j +1 } , j = 0 , ± , ± , ... where T > L E := { z : e i arg z ∈ e iE } . Recall that e iE is a set rarefied atevery point of the unit circle. Let Q be the set of rational numbers on the interval(1 , T ) . Set S Q := { z : | z | ∈ Q } , T j S Q := { zT j : z ∈ S Q } , A j := L E ∩ T j S Q , j = 0 , ± , ± , ... Lemma 5.7.7.1. There exists v ∈ U [ ρ ] such that (5.7.7.1) v ( z ) = −∞ for z ∈ A and (5.7.7.2) µ v ( e ) = 0 , ∀ e ⊂ C \ B . Proof. The set E is rarefied at its every point, hence it is polar ([Br,Ch.7, § { z : | z | = r } ∩ L E is polar (see [Br,Ch.3, § § A is polar.Hence there exists a positive measure µ concentrated on B for which the potential v ( z ) := R G p ( z/ζ ) dµ is equal to −∞ on A (see [Br,Ch.4, § 6, Applications]). It is easy to see that µ ∈ M ( ρ ) and hence v ∈ U [ ρ ] (see Th.3.1.4.2). (cid:3) Lemma 5.7.7.2. There exists ω ∈ U [ ρ ] such that the following conditions arefulfilled: (5.7.7.3) ω ( z ) = −∞ , z ∈ A := ∪ + ∞ j = −∞ A j ; ω ( T z ) = T ρ ω ( z ) Proof. Set for every E ⋐ C \ ν ( E ) := j =+ ∞ X j = −∞ T jρ µ v ( T − j E ∪ B )(compare Th.4.1.7.1). We have ν ∈ M ( ρ ) . Set ω ( z ) := Z G p ( z/ζ ) ν ( dξdη ) , ζ = ξ + iη. This ω satisfies (5.7.7.3). (cid:3) Exercise 5.7.7.1. Prove this using Th.4.1.7.1. Lemma 5.7.7.3. Let ω be a subharmonic function in C . Denote m ( φ ) := max { ω ( re iφ : r ∈ [1 , T ] } . Then there exists a constant C > −∞ such that m ( φ ) > C ∀ φ . Proof. If not, there exists a sequence φ n that we can assume to converge to φ ∞ suchthat m ( φ n ) → −∞ . By upper semicontinuity of ω we have ω ( z ) = −∞ , ze − iφ ∞ ∈ [1 , T ] . Thus ω ( z ) ≡ −∞ because the capacity of the segment in the plane is positiveand hence it is not polar for some subharmonic function. (cid:3) Recall that for v ∈ U [ ρ ] (see (4.1.3.1))(5.7.7.5) C ( v ) := D ′ − clos { v [ t ] : 0 < t < ∞} , (5.7.7.6) Ω( v ) := { v ′ ∈ U [ ρ ] : ( ∃ t k → ∞ )( v ′ = lim k →∞ v [ t k ] } (5.7.7.7) A ( v ) := { v ′ ∈ U [ ρ ] : ( ∃ τ k → v ′ = lim k →∞ v [ t k ] } By Theorems 4.1.3.4 and 4.2.1.2 if(5.7.7.8) A ( v ) ∩ Ω( v ) = ∅ , there exists u ∈ SH ( ρ ( r )) such that(5.7.7.9) Fr u = C ( v ) . Lemma 5.7.7.4. There exists v ∈ U [ ρ ] such that the following holds: (5.7.7.10) A ( v ) = Ω( v )(5.7.7.11) inf { v ( e iφ ) : v ∈ C ( v ) } = lim inf t → v ( te iφ ) = 0 , ∀ v ∈ C ( v ) , ∀ e iφ ∈ e iE , (5.7.7.12.) sup { v ( e iφ ) : v ∈ C ( v ) } 6 = inf { v ( e iφ ) : v ∈ C ( v ) } . Proof. Let ω ( z ) be constructed by Lemma 5.7.7.2. Set v ( z ) := ω ( z ) + D log + | z | . The condition (5.7.7.3) implies A ( ω ) = Ω( ω ) = { ω [ t ] : t ∈ [1 , T ] } because it is a Periodic Limit Set (see Th.4.1.7.1).Since (log + | z | ) [ t ] → , t → , t → ∞ the function v satisfies the condition A ( v ) = Ω( v ) = { ω [ t ] : t ∈ [1 , T ] } By Theorem 2.1.7.4 for the function v := v + we have A ( v ) = Ω( v ) = { ω +[ t ] : t ∈ [1 , T ] } . Note that v ( z ) = 0 for z ∈ A and since A is dense in L E (5.7.7.11) holds.Choosing D sufficiently large it is possible (using Lemma 5.7.7.3) to find on everyray { arg z = φ } a point z φ where v ( z φ ) > . Hence sup { v ( e iφ ) : v ∈ C ( v ) } > . Because of (5.7.7.11) and upper semicontinuity of inf { v ( e iφ ) : v ∈ C ( v ) } it is zerofor every e iφ . Thus (5.7.7.12) holds. (cid:3) Proof of Theorem 5.7.4.5. Let us construct by Theorems 4.1.3.4 and 4.2.1.2 a func-tion u ∈ SH ( ρ ( r ) such that Fr u = C ( v ) where v is taken from Lemma 5.7.7.4. Itdoes not belong to A reg,φ for any φ . The equality (5.7.5.9) holds for every φ ∈ E because of (5.7.7.11) by Theorem 5.7.4.6. (cid:3) The proof of Theorem 5.7.4.1 is a copy of the proof of Theorem 5.7.3.1. Exercise 5.7.8.1. Prove Theorem 5.7.4.1.Now we are going to prove Theorem 5.7.4.7 which implies (as it was shown inCorollary 5.7.4.8) Theorem 5.7.4.2. The main constructive element of the proof ofnecessity is Lemma 5.7.8.1. Let ǫ > , t > and φ ∈ [0 , π ) be fixed. Then there exists v ∈ U [ ρ ] with the following properties: (5.7.8.1) D ′ − lim t → v [ t ] = D ′ − lim t →∞ v [ t ] = 0 (5.7.8.2) v ( e iφ ) > v [ t ] ( e iφ ) , t ∈ (0 , ∩ (1 , ∞ ) and the inequality (5.7.8.3) v [ t ] ( e iφ ) − v ( e iφ ) ≥ − ǫ/ implies (5.7.8.4) t ∈ [ t , /t ]The last condition means that the function ψ ( t ) := v [ t ] ( e iφ ) can be more than ψ (1) − ǫ/ t = 1 . Proof. Consider the function(5.7.8.5) w ( z ) := log + | z | It is subharmonic and satisfies (5.7.8.1). Since the function ψ ( t ) := w [ t ] ( e iφ ) = t − ρ log + t has its only strict maximum in the point t max > , the function v ( z ) := w ( z/t max )has all the properties (5.7.8.1)-(5.7.8.4). (cid:3) After this lemma all the proof of Theorem 5.7.4.6 can be repeated with minimalchanges. Exercise 5.7.8.2. Prove Theorem 5.7.4.7. Now we are going to prove Theorem 5.7.4.3. Let us prove the following Lemma 5.7.9.1. Let Θ be a closed subset of [0 , π ) . Then for every σ > thereexists a π - periodic ρ -trigonometrically convex function h ( φ ) such that (5.7.9.1) h ( φ ) = σ for φ ∈ Θ and (5.7.9.2) h ( φ ) > σ for φ / ∈ Θ . Proof. We can suppose that 0 ∈ Θ otherwise we can shift it a little. The set[0 , π ) \ Θ is open and it can be represented as union of non-intersecting openintervals. If length of an interval is ≤ π/ρ we can construct a ρ -trigonometricalfunction that equals to σ on the ends of the interval. It is greater than σ in the allinner points of the interval because f ( φ ) ≡ σ is strictly ρ -trigonometrical function.If the length of the interval is greater than π/ρ , for example this is ( − l/ , l/ 2) with l > π/ ρ, we cover it by intersecting intervals of length less then π/ρ, construct h I ( φ ) as before for every interval I and set h ( φ ) = max I h I ( φ ) . It is obvious that h ( φ ) is greater than σ and it is ρ - trigonometrically convex. (cid:3) Theorem 5.7.4.3 is a corollary of Lemma 5.7.9.1 and the following Theorem 5.7.9.2. Let h and h be two ρ -trigonometrically convex functions.Then there exists a function f ∈ A ( ρ ( r )) such that h f ( φ ) = max( h ( φ ) , h ( φ )) , h f ( φ ) = min( h ( φ ) , h ( φ )) Proof. Consider the set(5.7.9.3) U := { v ( z ) = cr ρ h ( φ ) + (1 − c ) r ρ h ( φ ) : 0 ≤ c ≤ } It consists of invariant subharmonic functions,it is contained in U [ ρ ] and satisfiesthe condition of Theorem 4.1.4.1. Hence (Theorems 4.2.1.2, Corollary 5.3.1.5) thereexists a function f ∈ A ( ρ ( r )) such that(5.7.9.4) Fr f = U By formulae (3.2.1.1),(3.2.1.2) we obtain the assertion of the theorem, using (5.7.9.3). (cid:3) Exercise 5.7.9.1. Prove Theorem 5.7.4.3. The family of characteristics {F α , α ∈ A } is called independent if forevery subset A ′ ⊂ A (or subset in some class of subsets , for example, measurableor closed) there exists a function f = f A ′ ∈ A ( ρ ( r )) such that F α [ f ] = F α [ f ] , α ∈ A ′ F α [ f ] = F α [ f ] , α ∈ A \ A ′ It means that for every pointed subset of characteristics there exists a functionthat has regular growth with respect to this subset of characteristics and is not ofregular growth with respect to all other characteristics.Theorem 5.7.4.3 can be considered as an assertion of independence of the family(5.7.2.2). Theorem 5.7.10.1. The family χ F o (5.7.2.4) is independent, i.e., for every A ⊂ Z there exists f ∈ A ( ρ ( r ) such thatlim r →∞ r − ρ ( r ) Z π log | f ( re iφ ) g k ( φ ) dφ exists for all k ∈ A and does not exists for k ∈ Z \ A. For beginning we prove Lemma 5.7.10.2. There exist two ρ -trigonometrically convex functions h and h for which (5.7.10.11) Z π h ( φ ) g k ( φ ) dφ = Z π h ( φ ) g k ( φ ) dφ, k ∈ A (5.7.10.12) Z π h ( φ ) g k ( φ ) dφ = Z π h ( φ ) g k ( φ ) dφ, k ∈ Z \ A Proof. Let g ( φ ) ∈ C be a function, the Fourier coefficients of which with indices k ∈ A are equal to zero . We can represent it as a difference of ρ -trigonometricallyconvex functions in the following way. Suppose for simplicity that ρ is non-integer.Then take T ρ g = g ′′ + ρ g and consider h ( φ ) := 12 ρ sin πρ Z π ] cos ρ ( φ − ψ − π )( T ρ g ) + ( φ ) dφ ; h ( φ ) := 12 ρ sin πρ Z π ] cos ρ ( φ − ψ − π )( T ρ g ) − ( φ ) dφ. By Th.3.2.3.3 these functions are ρ trigonometrically convex and h − h = g. Hence(5.7.10.11) , (5.7.10.12) holds. (cid:3) Proof of Th.5.7.10.1. We consider a function f ∈ A ( ρ ( r )) with the limit set U := { v ( z ) = cr ρ h ( φ )+ (1 − c ) r ρ h ( φ ) : 0 ≤ c ≤ } with h , h from Lemma and exploitTheorem 5.7.1.4. (cid:3) Exercise 5.7.10.1. Do this in details. The point of depart on this topic is the following Theorem VT. [Va] , [Tich] Let f ∈ A ( ρ ) , ρ < have its zeros on the negative ray.If the limit lim r →∞ r − ρ log | f ( r ) | exists, then the limit lim r →∞ r − ρ n ( r ) exists. The latter means that f is a CRG-function.The general problem is the following. Let ρ be any non-integer number, f ∈ A ( ρ ( r )) and suppose all zeros of f lie on a finite system of rays(5.8.1.1) K S := { z = re iφ : 0 < r < ∞ , φ ∈ S } where(5.8.1.2) S := { e iθ j : j = 1 , , ..., m } We write n f ∈ M S . Let n j be a zero distribution on the ray { arg z = θ j } and all the limits(5.8.1.3) lim r →∞ r − ρ n j ( r ) := ∆ j exist. In such case we write n f ∈ M reg,S .Let K S be one more system of rays(5.8.1.4) S = { e iψ k : k = 1 , , ..., n } Some ψ k can coincide with some θ j . Suppose that f has regular growth on thissystem, i.e.,(5.8.1.5) h f ( φ ) = h f ( φ ) , e iφ ∈ S In such case we write f ∈ A reg,S . The problem is, what is the connection between S and S so that the implica-tion ( f ∈ A reg,S ) = ⇒ ( n f ∈ M reg,S ) holds.This problem can be reformulated in another way. For n f ∈ M reg,S if n f ∈M S it is necessary and sufficient that f is a CRG -function, because existence theangle density is equivalent to the existence of all the limits. So the problem canbe reformulated in the form: what is connection between S and S , so that theimplication ( f ∈ A reg,S ) = ⇒ ( f is CRG − f unction ) holds.Denote(5.8.1.6) G ( t, γ, ρ ) := G p ( e t − iγ ) e − ρt , p = [ ρ ]where G p is the Primary Kernel: G p ( z ) = log | − z | + ℜ p X k =1 z k /k Set ˆ G ( s, γ, ρ ) := ∞ Z −∞ G ( t, γ, ρ ) e − ist dt This is the Fourier transformation of G ( t, γ, ρ ). It can be computed (see, e.g,[Oz,Lemma3])ˆ G ( s, γ, ρ ) = π cos( π + γ )( ρ + is )( ρ + is ) sin π ( ρ + is )Consider the matrix(5.8.1.7) ˆ G ( s, S − S ) := k ˆ G ( s, θ j − ψ k , ρ ) k . We are going to prove Theorem 5.8.1.1. The implication { f ∈ A reg,S } ∧ { n f ∈ M S } = ⇒ { f is a CRG − f unction } holds iff (5.8.1.5) rank ˆ G ( s, S − S ) = m, ∀ s ∈ ( −∞ , ∞ )As a corollary we obtain the following ([De]) Theorem 5.8.1.2.(Delange). Suppose that S and S consist of one ray,i.e., S = { e iθ } , S = { e iψ } . The implication (5.8.1.5) holds iff (5.8.1.6) θ − ψ = (1 − (2 k + 1) / ρ ) π, k = 1 , , ... A Fourier transformation for distribution ν on the real axes is a distributionin the standard space S ′ (see [H¨o, v.1,Ch.7, § F ν )( s ) := lim ǫ → ∞ Z −∞ e its e − ǫt ν ( dt ) . where the right side is understood in the sense of distributions.For example, if ν ( dt ) := e is t dt, we have F ν ( s ) = δ ( s − s ) where δ is the Diracfunction. Exercise 5.8.2.1. Check this.For distribution and a summable function one can b define the convolution ,for which the property F ( f ∗ ν )( s ) = F f ( s ) F ν ( s ) holds. Proof of Theorem 5.8.1.1.. Since f ∈ M S the limit set Fr n f is concentrated on K S . So every v ∈ Fr [ f ] can be represented in the form (see Th.3.1.5.1):(5.8.2.1) v ( z ) = j = m X j =1 ∞ Z G p ( z/re iθ j ) µ j ( dr )where µ j is concentrated on the ray { arg ζ = θ j } and belong to U [ ρ ] . After changingvariables: r = e t , | z | = e t we obtain from (5.8.2.1)(5.8.2.1) v ( te iφ ) = j = m X j =1 ∞ Z −∞ G ( t − τ, φ − θ j , ρ ) µ j ( dτ ) where(5.8.2.2) e ρτ µ ( dτ ) := µ j ( dr ) , v ( te iφ ) := v ( | z | e iφ ) e − ρ | z | . The equality (5.8.2.1) can be written as(5.8.2.3) v ( te iφ ) = j = m X j =1 [ G ( • , φ − θ j , ρ ) ∗ µ j ]( t )where * stands for convolution. Then f ∈ A reg,S with n f ∈ M S , iff every pair v , v ∈ Fr [ f ] satisfies the condition(5.8.2.4) v ( z ) = v ( z ) , z ∈ K S Denote by µ ,j , µ ,j the restriction of µ v µ v , to the ray { arg z = θ j } . Set ν j := µ ,j − µ ,j Using (5.8.2.3) we can rewrite (5.8.2.4) in the form(5.8.2.5) j = m X j =1 [ G ( • , φ k − θ j , ρ ) ∗ ν j ]( t ) ≡ , k = 1 , , ..., n. Applying Fourier transforms we obtain a system of linear equations:(5.8.2.6) j = m X j =1 [ ˆ G ( • , ψ k − θ j , ρ ) · ˆ ν j ]( t ) ≡ , k = 1 , , ..., n. Suppose now that rank ˆ G ( s, S − S ) = m for every s ∈ R . The system (5.8.2.6)has only the trivial solution for every s . Thus ˆ ν j ( s ) ≡ , for j = 1 , , ..., m. Thisimplies ν j ( t ) ≡ j = 1 , , ..., m and ν j ≡ j = 1 , , ..., m. Thus µ v = µ v ,i.e, (by (5.8.2.3) ) Fr [ f ] consists of one function v ∈ U [ ρ ] . Thus f is a CRG-function.Conversely, suppose that rank ˆ G ( s, S − S ) < m for some s Then there exists a nontrivial solution ( b , ...b m ) that satisfies the correspondingsystem. We obtain that { ˆ ν j b j δ ( s − s ) , j = 1 , ...m } is a solution of (5.8.2.6) forall s ∈ R and hence the ν j ( dt ) = b j e its dt, j = 1 , , ...m Since ν j have bounded densities dν j /dt, we can find a constant C such thatsup {| dν j /dt | : 0 < t < ∞ , j = 1 , ...m } ≤ C. Set(5.8.2.7) µ ,j ( dt ) = Cdt + ν j ( dt ); µ ,j = Cdt. Both of these are measures. Now we pass to m ,j , m ,j via (5.8.2.2).It is easy tocheck that m ,j , m ,j ∈ M ( ρ ) . Exercise 5.8.2.2. Check this.Consider µ , µ ∈ M ( ρ ) which are defined uniquely by their restrictions µ ,j , µ ,j respectively on K S . Set v ( z ) := Z C G p ( z/ζ ) µ ( dξdη ); v ( z ) := Z C G p ( z/ζ ) µ ( dξdη ); ζ = ξ + iη. It is easy to check that the equality(5.8.2.8) v ( z ) = v ( z ) , z ∈ K S holds. Exercise 5.8.2.3. Check this.Since µ and µ is a finite sum of trigonometrical functions, for v and v thecondition (4.1.3.3) is satisfied. Thus by Theorem 4.3.6.1 there exists a function f ∈ A ( ρ ( r )) for which Fr [ f ] = [ ≤ c ≤ C ( cv + (1 − c ) v )Since for v ∈ C ( cv + (1 − c ) v ) (5.8.2.8) also holds, the same holds for v ∈ Fr [ f ]and this function is not a CRG-function. (cid:3) 6. Application to the completeness of exponential systemsin convex domains and the multiplicator problem The completeness of exponential systems in convex domains is intimately con-nected to the multiplicator problem. Considering a special form of exponent systemis related to the study of special subharmonic functions, that determine the peri-odic limit set, so called automorphic subharmonic functions. The next §§ Let Φ ∈ A ( ρ ( r )) and let H ( φ ) be a ρ -trigonometrically convex function. Afunction g ∈ A ( ρ ( r )) is called an H -multiplicator of Φ if the indicator h g Φ of theproduct g Φ satisfies the inequality h g Φ ( φ ) ≤ H ( φ ) , ∀ φ. In some questions (see § . Define H ( z ) := r ρ H ( φ ) , z = re iφ . Let v ∈ U [ ρ ] (see (3.1.2.4)). Consider the function m ( z, v, H ) := H ( z ) − v ( z ) . As will be proved in Corollary 6.1.9.3, the maximal subharmonic minorant of m ( z, v, H ) exists and is continuous. The maximal subharmonic minorant of m (m.s.m.) belonging to U [ ρ ] will be denoted by G H v, while the domain of definitionof the operator G H will be denoted by D H . Though m (0 , • , • ) = 0, the m.s.m. of m can differ from zero (as was remarked by A.E.Eremenko and M.L.Sodin), but ifthe m.s.m. of m equals zero at zero , then it belongs to U [ ρ ] . Exercise 6.1.1.1. Consider the function w ( z ) = (cid:26) | z | log | z | , if | z | ≤ | z | − , if | z | ≥ . It is subharmonic and belongs to U [1] . Show that the maximal subharmonic mino-rant of K | z | − w ( z ) is different from zero in 0 for every K > . Theorem 6.1.1.1. Φ ∈ A ( ρ ( r )) has an H -multiplicator iff (6.1.1.1) Fr [Φ] ⊂ D H . Proof of necessity. Let g be a multiplicator of Φ, i.e.,(6.1.1.2) h g Φ ( φ ) ≤ H ( φ )and let v ∈ Fr [Φ] . We can choose v g Φ ∈ Fr [ g Φ] and v g ∈ Fr [ g ] such that v g Φ = v + v g (see Th.3.1.2.4, fru1)). Exercise 6.1.1.2. Prove this directly.By definition of indicator (3.2.1.1) and (6.1.1.2) we have v g Φ ( z ) ≤ H ( z ) or v g ( z ) ≤ m ( z, v, H ) . Since v g ∈ U [ ρ ] , v ∈ D H . (cid:3) For proving sufficiency we need the following Theorem 6.1.1.2. The operator G H is1. upper semicontinuous in the D ′ -topology, 6.1.1.5i.e., ( v j → v ) ∧ ( G H v j → w ) = ⇒ ( w ∈ U [ ρ ]) ∧ ( w ( z ) ≤ G H ( z ) , z ∈ C ); ( G H v ) [ t ] = G H v [ t ] ; (see (3.1.2.4a) for P ≡ I ; )3.concave: ( ∀ v , v ∈ D H , c ∈ [0; 1]) = ⇒ ( v c := cv + (1 − c ) v ∈ D H ) and G H ( v c ) ≥ c G H ( v ) + (1 − c ) G H ( v ) . Proof. Let us prove 1). Suppose v j ∈ U [ ρ ] → v and G H v j → w. Then(6.1.1.3) G H v j ≤ H ( z ) − v j ( z ) , z ∈ C , Applying ( • ) ǫ from (2.6.2.2) and Th.2.3.4.5 reg 3), we obtain w ǫ ≤ ( H ) ǫ ( z ) − ( v ) ǫ ( z ) , z ∈ C w ǫ (0) ≥ Passing to the limit as ǫ ↓ w ( z ) ≤ H ( z ) − v ( z ) = m ( z, H, v ) , z ∈ C . Since 0 ≤ w (0) ≤ m (0 , H, v ) = 0 we have w (0) = 0 and , hence, w ∈ U [ ρ ] . Thus v ∈ D H and w ( z ) ≤ G H v ( z ).Let us prove 2). Since H ( z ) is invariant with respect to ( • ) [ t ] , ( G v ) [ t ] ≤ H ( z ) − v [ t ] . Hence,(6.1.1.4) ( G v ) [ t ] ( z ) ≤ ( G ( v [ t ] ))( z ) , because G ( v [ t ] ) is m a x i m a l subharmonic minorant. We can replace v with v [1 /t ] and obtain ( G v [1 /t ] ) [ t ] ( z ) ≤ G v ( z ) . Applying ( • ) [ t ] to the two sides of the inequality,we obtain G v [1 /t ] ( z ) ≤ ( G v ( z )) [1 /t ] . Now we can replace 1 /t with t and obtain thereverse inequality to (6.1.1.4), which, together with (6.1.1.4), proves 2).3). Let v , v ∈ D H and c ∈ [0; 1] . One has G v i ( z ) ≤ H ( z ) − v i ( z ) , i = 1 , , ∀ z. Then [ c G v + (1 − c ) G v ]( z ) ≤ H ( z ) − [ cv + (1 − c ) v ]( z ) . Thus [ c G v + (1 − c ) G v ]( z ) ≤ G [ cv + (1 − c ) v ]( z ) . (cid:3) Proof of sufficiency in Theorem 6.1.1.1. Assume that Fr [Φ] ⊂ D H and considerthe set(6.1.1.5) U := { ( v ′ , v ′′ ) : v ′′ ≤ G v ′ , v ′ ∈ Fr [Φ] } . Then U is non-empty, because of (6.1.1.1), closed, because of Th. 6.1.1.2, 1), andinvariant, because of Th.6.1.1.2, 2). Every fiber U ′′ = { v ′′ : v ′′ ≤ G v ′ } is convex because of Th.6.1.1.2, 3). By Th.4.4.1.2 there exists u ′′ ∈ U ( ρ ( r )) such that for the curve u := ( u ′ , u ′′ )(6.1.1.6) Fr [ u ] = U . By the Th.5.3.1.4 (Approximation Theorem) the function u ′′ can be replacedwith log | g | , where g ∈ A ( ρ ( r )) , retaining the property (6.1.1.6).Let us prove that g is an H -multiplicator of Φ . Indeed, set Π := g Φ . It isenough to prove that for every v Π ∈ Fr [Π](6.1.1.7) v Π ( z ) ≤ H ( z )Note that every v Π has the form v Π = v g + v, where ( v, v g ) ∈ U . Thus, because ofthe definition (6.1.1.5) v Π satisfies (6.1.1.7). (cid:3) Let us note that the pair ( v, G H v ) ∈ U because of closeness of U . Hence thefollowing assertion holds Proposition 6.1.1.3. Every Φ ∈ A ( ρ ) that satisfies (6.1.1.1) has a multiplicator g ∈ A ( ρ ) such that (6.1.1.8) v + G H v ∈ Fr [ g Φ] . Exercise 6.1.1.3. Check this in details.Although v ∈ U [ ρ ] is in general an upper semicontinuous function, Theorem 6.1.1.4. The function G H v ( z ) , v ∈ U [ ρ ] , is a continuous function thatis harmonic outside the set E = { z : G H v ( z ) = m ( z, v, H ) } . Proof. G G v ( z ) is continuous because of Corollary 6.1.9.3. If G H v ( z ) < v ( z ) and if G H v ( z ) is not harmonic in a neighborhood of z , we can make sweeping of massesfrom a small disc {| z − z | < ǫ } (see Th.2.7.2.1). The obtained subharmonic functionwill be grater than G H v ( z ) , contradicting maximality. (cid:3) Suppose that some H -multiplicator g = g ( z, Φ , H ) of the function Φ isfound. We examine the function Π = g Φ.The structure of its limit set is describedby the following statement: Proposition 6.1.2.1. Every v Π ∈ Fr [ g Φ] can be written as v Π = v + w , where v ∈ Fr [Φ] and w ∈ U [ ρ ] with the condition (6.1.2.1) w ( z ) ≤ G H ( z ) , ∀ z ∈ C , and, conversely, for every v ∈ Fr [Φ] there exists a v g , v g ( z ) ≤ G H v ( z ) , such that v + v g ∈ Fr [ g Φ] . . Exercise 6.1.2.1. Prove this, like in Exercise 6.1.1.1.An H -multiplicator G of the function Φ will be called ideally complementing if it satisfies the condition Fr [ G Φ] = { v Π = v + G H v : v ∈ Fr [Φ] } . If a multiplicator is ideally complementing then equality is achieved in (6.1.2.1)for all v ∈ Fr [Φ] . This make the multiplicator optimal in another respect. Recallthat an entire function f is of minimal type with respect to a proximate order ρ ( r ) , ρ ( r ) → ρ if (see (2.8.1.6)) σ f := lim sup r →∞ log M ( r, f ) r − ρ ( r ) = 0 . Proposition 6.1.2.2. Let G = G ( • , Φ , H ) be an ideally complementing H -multiplicatorof a function Φ .Then each H -multiplicator of the function Π = G Φ is of minimaltype. This proposition is proved in § H the condition(6.1.1.1) implies that Φ has an ideally complementing multiplicator. For instance,if Φ is a function of completely regular growth (see § Exercise 6.1.2.2. Prove this. Theorem 6.1.2.3. Every function with periodic limit set is ideally complementable. This theorem is proved in § C ⊂ R l be an l -dimensional connected compact and let { h ( φ, c ) : c ∈ C } be a set of ρ -t.c. functions that is continuous with respect to c ∈ C. For example, c ∈ [0 , 1] and h ( φ, c ) = ch ( φ ) + (1 − c ) h ( φ ) . The set(6.1.2.2) U ind := { v ( re iφ s ) = r ρ h ( φ, c ) : c ∈ C } is the limit set of an entire function. Exercise 6.1.2.3 Prove this using Theorem 4.3.6.1.Such a set is called a set of indicators . Entire functions with such limit setscan be also considered as a generalization of CRG-functions. Theorem 6.1.2.4. Every function Φ whose limit set is a set of indicators is ideallycomplementable. This theorem is proved in § H - multiplicator depends, of course, both on Φ ∈ A ( ρ ( r )) (or, more precisely, on its limit set Fr [Φ]) and on H. Theorem 6.1.2.5. Let Φ and H be such that the condition (6.1.1.1) is satis-fied.The function Φ has an ideally complementing H -multiplicator if and only if theoperator G H is continuous on Fr Φ . This theorem is proved in § G H . Weshall say that the maximum principle for U [ ρ ] is valid in the domain G, (which is,generally speaking, unbounded), if the conditions w ∈ U [ ρ ] , w ( z ) = 0 for z / ∈ G imply w ( z ) ≡ . Let us denote by H w a region of harmonicity of w ∈ U [ ρ ], i.e., a region wherethe conditions “ w is harmonic in G ” and “ G ⊃ H w ” imply G = H w . We remark that H w is a connected component of the open set on which w isharmonic. Generally it is not unique. The image of U ∈ U [ ρ ] will be denoted by G H U , while its closure in the D ′ -topology will be denoted by clos G H U. Theorem 6.1.2.6. Suppose for every w ∈ clos G H U and every H w the maximumprinciple for U [ ρ ] holds. Then G H is continuous on U. This theorem is proved in § § H such that the operator G H is not continuous on Fr [Φ] . This is also an example of the function that has notideally complementing multiplicator. Proof of Proposition 6.1.2.2. Let g be an ideally complementing multiplicator ofthe function Π = G Φ . We denote(6.1.3.1) θ ( z ) := ( gG Φ)( z ) . Let v g ∈ Fr [ g ] . Let us choose t j → ∞ such that:(log | g | ) t j → v g ; (log | Π | ) t j → v Π ∈ Fr [Π]; (log | θ | ) t j → v θ ∈ Fr [ θ ] . It follows from (6.1.3.1) that v θ = v g + v Π . Since g is a multiplicator of Π , we have(6.1.3.2) v θ ( z ) = v g ( z ) + v Π ( z ) ≤ H ( z ) . Since G is an ideally complementing multiplicator, v Π = v + G H v . So for all z ∈ C (6.1.3.2) implies ( v g + G H v )( z ) ≤ ( H − v )( z ) . Since G H v is the maximal subharmonic minorant, v g ( z ) ≤ v g ( z ) ≡ . Thus (see (3.2.1.1)) we have h g ( φ ) ≡ σ g = max ≤ φ ≤ π h g ( φ ) = 0 (cid:3) In order to prove Theorem 6.1.2.6 we need a number of auxiliary statements. Lemma 6.1.4.1. Let the maximum principle be valid in G for U [ ρ ] and for somecontinuous functions w , w ∈ U [ ρ ] satisfy:a)w is harmonic in G ; b) w ( z ) = w ( z ) outside of G. Then (6.1.4.1) w ( z ) ≤ w ( z ) , z ∈ G. Proof. We set w ( z ) := (cid:26) ( w − w ) + ( z ) , z ∈ G , z / ∈ G. This function is continuous in C and, evidently, subharmonic both in G and in C \ G. Since w ( z ) ≥ , the inequality for the mean values0 = w ( z ) ≤ π π Z w ( z + ǫre iφ ) dφ, z ∈ ∂G, implies the subharmonicity on ∂G. Since G satisfies the maximum principle for U [ ρ ], we have w ≡ 0, which is equivalent to (6.1.4.1). (cid:3) Now we shall dwell on some properties of maximal subharmonic minorantsand, in particular, of w = G H v . Let(6.1.4.2) E v := { z ∈ C : G H v ( z ) = m ( z, v, H ) } We remark that m ( z, v, H ) is a δ -subharmonic function in C whose charge will bedenoted by ν ( • , v ), its positive and negative parts will be denoted by ν + and ν − .Let us denote by µ H the measure of H ( z ) , It is decomposed into the productof measures (see § µ h = ∆ H ⊗ ρr ρ − dr, where ∆ H is the measure on the unit circle and ρr ρ − dr is the measure on the ray.It is obvious that(6.1.4.4) ν + ( • , v ) ≤ µ H ( • ) . We shall denote the mass distribution of w ∈ U [ ρ ] by µ w . The modulus of continuity of w (if w is continuous) will be denoted ω w ( z, h ) , z ∈ C , h > w ∈ G H U, U ⊂ U [ ρ ] which willbe useful in the sequel: Lemma 6.1.4.2. Let w ∈ G H U .Then1. w ∈ U [ ρ, σ ] where σ = 4 · ρ [max { H ( e iφ ) : φ ∈ [0 , π ] } + 2 σ ] σ = max { v ( z ) | z | − ρ : z ∈ C , v ∈ U } ; ν ( • , v ) | E v to E v is nonnegative, i.e., ν ( • , v ) | E v = ν + ( • , v ) | E v ; 3. outside E v the function w is harmonic, i.e., µ w | C \ E v = 0; µ w is bounded from above by ν + ( • , v ) , i.e., µ w ≤ ν + ( • , v ); G H U is equicontinuous on each compact set, i.e., ω w ( z, h ) ≤ C ( R, σ, ρ ) √ h log(1 /h ) , | z | ≤ R, where C ( R, σ, ρ ) is independent of w ∈ G H U. Proof. Let us prove property 1. We have T ( r, w ) := 12 π π Z w + ( re iφ ) dφ ≤ π r ρ π Z H + ( e iφ ) dφ + π Z v + ( re iφ ) dφ + π Z v − ( re iφ ) dφ . Since v (0) = 0, we have π Z v − ( re iφ ) dφ ≤ π Z v + ( re iφ ) dφ. Therefore T ( r, w ) ≤ [max { H ( e iφ ) : φ ∈ [0 , φ ] } + 2 σ ] r ρ . It is known (see Theorem 2.8.2.3, (2.8.2.5)) that M ( r ) ≤ T (2 r ). So we concludethat w ( z ) ≤ · ρ [max { H ( e iφ ; φ ∈ [0 , π ] } + 2 σ ] | z | ρ = σ | z | ρ . Let us prove property 2. To this end we shall use the following theorem (Grishin’sLemma)[Gr] Theorem A.F.G. Let g be a nonnegative δ -subharmonic function, and let ν g beits charge. Then the restriction ν g | E to the set E = { z : g ( z ) = 0 } is a measure. Applying this theorem to the function g := m ( z, v, H ) − G H v ( z ), we get(6.1.4.5) ν ( • , v ) | E v ≥ µ w | E v , hence, we obtain property 2.Let us prove property 3. Since w and v are upper semicontinuous, and H iscontinuous (see Th.3.2.5.5) , the set { z : ( w + v )( z ) − H ( z ) < } is open.Let us take a neighborhood of an arbitrary point of this set and replace thefunction w within it with the Poisson integral constructed using this function, i.e.,let us sweep out the mass from this neighborhood. The subharmonic function ob-tained would be strictly greater than the initial one, if the latter were not harmonic.This means that the initial w was not the maximal minorant. We have arrived ata contradiction, which proves property 3.Property 4. immediately follows from property 3. and (6.1.4.5).In order to prove property 5. we shall need auxiliary statement which will bestated as lemmas. Let P ( z, φ, R ) := 12 π R − | z | | z − Re iφ | be the Poisson kernel in the disc K R = { z : | z | < R } .Below C ′ s with indices will denote constants. Lemma 6.1.4.3. In the disc K R/ , we have | grad z P ( z, φ, R ) | ≤ C ( R ) , where C ( R ) depends only on R. Exercise 6.1.4.1. Prove this.We shall introduce the notation for the Green function for the Laplace operatorin the disc K R : G ( z, ζ, R ) := log (cid:12)(cid:12)(cid:12)(cid:12) R − ζzR ( z − ζ ) (cid:12)(cid:12)(cid:12)(cid:12) . The disc { ζ : | ζ − z | < t } will be denoted by K z,t . Lemma 6.1.4.4. Let z ∈ K R/ \ K ζ, √ h . Then for a small h | grad z G ( z, ζ, r ) | ≤ C ( R ) / √ h Exercise 6.1.4.2. Prove this.Let us denote µ ( z, t ) := µ ( K z,t ) . Lemma 6.1.4.5. For z ∈ K R/ , < t < R/ , we have µ H ( z, t ) ≤ C ( σ, R ) t. Proof. Applying Th.2.6.5.1 (Jensen-Privalov) to the function H ( z ) , we obtain M H = max { H ( e iφ : φ ∈ [0; 2 π ] } = ∆ H ( T ) /ρ where T is the unit circle.Now µ ( z, t ) ≤ ∆ H ( T ) | z | + t Z | z |− t r ρ − dr ≤ ρ M H R ρ − t ≤ σC ( ρ ) R ρ − t where C ( ρ ) is a constant depending only on ρ . This proves the lemma. (cid:3) Lemma 6.1.4.6. Let h < and suppose that a monotonic function µ ( t ) satisfiesthe condition (6.1.4.6) µ ( t ) < ct Then √ h Z log(1 /t ) µ ( dt ) ≤ (3 / c √ h log h Exercise 6.1.4.3. Prove this integrating by parts and using (6.1.4.6). Lemma 6.1.4.7. Let z ∈ K R/ and ζ ∈ K R . Then | log | ( R − ζz/R || ≤ C ( R ) Exercise 6.1.4.4. Prove this.Now we pass to the proof of assertion 5. from Lemma 6.1.4.2. According tothe F.Riesz theorem (Th.2.6.4.3) we represent w in the circle as(6.1.4.7) w ( z ) = H ( z, w ) − Z K R G ( z, ζ, R ) µ w ( dξdη ) , ζ = ξ + iη, where H ( z, w ) = 12 π π Z P ( z, φ, R ) w ( Re iφ ) dφ. It follows from Lemma 6.1.4.3 and 1. of Lemma 6.1.4.2 that(6.1.4.8) | grad z H ( z, w ) | ≤ C ( R ) 12 π π Z | w | ( Re iφ ) dφ ≤ C ( R )2 σR ρ . We split the integral in (6.1.4.7) into three terms ψ ( z, h ) := Z K r \ K z , √ h G ( z, ζ, R ) µ w ( dξdη ) ,ψ ( z, h ) = Z K z , √ h log | ( R − zζ ) /R | µ ( dξdη ) ψ ( z, h ) Z K z , √ h log | ζ − z | µ ( dξdη ) , where z is an arbitrary fixed point in K R/ . Combining property 4. and inequality (6.1.4.4) we have(6.1.4.9) µ w ( E ) ≤ µ H ( E ) , ∀ E ⊂ K R . For all z ∈ K z , √ h/ Lemma 6.1.4.4 yields(6.1.4.10) | grad ψ ( z, h ) | ≤ C ( r ) σR ρ / √ h. Combining Lemmas 6.1.4.5 and 6.1.4.7 with inequality (6.1.4.9), we get(6.1.4.11) | ψ ( z, h ) | ≤ C ( R ) C ( σ, R ) √ h. Further, from Lemmas 6.1.4.5 and 6.1.4.6, taking into account the factthat log | ζ − z | < , we obtain(6.1.4.12) | ψ ( z, h ) | ≤ (3 / C ( σ, R ) √ h log h. Now consider the difference∆ w := w ( z + ∆ z ) − w ( z ) , | ∆ z | < h < √ h/ . It can be represented as(6.1.4.13) ∆ w = ∆ ψ + ∆ ψ + ∆ ψ + ∆ H ( z, w ) . Choosing h small enough, one may assume that z +∆ z ∈ K √ h/ ,z . Thus, accordingto (6.1.4.11),(6.1.4.14) | ∆ ψ ( z , h ) | ≤ | ψ ( z , h ) | + | ψ ( z + h, h ) | ≤ C ( σ, R ) √ h. Likewise (6.1.4.12) yields(6.1.4.15) | ∆ ψ ( z , h ) | ≤ | ψ ( z , h ) | + | ψ ( z + h, h ) | ≤ C ( σ, R ) √ h log h. Finally, from (6.1.4.10) and (6.1.4.8), respectively, we obtain(6.1.4.16) | ∆ ψ | ≤ C ( σ, R ) √ h, | ∆ H ( z , w ) | ≤ C ( σ, R ) h. Substituting (6.1.4.14)-(6.1.4.16) into (6.1.4.13), we obtain relation 5. of Lemma6.1.4.2. (cid:3) The lemma is proved. In this item we are going to prove Theorem 6.1.2.6. However, before that,we prove Lemma 6.1.5.1. Let w n = G H v n , v n D ′ → v and w n D ′ → w ∞ . Set E ∞ := { z : w ∞ ( z ) = H ( z ) − v ( z ) } . Then w ∞ is harmonic in C \ E ∞ . Let us note that w ∞ , in general, is not the maximal subharmonic minorantbecause the operator G H can be only upper semicontinuous, as will be demonstratedby example in § w ∞ is a minorant of H − v because of Th. 6.1.1.2,1. Proof. Let z / ∈ E ∞ . Then there exists a δ > w ∞ ( z ) + v ( z ) ≤ H ( z ) − δ Since the function b ( z ) := w ∞ + v ( z ) − H ( z ) is upper semicontinuous, there existsan ǫ = ǫ ( δ ) such that b ( z ) < − δ for all z ∈ {| z − z | < ǫ } . Let ( • ) ǫ be a smoothing operator from (2.6.2.3). If w n D ′ → w then ( w n ) ǫ → w uniformly on every compact set (Th.2.3.4.5, reg3) and for every subharmonicfunction v the sequence v ǫ ( z ) ↓ v ( z ),when ǫ ↓ b ) ǫ ( z ) < − δ, for | z − z | < ǫ or ( w ∞ ) ǫ ( z ) + ( v ) ǫ ( z ) ≤ ( H ) ǫ ( z ) − δ. Thefunction H is continuous, hence uniformly continuous on the circle { z : | z − z | ≤ ǫ } .Thus we can replace ( H ) ǫ in the last inequality with H and δ with δ/ . So, we have(6.1.5.1) ( w ∞ ) ǫ ( z ) + v ǫ ( z ) ≤ H ( z ) − δ/ , | z − z | < ǫ. Since ( • ) ǫ is monotonic on subharmonic functions, we can replace ǫ in (6.1.5.1) withany ǫ < ǫ. So we obtain(6.1.5.2) ( w ∞ ) ǫ ( z ) + v ǫ ( z ) ≤ H ( z ) − δ/ , | z − z | < ǫ Since ( w n ) ǫ → ( w ∞ ) ǫ uniformly in the disc | z − z | ≤ ǫ we can replace in (6.1.5.2) w ∞ with w n and respectively v with v n , changing δ/ δ/ . After that we canpass to the limit as ǫ ↓ n. So we obtain w n ( z ) + v n ( z ) ≤ H ( z ) − δ/ , | z − z | < ǫ It means that the disc {| z − z | < ǫ } ⊂ C \ E v n . Because of Lemma 6.1.4.2, 3. w n is harmonic in this disc for all large n. Thus w ∞ is also harmonic, as the D ′ -limitof w n . (cid:3) Proof of Theorem 6.1.2.6. Let v n D ′ → v . Then the set w n = G H v n is equicontinuousby Lemma 6.1.4.2, 5., and we can choose from it a subsequence uniformly convergingto a continuous function w ∞ . Let w = G H v, E = E w , E ∞ being defined in Lemma6.1.5.1.Since ( w ∞ + v )( z ) ≤ H ( z ) , ( w + v )( z ) ≤ H ( z ) , ∀ z ∈ C and v is upper semicontinuous, whereas w and H are continuous, the sets E and E ∞ are closed.Since w ∞ ( z ) ≤ H ( z ) − v ( z ) we have(6.1.5.3) w ∞ ( z ) ≤ w ( z ) , ∀ z ∈ C , and therefore E ∞ ⊂ E. The function w is subharmonic in C \ E ∞ , and w ∞ is harmonic in C \ E ∞ by Lemma 6.1.5.1. They take the same values on E ∞ . As the maximum principleholds in H w ∞ by assumption we have, according to Lemma 6.1.4.1 the inequality(6.1.5.4) w ( z ) ≤ w ∞ ( z ) , ∀ z ∈ C . The inequalities (6.1.5.4) and (6.1.5.3) imply that w ( z ) = w ∞ ( z ) , i,e., G H is con-tinuous. (cid:3) Proof of Theorem 6.1.2.5. Sufficiency. We exploit the following criterion for exis-tence of a limit set that follows from Theorems 4.2.1.1, 4.2.1.2, 4.3.1.2 and Corollary5.3.1.5: Proposition 6.1.6.1. In order that U ⊂ U [ ρ ] be a limit set of an entire function f ∈ A ( ρ ( r )) it is necessary and sufficient that there exists a piecewise continuous, ω -dense in U asymptotically dynamic pseudo-trajectory (a.d.p.t) v ( •| t ) . Exercise 6.1.6.1. Check this.Let v Φ ( •| t ) be an a.d.p.t. corresponding to Fr Φ . Consider the pseudo-trajectory v g ( •| t ) := G H v Φ ( •| t ) . It exists because of (6.1.1.1). Prove that this pseudo-trajectoryis asymptotically dynamical, i.e., (4.3.1.1) is fulfilled. Recall that T τ • = ( • ) [ e τ ] . Using the property of invariance of G H (Th 6.1.1.2, 2.) we have T τ v g ( •| e t ) − v g ( •| e t + τ ) = G H [ T τ v Φ ( •| e t ) − v Φ ( •| e t + τ )] . Thus (4.3.1.1) is fulfilled because of continuity of G H . Also the condition of ω -denseness (4.3.1.4) is fulfilled and { w ∈ U [ ρ ] : ( ∃ t j → ∞ ) w = D ′ − lim v g ( •| e t j ) } = G H ( Fr Φ)The corresponding entire function g ∈ A ( ρ ( r )) with the limit set U g = G H ( Fr Φ) isan ideally complementing multiplicator, because Fr [ g Φ] = { v + G H v : v ∈ Fr [Φ] } . Exercise 6.1.6.2. Check this.Necessity. Let G be an ideally complementing multiplicator of Φ . Let us showthat G H is continuous on Fr [Φ] . Assume this is not the case, i.e., there existsa sequence v j → v such that G H v j → W and W = G H v. Since the limit set Fr [ G Φ] is closed, we have v j + G H v j → v + G H v, v j ∈ Fr [Φ]. On the other hand , v j + G H v j → v + W. Thus, W = G H v that is a contradiction. (cid:3) Proof of Theorem 6.1.2.3. Let Fr [Φ] be a periodic limit set, that is Fr [Φ] = C ( v ) = { v [ t ] : 1 ≤ t ≤ e P } , where v ∈ U [ ρ ] . We shall show that G H is continuous on U [ ρ ] . By Theorem 6.1.1.2,2) the equality ( G H v ) [ t ] = G H v [ t ] holds. Since the operation ( • ) [ t ] is continuous forall t, G H is continuous on C ( v ) . (cid:3) Now we are going to prove Theorem 6.1.2.4. However, we need some prepa-rations.Let h ( φ ) , φ ∈ [0 , π ) be a 2 π -periodic ρ -t.c.function, satisfying the conditionmax φ ∈ [0 , π ] h ( φ ) = σ We denote this class as T C [ ρ, σ ] and denote T C [ ρ ] := [ σ> T C [ ρ, σ ] . The class of function w = h − h where h , h ∈ T C [ ρ, σ ] will be denoted as δT C [ ρ, σ ] and denote also δT C [ ρ ] := [ σ> δT C [ ρ, σ ] . From properties of ρ -t.c.function (see §§ δ − ρ t.c.functions: Proposition 6.1.7.1. For w ∈ δT C [ ρ ] the following holds:1. w ′ ( φ − and w ′ ( φ + 0) exist at each point and are bounded in [0; 2 π ]; w ′ ( φ − 0) = w ′ ( φ + 0) for all φ ∈ [0; 2 π ] , except, perhaps, a countable set;3.the charge ∆ w generated by the function ∆ w := w ′ ( φ ) + ρ Z φ w ( θ ) dθ has bounded variation | ∆ w | ; the variation | ∆ w | ( α, β ) of the charge on the interval ( α ; β ) and the variation of the charge generated by derivative | w ′ | ( α ; β ) on the sameinterval satisfy the relation: | ∆ w | ( α, β ) ≥ | w ′ | ( α ; β ) + ρ ( β − α ) . w ∈ δT C [ ρ, σ ]max( | w ′ ( φ − | , | w ′ ( φ + 0) | ) ≤ C ( ρ, σ ) , φ ∈ [0; 2 π ]; r ρ w n D ′ → r ρ w and w n ∈ δT C [ ρ, σ ] , then w n → w uniformly on [0; 2 π ] . Exercise 6.1.7.1. Prove this using properties of ρ -t.c.functions.We also need a technical Lemma 6.1.7.2. Let M n ( φ ) be a sequence of functions which satisfies the condi-tions:1. M n ≥ M n (0) = 0; M n converges uniformly to M ∞ ( φ ) ≥ A sin ρφ, A > M ′ n ( φ − , M ′ n ( φ + 0) exist at every point, and they coincide almost every-where;4.there exists a sequence φ n ↓ such that for each arbitrarily small ǫ > andarbitrarily large n ∈ N there exists n > n for which the inequality M n ( φ n ) < ǫφ n holds.Then there exists a sequence ( ζ n , η n ) of disjoint intervals and a subsequence M k n such that (6.1.7.1) M ′ k n ( ζ n ) − M ′ k n ( η n ) ≥ Aρ/ . Proof. Set ǫ = 1 / , η = π/ ǫ n , η n , ζ n be already chosen. Set ǫ n +1 = ǫ n / , find φ n +1 < η n and choose k = k ( n )so that for k > k M k ( φ n +1 ) − Aρφ n +1 > − ǫ n +1 φ n +1 . This is possible because of 2. and sin ρφ ∼ φ, φ → . So we have(6.1.7.2) M k ( φ n +1 ) φ n +1 > Aρ − ǫ n +1 . Now, choose ψ n +1 < φ n +1 and k n +1 > k so that(6.1.7.3) M k n +1 ( ψ n +1 ) < ǫ n +1 ψ n +1 This is possible by 4. Thus for small ǫ n +1 from (6.1.7.2) and (6.1.7.3) we obtain(6.1.7.4) M k n +1 ( φ n +1 ) − M k n +1 ( ψ n +1 ) φ n +1 − ψ n +1 > (2 / Aρ On the other hand(6.1.7.5) M k n +1 ( ψ n +1 ) − M k n +1 (0) ψ n +1 − < ǫ n +1 On the interval ( ψ n +1 , φ n +1 ) there is a point η n +1 where the derivative exists andthe inequality(6.1.7.6) M ′ k n +1 ( η n +1 ) ≥ M k n +1 ( φ n +1 ) − M k n +1 ( ψ n +1 ) φ n +1 − ψ n +1 is valid. Also there is a point ζ n +1 ∈ (0 , ψ n +1 ) where derivative exists and theinequality(6.1.7.7) M ′ k n +1 ( ζ n +1 ) ≤ M k n +1 ( ψ n +1 ) − M k n +1 (0) ψ n +1 − (cid:3) Proof of Theorem 6.1.2.4. Denote by ˆ G H h the maximal ρ -t.c.minorante of H ( e iφ ) − h ( φ ) . It follows from Th.6.1.1.2, 2. that G H ( r ρ h ( φ ))( re iφ ) = r ρ ˆ G H h ( φ ) Exercise 6.1.7.2. Prove this.So taking in consideration Proposition 6.1.7.1, 5., one must prove Proposition 6.1.7.3. The operator ˆ G H is continuous on the set ˆ U ind := { h ( φ, c ) : c ∈ C } in the uniform topology.Proof. Let h n → h, h n , h ∈ ˆ U ind . Denote ˆ w n = ˆ G H h n , ˆ w = ˆ G H h, ˆ w ∞ = lim n →∞ ˆ w n . We set also ˆ M n = H − h n − ˆ w n , ˆ M ∞ = H − h − ˆ w ∞ , ˆ M = H − h − ˆ w. Let ( α n ; β n )be a maximum interval where ˆ M n ( φ ) > . We shall show that β n − α n ≤ π/ρ. Indeed, for a fixed n let us consider the functionˆ W n := ˆ w n + ǫ n L ( φ − ( α n + β n ) / L ( φ ) = (cid:26) cos | φ | , φ ∈ ( − π/ ρ ; π/ ρ );0 , φ ∈ [ − π ; π ] \ ( − π/ ρ ; π/ ρ ) , and ǫ n is small enough. If β n − α n > π/ρ , then ˆ W n is also a ρ -t.c. minorant of H − h n , i.e., ˆ w n is not maximal.If β n − α n = π/ρ, then, to ensure that ˆ w n is a maximal minorant, at least oneof the conditions(6.1.7.8) lim inf φ → α n +0 ˆ M n ( φ ) φ − α n = 0 , lim inf φ → β n − ˆ M n ( φ ) β n − φ = 0 . must be satisfied.Let us choose (and preserve the previous notation) a subsequence ˆ M n ( φ ) forwhich α n → α, and β n → β .If β − α < π/ρ, then the maximum principle for ρ -t.c.functions is valid. Re-peating arguments of proof of Theorem 6.1.2.6, we obtain ˆ w ∞ = ˆ w for all φ , whichproves Proposition 6.1.7.3 for the case considered. Exercise 6.1.7.3. Repeat them.Consider the case when β − α = π/ρ. Set q ( φ ) = ( w − w ∞ )( φ ) . The function q is ρ -trigonometric on the interval( α ; β ) since ˆ w and ˆ w ∞ are ρ -trigonometric, i.e., have the form A sin ρφ + B cos ρφ. Exercise 6.1.7.4. Explain this.Besides, we have q ≥ q ( α ) = q ( β ) = 0 . It is easy to see that q has theform(6.1.7.9) q ( φ ) = A sin ρ ( φ − α ) , A > . Exercise 6.1.7.5. Prove this.Since ˆ M ( φ ) ≥ , we have ˆ M ∞ ( φ ) = ˆ M ( φ ) + ( ˆ w − ˆ w ∞ )( φ ) ≥ ( ˆ w − ˆ w ∞ )( φ ) , ∀ φ, whence(6.1.7.10) ˆ M ∞ ( φ ) ≥ A sin ρ ( φ − α ) , A > . Since β n − α n ≤ π/ρ, the segment [ α, β ] contains the infinite sequence α n or β n . Let us single out a subsequence, let it be, for example, α n → α + 0 , α n ∈ [ α ; β ].Consider the sequence M n ( φ ) = ˆ M n ( φ − a ) . From the definition of M n and fromrelation 6.1.7.10 it follows that conditions 1. and 2. of Lemma 6.1.7.1 are fulfilled. Condition 3. is fulfilled because of property 1. of Proposition 6.1.7.1. Further, if α n α , then condition 4. of Lemma 6.1.7.1 is trivially true,since M n ( α n − α ) = 0;otherwise, if α n ≡ α , condition 4. follows from (6.1.7.8).Applying Lemma 6.1.7.1, we obtain the union of intervals satisfying (6.1.7.7).The equality H ( φ ) = ˆ M n − h n − ˆ w n yields the following inequality for the measure∆ H :∆ H (( η n ; ζ n )) ≥ Aρ/ . Summing this inequality and taking into account thefact that the intervals do not intersect, we obtain ∆ H ( ∪ n ( η n , ζ n )) = ∞ , which isimpossible. So, Proposition 6.1.7.3 is proved. (cid:3) Hence, Theorem 6.1.2.4 is proved. (cid:3) In this item we show an example of H and an entire function without anideally complementing H - multiplicator.According to Th.6.1.2.5 , to construct such an example it is sufficient to con-struct a limit set on which G H is not continuous.We set L ( η ) = (cid:26) cos | η | , η ∈ ( − π/ ρ ; π/ ρ );0 , η ∈ [ − π ; π ] \ ( − π/ ρ ; π/ ρ ) . Let us define X ∈ C ∞ so that X ( ξ ) = 1 for ξ < X = 0 for ξ > α. We set(6.1.8.1) κ := (1 /ρ ) max ( −∞ ;+ ∞ ) [2 ρX ′ + X ′′ ]( ξ ) , H ( η ) := L ( η ) + κ. We also set v ( ζ, c ) := [ H − X ( ξ − c ) L ( η )] e ρξ , ζ = ξ + iη, where H and L have been periodically extended from the interval [ − π ; π ] to( −∞ , + ∞ ) . As H ( z ) we take H ( z ) := H ( φ ) r ρ Lemma 6.1.8.1. We have (6.1.8.2) v (log z, c ) ∈ U [ ρ, σ ] , σ = 1 + κ, (6.1.8.3) G H v ( • , c ) ≡ , (6.1.8.4) lim c →∞ v (log z, c ) = κr ρ uniformly with respect to z ∈ K ⋐ C , and (6.1.8.5) G H ( κr ρ ) = L ( φ ) r ρ . Proof. For the Laplace operators in ζ and z it is true that ∆ ζ = ∆ z / | ζ | . Let uscheck that v ( ζ, c ) is subharmonic in ζ. We have∆ ζ v ( ζ, c ) = { [1 − X ( ξ − c )]( L ′′ + ρ L )( η ) + [ ρ κ − L ( ξ )[ X ′′ ( x − c ) + 2 ρX ′ ( x − c )] } e ρξ Exercise 6.1.8.1. Check this computation.Since X ( ξ ) ≤ L ( η ) is ρ -t.c.[1 − X ( ξ − c )]( L ′′ + ρ L )( η ) ≥ L ( ξ ) ≤ X ′′ ( x − c ) + 2 ρX ′ ( x − c )] ≤ κρ we have[ ρ κ − L ( ξ )[ X ′′ ( x − c ) + 2 ρX ′ ( x − c )] ≥ . Thus v (log z, c ) is subharmonic. Exercise 6.1.8.2. Prove that v (log z, c ) ∈ U [ ρ, σ ] for σ = 1 + κ. Let us prove (6.1.8.3). We have H ( z ) − v (log z, c ) = X (log r − c ) L ( φ ) r ρ . Since X = 0 for r > e c + α , the maximal subharmonic minorant of H − v is zero bythe maximum principle.Relation (6.1.8.4) is obvious, since X (log r − c ) converges to 1 uniformly onevery disc {| z | ≤ R } . Relation 6.1.8.5 follows from the equality H ( z ) − κr ρ = L ( φ ) r ρ ,since L ( φ ) r ρ ∈ U [ ρ ] . (cid:3) Now we pass to the construction of the example. Examine the set U := clos { v (log z, c ) : c ∈ [0; ∞ ) } . It contains the function D ′ − lim c →∞ v (log z, c ) = κr ρ . Let us consider the minimal convex ( • ) [ t ] -invariant set U containing U .The set iscontained in U [ ρ, κ ] . It is a limit set for a certain entire function Φ . Let us showthat G H is not continuous on Fr [Φ]. We take an arbitrary sequence c j → ∞ and set v j ( z ) := v (log z, c j ) ∈ U . Now D ′ − lim j →∞ v j = κr ρ by (6.1.8.4) and G H v j ( z ) = 0 , so D ′ − lim j →∞ G H v j = 0 but G H (lim v j ) = G H ( κr ρ ) = L ( φ ) r ρ Here we prove existence and continuity of maximal subharmonic minorantfor some classes of functions m ( z ) . Theorem 6.1.9.1. Let m ( z ) be a continuous function such that the set of sub-harmonic minorants is nonempty. Then the maximal subharmonic minorant of m exists and is continuous.Proof. The set of subharmonic minorants is not empty and partially ordered. In-deed, for every subset { u α , α ∈ A } of subharmonic minorants there exists u A =(sup { u α : α ∈ A } ) ∗ which is subharmonic and is a minorant of m , because m iscontinuous. Exercise 6.1.9.1. Explain this in details.Thus there exists a uniquely maximal element m.s.m.( z, m ), which is a sub-harmonic minorant of m. Let us prove that it is continuous at every point z . Since m.s.m.( z, m ) is uppersemicontinuous, m.s.m ( z, m ) > m.s.m ( z , m ) − ǫ for | z − z | < δ for arbitrary small ǫ and corresponding δ = δ ( ǫ ) . So we need toprove the inequality m.s.m ( z, m ) < m.s.m ( z , m ) + ǫ for arbitrary small ǫ and corresponding δ = δ ( ǫ ) . Perform sweeping m.s.m ( z, m ) from the disc | z − z | < δ such that the result u ( z, δ ) satisfies the inequality m.s.m ( z, m ) < u ( z, δ ) < m.s.m ( z, m ) + ǫ < m ( z ) + ǫ. Thus u ( z, δ ) − ǫ < m ( z ) . Hence m.s.m ( z, m ) > u ( z, δ ) − ǫ for all z. Since u ( z, δ )is continuous u ( z, δ ) > u ( z , δ ) − ǫ in the disc {| z − z | < δ } . So m.s.m.( z, m ) >u ( z , δ ) − ǫ > m.s.m ( z , m ) − ǫ (cid:3) Theorem 6.1.9.2. Let m = m − m , where m , m are subharmonic functions.Then the maximal subharmonic minorant of m exists. If m is continuous, thenthe maximal subharmonic minorant is continuous.Proof. Set M ǫ ( z, m ) := M ǫ ( z, m ) − M ǫ ( z, m ) , where M ǫ ( z, m i ) , i = 1 , M ǫ ( z, m ) is continuous (see Theorem 2.6.2.3 (Smooth ap-proximation)), there exists m.s.m.( z, M ǫ ( z, m )) . We have u ( z, m ) := lim sup ǫ → m.s.m ( z, M ǫ ( • , m ) ≤ lim ǫ → M ǫ ( z, m ) = m − m ( z ) = m ( z )Now we prove that the upper semicontinuous regularization u ∗ ( z, m ) also satisfiesthe inequality u ∗ ( z, m ) ≤ m ( z ) . Indeed, m + u ( z, m ) ≤ m ( z ) . Hence, M ǫ ( z, m ) + M ǫ ( z, u ( • , m )) ≤ M ǫ ( z, m ) . Passing to limit we obtain three subharmonic function and inequality m ( z ) + u ∗ ( z, m ) ≤ m ( z ) . We prove that u ∗ ( z, m ) is the m.s.m.( z, m ) . If not, there would exist a subharmonicfunction u which exceeds u ∗ ( z, m ) on a set of positive measure (otherwise theycoincide);thus we would have for some z and ǫu ∗ ( z, m ) < M ǫ ( z, u ) ≤ m.s.m. ( z, M ǫ ( z, m )) This contradicts the definition of u ∗ ( z, m ) . Now suppose that m is continuous at a point z . From Th.2.6.5.1 (Jensen-Privalov) we obtain that it is equivalent to Z ǫ µ m ( { z : | z − z | < t } ) t dt = o (1) , ǫ → µ m.s.m ( z,m ) ≤ µ m . Hence m.s.m. ( z, m ) is also continuous. (cid:3) Exercise 6.1.9.2. Prove continuity in details. Corollary 6.1.9.3. For m = m ( z, v, H ) the function G H v ( z ) := m.s.m. ( z, m ) exists and is continuous. Exercise 6.1.9.3. Prove Corollary 6.1.9.3. ρ -trigonometric convexity.6.2.1. One of the important and useful kinds of limit sets is periodic limit sets.They are determined by one subharmonic function v ∈ U [ ρ ] that satisfies the con-dition(6.2.1.1) v ( T z ) = T ρ v ( z ) , z ∈ C Such a function is called automorphic . They generate the class of so called L ρ -subfunctions, that is a generalization of ρ -trigonometrically convex functions. Inthis part we are going to review of properties of such functions from different pointof view, that will be useful for applications (see [ADP]).In connection with property (6.2.1.1) it is natural to consider so called T - homogeneous domains in C , i.e., such domains G that satisfy the condition { T z : z ∈ G } = G or shortly T G = G. As we can see they are invariant with respectdilation by T. For example, every component of an open set of harmonicity of anautomorphic function is a T -homogeneous domain.Let v satisfy (6.2.1.1).Then the function(6.2.1.2) q ( z ) := v ( e z ) e − ρx is 2 π periodic function in y and P -periodic in x, where P = log T. The function q can be considered as a function on a torus T P , obtained byidentifying the opposite sides of the rectangle Π = (0 , T ) × ( − π, π ) . The homology group of T P is nontrivial, and generated by the cycles γ x , γ y ,where γ x = T P ∩ { y = 0 } , γ y = T P ∩ { x = 0 } .Let π be the covering map of C onto T P , then φ = π ◦ log is a well-definedcovering map of C \ { } onto T P , where the group of deck transformations is givenby the dilations by T m for m ∈ Z . So if G is a given T -homogeneous domain, then(6.2.1.3) D = π ◦ log G = φ ( G )is a domain in T P . On the other hand, not every domain in T P has a T -homogeneousdomain as its preimage under φ . The preimage φ − ( D ) under φ is a possibly disconnected set which is invariant under dilations by T m for m ∈ Z . An intrinsicdescription is given by the next proposition. Proposition 6.2.1.1. Let γ be a closed curve in a domain D ⊂ T P that is homol-ogous in T P to a cycle γ = n x γ x + n y γ y , n x , n y ∈ Z . Then1. If n x = 0 for every such γ in D, , then φ − ( D ) = ∪ ∞ j = −∞ G j , where G j = T j G , G is an arbitrary connected component of φ − ( D ) , and G j ∩ G l = ∅ for j = l. 2. If there exists a curve γ as above with n x = 0 , then φ − ( D ) = ∪ k − q =0 G q , where k = min | n x | with the minimum taken over all such curves γ ; G is anarbitrary component of φ − ( D ); G j , j = 0 , , ..., k − , are disjoint T k -homogeneousdomains, and for every m ∈ Z , T m G = G q , provided m = lk + q, for some q ∈ Z , ≤ q ≤ k − , l ∈ Z . We call domains as in part 2 of Proposition 6.2.1.1 connected on spirals . Inparticular, this proposition shows that for every D connected on spirals, we canfind a connected T k - homogeneous domain that relates to D by (6.2.1.3).Let us give some examples. The domain D ′ = T P ∩ {| x − P/ | < P/ } is notconnected on spirals, whereas D ′′ = T P ∩ {| y | < π/ } is. It follows that D ′ ∩ D ′′ isnot connected on spirals whereas D ′ ∪ D ′′ is.The situation can be more complicated. Set x ′ ( x, y, α ) := x cos α + y sin α ; y ′ ( x, y, α ) := − x sin α + y cos α ; 0 ≤ α < π/ P := (1 / | x ′ ( P, π, − α ) | ; P := (1 / | y ′ ( P, π, − α ) | . Then R ′ = { z ′ = x ′ + i ′ y ′ : − P < x ′ < P ; − P < y ′ < P } is a fundamentalrectangle for T P in the corresponding coordinates. Set f ( y ′ ) := ( P − y ′ ) − − ( y ′ + P ) − and D , := { z ′ : − P < y ′ < P ; f ( y ′ ) < x ′ < f ( y ′ ) + d } where 0 < d < P . Then the domains D l,m := D , + 2 P l + 2 P mi ′ , l, m ∈ Z are disjoint, and theirunion D determines a domain ˆ D ⊂ T P . This ˆ D is determined completely by theintersection of D with the rectangle R = (0 , P ) × ( − π, π ) . The domain ˆ D is notconnected on spirals.One more example. Consider the family of lines L l := { z = x + iy : y = π/ ( kP ) x + lπ/k, x ∈ R } , l ∈ Z . It determines a closed curve (spiral) γ on T P with n = k. The open set D k = { z : | z − ζ | < ǫ, ζ ∈ L l , l ∈ Z } , < ǫ < P/ √ π + k , determines a domain ˆ D k on T P that is connected on spirals, and such that φ − ( ˆ D k )consists of k components, every one of them T k -homogeneous.Since the function v in (6.2.1.1) is subharmonic, the function q of (6.2.1.2) isupper semicontinuous and in the D ′ topology on T P satisfies the inequality L ρ q ≥ L ρ := ∆ + 2 ρ ∂∂x + ρ . Such functions q are called subfunctions with respect to L ρ , or L ρ - subfunctions . L ρ q is a positive measure on T P . The operator L ρ arises naturally by changing variables z log z in the Laplaceoperator ∆ ζ . Exercise 6.2.1.1 Check this. Set ζ = e z . Let us note that if q depends only on the variable y, it is a 2 π -periodic ρ -trigonometric convex function because L ρ turns into T ρ = ( • ) ′′ + ρ ( • ) (c.f. § Consider the solution of the homogeneous boundary problem(6.2.2.1) L ρ q = 0 in D ; q (cid:12)(cid:12) ∂D = 0 , where D is a domain in T P and q is bounded in a neighborhood of ∂D with boundaryvalue zero quasi-everywhere. This is a spectral problem for a pencil of differentialoperators ( [Ma]).A solution of this problem can be defined for an arbitrary domain D ⊂ T P with a boundary of positive capacity. The spectrum of the problem (6.2.2.1) consists of those (complex) ρ for which(6.2.2.1) holds for some function q 0. The minimal positive point of the spectrum ρ ( D ) exists iff there exists the spectrum. The spectrum exists iff the domain D is connected on spirals. In this case ρ ( D ) is the order of the minimal harmonicfunction in everyone of the domains G i that corresponds to D by Prop.6.2.1.1.The quantity ρ ( D ) is strictly monotonic . It means that if two domains D , D ∈ T P are such that D ⊂ D and the capacity of D \ D is positive, then ρ ( D ) <ρ ( D ) . For example, thus is the case of D = {| y | < d, d < π } and D is the samestrip without the segment { it : 0 ≤ t ≤ d } .In connection with the problem of multiplicator we considered the maximalsubharmonic minorant of a function m = H − v where v is a T - automorphicfunction. From Th.6.1.1.2, 2. we can obtain that if v is a T - automorphic function,then G H v is also T - automorphic. Exercise 6.2.2.1 Check this.Thus for this case finding D H in Th. 6.1.1.1 is reduced to finding a maximal L ρ -subfunction q that satisfies the inequality(6.2.2.2) q ( z ) ≤ m ( z ) := [ H ( e z ) − v ( e z )] e − ρx , z ∈ T P . We say that m ( z ) has an L ρ − subminorante. The idea of ρ ( D ) give possibility for the Theorem 6.2.2.1. If m has a non-zero L ρ -subminorant, then ρ ( D ) ≤ ρ for somecomponent D of the open set M + := { z : m ( z ) > } . Conversely, if ρ ( D ) < ρ (strict inequality) for some component D of the set M + , and m ( z ) ≥ for all z ∈ T P , then m has a non-zero L ρ -subminorant. Exercise 6.2.2.2 Prove that M + is open. If ρ / ∈ Z , the operator L ρ has a fundamental solution E ρ ( • − ζ ) in T P , where ζ is a shift by the torus, i.e., by the modulus P + i π. It means that L ρ E ρ ( • − ζ ) = δ ζ , in D ′ ( T P ) , where δ ζ is the Dirac function, concentrated at ζ. If ρ ∈ Z , there exists, like for operator T ρ and spherical operator (see Th.3.2.4.2,Th.3.2.6.3), a generalized fundamental solution E ′ ρ that satisfies the equation L ρ E ′ ρ ( • − ζ ) = δ ζ − cos ρ ( y − η ) , ζ = ξ + iη in D ′ ( T P ) . Theorem 6.2.3.1. Let ρ > , ρ / ∈ Z . Then every L ρ -subfunction on T P can berepresented in the form (6.2.3.1) q ( z ) = Z T P E ρ ( z − ζ ) ν ( dζ ) , where ν = L ρ q. This theorem is the counterpart of Th.’s 3.2.3.3, 3.2.6.2. Theorem 6.2.3.2. Let ρ > , ρ ∈ Z . Then the mass distribution ν = L ρ v satisfiesthe condition (6.2.3.2) Z T P e ± iρy ν ( dz ) = 0 , and the representation (6.2.3.3) q ( z ) = ℜ ( Ce iρy ) + Z T P E ′ ρ ( z − ζ ) ν ( dζ ) holds with C that is a complex scalar. This theorem is the counterpart of Th.’s 3.2.4.2, 3.2.6.2.Let D ⊂ T P and ρ ( D ) > ρ. Then the operator L ρ has in D the Green function − G ρ ( z, ζ, D ) . Thus for every q that is a L ρ -subfunction in D and bounded fromabove in D we have the representation(6.2.3.4) q ( z ) = g ( z ) − Z D G ρ ( z, ζ, D ) ν ( dζ ) , in which ν = L ρ q and g is the minimal majorant on ∂D of the function q , satisfying L ρ g = 0 in D. This is the counterpart of Th.2.6.4.3 (F.Riesz representation) and Th.3.2.5.1.From (6.2.3.4) one can easily obtain Theorem 6.2.3.3. (Maximum principle). If ρ ( D ) > ρ and q ( z ) is an L ρ -subfunction such that q ( z ) ≤ , z ∈ ∂D, then q ( z ) ≤ , z ∈ D. Exercise 6.2.3.1 Prove this. Theorem 6.2.3.4. An L ρ -subfunction in T P can not attain zero maximum if it isnot zero identically. Exercise 6.2.3.1 Prove this exploiting (6.2.1.2) and properties of subharmonicfunctions. Theorem 6.2.3.5. Let q be an L ρ -subfunction in T P . If q ( z ) ≤ for z ∈ T P then q ( z ) ≡ . Exercise 6.2.3.2 Prove this using Th.3.1.4.7 (**Liouville). Proposition 6.2.3.6. Let q D be the solution of the problem (6.2.2.1) in a domain D with a smooth boundary, corresponding to ρ = ρ ( D ) . Suppose that q D ( z ) = 1 for some z ∈ D. Then ∂q D ∂n > , ∀ z ∈ ∂D. Exercise 6.2.3.3 Prove this, using properties of positive harmonic functions. In the part devoted to completeness of exponential system ( § U [ ρ ] . A function v ∈ U [ ρ ] is called minimal if the function v − ǫr ρ has no subharmonic minorant forarbitrary small ǫ > . If v is T - automorphic, the corresponding L ρ -subfunction q iscalled minimal if the function q − ǫ has no L ρ -subminorant in T P . We formulate onesufficient condition for minimality and one sufficient condition for nonminimality. Theorem 6.2.4.1. Let H ρ ( q ) be the maximal open set on which L ρ q = 0 . If thereexists a connected component M ⊂ H ρ ( q ) such that ρ ( M ) < ρ, then q is a minimal L ρ -subfunction. For example, q ≡ Proposition 6.2.4.2. The function q is nonminimal if q ( z ) ≥ c or L ρ q − c > for some positive c for all z ∈ T P . For example, q ≡ c > Let Λ := { λ k } , k = 1 , , ... be a set of points in the complex plane C , satisfying the condition λ k = 0 and λ j = λ k , if k = j. Consider the canonical product(6.3.1.1) Φ Λ ( λ ) := Y k (1 − λ/λ k ) exp λ/λ k We suppose in this § that Φ Λ ( λ ) is an entire function of order one and normaltype, i.e., a function of exponential type (see [Levin, Ch.1, § Exercise 6.3.1.1 Formulate this theorem for entire functions of order one andnormal type under assumption that ρ ( r ) ≡ . We will suppose that the upper density of zeros (see § § Λ > . Let G ⊂ C be a convex bounded domain containing zero . This last requestdoes not restrict any of the further considerations connected to completeness, be-cause exp Λ := { e λ j z : λ j ∈ Λ } can be replaced by the system { e λ j ( z − z ) : λ j ∈ Λ } and e λ j ( z − z ) = C j e λ j z . Let A ( G ) be the space of holomorphic functions in G withthe topology of uniform convergence on compacts. We will study the completenessof the exponential systems(6.3.2.1) exp Λ := { e λ j z : λ j ∈ Λ } in A ( G ) . We will be interested in the following questions:1. completeness of exp Λ in A ( G ) . maximality of G for exp Λ , which is complete in A ( G );3. extremal overcompleteness of exp Λ in A ( G ) for a maximal G. Let us give a precise definitions of maximality and extremal overcompleteness.The completeness means that every function f ∈ A ( G ) can be approximated on ev-ery compact set K ⋐ G with arbitrary precision by linear combinations of functionsfrom exp Λ . A convex domain G is called maximal for a system exp Λ , which is complete in A ( G ) if for every domain G such that G ⋐ G exp Λ is not complete in A ( G ) . A system exp Λ is called extremely overcomplete in A ( G ) for a maximal G, iffor every sequence Λ := { λ j } such that Λ ∩ Λ = ∅ and ∆ Λ > G isnot maximal for the system exp Λ ∪ Λ . Another words, every essential enlargement of an extremely overcomplete sys-tem enlarges also the maximal domain of completeness. Let h Λ ( φ ) := lim sup r →∞ log | Φ Λ ( re iφ ) | r − be the indicator of Φ Λ . It is 1-trigonometrically convex function or simply trigono-metrically convex function (t.c.f). Let G Λ be the conjugate indicator diagram ofΦ Λ , i.e., a convex domain of the form G Λ := { z : max z ∈ G Λ ℜ ( ze iφ ) ≤ h Λ ( φ ) } . Let us describe conditions for completeness, maximality and extremal over-completeness when Λ is a regular set (see § Λ is a CRG - function (see § G Λ is enclosed in G if it can be enclosed in G by parallel translation,it is enclosed with sliding , if it can be moved after enclosing only in one direction, enclosed rigidly if it is impossible to move after enclosing, freely enclosed in everyother case of enclosing. Theorem 6.3.3.1. Let Λ be a regular set.Then the following holds:1. { exp Λ is not complete in A ( G ) } ⇐⇒ { G Λ is freely enclosed to G } ;2. { G is maximal for exp Λ } ⇐⇒ { G Λ is not freely enclosed to G } ;3. { exp Λ is extremely overcomplete in A ( G ) } ⇐⇒ { G Λ is enclosed rigidly in G } . Let us note that G is maximal for exp Λ but not extremely overcomplete if andonly if G Λ is enclosed with sliding in G. This theorem is a corollary of the more general Theorem 6.3.4.1, but will beproved independently in § If Λ is not regular, it is natural to exploit the notion of limit set (see § . Suppose the limit set of Φ Λ has the form Fr [Φ Λ ] := { v ( λ ) = | λ | ( ch + (1 − c ) h )(arg λ ) : c ∈ [0; 1] } where h , h are t.c.f.Such limit set is a particular case of U ind (6.1.2.2). It is called indicator limitset and it is indeed a limit set of an entire function (see Exercise 6.1.2.4)).The asymptotic behavior of the set Λ (i.e., the limit set of the correspondingmass distribution) can be described completely using Th.3.1.5.2. Exercise 6.3.4.2. Do that.We will call such Λ an indicator set . Denote by G , G the conjugate diagramof h , h . Since G , G are convex, the set αG + βG := { αz + βz : z ∈ G , z ∈ G } , α, β > h := αh + βh . Theorem 6.3.4.1. Let a set Λ be an indicator set.Then the following holds:1. { exp Λ is not complete in A ( G ) } ⇐⇒ { G and G are freely enclosed to G } ;2. { G is maximal for exp Λ } ⇐⇒ { G and G are enclosed in G and at leastone of them is not freely enclosed to G } ;3. { exp Λ is extremely overcomplete in A ( G ) } ⇐⇒ { cG + (1 − c ) G is enclosedrigidly in G ∀ c ∈ [0; 1] } . This theorem is proved in § h Λ = max( h , h )Thus the conjugate diagram G Λ of the function h Λ is the convex hull of G and G . Let us note that the indicator h Λ does not determine the completeness of thesystem exp Λ if Λ is not regular set, as the following example shows Example 6.3.4.1 . Let G := { z = x + iy : x = 1; − ≤ y ≤ } ,G := { z = x + iy : x = − − ≤ y ≤ } , and G = { z : | z | < ǫ } with a small ǫ. Exercise 6.3.4.3 Prove that G and G are freely enclosed to G and theirconvex hull is not enclosed.Let Λ be a set such that the interior of G Λ coincides with G. If Λ is regularset then exp Λ is complete in A ( G ), G is maximal for exp Λ and exp Λ is extremelyovercomplete in A ( G ) . If Λ is an indicator set, then the first two assertions hold but exp Λ can be notextremely overcomplete: Example 6.3.4.2 Set G := { z = x + iy : − ≤ x ≤ y = 0 } ; G := { z = x + iy : x = 1; − ≤ y ≤ } . Here G Λ is triangle in which G is freely enclosed and G is rigidly enclosed, but cG + (1 − c ) G is free enclosed for all c : 0 < c < . Exercise 6.3.4.4 Check this. Example 6.3.4.3 Set G := { z = x + iy : x = − y ∈ [ − 1; 1] } ; G := { z = x + iy : x = 1; y ∈ [ − 1; 1] } . Exercise 6.3.4.5 Check that G and G are inclosed with sliding in G Λ . If G and G are rigidly enclosed in G it does not imply in general that cG +(1 − c ) G are rigidly enclosed for all c ∈ [0; 1] . Example 6.3.4.4 Let G be an equilateral triangle inscribed in the circle | z | = 1 , let G be the same triangle rotated by the angle π/ , and let G be theunit disc. Exercise 6.3.4.6 Show that ( G + G ) is freely enclosed in G. If G ∩ G is rigidly enclosed in G then cG + (1 − c ) G is rigidly enclosed for c ∈ [0; 1] . Exercise 6.3.4.7 Check this.However this is not a necessary condition. Example 6.3.4.5 Set G := { z = x + iy : | x | < | y | < } ; G := { z = x + iy : x ∈ ( − , − x > y > − } ; G := { z = x + iy : x ∈ ( − , − < y < x } . Exercise 6.3.4.8 Check that every triangle cG + (1 − c ) G is rigidly enclosedin G and G ∩ G is freely enclosed. Consider in more details the conditions for extremal overcompleteness inthe case when Λ is indicator set and G Λ = G or, in other words, if(6.3.5.1) h Λ = h G . We can suppose that h and h are linearly independent, otherwise we exploitTheorem 6.3.3.1. If, for example, the inequality h ( φ ) ≤ h ( φ ) , ∀ φ, holds, theextremal overcompleteness is in the case when G is rigidly enclosed in G because G ∩ G = G , and this case was mentioned above (Exercise 6.3.4.7).Consider the general case. Denote g ( φ ) := | h − h | ( φ ), and set Θ Λ := { φ : g ( φ ) > } . This is an open set on the unit circle. Denote as I Λ := ( α , α )the maximal interval contained in Θ Λ and denote by d Λ its length. Since g ( φ ) iscontinuous(6.3.5.1) g ( α j ) = 0 , j = 1 , . If also at least one of the conditions:lim inf φ ∈ I Λ ,φ → α j g ( φ ) φ − α j = 0 , j = 1 , , is fulfilled we say g is zero with tangency on ∂I Λ . Theorem 6.3.5.1. Suppose Λ is an indicator set that satisfies (6.3.5.1). In orderthat exp Λ be extremely overcomplete in A ( G ) it is necessary and sufficient that atleast one of the following condition holds:1. d Λ < π ; d Λ = π and g is zero with tangency on ∂I Λ . This theorem is proved in § We call Λ periodic if Fr [Φ Λ ] is a periodic limit set (see Th.4.1.7.1). In suchcase all the limit set is determined by one subharmonic function v ∈ U [1] (see(4.1.3.1)). Let us characterize the system exp Λ for periodic Λ . Set(6.3.6.1) h G ( φ ) := max {ℜ ( ze iφ ) : z ∈ G } (6.3.6.2) m ( λ, G, v ) := | λ | h G (arg λ ) − v ( λ )Denote as G G v -the maximal subharmonic minorant of the function m ( λ, G, v )A function w ∈ U [1] is called minimal if the function w − ǫ | λ | has no subharmonicminorant in U [1] for every small ǫ > 0. The harmonic function of the form(6.3.6.3) H ( λ ) := | λ | ( A cos(arg λ ) + B sin(arg λ )) , for example, is minimal.We will denote as HARM the set of the functions of the form (6.3.6.3) . Theorem 6.3.6.1. Let Λ be a periodic set. The following holds:1. { exp Λ is not complete in A ( G ) } ⇐⇒ {G G v exists and is non minimal } ;2. { G is maximal for exp Λ } ⇐⇒ {G G v exists and is minimal } ;3. { exp Λ is extremely overcomplete in A ( G ) } ⇐⇒ {G G v ∈ HARM } . This theorem is proved in § Let us characterize the completeness of exp Λ for periodic Λ in other terms.For this we need the information that was presented in § ρ = 1 . Denote(6.3.7.1) q Λ ( z ) := v Λ ( e z ) e − x (compare with (6.2.1.2)). As it was explained in § L -subfunction on the torus T P . Set m ( z, G, q Λ ) = h G ( y ) − q Λ , D ( G, Λ) := { z : m ( z, G, q Λ ) > } ⊂ T P The set D ( G, Λ) is open because − m is an upper semicontinuous function (seeTh.2.1.2.4), denote ρ (Λ , G ) := min ρ ( M )where the minimum is taken over all components M of D ( G, Λ) , and it is attainedon one of the components because they are not intersecting and T P is compact. Exercise 6.3.7.1 Explain this in details, using properties of ρ ( D ) ( § Theorem 6.3.7.1. If (6.3.7.2) ρ (Λ , G ) ≥ then exp Λ is complete in G. This theorem is proved in § w := g G q Λ ( z ) be the maximal L -subminorant of q Λ . Denote by H Λ theopen set in T P where L w = 0 . Theorem 6.3.7.2. If there exists a component M of H Λ such that ρ ( M ) < then w is minimal , and, hence, G is maximal for exp Λ . This theorem follows directly from Th.6.2.3.1.It is not known if the condition (6.3.7.2) is necessary. Consider in details the situation, in which the domain G coincides with G Λ ,the conjugated indicator diagram of h Λ , i.e., we suppose that(6.3.7.3) h G ( φ ) = h Λ ( φ ) , ∀ φ. In this case m ( z, G, q Λ ) ≥ Theorem 6.3.7.3. In order that exp Λ be complete in G Λ it is necessary andsufficient that (6.3.7.4) ρ (Λ , G Λ ) ≥ . This theorem is proved in § h Λ ( y ) = max { q Λ ( x + iy ) : x ∈ [0; P } , the function m ( z, G Λ , q Λ ) has a zero in x for every fixed y .Thus the set D ( G, Λ) does not contain any curve y = const on the torus. Theorem 6.3.7.4. Let G be a strictly convex domain and let D ⊂ T P be suchthat T P \ D intersect every line { y = y } , y ∈ [0 , π ] . Then there exists a periodic Λ such that (6.3.7.6) G Λ = G , D ( G Λ , Λ) = D . This theorem is proved in § Example 6.3.7.1 Let D be the complement in T P to the set(6.3.7.7) M := { z = x + iy : x = f ( y ) , y ∈ [0; 2 π ] } where f ( y ) is a continuous 2 π -periodic function satisfying the condition0 < f ( y ) < P. Then ρ ( D ) = ∞ , because this domain is not connected on spirals (see § G there exists a periodic Λ such that G Λ = G and exp Λ is extremely overcomplete in G . Example 6.3.7.2. Let D be the complement to the set M := { z = x + iy : x = P π y, ≤ y ≤ π } Then(6.3.7.8) ρ ( D ) = 12 (cid:0) π/P ) (cid:1) (see § P , and using Th.6.3.7.4,it is possible make exp Λ complete ornon-complete in G (= G Λ ) for every strictly convex domain G . Now pass to generalizations. Denote by D G the natural domain of definitionof the operation G G , i.e. the set of v ∈ U [1] for which m ( λ, G, v ) (see (6.3.6.2)) hasa subharmonic minorant belonging to U [1] . Let Φ Λ be defined by the equality (6.3.1.1).The condition that for every v ∈ Fr [Φ Λ ] the function m ( λ, G, v ) has a subharmonic minorant belonging to U [1] ispossible to express by the relation(6.3.8.1) Fr [Φ Λ ] ⊂ D G (compare with (6.1.1.1)).We call the set U ⊂ U [1] minimal ( U ∈ M IN ) if for arbitrary small ǫ > w = w ǫ ∈ U such that the function w ǫ − ǫ | λ | has no subharmonic minorant,belonging to U [1] . Let us note that if U contains a minimal function in the sense of ( § U ∈ M IN. We denote the image of Fr [Φ Λ ] under the mapping by the operator G G as J G (Λ) . Theorem 6.3.8.1. The follow holds:1. { exp Λ is not complete in A ( G ) }⇐⇒ { (6.3.8.1) holds ∧ J G / ∈ M IN ; } { G is maximal for exp Λ }⇐⇒ { (6.3.8.1) holds ∧ J G ∈ M IN ; } { exp Λ is extremely overcomplete for maximal G } ⇐⇒ { (6.3.8.1) holds ∧ J G ∈ HARM ; } In the proof of Theorem 6.3.8.1 that we are going to prove now we exploit Theorem 6.3.9.1. (A.I.Markushevich) [see,Lev,,Ch.4, § . Let A ( C \ G ) be aclass of functions ψ which are holomorphic in C \ G and equal to zero in infinity.In order that the system exp Λ be complete in A ( G ) , it is necessary and sufficientthat the function (6.3.9.1) Φ( λ ) := Z L ψ e λz ψ ( z ) dz, where ψ ∈ A ( C \ G ) , and L ψ ⋐ G is a rectifiable closed curve,has the followingproperty: the condition (6.3.9.2) Φ( λ k ) = 0 , ∀ λ k ∈ Λ implies Φ( λ ) ≡ . Proof Theorem 6.3.8.1, 1.. Necessity. Let exp Λ is not complete . By Theorem6.3.9.1 Φ( λ k ) = 0 , but Φ( λ ) . The function g ( λ ) := Φ( λ ) / Φ Λ ( λ ) , where Φ Λ isfrom (6.3.1.1), is an entire function and it has order one and normal or minimaltype by Th.2.9.3.1. Set u g := log | g ( λ ) | ; u Φ ( λ ) := log | Φ( λ ) | ; u Λ ( λ ) := log | Φ( λ ) | . We have from (6.3.9.1) u Φ ( λ ) ≤ max {ℜ ( λz ) : z ∈ L ψ } + C ψ } , where C ψ is aconstant, depending possibly on ψ. This implies that(6.3.9.3) u Φ ( λ ) ≤ h G ( φ ) r + C ψ , λ = re iφ , for some convex domain G ⋐ G. Let v ∈ Fr [Φ Λ ] . Choose a sequence t j → ∞ for which ( u Λ ) t j → v, and thesequences ( u Φ ) t j and ( u g ) t j also converges to v Φ and v g respectively. From theequality u g ( λ ) = u Φ ( λ ) − u Λ ( λ ) we obtain v g ( λ ) = v Φ ( λ ) − v ( λ ) where v g ∈ Fr [ g ] , v Φ ∈ Fr [Φ] . Since (6.3.9.3) implies v Φ ( λ ) ≤ h G ( φ ) r (6.3.9.4) v g ( λ ) ≤ h G ( φ ) r − v ( λ )and it means that for every v ∈ Fr [Φ Λ ] G G v and hence G G v exist, i.e., the condition(6.3.8.1) holds.Let us show that the condition J G (Λ) / ∈ M IN is satisfied. We have for some δ > h G ( φ ) − h G ( φ ) ≤ − δ. From (6.3.9.4) we obtain(6.3.9.5) v g ( λ ) + δr ≤ m ( λ, G, v )The left hand side of the inequality (6.3.9.5) belongs to U [1] . Thus w v := G G v satisfies the condition v g ( λ ) + δr ≤ w v ( λ ) for every v ∈ Fr [Φ Λ ] . It means that J G (Λ) / ∈ M IN. Necessity is proved. (cid:3) For proving sufficiency we exploit the following assertion Theorem 6.3.9.2. (I.F.Krasichkov-Ternovskii). Suppose there exists an en-tire function g such that (6.3.9.6) h g Φ Λ ( φ ) < h G ( φ ) , ∀ φ. Then the system exp Λ is not complete for some convex domain G ⋐ G. This theorem connects the problem of completeness to the multiplicator prob-lem. Proof of Th.6.3.9.2. Let g ( λ ) satisfy (6.3.9.6). Denote by ψ ( z ) the Borel transfor-mation for Φ( λ ) := g ( λ )Φ Λ ( λ ) . By P´olya Theorem (see, for example, [Lev.,Ch.1, § ψ are contained in a convex domain G Φ which is the conju-gate diagram of the indicator h Φ ( φ ) . Thus the representation (6.3.9.1) holds with L ψ that embraces G Φ . It follows from (6.3.9.6) that G Φ ⋐ G. Thus it is possible tochoose L ψ between ∂G Φ and ∂G. Since (6.3.9.2)for Φ is fulfilled and Φ( λ ) , exp Λis non-complete in some convex G ⋐ G such that L ψ ⋐ G by Th. 6.3.9.1. (cid:3) Now we can prove sufficiency in Th.6.3.8.1, 1. From the condition J G (Λ) / ∈ M IN it follows that one can choose δ > ∀ v ∈ Fr [Φ Λ ] the functions w v − δr where w v := G G , have subharmonic minorants. As we already said in § z . Thus we can suppose that0 ∈ G and, hence, h G ( φ ) > φ. Let γ < δ be such that h G ( φ ) − γ > G ⋐ G satisfy(6.3.9.7) h G ( φ ) − γ/ > , h G ( φ ) − h G ( φ ) ≤ γ/ D G ⊃ Fr [Φ Λ ] , Indeed, for v ∈ Fr [Φ Λ ] we have m ( λ, G , v ) := h G ( φ ) r − v ( λ ) ≥ h G ( φ ) r − ( γ/ r − v ( λ ) ≥ (6.3.9.9) h G ( φ ) r − v ( λ ) − δr ≥ w v − δr Since the right hand side of (6.3.9.9) has a subharmonic minorant from U [1] , then(6.3.9.8) is proved. By Theorem 6.1.1.1 there exists a multiplicator g ( z ) ∈ A (1)such that(6.3.9.10) h g Φ ( φ ) ≤ h G ( φ ) < h G ( φ ) . From Th.6.3.9.2 we obtain that exp Λ is non-complete in G. Proof Theorem 6.3.8.1, 2.. Necessity. Let G j , j = 1 , , ... be a sequence of convexdomains, satisfying the conditions G j ⋑ G, G j ↓ G. Since exp Λ is not complete inevery A ( G j ) , D G j ⊃ Fr [Φ Λ ] by Th.6.3.8.1, 1.The sequence w j := G G j v satisfies w j ( λ ) ≤ h G j ( φ ) r − v ( λ ) , λ ∈ C . Since { w j } is compact and h G j → h G , one can find a subsequence with the limit w ∈ U [1] . Then w ( λ ) ≤ h G ( φ ) r − v ( λ ) . Hence G G v exists.If J G ∈ M IN would not hold, then, by Th.6.3.8.1, 1., exp Λ were non-completein A ( G ) , which contradicts maximality.Necessity is proved. Let us prove sufficiency.Completeness of exp Λ in A ( G ) follows from Th.6.3.8.1, 1.. We will prove thatexp Λ is non-complete in A ( G ) for every G ⋑ G under the condition D G ⊃ Fr [Φ Λ ].Set δ := min φ [ h G ( φ ) − h G ( φ )] > ∀ v ∈ Fr [Φ Λ ] G G v + δr ≤ h G ( φ ) r − v ( λ ) , λ ∈ C . This means that G G v ≥ G G v + δr. Hence J G (Λ) / ∈ M IN and, by Th.6.3.8.1, 1.,exp Λ is non-complete in A ( G ) . (cid:3) Proof of Th.6.3.8.1, 3.. Necessity. By Th.6.3.8.1, 2. from maximality G (6.3.8.1)follows. We will prove that G G v ∈ HARM ∀ v ∈ Fr [Φ Λ ] . Suppose it is not fulfilled,i.e., there exists v ∈ Fr [Φ Λ ] such that the mass distribution ν of the function w = G G v is not zero. By Proposition 6.1.1.3 there exists a multiplicator g suchthat v + w ∈ Fr [ g Φ Λ ] . Let Λ be the set of zeros of g. Since ν ∈ Fr Λ , ∆(Λ ) > , because ν = 0 and by the definitions in § g and suppose without lack of generality that theyare simple and Λ ∩ Λ = ∅ . The condition for a multiplicator gives the inequality: h g Φ Λ ( φ ) ≤ h G ( φ ) , ∀ φ. It implies m ( λ, G, v Π ) = rh G ( φ ) − v Π ≥ v Π ∈ Fr [ g Φ Λ ] . It means that m ( λ, G, v Π ) has zero as a minorant ∀ v Π ∈ Fr [ g Φ Λ ], i.e., D G ⊃ Fr [ g Φ Λ ] . So the domain G maximal although the system exp Λis replaced with the system exp(Λ ∪ Λ ) . This contradicts to extremal overcomplete-ness. Hence, ν ≡ w = G G v ∈ HARM .Necessity is proved. Let us prove sufficiency.Let the condition G G v ∈ HARM ∀ v ∈ Fr [Φ Λ ] hold. Suppose that there existsΛ such that ∆ Λ > G is maximal for the system exp(Λ ∪ Λ ) . Theorem 6.3.8.1,, 2. implies(6.3.9.11) D G ⊃ Fr [Φ Λ ] , where Λ = Λ ∪ Λ . For every v ∈ Fr [Φ Λ ] one can find v ∈ Fr [Φ Λ ] such that v := v + v ∈ Fr [Φ Λ ]The condition ∆ Λ > v for which the Riesz measure ν 0. For w = G G v one has the inequality w ≤ rh G − v by (6.3.9.11), so w + v ≤ rh G − v holds. Hence w v := G G v satisfies the inequality(6.3.9.12) ( w + v )( λ ) ≤ w v ( λ ) , ∀ λ ∈ C . Let us show that (6.3.9.12) is impossible.Indeed, since w v ∈ HARM w := w + v − w v ≤ w ∈ U [ ρ ]. Thus w ≡ . However the Riesz measure ν w ≥ ν , hence w . This contradiction proves sufficiency. (cid:3) Now we prove Theorems 6.3.3.1, 6.3.4.1 and 6.3.5.1. We need some auxil-iary assertions. Lemma 6.3.10.1. Let v := rh ( φ ) and G be the conjugated diagram of h . Thenthe following holds:1. { G is freely enclosed in G } ⇐⇒ {G G v is non-minimal } ; { G is enclosed to G but not free enclosed }⇐⇒ {G G v is minimal } ;3. { G is rigidly enclosed in G } ⇐⇒ {G G v ∈ HARM } ;4. { G is not enclosed in G } ⇐⇒ {G G v does not exist } ; To prove this lemma we need the following two: Lemma 6.3.10.2. Let v := rh ( φ ) . Then G G v = rh ( φ ) where h is the maximaltrigonometrically convex minorant of the function m ( φ, G, h ) := h G ( φ ) − h ( φ ) .Proof. Let v = G G v. Since v [ t ] = v for all t > v ) [ t ] = G G v [ t ] = G G v by Th.6.1.1.2, 2.Thus the function ˆ v := (cid:18) sup t ( v ) [ t ] (cid:19) ∗ ( λ ) ≥ v ( λ )and is also a subharmonic minorant belonging to U [1]. Thus v = ˆ v . However,the function ˆ v is invariant with respect to the transformation ( • ) [ t ] . Hence it hasthe form rh ( φ ) . The maximality of h ( φ ) follows from the maximality v . (cid:3) Lemma 6.3.10.3. In order that v := rh be a minimal function it is necessaryand sufficient that G , the conjugate diagram of h , be a segment (in particular, apoint).Proof. Let v = rh be minimal and let G be the conjugate diagram of h . If G is not segment, then it contains some disc of radius δ > . Hence there exists atrigonometric function A cos( φ − φ ) such that δ + A cos( φ − φ ) ≤ h ( φ ) . Multiplying this inequality by r, we obtain that v − δr has a harmonic (and hencesubharmonic) minorant. This contradicts to minimality. Inversely, Suppose v is not minimal. Then there exists δ > h ( φ )such that(6.3.10.1) h ( φ ) ≤ h ( φ ) − δ. For every t.c.f. h there exists a trigonometric function A cos( φ − φ ) such that(6.3.10.2) h ( φ ) + A cos( φ − φ ) ≥ δ − A cos( φ − φ ) ≤ h ( φ ) , which means that G contains some disc of radius δ > . So it is not a segment. (cid:3) Proof of Lemma 6.3.10.1. G is freely enclosed iff the following assertion holds: thereexists δ > A cos( φ − φ ) such that the inequality(6.3.10.3) h ( φ ) + δ − A cos( φ − φ ) ≤ h G ( φ ) . holds. Exercise 6.3.10.1. Prove this.Let G G v be non-minimal. By Lemma 6.3.10.2 it has the form w = rh , where h is the maximal trigonometrically convex minorant of m ( φ, G, h ) . There exists δ > w − δr has the maximal subharmonic minorant v = rh ( φ ) . Let A cos( φ − φ ) be a trigonometric function for which h ( φ ) + A cos( φ − φ ) ≥ . In addition, h ( φ ) ≤ h ( φ ) − δ, h ( φ ) ≤ h G − h ( φ )From this we obtain (6.3.10.3) and hence that G is free enclosed.Inversely, let G is freely enclosed in G. From (6.3.10.3) it follows that(6.3.10.4) δ − A cos( φ − φ ) ≤ h G ( φ ) − h ( φ ) . Multiplying (6.3.10.4) by r, we obtain that m ( λ, G, v ) has a minorant v = r ( δ − A cos( φ − φ )) which obviously is non-minimal. Hence, G G v is non-minimal. G is enclosed in G with sliding, hence there does not exists δ > E ( φ ) = B | sin φ | + A cos( φ − φ ) , such that the inequality(6.3.10.5) h ( φ ) + E ( φ ) ≤ h g ( φ )holds. Exercise 6.3.10.2 Prove this.Let G G v be minimal. By Lemma 6.3.10.2 it has the form w = rh and byLemma 6.3.10.3 h = E ( φ ) . Thus E ( φ ) ≤ ( h G − h )( φ ) , which is equivalent to(6.3.10.5).Prove 2., suppose G is not freely enclosed and hence only (6.3.10.5) is possible.If G G v were non-minimal, (6.3.10.3) would follow, as it was proved above. Thiscontradicts the supposition.The rigid enclosure is equivalent only to the inequality of the form h ( φ ) − A cos( φ − φ ) ≤ h G ( φ ) ∀ φ, and impossibility of enclosure is equivalent to the impossibility of even such aninequality. Thus all other assertions of the lemma can be proved analogously. Exercise 6.3.10.3 Do this in details. (cid:3) Proof Theorem 6.3.3.1. Regularity of Λ means that F r [Φ Λ ] = { v } where v = rh Λ . Thus J G (Λ) = {G G v } and all the assertions of Theorem 6.3.3.1 follows from Th.6.3.8.1 and Lemma 6.3.10.1. (cid:3) Exercise 6.3.10.4 Check this in details.For proving Theorem 6.3.4.1 we need an additional Lemma 6.3.10.4. Let Λ an indicator set, v = rh , v = rh . Then { J G (Λ) / ∈ M IN } ⇐⇒ {G G v and G G v are non − minimal } . Proof. Suppose w := G G v and w := G G v are not minimal, i.e., w − δr and w − δr have subharmonic minorants g and g . Then cg + (1 − c ) g is a minorant of the function cw + (1 − c ) w − δr , i.e., J G (Λ) / ∈ M IN. The inverse implication is trivial. (cid:3) Exercise 6.3.10.5 Prove this. Proof of Theorem 6.3.4.1. Suppose exp Λ is not complete. By theorem 6.3.8.1 J G / ∈ M IN. By Lemma 6.3.10.4 G G v and G G v are not minimal. Hence G and G are freely enclosed in G by Lemma 6.3.10.1. Since every one of these assertionsis reversible, the inverse implication also holds. Analogously the other cases areproved. (cid:3) Exercise 6.3.10.6 Prove all this in details. To prove Theorem 6.3.5.1 we need some auxiliary assertions. Lemma 6.3.11.1. Let φ be a maximum point of t.c.f. h ( φ ) and h ( φ ) ≥ . Then (6.3.11.1) h ( φ ) ≥ h ( φ ) cos( φ − φ ) , | φ − φ | ≤ π/ . Proof. Denote y ( φ ) := h ( φ ) cos( φ − φ ) . We have y ( φ ) = h ( φ ) and y ( φ ) is atrigonometric function. If y ( φ ) = h ( φ ) for some φ such that | φ − φ | < π/ y ( φ ) does not intersect h ( φ ) , this contradicts toProp.3.2.5.6. (cid:3) Lemma 6.3.11.2. Let H ( φ ) be a trigonometric function on the interval I = ( α, β ) of length ≤ π, such that H ( φ ) = 0 at one of the ends of I . Then every of theconditions1. H ( φ ) = 0 , φ ∈ ( α ; β ); H ( φ ) is zero on ∂I with tangency;implies H ( φ ) ≡ , φ ∈ I. Exercise 6.3.11.1 Prove this. Lemma 6.3.11.3. Let g ≥ be a continuous periodic function, and let Θ Λ , I Λ , d Λ be defined as in Theorem 6.3.5.1. In order that its maximal t.c. minorant be atrigonometrical function, it is necessary and sufficient satisfying at least one of theconditions:1. d Λ < π ; d Λ = π and g ( φ ) is zero with tangency on ∂I. Proof. Necessity. Suppose d Λ > π. Without loss of generality we can suppose that I Λ = ( α ; − α ), where α > π/ + φ := max(cos φ, , (6.3.11.2) a = inf (cid:18) g ( φ )cos + φ : φ ∈ ( − α ; α ) (cid:19) . We have a > . Set(6.3.11.3) (cid:26) a cos φ, | φ | ≤ π/ | φ | > π/ , where a ≤ a. The function h ( φ ) is a t.c.minorante of g ( φ ) and it is not a trigonometricfunction, which contradicts the supposition. Thus d Λ ≤ π. Suppose d Λ = π and the condition to be zero with tangency on ∂I does nothold. Then for a defined by (6.3.11.2) the condition minoranta > h ( φ )defined by (6.3.11.3) is a non-trigonometric of g. Sufficiency. The first condition holds and I = ( α ; β ) be an arbitrary intervalbelonging to Θ Λ ; let h ( φ ) be the maximal t.c.minorante of g ( φ ) . Set H ( φ ) := h ( φ ) cos( φ − φ ) , where φ is the maximum point of h ( φ ) on I. From inequality (6.3.11.1) and theconditions g ( α ) = g ( β ) = 0 follows H ( α ) = H ( β ) = 0 . Then, by Lemma 6.3.11.2,we obtain H ( φ ) ≡ . Thus h ( φ ) = 0 and h ( φ ) ≡ φ ∈ ( α ; β ), i.e., h ( φ ) istrigonometric.Let the second condition be fulfilled. Lemma 6.3.11.1 implies that H ( φ ) is zerowith tangency on ∂I. By Lemma 6.3.11.2 we obtain that h ( φ ) ≡ . (cid:3) Proof of Theorem 6.3.5.1. Necessity. Let us note that if v ∈ Fr [Φ Λ ] then for c ∈ [0; 1] we have the equality(6.3.11.4) m ( λ, G, v ) = r (cid:0) c ( h − h ) + + (1 − c )( h − h ) − (cid:1) ( φ ) := rm ( φ, c )Let exp Λ be extremely overcomplete in A ( G ). By Th. 6.3.8.1 J G ⊂ HARM, i.e.,for every c ∈ [0; 1] the maximal t.c.minorant of the function m ( φ, c ) is trigonometric.Since ∀ c ∈ [0; 1] , Θ Λ = { φ : m ( φ, c ) > } the necessity follows from Lemma 6.3.11.3.Sufficiency. It follows directly from Lemma 6.3.11.3. (cid:3) Exercise 6.3.11.2 Explain this. To prove Theorem 6.3.6.1 we need Lemma 6.3.12.1. If w ∈ U [1] is non-minimal then C ( w ) := { w [ t ] : 1 ≤ t ≤ e P } / ∈ M IN It follows from Th.6.1.1.2, 2. Exercise 6.3.12.1 Explain this in details. Proof of Theorem 6.3.6.1. By Theorem 6.1.1.2, 2. J G (Λ) = C ( G G v ) . Thus Lemma6.3.12.1 implies that J G (Λ) / ∈ M IN if and only if G G v is not minimal. ThusTheorem 6.3.8.1 implies Th. 6.3.6.1, 1. and 2. Suppose J G (Λ) ⊂ HARM . Hence, G G v = rH ( φ ) , where H is trigonometric.Inversely, Lemma 6.3.12.1 implies J G (Λ) = { rH ( φ ) } . (cid:3) Proof of Theorem 6.3.7.1. Let ρ (Λ , G ) > . Suppose w q := g G q exists. By definitionof ρ (Λ , G ) we have w q ( z ) ≤ z ∈ ∂D (Λ , G ) . By Th.6.2.3.3 (Maximum principle w q ≤ z ∈ D ( G, Λ) . Also w q ≤ z ∈ T P \ D ( G, Λ) by definition of D (Λ , G ) . By Theorem 6.2.3.4 w q ≡ exp Λ is completeby Th.6.3.6.1. If ρ (Λ , G ) = 1 , then the system exp Λ is complete for every G n ⋑ G, because of strict monotonicity of ρ ( • ) (see § G is the maximal domain. (cid:3) For proof of Theorem 6.3.7.3 we need an auxiliary assertion. We suppose that D is an image on T P by the map (6.2.1.3) of the domain G with a smooth boundary. Theorem 6.3.12.2. Let D ⊂ T P and ρ ( D ) ≤ . Then ρ ( T P \ D ) > , if D Λ = {ℜ z > } . For the proof we need the following assertion which was proved originally byA.Eremenko and M.Sodin: Theorem 6.3.12.3(Eremenko,Sodin). Let Γ be a Jordan curve, connecting and ∞ , T Γ = Γ for some T > . Let D + , D − be domains, into which Γ divides theplane, and let ρ , ρ be the orders of the minimal harmonic functions in D + and D − respectively.Then ρ + 1 ρ ≤ , and equality is attained only if Γ consists of two rays. We will give prove this theorem in § Proof of Theorem 6.3.12.2. Let ρ = ρ ( D ) , and suppose q ( z ) is a solution ofboundary problem (6.2.2.1), ρ = ρ ( T P \ D ) , q ( z ) is a solution of correspond-ing boundary problem. Then the image of the boundary under the map λ = e z (we denote it as Γ) satisfies the conditions of Theorem 6.3.12.3 and the functions v ( λ ) := q (log λ ) | λ | ρ and v ( λ ) := q (log λ ) | λ | ρ are positive harmonic functions in D + , D − with orders ρ and ρ respectively. By Theorem 6.3.12.3 we obtain1 /ρ ( D ) + 1 /ρ ( T P \ D ) ≤ . and equality holds only if Γ is a pair of rays, i.e., D Λ = {ℜ z > } . (cid:3) Proof Theorem 6.3.7.3. Necessity. Suppose ρ (Λ , G Λ ) < . Let us prove that exp Λ isnot complete. To this end we construct an L -minorant of m ( z, G Λ , Λ) and provethat it is not minimal.Let D ⋐ D ( G Λ , Λ) be a domain with smooth boundary for which ρ ( D ) = 1 . This is possible because of strict monotonicity ρ ( D ) ( § q be a solutionof the problem (6.2.2.1) satisfying the condition0 < max { q ( z ) : z ∈ D } ≤ { m ( z, G Λ , Λ); z ∈ D } − ǫ for sufficiently small ǫ. By Theorem 6.3.12.2 ρ ( T P \ D ) > . Thus the potentialΠ( z ) = − Z D G ρ ( z, ζ, D ) ν ( dζ )exists and ν can be chosen in such way that supp ν ⋐ T P \ D . By Proposition6.2.3.6 ∂q ∂n > , z ∈ D Thus ν can be chosen in such way that − ∂ Π ∂n < min ∂q ∂n , z ∈ ∂D . Then the function q ( z ) = (cid:26) q ( z ) , z ∈ D Π( z ) , z ∈ T P \ D , is an L -subfunction on T P . Exercise 6.3.12.2 Explain this in details, exploiting Th.2.7.2.1.The function q ( z ) satisfies the condition q ( z ) ≤ m ( z, G Λ , Λ) − ǫ, f orallz ∈ T P , because of negative potential. Hence, q ( z ) := q ( z ) + η for some η > m ( z, G Λ , Λ) and it is not minimal. Necessityis proved. Sufficiency follows from Th.6.3.7.1. (cid:3) Now we pass to the proof of Theorem 6.3.7.4 and construction of Example6.3.7.2. Proof of Theorem 6.3.7.4. The set T P \ D is closed. Let φ ( z ) be an infinitelydifferentiable function equal to zero on T P \ D and positive on D . Set q ( z ) := h ( y ) − ǫφ ( z ) , where h ( y ) is a t.c.f., corresponding to G , and let ǫ be small enough to satisfy L q ( z ) > , z ∈ T P . It is possible, because L h ( y ) > m ( z, G , q ) = ǫφ ( z ) . hence { z : m ( z, G , q ) > } = D . Take v ( λ ) := | λ | q (log λ )and construct an entire function Φ Λ for which Fr [Φ Λ ] = { v [ t ] : 1 ≤ t ≤ e P } It is easy to check that the zero distribution of this function has all the propertiesdemanded by Theorem 6.3.7.4. (cid:3) Exercise 6.3.13.1 Check this. Proof of (6.3.7.8). Consider the problem(6.3.13.1) L q ( z ) = 0 , q | x =(2 π/P ) y = 0Let us pass in the equation to new coordinates (cid:26) ξ = x cos α + y sin αη = − x sin α + y cos α, tan α = 2 π/P . Then the equation takes the form: (cid:20) ∂ ∂ξ + ∂ ∂η + 2 ρ (cid:18) cos α ∂∂ξ − sin α ∂∂η (cid:19) r (cid:21) R ( ξ, η ) = 0The condition of being zero on D is R ( ξ, πl cos α ) = 0 . l ∈ Z . The condition of periodicity gives R ( ξ + ( P/ cos α ) k, η ) = R ( ξ, η ) , k ∈ Z . We search for a solution that does not depend of ξ. We have R ′′ ( η ) − ρ sin αR ′ ( η ) + ρ R ( η ) = 0 , R (0) = R (2 π cos α ) = 0Further, R ( η ) = C e ( ρ sin α ) η cos(( ρ cos α ) η ) + C e ( ρ sin α ) η sin(( ρ cos α ) η )Exploiting the boundary condition, we have ρ min = (2 cos α ) − = 12 " (cid:18) πP (cid:19) . The corresponding eigenfunction R = exp ( ρ min sin α ) η sin(( ρ min cos α ) η ) . It is zero on T P \ D and positive in D , so it determined up to a constant multi-ple. (cid:3) We are going to prove Theorem 6.3.12.3. Actually we prove Theorem 6.3.14.1. Let Γ , Γ , ..., Γ n be Jordan curve, such that1. Γ i , i = 1 , , ..., n connect and ∞ ; T, | T | > (not necessarily real) for which T Γ i =Γ i , i = 1 , , ..., n. Let D i , i = 1 , , ..., n be domains into which the plane is divided, and let ρ i bethe order of the minimal harmonic function in D i . Then (6.3.14.1) X i /ρ i ≤ and equality holds if and only if Γ i are a logarithmic spirals (or rays, when T ∈ R + ).Proof. Denote by H i the minimal harmonic function in D i . Then H i = ℑ φ i where φ i : D i Π + is a conformal map of D i to upper half plane, φ (0) = 0 . The maps g i := φ i ( T φ − i ) : Π + Π + does continued by isomorphism to C , and g i (0) = 0 . Thus g i ( z ) = σ i z , where σ i > . Hence, φ i ( T z ) = σ i φ i ( z ) or T h i ( z ) = h i ( σ i z ) , h i := φ − i : Π + D i . Now we exploit the following inequality from [LevG](6.3.14.2) n X i =1 σ i ≤ T | log T | ≤ T . Notation2.1. R m , M ( f, x, ε ) , f ∗ ( x ) , C + ( E ) , C − ( E ) , χ G , χ F , Γ A , F A , K, M ( f, K ) ,K n , K max . σ ( G ) , µ, G ( µ ) , supp µ, M ( G ) , µ F ( E ) , ν, M d , ν + , ν − , | ν | , supp φ, ∗ → , ◦ E, E, σ ( R m × R m ) , Φ ⊗ Φ , µ ⊗ µ . ϕ n D → ϕ, α ( t ) , α ε ( x ) , ψ ε ( x ) , < f, ϕ >, < δ x , ϕ >, < δ ( n ) x , ϕ >, < µ, ϕ >, <αf, ϕ >, < f + f , ϕ >, < ∂∂x k f, ϕ >, f ǫ ( x ) , f | G , ] cos ρ ( φ ) . ∆ x , E m ( x ) , θ m , G ( x, y, Ω) , G ( x, y, K a,R ) Π( x, µ, D ) , G N ( x, y ) , Π N ( x, µ, D ) , Π( x, µ ) , Π( z, µ ) , U [ ρ ] cap G ( K, D ) , cap m ( K ) , cap m ( D ) , cap m ( E ) , cap m ( E ) , cap l ( K ) M ( x, r, u ) , N ( x, r, u ) , E ǫ , D − ǫ , u ǫ ( x ) , K R , M ( r, u ) , µ ( r, u ) , M ( r, u ) , N ( r, u ) ,M ( z ) . ˜ u ( x ) , µ x ( t ) , E ( α, α ′ , ǫ, µ ) , E n,δ . a ( r ) , ρ [ a ] , σ [ a ] , ρ ( r ) , σ [ a, ρ ( r )] , V ( r ) , L ( r ) , δSH ( R m ) , T ( r, u ) , ρ T [ u ] , σ T [ u ] ,σ T [ u, ρ ( r )] , ρ M [ u ] , σ M [ u ] , σ M [ u, ρ ( r )] , ρ [ µ ] , ¯∆[ µ ] , ¯∆[ µ, ρ ( r )] , N ( r, µ ) , ρ N [ µ ] , δ M ( R m ) . H ( z, cos γ, m ) , G ( x, y, R m ) , D k ( x, y ) , H ( z, cos γ, m, p ) , G p ( x, y, m ) , G p ( z, ζ, , Π( x, µ, p ) , δSH ( ρ ) , Π R< ( x, ν, ρ − , Π R> ( x, ν, ρ ) , δ R ( x, ν, ρ ) , δ R ( z, ν, ρ ) , δ R ( x, u, ρ ) ,M ( r, δ ) , ¯∆ δ [ u, ρ ] , Ω[ u, ρ ( r )] , T ( r, λ, > ) , T ( r, λ, < ) . V t , P t , SH ( R m , ρ, ρ ( r )) , SH ( ρ ( r )) , u t ( x ) , Fr [ u, ρ ( r ) , V • , R m ] , U [ ρ, σ ] , U [ ρ ] ,v [ t ] , M ( R m , ρ ( r )) , µ ∈ M ( ρ ( r )) , Fr [ µ, ρ ( r ) , V • , R m ] , Fr [ µ ] , M [ ρ, ∆] , M [ ρ ] , ν [ t ] . h ( x, u ) , h ( x, u ) , l x , x ( x ) , T ρ , G I ( φ, ψ ) , Π I ( φ, ds ) , T C ρ , Co Ω . ∆( G, µ ) , ∆( E, µ ) , ∆( K, µ ) , ∆( E, µ ) , Co Ω ( I ) , ∆ cl ( E ) , ∆ cl ( E ) , Ω G ( ǫ ) , Ω K ( ǫ ) . T t , ( T • , M ) , d ( • , • ) , Ω( T • ) , C ( m ) , Ω( m ) , A ( m ) , T t v. U , β ( x ) , b , k ( s ) , R ǫ v ( x ) , Str ( δ ) , v ( x | t ) , v ( • , t ) w ( •| t ) , w ( •|• ) . , ( u ) t , Fr [ u ] , U [ ρ ] . M ( r, f ) , T ( r, f ) , ρ T [ f ] , ρ M [ f ] , σ T [ f, ρ ( r )] , σ M [ f, ρ ( r )] , n ( K r ) , n ( r ) , ρ [ n ] , ∆[ n ] ,N ( r, n ) , ρ N [ n ] , ∆ N [ n ] , p [ n ] , Fr [ f ] , Fr [ n ] , M er ( ρ, ρ ( r )) , T ( r, f ) , ρ T [ f ] , σ T [ f, ρ ( r )] . α − mes C, C α , C . k g k p . h ( φ, f ) , h ( φ, f ) , h ( φ, f ) . N ( δ, X ) , ( X ) R f dδ, ( X ) R E f dδ, δ (Θ F ) , D r, Θ , δ z ( D r, Θ ) , A cl ( δ, χ Θ ) F ( u ) , H φ ( u ) , T ( u ) , M α ( u ) , M ( u ) , I αβ ( u ) , I ( u, g ) , F [ f ] , F [ f ] , χ H , χ I , χ F o . K S , S , G ( t, γ, ρ ) , ˆ G ( s, S − S ) , ( F ν )( s ) . H ( z ) , m ( z, v, H ) , G H , D H , U ind , ˆ U ind . T P , D ′ ( T P ) , q ( z ) , L ρ , E ρ ( • − ζ ) , E ′ ρ ( • − ζ ) , q D , G ρ ( z, ζ, D ) , H ρ ( q ) . Λ , Φ Λ ( λ ) , exp Λ , A ( G ) , h Λ ( φ ) , G Λ , αG + βG , Θ Λ , I Λ , , d Λ h G ( φ ) , m ( λ, G, v ) ,H ( λ ) , q Λ ( z ) , D ( G, Λ) , ρ (Λ , G ) , g G q, G G , D G , M IN, J G (Λ) , HARM, m ( φ, G, h ) , E ( φ ) . List of Terms2.1. µ µ is concentrated on E ∈ σ ( G )2.2. restriction of µ onto F ∈ σ ( G ).2.2. charge2.2. positive and negative , respectively, variations of ν full variation of ν Borel function restriction of µ on the set E topological space2.3. linear continuous functional on D distribution the n-th derivative of the Dirac delta-function2.3. regular distribution2.3. positive distribution2.3. product of a distribution f by an infinitely differentiable function α ( x )2.3. sum of distributions f and f partial derivative of distribution2.3. sequence of distributions f n converges to a distribution f regularization of the distribution f restriction of distribution f ∈ D ′ ( G ) to G ⊂ G fundamental solution of L at the point y sequence of distributions f n converges to a distribution f f f ∈ D ′ ( G ) to G ⊂ G ρ µ relative to D K relative to the domain D. E u ( x ) on the sphere S x,r := { y : | y − x | = r } u in K u f n of locally summable functions converges in L loc u n converges to a function u relative to α - Carlesonmeasure2.7. a point x ∈ R m ( α, α ′ , ǫ ) -normal with respect to the measure µ a ( r ) a ( r ) a ( r ) of minimal type2.8. a ( r ) of normal type2.8. a ( r ) of maximal type2.8. convergence exponent for the sequence { r j } ρ u ( x ) with respect to T ( r ) . ρ M [ u ] , σ M [ u ] , σ M [ u, ρ ( r )] µ µ. µ µ µ u ( x ) µ u ρ -subspherical function3.2. ρ -trigonometrically convex ( ρ -t.c.)3.2.fundamental relation of indicator. µ ∆( E, • ) ∆( E, • ) : E ∈ R m . ( ǫ, s ) -chain from m to m ′ T • (a.d.p.t.)4.3.piecewise continuous pseudo-trajectory w ( •|• ) ω –dense pseudo-trajectory ρ and normal type with respect to proximateorder ρ ( r ) ρ and normal type with respect to a proxi-mate order ρ ( r ) α -measure ρ -trigonometricity5.4.strictly ρ -t.c.f.5.4. concordant h and g ( X ) − integral with respect to a nonnegative measure δ. ρ H -multiplicator6.1.entire function is of minimal type with respect to a proximate order ρ ( r ) , ρ ( r ) → ρ U [ ρ ] is valid in the domain G v ∈ U [ ρ ] exp Λ G Λ is enclosed in G Λ is periodic6.3. w ∈ U [1] is minimal6.3. U ⊂ U [1] is minimal References [An] Anosov D.V. at al,, Ordinary differential equations and smooth dynamical systems , Dynam-ical systems I, Ency.Math.Sci,1, Springer-Verlag, 1997.[Ar]Arakeljan,N.U., Uniform approximation by entire functions on unbouded continua an estimateof the rate of their growth , Acad.N.Armjan.SSR Dokl (1962), 145-149. 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