Bounds on the growth of subharmonic frequently oscillating functions
BBounds on the growth of subharmonic frequently oscillating functions
Adi Gl¨ucksam ∗ April 15, 2020
Abstract
We present a Phragm´enLindel¨of type theorem with a flavor of Nevanlinna’s theorem for subharmonicfunctions with frequent oscillations between zero and one. We use a technique inspired by a paper of Jonesand Makarov.
In this note we introduce a Phragm´enLindel¨of type theorem for subharmonic functions which are bounded ona set, possessing a special geometric structure. The bounds we obtain depend heavily on the structure of theset, and are proven to be optimal. A different but valid angle on the matter, would be to say we present a moreelaborate version of Nevanlinna’s theorem for a special class of subharmonic functions with a ‘well distributed’zero set.To formally state our results, we will need the following definitions:
A cube I is called a basic cube if I = d (cid:81) j =1 [ n j , n j + 1) for n , · · · , n d ∈ Z , i.e I is a half open half closed cube withedge length one, whose vertices lie on the lattice Z d . Given a subharmonic function u and a basic cube I , weconsider the properties:(P1) sup x ∈ I u ( x ) ≥ λ d − ( I ∩ Z u ) ≥ ε d for some constant ε d >
0, where Z u := { u ≤ } and λ d − denotes the ( d − I satisfies both properties, we say that the function u oscillates in I . Otherwise, we say I is a roguebasic cube . ∗ Supported in part by Schmidt Futures program a r X i v : . [ m a t h . C V ] A p r o oscillations have any effect on the growth of a subharmonic function? The following observation suggeststhat the answer is yes: Observation 1.1
There exists a constant α d , which depends on the dimension alone, so that for every sub-harmonic function u defined in a neighbourhood of the unit ball in R d , B (0 , , if λ d − ( Z u ∩ B (0 , )) > ε > then sup y ∈ B (0 , u ( y ) ≥ u (0) e α d · ε . To prove this observation we rely on the following claim:
Claim 1.2
Let E ⊂ B (cid:0) , (cid:1) be a compact set with λ d − ( E ) > . Then ω (0 , E ; B (0 , \ E ) (cid:38) d λ d − ( E ) . ∗ This is definitely known and used by experts in the field. In two words, it is true is since the harmonicmeasure is bounded from bellow by a constant over the energy of the equilibrium measure of the set E , whichin turn is bounded from bellow by a constant times the ( d − E . For the reader’sconvenience, we include a proof of this claim in Appendix A. Proof of Observation 1.1.
Let E ⊆ Z u ∩ B (cid:0) , (cid:1) be a compact set with λ d − ( E ) > λ d − ( Z u ∩ B (0 , )). Then,following Claim 1.2, we know that ω (0 , E ; B (0 , \ E ) ≥ α d · ε for some uniform constant α d , which depends onthe dimension alone. By definition of harmonic measures, and since u | E ≡ u (0) ≤ (cid:90) ∂ ( B (0 , \ E ) u ( y ) dω (0 , y ; B (0 , \ E ) ≤ sup y ∈ B (0 , u ( y ) · ω (0 , ∂B (0 , \ E ; B (0 , \ E )= sup y ∈ B (0 , u ( y ) (1 − ω (0 , E ; B (0 , \ E )) ≤ sup y ∈ B (0 , u ( y ) (1 − α d · ε ) ≤ sup y ∈ B (0 , u ( y ) e − α d · ε , concluding the proof.It seems like every oscillation imposes an increment on the function. It is therefore interesting to investigatewhat is the relationship between the number of basic cubes where the function u oscillates and its minimalpossible growth.To describe the growth-rate of a non-negative subharmonic function u , we will abuse the notation M u to denotethe function from subsets of R d to R + defined by M u ( A ) := sup z ∈ A u ( z ) , as well as the function from R + to itselfdefined by M u ( R ) := M u ( B (0 , R )). ∗ Here, and anywhere else in this paper, we write A (cid:46) B if there exists α > A ≤ α · B . We write A ∼ B if A (cid:46) B and B (cid:46) A . If the constant depends on the dimension d , we will add a subscript d to each notation, i.e A (cid:46) d B and A ∼ d B .
2n this paper we produce a Phragm´enLindel¨of type theorem to describe bounds on the minimal possible growthof subharmonic functions with frequent oscillations between zero and one. We show that every subharmonicfunction u defined in a neighbourhood of [ − N, N ] d , and oscillating in all but a N basic cubes, has a lower bound M u (cid:0) [ − N, N ] d (cid:1) ≥ exp c d N (cid:0) a N N (cid:1) d − log dd − (cid:16) a N N (cid:17) , as long as N is large enough, and a N ≤ c · N d . The constants c d and c depend on the dimension d . We alsoshow that this bound is optimal.To describe the asymptotic behavior of the number of rogue basic cubes, those who do not satisfy eitherproperty (P1) or property (P2), we use a comparison function:Given a monotone non-decreasing function f ( t ) ≤ t d , a subharmonic function u , and a cube Q ⊂ R d with edgelength (cid:96) ( Q ), we let γ uf ( Q ) = γ ( Q ) = { I ⊂ Q, I is a rogue basic cube } f ( (cid:96) ( Q )) . We say u is f -oscillating in Q if u is defined in a neighbourhood of Q , and γ ( Q ) <
1. We say u is f -oscillating iflim sup N →∞ γ ([ − N, N ] d ) < . Lastly, we remind the reader the definition of regularly varying functions. A function f is called regularly varyingof index a ≥ f ( t ) = t a · g ( t ) and the function g satisfies that for every λ > t →∞ g ( λ · t ) g ( t ) = 1 . The function g is called slowly varying . For more information, see, for example, page 18 in [2]. Let us begin with a simple case: What can be said about the minimal possible growth of a non-negative subhar-monic function which oscillates in every basic cube, i.e an f -oscillating function for f ( t ) = 1?In light of Observation 1.1, a lower bound is not hard to obtain by noting that every ball of radius 2 √ d includesa basic cube as a subset. We may therefore apply Observation 1.1 with ε = ε ( d ) to find a sequence of points, { x n } ∈ R d , satisfying that | x n | − | x n − | = 2 √ d while u ( x n ) ≥ e δ d u ( x n − ) ≥ · · · ≥ e δ d · n for some δ d >
0. Thisimplies that lim inf r →∞ log ( M u ( B d (0 , r ))) r ≥ lim inf n →∞ log ( u ( x n ))2 n √ d ≥ δ d √ d > . In fact, most of the proofs of lower bounds we will mention, basically count the number of times we can applyObservation 1.1 to obtain a lower bound on the growth of the function.3o see a construction of an example with this optimal growth, one could extend Theorem 2 in [4] to higherdimensions, which we shall do in Section 3.It is interesting to ask what happens if we relax the condition that u oscillates in every basic cube. The firstresult in this direction, which already appears in [4], was for the case d = 2 where we allow u to have many roguebasic cubes, their number proportional to the area. Here is a restatement of it: Lemma 1.3 [4, Lemma 1] Let Q be a square of edge length (cid:96) > , and let f ( t ) = αt for some α ∈ (0 , . Thenevery non-negative subharmonic function u which is f -oscillating in Q , must satisfy M u ( Q ) ≥ e c α ( log (cid:96) log log (cid:96) ) . In [4], we also indirectly show a lower bound. We constructed a subharmonic function in C which is f -oscillating for f ( t ) = αt (for some α ∈ (0 , r →∞ log ( M u ( r ))log ε ( r ) < ∞ . These functions arise when studying the support of translation invariant probability measures on the space ofentire functions, i.e probability measures which are invariant under the action of the group C acting on the spaceof entire functions by translations: ( T w f )( z ) = f ( z + w ). For more information, see [4].The examples we discussed are two extremes. One discusses the case f ( t ) = 1, i.e there are no rogue basic cubes.The other deals with the situation where f ( t ) = αt for α ∈ (0 , Let f be a regularly varying function. What can be said about the minimal possible growth of f -oscillatingsubharmonic functions? Theorem 1.4 (A) Every f -oscillating subharmonic function u satisfies lim inf R →∞ log( M u ( R )) R ( f ( R ) R ) d − log dd − (cid:16) f ( R ) R (cid:17) > , provided that f ( t ) ≤ c · t d for all t large, where c is a constant, which depends on the dimension alone.(B) Let f be a regularly varying function of index α ∈ [0 , d ] . Assume that either α < d or f ( t ) = t d · g ( t ) for some monotone non-increasing function g . Then there exists an f -oscillating subharmonic function u atisfying lim sup R →∞ log( M u ( R )) R ( f ( R ) R ) d − log dd − (cid:16) f ( R ) R (cid:17) < ∞ . In particular, if f ( t ) = t α then the theorem above shows that the minimal possible growth for f -oscillatingsubharmonic functions is log ( M u ( R )) ≥ c d · R , α ≤ c d,α (cid:16) R d − α log d ( R ) (cid:17) d − , α > , and this bound is optimal.This theorem implies that in Lemma 1.3, the log log component in the numerator of the exponent’s power isnot essential. In fact, the following corollary can now be proved- Corollary 1.5
Let λ be a non-trivial translation-invariant probability measure on the space of entire functions.Then, for λ -a.e. function f , lim inf R →∞ log log M f ( R )log ( R ) > . To prove this corollary, one may use the original proof of Theorem 1 part (A) in [4], and replace the use ofLemma 1 with part (A) of Theorem 1.4 above.Never the less, the construction, described in Theorem 1.4 part (B) above, cannot be used to construct atranslation invariant probability measure with a similar upper bound. For more details see the end of Section 3.
Given a subharmonic function u , there is some connection between bounds on the harmonic measure of its zeroset in a large ball, ω (0 , Z u , B (0 , R ) \ Z u ), and estimates on the growth of the function u in B (0 , R ) (for definitionsand basic properties of harmonic measures see for example [6] chapter 3.6). Though this connection is mostlyconceptual, it is not surprising that some of the tools, used in this paper, are also used to estimate harmonicmeasures.We use a stopping time argument inspired by the one appearing in a paper by Jones and Makarov (see [7]).Jones and Makarov’s work in [7] is very influential in modern geometric function theory. For example, A.Baranovand H.Hedenmalm use some of the techniques in this paper to study integral means spectrum of conformalmappings (see [1]).The proof of Theorem 1.4 part (A) is based on a main lemma, which essentially produces a lower bound onthe quotient M u ([ − N,N ] d ) M u ( I ) for every N large enough and for at least half of the basic cubes I ⊂ [ − N, N ] d .5he main idea of the proof is as follows- for every cube I welook at the growing sequence of cubes centered at the samepoint as I , and contained in [ − N, N ] d (see Figure 1 to theright). We then choose a subsequence of this sequence of cubes,say { Q j } , satisfying that for some δ d >
0, for all j , for every ξ ∈ ∂Q j there exists r ξ so that on one hand B ( ξ, r ξ ) ⊂ Q j +1 and on the other handinf j inf ξ ∈ ∂Q j λ d − ( Z u ∩ B d ( ξ, r ξ )) r dξ ≥ δ d . Using a Observation 1.1, there exists a constant α d so that if x j ∈ ∂Q j satisfies that u ( x j ) = M u ( Q j ), and r j = r x j then I N Figure 1:
In this figure one can see the layersabout the cube I which are colored alternately. M u ( Q j ) = u ( x j ) ≤ M u ( B ( x j , r j )) · e − α d δ d ≤ M u ( Q j +1 ) · e − α d δ d , where the last inequality is by inclusion. Doing so recursively we conclude that M u ( I ) ≤ e − α d δ d M u ( Q ) ≤ · · · ≤ e − { Q j } α d δ d M u ([ − N, N ] d ) . It is therefore left the bound from bellow the number of elements in the subsequence { Q j } to conclude the proof.Here various combinatorial methods are used.It is not the particular statement of this lemma, but the methods used in the proof that are interesting on theirown and may find additional applications in many fields. The proof is completely elementary and assumes verybasic knowledge of potential theory. The proof of the lower bound, Theorem 1.4 part (A), can be found in Section2, and the formal statement of the lemma and its proof can be found in Subsection 2.2. We would have loved to use a construction which is similar to the one presented in [4]. Never the less, such aconstruction would have two main issues. The first is that the ‘self similarity’ of the function leads to accumulatingrogue basic cubes from smaller scales. This accumulation grows like the volume measure and can never work for f ( t ) (cid:28) t d (in fact, even for f ( t ) = t d we do not get the optimal bound using this method). The second issueis that this method uses hyperplanes to separate the similar copies of the function. This will not work for anyfunction t (cid:28) f ( t ) (cid:28) t d − in dimensions higher than two.6e will first show that for every R > u , such that u oscillates in all butat most f (2 R ) basic cubes in [ − R, R ] d . This is only possible on bounded domains, as we use a parameter, whichmust be positive, while it tends to zero as the diameter of the set tends to infinity. ‘Glueing’ different functionssolves the issue of accumulating rogue cubes. To solve the second issue, we will use functions which are non-zeroonly on tube-like sets, and the ‘glueing’ will be done along tubes as well. The construction can be found inSection 3. We bring here another proof of Theorem 1.4 part (A). This proof uses Jones and Makarov’s main lemma in [7]as a black box, instead of using the ideas presented in it. It was suggested to the author by M. Sodin, and usespotential theory. The proof can be found in Appendix B.
The author is lexicographically grateful to Ilia Binder, Persi Diaconis, Eugenia Malinnikova, and Mikhail Sodinfor several very helpful discussions. In particular, I would like to extend my gratitude to Chris Bishop whoinsisted on the connection between the problem presented here and the paper by Jones and Makarov, [7].
We begin this section with the statement of the main lemma. This lemma is the essence of the proof of the lowerbound:
Lemma 2.1
Let Q := (cid:2) − N , N (cid:3) d for some N (cid:29) , and let u be a subharmonic function defined in a neighbourhoodof Q . Denote by G the collection of all basic cubes in Q , and let E = E u denote sub-collection of cubes that donot satisfy property (P2), E = E u := { I ∈ G ; λ d − ( I ∩ Z u ) < ε d } . Then, there exists a constant c , depending on the dimension alone, so that if E ≤ c t d , then at least half ofthe basic cubes I ∈ G satisfy: M u ( I ) M u ( Q ) ≤ exp − c d N (cid:16) E N (cid:17) d − log dd − (cid:18) E N (cid:19) , where c d is a constant which depends on the dimension alone as well. We will first show how the main lemma implies the proof of Theorem 1.4 part (A):7 .1 The proof of Theorem 1.4 part (A)
Let c be the constant defined in the Main Lemma, Lemma 2.1, let f be a monotone increasing functionsatisfying that f ( t ) ≤ c t d for all large enough t , and let u be an f -oscillating function. Then, using the notationof the lemma, for all N large enough E = (cid:26) I ∈ G , I is a basic cube, λ d − ( I ∩ Z u ) < ε d (cid:27) = (cid:26) I ∈ G , I is a basic cube whichdoes NOT satisfy property (P2) (cid:27) ≤ { I ∈ G , I is a rogue basic cube } = f ( N ) · γ ( Q ) < f ( N ) ≤ c N d . We may therefore apply the Main Lemma to conclude that at least half of the basic cubes in Q satisfy M u ( I ) M u ( Q ) ≤ exp − c d N (cid:16) E N (cid:17) d − · log dd − (cid:18) E N (cid:19) . To get a similar bound with f ( N ) instead of E , we note that the function ψ defined by ψ ( x ) := 11 + x d − · log dd − (2 + x )is monotone decreasing in some neighbourhood of ∞ , which depends on the dimension. Combining this with thefact that E ≤ f ( N ), we see that for every N large enough M u ( I ) M u ( Q ) ≤ exp − c d N (cid:16) E N (cid:17) d − · log dd − (cid:18) E N (cid:19) = exp (cid:18) − c d N · ψ (cid:18) E N (cid:19)(cid:19) ≤ exp (cid:18) − c d N · ψ (cid:18) f ( N ) N (cid:19)(cid:19) = exp − c d N (cid:16) f ( N ) N (cid:17) d − · log dd − (cid:18) f ( N ) N (cid:19) . On the other hand, as without loss of generality c < , more than half of the basic cubes in Q satisfy property(P1), that is M u ( I ) ≥
1. We conclude that there exists at least one cube satisfying Property (P1) and theinequality above, implying that M u ( Q ) ≥ exp c d N (cid:16) f ( N ) N (cid:17) d − log dd − (cid:18) f ( N ) N (cid:19) M u ( I ) ≥ exp c d N (cid:16) f ( N ) N (cid:17) d − log dd − (cid:18) f ( N ) N (cid:19) . As this holds for every N ∈ N large enough,lim inf R →∞ log ( M u ( R )) R ( f ( R ) R ) d − · log dd − (cid:16) f ( R ) R (cid:17) > .2 The proof of Main Lemma Let K denote the approximation from aboveof the essential part of the set Z u ∩ Q by ele-ments of G \ E , K := (cid:91) I ∈G I (cid:54)∈E I (see Figure 2 to the right), and define thefunction ρ ( x ) := max (cid:110) √ d, r ( x ) (cid:111) , for Figure 2:
The gray area is the set Z u , the cubes with boldboundary are the components of K . The cubes marked with X arebasic cubes, which intersect Z u but do not satisfy (P2) for ε d = . r ( x ) := inf (cid:26) t > , m d (cid:18) K ∩ B (cid:18) x, t (cid:19)(cid:19) ≥ δ · m d ( B (0 , · t d (cid:27) , where m d denotes Lebesgue’s measure on R d , and δ is some constant, which depends on the dimension alone. Observation 2.2
There exists a constant δ d ∈ (0 , , which depends on the dimension d , so that for every xλ d − ( Z u ∩ B ( x, ρ ( x ))) ρ ( x ) d ≥ δ d . Proof. If ρ ( x ) = 2 √ d , then there exists t ∈ (cid:16) , √ d (cid:17) so that m d (cid:18) K ∩ B (cid:18) x, t (cid:19)(cid:19) ≥ δ · m d ( B (0 , · t d , and in particular B (cid:0) x, t (cid:1) intersects K , implying that there exists I ∈ G \ E so that I ⊂ B (cid:18) x, t √ d (cid:19) ⊆ B (cid:16) x, √ d (cid:17) , since t ≤ √ d ⇒ t + √ d ≤ √ d . As I (cid:54)∈ E , it satisfies property (P2) and therefore λ d − (cid:16) Z u ∩ B (cid:16) x, √ d (cid:17)(cid:17) ≥ λ d − (cid:18) Z u ∩ B (cid:18) x, t √ d (cid:19)(cid:19) ≥ λ d − ( Z u ∩ I ) ≥ ε d , implying that λ d − ( Z u ∩ B ( x, ρ ( x ))) ρ ( x ) d ≥ ε d d d d . If ρ ( x ) > √ d , then m d (cid:16) K ∩ B (cid:16) x, ρ ( x )2 (cid:17)(cid:17) ≥ δ · m d ( B (0 , · ρ ( x ) d , implying that B (cid:16) x, ρ ( x )2 (cid:17) intersects atleast (cid:6) δ · m d ( B (0 , · ρ ( x ) d (cid:7) elements in G \ E , by the pigeon-hole principle. On the other hand, ρ ( x ) > √ d implying that ρ ( x )2 + √ d ≤ ρ ( x ). Combining these two observations together, we conclude that λ d − ( Z u ∩ B ( x, ρ ( x ))) ≥ λ d − (cid:18) Z u ∩ B (cid:18) x, ρ ( x )2 + √ d (cid:19)(cid:19) ≥ ε d · (cid:26) I (cid:54)∈ E , I ∩ B (cid:18) x, ρ ( x )2 (cid:19) (cid:54) = ∅ (cid:27) ≥ ε d · (cid:6) δ · m d ( B (0 , · ρ ( x ) d (cid:7) ≥ ε d · δ · m d ( B (0 , · ρ ( x ) d .
9e conclude the proof by choosing δ d := ε d · min (cid:110) − d · d − d , δ · m d ( B (0 , (cid:111) . For every cube I we denote by A ( I, k ) the k th layer of basic cubes surrounding I (see for example thealternating gray and black cubed annuli in Figure 1 in the introduction). Formally, if I = d (cid:81) j =1 [ n j , n j + 1), then A ( I, k ) = d (cid:89) j =1 [ n j − k, n j + 1 + k ) \ d (cid:89) j =1 [ n j − ( k − , n j + k ) . For example in dimension d = 2 if I = [0 , , then A ( I,
1) = [ − , \ [0 , .We will construct a monotone non-decreasing function M : N → N , so that for every k fixed it satisfies(M) { I ∈ G , k ∈ K I } ≥ N d , where K I is a set defined for every basic cube I ∈ G by K I := (cid:26) k ∈ (cid:26) , · · · , N d (cid:27) , ∀ x ∈ Q ∩ A ( I, k ) , ρ ( x ) ≤ M ( k ) (cid:27) . For now, we assume such a function exists.For every cube I ∈ G , define the monotone increasing sequence { κ j } = { κ j ( I ) } in the following way: κ := min { k ∈ K I } κ j := min { k ∈ K I , k > κ j − + M ( κ j − ) } . We choose a cube I so that A (cid:0) I, N d (cid:1) ⊂ Q , and let γ j denote the outer boundary of A ( I, κ j ), and Q j denote theclosed cube whose boundary is γ j . By the way the set K I was defined, for every x ∈ γ j , ρ ( x ) ≤ M ( κ j ) ≤ dist ( γ j , γ j +1 ) ⇒ B ( x, ρ ( x )) ⊂ Q j +1 . By using a rescaled version of Observation 1.1, we deduce that, maybe for a different constant δ d , we have u ( x ) ≤ M u ( B ( x, ρ ( x ))) · e − δ d ≤ M u ( γ j +1 ) e − δ d = M u ( Q j +1 ) e − δ d . Applying this argument inductively and using the maximum principle, if Q denotes the cube I , then M u ( I ) M u ( Q ) = { κ j }− (cid:89) j =0 M u ( Q j ) M u ( Q j +1 ) ≤ exp ( − { κ j } · δ d ) . To conclude the proof, we need to bound from bellow { κ j } for all but at most half of the basic cubes I ∈ G .Note that if the function M is too big, the sequence { κ j } will be a very short sequence, for most cubes. On theother hand, the smaller the function M is, the harder it becomes to satisfy property (M).We shall conclude the proof in three steps: 10tep 1: Show that for at least N d of the basic cubes I ∈ G ( § ) { κ j ( I ) } ≥ N d (cid:88) k =1 M ( k ) . Step 2: Define the function M and prove that it satisfies condition (M).Step 3: Bound the sum N d (cid:80) k =1 1 M ( k ) from bellow. N d of the basic cubes I ∈ G , ( § ) holds. For every basic cube I ∈ G let B ( I ) := (cid:88) k ∈ K I M ( k ) , and define the collection X := I, B ( I ) ≥ N d (cid:88) k =1 M ( k ) . We will first show that every basic cube I ∈ X satisfies ( § ). Since M is a monotone non-decreasing function with M (1) ≥
1, and by the way the sequence { κ j } was defined, for every I ∈ X N d (cid:88) k =1 M ( k ) ≤ B ( I ) = 12 (cid:88) k ∈ K I M ( k ) = 12 { κ j } (cid:88) j =1 (cid:88) k ∈ KIκj ≤ k ≤ κj + M ( κj ) M ( k ) ≤ { κ j } (cid:88) j =1 κ j + M ( κ j ) (cid:88) k = κ j M ( k ) ≤ { κ j } (cid:88) j =1 M ( κ j ) + 1 M ( κ j ) ≤ { κ j } . To conclude the proof it is left to show that X ≥ N d .Following property (M),(Σ) (cid:88) I ∈G B ( I ) = (cid:88) I ∈G (cid:88) k ∈ K I M ( k ) = N d (cid:88) k =1 M ( k ) { I ∈ G , k ∈ K I } ≥ N d N d (cid:88) k =1 M ( k ) . On the other hand, by the way the set X was defined, B ( I ) ≤ N d (cid:80) k =1 1 M ( k ) , I (cid:54)∈ X N d (cid:80) k =1 1 M ( k ) , I ∈ X .
Then (cid:88) I ∈G B ( I ) = (cid:88) I ∈ X B ( I ) + (cid:88) I (cid:54)∈ X B ( I ) ≤ X N d (cid:88) k =1 M ( k ) + (cid:0) N d − X (cid:1) N d (cid:88) k =1 M ( k )= (cid:18) X + 112 N d (cid:19) N d (cid:88) k =1 M ( k ) by (Σ) ≤ N d (cid:18) X + 112 N d (cid:19) (cid:88) I ∈G B ( I ) = (cid:18) XN d + 111 (cid:19) (cid:88) I ∈G B ( I ) ⇒ X ≥ N d , concluding the proof. 11 .2.2 Step 2: constructing the function M : The definition of the function M will heavily depend on acollection of sets, M , with special properties. To definethe collection M we will need the following definition:Given k an interval I is called a dyadic interval of order k if there exists j so that I = ± j · k + (cid:2) , k (cid:1) . A cube C is called a dyadic cube of order k if there are d dyadicintervals of order k , I , I , · · · , I d so that C = d (cid:81) j =1 I j . Wesay a cube C is a dyadic cube if there exists k so that C is a dyadic cube of order k (see Figure 3 to the right). = 1 2 = 22 = 42 = 8 2 = 16 Figure 3:
This figure depicts dyadic cubes. The gray cubeis not a dyadic cube even though it has edge-length 4.
Note that a dyadic cube of order k is a disjoint union of 2 d ( k − j ) dyadic cubes of order j , and that every twodyadic cubes of the same order are disjoint. This means we can define a partial order on the collection of dyadiccubes.We assume without loss of generality that N = 2 N , since for every other N we can find N such that2 N < N ≤ N +1 , and the proof follows from this case with maybe a different constant c d . For every I ∈ G weassign ρ ( I ) = sup x ∈ I ρ ( x ). Then ρ , defined on elements of G , assumes a finite number of values, and one may orderthe cubes I ∈ G as I , · · · , I N d in an ascending ρ order, that is so that ρ ( I k ) ≥ ρ ( I k +1 ). For every I k ∈ G we finda dyadic cube J k so that:1. I k ⊂ J k .2. 2 ρ ( I k ) ≤ (cid:96) ( J k ) < ρ ( I k ).We define the collection M to be the collection of maximal elements { J k } , that is J k ∈ M ⇐⇒ I k (cid:54)⊂ k − (cid:91) i =1 J i , for if I k ⊂ J i for some i < k , then since ρ ( I i ) ≥ ρ ( I k ), J k ⊆ J i . M forms a cover for Q and it is uniquely defined.Let n (cid:96) denote the number of elements in M with edge length 2 (cid:96) . We will define the function M to be a stepfunction, with steps s m := min αN d (cid:80) (cid:96) ≥ m (cid:96) n (cid:96) d − , N d , for some numerical constant α which we shall choose later. It is not hard to see that this is a monotone non-decreasing sequence. We denote the index of the first element satisfying s m = N d by ¯ m + 1 and stop defining the12equence there. Let m be a constant, which will depend on the dimension, and determined later by Claim 2.3,and define the function M ( k ) := m , k ≤ s m m , s m − < k ≤ s m , m ∈ { m + 1 , · · · , ¯ m + 1 } . Though the definition of M may seem forced and unnatural now, we will soon see it is in fact very natural. Toshow that the function M satisfies property (M), we relay on two key claims. The first claim bounds the totalnumber of elements in G with large ρ . We will show that elements in M with large edge length do not cover abig part of Q : Claim 2.3
There exist a numerical constant C and an index m , which depend on the dimension d , so that (cid:88) m ≥ m m · d n m ≤ C E . Proof.
Recall that the maximal function is defined by M f ( x ) := sup B some ballwith x ∈ B m d ( B ) (cid:90) B f ( y ) dm d ( y ) . Let x be so that ρ ( x ) > √ d , then by the way ρ is defined m d (cid:16) K ∩ B (cid:16) x, ρ ( x )4 (cid:17)(cid:17) m d (cid:16) B (cid:16) x, ρ ( x )4 (cid:17)(cid:17) < δ , and, by definition, the maximal function M Q \ K satisfies M Q \ K ( x ) = sup B some ballwith x ∈ B m d ( B ) (cid:90) B Q \ K dm d = sup B some ballwith x ∈ B (cid:18) − m d ( B ∩ K ) m d ( B ) (cid:19) ≥ − m d (cid:16) K ∩ B (cid:16) x, ρ ( x )4 (cid:17)(cid:17) m d (cid:16) B (cid:16) x, ρ ( x )4 (cid:17)(cid:17) ≥ − δ ≥ , for δ appropriately chosen. We note that the same holds for every y ∈ B (cid:16) x, ρ ( x )4 (cid:17) . We conclude that B (cid:16) x, ρ ( x )4 (cid:17) ⊂ (cid:8) M Q \ K ≥ (cid:9) . Using the maximal function theorem,( † ) m d (cid:91) { x, ρ ( x ) > √ d } B (cid:18) x, ρ ( x )4 (cid:19) ≤ m d (cid:18)(cid:26) M Q \ K ≥ (cid:27)(cid:19) (cid:46) m d ( Q \ K ) = E , where the last equality is by the definition of the set K .Let J ∈ M be so that (cid:96) ( J ) > √ d . Then there exists x J ∈ J with ρ ( x J ) ≥ (cid:96) ( J )4 > √ d , implying that J ⊂ B ( x J , √ d · ρ ( x J )). We conclude that 116 √ d J ⊂ B (cid:18) x J , ρ ( x J )4 (cid:19) , √ d J is the cube concentric with J , having edge-length √ d · (cid:96) ( J ).We choose m to be the first integer satisfying that 2 m > √ d . By using the inequality above, since theelements in M are disjoint, ∞ (cid:88) m = m d · m n m = ∞ (cid:88) m = m (2 m ) d n m = (cid:88) J ∈M (cid:96) ( J ) ≥ m m d ( J ) = 16 d · d d (cid:88) J ∈M (cid:96) ( J ) ≥ m m d (cid:18) √ d J (cid:19) = 16 d · d d m d (cid:91) J ∈M (cid:96) ( J ) ≥ m √ d J (cid:46) m d (cid:91) J ∈M (cid:96) ( J ) ≥ m B (cid:18) x J , ρ ( x J )4 (cid:19) ≤ m d (cid:91) { x, ρ ( x ) > √ d } B (cid:18) x, ρ ( x )4 (cid:19) (cid:46) E , by ( † ), concluding the proof.The second claim gives a softer condition to guarantee that property (M) holds. Claim 2.4
There exists a numerical constant C so that for every c small enough if α = C − c C > , thenfor all m ≥ m , for every k ≤ (cid:32) α · N d (cid:80) (cid:96) ≥ m (cid:96) n (cid:96) (cid:33) d − at least of the basic cubes I ∈ G satisfy that ρ | A ( I,k ) ∩ Q ≤ m . It becomes clear now why the sequence { s m } was defined in that particular way: Fix k and let m be so that s m − < k ≤ s m , then by definition M ( k ) = 2 m . Next, k ∈ K I ⇐⇒ ρ | A ( I,k ) ∩ Q ≤ M ( k ) = 2 m . On the other hand, k ≤ s m = α · N d (cid:80) (cid:96) ≥ m (cid:96) n (cid:96) d − . Following the claim above, at least of the basic cubes I ∈ G satisfy that ρ | A ( I,k ) ∩ Q ≤ m , or, in other words, { I, k ∈ K I } ≥ N d , and so the function M satisfies property (M). Proof of Claim 2.4.
For every J ∈ M , for every x ∈ J , ρ ( x ) < (cid:96) ( J )2 , by the way the collection M was defined.In particular, if (cid:96) ( J ) < m , then for every x ∈ J , ρ ( x ) < m as well. We conclude that if A ( I, k ) only intersects J ∈ M with (cid:96) ( J ) < m , then every x ∈ A ( I, k ) must satisfy ρ ( x ) < m , as M forms a cover for Q . Overall, { I, ∃ x ∈ A ( I, k ) , ρ ( x ) > m } ⊆ (cid:91) J ∈M (cid:96) ( J ) > m { I, A ( I, k ) ∩ J (cid:54) = ∅} := (cid:91) J ∈M (cid:96) ( J ) > m B ( J, k ) . We will bound the number of elements in B ( J, k ), i.e given J ∈ M and k ∈ (cid:8) , · · · , N d (cid:9) , how many elements I ∈ G satisfy that J ∩ A ( I, k ) (cid:54) = ∅ ? 14o bound this quantity we will look at two cases. The first and simpler case is when (cid:96) ( J ) > k + 1. Since (cid:96) ( J ) > k + 1, the cube J must contain at least one outer vertex of the set A ( I, k ). We begin by choosing someorder on the outer vertices of A ( I, k ). Then we can associate each basic cube I ∈ B ( J, k ) with the location ofthe first outer vertex of A ( I, k ) lying inside J (see Figure 4b bellow). Because J is a union of basic cubes andso is A ( I, k ), the number of sites in J , in which such a vertex could be found, is equal to the number of basiccubes in J . We can therefore bound from above the number of elements in B ( J, k ), in this case, by the numberof possible vertices in A ( I, k ) times the number of basic cubes in J , that is by 2 d m d ( J ) = 2 d · (cid:96) ( J ) d .The second case is when (cid:96) ( J ) ≤ k + 1. In this case, if J ∩ A ( I, k ) (cid:54) = ∅ , then the boundary of J must intersect A ( I, k ). In particular, there are two adjacent vertices of J one inside (or on) A ( I, k ) and the other outside (oron) A ( I, k ). We can identify every basic cube I in B ( J, k ) by indicating the basic cube on A ([0 , d , k ) wherethe intersection of A ( I, k ) and J occurs, and the basic cube on the one dimensional facet connecting the twoadjacent vertices of J mentioned earlier (see Figure 4a bellow). If more than two vertices satisfy this, we choosean order on the collection of one dimensional facets and use the first one intersecting A ( I, k ). We concludethat the number of elements in B ( J, k ), in this case, is bounded by the number of basic cubes on all the onedimensional facets of J times the number of basic cubes in A ([0 , d , k ), which is bounded by2 d · (cid:96) ( J ) · d · (2 k + 1) d − (cid:46) (cid:96) ( J ) · k d − . A ( I, k ) JI (a) The case (cid:96) ( J ) ≤ k + 1. J IA ( I,k ) (b) The case (cid:96) ( J ) > k + 1. Figure 4:
This figure describes what happens in dimension d = 2. In both figures, the black cube is the basic cube we choose toindicate I ∈ B ( J, k ). { I, ∃ x ∈ J ∩ A ( I, k ) with ρ ( x ) > m } ≤ (cid:88) J ∈M (cid:96) ( J ) > m B ( J, k ) (cid:46) (cid:88) J ∈M (cid:96) ( J ) > m (cid:0) (cid:96) ( J ) d + k d − (cid:96) ( J ) (cid:1) (cid:46) (cid:88) J ∈M (cid:96) ( J ) > m m d ( J ) + k d − (cid:88) J ∈M (cid:96) ( J ) > m (cid:96) ( J ) = (cid:88) (cid:96)>m (cid:96) · d n (cid:96) + k d − (cid:88) (cid:96)>m (cid:96) n (cid:96) . Combining this with Claim 2.3, we see that if k d − ≤ αN d (cid:80) (cid:96) ≥ m (cid:96) n (cid:96) and E ≤ c N d , then there exists a constant C ,which depends on the dimension alone, so that { I, ∃ x ∈ Q ∩ A ( I, k ) with ρ ( x ) > m } ≤ · · · ≤ C (cid:88) (cid:96)>m (cid:96) · d n (cid:96) + C k d − (cid:88) (cid:96) ≥ m (cid:96) n (cid:96) ≤ C · C E + C α · N d ≤ N d ( C C c + C α ) = N d , by the way we defined α . We therefore get that { I, ∀ x ∈ Q ∩ A ( I, k ) with ρ ( x ) ≤ m } = N d − { I, ∃ x ∈ Q ∩ A ( I, k ) with ρ ( x ) > m } ≥ N d − N d
12 = 1112 N d , concluding the proof. N d (cid:80) k =1 1 M ( k ) from bellow: If s m = N d , then by the way the function M is defined, for all k we have M ( k ) = 2 m , implying that for someconstant which depends on the dimension alone, N d (cid:88) k =1 M ( k ) = N d (cid:88) k =1 m = c d · N, which yields a smaller bound (up to multiplication by a constant) than the one indicated in the lemma. Otherwise,assume that s m < N d . Since the function M was defined as a step function, N d (cid:88) k =1 M ( k ) = s m m + ¯ m +1 (cid:88) m = m +1 − m ( s m − s m − ) = s m m + ¯ m +1 (cid:88) m = m +1 − m s m − ¯ m +1 (cid:88) m = m +1 − m s m − = ¯ m +1 (cid:88) m = m − m s m − ¯ m (cid:88) m = m − m s m = 12 ¯ m (cid:88) m = m − m s m + 2 − ¯ m − s ¯ m +1 ∼ N · − ¯ m + ¯ m (cid:88) m = m s m m . To bound ¯ m (cid:80) m = m s m m from bellow, we use Jensen’s inequality with the convex function g ( t ) = t − d − : ¯ m (cid:88) m = m s m m = ¯ m (cid:88) m = m g (cid:18) m ( d − s d − m (cid:19) = ( ¯ m − m + 1) ¯ m (cid:80) m = m g (cid:16) m ( d − s d − m (cid:17) ¯ m − m + 1 ≥ ( ¯ m − m + 1) g ¯ m (cid:80) m = m m ( d − s d − m ¯ m − m + 1 = ( ¯ m − m + 1) d − (cid:18) ¯ m (cid:80) m = m m ( d − s d − m (cid:19) d − .
16o bound ¯ m (cid:80) m = m m ( d − s d − m from above, we note that by definition s m = α · N d (cid:80) (cid:96) ≥ m (cid:96) n (cid:96) d − . This implies that ¯ m (cid:88) m = m m ( d − s d − m = 1 αN d ¯ m (cid:88) m = m m ( d − (cid:88) (cid:96) ≥ m (cid:96) n (cid:96) = 1 αN d (cid:88) (cid:96) ≥ m (cid:96) n (cid:96) (cid:96) (cid:88) m = m m ( d − ∼ αN d (cid:88) (cid:96) ≥ m d · (cid:96) n (cid:96) ∼ E N d , following Claim 2.3. Then ¯ m (cid:88) m = m s m m ≥ ( ¯ m − m + 1) d − (cid:18) ¯ m (cid:80) m = m m ( d − s d − m (cid:19) d − (cid:38) ¯ m d − · (cid:18) N d E (cid:19) d − , implying that N d (cid:88) k =1 M ( k ) ∼ − ¯ m · N + ¯ m (cid:88) m = m s m m (cid:38) − ¯ m · N + ¯ m d − · (cid:18) N d E (cid:19) d − . We would like our lower bound to hold for every ¯ m . To find the optimal inequality, we define the function ϕ ( x ) := 2 − x + x d − · (cid:18) N E (cid:19) d − . As we know nothing about ¯ m , we will look for the absolute minimum of ϕ in (cid:2) , N d (cid:3) and use whatever inequalitythis minimum satisfies: ϕ (cid:48) ( x ) = − − x log(2) + (cid:18) d − (cid:19) (cid:18) N E (cid:19) d − · x d − = 0 ⇐⇒ x d − x ∼ (cid:18) E N (cid:19) d − and the later is a minimum point since ϕ (cid:48)(cid:48) ≥
0. The lower bound this minimum produces isexp − c d · N (cid:16) E N (cid:17) d − log dd − (cid:18) E N (cid:19) , concluding the proof of the Main Lemma. If f ( t ) (cid:46) t , then the upper bound we are looking to prove is exponential. Let W ( x , · · · , x d ) := cos(2 πx ) d (cid:89) j =2 cosh (cid:18) πx j √ d − (cid:19) { | x |≤ } ( x ) . (cid:8) | x | < (cid:9) . By taking the maximum overtranslations and rotations of this function, we create a subharmonic function u so that for every basic cube, I ,we have m d ( Z u ∩ I ) ≥ , while M u ( I ) ≥
1, i.e u oscillates in every basic cube. On the other hand, the growthof the function W is exponential, implying that the growth of the function u is exponential as well, concludingthe proof of the case f ( t ) (cid:46) t . From now on we will assume that f ( t ) (cid:29) t .Though the construction, which we will eventually present in this section, will be applicable for all thedimensions, there is an essential difference between the case of the plane and higher dimensions.We would have loved to use a construction similar to the one presented in [4] of a ‘self similar’ subharmonicfunction. Two issues arise when one tries this approach. The first issue lies in the very essence of ‘self similarity’.The whole idea of ‘self similarity’ means that for every compact set K and for every cube [ − (cid:96), (cid:96) ] d , there are ∼ (cid:96) d disjoint copies of the set K where the function u is defined in the same way, i.e there is some uniform δ > { ω j } δ · (cid:96) d j =1 ⊂ [ − (cid:96), (cid:96) ] d so that ω j + K ∩ ω m + K = ∅ whenever j (cid:54) = m , while for every z ∈ K and j we have u ( z ) = u ( z + ω j ). This implies that if there is one rogue basic cube, then for every (cid:96) large enough there are ∼ (cid:96) d rogue basic cubes in [ − (cid:96), (cid:96) ] d . For every function f , satisfying f ( t ) (cid:28) t d , we accumulate too many roguebasic cubes using this method, and even for f ( t ) ∼ t d we do not get the optimal bound on the growth using thismethod.The second issue arises only in dimensions higher than 3. The construction in [4] uses functions similar to thefunction W , presented in the proof of the case f ( t ) (cid:46) t . The support of these functions is of the form hyperspace × relatively small interval. We will refer to these sets as ‘walls’. These ‘walls’ are used to weld together copies ofthe function u on some compact sets, to get the ‘self similarity’ property. The number of basic cubes lying in theintersection of such a ‘wall’ and a cube of edge length N is N d − × (the length of the relatively small interval).As we are considering the case f ( t ) (cid:29) t , this does not form a problem for the plane. In higher dimensions, thisbecomes a problem whenever f ( t ) (cid:28) t d − , and in fact for any function f ( t ) (cid:28) t d this method will not producethe optimal bound on the growth.We will define a subharmonic function whose support looks like an upside down tree. Instead of using these‘wall’ sets, we will use tube-like sets. The entire support of the function u will be a union of tubes of differentdiameters. This solves the second issue we discussed. To address the first issue, instead of duplicating the samefunction over and over, we will create similar but different functions, which have less and less rogue basic cubes.We formally describe the support of this function bellow. The idea of trees (explained in Subsection 3.2 bellow)was inspired by numerous examples in Jones and Makarov’s paper, [7, Section 6].18 .2 The set We begin our discussion with a description of the set, on which the subharmonic function we create, is supported(in fact, it will be supported on an extremely similar set, but not exactly this set). This set is a union of 2 d disjoint sets, each of them will be kind of an upside down tree described from its tree-top to the trunk.Let { v j } denote the vertices of the cube [ − , d . For every scalar α we write α · v j to describe the rescalingof the point v j by α . For example, let v = (1 , , · · · , α · v = ( α, α, · · · , α ). Let R j denote the linearmap taking the cube v + [ − , d to v j + [ − , d with R j (0) = 0 and R j ( v ) = v j . The set we describe here willbe a union of disjoint copies of one set T ⊂ (0 , ∞ ) d , that is (cid:101) T := (cid:83) d − j =1 R j ( T ). It is therefore enough to describethe set T , which we shall do using a recursive description. Basic terminology and the first steps A tube of diameter δ is any translation or rotation of the set (cid:26) x = ( x , · · · , x d ) , ∀ j ≥ , | x j | < δ , and x ∈ [ a, b ] (cid:27) where a < b are both real numbers. Given a cube Q we will denoteby (cid:96) ( Q ) the edge length of Q .For every basic cube lying inside [0 , d , we connect the center ofthe basic cube, v ∗ , with v by a tube of diameter ε . We placethe tube so that the centerline of the tube is the line connecting v and v ∗ . We will call this structure a basic subtree of leavesdiameter ε and denote it T . For illustration in dimension d = 3see Figure 5 to the right. Figure 5:
A basic subtree in dimension d = 3,composed of 8 = 2 d tubes connecting the basiccubes surrounding v with v . On step 2, we note that [0 , d is composed of 2 d disjoint dyadic cubes of order 1, i.e of edge length 2. In oneof them, [0 , d , we already defined a basic subtree of leaves diameter ε . We shall call it the inner subtree ofrank 1 . In each of the other 2 d − ε in a similar manner tostep 1, for some ε . We will call these the outer subtrees of rank 1 . We then connect the center of each dyadiccube of order 1 to 2 v by a tube of relative diameter ε , i.e the diameter of the tube is 4 ε . These tubes will becalled brunches. More precisely, a brunch will be any tube of diameter bigger than 2 ε . We call this structure a subtree of rank 2 and denote it T . For illustration examples in dimensions 2 and 3, see Figures 6 and 7 bellow. Constructing an outer subtree of rank ( k + 1) In general, we would like the outer subtree of rank ( k + 1)to have two kinds of brunches diameters, corresponding to two stages of the tree.19 " " " " " Figure 6:
This figure depicts the first two steps in dimension d = 2. The figure to the left depicts the first step and the figure tothe right depicts the second step. The gray lines indicate the boundary of the basic cubes. The numbers next to the tubes indicatetheir diameter. (a) Closeup on a subtree of rank two. (b)
A subtree of rank two structure. (c)
A collection of basic subtrees for step two.
Figure 7:
This figure depicts the first 2 steps in dimension d = 3. The basic subtrees in Figure (7c) describes all the basic subtreesthat participate in the second step. They are enumerated to indicate they have different diameters. The first one is the basicsubtree from the first step appearing in Figure 5. It has diameter ε while the others have diameter ε . Figure (7a) zooms in onhow the tubes which are considered ‘brunches’ (when they are wide enough) are connected to the subtrees of rank 1. Given k we will describe the outer subtree of rank ( k + 1) centered at 2 k v . For all the other outer trees ofrank ( k + 1) will be translations and rotations of this structure. Define ε k := 2 s k − k for an integer s k satisfying20hat (cid:18) f (2 k )2 k (cid:19) d − ≤ s d − k s k ≤ (cid:18) f (2 k )2 k (cid:19) d − . We will construct the outer subtree of rank ( k + 1), (cid:93) T k +1 , starting from the outside in, i.e starting from thebig brunches. The cube 2 k v + [ − k , k ) d is composed of 2 d dyadic cubes of order k , i.e of edge length 2 k . Weconnect the center of each of these cubes to 2 k v by tubes of relative diameter ε k , i.e tubes of diameter 2 k ε k . Weplace these tubes so that their centerlines are the lines connecting 2 k v with the center of these dyadic cubes.Next, every dyadic sub-cube of order k is composed of 2 d dyadic sub-cubes of order k −
1. We connect the centerof each such sub-cube with the center of the dyadic cube of order k containing it by a tube of relative diameter ε k . We continue this process for s k steps. We continue to connect dyadic cubes until we reach the cubes of order0, these are basic cubes. Unlike the first s k steps, on the consecutive steps we use tubes of absolute diameter ε .We denote this structure by (cid:93) T k +1 . The general step
To create the subtree of degree k + 1 centered at 2 k v , we take the inner subtree of rank k we already created, T k , and 2 d − (cid:102) T k constructed above as the outer trees ofdegree k centered at 2 k v + 2 k − v j for j ∈ (cid:8) , · · · , d − (cid:9) . We then connect the center of each subtree to 2 k v using a tube of relative diameter δ k := (cid:16) f (2 k )2 k · d (cid:17) d − . We let T denote the limiting set, i.e T = (cid:83) ∞ k =1 T k . Properties of the set T Given a basic cube, I , we say the set T is sparse in I if m d ( I \ T ) > − √ d · (2 ε ) d − > , (the latter holds for ε small enough). We point out that every basic cube intersects the tree T , for if I intersectsa brunch then in particular it intersects T . Otherwise, it intersects a basic subtree, and those have leaves ofdiameter ε , meaning in this case, T is sparse in I . Claim 3.1
For every k (cid:26) basic cubes in [0 , k ) d in which T is NOT sparse (cid:27) (cid:46) f (2 k ) . Proof.
We note that T is sparse in every basic cube which does not intersect a brunch. To bound the number ofbasic cubes in which T is not sparse, we can therefore bound the number of basic cubes which intersect brunches.In addition, by the way the tree T is constructed, it is enough to to look at T k +1 to make such estimates for thecube [0 , k ) d .We begin by bounding the number of basic cubes intersecting brunches in (cid:102) T j , an outer subtrees of rank j . By21he way (cid:102) T j was constructed, all the brunches originate from the first s j steps of the construction: (cid:110) basic cubes intersecting brunches in (cid:102) T j (cid:111) ∼ d m d (cid:16)(cid:110) brunches in (cid:101) T j (cid:111)(cid:17) = s j − (cid:88) (cid:96) =0 (cid:110) dyadic cubes of order ( j − (cid:96) ) in (cid:102) T j (cid:111) · (tube length on step (cid:96) ) · (tube diameter on step (cid:96) ) d − ∼ s j − (cid:88) (cid:96) =0 d · (cid:96) · j − (cid:96) · (cid:0) ε j · j − (cid:96) (cid:1) d − = 2 j · d ε d − j · s j = 2 j · d · (cid:18) s j j (cid:19) d − · (cid:18) s d − j (cid:19) d − = 2 j (cid:18) s j · s d − j (cid:19) d − ≤ j · d − · f (2 j )2 j = 4 d − f (2 j ) , by the way the sequence { s j } was chosen. Now, f is regularly varying, therefore there exists a slowly varyingfunction g so that f ( t ) = t α g ( t ), where α ∈ [1 , d ], as f ( t ) (cid:29) t . This implies that for every t ≥ t large enough f ( t ) f (2 t ) = 12 α · g ( t ) g (2 t ) ∈ (cid:18) d +1 , (cid:19) . We will now bound the number of basic cubes in [0 , k ) d in which T k +1 is not sparse. Let j be the first integersatisfying 2 j > t , and let b j denotes the number of basic cubes intersecting a brunch in T j . Using the estimatewe showed for outer trees, b k = b k − + (cid:26) basic cubes in the ‘handle’connecting the subtrees (cid:27) + (cid:0) d − (cid:1) (cid:26) basic cubes intersecting brunchesin an outer subtree of rank ( k − (cid:27) (cid:46) d b k − + 2 k · (cid:0) k · δ k (cid:1) d − + f (2 k − ) = b k − + 2 k · d · f (2 k )2 k · d + f (2 k − ) ≤ b k − + 2 f (2 k ) (cid:46) k (cid:88) j =1 f (2 j )= j (cid:88) j =1 f (cid:0) j (cid:1) + f (cid:0) k (cid:1) k (cid:88) j = j +1 f (2 j ) f (2 k ) = j (cid:88) j =1 f (cid:0) j (cid:1) + f (cid:0) k (cid:1) k (cid:88) j = j +1 k − (cid:89) (cid:96) = j f (2 (cid:96) ) f (2 (cid:96) +1 ) ≤ j (cid:88) j =1 f (2 j ) + f (cid:0) k (cid:1) k (cid:88) j = j +1 (cid:18) (cid:19) k − j ∼ f (cid:0) k (cid:1) , since we saw that f ( t ) f (2 t ) ≤ . This concluds the proof, since the number of basic cubes in [0 , k ) d , in which T k +1 is not sparse, is bounded by b k plus the basic cubes intersecting the ‘handle’ connecting T k to T k +1 , whichintersects (cid:46) d f (2 k +1 ) ∼ d f (2 k ) basic cubes. In this subsection we shall construct the desired subharmonic function. The support of this function will beclosely related to the ‘tree set’ described in subsection 3.2 above. The first step is to construct a subharmonicfunction which is supported on half tubes.For every ε > T ε : R d → R by T ε ( x ) = T ε ( x , x , · · · , x d ) := cosh (cid:18) π √ d − ε · x (cid:19) d (cid:89) j =2 cos (cid:16) πε · x j (cid:17) · { | x j | < ε } ( x ) , x = ( x , · · · , x d ) ∈ R d . The set { T ε (cid:54) = 0 } is an infinite tube of diameter ε whose centerline is the x -axis.The function T ε is subharmonic as its laplacian is 0 on R d \ (cid:83) dj =2 (cid:8) | x j | = ε (cid:9) and along the partial hyper-planescomposing the boundary of this set, it is continuous and satisfies the mean value property.We define the function L ε ( x ) = max { T ε ( x ) − , } , x ≥ , otherwise . Figure 8 bellow shows the support of the function L π , i.e the set (cid:8) L π > (cid:9) . Figure 8:
This figure shows the surface (cid:110) L π > (cid:111) in dimension 3. Properties of the function L ε
1. The function L ε is continuous and subharmonic as locally it is either zero or the maximum between twosubharmonic functions.2. Define the set G ε := d (cid:92) j =2 (cid:110) | x j | ≤ ε (cid:111) ∩ (cid:26) | x | ≥ ε · d log 2 π √ d − (cid:27) . The set G ε is a closed set, and since for every | t | < ε , cos (cid:0) πε · t (cid:1) ≥ cos (cid:0) πε · ε (cid:1) = , every x ∈ G ε satisfies L ε ( x ) ≥ − ( d − cosh (cid:18) π √ d − ε · x (cid:19) > − ( d − exp (cid:16) π √ d − ε · | x | (cid:17) > − d exp (cid:0) log (cid:0) d (cid:1)(cid:1) = 1 .
3. The set { L ε > } is contained in a tube of diameter ε interseced with { x > } .4. Given a bounded set E ⊂ R d , M L ε ( E ) := sup x ∈ E L ε ( x ) ≤ exp (cid:18) π √ d − ε · sup x ∈ E | x | (cid:19) .We are now ready to construct the function. Our function will be a local uniform limit of a sequence of functions.The idea is that every brunch of diameter δ > T , constructed in Subsection 3.2, will correspond toa rotation and/or translation of the function L δ . Connecting subtrees will be done by taking maximum of somebig constant times L δ ∗ where δ ∗ is the diameter of the brunch connecting the subtrees in T .23 function corresponding to an outer subtree of rank ( k + 1) We first point out that composing L ε ona translation and/or a rotation does not change its growth-rate, only instead of taking supremum over | x | weshould measure the distance along the translated (and/or rotated) x -axis from where the origin was translated.We will construct the function corresponding to the outer subtree, which is centered at 2 k v , in k steps.For every 1 ≤ j ≤ d we keep the notation used in the previous subsection for vertices of the cube [ − , d , v j ,and let L j be the function L k ε k translated and rotated so that the x -axis is aligned with the line connecting2 k v + 2 k − v j with 2 k v , and the point x = (cid:16) ε k k · d log(2) π √ d − , · · · , (cid:17) , is translated to the point 2 k v + 2 k − v j , whilealong the line described by the parametrisation (1 − t ) · (cid:0) k v + 2 k − v j (cid:1) + t · k v the function L j increases with t . Let R ( x ) be the rotation map taking the line connecting 2 k v and its ‘parent vertex’, 2 k +1 v , to the x -axis,and define L ∗ ( x ) = L k ε k ( R ( x )). Define the function v ( x ) = max { L ∗ ( x ) , p · L j ( x ) } , L j ( x ) > x (cid:54)∈ R − (cid:0) G k ε k (cid:1) L ∗ ( x ) , otherwise . We will show that if p is chosen correctly, then v is well defined and subharmonic. It is enough to show that p can be chosen so that on ∂R − (cid:0) G k ε k (cid:1) already L ∗ ( x ) > p · L j ( x ) for all 1 ≤ j ≤ d . Using properties 2 and4 of the function L ε we see thatinf x ∈ ∂R − (cid:16) G kεk (cid:17) L ∗ ( x ) = inf x ∈ ∂G kεk L k ε k ( x ) > p · sup x ∈ ∂R − (cid:16) G kεk (cid:17) L j ( x ) , if we choose 1 p := exp (cid:18) πdε k (cid:19) ≥ sup x ∈ ∂R − (cid:16) G kεk (cid:17) L j ( x ) , since the distance between 2 k v + 2 k − v j and 2 k v is bounded by √ d · k − . We continue to define the functions v , · · · , v s k in a similar manner with L j being translations and/or rotations of the function L k − m ε k and a sequenceof constants p m satisfying that p m p m +1 = exp (cid:18) πdε k (cid:19) . We point out that on step m we have 2 d ( k − m ) independent locations where we perform this ‘glueing procedure’,but it is the same in every location (up to rotations and/or translations). We are therefore able to use the sameconstants, { p m } . After s k steps, we use the same construction but now with L ε and G ε instead of L k − m ε k and G k − m ε k . Then for all j > s k we choose p j p j +1 = exp (cid:18) πd · j ε (cid:19) . We then define τ k +1 ( x ) := p k v k ( x ). 24or every basic cube, I , in which the tree T is sparse, M τ k +1 ( I ) ≥ τ k +1 ≥
1. This makes every basic cube which does NOT intersect a brunch, a cube which satisfies both Property(P1) and Property (P2), i.e τ k +1 oscillates in every basic cube which does not intersect a brunch.To bound the growth of the function τ k +1 we use property 4 of the function L ε for ε = 2 k ε k = 2 s k . Let Q k := 2 k v + [ − k , k ) d , thensup x ∈ Q k τ k +1 ( x ) = 1 p k sup x ∈ Q k L ∗ ≤ p · k − (cid:89) j =1 p j p j +1 · exp (cid:18) π √ d − k ε k · √ d · k (cid:19) ≤ exp πd s k ε k + 1 ε k − s k (cid:88) j =1 j ≤ exp (cid:18) πd (cid:18) s k ε k + 1 ε k − s k (cid:19)(cid:19) = exp (cid:18) πdε k (cid:18) s k + 1 ε (cid:19)(cid:19) ≤ exp (cid:18) πd · s k ε k (cid:19) , for all k large enough, by the maximum principle. Now, s k was chosen so that 1 ≤ s d − k sk (cid:16) f (2 k )2 k (cid:17) d − ≤
4, implying thatsup x ∈ Q k τ k +1 ( x ) ≤ · · · ≤ exp (cid:18) πd · s k ε k (cid:19) = exp (cid:18) πd · s k · k s k (cid:19) ≤ exp πd · s k · k (cid:16) f (2 k ) s k · k (cid:17) d − = exp πd · s dd − k · (cid:0) k (cid:1) dd − f (2 k ) d − ≤ exp (cid:32) πd · k · dd − f (2 k ) d − log dd − (cid:18) f (2 k )2 k (cid:19)(cid:33) := M k . Lastly, we point out that the support of the function τ k +1 is almost the same as an outer subtree of rank k unionwith the support of the ‘handle’, L ∗ . The difference is, we have additional small tips at the beginning of everytube, but the number of additional rogue basic cubes in these tips is bounded by their number in the entirebrunch. Using an argument similar to the one done in Claim 3.1, (cid:8) Rogue basic cubes in [0 , k +1 ) d (cid:9) ≤ c τ f (2 k +1 ) , for some uniform constant c τ . Connecting outer subtrees of rank k to the subtree T k We will construct the desirable function induc-tively, as a local uniform limit of a sequence of functions { u k } satisfying that1. u k is subharmonic.2. u k is supported on [0 , k ) d and a translation and rotation of the function L k δ k (like a handle sticking out),for δ k := (cid:16) f (2 k )2 k · d (cid:17) d − , as defined in the previous subsection.3. For every j < k we have u k | [0 , j ) d ≡ u j . 25. sup [0 , k ) d u k ≤ M k := exp (cid:18) πd · k · dd − f (2 k ) d − log dd − (cid:16) f (2 k )2 k (cid:17)(cid:19) .5. There exists a uniform constant γ so that γ ([0 , k ) d ) ≤ γ · f (2 k ).Let τ be the function constructed above for k = 1, and define u ( x ) = τ ( x ). This function is subharmonicas τ is, and properties 2,3,4 and 5 hold by default. Assume that u k was defined so that it satisfies properties1,2,3,4 and 5 and let us construct the function u k +1 . We note that for every k , the function τ k is supportedon the outer subtree of rank k union with the support of the function L ∗ , the latter being exactly translationsand rotations of the function L k ε k . Informally, to define u k +1 we will ‘wrap’ the support of u k with 2 d − τ k , which we denote by τ jk for 1 ≤ j ≤ d −
1. Wewill only use translations and/or rotations of τ k which are not the opposite of the direction of the component, asif not to have overlaps between the support of u k and the support of τ jk . We then connect these functions with L k +1 δ k +1 in the ‘right direction’.Formally, without loss of generality assume that v d = ( − , − , · · · , − τ jk ( x ) = τ k ◦ R − j ( x − k +1 v ),where R j was the linear mapping taking v +[ − , d to v j +[ − , d so that R j (0) = 0 and R j ( v ) = v j . We wouldlike to ‘glue’ together the definition of the function u k with the functions τ jk . Let L k +1 denote the compositionof L k +1 δ k +1 with a translation and/or rotation which maps the set (cid:110) x = ( x , , · · · , ∈ R d , x ≥ ε k k · d log(2) π √ d − (cid:111) to the line connecting 2 k v with 2 k +1 v . We will denote by B k +1 the set R d \ G k +1 ε k +1 translated and rotatedin the same manner as L k +1 . Let u k +1 ( x ) = max (cid:110) τ jk ( x ) , M k · L k +1 ( x ) (cid:111) , x ∈ B k +1 M k · L k +1 ( x ) , otherwise . To conclude that u k +1 is subharmonic, it is enough show that on ∂B k +1 already τ jk ( x ) < M k · L k +1 ( x ) and soas subharmonicity is a local property, this implies that u k +1 is subharmonic. A computation, similar to the onedone in the construction of a function corresponding to an outer subtree of rank k , shows thatinf x ∈ ∂B k +1 M k · L k +1 ( x ) ≥ M k d − · cosh (cid:18) π √ d − k +1 · δ k +1 · k +1 δ k +1 · d log(2) π √ d − (cid:19) ≥ M k · − d exp (cid:0) log (cid:0) d (cid:1)(cid:1) = M k , M k ≥ sup x ∈ ∂B k +1 max (cid:110) τ k , · · · , τ d − k , u k (cid:111) , based on the induction assumption and the bounds proved for the function τ k in the previous paragraph. Toconclude the proof we need to show that properties 4 and 5 hold for u k +1 , since the rest of the properties areobvious by the construction.To see property 5 holds, we note that if c τ is a constant, so that the number of rogue basic cubes in the outer26rees of rank k is bounded by c τ f (2 k ), then by the induction assumption γ u k +1 f (cid:16)(cid:2) , k +1 (cid:1) d (cid:17) = γ u k f (cid:16)(cid:2) , k (cid:1) d (cid:17) + 2 k +1 (cid:0) k +1 δ k +1 (cid:1) d − + (cid:0) d − (cid:1) γ τ k f (cid:16)(cid:2) , k (cid:1) d (cid:17) ≤ γ · f (2 k ) + f (cid:0) k +1 (cid:1) + c τ (cid:0) d − (cid:1) f (2 k ) = f (cid:0) k +1 (cid:1) (cid:18) f (2 k ) f (2 k +1 ) (cid:0) γ + c τ (cid:0) d − (cid:1)(cid:1)(cid:19) ≤ f (cid:0) k +1 (cid:1) (cid:18) (cid:0) γ + c τ (cid:0) d − (cid:1)(cid:1)(cid:19) , as we saw that for k large enough f (2 k ) f (2 k +1 ) ≤ . If we choose γ so that γ ≥ (cid:0) γ + c τ (cid:0) d − (cid:1)(cid:1) ⇐⇒ γ ≥ (cid:18) c τ (cid:0) d − (cid:1)(cid:19) , then γ u k +1 f (cid:16)(cid:2) , k +1 (cid:1) d (cid:17) ≤ f (cid:0) k +1 (cid:1) (cid:18) (cid:0) γ + c τ (cid:0) d − (cid:1)(cid:1)(cid:19) ≤ γ · f (cid:0) k +1 (cid:1) and property 5 holds.It is left to prove the upper bound on u k +1 , i.e to show that property 4 holds. Following the maximumprinciple and using property 4 of the function L k +1 we see thatsup [0 , k +1 ) d u k +1 = M k · sup [0 , k +1 ) d L k +1 ( x ) ≤ M k · exp (cid:18) π √ d − k +1 δ k +1 · k +1 (cid:19) ≤ M k exp (cid:18) πdδ k +1 (cid:19) . To conclude the proof of the upper bound, we need to show that the latter is bounded by M k +1 , i.e( ♣ ) M k · exp (cid:18) πdδ k +1 (cid:19) ≤ M k +1 . To simplify of the notation, we let ϕ ( t ) := log dd − (cid:16) f ( t ) t (cid:17) . Then M k := exp (cid:32) πd · k · dd − f (2 k ) d − log dd − (cid:18) f (2 k )2 k (cid:19)(cid:33) = exp (cid:32) πd · k · dd − f (2 k ) d − ϕ (2 k ) (cid:33) , and therfore M k · exp (cid:18) πdδ k +1 (cid:19) ≤ M k +1 ⇐⇒ πd · k · dd − f (2 k ) d − ϕ (2 k ) + πd ( k +1) · dd − f (2 k +1 ) d − ≤ πd · ( k +1) · dd − f (2 k +1 ) d − ϕ (cid:0) k +1 (cid:1) ⇐⇒ (cid:18) f (2 k +1 )2 d f (2 k ) (cid:19) d − · ϕ (2 k ) + 14 ≤ ϕ (cid:0) k +1 (cid:1) . Now, f is regularly varying and so f ( t ) = t a g ( t ), where a ∈ [1 , d ] (as f ( t ) (cid:29) t ) and g is slowly varying. Inparticular, f (2 k +1 )2 d f (2 k ) = 2 a ( k +1) · g (cid:0) k +1 (cid:1) d · a · k g (2 k ) = 2 a − d · g (cid:0) · k (cid:1) g (2 k ) ≤ a − d (1 + o (1)) , a < d , a = d g is slowly varying in general, and monotone decreasing in the case a = d . Then ( ♣ ) holds if C a ϕ (2 k ) + 14 ≤ ϕ (cid:0) k +1 (cid:1) , for the constant C a := a − dd − (1 + o (1)) , a < d , a = d . To conclude the proof we will use the following inequality: for every x, y ≥ α ≥ x α + y α ≤ ( x + y ) α . To see this inequality holds, fix y and let ψ ( t ) := ( t + y ) α − ( t α + y α ). Then ψ (cid:48) ( t ) = α (cid:16) ( t + y ) α − − t α − (cid:17) ≥ y ≥ α ≥
1, which implies that ψ is monotone non-decreasing. As ψ (0) = 0, we conclude that ψ ( t ) ≥ t ≥
0, and the inequality follows.Using this inequality, we see that ( ♣ ) holds if C d − d a log (cid:18) f (2 k )2 k (cid:19) + 14 d − d ≤ C d − d a log (cid:18) f (2 k )2 k (cid:19) + 12 ≤ log (cid:32) f (cid:0) k +1 (cid:1) k +1 (cid:33) ⇐⇒ (cid:16) C d − d a − (cid:17) log (cid:18) f (2 k )2 k (cid:19) + 12 ≤ log (cid:32) f (cid:0) k +1 (cid:1) k +1 (cid:33) − log (cid:18) f (2 k )2 k (cid:19) = log (cid:32) f (cid:0) k +1 (cid:1) f (2 k ) (cid:33) . ( ♣♣ )To show that ( ♣♣ ) holds and conclude the proof, we will look into two different cases: • If a < d , then C d − d a − a − dd (1 + o (1)) − ≤ a − d d − < k large enough. On the other hand, since f ( t ) (cid:29) t , then log (cid:16) f (2 k )2 k (cid:17) → ∞ . We conclude that (cid:16) C d − d a − (cid:17) log (cid:18) f (2 k )2 k (cid:19) + 12 ≤ (cid:16) a − d d − (cid:17) log (cid:18) f (2 k )2 k (cid:19) + 12 → −∞ as k → ∞ . In particular, the left hand side of ( ♣♣ ) is bounded from above by ( −
1) for all k large enough.We bound the right hand side of ( ♣♣ ) from bellow by ( −
1) since a ≥ (cid:32) f (cid:0) k +1 (cid:1) f (2 k ) (cid:33) = log (cid:0) a − (cid:1) + log (cid:18) g (2 · k ) g (2 k ) (cid:19) → log (cid:0) a − (cid:1) ≥ k → ∞ , showing that if a < d , then ( ♣♣ ) holds. • If a = d , then the left hand side of ( ♣♣ ) is , and since g is slowly varyinglog (cid:32) f (cid:0) k +1 (cid:1) f (2 k ) (cid:33) = log (cid:0) d − (cid:1) + log (cid:18) g (2 · k ) g (2 k ) (cid:19) = log (cid:0) d − (cid:1) − o (1) ≥ , for all k large enough, showing that if a = d , then ( ♣♣ ) holds as well.28 efining a function on R d Let u denote the local uniform limit of this sequence of functions. To define thefunction u we let u ( x ) := d − (cid:88) j =0 u ◦ R − j ( x ) , for the linear mappings R j taking v to v j and the origin to itself.It is left to bound the number of rogue basic cubes in every large cube centered at the origin. For every kγ uf (cid:0) [ − k , k ] d (cid:1) = 2 d γ u f (cid:0) [0 , k ] d (cid:1) = 2 d γ u k +1 f (cid:0) [0 , k ] d (cid:1) = 2 d · (cid:8) Rogue basic cubes in [0 , k ] d (cid:9) f (2 k ) ≤ d · γ. For a general cube Q = [ − N, N ] d there exists k so that 2 k ≤ N < k +1 and as long as N is large enough γ ( Q ) ≤ f (2 k +2 ) γ ([ − k +1 , k +1 ] d ) f (2 k +1 ) (cid:46) f (2 k +2 ) f (2 k +1 ) = 2 α ( k +2) g (cid:0) · k +1 (cid:1) α ( k +1) g (2 k +1 ) ≤ d +1 . Doing the same construction with f rescaled so that the latter will be less than 1, will only effect the growth bychanging the constants, and so on one hand it will create an f -oscillating subharmonic function, on the otherhand it will have the optimal growth as needed.We conclude this section with two remarks: Remark 3.2
One may wonder what happens if we require some oscillation, without bounding its magnitude frombellow, i.e replace property (P1) with the property ( (cid:102) P ) λ d − ( { u > } ∩ I ) ≥ (cid:101) δ d . Using the same construction described above, with τ k = v k , i.e without rescaling v k by p k , generates an examplewhere all but at most f ( N ) basic squares in (cid:2) − N , N (cid:3) d satisfy properties ( (cid:102) P ) and ( P , while the growth of thefunction is bounded by a polynomial. This shows that a bound from bellow on the magnitude of the oscillation ineach basic cube is essential. Remark 3.3
We cannot use this construction to generate a translation invariant probability measure on thespace of entire functions using the means described in [4]. The reason is, our function does not satisfy condition(10) in Lemma 7, that there exist a sequence of squares { S k } (cid:37) C and a sequence of constants {M k } so that lim k →∞ lim inf n →∞ m (cid:18)(cid:26) w ∈ S n , max z ∈ w + S k u ≤ M k (cid:27)(cid:19) m ( S n ) = 1 . Appendices
The proof we present here is a concatenation of two inequalities. To state and prove these inequalities we willneed the following definitions: Define the function k d ( t ) := log ( t ) , d = 2 − t d − , d ≥ . Following [3], for every measure ν we let p ν ( x ) := (cid:90) R d k d ( | x − y | ) dν ( y ) I ( ν ) := (cid:90) R d p ν ( x ) dν ( x ) = (cid:90) R d (cid:90) R d k d ( | x − y | ) dν ( y ) dν ( x ) . The first is called the potential of the measure ν and the second is called the energy of the measure ν . It is knownthat if the set E is compact, then there exists a probability measure ν so that I ( ν ) = sup ν I ( ν )where the supremum is takes over all the Borel probability measures ν which are supported on E . The measure ν is called the equilibrium measure of E . For more information see, for example, Theorem 5.4 on page 209 of[6]. We are now ready to state these inequalities: Claim 4.1
Let E ⊂ B (cid:0) , (cid:1) be a compact set with λ d − ( E ) > . If ν is the equilibrium measure of E , then λ d − ( E ) (cid:46) d − I ( ν ) . The proof of this claim is a variation of the proof on p.94 in [3]. For the sake of completeness, and since thestatement is not exactly the same, we bring here the full proof.We will use the following result:
Lemma 4.2 (Frostman’s Lemma)
Let ϕ be a gauge function, i.e a positive, increasing function on [0 , ∞ ) with ϕ (0) = 0 . Let K ⊂ R d be a compact set with positive ϕ -Hausdorff content, λ ϕ ( K ) > . Then there is a positiveBorel measure µ on K satisfying that for every ball of radius r , B , µ ( B ) ≤ C d · ϕ ( r ) while µ ( K ) ≥ λ ϕ ( K ) . Theconstant C d depends on the dimension alone. This lemma, and its proof can be found for example as Lemma 3.1.1, p.83 in [3].30 roof of Claim 4.1.
Let µ be the measure we obtain by using Lemma 4.2 with ϕ ( t ) = t d − and K = E . Webegin by bounding from bellow the potential of the measure µ . Then since − k d is a monotone decreasing positivefunction on [0 ,
1] and diam ( E ) ≤ − p µ ( x ) := (cid:90) E − k d ( | x − y | ) dµ ( y ) ≤ ∞ (cid:88) n =0 (cid:90) | x − y | < − n | x − y | < − n − − k d ( | x − y | ) dµ ( y ) ≤ ∞ (cid:88) n =0 − k d (cid:0) − n − (cid:1) (cid:2) µ (cid:0)(cid:8) | x − y | < − n (cid:9)(cid:1) − µ (cid:0)(cid:8) | x − y | < − n − (cid:9)(cid:1)(cid:3) ≤ C d ∞ (cid:88) n =0 − n ( d − (cid:16) − k d (cid:16) − ( n +1) (cid:17)(cid:17) = 2 d − · C d ∞ (cid:88) n =0 − ( n +1)( d − (cid:16) − k d (cid:16) − ( n +1) (cid:17)(cid:17) = 2 d − · C d ∞ (cid:88) n =1 − n ( d − (cid:0) − k d (cid:0) − n (cid:1)(cid:1) = d − · C d ∞ (cid:80) n =1 log (2 n ) 2 − n , d = 22 d − · C d ∞ (cid:80) n =1 2 n ( d − n ( d − , d ≥ − p µ ( x ) ≤ d C d . Next, define the measure dν = dµµ ( E ) . Then ν is a probability measure supported on E , while I ( ν ) = 1 µ ( E ) I ( µ ) = − µ ( E ) (cid:90) R d − p µ ( x ) dµ ( x ) ≥ − d C d µ ( E ) ≥ − d C d λ d − ( E ) . By definition of equilibrium measure we see that I ( ν ) = sup µ I ( µ ) ≥ I ( ν ) ≥ − d C d λ d − ( E ) ⇒ λ d − ( E ) ≤ − d C d I ( ν ) , as I ( ν ) <
0, concluding our proof.
Claim 4.3
Let E ⊂ B (cid:0) , (cid:1) be a compact set with | I ( ν ) | < ∞ . Then ω (0 , E ; B (0 , \ E ) (cid:38) d − I ( ν ) . Proof.
Define the function w : B (0 , → R by w ( x ) = ω ( x, E, B (0 , \ E ) ξ ∈ B (0 , \ E . This function is super-harmonic, and by Poisson-Jensen formula for every x ∈ B (0 , w | ∂B (0 , ≡ (cid:63) ) − w ( x ) = − (cid:90) B (0 , G B ( y, x ) dµ w ( y ) , for µ w the Reisz measure of w , defined by dµ w = ∆ wσ ( ∂B (0 , dm d , where σ d is the surface measure in R d . As E ⊆ B (cid:0) , (cid:1) , there exists a constant c d > G B (0 , y ) > c d on E , implying that( (cid:63)(cid:63) ) ω (0 , E, B (0 , \ E ) = w (0) by ( (cid:63) ) = (cid:90) E G B (0 , y ) dµ w ( y ) ≥ c d · (cid:90) E dµ w ( y ) = c d · || µ w || . ν denote the equilibrium measure of the set E , which is compact. For every measure µ supported on E , by Frostman’s theorem: || µ || = µ ( E ) = − I ( ν ) (cid:90) R d (cid:90) R d − k d ( | x − y | ) dν ( x ) dµ ( x ) (cid:38) d − I ( ν ) (cid:90) E (cid:90) E G B ( x, y ) dµ ( x ) dν ( y ) , for G B Green’s function on B = B (0 , µ = µ w we get that || µ w || ≥ − I ( ν ) (cid:90) E (cid:90) E G B ( x, y ) dµ w ( y ) dν ( x ) by ( (cid:63) ) = − I ( ν ) (cid:90) E w ( x ) dν ( x ) ( a ) = − I ( ν ) (cid:90) E dν ( y ) ( b ) = − I ( ν ) , where ( a ) is since w | E ≡
1, and ( b ) is since ν is a probability measure supported on E .Combining this with ( (cid:63)(cid:63) ), we see that ω (0 , E, B (0 , \ E ) (cid:38) d || µ w || ≥ − I ( ν ) , concluding the proof.Combining these two claims we see that ω (0 , E ; B (0 , \ E ) ≥ − C I ( ν ) ≥ C · C · λ d − ( E ) , concluding the proof of Claim 1.2. It is possible to prove the Main Lemma, Lemma 2.1, by using the Main Lemma in [7] with out repeating Jonesand Makarov’s ingenious idea. This is done using the following lemma, suggested to the author by M.Sodin:
Lemma 4.4
Let u be a subharmonic function defined in a neighbourhood of Q := [ − N, N ] d for some N (cid:29) and define Q = (cid:104) − N √ d , N √ d (cid:105) d . Let E ⊆ Z u ∩ Q be a closed set. Then for every basic cube I ⊂ Q satisfying that λ d − ( E ∩ I ) > ε d > M u ( I ) M u ( Q ) (cid:46) ω ( I ∗ ) ω ( E ) , where ω ( · ) = ω ( ∞ , · ; R d \ E ) and I ∗ is a cube concentric with I having edge length which depends on thedimension alone. In light of this lemma, and using the Main Lemma in [7], we get another proof for Lemma 2.1. For complete-ness, we bring here a reformulation of the Main Lemma in [7]:32 emma 4.5 (The Main Lemma in [7])
Let E ⊂ B (0 , be a compact set , and let ω ( · ) = ω ( ∞ , · ; R d \ E ) . Wesubdivide [0 , d into N d cubes I with side length N and denote by G the whole collection of cubes. We define asubset E ⊂ G (”empty” squares), as follows: I ∈ E if λ d − ( E ∩ I ) ≤ ε d N d − , where λ d − is the ( d − -dimensional dyadic Hausdorff content. If E ≤ c d · N d for some absolute constant c d ,then for at least half of the cubes I ∈ G the following inequality holds: ω ( I ) ω ([0 , d ) < exp (cid:32) − α d (cid:18) N d N + E log d (cid:18) E N (cid:19)(cid:19) d − (cid:33) , where α d is an absolute constant.Alternative proof of Lemma 2.1. Following Lemma 4.4, it is enough to bound ω ( ∞ , I ∗ , R d \ E ) from above.Let I ∈ G \ E be a basic cube and let I ∗ be the corresponding cube from Lemma 4.4. We rescale the set E by (cid:96) ( I ∗ ) N so that rescaling I ∗ , we get a cube of edge-length N , and E ⊂ B (0 , Let g denote Green’s function for the domain Ω := R d \ E with pole at ∞ . Since E ⊆ Q ⊂ B (cid:0) , N (cid:1) and Green’sfunctions satisfy the subordination principle, for every | x | ≥ Ng ( x ) ≥ g R d \ B ( , N )( ∞ , x ) ≥ min | x | = N g R d \ B ( , N )( ∞ , x ) = min | x | =2 g R d \ B (0 , ( ∞ , x ) := c , where c is a constant which depends on the dimension, but independent of N . By the maximum principle, forevery subharmonic function, u , with E ⊂ Z u , u ( x ) M u ( Q ) ≤ ≤ g ( x ) c , implying that M u ( I ) M u ( Q ) (cid:46) M g ( I ) . To conclude the proof we shall bound M g ( I ) from above.Let u be a subharmonic function and let I be a basic cube satisfying property (P2), and let x I ∈ ∂I be apoint satisfying that M u ( I ) = u ( x I ). By taking a square double in edge length, we may assume without lossof generality that dist ( x I , E ∩ I ) > . We note that following Nevanlinna’s first fundamental theorem (see forexample [6, Theorem 3.19]), for every r > √ dM u (cid:16) B (cid:16) x I , √ d (cid:17)(cid:17) ≤ r d − ( r + √ d )( r − √ d ) d − (cid:18) u ( x I ) + max y ∈ E ∩ I G B ( x I ,r ) ( x I , y ) · µ u ( B ( x I , r )) (cid:19) , G B ( x I ,r ) is Green’s function for a ball of radius r centered at x I , and µ u is Riesz measure, defined by dµ u ( x ) = ∆ u ( x ) σ d ( ∂B (0 , dm d ( x ), for σ d the surface area measure in R d .If r = α · √ d , then for every d fixed r d − ( r + √ d )( r − √ d ) d − = α d − d d − · √ d ( α + 1) d d − ( α − d − = α d − ( α + 1)( α − d − = 1 + α (cid:0) − α (cid:1) d − (cid:38) , as α → ∞ . We conclude that for every δ > α so that( (cid:63) ) M u (cid:16) B (cid:16) x I , √ d (cid:17)(cid:17) ≤ (1 + δ ) (cid:18) u ( x I ) + max y ∈ E ∩ I G B ( x I ,α √ d ) ( x I , y ) · µ u ( B ( x I , α √ d )) (cid:19) . Next, since the basic cube I satisfies that λ d − ( E ∩ I ) > ε d >
0, following Observation 1.1, there exists δ d > u ( x I ) < M u (cid:16) B (cid:16) x I , √ d (cid:17)(cid:17) (1 − δ d ) . Combining this with ( (cid:63) ), we note that if we choose α > d , and, inparticular, on δ d ) M u (cid:16) B (cid:16) x I , √ d (cid:17)(cid:17) ≤ (cid:18) δ d (cid:19) (cid:18) u ( x I ) + max y ∈ E ∩ I G B ( x I ,α √ d ) ( x I , y ) · µ u ( B ( x I , α · √ d )) (cid:19) ≤ (cid:18) δ d (cid:19) (cid:16) (1 − δ d ) M u (cid:16) B (cid:16) x I , √ d (cid:17)(cid:17) + c d · µ u ( B ( x I , α · √ d )) (cid:17) , for some constant c d , which depends on the dimension alone.This implies that if I ∗ is a cube centered at x I with edge length 2 α √ d then M u ( I ) ≤ M u (cid:16) B (cid:16) x I , √ d (cid:17)(cid:17) ≤ c d δ d · µ u ( B ( x I , α · √ d )) ≤ c d δ d · µ u ( I ∗ ) . In particular, for u = g we see that M g ( I ) (cid:46) µ g ( I ∗ ) . To conclude the proof of the lemma, we use the following claim:
Claim 4.6
Let Ω be a domain with piecewise analytic boundary. Let g denote Green’s function of Ω with pole at x ∈ Ω . Then for every cube J , µ g ( J ) ∼ d ω ( x , J ; Ω) . Using this claim we conclude the proof by observing that M u ( I ) M u ( Q ) (cid:46) M g ( I ) (cid:46) µ g ( I ∗ ) ∼ d ω ( ∞ , I ∗ , R d \ E ) = ω ( I ∗ ) ω ( E ) . roof of Claim 4.6: By the way harmonic measure is defined we know that ω ( x , J ; Ω) = (cid:90) J dω ( x , · ; Ω) = − σ d ( ∂B (0 , (cid:90) J ∂g∂n y ( y ) dσ d ( y ) , where n y is the outer normal of Ω at the point y ∈ ∂ Ω, and σ d is the surface area measure in R d . Let g + ( x ) = g ( x ) , x ∈ Ω0 , x (cid:54)∈ ΩWe note that g + is continuous in R d and smooth on R d \ ∂ Ω, while µ g = µ g + where both are defined. As µ g + isa distribution, to conclude the proof we need to show that for every compactly supported smooth function ϕ (cid:90) R d g + ( y ) · ∆ ϕ ( y ) dm d ( y ) = − (cid:90) ∂ Ω ϕ ( y ) · ∂g∂n y ( y ) dσ d ( y ) . Let ϕ be such a function. Since on Ω both ϕ and g + are twice continuously differentiable, we may apply Green’ssecond identity, which can be derived from the divergence theorem in any dimension, to conclude that (cid:90) Ω g + ∆ ϕdm d ∆ g | Ω =0 = (cid:90) Ω ( g + ∆ ϕ − ϕ ∆ g + ) dm d By Green’ssecond Id = (cid:90) ∂ Ω (cid:18) g + · ∂ϕ∂n y − ϕ · ∂g + ∂n y (cid:19) dσ d ( y ) g | ∂ Ω =0 = − (cid:90) ∂ Ω ϕ · ∂g + ∂n y dσ d ( y ) g + | Ω = g = − (cid:90) ∂ Ω ϕ · ∂g∂n y dσ d ( y ) . Over all, we see that since g + | R d \ Ω ≡ (cid:90) R d g + ( y ) · ∆ ϕ ( y ) dm d ( y ) = (cid:90) R d \ Ω g + ( y ) · ∆ ϕ ( y ) dm d ( y ) + (cid:90) Ω g + ( y ) · ∆ ϕ ( y ) dm d ( y )= (cid:90) R d \ Ω dm d + (cid:90) Ω g + ∆ ϕdm d = − (cid:90) ∂ Ω ϕ · ∂g∂n y dσ d ( y ) , concluding the proof. [1] A. Baranov and H. Hedenmalm. Boundary properties of Green functions in the plane. Duke Math. J. ,145:1-24,2008.[2] N.H Bingham, C.M Goldie, and J.L Teugels. Regular Variation.
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