Growth of p th means of analytic and subharmonic functions in the unit disc and angular distribution of zeros
aa r X i v : . [ m a t h . C V ] S e p Growth of p th means of analytic andsubharmonic functions in the unit disc andangular distribution of zeros Igor ChyzhykovOctober 9, 2018
In memory of Professor Anatolii Grishin
Abstract
Answering a question of A.Zygmund in [22] G.MacLane and L.Rubeldescribed boundedness of L -norm w.r.t. the argument of log | B | , where B is a Blaschke product. We generalize their results in several directions.We describe growth of p th means, p ∈ (1 , ∞ ), of subharmonic functionsbounded from above in the unit disc. Necessary and sufficient conditionsare formulated in terms of the complete measure (of a subharmonic func-tion) in the sense of A.Grishin. We also prove sharp estimates of thegrowth of p th means of analytic and subharmonic functions of finite orderin the unit disc. In the present paper we investigate an interplay between zero distribution andgrowth of analytic functions in the unit disc D = { z ∈ C : | z | < } . Especiallywe are interested in growth of logarithmic means.Given a sequence ( a n ) in D such that P n (1 − | a n | ) < ∞ , we consider theBlaschke product B ( z ) = ∞ Y n =1 a n ( a n − z ) | a n | (1 − za n ) . (1)It was A. Zygmund (see [22]) who asked to describe those sequences ( a n ) in D that I ( r ) = 12 π Z π − π (log | B ( re iθ ) | ) dθ is bounded. In [22] G. Maclane and L. Rubel answered this question usingFourier series method. 1 heorem A ([22, Theorem 1]) . A necessary and sufficient condition that I ( r ) be bounded is that J ( r ) be bounded, where J ( r ) = ∞ X k =1 k (cid:12)(cid:12)(cid:12) ( r k − r − k ) X | a n |≤ r ¯ a kn + r k X | a n | >r (¯ a kn − a − kn ) (cid:12)(cid:12)(cid:12) . Since it was difficult to check boundedness of J ( r ) they gave also the follow-ing sufficient condition.Let n ( r, B ) be the number of zeros in the closed disc D (0 , r ); here and inwhat follows D ( z, r ) = { ζ ∈ C : | ζ − z | < r } . Theorem B ([22]) . If n ( r, B ) = O ((1 − r ) − ) , r ∈ (0 , , (2) then I ( r ) is bounded. They also noted that (2) is equivalent to the condition X | a n | >r (1 − | a n | ) = O ( √ − r ) . MacLane and Rubel also proved that (2) becomes necessary if all zeros lie on afinitely many rays emanating the origin, but it is not the case in general. Afterthat C. N. Linden ([19, Corollary 1]) generalized this showing that it is sufficientto require that the zero sequence is contained in a finite number of Stolz angleswith vertices on ∂ D . The last assertion is a consequence of the following result.Let R ( re iϕ , σ ) = n ζ : r ≤ | ζ | ≤ r , | arg ζ − ϕ | ≤ σ o . Theorem C ([19, Theorem 1]) . If I ( r ) < M , < r < . Then { a n ∈ R ( re iϕ , κ (1 − r ) γ ) } ≤ C √ M (1 + √ κ ) r (1 − r ) , γ ≥ ,C √ M (1 + √ κ ) r (1 − r ) − γ , ≤ γ < . (3)Results of MacLane and Rubel show that the order of magnitude of the firstestimate (3) is the best possible. Linden ([19]) also established sharpness of theestimate for γ ∈ [0 , p th integralmean of log | f | , where f is analytic in D . Since our approach is natural for sub-harmonic functions we introduce means for the class of subharmonic functionsin D . Note that log | f | is subharmonic provided that f is analytic. Characteri-zation of zeros of analytic functions f with log | f | ∈ L p ( D ) is obtained in [2].For a subharmonic function u in D and p ≥ m p ( r, u ) = (cid:18) π Z π | u ( re iθ ) | p dθ (cid:19) p , < r < ,ρ p [ u ] = lim sup r ↑ log + m p ( r, u ) − log(1 − r ) . m p ( r, log | f | ) was studied in many papers, for instance [20],[21], [23], [26], [12], [27], [3], [7]. Nevertheless, to the best of our knowledge,only one paper, namely [23], contains criteria of boundedness of p th means when u = log | B | . Unfortunately, proofs have not been published yet.In [23] Ya.V. Mykytyuk and Ya.V. Vasyl’kiv introduced two auxiliary func-tions defined on ∂ D by ( a n ): ψ r ( ζ ) = X r ≤| a n | < (1 − | a n | ) | ζ − a n | , ζ ∈ ∂ D , r ∈ [0 , . and ϕ ( ζ ), which satisfies the relation ϕ ( ζ ) ≍ Φ( ζ ) := { a n : | − a n ¯ ζ | < − | a n | ) } , i.e. the number of zeros in the Stolz angle with the vertex ζ . They establishedthat ψ and Φ belong to the same classes L p ( ∂ D ), p ∈ [1 , ∞ ), and ψ log | ψ | and Φ log | Φ | belong to L ( ∂ D ), simultaneously. Moreover, for a branch of log B in D with the radial cuts [ a k , a k | a k | ) the following statement holds. Theorem D ([23]) . Let B be a Blaschke product, and p ∈ (1 , ∞ ) . Then:1) m p ( r, log B ) is bounded on [0 , if and only if ψ ∈ L p ( ∂D ) .2) m ( r, log B ) is bounded if and only if ψ log + ψ ∈ L ( ∂ D ) .3) m p ( r, log | B | ) is bounded on [0 , if and only if sup Let u ∈ SH ∞ , γ ∈ (0 , , p ∈ (1 , ∞ ) . Let λ be the completemeasure of u . Necessary and sufficient that m p ( r, u ) = O ((1 − r ) γ − ) , r ↑ , (6) hold is that (cid:18)Z π λ p ( C ( ϕ, δ )) dϕ (cid:19) p = O ( δ γ ) , < δ < . (7)4 heorem 2. Let f ∈ H ∞ , γ ∈ (0 , , p ∈ (1 , ∞ ) . Let λ be the complete measureof log | f | . Necessary and sufficient that m p ( r, log | f | ) = O ((1 − r ) γ − ) , r ↑ , (8) hold is that (7) . Remark. It was proved in [3] that if supp λ ⊂ ∂ D , i.e., u is harmonic, γ ∈ (0 , 1) then (7) is equivalent to (10). Remark. Though Theorems 1 and 2 look like Carleson type results wecannot use standard tools (e.g. [11, Chap.9]) here, because u and log | f | havelogarithmic singularities.The crucial role in the proof of sufficiency plays Lemma 1. In order toprove necessity of Theorems 1 and 2 we essentially use the fact that kernels inrepresentation (4) preserve the sign. The method allows to spread the sufficientpart of Theorems 1 and 2 to functions of finite order of growth (see Theorems4, 5 below).Under additional assumptions on zero location of a Blaschke product (sup-port of the Riesz measure of the Green potential) ( ?? ) could be simplified. Theorem 3. Let u ( z ) = Z D log | z − ζ || − z ¯ ζ | dµ u ( ζ ) , Z D (1 − | ζ | ) dµ u ( ζ ) < ∞ , (9) α ∈ [0 , , p ∈ (1 , ∞ ) , α + p . Suppose that supp µ u is contained in a finitenumber of Stolz angles with vertices on ∂ D . Necessary and sufficient that m p ( r, u ) = O ((1 − r ) − α )) , r ↑ , (10) hold is that n ( r, u ) := µ u ( D (0 , r ) = O ((1 − r ) − α − p ) , r ↑ . (11)Similar to the case α = 0, the growth condition (11) is appeared to besufficient for (10) when u is of finite order (see Theorem 6 below). Remark. Taking u = log | B | , we obtain a generalization of MacLane andRubel, and Linden’s results mentioned in Subsection 1.1. Remark. If α + p ≥ In order to formulate results on angular distribution for unbounded analyticfunctions we need some growth characteristics. The standard characteristics arethe maximum modulus M ( r, f ) = max {| f ( z ) | : | z | = r } , and the Nevanlinnacharacteristic ([16]) T ( r, f ) = π R π log + | f ( re iθ ) | dθ , x + = max { x, } . Notethat both of them are bounded for f = B . Note that the order defined by T ( r, f ) coincides with ρ [log | f | ].It follows from results of C.Linden [18] that M ( r, f ) does take into accountthe angular distribution of the zeros when it grows sufficiently fast, namely,when the order of growth ρ M [ f ] = lim sup r ↑ log + log + M ( r, f ) − log(1 − r ) ≥ . 5o be more precise, consider the canonical product P ( z, ( a k ) , s ) := ∞ Y k =1 E ( A ( z, a k ) , s ) , where E ( w, s ) = (1 − w ) exp { w + w / · · · + w s /s } , s ∈ Z + , is the Weierstrass primary factor, A ( z, ζ ) = 1 − | ζ | − z ¯ ζ , z ∈ D , ζ ∈ D .Let ˜ (cid:3) ( re iϕ ) := R (cid:16) re iϕ , − r (cid:17) ,ν ( re iϕ ) be the number of zeros of P in ˜ (cid:3) ( re iϕ ). We define ν ( ϕ ) = lim sup r ↑ log + ν ( re iϕ , P ) − ln(1 − r ) , ν [ P ] = sup ϕ ν ( ϕ ) . (12)With the notation above we have ([18, Theorem V]) ρ M [ P ] ( = ν [ P ] , ρ M [ P ] ≥ , ≤ ν [ P ] ≤ , ρ M [ P ] < . (13)Given a Borel measure µ on D satisfying Z D (1 − | ζ | ) s +1 dµ f ( ζ ) < ∞ , s ∈ N ∪ { } , (14)define the canonical integral as U ( z ; µ, s ) := Z D log | E ( A ( z, ζ ) , s ) | dµ ( ζ ) . (15)Let ( q > − S α ( z ) = Γ(1 + q ) (cid:16) − z ) q +1 − (cid:17) , P q ( z ) = Re S q ( z ) , S q (0) = Γ( q + 1) . Note that S and P are the Schwarz and Poisson kernels, respectively.Let u be a subharmonic function in D of the form u ( z ) = U ( z ; µ, s ) − π π Z P s ( ze − iθ ) dψ ∗ ( θ ) + C, (16)where ψ ∗ ∈ BV [0 , π ], µ is the Riesz measure of u satisfying (14). Note thatevery subharmonic function u of finite order in D , i.e. satisfying log max { u ( z ) : | z | = r } = O (log − r ) ( r ↑ s ∈ N ∪ { } ([17], [9, Chap.9]).Let M be Borel’s subset of D such that M ∩ ∂ D is measurable with respectto the Lebesgue measure on ∂ D . Let u be a subharmonic function in D of theform (16). We set λ u ( M ) = Z D ∩ M (1 − | ζ | ) s +1 dµ u ( ζ ) + ψ ( M ∩ ∂ D ) , (17)6here µ u is the Riesz measure of u , ψ is the (signed) Stieltjes measure associatedwith ψ ∗ . Note that, in the case u = log | f | we have µ log | f | ( ζ ) = P n δ ( ζ − a n ),where ( a n ) is the zero sequence of f .Let | λ | denote the total variation of λ . Theorem 4. Let u be a subharmonic function in D of the form (16) , γ ∈ (0 , s + 1] , p ∈ (1 , ∞ ) . Let λ be defined by (17) . If (cid:18)Z π | λ | p ( C ( ϕ, δ )) dϕ (cid:19) p = O ( δ γ ) , < δ < , (18) holds, then m p ( r, u ) = O ((1 − r ) γ − s − ) , r ↑ . (19) Theorem 5. Let f be of the form f ( z ) = C q z ν P ( z, ( a k ) , q ) exp n π π Z S q ( ze − iθ ) dψ ∗ ( θ ) o , (20) where ψ ∗ ∈ BV [0 , π ] , ( a k ) is the zero sequence of f such that P k (1 −| a k | ) q +1 < + ∞ , ν ∈ Z + , C q ∈ C . Let γ ∈ (0 , s + 1] , p ∈ (1 , ∞ ) . Let λ be defined by (17) for u = log | f | . If (cid:18)Z π | λ | p ( C ( ϕ, δ )) dϕ (cid:19) p = O ( δ γ ) , < δ < , (21) holds, then m p ( r, log | f | ) = O ((1 − r ) γ − s − ) , r ↑ . (22) Remark. Suppose that µ is 1-periodic measure on R finite on the compactBorel sets, and p ≥ 1. Then (cid:18)Z (cid:16) µ (( x − δ, x + δ )) (cid:17) p dx (cid:19) p = O ( δ p ) , δ ∈ (0 , . In fact, assume that µ ([0 , C . Then by Fubini’s theorem Z (cid:16) µ (( x − δ, x + δ )) (cid:17) p dx ≤ (2 C ) p − Z µ (( x − δ, x + δ )) dx == (2 C ) p − Z Z ( x − δ,x + δ ) dµ ( y ) dx ≤≤ (2 C ) p − Z δ − δ dµ ( y ) Z ( y − δ,y + δ ) dx ≤ δµ (( − δ, δ )) ≤ C ) p δ. It follows from representation (16) that ρ [ u ] ≤ s . Then (19) implies ρ p [ u ] ≤ s + 1 − p . It is known that this is a sharp inequality ([20, 21]), in general. However,Theorems 4 and 5 characterize classes where ρ p [ u ] takes a particular value.Examples in Section 4 show that the assertion of Theorem 5 is sharp.The following theorem provides a sharp estimate for means of canonicalintegrals or products in terms of growth of their Riesz measures.7 heorem 6. Suppose that u is of the form (15) , s ∈ N ∪ { } , p ∈ (1 , ∞ ) , and α > are such that α + p < s + 1 . If (11) holds, then (10) is valid. K s ( z, ζ ) and representation of func-tions of finite order We define K ( z, ζ ) = G ( z, ζ )1 − | ζ | = 11 − | ζ | log (cid:12)(cid:12)(cid:12) − z ¯ ζz − ζ (cid:12)(cid:12)(cid:12) , z ∈ D , ζ ∈ D , z = ζ, where G ( z, ζ ) is the Green function for D . We have the following properties of K ( z, ζ ), z = re iϕ , ζ = ρe iθ . Proposition 1. The following hold:a) K ( z, 0) = − log | z | .b) ≤ K ( z, ζ ) ≤ −| z | | z − ζ | .c) If D ⋐ D , then uniformly in z ∈ D lim ρ ↑ K ( z, ρe iθ ) = 1 − | z | | ρe iθ − z | = P ( ze − iθ ) . d) | K ( z, ζ ) | ≥ 112 1 − | z | | z − ζ | , − | ζ | ≤ 12 (1 − | z | ) . Proof of the proposition. b) We have0 ≤ K ( re iϕ , ρe iθ ) = 12(1 − ρ ) log 1 − rρ cos( ϕ − θ ) + r ρ r − rρ cos( ϕ − θ ) + ρ == 12(1 − ρ ) log (cid:16) − r )(1 − ρ ) r − rρ cos( ϕ − θ ) + ρ (cid:17) ≤≤ − ρ ) (1 − r )(1 − ρ ) r − rρ cos( ϕ − θ ) + ρ ≤ − r | re iϕ − ρe iθ | . c) The assertion easily follows from the equality K ( z, ζ ) = 12(1 − | ζ | ) log (cid:16) − | z | )(1 − | ζ | ) | z − ζ | (cid:17) see b).d) It is proved in [7].Due to d), we set K ( z, e iθ ) := P ( ze − iθ ) preserving continuity of K on ∂ D with respect to the second variable.Let s ∈ N . We write K s ( z, ζ ) = − log | E ( A ( z, ζ ) , s ) | (1 − | ζ | ) s +1 , ζ ∈ D , z ∈ D , z = ζ, i.e. K ( z, ζ ) = K ( z, ζ ), we set K s ( z, z ) = −∞ , z ∈ D .Let D ∗ ( z, σ ) = { ζ : (cid:12)(cid:12) z − ζ − z ¯ ζ (cid:12)(cid:12) < σ } be the pseudohyperbolic disc with thecenter z and radius σ ∈ (0 , roposition 2. The following hold:i) | K s ( z, ζ ) | ≤ C ( s ) | − z ¯ ζ | s +1 , ζ D ∗ ( z, 17 ) . (23) ii) | K s ( z, ζ ) | ≤ C (1 − | ζ | ) s +1 log (cid:12)(cid:12) − z ¯ ζz − ζ (cid:12)(cid:12) , ζ ∈ D ∗ ( z, 17 ) . (24) iii) If z ∈ D ⋐ D , then K s ( z, ρe iθ ) ⇒ s +1 s + 1 Re 1(1 − ze − iθ ) s +1 = 2 s P s ( ze − iθ )( s + 1)! + C ( s ) , ρ ↑ . Proof of the proposition. The upper estimate for K s ( z, ζ ) K s ( z, ζ ) ≤ s +2 (1 − | ζ | ) s +1 | A ( z, ζ ) | s +1 ≤ s +3 | − z ¯ ζ | s +1 , z ∈ D , ζ ∈ D , (25)follows from the known estimate of the primary factor ([28, Chap. V.10]). Also, K s ( z, ζ ) = Re P ∞ j =1 1 j ( A ( z, ζ )) j (1 − | ζ | ) s +1 , (26)provided that | A ( z, ζ ) | < , so | K s ( z, ζ ) | ≤ | A ( z, ζ ) | s +1 ( s + 1)(1 − | ζ ) s +1 ≤ s +2 s + 1 1 | − z ¯ ζ | s +1 . Hence, it remains to consider the case when | A ( z, ζ ) | ≥ .Since for all z ∈ D , ζ ∈ D | A ( z, ζ ) | ≤ 2, we have for ζ D ∗ ( z, ) K s ( z, ζ )(1 − | ζ | ) s +1 = log (cid:12)(cid:12)(cid:12) z − ζ − z ¯ ζ (cid:12)(cid:12)(cid:12) + Re s X j =1 A ( z, ζ ) j ≥ − log 7 − s X j =1 j j == − C ( s ) ≥ − C ( s )2 s +1 | A ( z, ζ ) | s +1 . Hence, K s ( z, ζ ) ≥ − C ( s ) | − z ¯ ζ | s +1 , ζ D ∗ ( z, 17 ) , (27)and i) follows.Similar arguments give ii).Let us prove iii). If z ∈ D ⋐ D , then it follows from the representation (26)that K s ( z, ρe iθ ) ⇒ s +1 s + 1 Re 1(1 − ze − iθ ) s +1 = 2 s P s ( ze − iθ )( s + 1)! + C ( s ) , ρ ↑ . K s ( z, ζ ) the representation (16) could be rewritten inthe form (cf. [14], [13, Part II]) u ( z ) = − Z D K s ( z, ζ ) dλ ( ζ ) + C, (28)where s ( s +1)! dλ ( e iθ ) = π dψ ∗ ( θ ), and λ = λ u is defined by (17).Similarly, for any u ∈ SH ∞ we have the following representation (cf. (4)) u ( z ) = − Z ¯ D K ( z, ζ ) dλ u ( ζ ) + C. (29) Remark. The idea of such representation goes back to results of Martin [1,Chap. XIV]. Sufficiency of Theorem 1. We write u ( z ) = − Z D ∗ ( z, ) K ( z, ζ ) dλ ( ζ ) , u ( z ) = − Z ¯ D \ D ∗ ( z, ) K ( z, ζ ) dλ ( ζ ) . Let us estimate I = R π − π | u ( re iϕ ) | p dϕ .By the H¨older inequality | u ( re iϕ ) | = Z D ∗ ( re iϕ , ) log (cid:12)(cid:12)(cid:12) − re iϕ ¯ ζre iϕ − ζ (cid:12)(cid:12)(cid:12) dµ ( ζ ) ≤≤ (cid:18) Z D ∗ ( re iϕ , ) (cid:16) log (cid:12)(cid:12)(cid:12) − re iϕ ¯ ζre iϕ − ζ (cid:12)(cid:12)(cid:12)(cid:17) p dµ ( ζ ) (cid:19) p (cid:16) µ (cid:16) D ∗ (cid:16) re iϕ , (cid:17)(cid:17)(cid:17) p − p , hence I ≤ Z π − π (cid:18) Z D ∗ ( re iϕ , ) (cid:12)(cid:12)(cid:12) log (cid:12)(cid:12)(cid:12) − re iϕ ¯ ζre iϕ − ζ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p dµ ( ζ ) µ p − (cid:16) D ∗ (cid:16) re iϕ , (cid:17)(cid:17)(cid:19) dϕ. Since D ∗ ( z, h h ) ⊂ D ( z, (1 − | z | ) h ) ([5]), with h = we get D ∗ (cid:16) z, (cid:17) ⊂ D (cid:16) z, (1 − | z | ) 13 (cid:17) ⊂ (cid:3) (cid:16) z, 13 (1 − | z | ) (cid:17) , where (cid:3) ( re iϕ , σ ) = { ρe iθ : | ρ − r | ≤ σ, | θ − ϕ | ≤ σ } . Therefore, using Fubini’s10heorem, we deduce I ≤ Z π − π (cid:18)Z (cid:3) ( z, (1 − r )) (cid:16) log (cid:12)(cid:12)(cid:12) − re iϕ ¯ ζre iϕ − ζ (cid:12)(cid:12)(cid:12)(cid:17) p (cid:19) µ p − (cid:16) (cid:3) (cid:0) z, 13 (1 − r ) (cid:1)(cid:17) dµ ( ζ ) dϕ ≤≤ Z π − π (cid:18)Z (cid:3) ( z, (1 − r )) (cid:16) log (cid:12)(cid:12)(cid:12) − re iϕ ¯ ζre iϕ − ζ (cid:12)(cid:12)(cid:12)(cid:17) p (cid:19) µ p − (cid:16) (cid:3) (cid:0) re i arg ζ , 23 (1 − r ) (cid:1)(cid:17) dµ ( ζ ) dϕ = ≤ Z Z − π − − r ≤ θ ≤ π + − r | ρ − r |≤ − r | θ − ϕ |≤ − r (cid:16) log (cid:12)(cid:12)(cid:12) − rρe i ( ϕ − θ ) re iϕ − ρe iθ (cid:12)(cid:12)(cid:12)(cid:17) p µ p − (cid:16) (cid:3) (cid:0) re iθ , 23 (1 − r ) (cid:1)(cid:17) dµ ( ρe iθ ) dϕ = ≤ Z || ζ |− r |≤ (1 − r ) µ p − (cid:16) (cid:3) (cid:0) re i arg ζ , 23 (1 − r ) (cid:1)(cid:17) Z π − π (cid:16) log (cid:12)(cid:12)(cid:12) − re iϕ ¯ ζre iϕ − ζ (cid:12)(cid:12)(cid:12)(cid:17) p dϕ dµ ( ζ ) . We know that ([6]) for any a, b ∈ C , and p > Z π − π (cid:12)(cid:12)(cid:12) log (cid:12)(cid:12)(cid:12) a − e iθ b − e iθ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p dθ ≤ C ( p ) | a − b | holds. Using this inequality we obtain ( r ∈ ( , I ≤ C ( p )(1 − r ) Z || ζ |− r |≤ (1 − r ) µ p − (cid:16) (cid:3) (cid:0) re i arg ζ , 23 (1 − r ) (cid:1)(cid:17) dµ ( ζ ) . (30)In order to proceed we need the following lemma. Lemma 1. Let ν be a π periodic positive Borel measure on R , p ≥ , δ ∈ (0 , π ) .Then Z [ − π,π ) ν p − (( θ − δ, θ + δ )) dν ( θ ) ≤ p +1 δ Z [ − π,π ) ν p (( θ − δ, θ + δ )) dθ. (31) Proof of Lemma 1. First, we prove (31) for p = 1 .We have Z [ − π,π ) dν ( θ ) = Z [ − π,π ) δ Z θ + δ θ − δ dxdν ( θ ) ≤ Z [ − π − δ ,π + δ ) dx Z [ x − δ ,x + δ ) δ dν ( θ ) == Z [ − π − δ ,π + δ ) ν ([ x − δ , x + δ )) δ dx ≤ Z [ − π,π ) ν ([ x − δ , x + δ )) δ dx ≤ (32) ≤ Z [ − π,π ) ν (( x − δ, x + δ )) δ dx. (33)We now consider arbitrary finite p > 1. Applying (32) with dν ( θ ) = The author thanks Prof. Sergii Favorov for the idea of the proof of this lemma. p − (( θ − δ, θ + δ )) dν ( θ ), we get Z [ − π,π ) ν p − (( θ − δ, θ + δ )) dν ( θ ) = Z [ − π,π ) dν ( θ ) ≤ Z [ − π,π ) ν ([ x − δ , x + δ )) δ dx == 2 Z [ − π,π ) Z [ x − δ ,x + δ ) ν p − (( θ − δ, θ + δ )) dν ( θ ) dx ≤≤ Z [ − π,π ) ν p − (( x − δ , x + δ )) ν ([ x − δ , x + δ )) δ dx ≤≤ Z [ − π,π ) ν p (( x − δ , x + δ )) δ dx ≤≤ Z [ − π,π ) ν p (( x − δ , x ) ∪ [ x, x + δ )) δ dx ≤≤ p Z [ − π,π ) ν p (( x − δ , x )) δ dx + 2 p Z [ − π,π ) ν p ([ x, x + δ , x )) δ dx ≤≤ p +1 Z [ − π,π ) ν p (( x − δ, x − δ )) δ dx. The lemma is proved.Let us continue the proof of the sufficiency.We denote the nondecreasing function N r ( θ ) = λ ( { ρe iα : | r − ρ | ≤ 23 (1 − r ) , − π ≤ α ≤ θ } ) , θ ∈ [ − π, π ) . We extend it on the real axis preserving monotonicity by N r ( x + 2 π ) − N r ( x ) = N r (2 π ) − N r (0), x ∈ R . Let ν r be the corresponding Stieltjes measure on R .Estimate (30) can be written in the form I ≤ C (1 − r ) p − Z || ζ |− r |≤ (1 − r ) λ p − (cid:16) (cid:3) (cid:0) re i arg ζ , 23 (1 − r ) (cid:1)(cid:17) dλ ( ζ ) == C (1 − r ) p − Z π − π ν p − r (cid:16)h θ − 23 (1 − r ) , θ + 23 (1 − r ) i(cid:17) dν r ( θ ) ≤≤ p +1 C − r ) p Z π − π ν pr (cid:16)h θ − 23 (1 − r ) , θ + 23 (1 − r ) i(cid:17) dθ ≤≤ C ( p )(1 − r ) p Z π − π λ p (cid:16) (cid:3) (cid:0) re iθ , 23 (1 − r ) (cid:1)(cid:17) dθ ≤ C (1 − r ) p (1 − r ) pγ . We have used Lemma 1 and the assumption of the theorem on the completemeasure.Thus, we have (cid:16)Z π − π | u ( re iϕ ) | p dϕ (cid:17) p ≤ C ( p )(1 − r ) γ − . (34)Let us estimate u ( z ) = − R ¯ D K ( z, ζ ) d ˜ λ ( ζ ), where d ˜ λ ( ζ ) = χ ¯ D \ D ∗ ( z, ) ( ζ ) dλ ( ζ ).12ince supp ˜ λ ∩ D ∗ ( z, ) = ∅ , for ζ D ∗ ( z, ) we have by Proposition 1 that K ( z, ζ ) ≤ − | z | ) | − z ¯ ζ | . (35)Let E n = E n ( re iϕ ) = C ( ϕ, n (1 − r )), n ∈ N , E = ∅ . Then for ζ = ρe it ∈ D \ E n ( z ), n ≥ 1, we have | − ρre i ( ϕ − t ) | ≥ | − ρe i ( ϕ − t ) | − ρ (1 − r ) ≥ n (1 − r ) − (1 − r ) ≥ n − (1 − r ) . and | − ρre i ( ϕ − t ) | ≥ − rρ ≥ − r for ζ ∈ E ( z ). Therefore ( p + p ′ = 1) | u ( re iϕ ) | p ≤ (cid:18)(cid:18) [log − r ] X n =1 Z E n +1 \ E n + Z E (cid:19) − r ) | − re iϕ ¯ ζ | d ˜ λ ( ζ ) (cid:19) p ≤≤ p (cid:18) [log − r ] X n =1 Z E n +1 \ E n − r )(2 n − (1 − r )) d ˜ λ ( ζ ) + Z E − r d ˜ λ ( ζ ) (cid:19) p < ≤ (cid:16) − r (cid:17) p [log − r ]+1 X n =1 ˜ λ p ( E n ( z ))2 np (cid:18) ∞ X n =1 np ′ (cid:19) pp ′ ≤ C ( p )(1 − r ) p ∞ X n =1 ˜ λ p ( E n ( z ))2 np . (36)It follows from the latter inequalities and the assumption of the theoremthat ( r ∈ [ , Z π | u ( re iϕ ) | p dϕ ≤ C ( p )(1 − r ) p ∞ X n =1 Z π ˜ λ p ( E n ( re iϕ ))2 np dϕ ≤≤ C ( p )(1 − r ) p ∞ X n =1 (2 n (1 − r )) pγ np = C ( p )(1 − r ) p (1 − γ ) ∞ X n =1 np ( γ − ) = C ( p, γ )(1 − r ) p (1 − γ ) . (cid:16)Z π | u ( re iϕ ) | p dϕ (cid:17) p ≤ C ( γ, p )(1 − r ) − γ , r ∈ [0 , . Sufficiency of Theorem 1 is proved. Necessity. Using property d) of Proposition 1, we obtain | u ( re iθ ) | ≥ Z C ( ϕ, − r ) K ( re iϕ , ζ ) dλ ( ζ ) ≥ Z C ( ϕ, − r ) − r | re iϕ − ζ | dλ ( ζ ) . Elementary geometric arguments show that | re iϕ − ρe iθ | ≤ | re iϕ − e iθ | for 1 >ρ ≥ r ≥ 0. It then follows that | u ( re iθ ) | ≥ Z C ( ϕ, − r ) − r | re iϕ − e iθ | dλ ( ρe iθ ) ≥≥ π + 1) 1 − r (1 − r ) Z C ( ϕ, − r ) dλ ( ρe iθ ) ≥ λ ( C ( ϕ, − r ))3( π + 1)(1 − r ) . By the assumption of the theorem we deduce that C (1 − r ) (1 − γ ) p ≥ Z π | u ( re iϕ ) | p dϕ ≥ C R π λ p ( C ( ϕ, − r )) dϕ (1 − r ) p . R π λ p ( C ( ϕ, − r )) dϕ = O ((1 − r ) γp ) as r ↑ 1. This completes the proofof necessity. Proof of Theorem 3. Due to (7) we write u ( z ) = − Z ¯ D K s ( z, ζ ) dλ ( ζ ) == − Z D ∗ ( z, ) K s ( z, ζ ) dλ ( ζ ) − Z ¯ D \ D ∗ ( z, ) K s ( z, ζ ) dλ ( ζ ) ≡ u + u . According to (24) | u ( z ) | ≤ C ( s ) Z D ∗ ( z, ) log (cid:12)(cid:12)(cid:12) − z ¯ ζz − ζ (cid:12)(cid:12)(cid:12) dµ ( ζ ) . Its estimate repeats that for the case s = 0.Let us estimate p th means of u ( z ). Using Proposition 2 we deduce (cf.proof of Theorem 1) | u ( re iϕ ) | p ≤ (cid:18)(cid:18) [log − r ] X n =1 Z E n +1 \ E n + Z E (cid:19) C ( s ) | − re iϕ ¯ ζ | s +1 | d ˜ λ ( ζ ) | (cid:19) p ≤≤ C (1 − r ) ( s +1) p [log − r ]+1 X n =1 | ˜ λ | p ( E n ( z ))2 ( s + ) np (cid:18) ∞ X n =1 np ′ (cid:19) pp ′ ≤≤ C (1 − r ) ( s +1) p ∞ X n =1 | ˜ λ | p ( E n ( z ))2 ( s + ) np . It follows from the latter inequalities and the assumption of the theorem that Z π | u ( re iϕ ) | p dϕ ≤ C ( p )(1 − r ) ( s +1) p ∞ X n =1 Z π | ˜ λ | ( E n ( re iϕ ))2 ( s + ) np dϕ ≤≤ C ( p )(1 − r ) ( s +1) p ∞ X n =1 (2 n (1 − r )) pγ ( s + ) np = C ( p, γ )(1 − r ) p ( s +1 − γ ) , r ∈ h , (cid:17) . Finally, (cid:16)Z π | u ( re iϕ ) | p dϕ (cid:17) p ≤ C ( γ, p )(1 − r ) s +1 − γ , r ∈ h , (cid:17) . Proof of Theorem 3. Without loss of generality we assume that supp µ u ⊂ { z ∈ D : | − z | < − | z | ) } =: △ . Neccesity. Note that R (1 − δ, δ ) ⊂ C ( ϕ, δ ) for ϕ ∈ [ − δ, δ ]. Applying Theo-rem 1 we obtain (cid:18)Z δ − δ λ p ( R (1 − δ, δ )) dϕ (cid:19) p = O ( δ − α ) , < δ < , or µ u ( R (1 − δ, δ )) = O ( δ − α − p ) , < δ < . △ ⊂ D (0 , 12 ) ∪ ∞ [ n =1 R (1 − − n , − n ) (37)we deduce n (1 − − k , u ) ≤ C k X n =1 n ( α + p ) + C = O (2 k ( α + p ) ) , k ∈ N , and the assertion follows. Sufficiency. It follows from the assumptions that λ ( R ((1 − δ ) e iϕ , δ )) = O ( δ − α − p ) , δ ↓ . Then λ ( C ( ϕ, δ )) ≤ λ (cid:16) ∞ [ n =0 R (1 − δ n e iϕ , δ n ) (cid:17) ≤ C ∞ X n =0 (cid:16) δ n (cid:17) − α − p = O ( δ − α − p ) , δ ↓ . Since supp µ u ⊂ △ , we have Z π − π λ p ( C ( ϕ, δ )) dϕ = Z πδ − πδ λ p ( C ( ϕ, δ )) dϕ = O ( δδ p (1 − α − p ) ) = O ( δ p (1 − α ) ) , δ ↓ . It remains to apply Theorem 1. Proof of Theorem 6. We confine ourselves to the case s = 0. We keep thenotation from the proof of Theorem 1. It follows from estimate (30) that Z π − π | u ( re iϕ ) | p dϕ ≤ (1 − r ) n p − (cid:16) ( r + 23 (1 − r ) , u (cid:17) ) n (cid:16) ( r + 12 (1 − r ) , u (cid:17) ) = O (cid:16) (1 − r ) − αp (cid:17) . (38)Let us estimate p th mean of u . We use estimate (35), integral Minkowski’sinequality ([25, § A1]), standard estimates, and integration by parts (cid:16)Z π − π | u ( re iϕ ) | p dϕ (cid:17) p ≤ C (cid:16)Z π − π (cid:16)Z D − r | − re iϕ ¯ ζ | dλ ( ζ ) (cid:17) p dϕ (cid:17) p ≤≤ C Z D (cid:16)Z π − π (cid:16) − r | − re iϕ ¯ ζ | (cid:17) p dϕ (cid:17) p dλ ( ζ ) ≤ C Z D − r (1 − r | ζ | ) − p dλ ( ζ ) == C (1 − r ) Z (1 − t ) dn ( t, u )(1 − rt ) − p ≤ C (1 − r ) (cid:18)Z r dn ( t, u )(1 − t ) − p ++ Z r (1 − t ) dn ( t, u )(1 − r ) − p (cid:19) ≤ C (1 − r ) − p Z r n ( t, u ) dt = O ((1 − r ) − α ) , r ↑ . Taking into account (38), we obtain desired estimate.15 Examples Example 1. Following Linden [21, Lemma 1], given α ≥ β ∈ [0 , a k,m = (1 − − k ) e im − k , ≤ m ≤ [2 kβ ] (39)where each of numbers (39) is counted [2 αk ] times. Then for P ( z ) = P ( z, ( a k,m ) , s ),where s = min { q ∈ N : q > α + β − } we have (see [21]) n ( r, P ) ≍ (cid:16) − r (cid:17) α + β , ν ( r, P ) ≍ (cid:16) − r (cid:17) α , r ↑ . Therefore, by Theorem A [18] ρ M [ P ] = α . In [21] itis proved that ρ p [log |P| ] = α + β − p . We are going to prove that (cid:18)Z π λ p ( C ( ϕ, δ )) dϕ (cid:19) p ≥ C ( δ s +1 − α − β − p ) , δ ↓ . (40)It would imply that restriction (18) could not be weakened.We first assume that β ∈ (0 , δ ∈ (0 , δ ) we define ϕ δ = δ − β − πδ ,where δ is chosen such that ϕ δ > 0. Note that ϕ δ ∼ δ − β , δ ↓ 0. According tothe definition of C ( ϕ, δ ), a k,m ∈ C ( ϕ, δ ) if and only if1 − | a k,m | = 2 − k ≤ δ, ϕ − πδ ≤ m − k ≤ ϕ + πδ. (41)Let G ( ϕ, δ ) denote the set of ( k, m ) such that (41) is valid. It is easy to checkthat for ϕ ∈ (0 , ϕ δ ) the set G ( ϕ, δ ) is not empty. Let k ( ϕ ) = min { k : 2 − k [2 βk ] ≤ ϕ + πδ } , where ϕ ∈ (0 , ϕ δ ). Since k ( ϕ ) tends to infinity uniformly with respect to ϕ ∈ (0 , ϕ δ ) as δ ↓ 0, one can choose δ so small that for all δ ∈ (0 , δ ), ϕ ∈ (0 , ϕ δ ) and k ≥ k ( ϕ ) the inequality − βk (1 − β )(1 − βk ) log 2 ≤ k = k ( ϕ ) that | k ( ϕ + πδ ) − βk | < , − − βk ϕ + πδ < k (1 − β ) < − βk ϕ + πδ , (42) | k − − β log ϕ + πδ | < . It follows from the definition of ϕ δ and (42) that2 k > (1 − − k β ) − β δ > πδ , < ϕ < ϕ δ , (43)Then, according to (41), (43) for δ ∈ (0 , min { δ , δ } ) and ϕ ∈ ( ϕ δ , ϕ δ ), δ ↓ λ ( C ( ϕ, δ )) = X ( k,m ) ∈ G ( ϕ,δ ) [2 αk ]2 − k ( s +1) ≥ [2 k ( ϕ + πδ )] X m =[2 k ( ϕ − πδ )]+1 [2 αk ]2 − k ( s +1) ≥≥ [2 αk ]2 − k ( s +1) (2 k πδ − ≥ [2 αk ]2 − k s πδ ≥ πδ ( α − s ) k ∼ πδ (cid:16) ϕ (cid:17) α − s − β . (44)16t follows from the last estimate that (cid:18)Z π ( λ ( C ( ϕ, δ ))) p dϕ (cid:19) p ≥ πδ (cid:18)Z ϕ δ ϕ δ / ( ϕ s − α − β ) p dϕ (cid:19) p == π s − α − β p + 1) δϕ s − α − β + p (cid:12)(cid:12)(cid:12) ϕ δ ϕ δ / ∼ C ( s, α, p ) δ s − α − β − p , δ ↓ . In the case β = 1 the arguments could be simplified. By the choice of s , s > α . For 0 < ϕ ≤ , according to (41) we deduce λ ( C ( ϕ, δ )) = ∞ X k =[log δ ]+1 [2 k ( ϕ + πδ )] X m =[2 k ( ϕ − πδ )]+1 [2 αk ]2 − k ( s +1) ≥≥ ∞ X k =[log δ ]+1 [2 αk ]2 − k ( s +1) (2 k πδ − ≥ ∞ X k =[log δ ]+1 [2 αk ]2 − ks πδ ≥≥ πδ ∞ X k =[log δ ]+1 ( α − s ) k ≍ δ s − α . Hence, (cid:18)Z π ( λ ( C ( ϕ, δ ))) p dϕ (cid:19) p ≥ C ( s, α, p ) δ s − α , δ ↓ . If β = 0, then all zeros a k = 1 − − k are located on [0 , s > α − 1. For ϕ ∈ ( − πδ, πδ ) we then have according to (41) λ ( C ( ϕ, δ )) = ∞ X k =[log δ ]+1 [2 αk ]2 − k ( s +1) ≥ C ∞ X k =[log δ ]+1 ( α − s − k ≍ δ s − α . 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