On a Blaschke-type condition for subharmonic functions with two sets of singularities on the boundary
aa r X i v : . [ m a t h . C V ] J a n ON A BLASCHKE-TYPE CONDITION FORSUBHARMONIC FUNCTIONS WITH TWO SETS OFSINGULARITIES ON THE BOUNDARY
S. FAVOROV AND L. GOLINSKII
Abstract.
Given two compact sets, E and F , on the unit circle, westudy the class of subharmonic functions on the unit disk which cangrow at the direction of E and F (sets of singularities) at differentrate. The main result concerns the Blaschke-type condition for the Rieszmeasure of such functions. The optimal character of such condition isdemonstrated. To Victor Katsnelson on occasion of his 75-th anniversary
Introduction
In 1915, around a century ago, a seminal paper (6-pages note!) [2] byW. Blaschke came out. A condition widely known nowadays as the
Blaschkecondition for zeros of bounded analytic functions on the unit disk D (0.1) X ζ ∈ Z ( f ) (1 − | ζ | ) < ∞ was announced in this gem of Complex Analysis. Around 50 years agoboth the authors learned about the Blaschke condition from VK, being hisgraduate students.It is not our intention reviewing a vast literature with various refinementsand far reaching extensions of (0.1), which appeared since then. We mentiononly that in all such extensions the majorants of the (unbounded) functionsin question were radial , that is, they depended on the absolute value of theargument. In other words, the function was allowed to grow uniformly nearthe unit circle T .We came across functions with non-radial growth for the first time in aresult of Killip and Simon [12, Theorem 2.8], where this bound looked(0.2) log | L ( z, J ) | ≤ C | z − | , z ∈ D . In the spectral theory setting of this paper the function L (the perturbationdeterminant) turned out to belong to the Nevanlinna class, so its zerossatisfied (0.1). Date : January 10, 2019.
Key words and phrases. subharmonic functions; Riesz measure; harmonic majorant;the Green’s function; layer cake representation; harmonic measure.
The question arose naturally what one could say about the zeros of ageneric function which can grow at the directions toward some selectedcompact sets on T (we refer to these sets as the sets of singularities ). Forexample, in (0.2) this set is E = {± } . The study of such functions andtheir zero sets was initiated in [3, 4] for analytic functions, and in [6, 7]for subharmonic functions on D . To remain closer to the main subject ofour paper – functions with two sets of singularities on T – we mention tworesults from the preceding papers.Given a compact set F ⊂ T , denote by ρ F ( w ) the Euclidian distance froma point w ∈ C to the set F . Recall the following quantitative characteristicof F known as the Ahern–Clark type [1] α ( F ) := sup { α ∈ R : m (cid:0) ζ ∈ T : ρ F ( ζ ) < x (cid:1) = O ( x α ) , x → +0 } ,m ( A ) is the normalized Lebesgue measure of a set A .The first aforementioned result is a particular case of [4, Theorem 0.3]. Theorem A . Given a compact set F ⊂ T , let an analytic function f on D , | f (0) | = 1 , satisfy the growth condition log | f ( z ) | ≤ M (1 − | z | ) p ρ qF ( z ) , z ∈ D , M, p, q > . Then for each ε > there is a positive number C = C ( F, p, q, ε ) so that theBlaschke-type condition holds for the zero set Z ( f ) of f X ζ ∈ Z ( f ) (1 − | ζ | ) p +1+ ε ρ ( q − α ( F )+ ε ) + F ( ζ ) ≤ CM, ( x ) + := max( x, . As it was pointed out in [6], the natural setting of the problem in questionis the set of subharmonic functions of special growth. The analogue of theBlaschke condition involves then the Riesz measure (generalized Laplacian)of the corresponding function.The second result is a particular case n = 2 of [7, Theorem 5].Let E and F be two arbitrary compact sets on T . We define a class S p,q ( E, F ) of subharmonic on D functions v , which satisfy(0.3) v ( z ) ≤ Mρ pE ( z ) ρ qF ( z ) , M, p, q > . Theorem B . Given two disjoint compact sets
E, F ⊂ T , let a subhar-monic function v ∈ S p,q ( E, F ) . Then for each ε > the following Blaschke-type condition holds for the Riesz measure µ of v Z D (1 − | λ | ) ρ ( p − α ( E )+ ε ) + E ( λ ) ρ ( q − α ( F )+ ε ) + F ( λ ) µ ( dλ ) < ∞ . Both the above results actually deal with two sets of singularities, andeach case is extreme in a sense. Precisely, such sets are E = T and F inTheorem A, and the disjoint sets E and F in Theorem B. The goal of thispaper is to study the case of two generic compact sets which come up asthe sets of singularities of a subharmonic function v subject to some specialgrowth condition. LASCHKE-TYPE CONDITION 3
We impose certain restrictions on E and F in the form of “integrability”of the products(0.4) (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) = Z T m ( dζ ) ρ aE ( ζ ) ρ bF ( ζ ) < ∞ , a, b ≥ . Here is our main result. Theorem 0.1.
Given two compact sets E and F on T subject to (0.4) , leta subharmonic function v , v (0) ≥ , with the Riesz measure µ , belong to S p,q ( E, F ) . ( i ) . If both ≤ a < p and ≤ b < q hold, then for each ε > there is aconstant C = C ( p, q, a, b, ε ) so that (0.5) Z D ρ p − a + εE ( λ ) ρ q − b + εF ( λ )(1 − | λ | ) µ ( dλ ) ≤ CM (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . ( ii ) . If ≤ a < p , b ≥ q (0 ≤ b < q , a ≥ p ) , then for each ε > there is aconstant C = C ( p, q, a, ε ) ( C = C ( p, q, b, ε )) so that Z D ρ p − a + εE ( λ ) (1 − | λ | ) µ ( dλ ) ≤ CM (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) , (cid:16)Z D ρ q − b + εF ( λ ) (1 − | λ | ) µ ( dλ ) ≤ CM (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . (cid:17) (0.6)The procedure we suggest for solving the problem under consideration ispursued in three steps.Step 1. Given a function v ∈ S p,q ( E, F ), we find a domain Ω ⊂ D so that v has a harmonic majorant, i.e., the harmonic function U exists with v ≤ U on Ω. By the Riesz representation, see, e.g., [14, Theorem 4.5.4], which willfeature prominently in what follows,(0.7) v ( z ) = u ( z ) − Z Ω G Ω ( z, λ ) µ ( dλ ) , z ∈ Ω . Here u is the least harmonic majorant for v , µ the Riesz measure of v , G Ω the Green’s function for Ω G Ω ( z, λ ) := log 1 | z − λ | − h Ω ( z, λ ) , z, λ ∈ Ω ,h Ω is the solution to the Dirichlet problem on Ω for the boundary value h Ω ( z, ξ ) = log 1 | z − ξ | , ξ ∈ ∂ Ω . If Ω contains the origin, and v (0) ≥
0, we have from (0.7) with z = 0(0.8) Z Ω G Ω (0 , λ ) µ ( dλ ) ≤ u (0) ≤ U (0) . Step 2. We apply the lower bound for the Green’s function of the type G Ω (0 , λ ) ≥ c (1 − | λ | ) , λ ∈ Ω ′ ⊂ Ωto obtain Z Ω ′ (1 − | λ | ) µ ( dλ ) ≤ U (0) . The case of more general conditions on a function v and its associated measure wasconsidered in the papers [10], [11], but these conditions do not look as clear as ours. S. FAVOROV AND L. GOLINSKII
Step 3. To go over to the integration over the whole unit disk, we invoke anew two-dimensional version of the well-known “layer cake representation”(LCR) theorem, see Proposition 1.8.In the simplest case when Ω = D (see Theorem 2.1 below) the Green’sfunction is G D ( z, λ ) = log (cid:12)(cid:12)(cid:12)(cid:12) − ¯ λzz − λ (cid:12)(cid:12)(cid:12)(cid:12) , so we come to the Blaschke condition for µ of the form(0.9) Z D (1 − | λ | ) µ ( dλ ) ≤ Z D log 1 | λ | µ ( dλ ) ≤ U (0)in one step.We proceed as follows. In Section 1 we gather a collection of auxiliaryfacts on the harmonic measure and majorants, the bounds from below for theGreen’s function and LCR theorems. The main result is proved in Section 2.We also demonstrate its optimal character in Theorem 2.7.1. Preliminaries
Bounds for the harmonic measure.
Let γ = [ e iθ , e iθ ] be a closedarc on the unit circle T . For the harmonic measure of this arc with respectto the unit disk D the explicit expression is known [9, p. 26] ω ( λ, γ ; D ) = 2 α − ( θ − θ )2 π , λ ∈ D , where α is the angle subtended at λ by the arc γ .Let ζ ′ ∈ T , and 0 < t <
1. We put γ = γ t ( ζ ′ ) := { ζ ∈ T : | ζ − ζ ′ | ≤ t } , Γ = Γ t ( ζ ′ ) := { z ∈ D : | z − ζ ′ | = t } . (1.1)It is clear, that ω is constant on Γ (it is constant on each arc of a circle thatpasses through the endpoints of γ ). An elementary geometry provides theformula ω ( λ, γ t ( ζ ′ ); D ) = 12 − π arcsin t , λ ∈ Γ t ( ζ ′ ) . So, there is a uniform bound from below for the harmonic measure of γ on Γ(1.2) ω ( λ, γ t ( ζ ′ ); D ) ≥ , λ ∈ Γ t ( ζ ′ ) . To proceed further, given a compact set K ⊂ T , denote by ρ ( w ) = ρ K ( w ) := dist( w, K ) , w ∈ C , the Euclidian distance from w to K . Consider the sets on the unit circle(1.3) K t := { ζ ∈ T : ρ K ( ζ ) ≤ t } , K ′ t := { ζ ∈ T : ρ K ( ζ ) ≥ t } = T \ K t , and the set in D (1.4) Γ t ( K ) := { z ∈ D : ρ K ( z ) = t } . Note that K t and K ′ t are finite unions of disjoint closed arcs. LASCHKE-TYPE CONDITION 5
For each λ ∈ Γ t ( K ) there is ζ ′ ∈ K , such that | λ − ζ ′ | = ρ K ( λ ) = t , so λ ∈ Γ t ( ζ ′ ). If follows from relation (1.2) that ω ( λ, γ t ( ζ ′ ); D ) ≥ , λ ∈ Γ t ( K ) . But, by definition, K t ⊃ γ t ( ζ ′ ) for each ζ ′ ∈ K , so monotonicity of theharmonic measure yields ω ( λ, K t ; D ) ≥ ω ( λ, γ t ( ζ ′ ); D ) . Proposition 1.1.
Given a compact set K ⊂ T , let K t be its closed neigh-borhood (1.3) . Then (1.5) ω ( λ, K t ; D ) ≥ , λ ∈ Γ t ( K ) . Let us now turn to the upper bounds for the harmonic measure of K t .For a compact set K on T and 0 < t <
1, the open set(1.6) D t ( K ) := { w ∈ D : ρ K ( w ) > t } can be disconnected even for simple K . We denote by Ω t ( K ) the connectedcomponent of D t ( K ) that contains the origin. Clearly, Ω t ( K ) = ∅ for t ≥ t ( K ) ⊃ Ω τ ( K ) for τ > t . It is also important that(1.7) ∂ Ω t ( K ) ⊂ ∂D t ( K ) = Γ t ( K ) ∪ K ′ t . The following result will be helpful later on.
Proposition 1.2.
Given a compact set K ⊂ T , and s > , one has (1.8) { w ∈ D : ρ K ( w ) > s } ⊂ Ω s/ ( K ) . Proof.
Clearly, { w ∈ D : ρ K ( w ) > s } ⊂ n w ∈ D : ρ K ( w ) > s o , and we wish to show that the set on the left side is actually a subset ofthe connected component of the set on the right side that contains the ori-gin. The argument relies on a simple inequality, which we apply repeatedlythroughout the paper(1.9) ρ K ( z ) ≤ ρ K ( rz ) , z ∈ D , ≤ r ≤ . Indeed, by the triangle inequality ρ K ( z ) ≤ ρ K ( rz ) + | rz − z | , and so ρ K ( z ) ≤ ρ K ( rz ) + (1 − | rz | ) ≤ ρ K ( rz ) , as claimed.It follows from (1.9) that ρ K ( rz ) > s/ ≤ r ≤ ρ K ( z ) > s . In other words, the whole closed interval[0 , z ] ⊂ { ρ K ( w ) > s/ } , and so z ∈ Ω s/ ( K ), as needed. (cid:3) Proposition 1.3.
Given a number l ∈ (0 , , put k := πl − + 1 . Then thefollowing inequality holds for t < k − (1.10) ω ( λ, K t ; D ) ≤ lt (1 − | λ | ) , λ ∈ Ω kt ( K ) . S. FAVOROV AND L. GOLINSKII
Proof.
If 1 − | λ | > tl − , inequality (1.10) obviously holds. So we assume inwhat follows that(1.11) 1 − | λ | ≤ tl , | λ | ≥ − tl > − kl . For λ = | λ | e ı θ ∈ Ω kt ( K ), and ζ = e iϕ ∈ K t , the Poisson integral represen-tation for the harmonic measure reads ω ( λ, K t ; D ) = Z K t − | λ | | ζ − λ | m ( dζ ) = (1 − | λ | )2 π Z e iϕ ∈ K t dϕ (1 − | λ | ) + 4 | λ | sin ϕ − θ . Take ζ ∈ K such that ρ K ( e iθ ) = | e iθ − ζ | . Then, in view of (1.11), ρ K ( e iθ ) = | e iθ − ζ | = | λ − ζ + e iθ − λ |≥ ρ K ( λ ) − (1 − | λ | ) > kt − (1 − | λ | ) . (1.12)Take ζ ∈ K such that ρ K ( e iϕ ) = | e iϕ − ζ | ≤ t , so, by (1.12), | θ − ϕ | ≥ (cid:12)(cid:12) e iθ − e iϕ (cid:12)(cid:12) = (cid:12)(cid:12) e iθ − ζ + ζ − e iϕ (cid:12)(cid:12) ≥ ρ K ( e iθ ) − t > ( k − t − (1 − | λ | ) . Hence (1.11) implies π ≥ | θ − ϕ | ≥ ( k − t − tl = k t, k := k − − l . Going back to the Poisson integral, we see that ω ( λ, K t ; D ) ≤ − | λ | π | λ | Z k t ≤| ϕ − θ |≤ π dϕ sin ϕ − θ ≤ π (1 − | λ | )4 | λ | Z k t ≤| ϕ − θ |≤ π dϕ ( ϕ − θ ) , or, in view of (1.11), ω ( λ, K t ; D ) ≤ π (1 − | λ | )4 klkl − Z πk t dxx ≤ π (1 − | λ | )4 t klkl − k − − l − . An elementary calculation shows that π klkl − k − − l − < l, and (1.10) follows. (cid:3) Lower bounds for Green’s functions.
Under a
Green’s function ofthe domain Ω t ( K ) with singularity z we mean a nonnegative function of theform(1.13) G t ( z, λ ) = G Ω t ( K ) ( z, λ ) := log 1 | z − λ | − h t ( z, λ ) , z, λ ∈ Ω t ( K ) , where h t is the solution to the Dirichlet problem on Ω t ( K ) for the boundaryvalue(1.14) h t ( z, ξ ) = log 1 | z − ξ | , ξ ∈ ∂ Ω t ( K ) . Such function exists and is unique, as the boundary ∂ Ω t ( K ) is a non-polarset, see, e.g., [14]. The problem we address here is to obtain a lower boundfor G t (0 , · ) in a smaller domain Ω τ ( K ) with an appropriate τ > t . LASCHKE-TYPE CONDITION 7
Proposition 1.4.
The Green’s function G t (0 , · ) for the domain Ω t ( K ) withsingularity at the origin and < t < (12 π + 1) − admits the lower bound (1.15) G t (0 , λ ) ≥ − | λ | , λ ∈ Ω (12 π +1) t ( K ) . Proof.
Since1 − | ξ | ≤ ρ K ( ξ ) = t, | ξ | ≥ − t > , ξ ∈ Γ t ( K ) , one has h t (0 , ξ ) = log 1 | ξ | ≤ log 11 − t ≤ t, ξ ∈ ∂ Ω t ( K ) ∩ Γ t ( K ) . Next, h t (0 , ξ ) = 0 for ξ ∈ ∂ Ω t ( K ) ∩ K ′ t , so, by Proposition 1.1 and theMaximum Principle, h t (0 , λ ) ≤ tω ( λ, K t ; D ) , λ ∈ Ω t ( K ) . Now, the upper bound (1.10) with l = 1 /
12 and k = 12 π + 1 yields h t (0 , λ ) ≤ − | λ | , λ ∈ Ω (12 π +1) t ( K ) , and so G t (0 , λ ) ≥ log 1 | λ | − − | λ | ≥ − | λ | , λ ∈ Ω (12 π +1) t ( K ) , as needed. (cid:3) So far we have been dealing with one compact set K . Keeping in mindthe main topic of the paper, consider the intersection D t,s ( E, F ) := { w ∈ D : ρ E ( w ) > t, ρ F ( w ) > s } = D t ( E ) ∩ D s ( F ) , where E and F are compact sets on the unit circle, 0 < t, s <
1. Denoteby Ω t,s the connected component of this open set (or, that is the same, theconnected component of Ω t ( E ) ∩ Ω s ( F )) so that 0 ∈ Ω t,s . Clearly, Ω t,s = ∅ for max { t, s } ≥
1. It is not hard to check that ∂ Ω t,s ⊂ W ∪ W ∪ W , W := E ′ t ∩ F ′ s ,W := { w ∈ D : ρ E ( w ) = t, ρ F ( w ) ≥ s } ,W := { w ∈ D : ρ E ( w ) ≥ t, ρ F ( w ) = s } . (1.16)In particular,(1.17) ∂ Ω t,s ⊂ Γ t ( E ) ∪ Γ s ( F ) ∪ ( E ′ t ∩ F ′ s ) . The inclusion(1.18) D t,s ( E, F ) ⊂ Ω t/ ,s/ , can be verified in exactly the same way as (1.8) in Proposition 1.2.We complete with the lower bound for the Green’s function G t,s := G Ω t,s . Proposition 1.5.
The Green’s function G t,s (0 , · ) for the domain Ω t,s withsingularity at the origin and < t, s < (24 π + 1) − admits the lower bound (1.19) G t,s (0 , λ ) ≥ − | λ | , λ ∈ Ω (24 π +1) t, (24 π +1) s . S. FAVOROV AND L. GOLINSKII
Proof.
We follow the argument from the proof of Proposition 1.4. Write G t,s (0 , λ ) = log 1 | λ | − h ( λ ) , h ( ζ ) = log 1 | ζ | , ζ ∈ ∂ Ω t,s , so h ( ζ ) = 0 for ζ ∈ ∂ Ω t,s ∩ T . Since | ζ | ≥ − t > / , ζ ∈ Γ t ( E ) , | ζ | ≥ − s > / , ζ ∈ Γ s ( F ) , we have log 1 | ζ | ≤ − | ζ | ) ≤ t, ζ ∈ Γ t ( E ) , log 1 | ζ | ≤ − | ζ | ) ≤ s, ζ ∈ Γ s ( F ) . (1.20)In view of (1.17), (1.20) and Proposition 1.1, it follows from the MaximumPrinciple that h ( λ ) ≤ tω ( λ, E t ; D ) + 6 sω ( λ, F s ; D ) , λ ∈ Ω t,s . We apply the upper bound for the harmonic measure (1.10) tω ( λ, E t ; D ) + sω ( λ, F s ; D ) ≤ l (1 − | λ | ) , λ ∈ Ω t,s , so for l = 1 / k = 24 π + 1 we come to h ( λ ) ≤ − | λ | ≤
12 log 1 | λ | ⇒ G t,s (0 , λ ) ≥
12 log 1 | λ | ≥ − | λ | , as claimed. (cid:3) Harmonic majorant.
The result below concerns particular subhar-monic functions and their harmonic majorants.
Proposition 1.6.
Given two compact sets E and F on the unit circle, and a, b ≥ , assume that ρ − aE ρ − bF ∈ L ( T ) . Then the function (1.21) v a,b ( z ) := 1 ρ aE ( z ) ρ bF ( z ) , z ∈ D , is subharmonic and admits the harmonic majorant (1.22) v a,b ( z ) ≤ P a,b ( z ) := Z T − | z | | ζ − z | m ( dζ ) ρ aE ( ζ ) ρ bF ( ζ ) . Proof.
The case a = b = 0 is trivial, so let a + b >
0. By [14, Theorem 2.4.7],the function v a,b ( z ) = sup ξ ∈ E, η ∈ F | ( z − ξ ) − a ( z − η ) − b | is subharmonic. The inequality (1.9) implies(1.23) v a,b ( rζ ) ≤ a + b v a,b ( ζ ) , ζ ∈ T , ≤ r ≤ . The standard Maximum Principle states that v a,b ( rz ) ≤ Z T − | z | | ζ − z | v a,b ( rζ ) m ( dζ ) , z ∈ D , r < . The bound (1.22) is now immediate from the latter inequality as r → − (cid:3) LASCHKE-TYPE CONDITION 9
Remark 1.7.
As a matter of fact, P a,b is the least harmonic majorant for v a,b , see, e.g., [8, pp.36-37].1.4. Layer cake representation.
A key ingredient in our argument is thefundamental result in Analysis, known as the “layer cake representation”(LCR) see, e.g., [13, Theorem 1.13].
Theorem LCR . Let (Λ , ν ) be a measure space, and h ≥ c > Z Λ h c ( τ ) ν ( dτ ) = c Z ∞ x c − ν ( { τ : h ( τ ) > x } ) dx. In what follows we make use of the two-dimensional analogue of thisresult.
Proposition 1.8.
Let f, g ≥ be measurable functions on the measure space (Λ , σ ) , and α, β > . Then I := Z Λ f α ( τ ) g β ( τ ) σ ( dτ )= αβ Z ∞ Z ∞ x α − y β − σ ( { τ : f ( τ ) > x, g ( τ ) > y } ) dx dy. (1.25) Proof.
We apply the LCR (1.24) twice. Put ν ( dτ ) := g β σ ( dτ ), so I = Z Λ f α ( τ ) ν ( dτ ) = α Z ∞ x α − ν ( { τ : f ( τ ) > x } ) dx. Write Λ x := { τ : f ( τ ) > x } , and apply (1.24) once again ν (Λ x ) = Z Λ x g β ( τ ) σ ( dτ ) = β Z ∞ y β − ν ( { τ ∈ Λ x : g ( τ ) > y } ) dy = β Z ∞ y β − ν ( { τ ∈ Λ : f ( τ ) > x, g ( τ ) > y } ) dy, so Fubini’s theorem completes the proof. (cid:3) Problem with two compact sets
Let us go back to our main problem concerning the Blaschke-type condi-tion for the Riesz measure of the subharmonic function which can grow atthe direction of two sets of singularities on the unit circle.As a warm-up, we prove the following result.
Theorem 2.1.
Assume that E and F are two compact sets on T so that (0.4) holds with a = p , b = q . For each subharmonic function v ∈ S p,q ( E, F ) , v (0) ≥ , with the Riesz measure µ , the Blaschke condition holds (2.1) Z D (1 − | λ | ) µ ( dλ ) ≤ M k ρ − pE ρ − qF k . Proof.
By Proposition 1.6, v admits the harmonic majorant U = M P p,q with U (0) = M k ρ − pE ρ − qF k . Relation (0.9) completes the proof. (cid:3) The case when min( p, q ) = 0, so we actually have one compact set, waselaborated in [6].The main result of the paper, Theorem 0.1, concerns the rest of the valuesfor a and b , that is, either 0 ≤ a < p or 0 ≤ b < q . Proof of Theorem 0.1. (i). We proceed in three steps, following the procedure outlined in Introduc-tion.Step 1. Write the hypothesis (0.3) as v ( z ) ≤ v a,b ( z ) Mρ p − aE ( z ) ρ q − bF ( z ) , z ∈ D . In view of (1.16), Proposition 1.6, and the Maximum Principle, we come tothe bound v ( z ) ≤ U ( z ) = P a,b ( z ) Mt p − a s q − b , z ∈ Ω t,s . Step 2. Relation (0.8) now reads Z Ω t,s G t,s (0 , λ ) µ ( dλ ) ≤ u (0) ≤ U (0) = Mt p − a s q − b (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . By Proposition 1.5 with κ = 24 π + 1, one has Z Ω κt,κs (1 − | λ | ) µ ( dλ ) ≤ Mt p − a s q − b (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . By (1.18), D κt, κs ( E, F ) ⊂ Ω κt,κs , so putting ξ := 2 κt, η := 2 κs, ≤ t, s ≤ κ , we end up with the bound(2.2) Z D ξ,η ( E,F ) (1 − | λ | ) µ ( dλ ) ≤ κ ) p + q − a − b Mξ p − a η q − b (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . Step 3. The LCR theorem comes into play here. By Proposition 1.8 withΛ = D , σ = (1 − | λ | ) µ, f = ρ E , g = ρ F , α = p − a + ε, β = q − b + ε, we see that Z D ρ αE ( λ ) ρ βF ( λ ) σ ( dλ ) = αβ Z Z ξ α − η β − σ (cid:16) { λ : ρ E ( λ ) > ξ, ρ F ( λ ) > η } (cid:17) dξdη. But, due to (2.2), σ (cid:16) { λ : ρ E ( λ ) > ξ, ρ F ( λ ) > η } (cid:17) = Z D ξ,η ( E,F ) (1 − | λ | ) µ ( dλ ) ≤ CMξ p − a η q − b (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) , so, finally, Z D ρ αE ( λ ) ρ βF ( λ )(1 − | λ | ) µ ( dλ ) ≤ αβ CM (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) Z ξ ε − dξ Z η ε − dη, and the first statement is proved. LASCHKE-TYPE CONDITION 11 (ii). Assume now that 0 ≤ a < p and b ≥ q . The argument is the same butsimpler, as we appeal to the domain Ω t ( E ) and the standard one-dimensionalLCR theorem (1.24). Indeed, as in Step 1, we have v ( z ) ≤ U ( z ) = P a,b ( z ) 2 b − q Mt p − a , z ∈ Ω t ( E ) . Next, relation (0.8) provides Z Ω t ( E ) G t (0 , λ ) µ ( dλ ) ≤ b − q Mt p − a (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) , so, by Proposition 1.4 with κ = 12 π + 1, Z Ω κt ( E ) (1 − | λ | ) µ ( dλ ) ≤ b − q +1 Mt p − a (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . By (1.8), D κt ( E ) ⊂ Ω κt ( E ), and so for ξ = 2 κt we have Z D ξ ( E ) (1 − | λ | ) µ ( dλ ) ≤ (2 κ ) p − a b − q +1 Mξ p − a (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) . An application of LCR theorem in the form (1.24) withΛ = D , ν ( dλ ) = (1 − | λ | ) µ ( dλ ) , h = ρ E , c = p − a + ε leads to the first Blaschke-type condition in (0.6). The proof of the secondone is identical. ✷ The case a = b = 0 is important, for there are no integrability assumptionswhatsoever. Corollary 2.2.
Given two compact sets E , F on T , let a subharmonicfunction v , v (0) ≥ , belong to S p,q ( E, F ) . Then for each ε > there is aconstant C = C ( p, q, ε ) so that (2.3) Z D ρ p + εE ( λ ) ρ q + εF ( λ )(1 − | λ | ) µ ( dλ ) ≤ CM.
The results of Theorem 0.1 can be extended to the case of n compact setson the unit circle with no additional efforts. Theorem 2.3.
Let K , . . . , K n be compact subsets of T , and let v be asubharmonic function on D with Riesz measure µ such that v (0) ≥ and v ( z ) ≤ M ρ − p K ( z ) · · · ρ − p n K n ( z ) . z ∈ D . Suppose that ρ − a K ( ζ ) · · · ρ − a n K n ( ζ ) ∈ L ( T ) for some a < p , . . . , a k < p k , a k +1 = p k +1 , . . . , a n = p n , k ≤ n. Then for each ε > there is a constant C = C ( p , . . . , p n , a , . . . , a k , ε ) sothat Z D ρ p − a + εK ( λ ) · · · ρ p k − a k + εK k ( λ )(1 − | λ | ) dµ ( λ ) ≤ CM k ρ − a K ( z ) · · · ρ − a n K n ( z ) k . In view of further applications, let us mention a special case of subhar-monic functions v = log | f | with f analytic on the unit disk. Corollary 2.4.
Let an analytic function f , | f (0) | ≥ , satisfy the growthcondition (2.4) log | f ( z ) | ≤ Mρ pE ( z ) ρ qF ( z ) , M, p, q > , with two compact sets E , F on the unit circle. Assume that the relation (0.4) holds for some ≤ a < p and ≤ b < q . Then for each ε > there isa constant C = C ( p, q, a, b, ε ) so that ∞ X n =1 (1 − | λ n | ) ρ p − a + εE ( λ n ) ρ q − b + εF ( λ n ) ≤ CM (cid:13)(cid:13) ρ − aE ρ − bF (cid:13)(cid:13) , where { λ n } n ≥ are the zeros of f counting multiplicity. Next, we consider the situation where the integrability assumptions areimposed on ρ E and ρ F separately. At the moment the following partialresult is available. Proposition 2.5.
Let a subharmonic function v , v (0) ≥ , belong to S p,q ( E, F ) .Assume that (2.5) (cid:13)(cid:13) ρ − pE (cid:13)(cid:13) = Z T m ( dζ ) ρ pE ( ζ ) < ∞ , (cid:13)(cid:13) ρ − qF (cid:13)(cid:13) = Z T m ( dζ ) ρ qF ( ζ ) < ∞ . Let p ′ , q ′ ≥ be nonnegative constants such that p ′ + q ′ > max( p, q ) . Thenthere is a constant C = C ( p, q, p ′ , q ′ ) so that (2.6) Z D ρ p ′ E ( λ ) ρ q ′ F ( λ )(1 − | λ | ) µ ( dλ ) ≤ CM (cid:16)(cid:13)(cid:13) ρ − pE (cid:13)(cid:13) + (cid:13)(cid:13) ρ − qF (cid:13)(cid:13) (cid:17) . Proof.
We focus on two particular cases of Theorem 0.1, namely, a = 0 , b = q and a = p, b = 0. The corresponding conditions (0.4) agree with (2.5). Itfollows from (0.6) that(2.7) Z D (cid:16) ρ p + εE ( λ ) + ρ q + εF ( λ ) (cid:17) (1 − | λ | ) µ ( dλ ) ≤ CM (cid:16)(cid:13)(cid:13) ρ − pE (cid:13)(cid:13) + (cid:13)(cid:13) ρ − qF (cid:13)(cid:13) (cid:17) for arbitrary ε >
0. We choose this parameter from the condition(2.8) 0 < ε < p ′ + q ′ − max( p, q )2 . The argument below is quite elementary. Let 0 ≤ x, y ≤
2. If y ≤ x , wehave, by (2.8), x p ′ y q ′ = x p ′ + q ′ ≤ p ′ + q ′ − p − ε x p + ε . Similarly, for x ≤ y x p ′ y q ′ = y p ′ + q ′ ≤ p ′ + q ′ − q − ε y q + ε . So, for each 0 ≤ x, y ≤ x p ′ y q ′ ≤ C (cid:0) x p + ε + y q + ε (cid:1) , C = 2 p ′ + q ′ − min( p,q ) − ε . It remains only to put x := ρ E ( λ ), y := ρ F ( λ ) and make use of (2.7). Theproof is complete. (cid:3) LASCHKE-TYPE CONDITION 13
Remark 2.6.
In some instances the assumption v (0) ≥ −∞ < v (0) <
0, one can apply the above results to thefunction v ( z ) = v ( z ) − v (0), which belongs to the same class S p,q ( E, F ).But now the constant M depends on v , so we actually have quantitativeBlaschke-type conditions. For example,(2.9) Z D ρ p − a + εE ( λ ) ρ q − b + εF ( λ )(1 − | λ | ) µ ( dλ ) < ∞ holds in place of (0.5).If v (0) = −∞ , consider the Poisson integral in the disk | z | < / v h ( z ) := Z T − | z | | ζ − z | v ( ζ/ m ( dζ ) . Since v is upper semicontinuous, we see that lim z ′ → z h ( z ′ ) ≤ v ( z ) for each z with | z | = 1 / v ( z ) = (cid:26) max( v ( z ) , h ( z )) for | z | < / ,v ( z ) for | z | ≥ / D , and the restriction of its Riesz measure µ on the set { z ∈ D : | z | > / } agrees with µ . Therefore, Z D ρ p − aE ( λ ) ρ q − bF ( λ )(1 − | λ | ) (cid:16) µ ( dλ ) − µ ( dλ ) (cid:17) = O (1) . Since v (0) > −∞ , we again get the conclusions of the quantitative typesimilar to (2.9).We complete the paper with the result which demonstrates the optimalcharacter of the bound (0.5) in Theorem 0.1.Given a compact set K ⊂ T , define the value δ ( K ) := sup { d ≥ ρ − dK ∈ L ( T ) } . It is clear that 0 ≤ δ ( K ) ≤
1. The equality(2.10) Z T m ( dζ ) ρ dK ( ζ ) = 2 − d + dI ( d, K ) , d > , I ( d, K ) := Z m ( K t ) t d +1 dt follows easily from the LCR theorem (1.24), see [6, formula (15)]. Thecharacteristic I ( d, K ) appeared already in [5]. So, δ ( K ) = sup { d ≥ I ( d, K ) < ∞} . Choose two disjoint compact sets E and F with δ ( E ) > δ ( F ) >
0. Bythe definition, ρ − δ ( E )+ εE ∈ L ( T ) , ρ − δ ( F )+ εF ∈ L ( T ) , < ε < min( δ ( E ) , δ ( F )) , and so (0.4) holds with a = δ ( E ) − ε , b = δ ( F ) − ε ( E and F are disjoint).On the other hand, ρ − δ ( E ) − εE / ∈ L ( T ) , ρ − δ ( F ) − εF / ∈ L ( T ) , and, by (2.10), I ( δ ( E ) + ε, E ) = I ( δ ( F ) + ε, F ) = + ∞ . In notation (1.6) we take t , s small enough so that(2.11) D ct ( E ) ∩ D cs ( F ) = ∅ , D ct ( K ) := D \ D t ( K ) . Let p > δ ( E ), q > δ ( F ), and consider the function v ( z ) = v E ( z ) + v F ( z ) = 1 ρ pE ( z ) + 1 ρ qF ( z ) , v ∈ S p,q ( E, F ) . Denote by µ E ( µ F ) the Riesz measure of the subharmonic function v E ( v F ).The result in Theorem 0.1, (i), states that Z D ρ p − δ ( E )+2 εE ( λ ) ρ q − δ ( F )+2 εF ( λ ) (1 − | λ | ) µ ( dλ ) < ∞ , µ := µ E + µ F is the Riesz measure of v and ε > Theorem 2.7.
For < ε < min( p − δ ( E ) , q − δ ( F )) the relation holds (2.12) I := Z D ρ p − δ ( E ) − εE ( λ ) ρ q − δ ( F ) − εF ( λ ) (1 − | λ | ) µ ( dλ ) = + ∞ . Proof.
We bound the integral I from below in a few steps. Clearly, I ≥ Z D ct ( E ) ρ p − δ ( E ) − εE ( λ ) ρ q − δ ( F ) − εF ( λ ) (1 − | λ | ) µ ( dλ ) ≥ Z D ct ( E ) ρ p − δ ( E ) − εE ( λ ) ρ q − δ ( F ) − εF ( λ ) (1 − | λ | ) µ E ( dλ ) = I . By (2.11), one has ρ F ( λ ) > s as long as λ ∈ D ct ( E ), so(2.13) I ≥ s q − δ ( F ) − ε Z D ct ( E ) ρ p − δ ( E ) − εE ( λ ) (1 − | λ | ) µ E ( dλ ) . We apply [6, Theorem 2], which claims that now Z D ρ p − δ ( E ) − εE ( λ ) (1 − | λ | ) µ E ( dλ ) = Z D t ( E ) ρ p − δ ( E ) − εE ( λ ) (1 − | λ | ) µ E ( dλ )+ Z D ct ( E ) ρ p − δ ( E ) − εE ( λ ) (1 − | λ | ) µ E ( dλ ) = I + I c = + ∞ . But I < ∞ thanks to the property of the Riesz measure, so I c = + ∞ . Therelation (2.12) follows now from (2.13). (cid:3) References [1] P. Ahern, D. Clark, On inner functions with B p derivatives, Michigan Math. J. (1976), 107–118.[2] W. Blaschke, Eine Erweiterung des Satzes von Vitali ¨uber Folgen analytischer Funk-tionen, S.-B. S¨acks Akad. Wiss. Leipzig Math.-Natur. KI. (1915), 194–200.[3] A. Borichev, L. Golinskii, S. Kupin, A Blaschke-type condition and its application tocomplex Jacobi matrices, Bull. Lond. Math. Soc. (2009), 117–123.[4] A. Borichev, L. Golinskii, S. Kupin, On zeros of analytic functions satisfying non-radial growth conditions, Rev. Mat. Iberoam., , no. 3 (2018), 1153–1176.[5] L. Carleson, Sets of uniqueness for functions analytic in the unit disc, Acta Math., (1952), 325–345. LASCHKE-TYPE CONDITION 15 [6] S. Favorov, L. Golinskii, A Blaschke-Type condition for Analytic and SubharmonicFunctions and Application to Contraction Operators, Amer. Math Soc. Transl. (2009), 37–47.[7] S. Favorov, L. Golinskii, Blaschke-type conditions for analytic and subharmonic func-tions in the unit disk: local analogs and inverse problems, Comput. Methods Funct.Theory, (2012) no. 1, 151–166.[8] J. Garnett, Bounded analytic functions , Graduate Texts in Mathematics, ,Springer, New York, 2007.[9] J. Garnett, D. Marshall,
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