The Constant of Interpolation in Bloch Type Spaces

It is known that there exists a constant 0<Δ1<1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$0<\Delta _1 < 1$$\end{document} such that any Δ1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta _1$$\end{document}-separated sequence for the pseudohyperbolic distance in the open unit disk D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\textbf{D}$$\end{document} of C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {C}$$\end{document} is interpolating for the classical Bloch space B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {B}$$\end{document}. We will prove that 0.8114<Δ1<0.9785\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$0.8114< \Delta _1 < 0.9785$$\end{document} and we will also generalize this result for Bloch type spaces Bvp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {B}_{v_p}$$\end{document} for vp(z)=(1-|z|2)p\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$v_p(z)=(1-|z|^2)^p$$\end{document}. In particular, we will provide a construction to calculate an estimate of the lower and upper bounds for the corresponding constant of separation Δp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta _p$$\end{document} for these spaces. We also prove that Δp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta _p$$\end{document} tends to 1 when p→∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$p \rightarrow \infty $$\end{document}.


Introduction and Background
Let D be the open unit disk of the complex plane C. Recall that the classical Bloch space B is given by the space of analytic functions f : The study of interpolating sequences for the classical Bloch space B was started by Attele in [1] and Madigan and Matheson in [9].The study of interpolating sequences for Bloch type spaces B v was introduced in [10], where the weight 1−|z| 2 was substituted by a more general weight v(z).In this work, we will deal with weights v p (z) = (1−|z| 2 ) p for 1 ≤ p < ∞.
Recall that a sequence (z n ) ⊂ D is said to be an interpolating sequence for B v if for any (a n ) ∈ ∞ there exists f ∈ B v such that v(z n )f (z n ) = a n for any n ∈ N. The interpolating operator T : B v → ∞ is defined by T (f ) = (v(z n )f (z n )).Notice that T is clearly linear and (z n ) is interpolating for B v if and only if T is surjective or, equivalently, if there exists a mapping S : ∞ → B v such that T • S = Id ∞ .Given δ > 0, the sequence (z n ) is said

Lower Estimate
Let a > 1 and b > 0 and consider the sequence: K. Seip proved that Γ(a, b) is interpolating for B vp if and only if 2π b log a < 1 p .This is based on the study of the positive density D + of the sequence (see [11]).As Schuster observed [12], the constant of separation of Γ(a, b) is given by: Proposition 2.1.There exists a sequence (z n ) ⊂ D which is δ-separated for δ = 0.811458 and (z n ) is not interpolating for the Bloch space B. Hence, Δ 1 > 0.811458.
Proof.Consider the sequence Γ(a, b) in (2.1).Take b= 2π loga for a > 1 and b > 0.Then, we have 2π b log a = 1 so the sequence is not interpolating for B by the comments above.Notice that functions: log a ) 2 + 4 are non-decreasing and non-increasing, respectively.Hence, the highest value of S(a, b) will be got when: and bearing in mind that b = 2π log a , the function g(a) is clearly non-decreasing continuous with respect to a. Since lim a→1 + g(a) = −1 and lim a→+∞ g(a) = 1, there exists a unique a > 1 such that g(a) = 0. Indeed, a ≈ 9.60773 and b ≈ 2.77701, which yields: Hence, we have a sequence whose constant of separation is 0.811458 but fails to be interpolating for B.
We can easily generalize Proposition 2.1 to B vp spaces for 1 ≤ p < +∞ taking b = 2pπ/ log a and following the same pattern: Proposition 2.2.For any 1 ≤ p < +∞, there exist a > 1 and b > 0 such that the sequence Γ(a, b) is separated but fails to be interpolating for B vp .In particular, these values can be chosen such that the constant of separation tends to 1 when p → ∞.
and since p → ∞ then b → ∞ so the right term tends to 1.In order to get this, we need that the left term also tends to 1 which is only possible if a → ∞ (since b = 2pπ/ log a, we can take a bigger than p to get it and b also tends to infinity).Hence, S(a, b) → 1 and we are done.

Upper Estimate
Given a ∈ D and 0 < r < 1, the pseudohyperbolic open disks centered at a and radius r are denoted by D ρ (a, r) = {z ∈ D : ρ(z, a) < r}.A standard calculation (see for instance [14]) shows that D ρ (a, r) is the euclidean disk with center 1−r 2 1−r 2 |a| 2 a and radius 1−|a| 2 1−|a| 2 r 2 r.The following lemma is just an easy calculation: Proof.An easy calculation shows that w ∈ D ρ (z, r) if and only if ρ(z, w which is equivalent to: and we are done.Changing z with w and repeating the proof we obtain the second inequality. Then, D ρ (a, r δ ) ∩ D ρ (b, r δ ) = ∅.In addition, if ρ(a, b) = δ then the disks D ρ (a, r δ ) and D ρ (b, r δ ) are externally tangent in the complex plane C so the radius r δ is optimal to separate both disks.
Proof.It is an easy calculation that r δ is the solution of the equation: where last inequality is clear since given any 0 ≤ a < 1 the real function (x + a)/(1 + ax) is non-decreasing for 0 ≤ x < 1.This is a contradiction, so the disks are disjoint.Now notice that z ∈ D ρ (a, r δ ) ∩ D ρ (b, r δ ) if and only if ρ(a, z) ≤ r δ and ρ(b, z) ≤ r δ .Since ρ is invariant by automorphisms and ϕ a (a) = 0, this is equivalent to: ρ(0, ϕ a (z)) ≤ r δ and ρ(ϕ a (b), ϕ a (z)) ≤ r δ if and only if ϕ a (z) ∈ D ρ (0, r δ ) ∩ D ρ (ϕ a (b), r δ ).Hence z belongs or not to D ρ (a, r δ ) and D ρ (b, r δ ) if and only if ϕ a (z) belongs or not to D ρ (0, r δ ) and D ρ (ϕ a (b), r δ ) respectively.Notice that: denotes the corresponding euclidean disks, so we will prove that these disks are externally tangent.For this, it is sufficient to prove that that the euclidean distance between both centers equals to the sum of both radius, that is: and dividing by 1 + r 2 δ we have: and dividing by δ and solving for r δ we obtain: and we are done.
The following result is an easy consequence of the mean value for complex analytic functions f on D: Proposition 2.6.If f : D → C is analytic then for any a ∈ D and 0 < r < 1 we have: Proof.By Cauchy's integral formula, we have for any 0 < s < 1 that and using polar coordinates: Applying this inequality to the function h := (f • ϕ a )(ϕ a ) 2 , we obtain as a consequence of the change of variable formula and the Cauchy-Riemman equations that and we are done.
The following results are well-known (see Theorem 1.7 in [8] or Lemma 3.10 in [14]).The measure A will denote the classical Borel measure defined on C. Proposition 2.7.We have the following results: b) For any 0 ≤ x < 1 and s ∈ R, s neither 0 nor a negative integer, we have: is bounded on D.
Proof.Suppose that ρ(z n , z k ) ≥ δ > 0 for n = k and consider the radius r δ given in Lemma 2.5.Fix z ∈ D and define the function f z (w) := 1 (1−zw) q+1 , which is holomorphic on D. Applying Proposition 2.6 and evaluating at z n we have: ( being the second inequality a consequence of Lemma 2.4.By Proposition 2.7 a), we have: Therefore, for any z ∈ D, we have that: • Case r = 2.We obtain: and since the function Γ(x) is non-decreasing for x > 1.462, we have for any m ≥ 3: and bearing in mind that ), we obtain: 2 ) is bounded on D, we are done.
and since the function Γ(x) is non-decreasing for x > 1.462, we have for any m ≥ 3: So:

G(z)
where: and since r − 2 > 0, by Proposition 2.7 b), we have: Considering p ≥ r − 2, we have: where P 2 (|z|) is a polynomial depending on |z| so E(z) is uniformly bounded for z ∈ D.
Hence, we obtain a well-known result about separated sequences: Corollary 2.9.If (z n ) ⊂ D is a separated sequence then for every r > 1 we have: Proof.Take r > 1, p > 0 such that p ≥ r − 2 ≥ q−1 2 > −1.Apply Proposition 2.8 evaluated at z = 0 and we are done.
Our main result refines the Madigan and Matheson result to get an upper estimate for Δ 1 (see [9]).In addition, we generalize it for Bloch type spaces B vp and provide a procedure to estimate Δ p for any 1 ≤ p < +∞.
Proof.Let p ≥ 1 and consider r, q such that p ≥ r − 2 ≥ q−1 2 > −1 and p + r = q + 1.For any λ = (λ n ) ∈ ∞ , define: If z k0 = 0 for any k 0 (there could only be one term because of the separability) then substitute the corresponding k 0 -term of the previous series by λ k0 z.
Notice that T λ is well-defined since the series is absolutely and uniformly convergent on compact sets of D. To show this, set η := inf k∈N {|z n | | z k = 0}.Since (z k ) ⊂ D is separated it does not have accumulation points in D and thus η > 0. Now for any |z| ≤ s < 1, we have: because of Corollary 2.9.Adding λ k z for a possible z k = 0 does not affect to the convergence.
Thus, T λ is holomorphic for any λ ∈ ∞ and (1−z k z) q+1 , which implies together with Proposition 2.8 that: . By the comments above, it is clear that T is a well-defined linear bounded operator.Let L : ).We will prove that if Δ p is close enough to 1, then the operator L • T : ∞ → ∞ is invertible, so in particular L is surjective.We have that: and since r + p = q + 1 therefore: and thus: If we apply Proposition 2.6 to the function f n (z) := dw the last inequality being a consequence of Lemma 2.4.Since the disks D ρ (r k , r p ) are disjoint, we obtain: where r − 2 − α and q + 1 − 2α > 0. Bearing in mind Proposition 2.7 a) we obtain: Therefore, for any z ∈ D, we have that: and for any m ≥ 3, we also have: So: Calling: we have: and by Proposition 2.7 b) we have: Since p ≥ r − 2 we obtain: where Q(|z|) is the polynomial given by: C so we will be done if we prove that: and: Proof.Inequality 0.81 < Δ 1 is straightforward from Proposition 2.1.For the other inequality, use proof of Theorem 2.1.Take p = 1, r = 3 and q = 3.Notice that 0 ≤ |z n | < 1 and take 0 ≤ α < 1.We have: where the polynomial Q(|z n |) is given by: We will prove that for some 0 ≤ α <

ForRemark 2 . 3 .
some particular p, it is a straightforward calculation to determine the value of a and S(a, b) such that b = 2pπ log a which yields examples of more and more separated sequences which are not interpolating for B vp .For instance: There are Δ p -separated sequences (z n ) ⊂ D which fail to be interpolating for B vp and Δ p → 1 when p → ∞.To show it, we will prove that S(a, b) → 1 when p → ∞.Indeed, from:

1 ( 1 −
znz) q+1−2α at the point z = z k for k = n and taking r p =