---
abstract: 'We study thin interpolating sequences $\{\lambda_n\}$ and their relationship to interpolation in the Hardy space $H^2$ and the model spaces $K_\Theta = H^2 \ominus \Theta H^2$, where $\Theta$ is an inner function. Our results, phrased in terms of the functions that do the interpolation as well as Carleson measures, show that under the assumption that $\Theta(\lambda_n) \to 0$ the interpolation properties in $H^2$ are essentially the same as those in $K_\Theta$.'
address:
- |
    Pamela Gorkin, Department of Mathematics\
    Bucknell University\
    Lewisburg, PA USA 17837
- |
    Brett D. Wick, School of Mathematics\
    Georgia Institute of Technology\
    686 Cherry Street\
    Atlanta, GA USA 30332-0160
author:
- 'Pamela Gorkin$^\dagger$'
- 'Brett D. Wick$^\ddagger$'
title: Thin Sequences and Their Role in Model Spaces and Douglas Algebras
---

[^1]

[^2]

\[section\] \[thm\][Lemma]{} \[thm\][Corollary]{} \[thm\][Conjecture]{} \[thm\][Problem]{} \[thm\][Proposition]{} \[thm\][Definition]{} \[thm\][Remark]{}

Introduction and Motivation
===========================

A sequence $\{\lambda_j\}_{j=1}^\infty$ is an [interpolating sequence]{.nodecor} for $H^\infty$, the space of bounded analytic functions, if for every $w\in\ell^\infty$ there is a function $f\in H^\infty$ such that $$f(z_j) = w_j, ~\mbox{for all}~ j\in{\mathbb{N}}.$$ Carleson’s interpolation theorem says that $\{\lambda_j\}_{j=1}^\infty$ is an interpolating sequence for $H^\infty$ if and only if $$\label{Interp_Cond}
\delta = \inf_{j}\delta_j:=\inf_j \left\vert B_j(\lambda_j)\right\vert=\inf_{j}\prod_{k \ne j} \left|\frac{\lambda_j - \lambda_k}{1 - \overline{\lambda}_j \lambda_k}\right| > 0,$$ where $$B_j(z):=\prod_{k\neq j}\frac{-\overline{\lambda_k}}{{\ensuremath{\left\vert\lambda_k\right\vert}}}\frac{z-\lambda_k}{1-\overline{\lambda}_kz}$$ denotes the Blaschke product vanishing on the set of points $\{\lambda_k:k\neq j\}$.

In this paper, we consider sequences that (eventually) satisfy a stronger condition than . A sequence $\{\lambda_j\}\subset{\mathbb{D}}$ is *thin* if $$\lim_{j\to\infty}\delta_j:=\lim_{j\to\infty}\prod_{k\neq j}\left\vert\frac{\lambda_j-\lambda_k}{1-\overline{\lambda}_k\lambda_j}\right\vert=1.$$

Thin sequences are of interest not only because functions solving interpolation for thin interpolating sequences have good bounds on the norm, but also because they are interpolating sequences for a very small algebra: the algebra $QA =  VMO \cap H^\infty$, where $VMO$ is the space of functions on the unit circle with vanishing mean oscillation [@W].

Continuing work in [@CFT] and [@GPW], we are interested in understanding these sequences in different settings. This will require two definitions that are motivated by the work of Shapiro and Shields, [@SS], in which they gave the appropriate conditions for a sequence to be interpolating for the Hardy space $H^2$.

Considering more general Hilbert spaces will require the introduction of reproducing kernels: In a reproducing kernel Hilbert space $\mathcal{H}$ (see [@AM p. 17]) we let $K_{\lambda_n}$ denote the kernel corresponding to the point $\lambda_n$; that is, for each function in the Hilbert space we have that $f(\lambda_n)=\left\langle f, K_{\lambda_n}\right\rangle_{\mathcal{H}}$. If we have an $\ell^2$ sequence $a = \{a_n\}$, we define $$\|a\|_{N, \ell^2} = \left(\sum_{j \ge N} |a_j|^2\right)^{1/2}.$$ The concepts of interest are the following.

A sequence $\{\lambda_n\}\subset\Omega \subseteq \mathbb{C}^n$ is said to be [*an eventual $1$-interpolating sequence for a reproducing kernel Hilbert space $\mathcal{H}$*]{}, denoted $EIS_{\mathcal{H}}$, if for every $\varepsilon > 0$ there exists $N$ such that for each $\{a_n\} \in \ell^2$ there exists $f_{N, a} \in \mathcal{H}$ with $$f_{N, a}(\lambda_n) {\ensuremath{\left\|K_{\lambda_n}\right\|}}_{\mathcal{H}}^{-1}=f_{N, a}(\lambda_n) K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}} = a_n ~\mbox{for}~ n \ge N ~\mbox{and}~ \|f_{N, a}\|_{\mathcal{H}} \le (1 + \varepsilon) \|a\|_{N, \ell^2}.$$ A sequence $\{\lambda_n\}$ is said to be a [*strong asymptotic interpolating sequence for $\mathcal{H}$*]{}, denoted $AIS_{\mathcal{H}}$, if for all $\varepsilon > 0$ there exists $N$ such that for all sequences $\{a_n\} \in \ell^2$ there exists a function $G_{N, a} \in \mathcal{H}$ such that $\|G_{N, a}\|_\mathcal{H} \le \|a\|_{N,\ell^2}$ and $$\|\{G_{N, a}(\lambda_n) K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}}  - a_n\}\|_{N, \ell^2} < \varepsilon \|a\|_{N, \ell^2}.$$

Given a (nonconstant) inner function $\Theta$, we are interested in these sequences in model spaces; we define the model space for $\Theta$ an inner function by $K_\Theta = H^2 \ominus \Theta H^2$. The reproducing kernel in $K_\Theta$ for $\lambda_0 \in \mathbb{D}$ is $$K_{\lambda_0}^\Theta(z) = \frac{1 - \overline{\Theta(\lambda_0)}{\Theta(z)}}{1 - \overline{\lambda_0}z}$$ and the normalized reproducing kernel is $$k_{\lambda_0}^\Theta(z) = \sqrt{\frac{1 - |\lambda_0|^2}{1 - |\Theta(\lambda_0)|^2}} K_{\lambda_0}^\Theta(z).$$ Finally, note that $$K_{\lambda_0} = K_{\lambda_0}^\Theta + \Theta \overline{\Theta(\lambda_0)}K_{\lambda_0}.$$ We let $P_\Theta$ denote the orthogonal projection of $H^2$ onto $K_\Theta$.

We consider thin sequences in these settings as well as in Douglas algebras: Letting $L^\infty$ denote the algebra of essentially bounded measurable functions on the unit circle, a Douglas algebra is a closed subalgebra of $L^\infty$ containing $H^\infty$. It is a consequence of work of Chang and Marshall that a Douglas algebra $\mathcal{B}$ is equal to the closed algebra generated by $H^\infty$ and the conjugates of the interpolating Blaschke products invertible in $\mathcal{B}$, [@C; @M].

In this paper, we continue work started in [@GM] and [@GPW] investigating the relationship between thin sequences, $EIS_{\mathcal{H}}$ and $AIS_{\mathcal{H}}$ where $\mathcal{H}$ is a model space or the Hardy space $H^2$. In Section \[HSV\], we consider the notion of eventually interpolating and asymptotic interpolating sequences in the model space setting. We show that in reproducing kernel Hilbert spaces of analytic functions on domains in $\mathbb{C}^n$, these two are the same. Given results in [@GPW], this is not surprising and the proofs are similar to those in the $H^\infty$ setting. We then turn to our main result of that section. If we have a Blaschke sequence $\{\lambda_n\}$ in $\mathbb{D}$ and assume that our inner function $\Theta$ satisfies $|\Theta(\lambda_n)| \to 0$, then a sequence $\{\lambda_n\}$ is an $EIS_{K_\Theta}$ sequence if and only if it is an $EIS_{H^2}$ sequence (and therefore $AIS_{K_\Theta}$ sequence if and only if it is an $AIS_{H^2}$). In Section \[CMMS\] we rephrase these properties in terms of the Carleson embedding constants on the model spaces. Finally, in Section \[asip\_algebra\], we recall the definition of Douglas algebras and show that appropriate definitions and conditions are quite different in that setting.

Preliminaries
=============

Recall that a sequence $\{x_n\}$ in ${\mathcal{H}}$ is [*complete*]{} if $~\mbox{Span}\{x_n: n \ge 1\} = \mathcal{H}$, and [*asymptotically orthonormal*]{} ($AOS$) if there exists $N_0$ such that for all $N \ge N_0$ there are positive constants $c_N$ and $C_N$ such that $$\begin{aligned}
\label{thininequality} 
c_N \sum_{n \ge N} |a_n|^2 \le \left\|\sum_{n \ge N} a_n x_n\right\|^2_{{\mathcal{H}}} \le C_N \sum_{n \ge N} |a_n|^2,\end{aligned}$$ where $c_N \to 1$ and $ C_N \to 1$ as $N \to \infty$. If we can take $N_0 = 1$, the sequence is said to be an $AOB$; this is equivalent to being $AOS$ and a Riesz sequence. Finally, the Gram matrix corresponding to $\{x_j\}$ is the matrix $G = \left(\langle x_n, x_m \rangle\right)_{n, m \ge 1}$.

It is well known that if $\{\lambda_n\}$ is a Blaschke sequence with simple zeros and corresponding Blaschke product $B$, then $\{k_{\lambda_n}\}$, where $$k_{\lambda_n}(z)=\frac{(1-\left\vert \lambda_n\right\vert^2)^{\frac{1}{2}}}{(1-\overline{\lambda_n}z)},$$ is a complete minimal system in $K_B$ and we also know that $\{\lambda_n\}$ is interpolating if and only if $\{k_{\lambda_n}\}$ is a Riesz basis. The following beautiful theorem provides the connection to thin sequences.

\[Volberg\] The following are equivalent:

1.  $\{\lambda_n\}$ is a thin interpolating sequence;

2.  The sequence $\{k_{\lambda_n}\}$ is a complete $AOB$ in $K_B$;

3.  There exist a separable Hilbert space $\mathcal{K}$, an orthonormal basis $\{e_n\}$ for $\mathcal{K}$ and $U, K: \mathcal{K} \to K_B$, $U$ unitary, $K$ compact, $U + K$ invertible, such that $$(U + K)(e_n) = k_{\lambda_n} \text{ for all } n \in {\mathbb{N}}.$$

In [@F]\*[Section 3]{} and [@CFT]\*[Proposition 3.2]{}, the authors note that [@V]\*[Theorem 3]{} implies the following.

\[propCFT\] Let $\{x_n\}$ be a sequence in ${\mathcal{H}}$. The following are equivalent:

1.  $\{x_n\}$ is an AOB;

2.  There exist a separable Hilbert space $\mathcal{K}$, an orthonormal basis $\{e_n\}$ for $\mathcal{K}$ and $U, K: \mathcal{K} \to \mathcal{H}$, $U$ unitary, $K$ compact, $U + K$ left invertible, such that $$(U + K)(e_n) = x_n;$$

3.  The Gram matrix $G$ associated to $\{x_n\}$ defines a bounded invertible operator of the form $I + K$ with $K$ compact.

We also have the following, which we will use later in this paper.

\[prop5.1CFT\] If $\{\lambda_n\}$ is a sequence of distinct points in $\mathbb{D}$ and $\{k_{\lambda_n}^\Theta\}$ is an $AOS$, then $\{\lambda_n\}$ is a thin interpolating sequence.

\[theorem5.2CFT\] Suppose $\sup_{n \ge 1} |\Theta(\lambda_n)| < 1$. If $\{\lambda_n\}$ is a thin interpolating sequence, then either

\(i) $\{k_{\lambda_n}^\Theta\}_{n\ge1}$ is an $AOB$ or

\(ii) there exists $p \ge 2$ such that $\{k_{\lambda_n}^\Theta\}_{n \ge p}$ is a complete $AOB$ in $K_\Theta$.

Hilbert Space Versions {#HSV}
======================

Asymptotic and Eventual Interpolating Sequences {#asip}
-----------------------------------------------

Let $\mathcal{H}$ be a reproducing kernel Hilbert space of analytic functions over a domain $\Omega\subset{\mathbb{C}}^n$ with reproducing kernel $K_\lambda$ at the point $\lambda \in \Omega$. We define two properties that a sequence $\{\lambda_n\}\subset \Omega$ can have.

A sequence $\{\lambda_n\}\subset\Omega$ is an [eventual $1$-interpolating sequence for $\mathcal{H}$]{.nodecor}, denoted $EIS_{\mathcal{H}}$, if for every $\varepsilon > 0$ there exists $N$ such that for each $\{a_n\} \in \ell^2$ there exists $f_{N, a} \in \mathcal{H}$ with $$f_{N, a}(\lambda_n) {\ensuremath{\left\|K_{\lambda_n}\right\|}}_{\mathcal{H}}^{-1}=f_{N, a}(\lambda_n) K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}} = a_n ~\mbox{for}~ n \ge N ~\mbox{and}~ \|f_{N, a}\|_{\mathcal{H}} \le (1 + \varepsilon) \|a\|_{N, \ell^2}.$$

A sequence $\{\lambda_n\}\subset\Omega$ is a [strong asymptotic interpolating sequence for $\mathcal{H}$]{.nodecor}, denoted $AIS_{\mathcal{H}}$, if for all $\varepsilon > 0$ there exists $N$ such that for all sequences $\{a_n\} \in \ell^2$ there exists a function $G_{N, a} \in \mathcal{H}$ such that $\|G_{N, a}\|_\mathcal{H} \le \|a\|_{N,\ell^2}$ and $$\|\{G_{N, a}(\lambda_n) K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}}  - a_n\}\|_{N, \ell^2} < \varepsilon \|a\|_{N, \ell^2}.$$

We now wish to prove Theorem \[EISiffASI\] below. The proof, which is a modification of the proof of the open-mapping theorem, also yields a proof of the following proposition.

\[Banachspace\]

Let $X$ and $Y$ be Banach spaces and let $T: X \to Y$ be a bounded operator and $\varepsilon > 0$. If $$\sup_{\|y\| = 1} \inf_{\|x\| \le 1} \|Tx - y\| < \varepsilon < 1,$$ then for all $y \in Y$, there exists $x \in X$ such that $\|x\| \le \frac{1}{1 - \varepsilon} \|y\|$ and $Tx = y$.

Theorem \[EISiffASI\] follows from Proposition \[Banachspace\], but doing so requires dealing with several technicalities that obfuscate the underlying ideas, and so we present a direct proof of our desired implication. When we turn to Banach algebras, the corresponding implication (in Theorem \[main\_algebra\]) will be a direct consequence of Proposition \[Banachspace\]. We thank the referee for pointing out Proposition \[Banachspace\] to us.

\[EISiffASI\] Let $\mathcal{H}$ be a reproducing kernel space of analytic functions over the domain $\Omega\subset\mathbb{C}^n$ with reproducing kernel at the point $\lambda$ given by $K_\lambda$. Then $\{\lambda_n\}$ is an $EIS_{\mathcal{H}}$ sequence if and only if $\{\lambda_n\}$ is an $AIS_{\mathcal{H}}$.

If a sequence is an $EIS_{\mathcal{H}}$, then it is trivially $AIS_{\mathcal{H}}$, for given $\varepsilon > 0$ we may take $G_{N, a} = \frac{f_{N, a}}{(1 + \varepsilon)}$.

For the other direction, suppose $\{\lambda_n\}$ is an $AIS_{\mathcal{H}}$ sequence. Let $\varepsilon > 0$, $N := N(\varepsilon)$, and $\{a_j\}:=\{a_{j}^{(0)}\}$ be any sequence. First choose $f_0 \in \mathcal{H}$ so that for $n \ge N$ we have $$\|\{K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}} f_0(\lambda_n) - a_{n}^{(0)}\}\|_{N, \ell^2} < \frac{\varepsilon}{1+\varepsilon} \|a\|_{N, \ell^2}$$ and $$\|f_0\|_{\mathcal{H}} \le \|a\|_{N,\ell^2}.$$ Now let $a_{n}^{(1)} = a_{n}^{(0)} - K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}} f_0(\lambda_n)$. Note that $\|a^{(1)}\|_{N, \ell^2} < \frac{\varepsilon}{1+\varepsilon} \|a\|_{N, \ell^2}$. Since we have an $AIS_{\mathcal{H}}$ sequence, we may choose $f_1$ such that for $n \ge N$ we have $$\|\{f_1(\lambda_n)K_{\lambda_n}(\lambda_n)^{-\frac{1}{2}}  - a_{n}^{(1)}\}\|_{N, \ell^2} < \frac{\varepsilon}{1+\varepsilon} \|a^{(1)}\|_{N, \ell^2} <  \left(\frac{\varepsilon}{1+\varepsilon}\right)^2\|a\|_{N, \ell^2},$$ and $$\|f_1\|_{\mathcal{H}} \le \|a^{(1)}\|_{N, \ell^2}<\left(\frac{\varepsilon}{1+\varepsilon}\right)\|a\|_{N,\ell^2}.$$ In general, we let $$a_{j}^{(k)} = -f_{k - 1}(\lambda_j)K_{\lambda_j}(\lambda_j)^{-\frac{1}{2}} + a_{j}^{(k-1)}$$ so that $$\|a^{(k)}\|_{N, \ell^2} \le \frac{\varepsilon}{1+\varepsilon} \|a^{(k - 1)}\|_{N, \ell^2} \le \left(\frac{\varepsilon}{1+\varepsilon}\right)^2 \|a^{(k-2)}\|_{N, \ell^2} \le \cdots \le \left(\frac{\varepsilon}{1+\varepsilon}\right)^k \|a\|_{N, \ell^2}$$ and $$\|f_k\|_{\mathcal{H}} \le \|a^{(k)}\|_{N, \ell^2}<\left(\frac{\varepsilon}{1+\varepsilon}\right)^k\|a\|_{N,\ell^2}.$$ Then consider $f(z) = \sum_{k = 0}^\infty f_k(z)$. Since $f_k(\lambda_j) = \left(a_{j}^{(k)} - a_{j}^{(k+1)}\right)K_{\lambda_j}(\lambda_j)^{\frac{1}{2}}$ and $a_{j}^{(k)} \to 0$ as $k \to \infty$, we have for each $j \ge N$, $$f(\lambda_j) = a_{j}^{(0)} K_{\lambda_j}(\lambda_j)^{\frac{1}{2}} = a_jK_{\lambda_j}(\lambda_j)^{\frac{1}{2}}.$$ Further $\|f\|_{\mathcal{H}}\le \sum_{k = 0}^\infty \left(\frac{\varepsilon}{1+\varepsilon}\right)^{k} \|a\|_{N, \ell^2} = \frac{1}{1 - \frac{\varepsilon}{1+\varepsilon}} \|a\|_{N, \ell^2}=(1+\varepsilon)\|a\|_{N, \ell^2}$. This proves that $\{\lambda_n\}$ is an $EIS_{\mathcal{H}}$ sequence.

The Hardy and Model Spaces
--------------------------

We let $\Theta$ denote a nonconstant inner function and apply Theorem \[EISiffASI\] to the reproducing kernel Hilbert space $K_{\Theta}$. We also include statements and results about Carleson measures. Given a non-negative measure $\mu$ on ${\mathbb{D}}$, let us denote the (possibly infinite) constant $${\mathcal{C}}(\mu) =  \sup_{f \in H^2, f \neq 0} \frac{\|f\|^2_{L^2({\mathbb{D}}, \mu)}}{\|f\|^2_2}$$ as the Carleson embedding constant of $\mu$ on $H^2$ and $${\mathcal{R}}(\mu) =  \sup_{z\in\mathbb{D}} \frac{\|k_z\|_{L^2({\mathbb{D}}, \mu)}}{\|k_z\|_2}=\sup_{z} \|k_z\|_{L^2({\mathbb{D}}, \mu)}$$ as the embedding constant of $\mu$ on $k_z$, the normalized reproducing kernel of $H^2$. It is well-known that ${\mathcal{C}}(\mu)\approx {\mathcal{R}}(\mu)$, [@MR2417425; @nikolski].

\[main\] Let $\{\lambda_n\}$ be an interpolating sequence in $\mathbb{D}$ and let $\Theta$ be an inner function. Suppose that $\kappa:=\sup_{n} \left\vert \Theta(\lambda_n)\right| < 1$. The following are equivalent:

1.  $\{\lambda_n\}$ is an $EIS_{H^2}$ sequence\[eish2\];

2.  $\{\lambda_n\}$ is a thin interpolating sequence\[thin1\];

3.  \[aob\] Either

    1.  $\{k_{\lambda_n}^\Theta\}_{n\ge1}$ is an $AOB$, or

    2.  there exists $p \ge 2$ such that $\{k_{\lambda_n}^\Theta\}_{n \ge p}$ is a complete $AOB$ in $K_\Theta$;

4.  $\{\lambda_n\}$ is an $AIS_{H^2}$ sequence\[Aish2\];

5.  The measure $$\mu_N = \sum_{k \ge N} (1 - |\lambda_k|^2)\delta_{\lambda_k}$$ is a Carleson measure for $H^2$ with Carleson embedding constant ${\mathcal{C}}(\mu_N)$ satisfying ${\mathcal{C}}(\mu_N) \to 1$ as $N \to \infty$\[C1\];

6.  The measure $$\nu_N = \sum_{k \ge N}\frac{(1 - |\lambda_k|^2)}{\delta_k} \delta_{\lambda_k}$$ is a Carleson measure for $H^2$ with embedding constant ${\mathcal{R}}_{\nu_N}$ on reproducing kernels satisfying ${\mathcal{R}}_{\nu_N} \to 1$\[C2\].

    Further, and are equivalent to each other and imply each of the statements above. If, in addition, $ \Theta(\lambda_n) \to0$, then - are equivalent.

7.  $\{\lambda_n\}$ is an $EIS_{K_\Theta}$ sequence\[eis\];

8.  $\{\lambda_n\}$ is an $AIS_{K_\Theta}$ sequence\[ais\].

The equivalence between and is contained in Theorem \[EISiffASI\]. Similarly, this applies to and . In [@GPW]\*[Theorem 4.5]{}, the authors prove that , and are equivalent. The equivalence between , , and is contained in [@GPW]. That implies is Theorem \[theorem5.2CFT\]. That implies also follows from results in [@CFT], for if a sequence is an $AOB$ for some $p \ge 2$ it is an $AOS$ for $p \ge 2$ and hence thin by Proposition \[prop5.1CFT\] for $p \ge 2$. This is, of course, the same as being thin interpolating. Thus, we have the equivalence of equations , , , , , and , as well as the equivalence of and .\
Now we show that and are equivalent under the hypothesis that $\Theta(\lambda_n)\to 0$.\
$\Rightarrow$. Suppose that $\{\lambda_n\}$ is an $EIS_{K_\Theta}$ sequence. We will prove that this implies it is an $EIS_{H^2}$ sequence, establishing .\
Let $\varepsilon>0$ be given. Choose $\varepsilon^\prime < \varepsilon$ and let $N_1 = N(\varepsilon^\prime)$ be chosen according to the definition of $\{\lambda_n\}$ being an $EIS_{K_\Theta}$ sequence. Recall that $$\kappa_m = \sup_{n \ge m} |\Theta(\lambda_n)| \to 0,$$ so we may assume that we have chosen $N_1$ so large that $$\frac{1 + \varepsilon^\prime}{(1 - \kappa_{N_1}^2)^{1/2}} < 1 + \varepsilon.$$ Define $\{\tilde{a}_n\}$ to be $0$ if $n < N_1$ and $\tilde{a}_n=a_n \left(1-{\ensuremath{\left\vert\Theta(\lambda_n)\right\vert}}^2\right)^{-\frac{1}{2}}$ for $n \ge N_1$. Then $\{\tilde{a}_n\} \in \ell^2$. Select $f_a\in K_\Theta\subset H^2$ so that

$$f_a(\lambda_n) \left(\frac{1-{\ensuremath{\left\vert\Theta(\lambda_n)\right\vert}}^2}{1-{\ensuremath{\left\vert\lambda_n\right\vert}}^2}\right)^{-\frac{1}{2}} = \tilde{a}_n = 
      a_n \left(1-{\ensuremath{\left\vert\Theta(\lambda_n)\right\vert}}^2\right)^{-\frac{1}{2}} \, \textrm{ if } n \ge N_1$$ and $$\|f_a\| \le (1 + \varepsilon^\prime) \|\tilde{a}\|_{N_1, \ell^2} \le \frac{(1 + \varepsilon^\prime)}{(1 - \kappa_{N_1}^2)^{1/2}}\|a\|_{N_1, \ell^2} < (1 + \varepsilon) \|a\|_{N_1, \ell^2}.$$ Since $f_a\in K_\Theta$, we have that $f_a\in H^2$, and canceling out the common factor yields that $f_a(\lambda_n)(1-{\ensuremath{\left\vert\lambda_n\right\vert}}^2)^{-\frac{1}{2}}=a_n$ for all $n\geq N_1$. Thus $\{\lambda_n\}$ is an $EIS_{H^2}$ sequence as claimed.\
$\Rightarrow$. Suppose that $\Theta(\lambda_n) \to 0$ and $\{\lambda_n\}$ is an $EIS_{H^2}$ sequence; equivalently, that $\{\lambda_n\}$ is thin. We want to show that the sequence $\{\lambda_n\}$ is an $EIS_{K_\Theta}$ sequence. First we present some observations.\
First, looking at the definition, we see that we may assume that $\varepsilon > 0$ is small, for any choice of $N$ that works for small $\varepsilon$ also works for larger values.\
Second, if $f\in H^2$ and we let $\tilde{f}=P_{K_\Theta}f$, then we have that ${\ensuremath{\left\|\tilde{f}\right\|}}_2\leq {\ensuremath{\left\|f\right\|}}_2$ since $P_{K_\Theta}$ is an orthogonal projection. Next, we have $P_{K_\Theta} = P_+ - \Theta P_+ \overline{\Theta}$, where $P_+$ is the orthogonal projection of $L^2$ onto $H^2$, so letting $T_{\overline{\Theta}}$ denote the Toeplitz operator with symbol $\overline{\Theta}$ we have $$\label{Toeplitz}
\tilde{f}(z)=f(z)-\Theta(z)T_{\overline{\Theta}}(f)(z).$$ In what follows, $\kappa_m := \sup_{n \ge m}|\Theta(\lambda_n)|$ and recall that we assume that $\kappa_m \to 0$.\
Since $\{\lambda_n\}$ is an $EIS_{H^2}$ sequence, there exists $N_1$ such that for any $a\in\ell^2$ there exists a function $f_0\in H^2$ such that $$f_0(\lambda_n)=a_n\left(\frac{1 - |\Theta(\lambda_n)|^2}{1-{\ensuremath{\left\vert\lambda_n\right\vert}}^2}\right)^\frac{1}{2}~\mbox{for all}~n\geq N_1$$ and $${\ensuremath{\left\|f_0\right\|}}_{2}\leq (1+\varepsilon){\ensuremath{\left\|\{a_k (1 - |\Theta(\lambda_k)|^2)^{\frac{1}{2}}\}\right\|}}_{N_1,\ell^2} \le (1 + \varepsilon){\ensuremath{\left\|a \right\|}}_{N_1,\ell^2}.$$ Here we have applied the $EIS_{H^2}$ property to the sequence $\{a_k(1-\left\vert \Theta(\lambda_k)\right\vert^2)^{\frac{1}{2}}\}\in\ell^2$. By we have that $$\begin{aligned}
\tilde{f}_0(\lambda_k) & = & f_0(\lambda_k)-\Theta(\lambda_k) T_{\overline{\Theta}}(f_0)(\lambda_k)\\
& = & a_k(1 - |\Theta(\lambda_k)|^2)^\frac{1}{2}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{-\frac{1}{2}}-\Theta(\lambda_k) T_{\overline{\Theta}}(f_0)(\lambda_k)\quad\forall k\geq N_1 \end{aligned}$$ and ${\ensuremath{\left\|\tilde{f}_0\right\|}}_2\leq{\ensuremath{\left\|f_0\right\|}}_2\leq (1+\varepsilon){\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}$. Rearranging the above, for $k \ge N_1$ we have $$\begin{aligned}
{\ensuremath{\left\vert\tilde{f}_0(\lambda_k)(1 - |\Theta(\lambda_k)|^2)^{-\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{\frac{1}{2}}-a_k\right\vert}} & = & {\ensuremath{\left\vert\Theta(\lambda_k) T_{\overline{\Theta}}(f_0)(\lambda_k)(1 - |\Theta(\lambda_k)|^2)^{-\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{\frac{1}{2}}\right\vert}}\\
& \leq &  \kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}} {\ensuremath{\left\|f_0\right\|}}_2\\
&\leq& (1+\varepsilon) \kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}}  {\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}.\end{aligned}$$ We claim that $\{a^{(1)}_n\}=\{\tilde{f}_0(\lambda_n)(1 - |\Theta(\lambda_n)|^2)^{-\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_n\right\vert}}^2)^{\frac{1}{2}} - a_n\}\in\ell^2$ and that there is a constant $N_2$ depending only on $\varepsilon$ and the Carleson measure given by the thin sequence $\{\lambda_n\}$ such that $$\label{a1} {\ensuremath{\left\|a^{(1)}\right\|}}_{N_2,\ell^2}\leq (1+\varepsilon)^2\kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}} {\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}.$$

Since the sequence $\{\lambda_n\}$ is thin and distinct, it hence generates an $H^2$ Carleson measure with norm at most $(1+\varepsilon)$; that is, we have the existence of $N_2 \ge N_1$ such that $\kappa_{N_2}(1 - \kappa_{N_2}^2)^{-\frac{1}{2}} \le \kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}}$ and $$\begin{aligned}
{\ensuremath{\left\|a^{(1)}\right\|}}_{N_2,\ell^2} & = & \left(\sum_{k\geq N_2} {\ensuremath{\left\vert\Theta(\lambda_k)\right\vert}}^2 {\ensuremath{\left\vertT_{\overline{\Theta}}(f_0)(\lambda_k)\right\vert}}^2 (1 - |\Theta(\lambda_k)|^2)^{-1}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)\right)^{\frac{1}{2}}\\
& \leq &   (1+\varepsilon) \kappa_{N_2}(1 - \kappa_{N_2}^2)^{-\frac{1}{2}} {\ensuremath{\left\|T_{\overline{\Theta}}f_0\right\|}}_2 \nonumber\\
& \leq &(1+\varepsilon) \kappa_{N_2}(1 - \kappa_{N_2}^2)^{-\frac{1}{2}}  {\ensuremath{\left\|f_0\right\|}}_2\nonumber\\
& \leq & (1+\varepsilon)^2 \kappa_{N_1} (1 - \kappa_{N_1}^2)^{-\frac{1}{2}} {\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}\nonumber<\infty,\end{aligned}$$ completing the proof of the claim.

We will now iterate these estimates and ideas. Let $\widetilde{a^{(1)}_n}=-\frac{a^{(1)}_n}{(1 + \varepsilon)^2\kappa_{N_1} (1 - \kappa_{N_1}^2)^{-\frac{1}{2}} }$ for $n \ge N_2$ and $\widetilde{a^{(1)}_n} = 0$ otherwise. Then from (\[a1\]) we have that ${\ensuremath{\left\|\widetilde{a^{(1)}}\right\|}}_{N_1,\ell^2} = {\ensuremath{\left\|\widetilde{a^{(1)}}\right\|}}_{N_2,\ell^2}  \le {\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}$. Since $\{\lambda_n\}$ is an $EIS_{H^2}$ we may choose $f_1\in H^2$ with $$f_1(\lambda_n)=\widetilde{a_n^{(1)}}(1 - |\Theta(\lambda_n)|^2)^{\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_n\right\vert}}^2)^{-\frac{1}{2}}~\mbox{for all}~n\geq N_1$$ and, letting $\widetilde{f}_1 = P_{K_\Theta}(f_1)$, we have $${\ensuremath{\left\|\tilde{f}_1\right\|}}_{2}\leq{\ensuremath{\left\|f_1\right\|}}_{2}\leq (1+\varepsilon){\ensuremath{\left\|\widetilde{a^{(1)}}\right\|}}_{N_1,\ell^2}\leq (1+\varepsilon){\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}.$$ As above, $$\begin{aligned}
\widetilde{f}_1(\lambda_k) & = & f_1(\lambda_k)-\Theta(\lambda_k) T_{\overline{\Theta}}(f_1)(\lambda_k)\\
& = & \widetilde{a_k^{(1)}}(1 - |\Theta(\lambda_k)|^2)^{\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{-\frac{1}{2}}-\Theta(\lambda_k) T_{\overline{\Theta}}(f_1)(\lambda_k)\quad\forall k\geq N_1. \end{aligned}$$ And, for $k \ge N_1$ we have $$\begin{aligned}
{\ensuremath{\left\vert\tilde{f}_1(\lambda_k)(1 - |\Theta(\lambda_k)|^2)^{-\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{\frac{1}{2}}-\widetilde{a_k^{(1)}}\right\vert}} & = & {\ensuremath{\left\vert\Theta(\lambda_k) T_{\overline{\Theta}}(f_1)(\lambda_k)(1 - |\Theta(\lambda_k)|^2)^{-\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{\frac{1}{2}}\right\vert}}\\
& \leq &  \kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}}  {\ensuremath{\left\|f_1\right\|}}_2\\
& \leq & (1+\varepsilon)\kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}}   {\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}.\end{aligned}$$ Using the definition of $\widetilde{a^{(1)}}$, for $k \ge N_2$ one arrives at $$\begin{aligned}
{\ensuremath{\left\vert\left((1+\varepsilon)^2\kappa_{N_1}(1 - \kappa_{N_1}^2)^{-\frac{1}{2}} \tilde{f}_1(\lambda_k)+\tilde{f}_0(\lambda_k)\right)(1 - |\Theta(\lambda_k)|^2)^{-\frac{1}{2}}(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{\frac{1}{2}}-a_k\right\vert}}\\
\leq (1+\varepsilon)^3\kappa_{N_1}^2 (1 - \kappa_{N_1}^2)^{-1}{\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}.\end{aligned}$$ We continue this procedure, constructing sequences $a^{(j)}\in \ell^2$ and functions $\tilde{f}_j\in K_{\Theta}$ such that $${\ensuremath{\left\|a^{(j)}\right\|}}_{N_1,\ell^2}\leq (1+\varepsilon)^{2j}\left(\frac{\kappa_{N_1}}{(1 - \kappa_{N_1}^2)^{\frac{1}{2}}}\right)^j{\ensuremath{\left\|a\right\|}}_{N_1,\ell^2},$$ $$\left\vert \frac{(1-{\ensuremath{\left\vert\lambda_k\right\vert}}^2)^{\frac{1}{2}}}{(1 - |\Theta(\lambda_k)|^2)^{\frac{1}{2}}} \left(\sum_{l=0}^{j} (1+\varepsilon)^{2l}\left(\frac{\kappa_{N_1}}{(1 - \kappa_{N_1}^2)^{\frac{1}{2}}}\right)^{l}\tilde{f}_l(\lambda_k)\right) -a_k \right\vert\leq \left(1+\varepsilon\right)^{2j+1}\left(\frac{\kappa_{N_1}}{(1 - \kappa_{N_1}^2)^{\frac{1}{2}}}\right)^{j+1}\left\Vert a\right\Vert_{N_1,\ell^2},$$ and $${\ensuremath{\left\|\tilde{f}_j\right\|}}_2\leq (1+\varepsilon){\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}~\mbox{for all}~j\in {\mathbb{N}}.$$ Define $$F=\sum_{j = 0}^{\infty} (1+\varepsilon)^{2j} \left(\frac{\kappa_{N_1}}{(1 - \kappa_{N_1}^2)^{\frac{1}{2}}}\right)^j \tilde{f}_j.$$ Then $F\in  K_{\Theta}$ since each $\tilde{f}_j\in K_{\Theta}$ and, since $\kappa_m \to 0$, we may assume that $$(1 + \varepsilon)^2\left(\frac{\kappa_{N_1}}{(1 - \kappa_{N_1}^2)^{\frac{1}{2}}}\right) < 1.$$ So,

$${\ensuremath{\left\|F\right\|}}_2\leq \frac{(1+\varepsilon)}{1 - (1+\varepsilon)^2\left(\frac{\kappa_{N_1}}{\left(1 - \kappa_{N_1}^2\right)^{\frac{1}{2}}}\right)} {\ensuremath{\left\|a\right\|}}_{N_1,\ell^2}.$$ For this $\varepsilon$, consider $\varepsilon_M < \varepsilon$ with $\frac{(1+\varepsilon_M)}{1 - (1+\varepsilon_M)^2\left(\frac{\kappa_{N_M}}{\left(1 - \kappa_{N_M}^2\right)^{\frac{1}{2}}}\right)}<1+\varepsilon$. Then, using the process above, we obtain $F_M$ satisfying $F_M \in K_\Theta, \|F_M\|_2 \le  (1 + \varepsilon) \|a\|_{M, \ell^2}$ and $F_M(\lambda_n)\|K_{\lambda_n}\|^{-1}_\mathcal{H} = a_n$ for $n \ge M$. Taking $N(\varepsilon) = M$, we see that $F_M$ satisfies the exact interpolation conditions, completing the proof of the theorem.

We present an alternate method to prove the equivalence between $(1)$ and $(7)$. As noted above, by Theorem \[EISiffASI\] it is true that $(7)\Leftrightarrow (8)$ and thus it suffices to prove that $(1)\Rightarrow (8)\Leftrightarrow (7)$. Let $\varepsilon>0$ be given. Select a sequence $\{\delta_N\}$ with $\delta_N\to 0$ as $N\to\infty$. Since $(1)$ holds, then for large $N$ and $a\in \ell^2$ it is possible to find $f_N\in H^2$ so that $$f_N(a_n)(1-\left\vert\lambda_n\right\vert^2)^{\frac{1}{2}}=\left\Vert a\right\Vert_{N,\ell^2}^{-1} a_n\quad n\geq N$$ with $\left\Vert f_N\right\Vert_{2}\leq 1+\delta_N$. Now observe that we can write $f_N=h_N+\Theta g_N$ with $h_N\in K_\Theta$. Since $h_N$ and $g_N$ are orthogonal projections of $f_N$ onto subspaces of $H^2$, we also have that $\left\Vert h_N\right\Vert_{2}\leq 1+\delta_N$ and similarly for $g_N$.

By the properties of the functions above we have that: $$h_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}=f_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}-\Theta(\lambda_n)g_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}.$$ Hence, one deduces that $$\begin{aligned}
\left\Vert \left\{h_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}-\frac{a_n}{\left\Vert a\right\Vert_{N,\ell^2}}\right\}\right\Vert_{N,\ell^2} & \leq & \left\Vert \left\{f_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}-\frac{a_n}{\left\Vert a\right\Vert_{N,\ell^2}}\right\}\right\Vert_{N,\ell^2}\\
&  & + \left\Vert \left\{\Theta(\lambda_n)g_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}\right\}\right\Vert_{N,\ell^2}\\
& \leq & \left\Vert \left\{\frac{a_n}{\left\Vert a\right\Vert_{N,\ell^2}}\left(\left(\frac{1}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}-1\right)\right\}\right\Vert_{N,\ell^2}\\
& & +\frac{\sup_{m\geq N}\left\vert\Theta(\lambda_m)\right\vert}{(1-\kappa_N^2)^{\frac{1}{2}}} \left\Vert \left\{g_N(\lambda_n)\left(1-\left\vert \lambda_n\right\vert^2\right)^{\frac{1}{2}}\right\}\right\Vert_{N,\ell^2}.\end{aligned}$$ Now for $x$ sufficiently small and positive we have that $\frac{1}{\sqrt{1-x}}-1=\frac{1-\sqrt{1-x}}{\sqrt{1-x}}\lesssim \frac{x}{\sqrt{1-x}}$. Applying this with $x=\sup_{m\geq N} \left\vert\Theta(\lambda_m)\right\vert$ gives that: $$\left\Vert \left\{h_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}-\frac{a_n}{\left\Vert a\right\Vert_{N,\ell^2}}\right\}\right\Vert_{N,\ell^2} \leq \frac{\sup_{m\geq N}\left\vert\Theta(\lambda_m)\right\vert}{(1-\kappa_N^2)^{\frac{1}{2}}} \left(1+\left\Vert \left\{g_N(\lambda_n)\left(1-\left\vert \lambda_n\right\vert^2\right)^{\frac{1}{2}}\right\}\right\Vert_{N,\ell^2}\right).$$ Define $H_N=(1+\delta_N)^{-1} \left\Vert a\right\Vert_{N,\ell^2} h_N$, and then we have $H_N\in K_{\Theta}$ and $\left\Vert H_N\right\Vert_{2}\leq \left\Vert a\right\Vert_{N,\ell^2}$. Using the last estimate and adding and subtracting the quantity $\frac{a_n}{(1+\delta_N)}$ yields that: $$\begin{aligned}
\left\Vert \left\{H_N(\lambda_n)\left(\frac{1-\left\vert \lambda_n\right\vert^2}{1-\left\vert \Theta(\lambda_n)\right\vert^2}\right)^{\frac{1}{2}}-a_n\right\}\right\Vert_{N,\ell^2}   \leq & &  
\\  \left(\frac{\sup_{m\geq N}\left\vert\Theta(\lambda_m)\right\vert}{(1+\delta_N)(1-\kappa_N^2)^{\frac{1}{2}}} \left(1+\left\Vert \left\{g_N(\lambda_n)\left(1-\left\vert \lambda_n\right\vert^2\right)^{\frac{1}{2}}\right\}\right\Vert_{N,\ell^2}\right)+\delta_N\right)\left\Vert a\right\Vert_{N,\ell^2}.\end{aligned}$$ Note that the quantity: $$\left(\frac{\sup_{m\geq N}\left\vert\Theta(\lambda_m)\right\vert}{(1+\delta_N)(1-\kappa_N^2)^{\frac{1}{2}}} \left(1+\left\Vert \left\{g_N(\lambda_n)\left(1-\left\vert \lambda_n\right\vert^2\right)^{\frac{1}{2}}\right\}\right\Vert_{N,\ell^2}\right)+\delta_N\right)\lesssim \delta_N+\sup_{m\geq N}\left\vert\Theta(\lambda_m)\right\vert.$$ Here we have used that the sequence $\{\lambda_n\}$ is by hypothesis an interpolating sequence and hence: $\left\Vert \left\{g_N(\lambda_n)\left(1-\left\vert \lambda_n\right\vert^2\right)^{\frac{1}{2}}\right\}\right\Vert_{N,\ell^2}\lesssim \left\Vert g_N\right\Vert_{2}\leq 1+\delta_N$. Since by hypothesis we have that $\delta_N+\sup_{m\geq N}\left\vert\Theta(\lambda_m)\right\vert\to 0$ as $N\to\infty$, it is possible to make this less than the given $\varepsilon>0$, and hence we get a function $H_N$ satisfying the properties for $\{\lambda_n\}$ to be $AIS_{K_\Theta}$.

The proof above also gives an estimate on the norm of the interpolating function in the event that $\sup_n |\Theta(\lambda_n)| \le \kappa < 1$, but $(1 + \varepsilon)$ is no longer the best estimate.

Carleson Measures in Model Spaces {#CMMS}
---------------------------------

From Theorem \[main\], and , we have a Carleson measure statement for thin sequences in the Hardy space $H^2$. In this section, we obtain an equivalence in model spaces.

We now consider the embedding constants in the case of model spaces. As before, given a positive measure $\mu$ on ${\mathbb{D}}$, we denote the (possibly infinite) constant

$${\mathcal{C}}_{\Theta}(\mu) =  \sup_{f \in K_{\Theta}, f \neq 0} \frac{\|f\|^2_{L^2({\mathbb{D}}, \mu)}}{\|f\|^2_2}$$ as the Carleson embedding constant of $\mu$ on $K_{\Theta}$ and $${\mathcal{R}}_{\Theta}(\mu) =  \sup_{z} \|k^{\Theta}_z\|_{L^2({\mathbb{D}}, \mu)}^2$$ as the embedding constant of $\mu$ on the reproducing kernel of $K_\Theta$ (recall that the kernels $k^{\Theta}_z$ are normalized). It is known that for general measure $\mu$ the constants ${\mathcal{R}}_{\Theta}(\mu)$ and ${\mathcal{C}}_{\Theta}(\mu)$ are not equivalent, [@NV]. The complete geometric characterization of the measures for which ${\mathcal{C}}_{\Theta}(\mu)$ is finite is contained in [@LSUSW]. However, we always have that $${\mathcal{R}}_\Theta(\mu) \le {\mathcal{C}}_\Theta(\mu).$$ For $N > 1$, let $$\sigma_N = \sum_{k \ge N} \left\Vert K_{\lambda_k}^{\Theta}\right\Vert^{-2}\delta_{\lambda_k}=\sum_{k \ge N} \frac{1-\left\vert \lambda_k\right\vert^2}{1-\left\vert \Theta(\lambda_k)\right\vert^2}\delta_{\lambda_k}.$$ Note that for each $f \in K_{\Theta}$ $$\label{munorm}
        \| f\|^2_{L^2({\mathbb{D}}, \sigma_N)} = \sum_{k=N}^\infty \frac{(1 - |\lambda_k|^2)}{(1-\left\vert \Theta(\lambda_k)\right\vert^2)} |f(\lambda_k)|^2 = \sum_{k=N}^\infty |\langle f, k^{\Theta}_{\lambda_k}\rangle|^2,$$ and therefore we see that $$\label{e:CETests}
1 \le  {\mathcal{R}}_\Theta(\sigma_N) \le {\mathcal{C}}_\Theta(\sigma_N).$$

By working in a restricted setting and imposing a condition on $\{\Theta(\lambda_n)\}$ we have the following.

\[thm:Carleson\] Suppose $\Lambda = \{\lambda_n\}$ is a sequence in $\mathbb{D}$ and $\Theta$ is a nonconstant inner function such that $\kappa_m := \sup_{n \ge m}|\Theta(\lambda_n)|\to 0$. For $N > 1$, let $$\sigma_N = \sum_{k \ge N} \left\Vert K_{\lambda_k}^{\Theta}\right\Vert^{-2}\delta_{\lambda_k}=\sum_{k \ge N} \frac{1-\left\vert \lambda_k\right\vert^2}{1-\left\vert \Theta(\lambda_k)\right\vert^2}\delta_{\lambda_k}.$$ Then the following are equivalent:

1.  $\Lambda$ is a thin sequence;

2.  $ {\mathcal{C}}_\Theta(\sigma_N) \to 1$ as $N \to \infty$;

3.  $ {\mathcal{R}}_\Theta(\sigma_N) \to 1$ as $N \to \infty$.

We have $(2)\Rightarrow (3)$ by testing on the function $f=k_z^{\Theta}$ for all $z\in\mathbb{D}$, which is nothing more then .

We next focus on $(1)\Rightarrow(2)$. Let $f \in K_{\Theta}$ and let the sequence $a$ be defined by $a_j = \left\|K_{\lambda_j}\right\|^{-1}f(\lambda_j)$. By , $\left\|a\right\|_{N, \ell^2}^2 = \left\| f\right\|^2_{L^2({\mathbb{D}}, \sigma_N)}$, and since $\{k_{\lambda_j}^{\Theta}\}$ is an $AOB$, there exists $C_N$ such that $$\begin{aligned}
\left\|a\right\|_{N,\ell^2}^2 & =  \sum_{j \ge N} \left\|K_{\lambda_j}^\Theta\right\|^{-2}_{K_{\Theta}} |f(\lambda_j)|^2 =  \left\langle f, \sum_{j \ge N} a_j k_{\lambda_j}^\Theta \right\rangle_{K_{\Theta}} \le  \left\| f\right\|_{2} \left\|\sum_{j \ge N} a_j k_{\lambda_j}^\Theta\right\|_{K_{\Theta}} \le  C_N \left\| f\right\|_{2} \left\|a\right\|_{N,\ell^2}.\end{aligned}$$ By (1) and [@CFT]\*[Theorem 5.2]{}, we know that $C_N \to 1$ and since we have established that $\|f\|_{L^2({\mathbb{D}}, \sigma_N)} \le C_N \|f\|_2$, (1) $\Rightarrow$ (2) follows.

An alternate way to prove this is to use Theorem \[main\], $(2)\Rightarrow(5)$, and the hypothesis on $\Theta$. Since it is possible to then show that $\frac{\mathcal{C}_{\Theta}(\sigma_N)}{\mathcal{C}(\mu_N)}\to 1$. Indeed, given $\varepsilon>0$, we have that $1\leq\mathcal{C}(\mu_M)$ for all $M$, and since $\{\lambda_n\}$ is thin there exists a $N$ such that $\mathcal{C}(\mu_M)<1+\varepsilon$ for all $M\geq N$. Hence, $1\leq \mathcal{C}(\mu_M)<1+\varepsilon$ for all $M\geq N$. These facts easily lead to: $$\frac{1}{1+\varepsilon}\leq\frac{\mathcal{C}_{\Theta}(\sigma_M)}{\mathcal{C}(\mu_M)}$$ Further, since $\Theta$ tends to zero on the sequence $\{\lambda_n\}$ there is an integer, without loss we may take it to be $N$, so that $\frac{1}{1-\left\vert \Theta(\lambda_n)\right\vert^2}<1+\varepsilon$ for all $n\geq N$. From this we deduce that: $$\frac{\mathcal{C}_{\Theta}(\sigma_M)}{\mathcal{C}(\mu_M)}< (1+\varepsilon)\frac{\sup\limits_{f\in K_{\theta}} \sum_{n\geq M} (1-\left\vert \lambda_m\right\vert^2)\left\vert f(\lambda_m)\right\vert^2}{\mathcal{C}(\mu_M)}\leq (1+\varepsilon)$$ in the last estimate we used that $K_\Theta\subset H^2$ and so the suprema appearing in the numerator is always at most the expression in the denominator. Combining the estimates we have that for $M\geq N$, that: $$\frac{1}{1+\varepsilon}\leq \frac{\mathcal{C}_{\Theta}(\sigma_M)}{\mathcal{C}(\mu_M)}<1+\varepsilon$$ which yields the conclusion about the ratio tending to $1$ as $N\to \infty$.

Now consider $(3)\Rightarrow (1)$ and compute the quantity $\mathcal{R}_\Theta(\sigma_N)$. In what follows, we let $\Lambda_N$ denote the tail of sequence, $\Lambda_N=\{\lambda_k: k\geq N\}$. Note that we have $\left\vert 1-\overline{a}b\right\vert\geq 1-\left\vert a\right\vert$. Using this estimate we see that: $$\begin{aligned}
\sup_{z\in\mathbb{D}} \|k^{\Theta}_z\|_{L^2({\mathbb{D}}, \sigma_N)}^2 & = & \sup_{z\in\mathbb{D}} \sum_{k\geq N} \frac{(1-\left\vert \lambda_k\right\vert^2)}{(1-\left\vert \Theta(\lambda_k)\right\vert^2)} \frac{(1-\left\vert z\right\vert^2)}{(1-\left\vert \Theta(z)\right\vert^2)}\frac{\left\vert 1-\Theta(z)\overline{\Theta(\lambda_k)}\right\vert^2}{\left\vert 1-z\overline{\lambda_k}\right\vert^2}\\
 & \geq & \sup_{z\in\mathbb{D}} \sum_{k\geq N} \frac{(1-\left\vert \lambda_k\right\vert^2)(1-\left\vert z\right\vert^2)}{\left\vert 1-z\overline{\lambda_k}\right\vert^2} \frac{(1-\left\vert \Theta(z)\right\vert)(1-\left\vert \Theta(\lambda_k)\right\vert)}{(1-\left\vert \Theta(z)\right\vert^2)(1-\left\vert \Theta(\lambda_k)\right\vert^2)}\\
 & = & \sup_{z\in\mathbb{D}} \sum_{k\geq N} \frac{(1-\left\vert \lambda_k\right\vert^2)(1-\left\vert z\right\vert^2)}{\left\vert 1-z\overline{\lambda_k}\right\vert^2} \frac{1}{(1+\left\vert \Theta(z)\right\vert)(1+\left\vert \Theta(\lambda_k)\right\vert)}\\
  & \geq & \sup_{z\in\Lambda_N} \sum_{k\geq N} \frac{(1-\left\vert \lambda_k\right\vert^2)(1-\left\vert z\right\vert^2)}{\left\vert 1-z\overline{\lambda_k}\right\vert^2} \frac{1}{(1+\left\vert \Theta(z)\right\vert)(1+\left\vert \Theta(\lambda_k)\right\vert)}\\
    & \geq & \frac{1}{(1+\kappa_N)^2}\sup_{z\in\Lambda_N} \sum_{k\geq N} \frac{(1-\left\vert \lambda_k\right\vert^2)(1-\left\vert z\right\vert^2)}{\left\vert 1-z\overline{\lambda_k}\right\vert^2}.\end{aligned}$$ By the Weierstrass Inequality, we obtain for $M \ge N$ that $$\begin{aligned}
\label{wi}
 \prod_{k \geq N, k \neq M}  \left|  \frac{\lambda_k - \lambda_M}{1 - \bar \lambda_k \lambda_M}\right|^2 
& = & \prod_{k \geq N, k \neq M} \left(  1-  \frac{(1 - |\lambda_k|^2)(1 - |\lambda_M|^2)}{|1 - \bar \lambda_k \lambda_M|^2} \right)\nonumber\\  
 & \ge & 1 - \sum_{k \geq N, k \neq M}  \frac{(1- |\lambda_M|^2)(1- |\lambda_k|^2)}{ | 1 - \bar \lambda_k \lambda_M|^2}.\end{aligned}$$ Thus, by we have for $M \ge N$, $$\begin{aligned}
\frac{1}{(1+\kappa_N)^2}\sup_{z\in\Lambda_N} \sum_{k\geq N} \frac{(1-\left\vert \lambda_k\right\vert^2)(1-\left\vert z\right\vert^2)}{\left\vert 1-z\overline{\lambda_k}\right\vert^2} 
& \ge  & \frac{1}{(1+\kappa_N)^2}\left(\sum_{k \geq N, k\neq M} \frac{(1-\left\vert \lambda_k\right\vert^2)(1-\left\vert \lambda_M\right\vert^2)}{\left\vert 1-\lambda_M\overline{\lambda_k}\right\vert^2} + 1\right)\\
& \ge & \frac{1}{(1+\kappa_N)^2}\left(1 - \prod_{k \geq N, k\neq M}  \left|  \frac{\lambda_k - \lambda_M}{1 - \bar \lambda_k \lambda_M}\right|^2 + 1\right).\end{aligned}$$ Now by assumption, recalling that $\kappa_N := \sup_{n \ge N}|\Theta(\lambda_n)|$, we have $$\lim_{N \to \infty}\sup_{z\in\mathbb{D}} \|k^{\Theta}_z\|_{L^2({\mathbb{D}}, \sigma_N)}^2 = 1~\mbox{ and }~\lim_{N \to \infty} \kappa_N = 0,$$ so $$1 = \lim_{N \to \infty}\sup_{z\in\mathbb{D}} \|k^{\Theta}_z\|_{L^2({\mathbb{D}}, \sigma_N)}^2
\ge \lim_{N \to \infty} \frac{1}{(1+\kappa_N)^2}\left(1 - \prod_{k \geq N, k\neq M}  \left|  \frac{\lambda_k - \lambda_M}{1 - \bar \lambda_k \lambda_M}\right|^2 + 1\right) \ge 1.$$ Therefore, for any $M \ge N$ $$\label{e:large}
\prod_{k \geq N, k\neq M}  \left|  \frac{\lambda_k - \lambda_M}{1 - \bar \lambda_k \lambda_M}\right| > 1 - \varepsilon~\mbox{as}~N \to \infty.$$ Also, for any $\varepsilon>0$ there is an integer $N_0$ such that for all $M> N_0$ we have: $$\label{e:bigk}
\prod_{k \geq N_0, k\neq M}  \left|  \frac{\lambda_k - \lambda_M}{1 - \bar \lambda_k \lambda_M}\right| >1-\varepsilon.$$ Fix this value of $N_0$, and consider $k<N_0$. Further, for $k \ne M$ and $k<N_0$, $$\begin{aligned}
 1- \rho(\lambda_M, \lambda_k)^2 & =  1- \left|\frac{\lambda_k - \lambda_M}{1 - \bar\lambda_k \lambda_M}\right|^2 = \frac{(1- |\lambda_M|^2)(1- |\lambda_k|^2)}{ | 1 - \bar\lambda_k \lambda_M|^2}\\ 
&=  (1 - |\lambda_k|^2)\frac{(1 - |\lambda_M|^2)}{(1 - |\Theta(\lambda_M)|^2)} \frac{1 - |\Theta(\lambda_M)|^2}{\left\vert 1 - \bar \Theta(\lambda_M) \Theta(\lambda_k)\right\vert^2} \left|\frac{1 - \bar\Theta(\lambda_M) \Theta(\lambda_k)}{1 - \lambda_k \bar\lambda_M}\right|^2\\
& =   \frac{1 - |\Theta(\lambda_M)|^2}{\left\vert 1 - \bar \Theta(\lambda_M) \Theta(\lambda_k)\right\vert^2}(1 - |\lambda_k|^2) |k_{\lambda_M}^\Theta(\lambda_k)|^2\\
& =   \frac{1 - |\Theta(\lambda_M)|^2}{\left\vert 1 - \bar \Theta(\lambda_M) \Theta(\lambda_k)\right\vert^2} (1 - |\Theta(\lambda_k)|^2) \frac{(1 - |\lambda_k|^2)}{1 - |\Theta(\lambda_k)|^2} |k_{\lambda_M}^\Theta(\lambda_k)|^2\\
& \le   \frac{1 - |\Theta(\lambda_M)|^2}{\left\vert 1 - \bar \Theta(\lambda_M) \Theta(\lambda_k)\right\vert^2} \left( \|k_{\lambda_M}^\Theta\|_{L^2(\mathbb{D}, \sigma_M)}^2 - 1\right) \\
& \leq  \frac{1}{(1-\kappa_M)^2}\left( \|k_{\lambda_M}^\Theta\|_{L^2(\mathbb{D}, \sigma_M)}^2 - 1\right)\to 0 ~\mbox{as}~ M \to \infty,\end{aligned}$$ since $1\leq \|k_{\lambda_N}^\Theta\|_{L^2(\mathbb{D}, \sigma_N)}^2\leq\sup_{z} \|k_{z}^\Theta\|_{L^2(\mathbb{D}, \sigma_N)}^2$ and, by hypothesis, we have that $\kappa_N \to 0$ and $\mathcal{R}_{\Theta}(\sigma_N)\to 1$. Hence, it is possible to choose an integer $M_0$ sufficiently large compared to $N_0$ so that for all $M>M_0$ $$\rho(\lambda_k,\lambda_{M})>\left(1-\varepsilon\right)^{\frac{1}{N_0}}\quad k<N_0$$ which implies that $$\label{e:smallk}
\prod_{k<N_0} \rho(\lambda_k,\lambda_{M})>1-\varepsilon.$$

Now given $\varepsilon>0$, first select $N_0$ as above in . Then select $M_0$ so that holds. Then for any $M>M_0$ by writing the product $$\prod_{k\neq M} \rho(\lambda_k,\lambda_M)=\prod_{k<N_0} \rho(\lambda_k,\lambda_{M})\prod_{k>N_0, k\neq M} \rho(\lambda_k,\lambda_M)>(1-\varepsilon)^2.$$ For the first term in the product we have used to conclude that it is greater than $1-\varepsilon$. And for $M$ sufficiently large, by , we have that the second term in the product is greater than $1-\varepsilon$ as well. Hence, $B$ is thin as claimed.

Algebra Version {#asip_algebra}
===============

We now compare the model-space version of our results with an algebra version. Theorem \[main\] requires that our inner function satisfy $\Theta(\lambda_n) \to 0$ for a thin interpolating sequence $\{\lambda_n\}$ to be an $AIS_{K_\Theta}$ sequence. Letting $B$ denote the Blaschke product corresponding to the sequence $\{\lambda_n\}$, denoting the algebra of continuous functions on the unit circle by $C$, and letting $H^\infty + C = \{f + g: f \in H^\infty, g \in C\}$ (see [@Sarason1] for more on this algebra), we can express this condition in the following way: $\Theta(\lambda_n) \to 0$ if and only if $\overline{B} \Theta \in H^\infty + C$. In other words, if and only if $B$ divides $\Theta$ in $H^\infty + C$, [@AG; @GIS].

We let $\mathcal{B}$ be a Douglas algebra; that is, a uniformly closed subalgebra of $L^\infty$ containing $H^\infty$. It will be helpful to use the maximal ideal space of our algebra. Throughout $M(\mathcal{B})$ denotes the maximal ideal space of the algebra $\mathcal{B}$; that is, the set of nonzero continuous multiplicative linear functionals on $\mathcal{B}$.

We now consider thin sequences in uniform algebras. This work is closely connected to the study of such sequences in general uniform algebras (see [@GM]) and the special case $B = H^\infty$ is considered in [@HIZ]. With the weak-$\star$ topology, $M(\mathcal{B})$ is a compact Hausdorff space. In interpreting our results below, it is important to recall that each $x \in M(H^\infty)$ has a unique extension to a linear functional of norm one and, therefore, we may identify $M(\mathcal{B})$ with a subset of $M(H^\infty)$. In this context, the condition we will require (see Theorem \[main\_algebra\]) for an $EIS_\mathcal{B}$ sequence to be the same as an $AIS_\mathcal{B}$ sequence is that the sequence be thin near $M(\mathcal{B})$. We take the following as the definition (see [@SW]):

An interpolating sequence $\{\lambda_n\}$ with corresponding Blaschke product $b$ is said to be [thin near $M(\mathcal{B})$]{.nodecor} if for any $0<\eta < 1$ there is a factorization $b = b_1 b_2$ with $b_1$ invertible in $\mathcal{B}$ and $$|b_2^\prime(\lambda_n)|(1 - |\lambda_n|^2) > \eta$$ for all $n$ such that $b_2(\lambda_n) = 0$.

We will be interested in two related concepts that a sequence can have. We first introduce a norm on a sequence $\{a_n\}\in \ell^\infty$ that is induced by a second sequence $\{\lambda_n\}$ and a set $\mathcal{O}\supset M(\mathcal{B})$ that is open in $M(H^\infty)$. Set $I_\mathcal{O}=\{n\in{\mathbb{Z}}: \lambda_n\in\mathcal{O}\}$. Then we define $${\ensuremath{\left\|a\right\|}}_{\mathcal{O},\ell^\infty}=\sup\{ {\ensuremath{\left\verta_n\right\vert}}: n\in I_\mathcal{O}\}.$$

A Blaschke sequence $\{\lambda_n\}$ is an [eventual $1$-interpolating sequence in a Douglas algebra $\mathcal{B}$]{.nodecor}, denoted $EIS_{\mathcal{B}}$, if for every $\varepsilon > 0$ there exists an open set $\mathcal{O}\supset M(\mathcal{B})$ such that for each $\{a_n\} \in \ell^\infty$ there exists $f_{\mathcal{O}, a} \in H^\infty$ with $$f_{\mathcal{O}, a}(\lambda_n) = a_n ~\mbox{for}~ \lambda_n\in\mathcal{O}  ~\mbox{and}~ \|f_{\mathcal{O}, a}\|_{\infty} \le (1 + \varepsilon) \|a\|_{\mathcal{O}, \ell^\infty}.$$

A Blaschke sequence $\{\lambda_n\}$ is a [strong asymptotic interpolating sequence in a Douglas algebra $\mathcal{B}$]{.nodecor}, denoted $AIS_{\mathcal{B}}$, if for all $\varepsilon > 0$ there exists an open set $\mathcal{O}\supset M(\mathcal{B})$ such that for all sequences $\{a_n\} \in \ell^\infty$ there exists a function $G_{\mathcal{O}, a} \in H^\infty$ such that $\|G_{\mathcal{O}, a}\|_{\infty} \le \|a\|_{\mathcal{O},\ell^\infty}$ and $$\|\{G_{\mathcal{O}, a}(\lambda_n)  - a_n\}\|_{\mathcal{O}, \ell^\infty} < \varepsilon \|a\|_{\mathcal{O}, \ell^\infty}.$$

\[EISiffASI\_algebra\] Let $\mathcal{B}$ be a Douglas algebra. Let $\{\lambda_n\}$ be a Blaschke sequence of points in ${\mathbb{D}}$. Then $\{\lambda_n\}$ is an $EIS_{\mathcal{B}}$ sequence if and only if $\{\lambda_n\}$ is an $AIS_{\mathcal{B}}$.

If a sequence is an $EIS_{\mathcal{B}}$, then it is trivially $AIS_{\mathcal{B}}$, for given $\varepsilon > 0$ we may take $G_{N, a} = \frac{f_{N, a}}{(1 + \varepsilon)}$.

For the other direction, suppose $\{\lambda_n\}$ is an $AIS_{\mathcal{B}}$ sequence. Let $\varepsilon > 0$ be given and let $\varepsilon^\prime < \frac{\varepsilon}{1 + \varepsilon}$. Let $\mathcal{O} \supset M(\mathcal{B})$ denote the open set we obtain from the definition of $AIS_{\mathcal{B}}$ corresponding to $\varepsilon^\prime$. Reordering the points of the sequence in $\mathcal{O}$ so that they begin at $n = 1$ and occur in the same order, we let $T: H^\infty \to \ell^\infty$ be defined by $T(g) = \{g(\lambda_{n})\}$. We let $y_\mathcal{O}$ denote the corresponding reordered sequence. Then $T$ is a bounded linear operator between Banach spaces, so we may use Proposition \[Banachspace\] to choose $f \in H^\infty$ so that $Tf = y_\mathcal{O}$ and $\|f\| < \frac{1}{1 - \varepsilon^\prime} \|y_\mathcal{O}\|_{\ell^\infty} < (1 + \varepsilon) \|y\|_{\mathcal{O}, \ell^\infty}$ to complete the proof.

Letting $\overline{B}$ denote the set of functions with conjugate in $B$, we mention one more set of equivalences. In [@SW Theorem 1] Sundberg and Wolff showed that an interpolating sequence $\{\lambda_n\}$ is thin near $M(\mathcal{B})$ if and only if for any bounded sequence of complex numbers $\{w_n\}$ there exists a function in $f \in H^\infty \cap \overline{B}$ such that $f(\lambda_n) = w_n$ for all $n$.

Finally, we note that Earl ([@E Theorem 2] or [@E2]) proved that given an interpolating sequence for the algebra $H^\infty$ satisfying $$\inf_n \prod_{j \ne n} \left|\frac{z_j - z_n}{1 - \overline{z_j} z_n}\right| \ge \delta > 0$$ then for any bounded sequence $\{\omega_n\}$ and $$\label{Earl}
M > \frac{2 - \delta^2 + 2(1 - \delta^2)^{1/2}}{\delta^2} \sup_n |\omega_n|$$ there exists a Blaschke product $B$ and a real number $\alpha$ so that $$M e^{i \alpha} B(\lambda_j) = \omega_j~\mbox{for all}~j.$$ Using the results of Sundberg-Wolff and Earl, we obtain the following theorem.

\[main\_algebra\] Let $\{\lambda_n\}$ in $\mathbb{D}$ be an interpolating Blaschke sequence and let $\mathcal{B}$ be a Douglas algebra. The following are equivalent:

1.  $\{\lambda_n\}$ is an $EIS_{\mathcal{B}}$ sequence; \[EIS\_Douglas\]

2.  $\{\lambda_n\}$ is a $AIS_{\mathcal{B}}$ sequence; \[AIS\_Douglas\]

3.  $\{\lambda_n\}$ is thin near $M(\mathcal{B})$;\[nearthin\]

4.  for any bounded sequence of complex numbers $\{w_n\}$ there exists a function in $f \in H^\infty \cap \overline{B}$ such that $f(\lambda_n) = w_n$ for all $n$.\[SW\]

The equivalence between and is contained in Theorem \[EISiffASI\_algebra\]. The equivalence of and is the Sundberg-Wolff theorem.

We next prove that if a sequence is thin near $M(\mathcal{B})$, then it is an $EIS_{\mathcal{B}}$ sequence. We let $b$ denote the Blaschke product associated to the sequence $\{\lambda_n\}$.

Given $\varepsilon>0$, choose $\gamma$ so that $$\left(\frac{1 + \sqrt{1 - \gamma^2}}{\gamma}\right)^2 < 1 + \varepsilon.$$ Choose a factorization $b = b_1^\gamma b_2^\gamma$ so that $\overline{b_1^\gamma} \in \mathcal{B}$ and $\delta(b_2) = \inf (1 - |\lambda|^2)|b_2^\gamma \,^\prime(\lambda)| > \gamma$. Since $|b_1^\gamma| = 1$ on $M(\mathcal{B})$ and $\gamma < 1$, there exists an open set $\mathcal{O} \supset M(\mathcal{B})$ such that $|b_1^\gamma| > \gamma$ on $\mathcal{O}$. Note that if $b(\lambda) = 0$ and $\lambda \in \mathcal{O}$, then $b_2(\lambda) = 0$.

The condition on $b_2^\gamma$ coupled with Earl’s Theorem (see ), gives rise to functions $\{f_k^\gamma\}$ in $H^\infty$ (!), and hence in $\mathcal{B}$ so that $$\label{estimate}
f_j^\gamma(\lambda_k) = \delta_{jk} \, ~\mbox{whenever}~ b_2^\gamma(\lambda_k) = 0 ~\mbox{and}~\sup_{z \in \mathbb{D}}\sum_{j}{\ensuremath{\left\vertf_j^\gamma(z)\right\vert}}\leq \left(\frac{1 + \sqrt{1 - \gamma^2}}{\gamma}\right)^2.$$

Now given $a\in\ell^\infty$, choose the corresponding P. Beurling functions (as in ) and let $$f^\gamma_{\mathcal{O}, a}=\sum_{j} a_j f_j^\gamma.$$ By construction we have that $f_{\mathcal{O},a}(\lambda_n)=a_n$ for all $\lambda_n\in\mathcal{O}$. Also, by Earl’s estimate , we have that $${\ensuremath{\left\|f_{\mathcal{O},a}^{\gamma}\right\|}}_{\infty} \leq (1+\varepsilon)\|a\|_\infty.$$ Thus, implies .

Finally, we claim implies . Suppose $\{\lambda_n\}$ is a $EIS_{\mathcal{B}}$ sequence. Let $0 < \eta < 1$ be given and choose $\eta_1$ with $1/(1 + \eta_1) > \eta$, a function $f \in H^\infty$ and $\mathcal{O} \supset M(\mathcal{B})$ open in $M(H^\infty)$ with $$f_{\mathcal{O}, n}(\lambda_m) = \delta_{nm}~\mbox{for}~\lambda_m \in \mathcal{O}~\mbox{and}~\|f\|_{\mathcal{O}, n} \le 1 + \eta_1.$$ Let $b_2$ denote the Blaschke product with zeros in $\mathcal{O}$, $b_1$ the Blaschke product with the remaining zeros and let $$f_{\mathcal{O}, n}(z) = \left(\prod_{j \ne n: b_2(\lambda_j) = 0} \frac{z - \lambda_j}{1 - \overline{\lambda_j}z}\right) h(z),$$ for some $h \in H^\infty.$ Then $\|h\|_{\infty} \le 1 + \eta_1$ and $$1 = |f_{\mathcal{O}, n}(\lambda_n)| = \left|\left(\prod_{j \ne n; b_2(\lambda_j) = 0} \frac{\lambda_n - \lambda_j}{1 - \overline{\lambda_j}\lambda_n}\right) h(\lambda_n)\right| \le (1 + \eta_1) \prod_{j \ne n} \left|\frac{\lambda_n - \lambda_j}{1 - \overline{\lambda_j}\lambda_n}\right|.$$ Therefore $$(1 - |\lambda_n|^2)|b_2^\prime(\lambda_n)| = \prod_{j \ne n: b_2(\lambda_j) = 0} \left|\frac{\lambda_n - \lambda_j}{1 - \overline{\lambda_j}\lambda_n}\right| \ge 1/(1 + \eta_1) > \eta.$$

Now because we assume that $\{\lambda_n\}$ is interpolating, the Blaschke product $b = b_1 b_2$ with zeros at $\{\lambda_n\}$ will vanish at $x \in M(H^\infty)$ if and only if $x$ lies in the closure of the zeros of $\{\lambda_n\}$, [@Hoffman]\*[p. 206]{} or [@Garnett]\*[p. 379]{}. Now, if we choose $\mathcal{V}$ open in $M(H^\infty)$ with $M(\mathcal{B}) \subset \mathcal{V} \subset \overline{\mathcal{V}} \subset \mathcal{O}$, then $b_1$ has no zeros in ${\mathcal{V}} \cap \mathbb{D}$ and, therefore, no point of $M(\mathcal{B})$ can lie in the closure of the zeros of $b_1$. So $b_1$ has no zeros on $M(\mathcal{B})$. Thus we see that $b_1$ is bounded away from zero on $M(\mathcal{B})$ and, consequently, $b_1$ is invertible in $\mathcal{B}$.

We note that we do not need the full assumption that $b$ is interpolating; it is enough to assume that $b$ does not vanish identically on a Gleason part contained in $M(\mathcal{B})$. Our goal, however, is to illustrate the difference in the Hilbert space and uniform algebra setting and so we have stated the most important setting for our problem.

[^1]: $\dagger$ Research supported in part by Simons Foundation Grant 243653

[^2]: $\ddagger$ Research supported in part by a National Science Foundation DMS grant \# 0955432.
