Traces of the Nevanlinna class on discrete sequences

TL;DR

This paper characterizes when a discrete sequence in the unit disk can be covered by a finite union of Nevanlinna class interpolating sequences, using divided differences and harmonic control.

Contribution

It provides a precise characterization of unions of Nevanlinna class interpolating sequences via divided differences and harmonic majorants.

Findings

Characterization of unions of $n$ interpolating sequences for $N$

Trace of $N$ on $ ext{Lambda}$ matches functions with controlled divided differences

Conditions involving positive harmonic functions for sequence interpolation

Abstract

We show that a discrete sequence $\Lambda$ of the unit disk is the union of $n$ interpolating sequences for the Nevanlinna class $N$ if and only if the trace of $N$ on $\Lambda$ coincides with the space of functions on $\Lambda$ for which the divided differences of order $n - 1$ are uniformly controlled by a positive harmonic function.

Equations

$$\mathcal{N}=\bigl{\{}f\in\operatorname{Hol}(\mathbb{D}):\lim_{r\to 1}\frac{1}{2\pi}\int_{0}^{2\pi}\log^{+}|f(re^{i\theta})|\;d\theta<\infty\bigr{\}}.$$

$$f({\lambda }_n)=v_n,n\in \mathbb{N}.$$

$$b_{\lambda }(z)=\frac{z-\lambda }{1-\bar{\lambda }z}\text{and}{B}_{\lambda }(z)=\frac{B(z)}{b_{\lambda }(z)}.$$

$$\rho (z,w)=\left|\frac{z-w}{1-\bar{z}w}\right|,$$

$$|B_{\lambda }(\lambda )|=(1-|\lambda |)|B^{\prime }(\lambda )|\ge e^{-H(\lambda )},\lambda \in \Lambda .$$

$$\mathcal{N}(\Lambda )=\left\{\{\omega (\lambda ){\}}_{\lambda \in \Lambda }:\exists H\in {\mathrm{Har}}_{+}(\mathbb{D}),{\log }_{+}|\omega (\lambda )|\le H(\lambda ),\lambda \in \Lambda \right\}.$$

$$\begin{array}{rl}{\mathcal{R}}_{\Lambda }:\mathcal{N}& ⟶\mathcal{N}(\Lambda )\\ f& \mapsto \{f(\lambda ){\}}_{\lambda \in \Lambda },\end{array}$$

$$\begin{array}{rl}{\Delta }^0\omega ({\lambda }_1)& =\omega ({\lambda }_1),\\ {\Delta }^j\omega ({\lambda }_1,\dots ,{\lambda }_{j+1})& =\frac{{\Delta }^{j-1}\omega ({\lambda }_2,\dots ,{\lambda }_{j+1})-{\Delta }^{j-1}\omega ({\lambda }_1,\dots ,{\lambda }_j)}{b_{{\lambda }_1}({\lambda }_{j+1})}j\ge 1.\end{array}$$

$${\Lambda }^n=\{({\lambda }_1,\dots ,{\lambda }_n)\in \Lambda \times \stackrel{\stackrel{n}{⌣}}{\cdots }\times \Lambda :{\lambda }_j\ne {\lambda }_k\text{if}j\ne k\},$$

$$\underset{({\lambda }_1,\dots ,{\lambda }_n)\in {\Lambda }^n}{\sup }|{\Delta }^{n-1}\omega ({\lambda }_1,\dots ,{\lambda }_n)|e^{-[H({\lambda }_1)+\cdots +H({\lambda }_n)]}<+\infty .$$

$$\underset{({\lambda }_1,\dots ,{\lambda }_{n+1})\in {\Lambda }^{n+1}}{\sup }\left|\frac{{\Delta }^{n-1}\omega ({\lambda }_2,\dots ,{\lambda }_{n+1})-{\Delta }^{n-1}\omega ({\lambda }_1,\dots ,{\lambda }_n)}{b_{{\lambda }_1}({\lambda }_{n+1})}\right|e^{-[H({\lambda }_1)+\cdots +H({\lambda }_{n+1})]}<\infty .$$

$$\begin{gathered} {\Delta }^{n-1}\omega ({\lambda }_1,\dots ,{\lambda }_n)=\frac{{\Delta }^{n-1}\omega ({\lambda }_1,\dots ,{\lambda }_n)-{\Delta }^{n-1}\omega ({\lambda }_1^0,{\lambda }_1,\dots ,{\lambda }_{n-1})}{b_{{\lambda }_1^0}({\lambda }_n)}{b}_{{\lambda }_1^0}({\lambda }_n)+ \\ +\frac{{\Delta }^{n-1}\omega ({\lambda }_1^0,{\lambda }_1,\dots ,{\lambda }_{n-1})-{\Delta }^{n-1}\omega ({\lambda }_2^0,{\lambda }_1^0,\dots ,{\lambda }_{n-2})}{b_{{\lambda }_2^0}({\lambda }_{n-1})}{b}_{{\lambda }_2^0}({\lambda }_{n-1})+\cdots + \\ \frac{{\Delta }^{n-1}\omega ({\lambda }_{n-1}^0,\dots ,{\lambda }_1^0,{\lambda }_1)-{\Delta }^{n-1}\omega ({\lambda }_n^0,\dots ,{\lambda }_1^0)}{b_{{\lambda }_n^0}({\lambda }_1)}{b}_{{\lambda }_n^0}({\lambda }_1)+{\Delta }^{n-1}\omega ({\lambda }_n^0,\dots ,{\lambda }_1^0) \end{gathered}$$

$$\begin{gathered} \left|{\Delta }^{n-1}\omega ({\lambda }_1,\dots ,{\lambda }_n)\right|\le e^{H({\lambda }_1^0)+H({\lambda }_1)\cdots +H({\lambda }_n)}\rho ({\lambda }_1^0,{\lambda }_n)+e^{H({\lambda }_1^0)+H({\lambda }_2^0)\cdots +H({\lambda }_{n-1})}\rho ({\lambda }_2^0,{\lambda }_{n-1})+ \\ +\cdots +e^{H({\lambda }_1^0)+\cdots +H({\lambda }_n^0)+H({\lambda }_1)}\rho ({\lambda }_n^0,{\lambda }_1)+{\Delta }^{n-1}\omega ({\lambda }_n^0,\dots ,{\lambda }_1^0) \\ \le K({\lambda }_1^0,\dots ,{\lambda }_n^0)e^{H({\lambda }_1)+\cdots +H({\lambda }_n)}, \end{gathered}$$

$$\frac{1-\rho (z,w)}{1+\rho (z,w)}\le \frac{H(z)}{H(w)}\le \frac{1+\rho (z,w)}{1-\rho (z,w)}.$$

$$|{\Delta }^{j-1}f(z_1,\dots ,z_j)|\le e^{H(z_1)+\cdots +H(z_j)}\text{for all (z1,…,zj)∈Dj}.$$

$${\Delta }^{j}f(z_1,\dots ,z_{j+1})=\frac{{\Delta }^{j-1}f(z_2,\dots ,z_{j+1})-{\Delta }^{j-1}f(z_1,\dots ,z_j)}{b_{z_1}(z_{j+1})}.$$

$$|\Delta^{j}f(z_{1},\ldots,z_{j+1})|\leq\frac{1}{\rho(z_{1},z_{j+1})}\bigl{(}e^{H(z_{2})+\cdots+H(z_{j+1})}+e^{H(z_{1})+\cdots+H(z_{j})}\bigr{)}.$$

$$|{\Delta }^{j}f(z_1,\dots ,z_{j+1})|\le 4e^{H(z_1)+\cdots +H(z_{j+1})},$$

$$|{\Delta }^{j}f(z_1,\dots ,z_{j+1})|$$

$$\le \underset{\xi :\rho (\xi ,z_{j+1})=1/2}{\sup }4e^{H(z_1)+\cdots +H(z_j)+H(\xi )}$$

$$\le 4e^{2[H(z_1)+\cdots +H(z_j)+H(z_{j+1})]}.$$

$$\left|{\Delta }^n\omega ({\lambda }_1,\dots ,{\lambda }_{n+1})\right|$$

$$\le e^{H_0({\lambda }_1)}\left(e^{H({\lambda }_2)+\cdots +H({\lambda }_{n+1})}+e^{H({\lambda }_1)+\cdots +H({\lambda }_n)}\right)$$

$$\underset{\lambda \in \Lambda }{\sup }\#[\Lambda \cap D(\lambda ,e^{-H(\lambda )})]\le n.$$

$${\Lambda }_a$$

$${\Lambda }_b$$

$$\begin{array}{rl}|{\Delta }^n\omega ({\lambda }_1,\dots ,{\lambda }_{n+1})|& \le \frac{|{\Delta }^{n-1}\omega ({\lambda }_2,\dots ,{\lambda }_{n+1})|+|{\Delta }^{n-1}\omega ({\lambda }_1,\dots ,{\lambda }_n)|}{\rho ({\lambda }_1,{\lambda }_{n+1})}\\ & \le e^{H({\lambda }_1)}\left(e^{H_0({\lambda }_2)+\cdots +H_0({\lambda }_{n+1})}+e^{H_0({\lambda }_1)+\cdots +H_0({\lambda }_n)}\right)\\ & \le 2e^{H({\lambda }_1)}{e}^{H_0({\lambda }_1)+\cdots +H_0({\lambda }_{n+1})}.\end{array}$$

$$\#[\Lambda \cap D(\lambda ,e^{-H(\lambda )})]>n.$$

$$Q_{k,j}=\bigl{\{}z=re^{i\theta}\in\mathbb{D}:1-2^{-k}\leq r<1-2^{-k-1}\ ,\ j\frac{2\pi}{k}\leq\theta<(j+1)\frac{2\pi}{k}\bigr{\}},$$

$$r_{k,j}=\inf \{r>0:\exists \lambda \in {\Lambda }_{k,j}:\#(\Lambda \cap \overline{D(\lambda ,r)})\ge n+1\}.$$

$$\underset{k,j}{\inf }\frac{r_{k,j}}{e^{-H({\alpha }_{k,j})}}=0.$$

$$\frac{r_{k,j}}{e^{-H({\alpha }_{k,j})}}\ge \eta .$$

Full text

Second and third authors supported by the Generalitat de Catalunya (grants 2014 SGR 289 and 2014SGR 75) and the Spanish Ministerio de Ciencia e Innovación (projects MTM2014-51834-P and MTM2014-51824-P)

1. Definitions and statement

This note deals with some properties of the classical Nevanlinna class consisting of the holomorphic functions in the unit disk $\mathbb{D}$ for which $\log_{+}|f|$ has a positive harmonic majorant. We denote by $\operatorname{Har}_{+}(\mathbb{D})$ the set of non-negative harmonic functions in $\mathbb{D}$. Equivalently,

[TABLE]

Definition**.**

A discrete sequence of points $\Lambda$ in $\mathbb{D}$ is called interpolating for $N$ (denoted $\Lambda\in\operatorname{Int}\mathcal{N}$) if the trace space $N|\Lambda$ is ideal, or equivalently, if for every $v\in\ell^{\infty}$ there exists $f\in\mathcal{N}$ such that

[TABLE]

Interpolating sequences for the Nevanlinna class were first investigated by Naftalevitch [6]. A rather complete study was carried out much later in [4]. Let $B$ denote the Blaschke product associated to a Blaschke sequence $\Lambda$. Let

[TABLE]

Let’s also consider the pseudohyperbolic distance in $\mathbb{D}$, defined as

[TABLE]

and the corresponding pseudohyperbolic disks $D(z,r)=\{w\in\mathbb{D}:\rho(z,w)<r\}$.

According to [4, Theorem 1.2] $\Lambda\in\operatorname{Int}\mathcal{N}$ if and only if there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

Moreover in such case the trace space is

[TABLE]

Other properties and characterizations of Nevanlinna interpolating sequences have been given recently in [3]. In these terms $\Lambda\in\operatorname{Int}\mathcal{N}$ when for every sequence $\omega(\Lambda)\in\mathcal{N}(\Lambda)$ there exists $f\in\mathcal{N}$ such that $f(\lambda)=\omega(\lambda)$, $\lambda\in\Lambda$. In terms of the restriction operator

[TABLE]

$\Lambda$ is interpolating when $\mathcal{R}_{\Lambda}(\mathcal{N})=\mathcal{N}(\Lambda)$.

Definition 1.1**.**

Let $\Lambda$ be a discrete sequence in $\mathbb{D}$ and $\omega$ a function given on $\Lambda$. The pseudohyperbolic divided differences of $\omega$ are defined by induction as follows

[TABLE]

For any $n\in\mathbb{N}$, denote

[TABLE]

and consider the set $X^{n-1}(\Lambda)$ consisting of the functions defined in $\Lambda$ with divided differences of order $n-1$ uniformly controlled by a positive harmonic function $H$ i.e., such that for some $H\in\operatorname{Har}_{+}(\mathbb{D})$,

[TABLE]

Lemma 1.2**.**

Let $n\in\mathbb{N}$. For any sequence $\Lambda\subset\mathbb{D}$, we have $X^{n}(\Lambda)\subset X^{n-1}(\Lambda)\subset\cdots\subset X^{0}(\Lambda)=\mathcal{N}(\Lambda)$.

Proof.

Assume that $\omega(\Lambda)\in X^{n}(\Lambda)$, that is,

[TABLE]

Then, given $(\lambda_{1},\dots,\lambda_{n})\in\Lambda^{n}$ and taking $\lambda_{1}^{0},\dots,\lambda_{n}^{0}$ from a finite set (for instance the $n$ first $\lambda^{0}_{j}\in\Lambda$ different of all $\lambda_{j}$) we have

[TABLE]

Since $\omega\in X^{n-1}(\Lambda)$ there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ and a constant $K(\lambda^{0}_{1},\dots,\lambda^{0}_{n})$ such that

[TABLE]

and the statement follows. ∎

The main result of this note is modelled after Vasyunin’s description of the sequences $\Lambda$ in $\mathbb{D}$ such that the trace of the algebra of bounded holomorphic functions $H^{\infty}$ on $\Lambda$ equals the space of pseudohyperbolic divided differences of order $n$ (see [7], [8]). Similar results hold also for Hardy spaces (see [1] and [2]) and the Hörmander algebras, both in $\mathbb{C}$ and in $\mathbb{D}$ [5]. The analogue in our context is the following.

Main Theorem**.**

The identity $\mathcal{N}|\Lambda=X^{n-1}(\Lambda)$ holds if and only if $\Lambda$ is the union of $n$ interpolating sequences for $\mathcal{N}$.

2. General properties

Throughout the proofs we will use repeatedly the well-known Harnack inequalities: for $H\in\operatorname{Har}_{+}(\mathbb{D})$ and $z,w\in\mathbb{\mathbb{D}}$,

[TABLE]

We shall always assume, without loss of generality, that $H\in\operatorname{Har}_{+}(\mathbb{D})$ is big enough so that for $z\in D(\lambda,e^{-H(\lambda)})$ the inequalities $1/2\leq H(z)/H(\lambda)\leq 2$ hold. Actually it is sufficient to assume $\inf\{H(z):z\in\mathbb{D}\}\geq\log 3$.

We begin by showing that one of the inclusions of the Main Theorem is inmediate.

Proposition 2.1**.**

For all $n\in\mathbb{N}$, the inclusion $\mathcal{N}|\Lambda\subset X^{n-1}(\Lambda)$ holds.

Proof.

Let $f\in\mathcal{N}$. Let us show by induction on $j\geq 1$ that there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

As $f\in\mathcal{N}$, there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that $|\Delta^{0}f(z_{1})|=|f(z_{1})|\leq e^{H(z_{1})}$, $z_{1}\in\mathbb{D}$.

Assume that the property is true for $j$ and let $(z_{1},\ldots,z_{j+1})\in\mathbb{D}^{j+1}$. Fix $z_{1},\ldots,z_{j}$ and consider $z_{j+1}$ as the variable in the function

[TABLE]

By the induction hypothesis, there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

If $\rho(z_{1},z_{j+1})\geq 1/2$ we get directly

[TABLE]

and choosing for instance $\tilde{H}=H+\log 4$ we get the desired estimate.

If $\rho(z_{1},z_{j+1})\leq 1/2$ we apply the maximum principle and Harnack’s inequalities

[TABLE]

Choosing here $\tilde{H}=2H+\log 4$ we get the desired estimate. ∎

Definition 2.2**.**

A sequence $\Lambda$ is weakly separated if there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that the disks $D(\lambda,e^{-H(\lambda)})$, $\lambda\in\Lambda$, are pairwise disjoint.

Remark 2.3**.**

If $\Lambda$ is weakly separated then $X^{0}(\Lambda)=X^{n}(\Lambda)$, for all $n\in\mathbb{N}$.

By Lemma 1.2, to see this it is enough to prove (by induction) that $X^{0}(\Lambda)\subset X^{n}(\Lambda)$ for all $n\in\mathbb{N}$.

For $n=0$ this is trivial.

Assume now that $X^{0}(\Lambda)\subset X^{n-1}(\Lambda)$ and take $\omega(\Lambda)\in X^{0}(\Lambda)$. Since $\rho(\lambda_{1},\lambda_{n+1})\geq e^{-H_{0}(\lambda_{1})}$ for some $H_{0}\in\operatorname{Har}_{+}(\mathbb{D})$ we have

[TABLE]

for some $H\in\operatorname{Har}_{+}(\mathbb{D})$. Taking $\tilde{H}=H+H_{0}$ we are done.

Lemma 2.4**.**

Let $n\geq 1$. The following assertions are equivalent:

• (a)

$\Lambda$* is the union of $n$ weakly separated sequences,*

• (b)

There exist $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

• (c)

$X^{n-1}(\Lambda)=X^{n}(\Lambda)$.

Proof.

(a) $\Rightarrow$(b). This is clear, by the weak separation.

(b) $\Rightarrow$(a). We proceed by induction on $j=1,\ldots,n$. For $j=1$, it is again clear by the definition of weak separation. Assume the property true for $j-1$. Let $H\in\operatorname{Har}_{+}(\mathbb{D})$ , $\inf\{H(z):z\in\mathbb{D}\}\geq\log 3$, be such that $\sup_{\lambda\in\Lambda}\#[\Lambda\cap D(\lambda,e^{-H(\lambda)})]\leq j$. We split the sequence $\Lambda=\Lambda_{a}\cup\Lambda_{b}$ where

[TABLE]

Now, for every $\lambda\in\Lambda_{b}$, we have $\#(\Lambda\cap D(\lambda,e^{-10H(\lambda)}))\leq j-1$, and by the induction hypothesis, $\Lambda_{b}$ splits into $j-1$ separated sequences $\Lambda_{1},\ldots,\Lambda_{j-1}$.

In the case $\lambda\in\Lambda_{a}$, there is obviously no point in the annulus $D(\lambda,e^{-H(\lambda)})\setminus D(\lambda,e^{-10H(\lambda)})$ which means that the $j$ points in $D(\lambda,e^{-10H(\lambda)}))$ are far from the other points of $\Lambda$. So we can add each one of these $j$ points in a weakly separated way to one of the sequences $\Lambda_{1},\ldots,\Lambda_{j-1}$, and the $j$-th point in a new sequence $\Lambda_{j}$ (which is of course weakly separated since the groups $\Lambda\cap D(\lambda,e^{-10H(\lambda)})$ appearing in $\Lambda_{a}$ are weakly separated).

(b)$\Rightarrow$(c). It remains to see that $X^{n-1}(\Lambda)\subset X^{n}(\Lambda)$. Given $\omega(\Lambda)\in X^{n-1}(\Lambda)$ and points $(\lambda_{1},\ldots,\lambda_{n+1})\in\Lambda^{n+1}$, we have to estimate $\Delta^{n}\omega(\lambda_{1},\ldots,\lambda_{n+1})$. Under the assumption (b), at least one of these $n+1$ points is not in the disk $D(\lambda_{1},e^{-H(\lambda_{1})})$. Note that $\Lambda^{n}$ is invariant by permutation of the $n+1$ points, thus we may assume that $\rho(\lambda_{1},\lambda_{n+1})\geq e^{-H(\lambda_{1})}$. Using the fact that $\omega(\Lambda)\in X^{n-1}(\Lambda)$, there exists $H_{0}\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

Taking $\tilde{H}=H_{0}+H+\log 2$ we get the desired estimate.

(c)$\Rightarrow$(b). We prove this by contraposition. Assume that for all $H\in\operatorname{Har}_{+}(\mathbb{D})$ there exists $\lambda\in\Lambda$ such that

[TABLE]

Consider the partition of $\mathbb{D}$ into the dyadic squares

[TABLE]

where $k\geq 0$ and $j=0,\dots 2^{k}-1$.

Let $\Lambda_{k,j}=\Lambda\cap Q_{k,j}$ and

[TABLE]

Take $\alpha_{k,j}\in\Lambda_{k,j}$ such that $\#(\Lambda\cap\overline{D(\alpha_{k,j},r_{k,j})})\geq n+1$.

Claim: For all $H\in\operatorname{Har}_{+}(\mathbb{D})$,

[TABLE]

To see this assume otherwise that there exist $H\in\operatorname{Har}_{+}(\mathbb{D})$ and $\eta>0$ with

[TABLE]

In particular, by Harnack’s inequalities,

[TABLE]

Let $\tilde{H}:=\log(2/\eta)+4H\in\operatorname{Har}_{+}(\mathbb{D})$. By the hypothesis (2) there exist $k_{0}\geq 0$, $j_{0}\in\{0,\dots,2^{k_{0}}-1\}$, $\lambda_{k_{0},j_{0}}\in\Lambda_{k_{0},j_{0}}$ such that

[TABLE]

In particular, by definition of $r_{k,j}$, we have $r_{k_{0},j_{0}}\leq e^{-\tilde{H}(\lambda_{k_{0},j_{0}})}$, that is

[TABLE]

which contradicts (3).

Now take a separated sequence $\mathcal{L}\subset\{\alpha_{k,j}\}_{k,j}$ for which the disks $D(\alpha,r_{\alpha})$, $\alpha\in\mathcal{L}$, are disjoint, where for $\alpha=\alpha_{k,j}\in\mathcal{L}$ we denote $r_{\alpha}=r_{k,j}$. Given $\alpha\in\mathcal{L}$, let $\lambda_{1}^{\alpha},\dots,\lambda_{n}^{\alpha}$ be its $n$ nearest (not necessarily unique) points, arranged by increasing distance. Notice that $\rho(\alpha,\lambda_{n}^{\alpha})=r_{\alpha}$.

In order to construct a sequence $\omega(\Lambda)\in X^{n-1}(\Lambda)\setminus X^{n}(\Lambda)$, put

[TABLE]

To see that $\omega(\Lambda)\in X^{n-1}(\Lambda)$ let us estimate $\Delta^{n-1}\omega(\lambda_{1},\ldots,\lambda_{n})$ for any given $(\lambda_{1},\ldots,\lambda_{n})\in\Lambda^{n}$. By the separation conditions on $\mathcal{L}$, we know that none of the $\lambda_{j}^{\alpha}$ is in $\mathcal{L}$. Hence, we may assume that at most one of the points is in $\mathcal{L}$. On the other hand, it is clear that $\Delta^{n-1}\omega(\lambda_{1},\ldots,\lambda_{n})=0$ if all the points are in $\Lambda\setminus\mathcal{L}$. Thus, taking into account that $\Delta^{n-1}$ is invariant by permutations, we will only consider the case where $\lambda_{n}$ is some $\alpha\in\mathcal{L}$ and $\lambda_{1},\ldots,\lambda_{n-1}$ are in $\Lambda\setminus\mathcal{L}$. In that case,

[TABLE]

as desired.

On the other hand, a similar computation yields

[TABLE]

The Claim above prevents the existence of $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

since otherwise, again by Harnack’s inequalities, we would have

[TABLE]

∎

It is clear from the characterization (1) of interpolating sequences for $\mathcal{N}$ that such sequences must be weakly separated. The previous result gives another way of showing it.

Corollary 2.5**.**

If $\Lambda$ is an interpolating sequence, then it is weakly separated.

Proof.

If $\Lambda$ is an interpolating sequence, then $\mathcal{N}|\Lambda=X^{0}(\Lambda)$. On the other hand, by Proposition 2.1, $\mathcal{N}|\Lambda\subset X^{1}(\Lambda)$. Thus $X^{0}(\Lambda)=X^{1}(\Lambda)$. We conclude by the preceding lemma applied to the particular case $n=1$. ∎

The covering provided by the following result will be useful.

Lemma 2.6**.**

Let $\Lambda_{1},\ldots,\Lambda_{n}$ be weakly separated sequences. There exist $H\in\operatorname{Har}_{+}(\mathbb{D})$, positive constants $\alpha,\beta$, a subsequence $\mathcal{L}\subset\Lambda_{1}\cup\cdots\cup\Lambda_{n}$ and disks $D_{\lambda}=D(\lambda,r_{\lambda})$, $\lambda\in\mathcal{L}$, such that

• (i)

$\Lambda_{1}\cup\cdots\cup\Lambda_{n}\subset\cup_{\lambda\in\mathcal{L}}D_{\lambda}$,

• (ii)

$e^{-\beta H(\lambda)}\leq r_{\lambda}\leq e^{-\alpha H(\lambda)}$* for all $\lambda\in\mathcal{L}$,*

• (iii)

$\rho(D_{\lambda},D_{\lambda^{\prime}})\geq e^{-\beta H(\lambda)}$* for all $\lambda,\lambda^{\prime}\in\mathcal{L}$, $\lambda\neq\lambda^{\prime}$.*

• (iv)

$\#(\Lambda_{j}\cap D_{\lambda})\leq 1$* for all $j=1,\dots,n$ and $\lambda\in\mathcal{L}$.*

Proof.

Let $H\in\operatorname{Har}_{+}(\mathbb{D})$ be such that

[TABLE]

We will proceed by induction on $k=1,\ldots,n$ to show the existence of a subsequence $\mathcal{L}_{k}\subset\Lambda_{1}\cup\cdots\cup\Lambda_{k}$ such that:

[TABLE]

Then it suffices to chose $\mathcal{L}=\mathcal{L}_{n}$, $\alpha=\alpha_{n}$, $\beta=\beta_{n}$, $r_{\lambda}=R_{\lambda}^{n}$. The weak separation and the fact that $r_{\lambda}<e^{-H(\lambda)}/3$ implies that $\#\Lambda_{j}\cap D(\lambda,r_{\lambda})\leq 1$, $j=1,\dots,k$, hence the lemma follows.

For $k=1$, the property is clearly verified with $\mathcal{L}_{1}=\Lambda_{1}$ and $R_{\lambda}^{1}=e^{-CH(\lambda)}$, with $C$ big enough so that $(iii)_{1}$ holds ($C=3$, for instance). Properties $(i)_{1}$, $(ii)_{1}$ follow immediately.

Assume the property true for $k$ and split $\mathcal{L}_{k}=\mathcal{M}_{1}\cup\mathcal{M}_{2}$ and $\Lambda_{k+1}=\mathcal{N}_{1}\cup\mathcal{N}_{2}$, where

[TABLE]

Now, we put $\mathcal{L}_{k+1}=\mathcal{L}_{k}\cup\mathcal{N}_{2}$ and define the radii $R_{\lambda}^{k+1}$ as follows:

[TABLE]

It is clear that $(i)_{k+1}$ holds:

[TABLE]

Also, by the induction hypothesis,

[TABLE]

Thus, to see $(ii)_{k+1}$ there is enough to choose $\alpha_{k+1},\beta_{k+1}$ such that

[TABLE]

for instance $\alpha_{k+1}=\alpha_{k}-1$, and

[TABLE]

that is $\beta_{k+1}H(\lambda)\geq\beta_{k}H(\lambda)+\log 8$. Assuming without loss of generality that $H(\lambda)\geq\log 8$, there is enough choosing $\beta_{k+1}\geq\beta_{k}+1$.

In order to prove $(iii)_{k}$ take now $\lambda,\lambda^{\prime}\in\mathcal{L}_{k+1}$, $\lambda\neq\lambda^{\prime}$. Notice that

[TABLE]

Split into four different cases:

1. $\lambda,\lambda^{\prime}\in\mathcal{L}_{k}$. Assume without loss of generality that $H(\lambda)\leq H(\lambda^{\prime})$. Then, by the definition of $R_{\lambda}^{k+1}$, we see that

[TABLE]

By inductive hypothesis

[TABLE]

Thus, by (5),

[TABLE]

2. $\lambda,\lambda^{\prime}\in\mathcal{N}_{2}$. Assume also $H(\lambda)\leq H(\lambda^{\prime})$. Condition (4) implies $\rho(\lambda,\lambda^{\prime})\geq e^{-H(\lambda)}$, hence

[TABLE]

If $\beta_{k}\geq 2$, by (5) we have

[TABLE]

3. $\lambda\in\mathcal{M}_{1},{\lambda^{\prime}}\in\mathcal{N}_{2}$ By definition of $\mathcal{M}_{1}$ there exists $\beta\in\mathcal{N}_{1}$ such that

[TABLE]

Then, using (4) on $\beta,\lambda^{\prime}\in\Lambda_{k+1}$, we have, by Harnack’s inequalities (if $\beta_{k}\geq 4$),

[TABLE]

4. $\lambda\in\mathcal{M}_{2},{\lambda^{\prime}}\in\mathcal{N}_{2}$. Taking into account the definition of $R_{\lambda}^{k+1},R_{\lambda^{\prime}}^{k+1}$ we have

[TABLE]

Since

[TABLE]

by inductive hypothesis and by (5)

[TABLE]

All together, it is enough to start with $C>n$, define $\alpha_{1}=\beta_{1}=C$, and then define $\alpha_{k}$, $\beta_{k}$ inductively by

[TABLE]

∎

3. Proof of Main Theorem. Necessity

Assume $\mathcal{N}|\Lambda=X^{n-1}(\Lambda)$, $n\geq 2$. Using Proposition 2.1, we have $X^{n-1}(\Lambda)=X^{n}(\Lambda)$, and by Lemma 2.4 we deduce that $\Lambda=\Lambda_{1}\cup\cdots\cup\Lambda_{n}$, where $\Lambda_{1},\ldots,\Lambda_{n}$ are weakly separated sequences. We want to show that each $\Lambda_{j}$ is an interpolating sequence.

Let $\omega(\Lambda_{j})\in\mathcal{N}(\Lambda_{j})=X^{0}(\Lambda_{j})$. Let $\cup_{\lambda\in\mathcal{L}}D_{\lambda}$ be the covering of $\Lambda$ given by Lemma 2.6. We extend $\omega(\Lambda_{j})$ to a sequence $\omega(\Lambda)$ which is constant on each $D_{\lambda}\cap\Lambda_{j}$ in the following way:

[TABLE]

We verify by induction that the extended sequence is in $X^{k-1}(\Lambda)$ for all $k\leq n$. It is clear that it belongs to $X^{0}(\Lambda)$.

Assume that $\omega\in X^{k-2}(\Lambda)$, $k\geq 2$, and consider $(\alpha_{1},\ldots,\alpha_{k})\in\Lambda^{k}$. If all the points are in the same $D_{\lambda}$ then $\Delta^{k-1}\omega(\alpha_{1},\ldots,\alpha_{k})=0$, so we may assume that $\alpha_{1}\in D_{\lambda}$ and $\alpha_{k}\in D_{\lambda^{\prime}}$ with $\lambda\not=\lambda^{\prime}$. Then we have, for some $H_{0}\in\operatorname{Har}_{+}(\mathbb{D})$,

[TABLE]

With this and the induction hypothesis it is clear that for some $H\in\operatorname{Har}_{+}(\mathbb{D})$,

[TABLE]

Taking for instance $\tilde{H}=H+\beta H_{0}+\log 2$ we get

[TABLE]

thus $\omega(\Lambda)\in X^{k-1}(\Lambda)$. By assumption there exist $f\in\mathcal{N}$ interpolating the values $\omega(\Lambda)$. In particular $f$ interpolates $\omega(\Lambda_{j})$.

4. Proof of the Main Theorem. Sufficiency

Assume $\Lambda=\Lambda_{1}\cup\dots\cup\Lambda_{n}$, where $\Lambda_{j}\in\operatorname{Int}\mathcal{N}$, $j=1,\dots,n$, and denote $\Lambda_{j}=\{\lambda_{k}^{(j)}\}_{k\in\mathbb{N}}$. Denote also by $B_{j}$ the Blaschke product with zeros on $\Lambda_{j}$. We will use the following property of the Nevanlinna interpolating sequences (see Theorem 1.2 in [3]).

Lemma 4.1**.**

Let $\Lambda\in\operatorname{Int}N$ and let $B$ the Blaschke product associated to $\Lambda$. There exists $H_{1}\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

According to Proposition 2.1 we only need to see that $X^{n-1}(\Lambda)\subset\mathcal{N}|\Lambda$. Let then $\omega(\Lambda)\in X^{n-1}(\Lambda)$ and split it

[TABLE]

where $\omega_{k}^{(j)}=\omega(\lambda_{k}^{(j)})$, $j=1,\dots,n$, $k\in\mathbb{N}$. By Lemma 1.2 and the hypothesis $\{\omega_{k}^{(1)}\}_{k\in\mathbb{N}}\in X^{0}(\Lambda_{1})$, hence there exists $f_{1}\in\mathcal{N}$ such that

[TABLE]

In order to interpolate also the values $\{\omega_{k}^{(2)}\}_{k}$ consider functions of the form

[TABLE]

Immediately $f_{2}(\lambda_{k}^{(1)})=f_{1}(\lambda_{k}^{(1)})=\omega_{k}^{(1)}$, $k\in\mathbb{N}$, and we will have $f_{2}(\lambda_{k}^{(2)})=\omega_{k}^{(2)}$ as soon as we find $g_{2}\in\mathcal{N}$ such that

[TABLE]

Since $\Lambda_{2}\in\operatorname{Int}\mathcal{N}$ such $g_{2}$ will exist as soon as the sequence in the right hand side is majorized by a sequence of the form $\{e^{H(\lambda_{k}^{(2)})}\}_{k}$.

Given $\lambda_{k}^{(2)}\in\Lambda_{2}$ pick $\lambda_{k}^{(1)}$ such that $\rho(\lambda_{k}^{(2)},\Lambda_{1})=\rho(\lambda_{k}^{(2)},\lambda_{k}^{(1)})$. There is no restriction in assuming that $\rho(\lambda_{k}^{(2)},\lambda_{k}^{(1)})\leq 1/2$. Then, by Lemma 4.1 there exists $H_{1}\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

Now, since $f_{1}(\lambda_{k}^{(1)})=\omega_{k}^{(1)}$ we have

[TABLE]

By hypothesis, and since $f_{1}\in\mathcal{N}$, there exists $H_{2}\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

and therefore, by Harnack’s inequalities,

[TABLE]

In general, assume that we have $f_{n-1}\in\mathcal{N}$ such that

[TABLE]

We look for a function $f_{n}\in\mathcal{N}$ interpolating the whole $\Lambda$ of the form

[TABLE]

We need then $g_{n}\in\mathcal{N}$ with

[TABLE]

Let us see that the sequence of values in the right hand side of this identity have a majorant of the form $\{e^{H(\lambda_{k}^{(n)})}\}_{k}$.

Pick $\lambda_{k}^{(j)}\in\Lambda_{j}$, $j=1,\dots,n-1$ such that $\rho(\lambda_{k}^{(n)},\Lambda_{j})=\rho(\lambda_{k}^{(n)},\lambda_{k}^{(j)})$. There is no restriction in assuming that $\rho(\lambda_{k}^{(n)},\lambda_{k}^{(j)})\leq 1/2$. Since $f_{n-1}(\lambda_{k}^{(j)})=\omega_{k}^{(j)}$, $j=1,\dots,n-1$, an immediate computation shows that

[TABLE]

Again by Lemma 4.1, there exists $H_{1}\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

Hence, by hypothesis and the fact that $f_{n-1}\in\mathcal{N}$ there exists $H\in\operatorname{Har}_{+}(\mathbb{D})$ such that

[TABLE]

Finally, by Harnack’s inequalities, this is bounded by $e^{2n(H(\lambda_{k}^{(n)})+H_{1}(\lambda_{k}^{(n)}))}$.