See pages 1-last of main.pdf


# Generalized Hilbert operators acting on weighted spaces of holomorphic
functions with sup-norms

María J. Beltrán-Meneu111Departament de Matemàtiques, Universitat Jaume I,
Av.Vicent Sos Baynat, s/n, Castelló de la Plana, E-12071 (Spain),
<EMAIL_ADDRESS>https://orcid.org/0000-0001-9645-096X., José Bonet222Instituto
Universitario de Matemática Pura y Aplicada IUMPA, Universitat Politècnica de
València, Camino de Vera, s/n, Valencia, E-46022 (Spain<EMAIL_ADDRESS>https://orcid.org/0000-0002-9096-6380. and Enrique Jordá333Instituto
Universitario de Matemática Pura y Aplicada IUMPA, Universitat Politècnica de
València, Camino de Vera, s/n, Valencia, E-46022 (Spain<EMAIL_ADDRESS>https://orcid.org/0000-0003-2980-1699.

###### Abstract

The behaviour of the generalized Hilbert operator associated with a positive
finite Borel measure $\mu$ on $[0,1)$ is investigated when it acts on weighted
Banach spaces of holomorphic functions on the unit disc defined by sup-norms
and on Korenblum type growth Banach spaces. It is studied when the operator is
well defined, bounded and compact. To this aim, we study when it can be
represented as an integral operator. We observe important differences with the
behaviour of the Cesàro-type operator acting on these spaces, getting that
boundedness and compactness are equivalent concepts for some standard weights.
For the space of bounded holomorphic functions on the disc and for the Wiener
algebra, we get also this equivalence, which is characterized in turn by the
summability of the moments of the measure $\mu.$ In the latter case, it is
also equivalent to nuclearity. Nuclearity of the generalized Hilbert operator
acting on related spaces, such as the classical Hardy space, is also analyzed.

Keywords: Generalized Hilbert operator, Weighted Banach spaces of analytic
functions, Boundness, Compactness, Nuclearity.

2020 Mathematics Subject Classification: Primary 47B38; Secondary 46E10,
46E15, 47A10, 47A16, 47A35, 47B35.

##  1 Introduction

Let $H(\mathbb{D})$ be the space of holomorphic functions on the unit disc
$\mathbb{D}$ of the complex plane $\mathbb{C}$ endowed with the topology
$\tau_{0}$ of uniform convergence on the compact subsets of $\mathbb{D}$. If
$\mu$ is a positive finite Borel measure on $[0,1)$, we denote by
$\mu_{n}:=\int_{0}^{1}t^{n}d\mu(t)$ the $n$-th moment of $\mu$,
$n\in\mathbb{N}_{0}$. Here
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}=\\{0,1,2,...\\}$. For $f\in
H(\mathbb{D}),$ $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n},$ $z\in\mathbb{D},$ we
consider its image by the generalized Hilbert operator

$H_{\mu}(f)(z):=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\mu_{n+k}a_{k}\right)z^{n},\
z\in\mathbb{D},$

in case it defines an analytic function in $\mathbb{D}$. The operator
$H_{\mu}$ is a generalization of the classical Hilbert operator $H$. Indeed,
if $\mu$ is the Lebesgue measure, then $H_{\mu}=H$ and
$\mu_{n}=\frac{1}{n+1},n\in\mathbb{N}_{0}.$

The following related operator will also be considered:

$I_{\mu}(f)(z):=\int_{0}^{1}\frac{f(t)}{1-tz}d\mu(t),$

whenever it defines an analytic function in $\mathbb{D}$ for $f\in
H(\mathbb{D})$. It is well known that, if
$\int_{0}^{1}\frac{d\mu(t)}{1-t}<\infty,$ i.e.,
$\sum_{n=0}^{\infty}\mu_{n}<\infty,$ then $I_{\mu}$ is well defined and
continuous on $H(\mathbb{D}).$

These two operators coincide on the polynomials. Therefore, if they are
continuous on a space containing the polynomials as a dense subset, then they
must coincide.

The Hilbert operator and the generalized Hilbert operators have received much
attention in the last decades and their behaviour have been investigated when
acting on different spaces of holomorphic functions on the disc. We refer the
reader to [2, 12, 13, 25] and the references therein for the Hilbert matrix
and to [9, 15, 16, 17, 18, 22, 31, 33] for the generalized Hilbert and other
Hankel operators.

The purpose of our article is to investigate the boundedness and compactness
of the generalized Hilbert operator $H_{\mu}$ and its companion operator
$I_{\mu}$ when they act on weighted spaces of holomorphic functions on
$\mathbb{D}$ defined by sup-norms, and in particular on the Korenblum type
growth Banach spaces $H_{v_{\gamma}}^{\infty}$ and $H_{v_{\gamma}}^{0}$,
$\gamma>0$. When $p<1$, the results that we present for the continuity and
compactness of $I_{\mu}:H_{v_{\gamma}}^{\infty}\to H_{v_{p}}^{\infty}$ are
analogous to those obtained for the Cesàro-type operator in [3]. When $p\geq
1$ the situation is completely different. In particular, we show that the
continuity of $I_{\mu}:H_{v_{1}}^{\infty}\to H_{v_{1}}^{\infty}$ is equivalent
to the compactness of the operator. Both conditions are equivalent to the
above mentioned summability of the sequence of moments $(\mu_{n})_{n}$
associated to $\mu$. In the last section we show that the summability
condition is in fact equivalent to the continuity and also to the compactness
of $I_{\mu}$ when it acts on the space of bounded holomorphic functions on the
disc $H^{\infty}$ and in the Wiener algebra $A(\mathbb{T})$. We get
$I_{\mu}=H_{\mu}$ in these cases. Moreover, we show that the continuity of
$I_{\mu}:A(\mathbb{T})\rightarrow A(\mathbb{T})$ is equivalent not only to the
compactness but also to nuclearity. The equivalence of the summability of
$(\mu_{n})_{n}$ and the continuity of
$I_{\mu}=H_{\mu}\in\mathcal{L}(H^{\infty})$ was obtained recently by Girela
and Merchán in [17].

We present some necessary definitions. A weight $v$ on $\mathbb{D}$ is a
radial ($v(z)=v(|z|),\ z\in\mathbb{D}$), continuous and non-increasing
function with respect to $|z|$ satisfying $\lim_{r\rightarrow 1^{-}}v(r)=0$.
Given a weight $v$, we define the weighted Banach spaces of holomorphic
funcions

$H_{v}^{\infty}:=\\{f\in H(\mathbb{D}):\
\sup_{z\in\mathbb{D}}v(z)|f(z)|<\infty\\}$

and

$H_{v}^{0}:=\\{f\in H_{v}^{\infty}:\ \lim_{|z|\rightarrow
1^{-}}v(z)|f(z)|=0\\}.$

Both are Banach spaces under the norm
$\|f\|_{v}:=\sup_{z\in\mathbb{D}}v(z)|f(z)|,\ f\in H_{v}^{\infty}.$ They
appear in the study of growth conditions of analytic functions and have been
investigated by several authors since the work of Shields and Williams (see
eg. [4], [5], [26], [32] and the references therein). The space $H_{v}^{0}$
coincides with the closure of the polynomials in $H_{v}^{\infty}$ and
$H_{v}^{\infty}$ is canonically isomorphic to the bidual $(H_{v}^{0})^{**}$
[6].

The constant weight $v\equiv 1$ does not satisfy our assumptions. In this
case, $H_{v}^{\infty}$ is the classical Banach space $H^{\infty}$ of bounded
holomorphic functions on $\mathbb{D}$, and $H_{v}^{0}$ reduces to $\\{0\\}$.
We will state and prove some results for $H^{\infty}$, but when we refer to a
weight in the statements, the constant weight is excluded, unless it is
explicitly mentioned. The so-called standard weights are defined by
$v_{\gamma}(r)=(1-r)^{\gamma},$ $r>0,$ $\gamma>0,$ the spaces
$H_{v_{\gamma}}^{\infty}$ and $H_{v_{\gamma}}^{0}$ are the classical Korenblum
type growth Banach spaces $A^{-\gamma}$ and $A^{-\gamma}_{0},$ respectively.
These spaces play an important role in interpolation and sampling of
holomorphic functions [19, Chapters 4 and 5]. Shields and Williams [32] proved
that $H_{v_{\gamma}}^{\infty}$ is isomorphic to $\ell^{\infty}$ and that
$H_{v_{\gamma}}^{0}$ is isomorphic to $c_{0}$; a result that was extended by
Lusky [26, Theorem 1.1]. A survey about weighted Banach spaces of analytic
functions of the type considered in this paper and operators between them can
be seen in [8].

A weight $v$ is called essential if there is a constant $c>0$ such that, for
each $z\in\mathbb{D}$,

$c/v(z)\leq\sup\\{|f(z)|\ ;\|f\|_{v}\leq 1\\}\leq 1/v(z).$

This terminology was introduced in [5], where this type of weights were
investigated. Every standard weight $v_{\gamma}$ is essential. The following
consequence of a deep result due to Abakumov and Doubtsov [1] (see also [8,
Theorem 2]) is useful later in the article.

###### Lemma 1

If the weight $v$ is essential, then there are $f_{v}\in H_{v}^{\infty}$ and
$D(v)>0$ such that $1/v(z)\leq D(v)f_{v}(|z|)$ for each $z\in\mathbb{D}$. In
particular $f_{v}(t)>0$ for each $t\in[0,1)$.

###### Proof.

By [1, Theorem 4] the weight $v$ is essential if and only if there is a
sequence $(a_{k})_{k=0}^{\infty}$ of non-negative numbers and there are
constants $c_{1},c_{2}>0$ such that, for each $0\leq r<1$,

$c_{1}\frac{1}{v(r)}\leq\sum_{k=0}^{\infty}a_{k}r^{k}\leq
c_{2}\frac{1}{v(r)}.$

Then the function $f_{v}(z):=\sum_{k=0}^{\infty}a_{k}z^{k},\ z\in\mathbb{D},$
belongs to $H(\mathbb{D})$, and, for each $z\in\mathbb{D}$,

$v(z)|f_{v}(z)|\leq v(z)\sum_{k=0}^{\infty}a_{k}|z|^{k}\leq c_{2}.$

This implies $f_{v}\in H^{\infty}_{v}$. Moreover,
$\frac{1}{v(r)}\leq(1/c_{1})f_{v}(r)$ for each $0\leq r<1$, and the conclusion
follows with $D(v)=1/c_{1}$. ∎

Given two functions $f$ and $g$ defined in an (possibly unbounded) interval
$I$ of real numbers, we write $f(x)\lesssim g(x)$ to denote the existence of
$C>0$ such that $f(x)\leq Cg(x)$ for each $x\in I$. In case $f(x)\lesssim
g(x)\lesssim f(x),$ we write $f(x)\cong g(x).$ If $a$ is an extreme point of
the interval $I$ (possible infinity), we shall use Landau’s notation
$f(x)=O(g(x))$ as $x\to a$ and $f(x)=o(g(x))$ as $x\to a$.

We recall that a positive finite Borel measure $\mu$ on $[0,1)$ is an
$s$-Carleson measure, $s>0,$ if

$\mu([t,1))=O((1-t)^{s}),\text{ as $t\rightarrow 1^{-}$},$

and a vanishing $s$-Carleson measure if

$\mu([t,1))=o((1-t)^{s})\text{ as $t\rightarrow 1^{-}$}.$

A $1$-Carleson measure and a vanishing $1$-Carleson measure will be simply
called a Carleson measure and a vanishing Carleson measure, respectively. By
[14, Lemma 2], $\mu$ is a Carleson measure if and only if
$\mu_{n}=O(\frac{1}{n})$ as $n\to\infty,$ and $\mu$ is a vanishing Carleson
measure if and only if $\mu_{n}=o(\frac{1}{n})$ as $n\to\infty.$

Our notation for concepts from functional analysis and operator theory is
standard. We refer the reader e.g. to [11] and [27]. Given a Banach space $X,$
we denote by ${X}^{*}$ its topological dual, that is, the space of all
continuous linear forms on $X,$ and given a continuous operator $T$ on $X,$ we
denote by ${T}^{*}$ its _transpose_.

##  2 $I_{\mu}$ as an operator with values in $H(\mathbb{D})$

In this section $(X,\|\cdot\|)$ is a general Banach space of analytic
functions, that is, a Banach space containing the polynomials such that the
inclusion $X\hookrightarrow(H(\mathbb{D}),\tau_{0})$ is continuous. We
analyze, for a general measure $\mu$, when the operator $I_{\mu}:X\to
H(\mathbb{D})$ is well defined and continuous. We also study the relation with
the operator $H_{\mu}:X\to H(\mathbb{D})$.

First, observe that, given $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in
H(\mathbb{D}),$ we get $I_{\mu}(f)(0)=\int_{0}^{1}f(t)d\mu(t)$. So, if
$I_{\mu}$ is well defined, then the restriction of $f$ to $[0,1)$ must belong
to $L([0,1),\mu).$ In the next proposition we prove that this condition is
also sufficient to get $I_{\mu}(f)\in H(\mathbb{D})$.

###### Proposition 2

Let $\mu$ be a positive finite Borel measure on $[0,1)$. If $f\in
H(\mathbb{D})$ satisfies that its restriction to $[0,1)$ belongs to
$L_{1}([0,1),\mu)$, then $I_{\mu}(f)\in H(\mathbb{D})$.

###### Proof.

Fix $f\in H(\mathbb{D})\cap L_{1}([0,1),\mu)$. For $0\leq r<1,$ put
$I^{r}_{\mu}(f)(z):=\int_{0}^{r}\frac{f(t)}{1-tz}d\mu(t)\in H(\mathbb{D}).$
Given $0<R<1,$ we get

$\sup_{z\in\mathbb{D},|z|\leq
R}|I_{\mu}(f)(z)-I_{\mu}^{r}(f)(z)|\leq\frac{1}{1-R}\int_{r}^{1}|f(t)|d\mu(t).$

Hence, as $\int_{0}^{1}|f(t)|d\mu(t)<\infty$, we get
$I_{\mu}^{r}(f)\rightarrow I_{\mu}(f)$ as $r\rightarrow 1$ in $\tau_{0}$.
Hence, $I_{\mu}(f)\in H(\mathbb{D})$. ∎

If we look now at the Hilbert-type operator, we see that
$H_{\mu}(f)(0)=\sum_{n=0}^{\infty}\mu_{n}a_{n}$. So, if $H_{\mu}$ is well
defined for $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H(\mathbb{D})$, then
$\sum_{n=0}^{\infty}\mu_{n}a_{n}$ must be convergent. In the next proposition
we prove that if we consider the stronger condition
$\sum_{n=0}^{\infty}\mu_{n}|a_{n}|<\infty$, we even get that $H_{\mu}(f)$ and
$I_{\mu}(f)$ are well defined and coincide.

###### Proposition 3

Let $X\hookrightarrow(H(\mathbb{D}),\tau_{0})$ be a Banach space of analytic
functions and let $\mu$ be a positive finite Borel measure on $[0,1)$. If
$\sum_{n=0}^{\infty}\mu_{n}|a_{n}|<\infty$ for each sequence $(a_{n})_{n}$
such that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in X$, $z\in\mathbb{D}$, then
$H_{\mu}:X\to H(\mathbb{D})$ and $I_{\mu}:X\to H(\mathbb{D})$ are well defined
and $I_{\mu}=H_{\mu}$.

###### Proof.

On the one hand, as $(\mu_{n})_{n\in\mathbb{N}_{0}}$ is decreasing, by
hypothesis $(\sum_{k=0}^{\infty}\mu_{n+k}a_{k})_{n\in\mathbb{N}_{0}}$ is
bounded, thus
$H_{\mu}(f)(z)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\mu_{n+k}a_{k}\right)z^{n}\in
H(\mathbb{D})$ for every $f\in X$, $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$,
$z\in\mathbb{D}$. On the other hand, the hypothesis yields that the
restriction of every $f\in X$ to $[0,1)$ belongs to $L_{1}([0,1),\mu)$.
Actually, the monotone convergence theorem gives
$\int_{0}^{1}|f(t)|d\mu(t)\leq\sum_{n=0}^{\infty}\mu_{n}|a_{n}|,$ hence
$I_{\mu}:X\to H(\mathbb{D})$ is well defined by Proposition 2. The coincidence
of $I_{\mu}$ and $H_{\mu}$ on $X$ follows by the dominated convergence theorem
using the Taylor series of $\displaystyle\frac{1}{1-tz}$. ∎

We need Abel summability of a series to write $I_{\mu}$ in terms of the
Hilbert-type operator below in Proposition 4.

###### Definition 1

A series $\sum_{n=0}^{\infty}a_{n}$, $a_{n}\in\mathbb{C},$ is said to be Abel
summable to $S\in\mathbb{C}$ if $\sum_{n=0}^{\infty}a_{n}r^{n}$ is convergent
for every $0<r<1$ and $\lim_{r\rightarrow
1^{-}}\sum_{n=0}^{\infty}a_{n}r^{n}=S$. In this case we write
$S=(A)$-$\sum_{n=0}^{\infty}a_{n}$.

It is known that, if $\sum_{n=0}^{\infty}a_{n}=S$, then
$(A)-\sum_{n=0}^{\infty}a_{n}=S$.

In order to state the following proposition we introduce some notation. For
$f\in H(\mathbb{D})$ and $0<r\leq 1$ we set $f_{r}(z):=f(rz)$. It is clear
that $\lim_{r\to 1^{-}}f_{r}=f$ in $(H(\mathbb{D}),\tau_{0}).$

###### Proposition 4

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let
$X\hookrightarrow(H(\mathbb{D}),\tau_{0})$ be a Banach space of analytic
functions. Assume that

  * (a)

there is $M>0$ such that for each $f\in X$ there is $g\in L_{1}([0,1),\mu)$
satisfying $|f_{r}(t)|\leq g(t)$ for every $t\in[0,1)$, $0<r\leq 1$ and
$\int_{0}^{1}g(t)d\mu(t)\leq M\|f\|$.

Then $I_{\mu}:(X,\|\ \|)\to(H(\mathbb{D}),\tau_{0})$ is a well defined
continuous operator with representation

$I_{\mu}(f)(z)=\displaystyle\sum_{n=0}^{\infty}\left((A)-\sum_{k=0}^{\infty}\mu_{n+k}a_{k}\right)z^{n},\
z\in\mathbb{D},$

for every $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in X.$ Whenever
$\sum_{k=0}^{\infty}\mu_{n+k}a_{k}$ converges for every $n\in\mathbb{N}_{0}$,
then $I_{\mu}(f)=H_{\mu}(f)$ on $X.$

###### Proof.

By Proposition 2 and the hypothesis on $X$, $I_{\mu}(f)\in H(\mathbb{D})$ for
every $f\in X$. We show that $I_{\mu}:X\to H(\mathbb{D})$ is continuous. Let
$(f_{n})_{n}\subseteq X$ be a sequence convergent to $f\in X$. Fix
$0<r_{0}<1$, $\varepsilon>0$ and let $n_{0}\in\mathbb{N}$ such that
$\|f_{n}-f\|<\frac{\varepsilon(1-r_{0})}{M}$ for each $n\geq n_{0}$. From the
hypothesis, for each $n\in\mathbb{N}$ there is $g_{n}\in L_{1}([0,1),\mu)$
such that $|f_{n}(t)-f(t)|\leq g_{n}(t)$ for each $t\in[0,1)$ and
$\int_{0}^{1}g_{n}(t)d\mu(t)\leq M\|f_{n}-f\|$. Hence, for $n\geq n_{0}$,

$\displaystyle\sup_{|z|\leq r_{0}}|I_{\mu}(f_{n})(z)-I_{\mu}(f)(z)|$
$\displaystyle\leq$ $\displaystyle\sup_{|z|\leq
r_{0}}\int_{0}^{1}\frac{|f_{n}(t)-f(t)|}{|1-tz|}d\mu(t)$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-r_{0}}\int_{0}^{1}|f_{n}(t)-f(t)|d\mu(t)\leq\frac{1}{1-r_{0}}\int_{0}^{1}g_{n}(t)d\mu(t)\leq$
$\displaystyle\leq$
$\displaystyle\frac{1}{1-r_{0}}M\|f_{n}-f\|\leq\varepsilon$

Therefore, $I_{\mu}:(X,\|\ \|)\to(H(\mathbb{D}),\tau_{0})$ is continuous.

Now, let us see that $I_{\mu}$ can be represented in terms of the Hilbert-type
operator. First, we prove that given $f\in X$, $\lim_{r\to
1^{-}}I_{\mu}(f_{r})=I_{\mu}(f)$ in $\tau_{0}$. Consider $g\in
L_{1}([0,1),\mu)$ satisfying (a) for $f$. Fix $0<R<1,$ $\varepsilon>0$ and let
$0<r_{0}<1$ be such that
$\int_{r_{0}}^{1}g(t)d\mu(t)\leq\frac{(1-R)\varepsilon}{4}$. Since $\lim_{r\to
1^{-}}f_{r}=f$ in the compact open topology, there is $0<r_{1}<1$ such that
$|f_{r}(z)-f(z)|\leq\frac{(1-R)\varepsilon}{2}$ for $r\geq r_{1}$, $|z|\leq
r_{0}$. Hence, for every $r\geq r_{1}$ we get

$\displaystyle\sup_{|z|\leq R}|I_{\mu}(f_{r})(z)-I_{\mu}(f)(z)|$
$\displaystyle\leq$ $\displaystyle\sup_{|z|\leq
R}\int_{0}^{1}\frac{|f_{r}(t)-f(t)|}{|1-tz|}d\mu(t)$ (2.1) $\displaystyle\leq$
$\displaystyle\frac{1}{1-R}\int_{0}^{r_{0}}|f_{r}(t)-f(t)|d\mu(t)+\frac{2}{1-R}\int_{r_{0}}^{1}g(t)d\mu(t)$
$\displaystyle\leq$ $\displaystyle\varepsilon.$

Given $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}\in X$ and $0<r<1,$ then
$f_{r}(z)=\sum_{k=0}^{\infty}a_{k}r^{k}z^{k}\in H^{\infty}.$ Observe that
$I_{\mu}(f_{r})=H_{\mu}(f_{r})$. Indeed, given $z\in\mathbb{D}$ we get

$\displaystyle I_{\mu}(f_{r})(z)$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{f(rt)}{1-tz}d\mu(t)=\int_{0}^{1}\frac{\sum_{k=0}^{\infty}a_{k}r^{k}t^{k}}{1-tz}d\mu(t)=\sum_{k=0}^{\infty}a_{k}r^{k}\int_{0}^{1}\frac{t^{k}}{1-tz}d\mu(t)$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}a_{k}r^{k}\int_{0}^{1}t^{k}\left(\sum_{n=0}^{\infty}(tz)^{n}\right)d\mu(t)=\sum_{k=0}^{\infty}a_{k}r^{k}\left(\sum_{n=0}^{\infty}\mu_{n+k}z^{n}\right)=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\mu_{n+k}a_{k}r^{k}z^{n}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\mu_{n+k}a_{k}r^{k}\right)z^{n}=H_{\mu}(f_{r})(z).$

We can perform all the permutations above because, for a fixed $0<r<1$ and
$0<R<1$ the series is absolutely convergent on $t\in[0,1]$ and $|z|\leq R$.

Since $I_{\mu}(f_{r})$ converges to $I_{\mu}(f)$ in $\tau_{0}$ as
$r\rightarrow 1^{-},$ we get $\lim_{r\rightarrow
1^{-}}\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\mu_{n+k}a_{k}r^{k}\right)z^{n}=I_{\mu}(f)(z)$
in $\tau_{0},$ and thus we get convergence on the coordinates, i.e.
$\lim_{r\rightarrow
1^{-}}\sum_{k=0}^{\infty}\mu_{n+k}a_{k}r^{k}=\frac{(I_{\mu}(f))^{(n)}(0)}{n!}$
for every $n\in\mathbb{N}.$ Therefore, for every $f\in X,$

$I_{\mu}(f)(z)=\sum_{n=0}^{\infty}\left((A)-\sum_{k=0}^{\infty}\mu_{n+k}a_{k}\right)z^{n},\
z\in\mathbb{D}.$

∎

###### Theorem 5

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let
$X\hookrightarrow(H(\mathbb{D}),\tau_{0})$ be a Banach space of analytic
functions. The operator $I_{\mu}:X\to H(\mathbb{D})$ is well defined and
continuous and admits the expression
$I_{\mu}(f)=\sum_{n=0}^{\infty}\left((A)-\sum_{k=0}^{\infty}\mu_{n+k}a_{k}\right)z^{n}$
for $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in X$ and $I_{\mu}(f)=H_{\mu}(f)$ if
$\sum_{k=0}^{\infty}\mu_{n+k}a_{k}$ is convergent for each
$n\in\mathbb{N}_{0}$ in the following cases:

  * (i)

When $X=H_{v}^{\infty}$ and $\frac{1}{v}\in L_{1}([0,1),\mu)$, $v$ being a
weight on $\mathbb{D}.$ If the weight $v$ is essential, then the condition
$\frac{1}{v}\in L_{1}([0,1),\mu)$ is also necessary in order to have
$I_{\mu}:X\to H(\mathbb{D})$ well defined.

  * (ii)

When $X=H^{1}$ and $\mu$ is a Carleson measure.

###### Proof.

(i) Just observe that we can apply Proposition 4 with
$M=\int_{0}^{1}\frac{1}{v(t)}d\mu(t)$ and $g=\frac{\|f\|_{v}}{v}\in
L_{1}([0,1),\mu)$, for a given $f\in H_{v}^{\infty}$. As
$\|f_{r}\|_{v}\leq\|f\|_{v}$ for each $f\in H_{v}^{\infty}$ and $0<r\leq 1$,
then $|f_{r}(t)|\leq\frac{||f_{r}||_{v}}{v(t)}\leq g(t)$ for $0\leq t<1$,
$0<r\leq 1$.

Now assume that $I_{\mu}:H_{v}^{\infty}\to H(\mathbb{D})$ is well defined and
that $v$ is essential. By Lemma 1 there are $f_{v}\in H_{v}^{\infty}$ and
$D(v)>0$ such that $1/v(t)\leq D(v)f_{v}(t)$ for each $t\in[0,1)$. We have

$I_{\mu}(f_{v})(0)=\int_{0}^{1}f_{v}(t)d\mu(t)\geq\frac{1}{D(v)}\int_{0}^{1}\frac{1}{v(t)}d\mu(t),$

and $1/v(t)\in L_{1}([0,1),\mu)$.

(ii) Let $C>0$ be such that $\mu_{n}\leq\frac{C}{n+1}$ for each
$n\in\mathbb{N}_{0}$. Given $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H^{1}$,
consider $g(t)=\sum_{n=0}^{\infty}|a_{n}|t^{n},$ $0\leq t<1$. The monotone
convergence theorem combined with the Hardy inequality give $g\in
L_{1}([0,1),\mu)$ and
$\int_{0}^{1}g(t)d\mu(t)=\sum_{n=0}^{\infty}|a_{n}|\mu_{n}\leq
C\sum_{n=0}^{\infty}\frac{|a_{n}|}{n+1}\leq C\pi\|f\|_{H^{1}},$ and then (a)
in Proposition 4 is satisfied, since $|f_{r}(t)|\leq g(t)$ holds for $0\leq
t<1$, $0<r\leq 1$. ∎

##  3 $I_{\mu}$ as an operator between weighted Banach spaces of analytic
functions.

Now we study the action of $I_{\mu}$ on weighted Banach spaces of holomorphic
functions defined by general weights as well as for $H^{\infty}$, and give a
characterization of the boundedness and compactness of the operator in the
case of standard weights. In what follows, $B_{v}^{\infty}$ denotes the closed
unit ball of $H_{v}^{\infty}.$

###  3.1 General weights

###### Lemma 6

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let $v$ be weight
on $\mathbb{D}$ such that $1/v\in L_{1}([0,1),\mu)$. Then
$I_{\mu}:(B_{v}^{\infty},\tau_{0})\to H(\mathbb{D})$ is continuous.

###### Proof.

The hypothesis and Theorem 5(i) imply that $I_{\mu}:H_{v}^{\infty}\to
H(\mathbb{D})$ is a well defined continuous linear operator. Let us see that
for every $(f_{n})_{n}\subseteq B_{v}^{\infty}$ convergent to $f\in
B_{v}^{\infty}$ in $\tau_{0},$ then $(I_{\mu}(f_{n}))_{n}$ converges to
$I_{\mu}(f)$ in $\tau_{0}$. If $(f_{n})_{n}\subseteq B_{v}^{\infty}$, then
$|f_{n}(t)|\leq\frac{1}{v(t)}$ for every $0\leq t<1,$ and every
$n\in\mathbb{N}.$ As $\frac{1}{v(t)}\in L_{1}([0,1),\mu),$ the conclusion
follows proceeding analogously as in (2.1) in Proposition 4, taking $g=1/v$
and considering $f_{n}$ instead of $f_{r}.$ ∎

In the next theorem we give sufficient conditions for the continuity of the
operator $I_{\mu}$ between weighted Banach spaces of holomorphic functions of
type $H_{v}^{\infty}$. These conditions are stronger than the assumption
$\int_{0}^{1}\frac{1}{v(t)}d\mu(t)<\infty$, which appeared in Theorem 5.

###### Theorem 7

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let $v,w$ be
weights on $\mathbb{D}.$ Consider the following statements:

  * (i)

$C(v,w):=\sup_{0\leq r<1}w(r)\int_{0}^{1}\frac{d\mu(t)}{v(t)(1-tr)}<\infty,$

  * (ii)

$I_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is well defined and continuous.

Then (i) always implies (ii). If we assume in addition that the weight $v$ is
essential, then (i) and (ii) are equivalent.

###### Proof.

If we assume (i) then $1/v\in L_{1}([0,1),\mu)$ (just take $r=0$). Thus,
Theorem 5(i) implies that $I_{\mu}:H_{v}^{\infty}\to H(\mathbb{D})$ is a well
defined continuous operator. Moreover, for each $f\in H_{v}^{\infty},$

$\sup_{0\leq r<1}w(r)\max_{|z|=r}|I_{\mu}(f)(z)|\leq\|f\|_{v}\sup_{0\leq
r<1}w(r)\int_{0}^{1}\frac{d\mu(t)}{v(t)(1-tr)},$

hence $I_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{\infty}$ and (ii) is a
consequence of the closed graph theorem.

For the converse, we assume (ii) and that the weight $v$ is essential. By
Lemma 1 there are $f_{v}\in H_{v}^{\infty}$ and $D(v)>0$ such that $1/v(t)\leq
D(v)f_{v}(t)$ for each $t\in[0,1)$. We have

$w(r)\int_{0}^{1}\frac{d\mu(t)}{v(t)(1-tr)}\leq
D(v)\sup_{z\in\mathbb{D},|z|=r}w(z)\left|\int_{0}^{1}\frac{f_{v}(t)}{1-tz}d\mu(t)\right|=D(v)\sup_{z\in\mathbb{D},|z|=r}w(z)|I_{\mu}(f_{v})(z)|.$
(3.1)

So, if we assume $I_{\mu}(f_{v})\in H_{w}^{\infty},$ we get that the condition
in (i) is also necessary. ∎

###### Proposition 8

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let $v,w$ be
weights on $\mathbb{D}$ such that $1/v\in L_{1}([0,1),\mu)$. The following
conditions are equivalent:

  * (i)

$I_{\mu}:H_{v}^{0}\to H_{w}^{0}$ is continuous.

  * (ii)

$I_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is continuous and $I_{\mu}(1)\in
H_{w}^{0}$.

If the equivalent conditions hold, then the bitranspose $I_{\mu}^{**}$ of
$I_{\mu}:H_{v}^{0}\to H_{w}^{0}$ coincides with $I_{\mu}:H_{v}^{\infty}\to
H_{w}^{\infty},$ and
$\|I_{\mu}\|_{\mathcal{L}(H_{v}^{\infty},H_{v}^{\infty})}=\|I_{\mu}\|_{\mathcal{L}(H_{v}^{0},H_{v}^{0})}.$

###### Proof.

Assume that (i) holds. Then $I_{\mu}(1)\in H_{w}^{0}$ immediately. We can
conclude by Lemma 6, proceeding analogously as in [3, Lemma 1].

If we assume that (ii) holds, for each $0<r<1$,

$\max_{|z|=r}|I_{\mu}(z^{k})|=\sum_{n=0}^{\infty}\mu_{n+k}r^{n}\leq\sum_{n=0}^{\infty}\mu_{n}r^{n}=\max_{|z|=r}|I_{\mu}(1)|,$
(3.2)

we get $I_{\mu}(z^{k})\in H_{w}^{0}$ for each $k\in\mathbb{N}$. Since the
polynomials $\mathcal{P}\subseteq H_{v}^{0}$ are norm dense in $H_{v}^{0}$, we
get $I_{\mu}(H_{v}^{0})\subseteq H_{w}^{0}$ and $I_{\mu}:H_{v}^{0}\to
H_{w}^{0}$ is continuous by the closed graph theorem. The last assertion
follows reasoning analogously as in [3, Lemma 1]. ∎

###### Proposition 9

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and $v$ a weight on
$\mathbb{D}$ such that $1/v\in L_{1}([0,1),\mu)$. Consider
$I_{\mu}:H_{v}^{\infty}\rightarrow H(\mathbb{D})$. Then:

  * (i)

If $\mu$ is a Carleson measure and $w$ is a weight such that
$w(r)|\log(1-r)|\rightarrow 0$ as $r\rightarrow 1^{-}$, then $I_{\mu}(1)\in
H_{w}^{0}.$ This condition is satisfied, for instance, in case $w$ is a
standard weight.

  * (ii)

If $\mu$ is an $s$-Carleson measure for $0<s<1$, then $I_{\mu}(1)\in
H_{v_{\gamma}}^{0}$ for each $\gamma>1-s$.

###### Proof.

By Theorem 5(i), the hypothesis implies $I_{\mu}:H_{v}^{\infty}\rightarrow
H(\mathbb{D})$ is well defined and continuous. As $I_{\mu}(1)=C_{\mu}(1)$,
where $C_{\mu}$ is the Cesàro-type operator associated with the Borel measure
$\mu,$ [3, Proposition 2] yields the conclusions. ∎

###### Proposition 10

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and $v,w$ be weights
on $\mathbb{D}$ such that $1/v\in L_{1}([0,1),\mu)$ and $I_{\mu}:H_{v}^{0}\to
H_{w}^{0}$ is continuous. Then $I_{\mu}:H_{v}^{0}\to H_{w}^{0}$ and
$I_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ are compact if and only if
$I_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{0}.$

###### Proof.

By Proposition 8, $I_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is continuous and
the operators $I_{\mu}^{**}$ and $I_{\mu}$ coincide on $H_{v}^{\infty}$.
Therefore $I_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{0}$ if and only if
$I^{**}_{\mu}((H_{v}^{0})^{**})\subseteq H_{w}^{0}$. This fact holds if and
only if $I_{\mu}:H_{v}^{0}\to H_{w}^{0}$ is weakly compact by [20, 17.2.7].
Now [28, Corollary 2.5] implies that this is equivalent to the compactness of
$I_{\mu}:H_{v}^{0}\to H_{w}^{0}$. Finally, Proposition 8 together with Banach-
Schauder theorem yield that $I_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is
compact if and only if $I_{\mu}:H_{v}^{0}\to H_{w}^{0}$ is compact. ∎

###### Theorem 11

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let $v,w$ be
weights on $\mathbb{D}.$ Assume that $1/v\in L_{1}([0,1),\mu)$. Consider the
following statements:

  * (i)

$\lim_{r\to 1^{-}}w(r)\int_{0}^{1}\frac{d\mu(t)}{v(t)(1-tr)}=0,$

  * (ii)

$I_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$ is compact.

  * (iii)

$I_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is compact and $I_{\mu}(1)\in
H_{w}^{0}.$

  * (iv)

$I_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$ is continuous.

Then (i) implies (ii), which imples (iii), and (iii) implies (iv). If in
addition $v$ is essential, then the four conditions are equivalent.

###### Proof.

Observe that the hypothesis (i) implies that $I_{\mu}:H_{v}^{\infty}\to
H_{w}^{\infty}$ is continuous by Theorem 7. Moreover, for each $f\in
B_{v}^{\infty}$ and $0<r<1,$

$w(r)\max_{|z|=r}|I_{\mu}(f)(z)|\leq
w(r)\int_{0}^{1}\frac{d\mu(t)}{v(t)(1-tr)}.$

Hence, $I_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{0}$. The compactness of
$I_{\mu}$ follows by Proposition 10, so we have that (i) implies (ii). (ii)
implies (iii) is trivial and (iii) implies (iv) by Propositions 8 and 10.

Now, let us assume the weight $v$ is essential. By Lemma 1 there are $f_{v}\in
H_{v}^{\infty}$ and $D(v)>0$ such that $1/v(t)\leq D(v)f_{v}(t)$ for each
$t\in[0,1)$. If condition (iv) hold, then (i) follows immediately from
$\lim_{r\to 1^{-}}w(r)|I_{\mu}(f_{v})(r)|=0.$ ∎

###### Corollary 12

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and let $v,w$ be
weights on $\mathbb{D}.$ Assume that $1/v\in L_{1}([0,1),\mu)$ and that
$w(r)\leq 1-r$ for every $0<r<1$. Then $I_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$
is compact.

###### Proof.

If we define $g_{r}(t):=\frac{w(r)}{v(t)(1-tr)}$, $t\in[0,1)$, $0<r<1$, we
have $|g_{r}(t)|\leq\frac{1}{v(t)}$ for every $0<r<1$, $0\leq t<1$ and
$\lim_{r\to 1^{-}}g_{r}(t)=0$ for all $0\leq t<1$. The dominated convergence
theorem permits us to conclude that condition (i) in Theorem 11 holds. ∎

###  3.2 Standard weights

In this section we consider weighted Banach spaces of holomorphic functions
determined by standard weights $v_{\gamma}(r)=(1-r)^{\gamma},$ $0\leq r<1,\
\gamma>0.$ In Theorems 15 and 16 we get that, for $\gamma>0$ and
$\delta\in(-\gamma,1-\gamma),$ the operator
$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$ is
continuous (compact) if and only if the Cesàro-type operator
$C_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$ is
so (see [3, Theorems 10 and 11]). First, we need two lemmata.

###### Lemma 13

Let $\mu$ be a positive finite Borel measure on $[0,1)$, $\gamma>0,$ and
$\delta\in(-\gamma,1-\gamma).$ The following assertions are equivalent:

  * (i)

$\sup_{0\leq
r<1}(1-r)^{\gamma+\delta}\int_{0}^{1}\frac{d\mu(t)}{(1-t)^{\gamma}(1-tr)}<\infty$

  * (ii)

$\sup_{0\leq
r<1}(1-r)^{\gamma+\delta}\int_{0}^{1}\frac{d\mu(t)}{(1-tr)^{\gamma+1}}<\infty$.

  * (iii)

$\mu$ is a $(1-\delta)$-Carleson measure.

###### Proof.

(ii) is equivalent to (iii) by [3, Theorem 10]. Since
$\frac{1}{1-tr}\leq\frac{1}{1-t}$ for every $r,t\in[0,1)$, we get immediately
that (i) implies (ii). We show that (iii) implies (i). By hypothesis, there
exists $C>0$ such that $\mu[t,1)\leq C(1-t)^{1-\delta}.$ Thus, using
integration by parts and [12, Lemma 2.5], we get

$\displaystyle\int_{0}^{1}\frac{d\mu(t)}{(1-t)^{\gamma}(1-tr)}$
$\displaystyle=$
$\displaystyle\mu([0,1))+\int_{0}^{1}\left(\frac{1}{(1-t)^{\gamma}(1-tr)}\right)^{\prime}\mu([t,1))dt$
$\displaystyle\leq$
$\displaystyle\mu([0,1))+C\int_{0}^{1}\frac{\gamma(1-tr)+r(1-t)}{(1-tr)^{2}(1-t)^{\gamma+\delta}}dt$
$\displaystyle\leq$
$\displaystyle\mu([0,1))+C(\gamma+r)\int_{0}^{1}\frac{dt}{(1-tr)(1-t)^{\gamma+\delta}}$
$\displaystyle\leq$
$\displaystyle\mu([0,1))+C^{\prime}(\gamma+r)(1-r)^{-(\gamma+\delta)}.$

∎

The proof of the next Lemma is completely analogous of that of Lemma 13, using
[3, Theorem 11].

###### Lemma 14

Let $\mu$ be a positive finite Borel measure on $[0,1)$, $\gamma>0,$ and
$\delta\in(-\gamma,1-\gamma).$ The following assertions are equivalent:

  * (i)

$\lim_{r\to
1^{-}}(1-r)^{\gamma+\delta}\int_{0}^{1}\frac{d\mu(t)}{(1-t)^{\gamma}(1-tr)}=0.$

  * (ii)

$\lim_{r\to
1^{-}}(1-r)^{\gamma+\delta}\int_{0}^{1}\frac{d\mu(t)}{(1-tr)^{\gamma+1}}=0$.

  * (iii)

$\mu$ is a vanishing $(1-\delta)$-Carleson measure.

###### Theorem 15

Let $\mu$ be a positive finite Borel measure on $[0,1)$, $\gamma>0,$ and
$\delta\in(-\gamma,1-\gamma).$ The following are equivalent:

  * (i)

$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$ is
continuous.

  * (ii)

$I_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is
continuous and $1/v_{\gamma}\in L_{1}([0,1),\mu)$.

  * (iii)

$\mu$ is a $(1-\delta)$-Carleson measure.

  * (iv)

$\mu_{n}=O(\frac{1}{n^{1-\delta}})$.

###### Proof.

The equivalence of (i) and (iii) follows from Theorem 7 and Lemma 13. The
equivalence between (iii) and (iv) is stated in [9, Proposition 1] without
proof and a proof can be found in [7, Corollary 4.6]. To see that these three
equivalent conditions are equivalent to (ii), by Proposition 8 we only need to
show that $I_{\mu}(1)\in H_{v_{\gamma+\delta}}^{0}$ when (i) holds. By the
equivalence between conditions (i) and (iii) and by Theorem 5(i),
$I_{\mu}(1)\in H_{v_{\gamma+\delta}}^{0}$ is satisfied for every $\gamma>0$ by
Proposition 9(i) when $-\gamma<\delta\leq 0$ and by Proposition 9(ii) when
$\delta\in(0,1-\gamma)$ (just take $s=1-\delta$ and consider
$H_{v_{\gamma+\delta}}^{0}$ instead of $H_{v_{\gamma}}^{0}$ in the
proposition). ∎

###### Theorem 16

Let $\mu$ be a positive finite Borel measure on $[0,1)$, $\gamma>0,$ and
$\delta\in(-\gamma,1-\gamma).$ The following are equivalent:

  * (i)

$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$ is
compact.

  * (ii)

$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is
compact.

  * (iii)

$I_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is compact
and $1/v_{\gamma}\in L_{1}([0,1),\mu)$.

  * (iv)

$\mu$ is a vanishing $(1-\delta)$-Carleson measure.

  * (v)

$\mu_{n}=o(\frac{1}{n^{1-\delta}})$.

###### Proof.

(i) is equivalent to (ii) by Theorem 15 and Proposition 10. (ii) and (iii) are
equivalent by Theorem 5(i) and Proposition 10. The equivalence with (iv)
follows by Theorem 11 and Lemma 14, and (iv) is equivalent to (v) by [3,
Theorem 11]. ∎

In the case $\delta\geq 1-\gamma$, $\gamma>0,$ we prove that the continuity
and compactness of the operator $I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow
H_{v_{\gamma+\delta}}^{\infty}$ are equivalent to the well definition of
$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H(\mathbb{D})$.

###### Theorem 17

Let $\mu$ be a positive finite Borel measure on $[0,1)$, $\gamma>0$. The
following are equivalent:

  * (i)

$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H(\mathbb{D})$ is well defined
(even continuous).

  * (ii)

$\int_{0}^{1}\frac{d\mu(t)}{(1-t)^{\gamma}}<\infty.$

  * (iii)

$\sum_{n=0}^{\infty}\mu_{n}n^{\gamma-1}<\infty.$

  * (iv)

$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{1}}^{\infty}$ is a compact
operator, i.e. $I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow
H_{v_{\gamma+\delta}}^{\infty}$ is so for every $\gamma+\delta\geq 1$.

  * (v)

$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{1}}^{\infty}$ is well
defined, i.e. $I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow
H_{v_{\gamma+\delta}}^{\infty}$ is so for every $\gamma+\delta\geq 1$.

In particular, $I_{\mu}:H_{v_{1}}^{\infty}\rightarrow H_{v_{1}}^{\infty}$ is
well defined if and only if it is compact if and only if
$\sum_{n}\mu_{n}<\infty.$ If (i)-(v) are satisfied, then
$I_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{1}}^{0}$ is a compact operator,
i.e. $I_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is so
for every $\gamma+\delta\geq 1$, and
$I_{\mu}(H_{v_{\gamma}}^{\infty})\subseteq H_{v_{1}}^{0}$ for every
$\gamma>0.$

###### Proof.

The equivalence between (i) and (ii) is a consequence of Theorem 5(i), since
$\frac{1}{v_{\gamma}(t)}=\frac{1}{(1-t)^{\gamma}}$. (ii) is equivalent to
(iii) by the fact that $f(z)=\frac{1}{(1-z)^{\gamma}},\ z\in\mathbb{D},$
satisfies $f(z)=\sum_{n=0}^{\infty}a_{n}(\gamma)z^{n},\ z\in\mathbb{D},$ with
$a_{n}(\gamma)>0,$ $a_{n}(\gamma)\cong n^{\gamma-1}$ (eg.[14, p.12]).
Corollary 12 ensures that (i) implies
$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{1}}^{\infty}$ is compact.
Since $H_{v_{\gamma_{1}}}^{\infty}\hookrightarrow H_{v_{\gamma_{2}}}^{\infty}$
continuously for every $\gamma_{2}>\gamma_{1}>0,$ (iv) holds. (iv) implies (v)
and (v) implies (i) are trivial. The assertion about
$I_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{1}}^{0}$ follows from the fact
that if (iii) is satisfied, then $\mu$ is a $\gamma$-Carleson measure, thus
Proposition 9 yields $I_{\mu}(1)\in H_{v_{1}}^{0}$ and Propositions 8 and 10
yield the conclusion. ∎

We get the following extension of [2, Theorem 2.1 (ii) and (iii)], where the
continuity of the operators is obtained for $\mu$ being the Lebesgue measure.

###### Corollary 18

If $\mu$ is a Carleson but not vanishing Carleson measure on $[0,1),$ then
$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma}}^{\infty}$ is
continuous if and only if $0<\gamma<1$. In this case,
$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma}}^{\infty}$ is never
compact.

###### Proof.

$\sum_{n=0}^{\infty}\mu_{n}<\infty$ implies $\lim_{n}n\mu_{n}=0$ [24, Theorem
3.3.1], hence if $\mu$ is not vanishing Carleson, then
$\sum_{n=0}^{\infty}\mu_{n}$ must diverge. Hence, if $\gamma\geq 1,$ then
$I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma}}^{\infty}$ is not
well defined by the equivalence between (iii) and (v) in Theorem 17. If
$0<\gamma<1$ then $I_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow
H_{v_{\gamma}}^{\infty}$ is continuous but not compact by Theorems 15 and 16.
∎

For $0<\gamma<1$, both the continuity and compactness of
$I_{\mu}:H_{v_{\gamma}}^{\infty}\to H_{v_{\gamma}}^{\infty}$ are equivalent to
those of the Cesàro type operator $C_{\mu}:H_{v_{\gamma}}^{\infty}\to
H_{v_{\gamma}}^{\infty}$ by [3, Theorems 10 and 11] and Theorems 15 and 16
above. The next example shows a difference between the behaviour of $I_{\mu}$
and $H_{\mu}$, and the behaviour of $C_{\mu}$ when $\gamma\geq 1$.

###### Example 19

For the positive finite Borel measure $\mu(t)=\frac{dt}{\log\frac{e}{1-t}}$,
$t\in[0,1),$ we get:

  * (i)

The operator $I_{\mu}:H_{v_{1}}^{\infty}\rightarrow H(\mathbb{D})$ is not well
defined.

  * (ii)

The operator $H_{\mu}:H_{v_{1}}^{\infty}\rightarrow H(\mathbb{D})$ is not well
defined.

  * (iii)

The Cesàro-type operator $C_{\mu}:H_{v_{1}}^{\infty}\rightarrow
H_{v_{1}}^{\infty}$ is compact.

###### Proof.

By [30, Lemma 2.1 (vi)], we get that $\mu$ is a positive finite Borel measure
with moments $\mu_{n}\cong\frac{1}{n\log n},$ $n\in\mathbb{N}_{0}$. So, it is
a vanishing Carleson measure but $\sum_{n=0}^{\infty}\mu_{n}$ diverges. (i) is
a consecuence of Theorem 17. (ii) follows from the fact that
$f(z)=\frac{1}{1-z}\in H_{v_{1}}^{\infty}$ but
$H_{\mu}(f)(0)=\sum_{n=0}^{\infty}\mu_{n}$. (iii) is immediate from [3,
Theorem 11]. ∎

Recall that if $H_{\mu}$ is well defined on $H_{v_{\gamma}}^{\infty}$, then
$\sum_{n=0}^{\infty}\mu_{n}a_{n}$ converges in $\mathbb{C}$ for each
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H_{v_{\gamma}}^{\infty}$. This is
satisfied if $\sum_{n=0}^{\infty}\mu_{n}|a_{n}|<\infty$ for each
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H_{v_{\gamma}}^{\infty}$. The following
result gives us information about the convergence of the last series for
$s$-Carleson measures.

###### Proposition 20

[29, Theorem 18], [21, Theorem 2.1] Let $\gamma>0$ and $s\geq 1.$ Then
$\sum_{n=0}^{\infty}\frac{|a_{n}|}{(n+1)^{s}}<\infty$ for every
$\sum_{n=0}^{\infty}a_{n}z^{n}\in H_{v_{\gamma}}^{\infty}$ if and only if
$s>\gamma+1/2.$ In particular:

  * (i)

If $0<\gamma<1/2,$ then $\sum_{n=0}^{\infty}\frac{|a_{n}|}{n+1}<\infty$ for
every $\sum_{n=0}^{\infty}a_{n}z^{n}\in H_{v_{\gamma}}^{\infty}$.

  * (ii)

If $\gamma\geq 1/2,$ then there exists $\sum_{n=0}^{\infty}a_{n}z^{n}\in
H_{v_{\gamma}}^{\infty}$ such that $\sum_{n=0}^{\infty}\frac{|a_{n}|}{n+1}$
diverges.

As a consequence of Proposition 20(ii), $H_{v_{\gamma}}^{\infty}\nsubseteq
H^{1}$ for $\gamma\geq 1/2,$ as a consequence of the Hardy inequality.

###### Corollary 21

Let $\mu$ be a positive finite Borel measure on $[0,1)$, $\gamma>0,$ and
$\int_{0}^{1}\frac{d\mu(t)}{v_{\gamma}(t)}<\infty.$ Then
$H_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H(\mathbb{D})$ is well defined and
continuous and $H_{\mu}=I_{\mu}$ on $H_{v_{\gamma}}^{\infty}$ in the following
cases:

  * (a)

If $\gamma<\frac{1}{2}$ and $\mu$ is a Carleson measure. In this case,
$H_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$
and $H_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ are
continuous for $\delta=0$ and compact for $\delta>0$.

  * (b)

If $\gamma\geq\frac{1}{2}$ and $\mu$ is an $s$-Carleson measure for
$s>\gamma+1/2$. In this case, $H_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow
H_{v_{\gamma+\delta}}^{\infty}$ and $H_{\mu}:H_{v_{\gamma}}^{0}\rightarrow
H_{v_{\gamma+\delta}}^{0}$ are continuous for $\delta=1-s$ and compact for
$\delta>1-s$.

###### Proof.

This is a consequence of Proposition 3 and Proposition 20, together with
Theorems 15, 16 and 17. ∎

##  4 The generalized Hilbert operator $H_{\mu}$ on solid hulls and cores

The solid hull and the solid core of $H_{v_{\gamma}}^{\infty}$ are also of
interest in the study of the well definition of $H_{\mu}$ on this space. In
this context we identify an analytic function
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H(\mathbb{D})$ with the sequence of its
Taylor coefficients $(a_{n})_{n}$. Given $A$ a vector space of complex
sequences containing the space of all the sequences with finitely many non-
zero coordinates, the _solid hull_ of $A$ is

$S(A):=\\{(c_{n})_{n}:\ \exists(a_{n})_{n}\in A\text{ such that
}|c_{n}|\leq|a_{n}|\ \forall n\in\mathbb{N}\\}\supseteq A,$

and the _solid core_ of $A$ is

$s(A):=\\{(c_{n})_{n}:\ (c_{n}a_{n})_{n}\in A\
\forall(a_{n})_{n}\in\ell_{\infty}\\}\subseteq A.$

In the next result we describe the solid hull and the solid core of the spaces
under consideration.

###### Proposition 22

[23, Theorems 8.2.1 and 8.3.4] For every $\gamma>0$,

$S(H_{v_{\gamma}}^{\infty})=\left\\{(a_{m})_{m}:\
\sup_{n\in\mathbb{N}_{0}}\left(\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|a_{m}|^{2}}{(m+1)^{2\gamma}}\right)^{1/2}<\infty\right\\},$

and

$s(H_{v_{\gamma}}^{\infty})=\left\\{(a_{m})_{m}:\
\sup_{n\in\mathbb{N}_{0}}\left(\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|a_{m}|}{(m+1)^{\gamma}}\right)<\infty\right\\}$

Both are Banach spaces of analytic functions for the norms
$\|f\|_{S(H_{v_{\gamma}}^{\infty})}$ and $\|f\|_{s(H_{v_{\gamma}}^{\infty})}$
defined by the supremum.

###### Proposition 23

Let $\gamma>0$ and let $\mu$ be an $s$-Carleson measure, for $s>0$.

  * (i)

If $s>\gamma+1/2$, then $H_{\mu}:S(H_{v_{\gamma}}^{\infty})\rightarrow
H(\mathbb{D})$ is a well defined continuous operator and $H_{\mu}=I_{\mu}$ on
$S(H_{v_{\gamma}}^{\infty})$. If in addition $\gamma\geq\frac{1}{2},$ then
$H_{\mu}:S(H_{v_{\gamma}}^{\infty})\rightarrow S(H_{v_{\gamma}}^{\infty})$ is
a well defined continuous operator.

  * (ii)

If $s>\gamma$, then $H_{\mu}:s(H_{v_{\gamma}}^{\infty})\rightarrow
H(\mathbb{D})$ is a well defined continuous operator and $H_{\mu}=I_{\mu}$ on
$s(H_{v_{\gamma}}^{\infty})$. If in addition $\gamma\geq 1,$ then
$H_{\mu}:s(H_{v_{\gamma}}^{\infty})\rightarrow s(H_{v_{\gamma}}^{\infty})$ is
a well defined continuous operator.

###### Proof.

(i) Given $f(z)=\sum_{m=0}^{\infty}a_{m}z^{m}\in S(H_{v_{\gamma}}),$ by
Proposition 22 there exists $C>0$ such that
$\displaystyle\left(\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|a_{m}|^{2}}{(m+1)^{2\gamma}}\right)^{1/2}<C\text{
for every }n\in\mathbb{N}_{0}.$ As $\|\ \|_{1}\leq d^{1/2}\|\ \|_{2}$ in
$\mathbb{C}^{d},$ with $d=2^{n}$, we get

$\displaystyle\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|a_{m}|}{(m+1)^{\gamma}}\leq
2^{n/2}C.$

Therefore, for every $\delta>1/2$,

$\sum_{n=0}^{\infty}\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|a_{m}|}{(m+1)^{\gamma+\delta}}\leq\sum_{n=0}^{\infty}\frac{1}{2^{n\delta}}\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|a_{m}|}{(m+1)^{\gamma}}\leq
C\sum_{n=0}^{\infty}\frac{1}{2^{n(\delta-1/2)}}<\infty.$ (4.1)

Hence, if $\mu$ is an $s$-Carleson measure for $s>\gamma+1/2,$ then
$\sum_{n=0}^{\infty}\mu_{n}|a_{n}|<\infty$. We conclude by Proposition 3 and
Proposition 4 that $H_{\mu}:S(H_{v_{\gamma}}^{\infty})\to H(\mathbb{D})$ is a
well defined continuous operator and $H_{\mu}=I_{\mu}$ on
$S(H_{v_{\gamma}}^{\infty})$. Indeed, to apply Proposition 4 take
$g(t)=\sum_{n=0}^{\infty}|a_{n}|t^{n}$ for
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in S(H_{v_{\gamma}}^{\infty})$. (4.1)
gives, for $M>0$ such that $\mu_{n}\leq\frac{M}{(n+1)^{s}}$,
$n\in\mathbb{N}_{0},$

$\int_{0}^{1}g(t)d\mu(t)=\sum_{n=0}^{\infty}\mu_{n}|a_{n}|\leq
M\sum_{n=0}^{\infty}\frac{1}{2^{n(s-\gamma-1/2)}}\|f\|_{S(H_{v_{\gamma}}^{\infty})}.$

In this case, let us see that $H_{\mu}(S(H_{v_{\gamma}}^{\infty}))\subseteq
S(H_{v_{\gamma}}^{\infty})$ if $\gamma\geq 1/2.$ Let
$M=\sum_{n=0}^{\infty}\mu_{n}|a_{n}|$ for
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in S(H_{v_{\gamma}}^{\infty})$. For every
$n\in\mathbb{N}_{0}$ we get

$\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|\sum_{k=0}^{\infty}\mu_{k+m}a_{k}|^{2}}{(m+1)^{2\gamma}}\leq
2^{n}\frac{M^{2}}{(2^{n}+1)^{2\gamma}}\leq M^{2}2^{n(1-2\gamma)}\leq M^{2}.$

The continuity holds by the closed graph theorem, since
$S(H_{v_{\gamma}}^{\infty})\hookrightarrow H(\mathbb{D})$ continuously.

(ii) Removing the factor $2^{n/2}$ in the proof of (i), we get that if $\mu$
is an $s$-Carleson measure for $s>\gamma,$ then there exists $M>0$ such that
$\sum_{k=0}^{\infty}\mu_{n+k}|a_{k}|<M$ for every $n\in\mathbb{N}_{0},$ thus,
$H_{\mu}(f)(z)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{\infty}\mu_{n+k}a_{k}\right)z^{n}\in
H(\mathbb{D})$ and $H_{\mu}=I_{\mu}:s(H_{v_{\gamma}}^{\infty})\to
H(\mathbb{D})$ is continuous. Let us see that
$H_{\mu}(s(H_{v_{\gamma}}^{\infty}))\subseteq s(H_{v_{\gamma}}^{\infty})$ if
$\gamma\geq 1$. Indeed, for every $n\in\mathbb{N}_{0},$

$\sum_{m=2^{n}}^{2^{n+1}-1}\frac{|\sum_{k=0}^{\infty}\mu_{k+m}a_{k}|}{(m+1)^{\gamma}}\leq
2^{n}\frac{M}{(2^{n}+1)^{\gamma}}\leq M2^{n(1-\gamma)}\leq M.$

The continuity holds again by the closed graph theorem. ∎

##  5 Hilbert-type operators associated to summable moment measures

In this section we deal with positive finite Borel measures $\mu$ on $[0,1)$
such that $\sum_{n=0}^{\infty}\mu_{n}<\infty$. This condition implies $\mu$ is
a vanishing Carleson measure [24, Theorem 3.3.1].

Proceeding as in the proof of [3, Theorem 7], we obtain that this condition
gives the equivalence of the boundedness and compactness of the Hilbert-type
operator on $H^{\infty}$. We include the proof for the sake of completeness.

###### Theorem 24

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\int_{0}^{1}\frac{d\mu(t)}{1-t}<\infty$, i.e.
$\sum_{n=0}^{\infty}\mu_{n}<\infty$. Then $I_{\mu}:H^{\infty}\rightarrow
H^{\infty}$ is compact and $H_{\mu}=I_{\mu}.$

###### Proof.

Let us assume without loss of generality that $\mu([0,1))=1$. By [17, Theorem
1.2], $I_{\mu}$ and $H_{\mu}$ are continuous on $H^{\infty}$ and
$H_{\mu}=I_{\mu}.$ By the proof of [3, Lemma 6] we only need to show that
$(\|I_{\mu}(f_{n})\|_{\infty})_{n}$ is convergent to 0 whenever
$(f_{n})_{n}\subseteq B_{H^{\infty}}$ is convergent to 0 in $\tau_{0}$. Let
$\varepsilon>0$. By hypothesis, there is $s_{0}\in(0,1)$ such that

$\int_{s_{0}}^{1}\frac{d\mu(t)}{1-t}<\frac{\varepsilon}{2}.$ (5.1)

We have that

$|I_{\mu}(f)(z)|=\left|\int_{0}^{1}\frac{f(t)}{1-tz}d\mu(t)\right|$ (5.2)

for all $f\in H^{\infty}$, $z\in\mathbb{D}$. For $z\in\mathbb{D}$,
$t\in[s_{0},1]$ and $f\in B_{H^{\infty}}$ we have
$|\frac{f(t)}{1-tz}|\leq\frac{1}{1-t}$. Hence, by (5.1), for each
$n\in\mathbb{N}$ we have

$\left|\int_{s_{0}}^{1}\frac{f_{n}(t)}{1-tz}d\mu(t)\right|<\frac{\varepsilon}{2}.$
(5.3)

Take $n_{0}\in\mathbb{N}$ such that, for every $n\geq n_{0},$
$|f_{n}(\omega)|<\frac{\varepsilon(1-s_{0})}{2}$ for every $|\omega|\leq
s_{0}.$ For $z\in\mathbb{D}$ and $n\geq n_{0}$ we also have

$\left|\int_{0}^{s_{0}}\frac{f_{n}(t)}{1-tz}d\mu(t)\right|\leq\int_{0}^{s_{0}}\frac{\max_{\omega\in
s_{0}\overline{\mathbb{D}}}|f_{n}(\omega)|}{1-s_{0}}d\mu(t)<\frac{\varepsilon}{2}.$
(5.4)

Putting together (5.2), (5.3) and (5.4) we conclude that
$\|H_{\mu}(f_{n})\|_{\infty}<\varepsilon$ for $n\geq n_{0}$, and so,
$H_{\mu}:H^{\infty}\rightarrow H^{\infty}$, is compact. ∎

Theorem 24 permits us to complement [17, Theorem 1.2] with more equivalent
conditions.

###### Theorem 25

Let $\mu$ be a positive finite Borel measure on $[0,1).$ The following
conditions are equivalent:

  * (i)

$H_{\mu}:H^{\infty}\rightarrow H^{\infty}$ is continuous.

  * (ii)

$H_{\mu}:H^{\infty}\rightarrow H^{\infty}$ is compact.

  * (iii)

$\int_{0}^{1}\frac{d\mu(t)}{1-t}<\infty.$

  * (iv)

$\sum_{n=0}^{\infty}\mu_{n}<\infty.$

  * (v)

$H_{\mu}:A(\mathbb{D})\rightarrow A(\mathbb{D})$ is continuous.

  * (vi)

$H_{\mu}:A(\mathbb{D})\rightarrow A(\mathbb{D})$ is compact.

If the assertions hold, then we get $H_{\mu}=I_{\mu}$ on $H^{\infty}$.

###### Proof.

The equivalence between (i), (iii) and (iv) is given in [17, Proof of Theorem
1.2]. Theorem 24 yields that (iv) implies (ii), and, since (ii) implies (i) we
have the equivalence of the first four statements. Since $A(\mathbb{D})$ is a
subspace of $H^{\infty}$, we get that (i) implies (v) if we show that (iv)
yields $H_{\mu}(A(\mathbb{D}))\subseteq A(\mathbb{D})$. For each
$k\in\mathbb{N}_{0}$, $H_{\mu}(z^{k})=\sum_{n=0}^{\infty}\mu_{n+k}z^{n}\in
A(\mathbb{D})$. We conclude since the polynomials $\mathcal{P}$ are dense in
$A(\mathbb{D})$. Finally, (v) implies (iv) since
$H_{\mu}(1)(1)=\sum_{n=0}^{\infty}\mu_{n}$. Trivially, (vi) implies (v), and
(v) and (ii) imply (vi). ∎

In what follows, we study the compactness and nuclearity of the Hilbert-type
operator on different spaces of holomorphic functions on which the operator is
well-defined under the assumption $\sum_{n=0}^{\infty}\mu_{n}<\infty.$ We
recall that given two Banach spaces $X$ and $Y$, we say that an operator
$T:X\rightarrow Y$ is _nuclear_ if there exist $(x_{n}^{*})_{n}\subseteq
X^{*}$ and $(y_{n})_{n}\subseteq Y$ such that
$\sum_{n=1}^{\infty}\|x_{n}^{*}\|\|y_{n}\|<\infty$ and
$T=\sum_{n=1}^{\infty}x_{n}^{*}\otimes
y_{n}=\sum_{n=1}^{\infty}x_{n}^{*}(\cdot)y_{n}.$ Nuclear operators are
compact.

In the rest of this section, we consider the following general spaces of
analytic functions:

$\ell^{p}_{A}:=\\{f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H(\mathbb{D}):\
\|f\|^{p}_{p}:=\sum_{n=0}^{\infty}|a_{n}|^{p}<\infty\\}\text{ for $1\leq
p<\infty.$}$ $\ell^{\infty}_{A}:=\\{f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in
H(\mathbb{D}):\ \|f\|_{\infty}:=\sup_{n}|a_{n}|<\infty\\}.$

These spaces have been thoroughly investigated in [10]. The most relevant
examples are the Wiener algebra (for $p=1$) $\ell^{1}_{A}=A(\mathbb{T})$, and
(for $p=2$) the classical Hardy space $\ell^{2}_{A}=H^{2}$. In the rest of the
paper we identify a function
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in\ell^{p}_{A},$ $1\leq p\leq\infty,$ with
its coefficients $(a_{n})_{n}$.

###### Proposition 26

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\sum_{n=0}^{\infty}\mu_{n}<\infty$. Then
$H_{\mu}:\ell^{\infty}_{A}\to\ell^{\infty}_{A}$ is well defined and
$H_{\mu}=I_{\mu}$ on $\ell^{\infty}_{A}$.

###### Proof.

If $M=\sum_{n=0}^{\infty}\mu_{n}$, then it is immediate that
$\|H_{\mu}((a_{n})_{n})\|_{\infty}\leq M\|(a_{n})_{n}\|_{\infty}$ for all
$(a_{n})_{n}\in l_{\infty}$. $I_{\mu}:\ell^{\infty}_{A}\to H(\mathbb{D})$ is
well defined by Proposition 4. Indeed, for
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ with $(a_{n})_{n}\in l_{\infty}$, we
consider $g(z)=\sum_{n=0}^{\infty}|a_{n}|z^{n}$. Then
$\int_{0}^{1}g(t)d\mu(t)\leq\|(a_{n})\|_{\infty}\sum_{n=0}^{\infty}\mu_{n}$. ∎

###### Lemma 27

Let $T:\ell^{\infty}_{A}\to\ell^{\infty}_{A}$, be given by
$x\mapsto\sum_{k=0}^{\infty}\langle y_{k},x\rangle e_{k}$, with
$y_{k}\in\ell^{1}_{A}$ for each $k\in\mathbb{N}_{0}$ and
$(\|y_{k}\|_{1})_{k\in\mathbb{N}_{0}}$ bounded. Then, for $1\leq p\leq\infty$
and $p^{\prime}:=\frac{p}{p-1}$ if $1<p<\infty$, $p^{\prime}=\infty$ if $p=1,$
and $p^{\prime}=1$ if $p=\infty$, we have:

  * (a)

If $(\|y_{k}\|_{p^{\prime}})_{k\in\mathbb{N}_{0}}\in l_{1}$ then
$T(\ell^{p}_{A})\subseteq\ell^{1}_{A}$, and the restriction
$T:\ell^{p}_{A}\to\ell^{1}_{A}$ is a nuclear operator.

  * (b)

If $(\|y_{k}\|_{p^{\prime}})_{k\in\mathbb{N}_{0}}\in l_{q},$ $q\neq\infty,$
then $T(\ell^{p}_{A})\subseteq\ell^{q}_{A}$ and the restriction
$T:\ell^{p}_{A}\to\ell^{q}_{A}$ is a compact operator.

###### Proof.

Observe that $(\ell^{p}_{A})^{*}=\ell^{p^{\prime}}_{A}$ if $1\leq p<\infty$
and $\ell^{1}_{A}\subseteq(\ell^{\infty}_{A})^{*}$ for the case $p=\infty$.
Assertion (a) follows from the very definition of a nuclear operator. Under
the assumptions of (b), it follows immediately that
$T(\ell^{p}_{A})\subseteq\ell^{q}_{A}$. Moreover, for the restriction
$T:\ell^{p}_{A}\rightarrow\ell^{q}_{A}$ we have
$T=\|\cdot\|-\lim_{n}\sum_{k=0}^{n}\langle y_{k},x\rangle e_{k}$, which
certainly implies the compactness of $T$. Actually, if $(y_{k})_{k}$ is a
sequence in $\ell^{p^{\prime}}_{A}$ such that $(\|y_{k}\|_{p^{\prime}})_{k}\in
l_{q}$, then for each $x\in B_{\ell^{p}_{A}}$ we get

$\left\|T(x)-\sum_{k=0}^{k_{0}}\langle y_{k},x\rangle
e_{k}\right\|_{q}\leq\left(\sum_{k=k_{0}+1}^{\infty}\|y_{k}\|_{p^{\prime}}^{q}\right)^{\frac{1}{q}}.$

∎

For a non increasing sequence $(\mu_{n})_{n\in\mathbb{N}_{0}}$ of positive
numbers such that $\sum_{n=0}^{\infty}\mu_{n}<\infty$, we consider the Hilbert
and Cesàro type operators defined on $\ell^{\infty}_{A}$, i.e.
$H_{\mu}:\ell^{\infty}_{A}\to\ell^{\infty}_{A}$,
$H_{\mu}((a_{n})_{n\in\mathbb{N}_{0}})=(\sum_{n=0}^{\infty}\mu_{n+k}a_{n})_{k\in\mathbb{N}_{0}}$,
$C_{\mu}:\ell^{\infty}_{A}\to\ell^{\infty}_{A}$,
$C_{\mu}((a_{n})_{n\in\mathbb{N}_{0}})=(\mu_{k}\sum_{n=0}^{k}a_{n})_{k\in\mathbb{N}_{0}}$.
These operators are well defined. We see that both satisfy the hypothesis of
Lemma 27. In fact, taking $y_{k}^{h}=(\mu_{n+k})_{n\in\mathbb{N}_{0}}$ for
$H_{\mu}$ and $y^{c}_{k}=(a^{k}_{j})_{j\in\mathbb{N}_{0}}$, where
$a^{k}_{j}=\mu_{k}$ if $j\leq k$ and $a_{j}^{k}=0$ for $j>k$, we get

$\|y_{k}^{h}\|_{1}\leq\sum_{n=0}^{\infty}\mu_{n}\quad\text{for all
}k\in\mathbb{N}_{0}$

and

$\|y_{k}^{c}\|_{1}\leq\sup\\{(n+1)\mu_{n}:\
n\in\mathbb{N}_{0}\\}<\infty\quad\text{for all }k\in\mathbb{N}_{0}.$

For the last estimate we have used the fact that when
$(\mu_{n})_{n\in\mathbb{N}_{0}}$ is a decreasing sequence of positive numbers
such that $\sum_{n=0}^{\infty}\mu_{n}<\infty,$ then $\lim_{n}n\mu_{n}=0$.

###### Proposition 28

Let $(\mu_{n})_{n\in\mathbb{N}_{0}}$ be a non increasing sequence of positive
numbers such that $\sum_{n=0}^{\infty}\mu_{n}<\infty$. The restrictions
$H_{\mu}^{1}:\ell^{1}_{A}\to\ell^{1}_{A}$ and
$C_{\mu}^{1}:\ell^{1}_{A}\to\ell^{1}_{A}$ of the Hilbert and Cesàro type
operators are nuclear operators. Moreover, the operator
$H_{\mu}:\ell^{\infty}_{A}\to\ell^{\infty}_{A}$ satisfies
$H_{\mu}=(H_{\mu}^{1})^{*}$, and it is also a compact operator.

###### Proof.

The result for the restrictions follows from Lemma 27 (a) for $p=1$,
$p^{\prime}=\infty$, and the equalities
$\|y_{k}^{h}\|_{\infty}=\|y_{k}^{c}\|_{\infty}=\mu_{k},$ $k\in\mathbb{N}_{0}.$
We observe to conclude that $(H_{\mu}^{1})^{*}=H_{\mu}$. In fact,
$(H_{\mu}^{1})^{*}$ is $w^{*}-w^{*}$ continuous,
$\langle(H_{\mu}^{1})^{*}(e_{j}),e_{n}\rangle=\langle
e_{j},H_{\mu}^{1}(e_{n})\rangle=\langle y_{j}^{h},e_{n}\rangle=\mu_{j+n}$ and
$\langle H_{\mu}(e_{j}),e_{n}\rangle=\langle
y_{n}^{h},e_{j}\rangle=\mu_{j+n}$. ∎

###### Proposition 29

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\sum_{n=0}^{\infty}\mu_{n}<\infty$ and with $1\leq p<\infty$ satisfying
$(\|(\mu_{n+k})_{n}\|_{p})_{k}\in l_{1}$. The Hilbert-type operator
$H_{\mu}:H^{\infty}\to A(\mathbb{T})$ is a well defined compact linear
operator.

###### Proof.

Since $H^{\infty}$ is a Grothendieck space and $A(\mathbb{T})$ is separable
and has the Schur property, we only need to show that $H_{\mu}$ is well
defined. If $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H^{\infty}$, then
$(a_{n})_{n}\in l_{q}$ for each $q>1$. In particular, $(a_{n})_{n}\in
l_{p^{\prime}}$. We have
$H_{\mu}(f)=\sum_{k=0}^{\infty}(\sum_{n=0}^{\infty}\mu_{n+k}a_{n})z^{k}$. For
each $k\in\mathbb{N}_{0}$,

$\left|\sum_{n=0}^{\infty}\mu_{n+k}a_{n}\right|\leq\|(a_{n}\mu_{n+k})_{n}\|_{1}\leq\|(a_{n})_{n}\|_{p^{\prime}}\|(\mu_{n+k})_{n}\|_{p},$

and we conclude from the hypothesis. ∎

###### Theorem 30

Let $\mu$ be a positive finite Borel measure on $[0,1)$. The following are
equivalent:

  * (i)

The operators $H_{\mu}:A(\mathbb{T})\to A(\mathbb{T})$ and
$C_{\mu}:A(\mathbb{T})\to A(\mathbb{T})$ are well defined.

  * (ii)

The operators $H_{\mu}:A(\mathbb{T})\to A(\mathbb{T})$ and
$C_{\mu}:A(\mathbb{T})\to A(\mathbb{T})$ are nuclear.

  * (iii)

$\sum_{n=0}^{\infty}\mu_{n}<\infty$.

###### Proof.

We use the identification $A(\mathbb{T})=\ell^{1}_{A}$. If the operators are
well defined, then $H_{\mu}(1)=C_{\mu}(1)=\sum_{n=0}^{\infty}\mu_{n}z^{n}$,
and we have (i) implies (iii). Conversely, if we assume (iii), we get (ii)
from Proposition 28. ∎

###### Proposition 31

Let $(\mu_{n})_{n\in\mathbb{N}_{0}}$ be a non increasing sequence of positive
numbers such that $\mu_{n}=O(\frac{1}{n^{r}})$ for some $r>1$. For each $1\leq
q<\infty$ there is $p_{0}$ such that
$H_{\mu}(\ell^{p}_{A})\subseteq\ell^{q}_{A}$ and
$C_{\mu}(\ell^{p}_{A})\subseteq\ell^{q}_{A}$ for each $1<p<p_{0}$. Moreover,
the corresponding restrictions $H_{\mu}:\ell^{p}_{A}\to\ell^{q}_{A}$ and
$C_{\mu}:\ell^{p}_{A}\to\ell^{q}_{A}$ are nuclear if $q=1$ and compact if
$q>1$. For the particular case $p=q=2$ we get
$H_{\mu}:\ell^{2}_{A}\to\ell^{2}_{A}$ and
$C_{\mu}:\ell^{2}_{A}\to\ell^{2}_{A}$ are well defined compact operators, and
they are nuclear if $r>\frac{3}{2}$.

###### Proof.

Let $M>0$ such that $\mu_{n}\leq\frac{M}{(n+1)^{r}},$ $n\in\mathbb{N}_{0}$.
For $H_{\mu}(x):=\sum_{k=0}^{\infty}\langle y_{k}^{h},x\rangle e_{k}$,
$y_{k}^{h}=(\mu_{n+k})_{n\in\mathbb{N}_{0}}$, $k\in\mathbb{N},$ and
$p^{\prime}=\frac{p}{p-1}$, $p>1,$ we get

$\|y_{k}^{h}\|_{p^{\prime}}\leq\left(\sum_{n=0}^{\infty}\frac{M^{p^{\prime}}}{(n+k+1)^{rp^{\prime}}}\right)^{\frac{1}{p^{\prime}}}\leq
M\left(\int_{k}^{\infty}\frac{1}{x^{rp^{\prime}}}dx\right)^{\frac{1}{p^{\prime}}}=C\frac{1}{k^{r-\frac{1}{p^{\prime}}}},$
(5.5)

where $C$ depends on $r$ and $p^{\prime}$ but not on $k$.

For the Cesàro operator we get, for some $C$ which does not depend on $k$,

$\|y_{k}^{c}\|_{p^{\prime}}\leq(k+1)^{\frac{1}{p^{\prime}}}\mu_{k}\leq
C\frac{1}{k^{r-\frac{1}{p^{\prime}}}}.$ (5.6)

Hence $(\|y_{k}^{h}\|_{p^{\prime}})_{k}\in l_{q}$ and
$(\|y_{k}^{c}\|_{p^{\prime}})_{k}\in l_{q}$ whenever

$q\left(r-\frac{1}{p^{\prime}}\right)=q\left(r-\frac{p-1}{p}\right)>1.$ (5.7)

Since $r>1$ and $q\geq 1$, the above inequality holds for $p>1$ close enough
to 1. Now the result follows directly from Lemma 27. For the statements about
$\ell^{2}_{A}$, i.e $p=p^{\prime}=2$, we observe that the estimate (5.7) is
satisfied for every $r>1$ if $q=2$. For $q=1$ the estimate is satisfied when
$r>\frac{3}{2}$. ∎

###### Theorem 32

Let $\mu$ be an $r$-Carleson measure on $[0,1)$ for some $r>1$.

  * (i)

$H_{\mu}:H^{\infty}\to A(\mathbb{T})$ and $C_{\mu}:H^{\infty}\to
A(\mathbb{T})$ are both well defined nuclear operators.

  * (ii)

$H_{\mu}:H^{2}\to H^{2}$ and $C_{\mu}:H^{2}\to H^{2}$ are both well defined
compact operators. $H_{\mu}$ is self-adjoint.

  * (iii)

If $r>\frac{3}{2},$ then $H_{\mu}:H^{2}\to A(\mathbb{T})$ and
$C_{\mu}:H^{2}\to A(\mathbb{T})$ are both well defined nuclear operators.

###### Proof.

From the continuous inclusion $H^{\infty}\hookrightarrow\ell^{p}_{A}$ for each
$p>1$, we get $(i)$, considering the composition of the inclusion and
$H_{\mu}:\ell^{p}_{A}\to\ell^{1}_{A}$ for some convenient $p$ close enough to
1 to ensure that $H_{\mu}$ is nuclear as a consequence of Proposition 31. The
identification $H^{2}=\ell^{2}_{A},$ together with Proposition 31, provides
(ii) and (iii). We see $H_{\mu}$ is self-adjoint proceeding as in the proof of
Proposition 28. ∎

Acknowledgments. The authors are grateful to Prof. A. Aleman for his help
concerning the convergence of the series defining the generalized Hilbert
operator and his suggestion to use Abel summability. They also want to thank
Prof. J.A. Peláez for providing them with Example 19.

The research of José Bonet and Enrique Jordá was partially supported by the
project PID2020-119457GB-100 funded by MCIN/AEI/10.13039/501100011033 and by
“ERDF A way of making Europe”.

Statements and Declarations

  * •

Competing Interests: The authors have no relevant financial or non-financial
interests to disclose.

  * •

Availability of data and materials: Data sharing not applicable to this
article as no datasets were generated or analysed.

  * •

Authors’ contributions: All authors contributed equally to the study
conception and design. All authors read and approved the final manuscript.

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