# Cesàro operators associated with Borel measures acting on weighted spaces of
holomorphic functions with sup-norms

María J. Beltrán-Meneu111Departament de Matemàtiques, Universitat Jaume I,
<EMAIL_ADDRESS>José Bonet222Departament de Matemàtica Aplicada, Universitat
Politècnica de València<EMAIL_ADDRESS>and Enrique Jordá333Departament de
Matemàtica Aplicada, Universitat Politècnica de València<EMAIL_ADDRESS>
###### Abstract

Let $\mu$ be a positive finite Borel measure on $[0,1).$ Cesàro-type operators
$C_{\mu}$ when acting on weighted spaces of holomorphic functions are
investigated. In the case of bounded holomorphic functions on the unit disc we
prove that $C_{\mu}$ is continuous if and only if it is compact. In the case
of weighted Banach spaces of holomorphic function defined by general weights,
we give sufficient and necessary conditions for the continuity and
compactness. For standard weights, we characterize the continuity and
compactness on classical growth Banach spaces of holomorphic functions. We
also study the point spectrum and the spectrum of $C_{\mu}$ on the space of
holomorphic functions on the disc, on the space of bounded holomorphic
functions on the disc, and on the classical growth Banach spaces of
holomorphic functions. All characterizations are given in terms of the
sequence of moments $(\mu_{n})_{n\in\mathbb{N}_{0}}$. The continuity,
compactness and spectrum of $C_{\mu}$ acting on Fréchet and (LB) Korenblum
type spaces are also considered.

Keywords: Cesàro-type operators, Weighted Banach spaces of analytic functions,
Korenblum type spaces, Compactness, Spectrum.

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

##  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 of
uniform convergence on the compact subsets of $\mathbb{D}$. If $\mu$ is a
positive finite Borel measure on $[0,1)$ and $f\in H(\mathbb{D}),$
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n},$ $z\in\mathbb{D},$ the Cesàro-type
operator associated with $\mu$ is defined by

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

where $\mu_{n}:=\int_{0}^{1}t^{n}d\mu(t)$ is the n-th moment of $\mu$,
$n\in\mathbb{N}_{0}$. This operator was studied by Galanopoulos et al. in
[18]. They proved, in particular, that $C_{\mu}:\ H(\mathbb{D})\rightarrow
H(\mathbb{D})$ is continuous and has the integral representation ([18,
Proposition 1]):

$C_{\mu}(f)(z)=\int_{0}^{1}\frac{f(tz)}{1-tz}d\mu(t),\ f\in H(\mathbb{D}).$

It is a generalization of the classical Cesàro operator $C$. Indeed, if $\mu$
is the Lebesgue measure, then $C_{\mu}=C,$ with
$\mu_{n}:=\int_{0}^{1}t^{n}dt=\frac{1}{n+1},\ n\in\mathbb{N}.$

Several authors have analyzed the continuity and compactness of Cesàro-type
operators on different Banach spaces of holomorphic functions on $\mathbb{D}.$
By the closed graph theorem, when dealing with Banach spaces $X$ continuously
embedded in $H(\mathbb{D}),$ to prove the continuity of $C_{\mu}$ it suffices
to show that the operator is well defined on $X.$ Galanopoulos et al. [18]
characterized the boundedness and compactness on the Hardy spaces $H^{p},$
$1\leq p\leq\infty,$ on the weighted Bergman spaces $A_{\alpha}^{p},$
$1<p<\infty$ and $\alpha>-1$, on BMOA, and on the Bloch space $\mathcal{B},$
obtaining conditions on the positive finite Borel measure $\mu.$ Measures of
Carleson type played a basic role, relating boundedness to Carleson measures
in the case of $H^{p}$ and $A_{\alpha}^{p},$ and to logarithmic Carleson
measures in the case of BMOA, and $\mathcal{B}.$ Compactness was related to
vanishing and vanishing logarithmic Carleson measures, respectively. In our
study we deal with $s$-Carleson and vanishing $s$-Carleson measures.

Galanopoulos et al. [18] also characterized the continuity of $C_{\mu}$ on the
Banach space of bounded holomorphic functions on the disc
$H^{\infty}(\mathbb{D})$, space on which the Cesàro operator is not well
defined [15], and asked whether $C_{\mu}(H^{\infty}(\mathbb{D}))\subseteq
BMOA$. Blasco [10] solved this question in the afirmative whenever $\mu$ is a
positive Carleson measure and studied the boundedness of $C_{\mu}$ on Hardy,
$BMOA$ and Bloch spaces for complex Borel measures $\mu.$ This question has
also been solved by Bao et al. [6]. Recently, Galanopoulos et al. continue
their study of the Cesàro-type operator on different Hilbert spaces of
holomorphic functions [19], on the Dirichlet space and on analytic Besov
spaces [17].

In our work, we study the action of Cesàro-like operators on weigthed spaces
of holomorphic functions on $\mathbb{D}$ defined by sup-norms. A weight $v$ on
$\mathbb{D}$ is a radial ($v(z)=v(|z|),\ z\in\mathbb{D}$), continuous and
decreasing function with respect to $|z|$ satisfying $\lim_{r\rightarrow
1^{-}}v(r)=0$. Given a weight $v$, we consider 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. [7], [8], [24], [30] 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 $(H_{v}^{0})^{**}$
[9]. Observe that the constant weight $v\equiv 1$ does not satisfy our
assumptions. In this case, $H_{v}^{\infty}$ is the classical Banach space
$H^{\infty}(\mathbb{D})$ of bounded holomorphic functions on $\mathbb{D}$, and
$H_{v}^{0}$ reduces to $\\{0\\}$ in this case. We will state and prove some
results for $H^{\infty}(\mathbb{D})$, 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 growth
Banach spaces $A^{-\gamma}$ and $A^{-\gamma}_{0}.$ These spaces play an
important role in interpolation and sampling of holomorphic functions [22,
Chapters 4 and 5]. Shields and Williams [30] proved that
$H_{v_{\gamma}}^{\infty}$ is isomorphic to $\ell^{\infty}$ and that
$H_{v_{\gamma}}^{0}$ is isomorphic to $c_{0}$; see also [24, Theorem 1.1].
Aleman and Persson obtained the continuity of the classical operator $C$ on
the spaces $H_{v_{\gamma}}^{\infty}$ and $H_{v_{\gamma}}^{0}$ and determined
their spectrum (see [27, Theorem 4.1] and [4, Theorems 4.1 and 5.1,
Corollaries 2.1 and 5.1]). Albanese et al. [1] complemented their work
determining the norm of the operator and proved that it is not weakly compact
on $H_{v_{\gamma}}^{\infty},$ neither on $H_{v_{\gamma}}^{0}.$

Given two functions $f$ and $g$ defined in an (unbounded) interval of real
numbers, we write $f(x)=O(g(x))$ as $x\to a$ ($x\rightarrow\infty$) if there
exist a constant $C>0$ (and $M>0$) such that $|f(x)|\leq C|g(x)|$ in a
neighbourhood of $a$ ( for every $x\geq M$), and $f(x)=o(g(x))$ as $x\to a$ if
$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=0$ (here $a=\infty$ is allowed). The
expression $f(x)\lesssim g(x)$ means $f(x)=O(g(x)).$ When $f(x)\lesssim
g(x)\lesssim f(x),$ we write $f(x)\cong g(x).$ We recall that a finite
positive Borel measure $\mu$ on $[0,1)$ is an $s$-Carleson measure, $s>0,$ if

$\mu([t,1))=O((1-t)^{s}),\ 0\leq t<1,$

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
[18, Lemma 2], $\mu$ is a Carleson measure if and only if
$\mu_{n}=O(\frac{1}{n})$, $n\in\mathbb{N},$ and $\mu$ is a vanishing Carleson
measure if and only if $\mu_{n}=o(\frac{1}{n}).$

Our notation for concepts from functional analysis and operator theory is
standard. We refer the reader e.g. to [13] and [25]. 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_. For weighted Banach spaces of analytic
functions of the type considered in this paper and operators between them, see
[11].

We briefly describe the content of the article. In section 2 we study the
continuity and the compactness of $C_{\mu}$ on different spaces of holomorphic
functions on the unit disc. In the case of $H^{\infty}(\mathbb{D})$ we prove
that $C_{\mu}$ is continuous if and only if it is compact, thus complementing
[18, Theorem 4]. In the case of weighted Banach spaces of holomorphic function
defined by general weights, we give sufficient and necessary conditions for
the continuity and compactness. For standard weights, we characterize when
$C_{\mu}$ is continuous and compact on classical growth Banach spaces of
holomorphic functions. Section 3 studies the spectrum and the point spectrum
of $C_{\mu}$ on $H(\mathbb{D}),$ on $H^{\infty}(\mathbb{D}),$ and on the
classical growth Banach spaces of holomorphic functions. All characterizations
are given in terms of the sequence of moments
$(\mu_{n})_{n\in\mathbb{N}_{0}}$. The continuity, compactness and spectrum of
$C_{\mu}$ acting on Fréchet and (LB) Korenblum type spaces are considered in
Section 4. Finally, we include an Appendix devoted to give concrete examples
of sequences $(\mu_{n})_{n\in\mathbb{N}_{0}}$ which are solutions of the
Hausdorff moment problem, that is, sequences $(\mu_{n})_{n\in\mathbb{N}_{0}}$
which are the moments of a positive finite Borel measure $\mu$ supported on
$[0,1)$. These examples play a role in Sections 2 and 3.

##  2 Boundedness and compactness of $C_{\mu}$ on spaces of holomorphic
functions

###  2.1 Weighted Banach spaces of holomorphic functions

In this section we study the Cesàro-type operator $C_{\mu}$ on weighted Banach
spaces of holomorphic functions defined by general weights as well as for
$H^{\infty}(\mathbb{D})$.

In the proof of next Lemma we use the following notation: for a weight $v$,
$B^{\infty}_{v}$ and $B^{0}_{v}$ denote the closed unit ball of
$H_{v}^{\infty}$ and $H_{v}^{0}$, respectively. It is easy to see by Montel’s
theorem that $B^{\infty}_{v}$ is a compact subset of $H(\mathbb{D})$ for the
topology $\tau_{0}$ of uniform convergence on the compact subsets of
$\mathbb{D}$. Moreover, [8, Lemma 1.1] implies that $B^{0}_{v}$ is
$\tau_{0}$-dense in $B^{\infty}_{v}$ and [9, Theorem 1.1 and Example 2.1]
yields that $H^{\infty}_{v}$ and $(H^{0}_{v})^{**}$ are isometrically
isomorphic. In fact, the compact open topology $\tau_{0}$, the pointwise
topology and the weak* topology $\sigma(H^{\infty}_{v},(H^{0}_{v})^{*})$
coincide on the closed unit ball $B^{\infty}_{v}$, by Alaoglu-Bourbaki’s
theorem, and consequently on all the bounded subsets.

###### Lemma 1

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

  * (i)

$C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is continuous and
$C_{\mu}(1)=\sum_{n=0}^{\infty}\mu_{n}z^{n}\in H_{w}^{0}$.

  * (ii)

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

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

###### Proof.

(i)$\Rightarrow$ (ii) If $C_{\mu}(1)=\sum_{n=0}^{\infty}\mu_{n}z^{n}\in
H_{w}^{0},$ then $C_{\mu}(z^{N})\in H_{w}^{0}$ for every $N\in\mathbb{N}.$
Indeed, $C_{\mu}(z^{N})=C_{\mu}(1)-\sum_{n=0}^{N-1}\mu_{n}z^{n}\in H_{w}^{0}$,
as the polynomials $\mathcal{P}$ are contained in $H_{w}^{0}.$ Moreover, since
the set of polynomials $\mathcal{P}$ is densely contained in $H_{w}^{0},$
$C_{\mu}(H_{w}^{0})=C_{\mu}(\overline{\mathcal{P}})\subseteq\overline{C_{\mu}(\mathcal{P})}\subseteq
H_{w}^{0},$ where the closure is taken with respect to $\|\ \|_{w}.$ So, the
closed graph theorem yields the continuity.
(ii)$\Rightarrow$ (i) If $C_{\mu}:H_{v}^{0}\to H_{w}^{0}$ is continuous, then
it is obvious that $C_{\mu}(1)\in H_{w}^{0}.$ Moreover, there is $M>0$ such
that $C_{\mu}(B^{0}_{v})\subseteq MB^{0}_{w}$. The continuity of
$C_{\mu}:H(\mathbb{D})\rightarrow H(\mathbb{D})$ implies (with closures taken
in $(H(\mathbb{D}),\tau_{0})$)

$C_{\mu}(B^{\infty}_{v})=C_{\mu}(\overline{B^{0}_{v}})\subseteq\overline{C_{\mu}(B^{0}_{v})}\subseteq
M\overline{B^{0}_{w}}\subseteq MB^{\infty}_{w}.$

This implies that $C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is well defined
and continuous.

Now assume that these two equivalent conditions are satisfied. By our comments
before the Lemma (see [9]), $C_{\mu}^{**}:H_{v}^{\infty}\to H_{w}^{\infty}$
and $C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ are two continuous linear
operators which coincide on $H^{0}_{v}$. To see that they also coincide on
$H^{\infty}_{v}$, take $f\in B^{\infty}_{v}$. There is a sequence
$(f_{n})_{n}\in B^{0}_{v}$ which converges to $f$ for the topology $\tau_{0}$.
Since $C_{\mu}:H(\mathbb{D})\rightarrow H(\mathbb{D})$ is continuous,
$(C_{\mu}(f_{n}))_{n}$ converges to $C_{\mu}(f)$ in
$(H(\mathbb{D}),\tau_{0})$. On the other hand, $C_{\mu}^{**}$ is weak*-weak*
continuous and $(f_{n})_{n}$ converges to $f$ for the topology
$\sigma(H^{\infty}_{v},(H^{0}_{v})^{*})$. Hence, the sequence
$C_{\mu}(f_{n})=C^{**}_{\mu}(f_{n}),n=1,2,...$ converges to $C^{**}_{\mu}(f)$
in $H^{\infty}_{w}$ for the topology $\sigma(H^{\infty}_{w},(H^{0}_{w})^{*})$.
Since this topology coincides with $\tau_{0}$ on the bounded subsets of
$H^{\infty}_{w}$, we get $C^{**}_{\mu}(f)=C_{\mu}(f)$. The coincidence that we
have just proved implies the assertion about the operator norms. ∎

###### Proposition 2

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

  * (i)

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

  * (ii)

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

###### Proof.

(i) By assumption, there exists $C>0$ such that $\mu_{n}\leq\frac{C}{n+1}$ for
every $n\in\mathbb{N}$ (see [7, Lemma 2]). So, for $z\in\mathbb{D},z\neq 0,$

$w(z)|C_{\mu}(1)(z)|\leq w(z)\sum_{n=0}^{\infty}\mu_{n}|z|^{n}\leq
Cw(z)\sum_{n=0}^{\infty}\frac{|z|^{n}}{n+1}=-Cw(z)\frac{\log(1-|z|)}{|z|}.$

Now, the hypothesis on $w$ implies $\lim_{|z|\rightarrow
1^{-}}w(z)|C_{\mu}(1)(z)|=0.$

(ii) Fix $\gamma>1-s$ and take $p\in\mathbb{R}$ such that $p>1/s$ and
$(p-1)/p<\gamma$. This choice is possible since
$\lim_{p\to\frac{1}{s}}\frac{p-1}{p}=1-s$. Let $q:=p/(p-1)$. Then
$(\mu_{n})_{n}\in l_{p}$, $(z^{n})_{n}\in l_{q}$ for any $z\in\mathbb{D}$ and
$\|(z_{n})_{n}\|_{q}=\frac{1}{(1-|z|^{q})^{\frac{1}{q}}}$. We apply Hölder
inequality to get

$|C_{\mu}(1)(z)|\leq\sum_{n=0}^{\infty}\mu_{n}|z|^{n}\leq\|(\mu_{n})_{n}\|_{p}\frac{1}{(1-|z|^{q})^{\frac{1}{q}}}.$

Since $\frac{1}{q}=\frac{p-1}{p}<\gamma$, this yields
$|C_{\mu}(1)(z)|=o\left(\frac{1}{(1-|z|)^{\gamma}}\right)$ as $|z|\rightarrow
1^{-}$.

∎

###### Example 1

Let $\mu$ be a positive finite Borel measure on $[0,1)$ with
$\sum_{n=0}^{\infty}\mu_{n}=\infty$. Then
$v(z):=1/\sum_{n=0}^{\infty}\mu_{n}|z|^{n},\ z\in\mathbb{D},$ is a weight on
$\mathbb{D}$. The function
$f(z):=\sum_{n=0}^{\infty}\mu_{n}z^{n}=C_{\mu}(1)(z),\ z\in\mathbb{D},$
belongs to $H(\mathbb{D})$ and has radius of convergence $1$. Moreover, $f\in
H_{v}^{\infty}\setminus H_{v}^{0}$, since
$\max_{|z|=r}|f(z)|=\sum_{n=0}^{\infty}\mu_{n}r^{n}$ for each $0\leq r<1$. On
the other hand, $C_{\mu}:H^{\infty}(\mathbb{D})\rightarrow H_{v}^{\infty}$ is
continuous, because, for each $f\in H^{\infty}(\mathbb{D})$ and
$z\in\mathbb{D}$, we have

$|C_{\mu}(f)(z)|\leq||f||_{\infty}\int_{0}^{1}\frac{d\mu(t)}{|1-tz|}\leq||f||_{\infty}\int_{0}^{1}(\sum_{n=0}^{\infty}t^{n}|z|^{n})d\mu(t)\leq||f||_{\infty}\frac{1}{v(|z|)}.$

###### Proposition 3

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

###### Proof.

By Lemma 1, the operators $C_{\mu}^{**}$ and $C_{\mu}$ coincide on
$H_{v}^{\infty}$. Therefore $C_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{0}$ if
and only if $C^{**}_{\mu}((H_{v}^{0})^{**})\subseteq H_{w}^{0}$. This fact
holds if and only if $C_{\mu}:H_{v}^{0}\to H_{w}^{0}$ is weakly compact by
[23, 17.2.7]. Now [26, Corollary 2.5] implies that this is equivalent to the
compactness of $C_{\mu}:H_{v}^{0}\to H_{w}^{0}$. Finally, Lemma 1 also yield
that $C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is compact if and only if
$C_{\mu}:H_{v}^{0}\to H_{w}^{0}$ is compact. ∎

###### Proposition 4

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

  * (i)

If $C(v,w):=\sup_{0\leq
r<1}w(r)\int_{0}^{1}\frac{d\mu(t)}{v(tr)(1-tr)}<\infty,$ then
$C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is continuous.

  * (ii)

If $\lim_{r\to 1^{-}}w(r)\int_{0}^{1}\frac{d\mu(t)}{v(tr)(1-tr)}=0,$ then
$C_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{0}$. In particular, the condition
implies that $C_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$ and $C_{\mu}:H_{v}^{0}\to
H_{w}^{0}$ are compact.

The conditions are also necessary if the function $f:[0,1)\to\mathbb{R}_{+}$,
$r\mapsto 1/v(r)$ admits holomorphic extension $\tilde{f}$ to $\mathbb{D}$
satisfying $|\tilde{f}(z)|\leq f(|z|)$ for every $z\in\mathbb{D}.$ In this
case, $\|C_{\mu}\|=C(v,w).$

###### Proof.

Let $f\in H_{v}^{\infty}$ with $\|f\|_{v}\leq 1.$ For $0\leq r<1,$

$\sup_{z\in\mathbb{D},|z|=r}w(z)\left|C_{\mu}f(z)\right|\leq\sup_{z\in\mathbb{D},|z|=r}w(|z|)\int_{0}^{1}\frac{|f(tz)|}{(1-t|z|)}d\mu(t)\leq
w(r)\int_{0}^{1}\frac{d\mu(t)}{v(tr)(1-tr)},$ (2.1)

so the condition in (i) yields the boundedness of $C_{\mu}:H_{v}^{\infty}\to
H_{w}^{\infty}$ with $\|C_{\mu}\|\leq C(v,w),$ and the condition in (ii), that
$C_{\mu}(H_{v}^{\infty})\subseteq H_{w}^{0}$, in particular, that
$C_{\mu}(1)\in H_{w}^{0}.$ This fact, together with (i) and Lemma 1 implies
the continuity of $C_{\mu}:H_{v}^{0}\to H_{w}^{0}.$ The compactness of
$C_{\mu}:H_{v}^{0}\to H_{w}^{0}$ and $C_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$
follows by Proposition 3.

For the converses, assume the function $f:[0,1)\to\mathbb{R}_{+}$, $r\mapsto
1/v(r)$ admits holomorphic extension $\tilde{f}$ to $\mathbb{D}$ satisfying
$|\tilde{f}(z)|\leq f(|z|)$ for every $z\in\mathbb{D}.$ As
$v(|z|)|\tilde{f}(z)|\leq v(|z|)f(|z|)$ and $v(r)f(r)=1,$ $0\leq r<1,$ we get
$\|\tilde{f}\|_{v}=1.$ Moreover,

$w(r)\int_{0}^{1}\frac{d\mu(t)}{v(tr)(1-tr)}\leq\sup_{z\in\mathbb{D},|z|=r}w(z)\left|\int_{0}^{1}\frac{\tilde{f}(tz)}{(1-tz)}d\mu(t)\right|=\sup_{z\in\mathbb{D},|z|=r}w(z)|C_{\mu}(\tilde{f})|.$
(2.2)

So, if we assume $C_{\mu}(\tilde{f})\in H_{v}^{\infty},$ we get the condition
in (i) is also necessary, and analogously for (ii) if we assume
$C_{\mu}(\tilde{f})\in H_{v}^{0}.$ The conclusion about the norm of the
operator follows from (2.1) and (2.2). ∎

###### Remark 1

Taking $\tilde{f}(z)=1/(1-z)^{\gamma}$, $z\in\mathbb{D},$ we get that the
necessity in Proposition 4 applies to standard weights
$v_{\gamma}(z)=(1-|z|)^{\gamma}$, $\gamma>0.$

###### Corollary 5

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and $v,w$ be weights
on $\mathbb{D}$ such that $w(|z|)\lesssim
v(|z|)(1-|z|)^{\delta},z\in\mathbb{D}.$

  * (i)

If $\delta>1,$ then $C_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$ and
$C_{\mu}:H_{v}^{0}\to H_{w}^{0}$ are continuous and compact.

  * (ii)

If $\delta=1,$ then $C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is continuous.

  * (iii)

If $0\leq\delta<1,$ then:

    * (a)

If $\sup_{0\leq
r<1}v(r)(1-r)^{\delta}\int_{0}^{1}\frac{d\mu(t)}{v(tr)(1-tr)}<\infty,$ then
$C_{\mu}:H_{v}^{\infty}\to H_{w}^{\infty}$ is continuous.

    * (b)

If $\lim_{r\rightarrow
1^{-}}v(r)(1-r)^{\delta}\int_{0}^{1}\frac{d\mu(t)}{v(tr)(1-tr)}=0,$ then
$C_{\mu}:H_{v}^{\infty}\to H_{w}^{0}$ and $C_{\mu}:H_{v}^{0}\to H_{w}^{0}$ are
compact.

###### Remark 2

Observe that the condition in Corollary 5(iii)(a) is satisfied, for instance,
if $\int_{0}^{1}\frac{d\mu(t)}{(1-t)^{1-\delta}}<\infty,$ or equivalently, if
$C_{\mu}((1-z)^{\delta})\in H^{\infty}(\mathbb{D}).$ If $\delta=0,$ the
boundedness of the integral is equivalent to $\sum_{n}\mu_{n}<\infty$ and to
$C_{\mu}(1)\in H^{\infty}(\mathbb{D})$ ([18, Theorem 4]), and thus, this
condition implies $C_{\mu}$ is continuous on $H_{v}^{\infty}$ and on
$H_{v}^{0}$ by Lemma 1.

Now we see the condition $\sum_{n}\mu_{n}<\infty$ in Remark 2 also implies
compactness of $C_{\mu}$ on $H_{v}^{\infty}$ and on $H_{v}^{0}$. It yields the
equivalence of the boundedness and compactness of $C_{\mu}$ on
$H^{\infty}(\mathbb{D})$ for the case $v\equiv 1.$ To do this, we need a
technical lemma, which generalizes the Weak Convergence Theorem appearing in
[29, Section 2.4], widely used in the context of composition operators.

###### Lemma 6

Let $F$ be a Fréchet Montel space, and let $X,Y$ Banach spaces continuously
embedded in $F$. Assume that $T:F\to F$ is continuous. If $T(X)\subseteq Y$
and $(T(f_{n}))_{n}$ is norm convergent to $0$ for each sequence $(f_{n})_{n}$
bounded in $X$ and convergent to $0$ in $F$, then $T:X\to Y$ is a compact
operator.

###### Proof.

Let $T$ be as in the hypothesis. The closed graph theorem yields that $T:X\to
Y$ is continuous. Assume $T(B_{X})$ is not precompact. Hence, there exists
$\varepsilon>0$ and a sequence $(f_{n})_{n}\subseteq B_{X}$ such that
$\|T(f_{n})-T(f_{m})\|>\varepsilon$ for all $m\neq n$. Since $B_{X}$ is
relatively compact in $F$, we can assume, by passing to a subsequence, that
$(f_{n})_{n}$ is convergent in $F$. Hence, $\frac{1}{2}(f_{n}-f_{n+1})$ is
convergent to 0 in $F$, and we get a contradiction with the assumption, as for
each $n\in\mathbb{N}$, it holds
$\left\|T\left(\frac{1}{2}(f_{n}-f_{n+1})\right)\right\|>\frac{\varepsilon}{2}.$
∎

###### Theorem 7

Let $\mu$ be a positive finite Borel measure on $[0,1).$ If
$\sum_{n=0}^{\infty}\mu_{n}<\infty,$ then $C_{\mu}:H_{v}^{\infty}\rightarrow
H_{v}^{\infty},$ $C_{\mu}:H_{v}^{0}\to H_{v}^{0}$ and
$C_{\mu}:H^{\infty}(\mathbb{D})\rightarrow H^{\infty}(\mathbb{D})$ are
compact.

###### Proof.

Let $v$ be a weight, including also the special case $v\equiv 1$ for which
$H_{v}^{\infty}=H^{\infty}(\mathbb{D}).$ Let us assume without loss of
generality that $\mu([0,1))=1$. By Remark 2 and [18, Theorem 4], $C_{\mu}$ is
continuous on the Banach spaces considered, so by Lemma 6 we only need to show
that $(\|C_{\mu}(f_{n})\|_{v})_{n}$ is convergent to 0 whenever
$(f_{n})_{n}\subseteq B_{H_{v}^{\infty}(\mathbb{D})}$ is convergent to 0 in
the compact open topology. Let $\varepsilon>0$. The hypothesis yields
$\int_{0}^{1}\frac{d\mu(t)}{1-t}<\infty$ ([18, Theorem 4]). So, there is
$s_{0}\in(0,1)$ such that

$\int_{s_{0}}^{1}\frac{d\mu(t)}{1-t}<\frac{\varepsilon}{2}.$ (2.3)

We have that

$|C_{\mu}(f)(z)|=\left|\int_{0}^{1}\frac{f(tz)}{1-tz}d\mu(t)\right|$ (2.4)

for all $f\in H_{v}^{\infty}$, $z\in\mathbb{D}$. For $z\in\mathbb{D}$,
$t\in[s_{0},1]$ and $f\in B_{H_{v}^{\infty}}$ we have
$v(tz)|\frac{f(tz)}{1-tz}|\leq\frac{1}{1-t}$. Hence, by (2.3), for each
$n\in\mathbb{N}$ we have

$v(z)\left|\int_{s_{0}}^{1}\frac{f_{n}(tz)}{1-tz}d\mu(t)\right|<\frac{\varepsilon}{2}.$
(2.5)

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

$v(z)\left|\int_{0}^{s_{0}}\frac{f_{n}(tz)}{1-tz}d\mu(t)\right|\leq\int_{0}^{s_{0}}\frac{\max_{\omega\in
s_{0}\overline{\mathbb{D}}}v(\omega)|f_{n}(\omega)|}{1-s_{0}}d\mu(t)<\frac{\varepsilon}{2}.$
(2.6)

Putting together (2.4), (2.5) and (2.6) we conclude that
$\|C_{\mu}(f_{n})\|_{v}<\varepsilon$ for $n\geq n_{0}$, and so,
$C_{\mu}:H^{\infty}(\mathbb{D})\rightarrow H^{\infty}(\mathbb{D})$,
$C_{\mu}:H_{v}^{\infty}\rightarrow H_{v}^{\infty}$ and $C_{\mu}:H_{v}^{0}\to
H_{v}^{0}$ are compact. ∎

Theorem 7 for $H^{\infty}(\mathbb{D})$ permits us to conclude the following
Corollary which gives compactness is equivalent to continuity for
$C_{\mu}:H^{\infty}(\mathbb{D})\rightarrow H^{\infty}(\mathbb{D})$. This
continuity was characterized with several conditions in [18, Theorem 4] which
we repeat in the statement.

###### Theorem 8

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

  * (i)

$C_{\mu}:H^{\infty}(\mathbb{D})\rightarrow H^{\infty}(\mathbb{D})$ is
continuous.

  * (ii)

$C_{\mu}:H^{\infty}(\mathbb{D})\rightarrow H^{\infty}(\mathbb{D})$ is compact.

  * (iii)

$C_{\mu}(1)\in H^{\infty}(\mathbb{D}).$

  * (iv)

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

  * (v)

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

Observe that a positive finite Borel measure $\mu$ on $[0,1)$ such that
$\sum_{n}\mu_{n}<\infty$ is a vanishing Carleson measure by [21, Theorem
3.3.1].

###### Lemma 9

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and $v,w$ weights on
$\mathbb{D}.$ If $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H_{v}^{\infty}$ is
such that $a_{n}\in\mathbb{R},a_{n}>0$ for every $n\in\mathbb{N}_{0},$ then:

  * (i)

If $C_{\mu}(f)\in H_{w}^{\infty},$ then there exists $C>0$ such that

$\sup_{n}\mu_{n}\|z^{n}\|_{w}\left(\sum_{k=0}^{n}\sum_{j=0}^{k}a_{j}\right)\leq
C.$

  * (ii)

If $C_{\mu}(f)\in H_{w}^{0},$ then for every $\varepsilon>0$ there exist
$n_{0}\in\mathbb{N}$ and $r_{0}\in[0,1)$ such that

$\sup_{n\geq n_{0}}\sup_{r_{0}\leq
r<1}\mu_{n}w(r)r^{n}\left(\sum_{k=0}^{n}\sum_{j=0}^{k}a_{j}\right)\leq\varepsilon.$

###### Proof.

Let $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}\in H_{v}^{\infty}$ be as in the
hypothesis and consider
$C_{\mu}(f)=\sum_{k=0}^{\infty}\mu_{k}\left(\sum_{j=0}^{k}a_{j}\right)z^{k}\in
H(\mathbb{D}).$ As $(\mu_{k})_{k}$ and $(r^{k})_{k},$ $0\leq r<1,$ are
decreasing, given $n\in\mathbb{N}$ and $0\leq r_{0}<1$ we have

$\displaystyle\sup_{r_{0}\leq|z|<1}$ $\displaystyle
w(z)\left|\sum_{k=0}^{n}\mu_{k}\left(\sum_{j=0}^{k}a_{j}\right)z^{k}\right|\geq\sup_{r_{0}\leq
r<1}w(r)\sum_{k=0}^{n}\mu_{k}\left(\sum_{j=0}^{k}a_{j}\right)r^{k}$
$\displaystyle\geq$ $\displaystyle\sup_{r_{0}\leq
r<1}\mu_{n}w(r)r^{n}\left(\sum_{k=0}^{n}\sum_{j=0}^{k}a_{j}\right).$

Thus, if $C_{\mu}(f)\in H_{w}^{\infty},$ then the first supremum is bounded
for $r_{0}=0$ and every $n\in\mathbb{N}.$ If $C_{\mu}(f)\in H_{w}^{0},$ for
every $\varepsilon>0$ there exists $n_{0}\in\mathbb{N}$ and $r_{0}\in[0,1)$
such that it is bounded by $\varepsilon$ for every $n\geq n_{0}.$ ∎

###  2.2 Classical growth Banach spaces of holomorphic functions

In the case of weighted Banach spaces of holomorphic functions determined by
standard weights $v_{\gamma}(r)=(1-r)^{\gamma},$ $0\leq r<1,\ \gamma>0,$ we
obtain a complete characterization of the boundedness and compactness of the
operator. The following theorem also provides a characterization of
$s$-Carleson measures in terms of the boundedness of a family of integrals.
The proof of (iii) $\Rightarrow$ (iv) was stated in [12, Proposition 1]
without proof and a proof can be found in [10, Corollary 4.6]. We include it
for completeness.

###### Theorem 10

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

  * (i)

$C_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$ is
a continuous operator

  * (ii)

$C_{\mu}\left(\frac{1}{(1-z)^{\gamma}}\right)\in
H_{v_{\gamma+\delta}}^{\infty}$

  * (iii)

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

  * (iv)

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

  * (v)

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

  * (vi)

$C_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is a
continuous operator.

###### Proof.

Let us see that the first five items are equivalent:
(i)$\Rightarrow$ (ii) is trivial, as $f(z)=\frac{1}{(1-z)^{\gamma}},\
z\in\mathbb{D},$ clearly belongs to $H_{v_{\gamma}}^{\infty}.$

(ii)$\Rightarrow$ (iii) It is known 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.[18, p.12]). Hence, if $C_{\mu}\left(\frac{1}{(1-z)^{\gamma}}\right)\in
H_{v_{\gamma+\delta}}^{\infty},$ by Lemma 9 there exists $C>0$ such that

$\sup_{n}\mu_{n}\|z^{n}\|_{v_{\gamma+\delta}}\left(\sum_{k=1}^{n}\sum_{j=1}^{k}j^{\gamma-1}\right)\leq
C.$

As
$\|z^{n}\|_{v_{\gamma+\delta}}=\frac{(\gamma+\delta)^{\gamma+\delta}}{(n+\gamma+\delta)^{\gamma+\delta}}\frac{n^{n}}{(n+\gamma+\delta)^{n}},$
we get that, for every $n\in\mathbb{N},$

$\displaystyle\mu_{n}\frac{(\gamma+\delta)^{\gamma+\delta}}{(n+\gamma+\delta)^{\gamma+\delta}}\frac{n^{n}}{(n+\gamma+\delta)^{n}}\left(\sum_{k=1}^{n}\sum_{j=1}^{k}j^{\gamma-1}\right)\leq
C$ (2.8)

Observe that, if $\gamma\geq 1$ then
$\sum_{j=1}^{m}j^{\gamma-1}\geq\int_{0}^{m}x^{\gamma-1}dx=\frac{m^{\gamma}}{\gamma},$
and if $0<\gamma<1,$ then
$\sum_{j=1}^{m}j^{\gamma-1}\geq\int_{0}^{m}(x+1)^{\gamma-1}dx=\frac{(m+1)^{\gamma}-1}{\gamma},$
$m\in\mathbb{N}.$ So, $\sum_{j=1}^{m}j^{\gamma-1}\gtrsim m^{\gamma}$ for every
$\gamma>0.$ Hence,

$\sum_{k=1}^{n}\sum_{j=1}^{k}j^{\gamma-1}\gtrsim\sum_{k=1}^{n}k^{\gamma}\gtrsim
n^{\gamma+1}$

As a consequence, we get that there exists $C^{\prime}>0$ such that, for every
$n\in\mathbb{N},$

$\mu_{n}\frac{n^{\gamma+1}}{(n+\gamma+\delta)^{\gamma+\delta}}\frac{n^{n}}{(n+\gamma+\delta)^{n}}\leq
C^{\prime}$

Equivalently,

$\mu_{n}\leq
C^{\prime}\left(\frac{n+\gamma+\delta}{n}\right)^{n+\gamma+\delta}\frac{1}{n^{1-\delta}}$

Since
$\lim_{n}\left(\frac{n+\gamma+\delta}{n}\right)^{n+\gamma+\delta}=e^{\gamma+\delta},$
we get $\mu_{n}=O(\frac{1}{n^{1-\delta}}).$

(iii)$\Rightarrow$(iv) Let us see that for every $s>0,$ if
$\mu_{n}=O(\frac{1}{n^{s}}),$ then $\mu$ is an $s$-Carleson measure. Assume
there exists $C>0$ such that $\mu_{n}\leq Cn^{-s}$ for every $n\in\mathbb{N}.$
It is enough to see that $\mu([t,1))\leq C(1-t)^{s}$ for every
$t\in[\frac{1}{2},1).$ Consider $\frac{1}{2}\leq t<1$ and take
$n\in\mathbb{N},$ $n\geq 2,$ such that $1-\frac{1}{n}\leq t<1-\frac{1}{n+1}.$
Thus, there exists $C^{\prime}>0$ such that
$t^{n}\geq\left(1-\frac{1}{n}\right)^{n}\geq C^{\prime}.$ Hence,

$\mu([t,1))\leq\mu[1-\frac{1}{n},1)\leq\frac{1}{C^{\prime}}\int_{1-\frac{1}{n}}^{1}t^{n}d\mu(t)\leq\frac{C}{C^{\prime}}n^{-s}\leq
D(n+1)^{-s}\leq D(1-t)^{s}$

for some $D>0,$ as we wanted to see.

(iv)$\Rightarrow$(v) By hypothesis, there exists $C>0$ such that $\mu[t,1)\leq
C(1-t)^{1-\delta}.$ Thus, using integration by parts we get

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

and the conclusion holds.

(v)$\Rightarrow$(i) is a particular case of Proposition 4.
To prove that the first five equivalent conditions are equivalent to (vi), by
Lemma 1 we only need to show that $C_{\mu}(1)\in H_{v_{\gamma+\delta}}^{0}$
when (i) holds. By the equivalence between conditions (i) and (iii), the
condition $C_{\mu}(1)\in H_{v_{\gamma+\delta}}^{0}$ is satisfied for every
$\gamma>0$ by Proposition 2(i) when $-\gamma<\delta\leq 0$ and by Proposition
2(ii) when $\delta\in(0,1)$ (just take $s=1-\delta$ and consider
$H_{v_{\gamma+\delta}}^{0}$ instead of $H_{v_{\gamma}}^{0}$ in the
proposition). ∎

The following theorem characterize the compactness of the operator and
provides also a characterization of vanishing $s$-Carleson measures in terms
of integrals, complementing the results in [12, Proposition 1], [18, Lemma 2]
and [10, Corollary 4.6].

###### Theorem 11

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

  * (i)

$C_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is a compact
operator

  * (ii)

$C_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}$ is
a compact operator

  * (iii)

$C_{\mu}(H_{v_{\gamma}}^{\infty})\subseteq H_{v_{\gamma+\delta}}^{0}$

  * (iv)

$C_{\mu}\left(\frac{1}{(1-z)^{\gamma}}\right)\in H_{v_{\gamma+\delta}}^{0}$

  * (v)

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

  * (vi)

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

  * (vii)

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

###### Proof.

(i) $\Rightarrow$ (ii) follows from Lemma 1, since the bitranspose of
$C_{\mu}:H_{v_{\gamma}}^{0}\rightarrow H_{v_{\gamma+\delta}}^{0}$ is
$C_{\mu}:H_{v_{\gamma}}^{\infty}\rightarrow H_{v_{\gamma+\delta}}^{\infty}.$
(ii) $\Rightarrow$ (iii) follows from Proposition 3 and the equivalence
between (i) and (vi) in Theorem 10.
(iii)$\Rightarrow$(iv) is trivial, as $\frac{1}{(1-z)^{\gamma}}\in
H_{v_{\gamma}}^{\infty}$.
(iv) $\Rightarrow$ (v) We fix $\varepsilon>0$. The function
$f(z)=\frac{1}{(1-z)^{\gamma}},\ z\in\mathbb{D},$ 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.[18, p.12]). By the
proof of Theorem 10 (ii) $\Rightarrow$ (iii), there exists $c>0$ such that
$\sum_{k=1}^{n}\sum_{j=1}^{k}j^{\gamma-1}\geq cn^{\gamma+1}.$ Let
$\varepsilon^{\prime}>0$ such that

$\frac{\varepsilon^{\prime}}{c(\gamma+\delta)^{\gamma+\delta}}\sup_{n}\left(\frac{n+\gamma+\delta}{n}\right)^{n+\gamma+\delta}<\varepsilon.$

If $C_{\mu}\left(\frac{1}{(1-z)^{\gamma}}\right)\in
H_{v_{\gamma+\delta}}^{0},$ by Lemma 9(ii), there is $n_{0}\in\mathbb{N}$ and
$r_{0}\in[0,1)$ such that, for every $n\geq n_{0}$,

$\mu_{n}\left(\sup_{r_{0}\leq
r<1}v_{\gamma+\delta}(r)r^{n}\right)\left(\sum_{k=0}^{n}\sum_{j=0}^{k}j^{\gamma-1}\right)\leq\varepsilon^{\prime}.$

The supremum $\sup_{0\leq
r<1}v_{\gamma+\delta}(r)r^{n}=\|z^{n}\|_{v_{\gamma+\delta}}=\frac{(\gamma+\delta)^{\gamma+\delta}}{(n+\gamma+\delta)^{\gamma+\delta}}\frac{n^{n}}{(n+\gamma+\delta)^{n}}$
is attained in $s_{n}=\frac{n}{n+\gamma+\delta}.$ So, as $\lim_{n}s_{n}=1,$
there exists $n_{1}\in\mathbb{N},$ $n_{1}\geq n_{0},$ such that $s_{n}\geq
r_{0}$ for every $n\geq n_{1}.$ Therefore, for $n\geq n_{1}$ we get

$\mu_{n}\frac{(\gamma+\delta)^{\gamma+\delta}}{(n+\gamma+\delta)^{\gamma+\delta}}\frac{n^{n}}{(n+\gamma+\delta)^{n}}n^{\gamma+1}\leq\frac{\varepsilon^{\prime}}{c},$

which implies

$n^{1-\delta}\mu_{n}\leq\frac{\varepsilon^{\prime}}{c(\gamma+\delta)^{\gamma+\delta}}\left(\frac{n+\gamma+\delta}{n}\right)^{n+\gamma+\delta}\leq\varepsilon,$

as we wanted to see.
(v)$\Rightarrow$(vi) Let us see that for every $s>0,$ if
$\mu_{n}=o(\frac{1}{n^{s}}),$ then $\mu$ is a vanishing $s$-Carleson measure.
By hypothesis, given $\varepsilon>0$ there exists $n_{0}\in\mathbb{N},$
$n_{0}\geq 2$ such that $\mu_{n}\leq e^{-1}\varepsilon(n+1)^{-s}$ for every
$n\geq n_{0}.$ Fix $t\in[1-{\frac{1}{n_{0}}},1)$ and take $n\geq n_{0}$ such
that $1-\frac{1}{n}\leq t<1-\frac{1}{n+1}.$ We get
$t^{n}\geq\left(1-\frac{1}{n}\right)^{n}\geq e^{-1},$ and thus,

$\mu([t,1))\leq\mu[1-\frac{1}{n},1)\leq
e\int_{1-\frac{1}{n}}^{1}t^{n}d\mu(t)\leq
e\mu_{n}\leq\varepsilon(n+1)^{-s}\leq\varepsilon(1-t)^{s},$

as we wanted to see.

(vi)$\Rightarrow$(vii) Given $\varepsilon>0$, take $\varepsilon^{\prime}>0$
such that $\varepsilon^{\prime}(\gamma+1)/(\gamma+\delta)<\varepsilon/3$ and
choose $t_{0}\in[0,1)$ such that
$\mu([t,1))\leq\varepsilon^{\prime}(1-t)^{1-\delta}$ for every $t\geq t_{0}.$
As

$\displaystyle\int_{0}^{1}\frac{1}{(1-tr)^{\gamma+1}}d\mu(t)\leq\frac{1}{(1-t_{0})^{\gamma+1}}\mu[0,1)+\int_{t_{0}}^{1}\frac{1}{(1-tr)^{\gamma+1}}d\mu(t),$

integrating by parts, we get

$\displaystyle\int_{t_{0}}^{1}\frac{d\mu(t)}{(1-tr)^{\gamma+1}}$
$\displaystyle=$
$\displaystyle\frac{\mu[t_{0},1)}{(1-rt_{0})^{\gamma+1}}+r(\gamma+1)\int_{t_{0}}^{1}\frac{\mu[t,1)}{(1-rt)^{\gamma+2}}dt$
$\displaystyle\leq$
$\displaystyle\frac{\mu(t_{0},1]}{(1-t_{0})^{\gamma+1}}+\varepsilon^{\prime}r(\gamma+1)\int_{0}^{1}\frac{(1-t)^{1-\delta}}{(1-rt)^{\gamma+2}}dt$
$\displaystyle\leq$
$\displaystyle\frac{\mu(0,1]}{(1-t_{0})^{\gamma+1}}+\varepsilon^{\prime}r(\gamma+1)\int_{0}^{1}\frac{dt}{(1-rt)^{\gamma+\delta+1}}$
$\displaystyle\leq$
$\displaystyle\frac{\mu(0,1]}{(1-t_{0})^{\gamma+1}}+\varepsilon^{\prime}\frac{(\gamma+1)}{\gamma+\delta}\frac{1}{(1-r)^{\gamma+\delta}}$
$\displaystyle\leq$
$\displaystyle\frac{\mu(0,1]}{(1-t_{0})^{\gamma+1}}+\frac{\varepsilon}{3}\frac{1}{(1-r)^{\gamma+\delta}}.$

If we choose $0<r_{0}<1$ such that
$(1-r)^{\gamma+\delta}\frac{\mu(0,1]}{(1-t_{0})^{\gamma+1}}\leq\frac{\varepsilon}{3}$
for each $r_{0}\leq r<1$, (vii) is satisfied.

(vii)$\Rightarrow$(i) follows by Proposition 4(ii). ∎

###### Corollary 12

Fix $\gamma>0$ and $0<p<\gamma+1$. Let $\mu_{n}=(n+1)^{-p}$ and consider $\mu$
the probability measure which is the solution of the Hausdorff moment problem
for $(\mu_{n})_{n\in\mathbb{N}_{0}}$. Then $C_{\mu}:H_{v_{\gamma}}\to
H_{v_{\gamma+1-p}}$ is continuous, not compact, and $C_{\mu}:H_{v_{\gamma}}\to
H_{v_{\gamma+1-t}}$ is not continuous for any $p<t<\gamma+1$.

###### Proof.

It follows immediately by applying Theorems 10 and 11 for $\delta=1-p$ and
$\delta=1-t$. ∎

The existence of the probability measure satisfying the conditions in
Corollary 12 is due to Example 4 in the appendix. We remark that this
corollary, for $p=1$ shows the non compactness of the classical Cesàro
operator $C$ in any $H_{v_{\gamma}}$, result which follows from the
description of the spectrum given in [27].

##  3 The spectrum of $C_{\mu}$ on spaces of holomorphic functions

We recall that given a continuous linear operator $T$ on a locally convex
space $E,$ the _spectrum_ of $T$ is the set
$\sigma(T):=\\{\lambda\in\mathbb{C}:\ \lambda I-T\text{ is not
invertible}\\}$. When the space $E$ is Fréchet, Banach or regular (LB), the
closed graph theorem implies that $\sigma(T):=\\{\lambda\in\mathbb{C}:\
\lambda I-T\text{ is not bijective}\\}$. The point spectrum of $T$ is defined
as $\sigma_{p}(T):=\\{\lambda\in\mathbb{C}:\ \lambda I-T\text{ is not
injective}\\}$. When $E$ is Banach and $T$ is compact, the Riesz theorem
ensures that $\sigma_{p}(T)$ is countable,
$\sigma(T)=\overline{\sigma_{p}(T)}$ and
$\overline{\sigma_{p}(T)}\setminus\sigma_{p}(T)\subseteq\\{0\\}$. The set
$\rho(T)=\mathbb{C}\setminus\sigma(T)$ is called the resolvent of $T.$ If we
want to distinguish the space, we write $\sigma(T,E),$ $\sigma_{p}(T,E)$ and
$\rho(T,E).$

###  3.1 The spectrum of $C_{\mu}:H(\mathbb{D})\to H(\mathbb{D}).$

###### Proposition 13

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and let $C_{\mu}:H(\mathbb{D})\to H(\mathbb{D})$. The point
spectrum satisfies $\sigma_{p}(C_{\mu},H(\mathbb{D}))=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}$ and the eigenspace associated to $\mu_{n}$ is generated
by the function

$f_{n}(z)=\sum_{k=n}^{\infty}a_{k}z^{k},\
a_{k}=\frac{\mu_{k}\mu_{n}^{k-n-1}}{\prod_{j=n+1}^{k}(\mu_{n}-\mu_{j})},\
k\geq n+1,\ a_{n}=1.$ (3.1)

###### Proof.

If $\lambda\in\mathbb{C}$ is an eigenvalue, then there exists
$f=\sum_{n=0}^{\infty}a_{n}z^{n}\in H(\mathbb{D}),$ $f\neq 0,$ such that

$\sum_{n=0}^{\infty}\mu_{n}(\sum_{k=0}^{n}a_{k})z^{n}=C_{\mu}(f)=\lambda
f=\sum_{n=0}^{\infty}\lambda a_{n}z^{n}.$

Thus, for every $n\in\mathbb{N}$ we get $\lambda
a_{n}=\mu_{n}\sum_{k=0}^{n}a_{k},$ or equivalently,
$a_{0}(\lambda-\mu_{0})=0,$
$a_{n}(\lambda-\mu_{n})=\mu_{n}\sum_{k=0}^{n-1}a_{k},$ $n\geq 1.$ If $a_{n}=0$
for every $n\leq n_{0}-1$ and $a_{n_{0}}\neq 0,$ then we get
$a_{n_{0}}(\lambda-\mu_{n_{0}})=0,$ and thus, $\lambda=\mu_{n_{0}}.$ This
yields $\sigma_{p}(C_{\mu},H(\mathbb{D}))\subseteq\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}.$ Moreover, as $(\mu_{n})_{n}$ is strictly decreasing by
Lemma 32(iii),

$a_{n}=\frac{\mu_{n}}{\mu_{n_{0}}-\mu_{n}}\sum_{k=0}^{n-1}a_{k},\ n\geq
n_{0}+1,$ (3.2)

which implies
$\sum_{k=0}^{n}a_{k}=a_{n}\frac{\mu_{n_{0}}-\mu_{n}}{\mu_{n}}+a_{n}=\frac{\mu_{n_{0}}}{\mu_{n}}a_{n}.$
Using this fact and (3.2), we get by induction that if $f_{n_{0}}$ is an
eigenfunction associated to $\mu_{n_{0}},$ it must be of the form

$f_{n_{0}}=\sum_{n=n_{0}}^{\infty}a_{n}z^{n},\text{ where
}a_{n}=\frac{\mu_{n}\mu_{n_{0}}^{n-n_{0}-1}}{\prod_{j=n_{0}+1}^{n}(\mu_{n_{0}}-\mu_{j})}a_{n_{0}},\
n\geq n_{0}+1,\ a_{n_{0}}\in\mathbb{C}.$

Now, observe that

$\frac{a_{n}}{a_{n+1}}=\frac{\mu_{n}}{\mu_{n+1}}\frac{\mu_{n_{0}}-\mu_{n+1}}{\mu_{n_{0}}}>0,$

hence, by Lemma 32 (i) and (ii) the radius of convergence of $f_{n_{0}}$ is

$R=\lim_{n}\frac{a_{n}}{a_{n+1}}=\lim_{n}\frac{\mu_{n}}{\mu_{n+1}}\geq 1,$

which implies $f_{n_{0}}\in H(\mathbb{D})$ for every $a_{n_{0}}\in\mathbb{C}$.
As this is satisfied for every $n_{0}\in\mathbb{N}_{0},$ we get
$\sigma_{p}(C_{\mu},H(\mathbb{D}))=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}.$ ∎

In the case of the Cesàro operator $C:H(\mathbb{D})\to H(\mathbb{D}),$ we get
that the eigenspace associated to the eigenvalue $\mu_{n}=\frac{1}{n+1}$ is
generated by the eigenfunction $\frac{z^{n}}{(1-z)^{n+1}},$ as obtained in
[27].

###### Proposition 14

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:H(\mathbb{D})\to H(\mathbb{D}).$ The
following are equivalent:

  * (i)

$0\notin\sigma(C_{\mu},H(\mathbb{D})).$

  * (ii)

$C_{\mu}:H(\mathbb{D})\to H(\mathbb{D})$ is surjective.

  * (iii)

$M_{(1/\mu_{n})_{n}}:H(\mathbb{D})\to H(\mathbb{D}),$
$\sum_{n=0}^{\infty}b_{n}z^{n}\mapsto\sum_{n=0}^{\infty}\frac{b_{n}}{\mu_{n}}z^{n}$
is continuous.

  * (iv)

For every $r\in(0,1)$ there exist $s\in(r,1)$ and $C>0$ such that
$\frac{1}{\mu_{n}}\leq C\left(\frac{s}{r}\right)^{n}$ for every
$n\in\mathbb{N}.$

  * (v)

For every $\varepsilon\in(0,1)$ there exists $C>0$ such that
$\frac{1}{\mu_{n}}\leq C(1+\varepsilon)^{n}$ for every $n\in\mathbb{N}.$

###### Proof.

(i) is equivalent to (ii) because $0\notin\sigma_{p}(C_{\mu},H(\mathbb{D}))$
by Proposition 13.
(ii) $\Leftrightarrow$ (iii) Given $f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\in
H(\mathbb{D}),$ consider the problem of finding
$g_{f}(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H(\mathbb{D})$ such that

$\sum_{n=0}^{\infty}\mu_{n}(\sum_{k=0}^{n}a_{k})z^{n}=C_{\mu}(g_{f})(z)=f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}.$

This yields $\sum_{k=0}^{n}a_{k}=\frac{b_{n}}{\mu_{n}}$ for every
$n\in\mathbb{N}_{0}.$ As we can assume without loss of generality that
$\mu_{0}=1,$ we obtain

$a_{n}=\sum_{k=0}^{n}a_{k}-\sum_{k=0}^{n-1}a_{k}=\frac{b_{n}}{\mu_{n}}-\frac{b_{n-1}}{\mu_{n-1}},\
n\in\mathbb{N},\ a_{0}=b_{0}.$

As a consequence, $C_{\mu}:H(\mathbb{D})\to H(\mathbb{D})$ is surjective if
and only if

$g_{f}(z)=b_{0}+\sum_{n=1}^{\infty}\left(\frac{b_{n}}{\mu_{n}}-\frac{b_{n-1}}{\mu_{n-1}}\right)z^{n}\in
H(\mathbb{D})$ (3.3)

for every $f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\in H(\mathbb{D}).$ And this
condition is equivalent to say that the operator

$M_{(1/\mu_{n})_{n}}:H(\mathbb{D})\to H(\mathbb{D}),\
f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\mapsto
h_{f}(z)=\sum_{n=0}^{\infty}\frac{b_{n}}{\mu_{n}}z^{n}$ (3.4)

is continuous. Indeed, $g_{f}(z)=(1-z)h_{f}$.
(iii) $\Leftrightarrow$ (iv) First, observe that
$M_{(1/\mu_{n})_{n}}:H(\mathbb{D})\to H(\mathbb{D})$ is continuous if and only
if for every $r\in(0,1)$ there exist $s\in(r,1)$ and $C>0$ such that
$\sup_{n}\frac{|b_{n}|}{\mu_{n}}r^{n}\leq C\sup_{n}|b_{n}|s^{n}$ for every
$\sum_{n=0}^{\infty}b_{n}z^{n}\in H(\mathbb{D}).$ And this condition is
equivalent to (iv) (by evaluation on the monomials).
(iv) $\Rightarrow$ (v) follows easily if we take $r=\frac{1}{1+\varepsilon}.$
For the converse, given $0<r<1,$ consider $0<\varepsilon<\frac{1-r}{r},$
$\epsilon<1,$ and $s\in[r(1+\varepsilon),1).$ Now, the hypothesis yields $C>0$
such that $\frac{1}{\mu_{n}}\leq C(1+\varepsilon)^{n}\leq
C\left(\frac{s}{r}\right)^{n}$ for every $n\in\mathbb{N}.$ ∎

###### Proposition 15

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:H(\mathbb{D})\to H(\mathbb{D}).$

  * (a)

For $\lambda\in\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\},$ the operator
$C_{\mu}-\lambda I:H(\mathbb{D})\to H(\mathbb{D})$ is not injective and not
surjective. In fact, $z^{k}\notin(C_{\mu}-\mu_{k}I)(H(\mathbb{D})),$
$k\in\mathbb{N}_{0}.$

  * (b)

For $\lambda\notin\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}\cup\\{0\\},$ the
following are equivalent:

    * (i)

$\lambda\notin\sigma(C_{\mu},H(\mathbb{D})).$

    * (ii)

$C_{\mu}-\lambda I:H(\mathbb{D})\to H(\mathbb{D})$ is surjective.

    * (iii)

For every $\sum_{k=0}^{\infty}b_{k}z^{k}\in H(\mathbb{D}),$

$\sum_{k=0}^{\infty}\mu_{k}\left(\sum_{n=0}^{k-1}\frac{b_{n}}{\prod_{j=n}^{k}\left(1-\frac{\mu_{j}}{\lambda}\right)}\right)z^{k}\in
H(\mathbb{D})$ (3.5)

###### Proof.

(a) First, take $k\in\mathbb{N}_{0}$ and consider
$C_{\mu}-\mu_{k}I:H(\mathbb{D})\to H(\mathbb{D}).$ By Proposition 13 we know
that $C_{\mu}-\mu_{k}I$ is not injective. Let us see that it is not surjective
either by showing that $z^{k}$ does not belong to the image of
$C_{\mu}-\mu_{k}I.$ Indeed, if $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in
H(\mathbb{D})$ is such that $(C_{\mu}-\mu_{k}I)(f)(z)=z^{k},$ then

$\displaystyle
z^{k}=\sum_{n=0}^{\infty}\left(\mu_{n}(\sum_{j=0}^{n}a_{j})-\mu_{k}a_{n}\right)z^{n}.$

Since $(\mu_{n})_{n\in\mathbb{N}_{0}}$ is a sequence of strictly positive
numbers, there is no $(a_{n})_{n\in\mathbb{N}_{0}}\subseteq\mathbb{C}$ such
that $\mu_{n}(\sum_{j=0}^{n}a_{j})-\mu_{k}a_{n}=0$ for $n\neq k$ and
$\mu_{k}(\sum_{j=0}^{k}a_{j})-\mu_{k}a_{k}=1$. Hence $C_{\mu}-\mu_{k}I$ is not
surjective.

(b) We take $\lambda\notin\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}\cup\\{0\\},$
$f(z)=\sum_{k=0}^{\infty}b_{k}z^{k}\in H(\mathbb{D}),$ and consider the
problem of finding $g_{f}(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$ such that

$\sum_{k=0}^{\infty}\mu_{k}(\sum_{n=0}^{k}a_{n})z^{k}-\lambda\sum_{k=0}^{\infty}a_{k}z^{k}=(C_{\mu}-\lambda
I)(g_{f})(z)=f(z)=\sum_{k=0}^{\infty}b_{k}z^{k}.$

This implies $\mu_{k}\sum_{n=0}^{k}a_{n}-\lambda a_{k}=b_{k}$ for every
$k\in\mathbb{N}_{0},$ that is,
$a_{k}(\mu_{k}-\lambda)=b_{k}-\mu_{k}\sum_{n=0}^{k-1}a_{n}.$ Assuming without
loss of generality that $\mu_{0}=1$ we get

$a_{k}=\frac{b_{k}}{\mu_{k}-\lambda}-\frac{\mu_{k}}{\mu_{k}-\lambda}\sum_{n=0}^{k-1}a_{n},\
k\in\mathbb{N},\ a_{0}=\frac{b_{0}}{\mu_{0}-\lambda}.$

Inductively, for every $k\in\mathbb{N},$ we get

$\displaystyle a_{k}$ $\displaystyle=$
$\displaystyle\frac{b_{k}}{\mu_{k}-\lambda}+\mu_{k}\sum_{n=0}^{k-1}(-1)^{k-n}\frac{\lambda^{k-n-1}b_{n}}{\prod_{j=n}^{k}(\mu_{j}-\lambda)}$
(3.6) $\displaystyle=$
$\displaystyle\frac{b_{k}}{\mu_{k}-\lambda}-\frac{\mu_{k}}{\lambda^{2}}\sum_{n=0}^{k-1}\frac{b_{n}}{\prod_{j=n}^{k}(1-\frac{\mu_{j}}{\lambda})}.$

Now, observe that $\sum_{k=0}^{\infty}\frac{b_{k}}{\mu_{k}-\lambda}z^{k}\in
H(\mathbb{D})$ since $|\mu_{k}-\lambda|>\delta$ for every $k\in\mathbb{N}$ for
some $\delta>0$. Therefore, $g_{f}(z)\in H(\mathbb{D})$ if and only if
$\sum_{k=0}^{\infty}\mu_{k}\left(\sum_{n=0}^{k-1}\frac{b_{n}}{\prod_{j=n}^{k}\left(1-\frac{\mu_{j}}{\lambda}\right)}\right)z^{k}\in
H(\mathbb{D})$. This gives the equivalence (ii)$\Leftrightarrow$(iii). Observe
that (i)$\Leftrightarrow$(ii) is trivial, since
$\lambda\notin\sigma_{p}(C_{\mu},H(\mathbb{D})).$ ∎

###### Remark 3

Condition (3.5) is satisfied, for instance, if
$\sum_{n=0}^{\infty}\mu_{n}<\infty.$ Indeed, in this case the infinite product
converges, and thus the radius of convergence of the power series is the one
of $\sum_{k=0}^{\infty}\mu_{k}(\sum_{n=0}^{k}|b_{n}|)z^{k}$, which belongs to
$H(\mathbb{D})$ because $f(z)=\sum_{k=0}^{\infty}|b_{k}|z^{k}\in
H(\mathbb{D})$ and $C_{\mu}$ is continuous. In the next proposition we prove
that this assumption can be relaxed.

###### Lemma 16

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

  * (i)

Fix $n\in\mathbb{N}.$ If $C>0$ satisfies that there exists
$j_{0}\in\mathbb{N},j_{0}\geq n$ such that $\mu_{j}\leq\frac{C}{j}$ for every
$j>j_{0},$ then for every $k\in\mathbb{N},$ $k\geq n$ and
$\lambda\in\mathbb{C},$ $\lambda\notin\\{\mu_{j}:\ j\geq n\\}\cup\\{0\\},$ we
get

$\displaystyle\prod_{j=n}^{k}\left|1-\frac{\mu_{j}}{\lambda}\right|\gtrsim\frac{1}{k^{\lceil
C|Re(1/\lambda)|\rceil}},$ (3.7)

where $\lceil x\rceil$ denotes the least integer greater than or equal to
$x>0$. Here, the constant giving the lower estimate can depend on $C$ and on
$n.$

  * (ii)

Fix $n\in\mathbb{N}.$ If $D>0$ satisfies that there exists
$j_{0}\in\mathbb{N}$ $j_{0}\geq n,$ satisfying $\mu_{j}\geq\frac{D}{j}$ for
every $j>j_{0},$ then for every $k\in\mathbb{N},$ $k\geq n$ and $a>\mu_{n},$
we get

$\prod_{j=n}^{k}\left(1-\frac{\mu_{j}}{a}\right)\lesssim\left(\frac{1}{k}\right)^{\lfloor
D/a\rfloor},$

where $\lfloor x\rfloor$ denotes the biggest integer smaller than or equal to
$x>0$. Here, the constant giving the upper estimate can depend on $D$ and on
$n.$

###### Proof.

(i) Fix $n\in\mathbb{N}$ and consider $\lambda,$ $C$ and $j_{0}\in\mathbb{N},$
$j_{0}\geq\max\\{n,\lceil C|Re(1/\lambda)|\rceil\\},$ as in the hypothesis.
Then, for each $k\geq j_{0}+1,$ we get

$\displaystyle\prod_{j=n}^{k}\left|1-\frac{\mu_{j}}{\lambda}\right|$
$\displaystyle\gtrsim$
$\displaystyle\prod_{j=j_{0}+1}^{k}\left|1-\frac{\mu_{j}}{\lambda}\right|\geq\prod_{j=j_{0}+1}^{k}\left|Re\left(1-\frac{\mu_{j}}{\lambda}\right)\right|$
$\displaystyle\geq$
$\displaystyle\prod_{j=j_{0}+1}^{k}\left(1-\frac{C}{j}\left|Re\left(\frac{1}{\lambda}\right)\right|\right)\geq\prod_{j=j_{0}+1}^{k}\frac{\left(j-\lceil
C|Re(1/\lambda)|\rceil)\right)}{j}$ $\displaystyle\geq$
$\displaystyle\frac{(k-\lceil
C\left|Re\left(1/\lambda\right)\right|\rceil)!}{k!}\geq\frac{1}{k^{\lceil
C\left|Re\left(1/\lambda\right)\right|\rceil}}.$

(ii) Fix $n\in\mathbb{N}$ and consider $a$, $j_{0}$ and $D$ as in the
hypothesis (consider the minimun $j_{0}\geq n$ satisfying the condition). For
$k\geq j_{0}+1$, we get

$\displaystyle\prod_{j=n}^{k}\left(1-\frac{\mu_{j}}{a}\right)$
$\displaystyle\lesssim$
$\displaystyle\prod_{j=j_{0}+1}^{k}\left(1-\frac{\lfloor
D/a\rfloor}{j}\right)=\prod_{j=j_{0}+1}^{k}\frac{j-\lfloor
D/a\rfloor}{j}=\frac{j_{0}!}{(j_{0}-\lfloor D/a\rfloor)!}\frac{(k-\lfloor
D/a\rfloor)!}{k!}$ $\displaystyle\leq$ $\displaystyle\frac{{j_{0}}^{\lfloor
D/a\rfloor}}{(k-\lfloor D/a|\rfloor)^{\lfloor
D/a\rfloor}}\lesssim\left(\frac{1}{k}\right)^{\lfloor D/a\rfloor},$

where the constant giving the upper estimate can depend on $D$ and $n.$ ∎

###### Remark 4

In Lemma 16(i), if we assume that $\mu_{j}\leq\frac{C}{j}$ for all
$j\in\mathbb{N}$, we get
$\prod_{j=n}^{k}\left|1-\frac{\mu_{j}}{\lambda}\right|\geq M\frac{1}{k^{\lceil
C|Re(1/\lambda)|\rceil}}$, where $M>0$ does not depend on $k,$ nor on $n$.

###### Proposition 17

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:H(\mathbb{D})\to H(\mathbb{D}).$ If $\mu$
is a Carleson measure, then
$\sigma(C_{\mu},H(\mathbb{D}))\setminus\\{0\\}=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}.$

###### Proof.

By Proposition 13 we get $\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}\subseteq\sigma(C_{\mu}).$ Let us see
$\sigma(C_{\mu})\setminus\\{0\\}\subseteq\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}.$
If $\mu$ is a Carleson measure, we can find $C>0$ such that
$\mu_{j}\leq\frac{C}{j}$ for every $j\in\mathbb{N}.$ Let
$\lambda\notin\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}\cup\\{0\\}.$ By Lemma 16(i)
and Remark 4, we get $M>0$ such that, for $n\in\mathbb{N}$ and $k\geq n,$

$\displaystyle\prod_{j=n}^{k}\left|1-\frac{\mu_{j}}{\lambda}\right|\geq\frac{M}{k^{\lceil
C|Re(1/\lambda)|\rceil}}.$ (3.8)

Therefore, for every $f(z)=\sum_{k=0}^{\infty}b_{k}z^{k}\in H(\mathbb{D})$ and
every $k\in\mathbb{N}$

$\mu_{k}\left|\sum_{n=0}^{k-1}\frac{b_{n}}{\prod_{j=n}^{k}\left(1-\frac{\mu_{j}}{\lambda}\right)}\right|\leq\frac{1}{M}k^{\lceil
C|Re(1/\lambda)|\rceil}\mu_{k}\sum_{n=0}^{k-1}|b_{n}|.$

Since $\tilde{f}(z)=\sum_{k=0}^{\infty}|b_{k}|z^{k}\in H(\mathbb{D})$, and
$C_{\mu}(\tilde{f})=\sum_{k=0}^{\infty}\mu_{k}\left(\sum_{n=0}^{k}|b_{n}|\right)z^{k}\in
H(\mathbb{D}),$ the radius of convergence formula yields

$\displaystyle\sum_{k=0}^{\infty}\mu_{k}\left(\sum_{n=0}^{k-1}\frac{b_{n}}{\prod_{j=n}^{k}\left(1-\frac{\mu_{j}}{\lambda}\right)}\right)z^{k}\in
H(\mathbb{D}).$

We conclude by Proposition 15. ∎

###### Examples 18

  * (i)

If $\mu$ is a positive finite Borel measure on $[0,1)$ such that
$\mu_{n}\cong\frac{1}{n^{s}},$ $s\geq 1,$ then
$\sigma(C_{\mu},H(\mathbb{D}))=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}.$ In
particular, this is satisfied for the Cesàro operator $C.$

  * (ii)

The positive mesure $\mu$ sastisfying $\mu_{n}=\frac{1}{2^{n}},$
$n\in\mathbb{N},$ is $s$-Carleson for every $s>0,$ satisfies
$\sum_{n}\mu_{n}<\infty$ and $\sigma(C_{\mu},H(\mathbb{D}))=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}\cup\\{0\\}.$

  * (iii)

If $d\mu(t)=\exp(-\frac{\alpha}{(1-t)^{\beta}})dt,$ $\alpha,\beta>0,$ then
$\sigma(C_{\mu},H(\mathbb{D}))=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}.$ Indeed,
$\mu_{n}\sim
n^{-\frac{\beta+2}{2(\beta+1)}}\exp\left(-Bn^{\frac{\beta}{\beta+1}}\right),$
with
$B=\alpha^{\frac{1}{\beta+1}}\left(\beta^{\frac{1}{\beta+1}}+\beta^{-\frac{1}{\beta+1}}\right)$,
by [5, Lemma 4.28] and [16, Lemma 1].

###  3.2 The spectrum of $C_{\mu}:H_{v_{\gamma}}^{\infty}\to
H_{v_{\gamma}}^{\infty}.$

###### Proposition 19

Let $X\subseteq H(\mathbb{D})$ be a Banach space of analytic functions and
$\mu$ a positive finite Borel measure on $[0,1)$ such that $\mu((0,1))>0$. If
$C_{\mu}$ is bounded on $X$, then:

  * (i)

$\sigma_{p}(C_{\mu},X)\subseteq\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}.$

  * (ii)

If $X$ contains the polynomials, then $\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}\cup\\{0\\}\subseteq\sigma(C_{\mu},X).$

  * (ii)

If $X$ contains the polynomials and $C_{\mu}$ is compact on $X,$ then
$\sigma_{p}(C_{\mu},X)=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}$ and
$\sigma(C_{\mu},X)=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}\cup\\{0\\}.$

###### Proof.

(i) follows by Proposition 13. (ii) is a consequence of Proposition 15(a) (see
[14] and [27] for the result on the classical Cesàro operator). By the
compactness of $C_{\mu},$ (iii) is a direct consequence of (i) and (ii). ∎

Theorem 7 and Proposition 19 yield the following result for
$H^{\infty}(\mathbb{D})$ and for general weighted Banach spaces of holomorphic
functions.

###### Proposition 20

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and $\sum_{n=0}^{\infty}\mu_{n}<\infty.$ The operators
$C_{\mu}:H^{\infty}(\mathbb{D})\to H^{\infty}(\mathbb{D}),$
$C_{\mu}:H^{\infty}_{v}\to H^{\infty}_{v}$ and $C_{\mu}:H^{0}_{v}\to
H^{0}_{v}$ satisfy $\sigma_{p}(C_{\mu})=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}$
and $\sigma(C_{\mu})=\\{\mu_{n}:\ n\in\mathbb{N}_{0}\\}\cup\\{0\\}.$

Analogously, by Theorem 11 and Proposition 19 we get the following corollary:

###### Proposition 21

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and $\mu$ is vanishing Carleson. Then:

  * (i)

$\sigma_{p}(C_{\mu},H_{v_{\gamma}})=\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{0})=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\},$

  * (ii)

$\sigma(C_{\mu},H_{v_{\gamma}})=\sigma(C_{\mu},H_{v_{\gamma}}^{0})=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}\cup\\{0\\}$.

In particular, this is satisfied if $\mu$ is an $s$-Carleson measure, for
$s>1$.

For non necessarily vanishing Carleson measures, we get inclusions for the
point spectrum in the spirit of that obtained in [27] and [4] for the
classical Cesàro operator.

###### Proposition 22

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and let $C>0$ such that $\mu_{n}\leq\frac{C}{n}$ eventually.
For $\gamma>0$ we have:

  * (i)

$\\{\mu_{n}:\lceil\frac{C}{\mu_{n}}\rceil\leq\gamma\\}=\\{\mu_{n}:\frac{1}{(1-z)^{\lceil\frac{C}{\mu_{n}}\rceil}}\in
H_{v_{\gamma}}^{\infty}\\}\subseteq\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{\infty})\subseteq\\{\mu_{n}:n\in\mathbb{N}_{0}\\}$

  * (ii)

$\\{\mu_{n}:\lceil\frac{C}{\mu_{n}}\rceil<\gamma\\}=\\{\mu_{n}:\frac{1}{(1-z)^{\lceil\frac{C}{\mu_{n}}\rceil}}\in
H_{v_{\gamma}}^{0}\\}\subseteq\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{0})\subseteq\\{\mu_{n}:n\in\mathbb{N}_{0}\\}$

Moreover, if there exists $D>0$ (consider the maximum) such that
$\mu_{n}\geq\frac{D}{n}$ eventually, then we even get

  * (iii)

$\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{\infty})\subseteq\\{\mu_{n}:\lfloor\frac{D}{\mu_{n}}\rfloor\leq\gamma\\}=\\{\mu_{n}:\frac{1}{(1-z)^{\lfloor\frac{D}{\mu_{n}}\rfloor}}\in
H_{v_{\gamma}}^{\infty}\\}$

  * (iv)

$\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{0})\subseteq\\{\mu_{n}:\lfloor\frac{D}{\mu_{n}}\rfloor<\gamma\\}=\\{\mu_{n}:\frac{1}{(1-z)^{\lfloor\frac{D}{\mu_{n}}\rfloor}}\in
H_{v_{\gamma}}^{0}\\}$

Here, $\lceil x\rceil$ $(\text{resp.}\lfloor x\rfloor)$ denotes the least
(greatest) integer greater (smaller) than or equal to $x.$

###### Proof.

As $H_{v_{\gamma}}^{0}(\mathbb{D})\subseteq
H_{v_{\gamma}}^{\infty}(\mathbb{D})\subseteq H(\mathbb{D})$, Proposition 19(i)
yields
$\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{0})\subseteq\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{\infty})\subseteq\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\},$ and if $\mu_{n}$ is an eigenvalue, by Proposition 13
the corresponding eigenspace is generated by the function
$f_{n}(z)=\sum_{k=n}^{\infty}a_{k}z^{k},$ with $a_{n}=1$ and

$a_{k}=\frac{\mu_{k}\mu_{n}^{-1}}{\prod_{j=n+1}^{k}(1-\frac{\mu_{j}}{\mu_{n}})},\
k\geq n+1.$

By Lemma 16(i) we get
$\prod_{j=n+1}^{k}\left(1-\frac{\mu_{j}}{\mu_{n}}\right)\gtrsim\frac{1}{k^{\lceil\frac{C}{\mu_{n}}\rceil}}.$
Thus, as $a_{k}\lesssim k^{\lceil\frac{C}{\mu_{n}}\rceil-1}$ for every
$k\in\mathbb{N}_{0},$ and $g_{\alpha}(z)=\frac{1}{(1-z)^{\alpha}},\
z\in\mathbb{D},$ satisfies
$g_{\alpha}(z)=\sum_{k=0}^{\infty}c_{k}(\alpha)z^{k},\ z\in\mathbb{D},$ with
$c_{k}(\alpha)\cong k^{\alpha-1}$ (eg.[18, p.12]), we have

$\sup_{|z|=r}|f_{n}(z)|v_{\gamma}(z)\lesssim\sum_{k=n}^{\infty}k^{\lceil\frac{C}{\mu_{n}}\rceil-1}r^{k}v_{\gamma}(r)\lesssim\sup_{|z|=r}\left|\frac{1}{(1-z)^{\lceil\frac{C}{\mu_{n}}\rceil}}\right|v_{\gamma}(z).$

This yields (i) and (ii). Analogously, for $n$ fixed and $D$ as in the
hypothesis, Lemma 16(ii) yields
$\prod_{j=n+1}^{k}\left(1-\frac{\mu_{j}}{\mu_{n}}\right)\lesssim\left(\frac{1}{k}\right)^{\lfloor
D/\mu_{n}\rfloor}.$ Now, as $c_{k}(\lfloor\frac{D}{\mu_{n}}\rfloor)\lesssim
k^{\lfloor\frac{D}{\mu_{n}}\rfloor-1}\lesssim a_{k}$ for every
$k\in\mathbb{N}_{0},$ we have

$\sup_{|z|=r}|f_{n}(z)|v_{\gamma}(z)\gtrsim\sum_{k=n}^{\infty}k^{\lfloor\frac{D}{\mu_{n}}\rfloor-1}r^{k}v_{\gamma}(r)\gtrsim\sup_{|z|=r}\left|\frac{1}{(1-z)^{\lfloor\frac{D}{\mu_{n}}\rfloor}}\right|v_{\gamma}(z)$

which concludes (iii) and (iv). ∎

As an immediate consequence of Proposition 22, we get the following:

###### Corollary 23

Let $\mu$ be a Carleson measure. For every $n_{0}\in\mathbb{N}_{0}$ there
exists $\gamma>0$ such that $\\{\mu_{n}:\ n\leq n_{0},\
n\in\mathbb{N}_{0}\\}\subseteq\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{0})\subseteq\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{\infty})$.

In the next example we show that multiples of the Cesàro operator are not the
only case satisfying the two conditions in Proposition 22.

###### Example 2

Let $\mu$ be a Borel probability on $[0,1)$ satisfying
$\mu_{n}=\frac{n+2}{2(n+1)^{2}},$ $n\in\mathbb{N}_{0}$ (see Example 5). Then
$\mu_{n}\leq\frac{1/2}{n}$ for every $n\in\mathbb{N}$ and
$\mu_{n}\geq\frac{t/2}{n}$ eventually for every $t<1.$ Moreover, $\lim_{t\to
1}\left\lfloor\frac{t(n+1)^{2}}{n+2}\right\rfloor=\left\lfloor\frac{(n+1)^{2}}{n+2}\right\rfloor$
for every $n\in\mathbb{N}_{0}.$ Fix $\gamma>0$ and let
$n_{0}=\max\\{n\in\mathbb{N}:\
\left\lceil\frac{(n+1)^{2}}{n+2}\right\rceil\leq\gamma\\}$. Then Proposition
22 gives $\\{\mu_{n}:\ n\leq
n_{0}\\}\subseteq\sigma_{p}(C_{\mu},H_{v_{\gamma}}^{\infty})\subseteq\\{\mu_{n}:\
n\leq n_{0}+1\\}$.

The example above shows that when $\mu$ is a Borel probability on $[0,1)$ with
$\lim n\mu_{n}=a>0$, then the point spectrum of $C_{\mu}$ on
$H_{v_{\gamma}}^{\infty}$ is “similar” as that of the classical Cesàro
operator (see [27, Theorem 4.1] and [4, Theorems 4.1 and 5.1, Corollaries 2.1
and 5.1]).

###### Remark 5

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0.$ If $\mu$ is vanishing Carleson, as $C_{\mu}$ is compact we
know by Proposition 21 that the point spectrum of $C_{\mu}$ on the weighted
Banach spaces of holomorphic functions for standard weights consists of all
the moments of the measure. Moreover, we have:

  * (a)

For $\gamma\geq 1,$ Proposition 22 gives a direct proof of Proposition 21.
Indeed, for each $n\in\mathbb{N}_{0}$, let $\varepsilon>0$ such that
$\lceil\varepsilon/\mu_{n}\rceil=1$. Since $\mu$ is vanishing Carleson, we can
apply Proposition 22(i) for $C=\varepsilon$ and we get $\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}=\sigma_{p}(C_{\mu},H_{v_{\gamma}})$. Now, the
compactness given by Theorem 11 yields the result on the spectrum.

  * (b)

For $\gamma<1,$ the set
$\\{\mu_{n}:\lceil\frac{C}{\mu_{n}}\rceil\leq\gamma\\}$ is empty for each
$C>0$, but in any case $\sigma_{p}(C_{\mu},H_{v_{\gamma}})=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}$.

##  4 The operator $C_{\mu}$ on Korenblum Fréchet and (LB) spaces

In this section we study the Cesàro-type operator $C_{\mu}$ when it acts on
Korenblum Fréchet and (LB) spaces.

For $0<\alpha<\beta,$ then $H_{v_{\alpha}}^{\infty}\subseteq
H_{v_{\beta}}^{\infty}$ with continuous inclusions. So, given $\gamma>0$ we
can consider the projective limit

$A_{+}^{-\gamma}:=\cap_{\mu>\gamma}H_{v_{\mu}}^{\infty}=\\{f\in
H(\mathbb{D}):\
\sup_{z\in\mathbb{D}}(1-|z|)^{\gamma+\frac{1}{k}}|f(z)|<\infty\text{ for every
}k\in\mathbb{N}\\},$

which is a Fréchet Schwartz space under the norms $\|\
\|_{v_{\gamma}+\frac{1}{k}},$ $k\in\mathbb{N}.$ Analogously, given $\gamma>0,$
and $k\geq k_{0}$ with $\gamma-\frac{1}{k_{0}}>0,$ we can consider the
inductive limit

$A_{-}^{-\gamma}:=\cup_{\mu<\gamma}H_{v_{\mu}}=\\{f\in H(\mathbb{D}):\
\sup_{z\in\mathbb{D}}(1-|z|)^{\gamma-\frac{1}{k}}|f(z)|<\infty\text{ for some
}k\in\mathbb{N}\\},$

endowed with the finest locally convex topology such that all inclusions
$H_{v_{\gamma-\frac{1}{k}}}^{\infty}\subseteq A_{-}^{-\gamma}$ are continuous.
With this topology, $A_{-}^{-\gamma}$ is a complete regular (DFS)-space. If
$\gamma=\infty,$ then $A_{-}^{-\gamma}$ is the Korenblum space [22] denoted
simply by

$A^{-\infty}:=\operatorname*{ind}_{n\in\mathbb{N}}H_{v_{n}}^{\infty}=\\{f\in
H(\mathbb{D}):\ \sup_{z\in\mathbb{D}}(1-|z|)^{n}|f(z)|<\infty\text{ for some
}n\in\mathbb{N}\\}.$

Albanese et al. [2] determined the spectrum of the Cesàro operator $C$ acting
on the Korenblum type spaces $A_{+}^{-\gamma},$ $A_{-}^{-\gamma}$ and
$A^{-\infty}$ and used it to prove the operator is never compact on them.

###### Proposition 24

Let $\mu$ be a positive finite Borel measure on $[0,1).$ The operator
$C_{\mu}:A^{-\infty}\rightarrow A^{-\infty}$ is continuous.

###### Proof.

For the weights $v_{n}(z)=(1-|z|)^{n},$ $z\in\mathbb{D},$ $n\in\mathbb{N},$
the operator $C_{\mu}:H_{v_{n}}\rightarrow H_{v_{n+1}}$ is continuous for
every $n\in\mathbb{N}$ by Corollary 5. So, the continuity follows from the
continuity of operators on inductive limits. ∎

###### Proposition 25

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

  * (i)

$\mu$ is an $\varepsilon$-Carleson measure for every $0<\varepsilon<1,$

  * (ii)

$C_{\mu}:A_{+}^{-\gamma}\rightarrow A_{+}^{-\gamma}$ is continuous for every
$\gamma\geq 0,$

  * (iii)

$C_{\mu}:A_{-}^{-\gamma}\rightarrow A_{-}^{-\gamma}$ is continuous for every
$\gamma>0.$

###### Proof.

(i) $\Leftrightarrow$ (ii) It is known that $C_{\mu}$ is continuous on
$A_{+}^{-\gamma}$ if and only if for every $k\in\mathbb{N},$ there exists
$l\in\mathbb{N},$ $l\geq k$ such that
$C_{\mu}:H_{v_{\gamma+\frac{1}{l}}}^{\infty}\rightarrow
H_{v_{\gamma+\frac{1}{k}}}^{\infty}$ is continuous (see eg. [3, Lemma 2.5]).
So, a necessary condition for the continuity is that $\mu$ is a
$(1-(\frac{1}{k}-\frac{1}{l}))$-Carleson measure. As this must be satisfied
for every $k\in\mathbb{N},$ we get $\mu$ must be an $\varepsilon$-Carleson
measure for every $0<\varepsilon<1.$ Let us see the suficiency. Given
$k\in\mathbb{N},$ select $\delta>0$ with $\delta<\frac{1}{k}.$ For
$\frac{1}{l}<\frac{1}{k}-\delta,$ as $\mu$ is a $(1-\delta)$-Carleson measure,
Theorem 10 yields

$C_{\mu}:H_{v_{\gamma+\frac{1}{l}}}^{\infty}\rightarrow
H_{v_{\gamma+\frac{1}{l}+\delta}}^{\infty}$

is continuous. Since $\gamma+\frac{1}{l}+\delta<\gamma+\frac{1}{k},$ the
continuity follows.

(i) $\Leftrightarrow$ (iii) By the Grothendieck factorization theorem,
$C_{\mu}$ is continuous on $A_{-}^{-\gamma}$ if and only if for every
$k\in\mathbb{N},$ there exists $l\in\mathbb{N},$ $l\geq k$ such that
$C_{\mu}:H_{v_{\gamma-\frac{1}{k}}}^{\infty}\rightarrow
H_{v_{\gamma-\frac{1}{l}}}^{\infty}$ is continuous. Therefore, $\mu$ must be a
$(1-(\frac{1}{k}-\frac{1}{l}))$-Carleson measure for every $k\in\mathbb{N},$
i.e. (i) must be satisfied. Assume now (i) holds. Given $k\in\mathbb{N},$
select $\delta>0$ with $\delta<\frac{1}{k}$ and $l\in\mathbb{N}$ such that
$\frac{1}{l}<\frac{1}{k}-\delta.$ As $\mu$ is a $(1-\delta)$-Carleson measure,
Theorem 10 yields

$C_{\mu}:H_{v_{\gamma-\frac{1}{k}}}^{\infty}\rightarrow
H_{v_{\gamma-\frac{1}{k}+\delta}}^{\infty}$

is continuous. Since $\gamma-\frac{1}{k}+\delta<\gamma-\frac{1}{l},$ we get
the continuity. ∎

Now we investigate the spectrum of the Cesàro-type operator acting on the
Korenblum space $A^{-\infty}.$ By Proposition 13 and Corollary 23 we get:

###### Proposition 26

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:A^{-\infty}\to A^{-\infty}.$ If $\mu$ is
a Carleson measure, then, $\sigma_{p}(C_{\mu},A^{-\infty})=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}.$

###### Lemma 27

[20, Lemma 9] A function $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\in H(\mathbb{D})$
belongs to $A^{-\infty}$ if and only if there exists $k\in\mathbb{N}$ such
that $\sup_{n}\frac{|a_{n}|}{(n+k)^{k}}<\infty$.

###### Proposition 28

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:A^{-\infty}\to A^{-\infty}.$ If $\mu$ is
a Carleson measure, then the following are equivalent:

  * (i)

$0\notin\sigma(C_{\mu},A^{-\infty}).$

  * (ii)

$C_{\mu}:A^{-\infty}\to A^{-\infty}$ is surjective.

  * (iii)

$M_{(1/\mu_{n})_{n}}:A^{-\infty}\to A^{-\infty},$
$f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\mapsto
h_{f}(z)=\sum_{n=0}^{\infty}\frac{b_{n}}{\mu_{n}}z^{n}$ is continuous.

  * (iv)

There exist $C,s>0$ such that $\frac{1}{\mu_{n}}\leq Cn^{s}$ for every
$n\in\mathbb{N}.$

###### Proof.

(i) $\Leftrightarrow$ (ii) by Proposition 26.
(ii) $\Leftrightarrow$ (iii) Proceeding as in the proof of Proposition 14, we
get that $C_{\mu}:A^{-\infty}\to A^{-\infty}$ is surjective if and only if

$g_{f}(z)=b_{0}+\sum_{n=1}^{\infty}\left(\frac{b_{n}}{\mu_{n}}-\frac{b_{n-1}}{\mu_{n-1}}\right)z^{n}\in
A^{-\infty}$ (4.1)

for every $f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\in A^{-\infty}.$ And this
condition is equivalent to say that the operator

$M_{(1/\mu_{n})_{n}}:A^{-\infty}\to A^{-\infty},\
f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\mapsto
h_{f}(z)=\sum_{n=0}^{\infty}\frac{b_{n}}{\mu_{n}}z^{n}$ (4.2)

is continuous. Indeed, $g_{f}(z)=(1-z)h_{f}(z)\in A^{-\infty}$ if and only if
$h_{f}\in A^{-\infty}.$
(iii) $\Leftrightarrow$ (iv) Lemma 27 implies that
$h_{f}(z)=\sum_{n=0}^{\infty}\frac{b_{n}}{\mu_{n}}z^{n}\in A^{-\infty}$ if and
only if there exist $k\in\mathbb{N}$ and $D>0$ such that
$\frac{|b_{n}|}{\mu_{n}}\leq Dn^{k}$ for every $n\in\mathbb{N}.$ So, if this
condition is satisfied and we consider $f(z)=\frac{1}{1-z}\in A^{-\infty},$
since $b_{n}=1$ for every $n\in\mathbb{N}$ we get $\frac{1}{\mu_{n}}\leq
Dn^{k}$ for every $n\in\mathbb{N}.$ On the other hand, if there exist $s,C>0$
such that $\frac{1}{\mu_{n}}\leq Cn^{s}$ for every $n\in\mathbb{N},$ the
conclusion follows easily by Lemma 27, as $f\in A^{-\infty}$ yields the
existence of $k\in\mathbb{N}$ and $D>0$ such that $b_{n}\leq Dn^{k}$ for every
$n\in\mathbb{N}.$ Now, Lemma 27 implies (iv). ∎

###### Examples 29

Let $\mu$ be a positive finite Borel measure on $[0,1)$ and consider
$C_{\mu}:A^{-\infty}\to A^{-\infty}.$

  * (i)

If $\mu$ satisfies $\mu_{n}\cong\frac{1}{n^{s}},$ $s>0,$ then
$0\notin\sigma(C_{\mu}).$

  * (ii)

If $\mu$ satisfies $\mu_{n}=\frac{1}{2^{n}},$ $n\in\mathbb{N},$ then
$0\in\sigma(C_{\mu}).$

###### Proposition 30

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:A^{-\infty}\to A^{-\infty}.$ If $\mu$ is
a Carleson measure and
$\lambda\notin\\{\mu_{n}:n\in\mathbb{N}_{0}\\}\cup\\{0\\},$ then the operator
$C_{\mu}-\lambda I:A^{-\infty}\to A^{-\infty}$ is surjective, i.e.,
$\lambda\notin\sigma(C_{\mu},A^{-\infty})$.

###### Proof.

Let $\lambda\notin\\{\mu_{n}:n\in\mathbb{N}_{0}\\}\cup\\{0\\}.$ By the proof
of Proposition 15, it is enough to show that for every
$f(z)=\sum_{n=0}^{\infty}b_{n}z^{n}\in A^{-\infty},$ the functions
$g_{f}(z)=\sum_{n=0}^{\infty}\frac{b_{n}}{\mu_{n}-\lambda}z^{n}\in
A^{-\infty}$ and

$h_{f}(z)=\sum_{n=0}^{\infty}\mu_{n}\left(\sum_{k=0}^{n-1}\frac{b_{k}}{\prod_{j=k}^{n}\left(1-\frac{\mu_{j}}{\lambda}\right)}\right)z^{n}\in
A^{-\infty}.$

By Lemma 27, there exists $s\in\mathbb{N}$ such that
$\sup_{n}\frac{|b_{n}|}{(n+s)^{s}}<\infty.$ Therefore, as there exists
$\delta>0$ with $|\mu_{n}-\lambda|\geq\delta$ for every $n\in\mathbb{N},$ we
get $\sup_{n}\frac{|b_{n}|}{|\mu_{n}-\lambda|(n+s)^{s}}<\infty$ and $g_{f}\in
A^{-\infty}.$ Now, take $m\in\mathbb{N},\ m\geq s+\lceil
C|Re(1/\lambda)|\rceil,$ where $C>0$ is such that $\mu_{n}\leq\frac{C}{n}$ for
every $n\in\mathbb{N}.$ Now, by Lemma 16(i) and Remark 4 we get

$\displaystyle\sup_{n}\frac{1}{(n+m)^{m}}\mu_{n}\left|\sum_{k=0}^{n-1}\frac{b_{k}}{\prod_{j=k}^{n}\left(1-\frac{\mu_{j}}{\lambda}\right)}\right|$
$\displaystyle\lesssim$ $\displaystyle\sup_{n}\frac{n^{\lceil
C|Re(1/\lambda)|\rceil-1}}{(n+m)^{m}}\sum_{k=0}^{n-1}|b_{k}|$
$\displaystyle\lesssim$ $\displaystyle\sup_{n}\frac{n^{\lceil
C|Re(1/\lambda)|\rceil-1}}{(n+m)^{m}}(n+s-1)^{s+1}<\infty.$

Therefore, Lemma 27 yields $h_{f}\in A^{-\infty}$. ∎

As a consequence of Propositions 26 and 30, the following corollary holds:

###### Corollary 31

Let $\mu$ be a positive finite Borel measure on $[0,1)$ such that
$\mu((0,1))>0$ and consider $C_{\mu}:A^{-\infty}\to A^{-\infty}.$ If $\mu$ is
a Carleson measure, then
$\sigma(C_{\mu},A^{-\infty})\setminus\\{0\\}=\\{\mu_{n}:\
n\in\mathbb{N}_{0}\\}.$

### Acknowledgments

The research of the three authors was partially supported by GVA-
AICO/2021/170. The research of the second and third authors was also partially
supported by the project PID2020-119457GB-100 funded by
MCIN/AEI/10.13039/501100011033 and by “ERFD A way of making Europe”.

##  Appendix A Hausdorff moment problem for Borel probabilities on $[0,1)$

This appendix collects a few classical results about Hausdorff moment problem
for Borel probabilities on $[0,1)$. They provide us with examples which are
relevant in connection with the theorems in the article.

We start with the following result, which is known to specialists. We state it
and give its short proof, since it is very useful in the article.

###### Lemma 32

Let $\mu$ be a positive finite Borel measure on $[0,1).$ Then:

  * (i)

$(\mu_{n})_{n}$ is a decreasing sequence satisfying $\lim_{n}\mu_{n}=0.$

  * (ii)

$\left(\frac{\mu_{n}}{\mu_{n+1}}\right)_{n}$ is a decreasing sequence bounded
below by $1$, so the limit $L=\lim_{n}\frac{\mu_{n}}{\mu_{n+1}}$ exists, with
$L\geq 1.$

  * (iii)

If $\mu((0,1))>0$, then $(\mu_{n})_{n}$ is a strictly decreasing sequence.

###### Proof.

(i) $(\mu_{n})_{n}$ is cleary decreasing, so $\lim_{n}\mu_{n}=0$ by the
Monotone Convergence Theorem.
(ii) By Hölder’s inequality, we get

$\int_{0}^{1}t^{n}d\mu\leq\left(\int_{0}^{1}(t^{\frac{n-1}{2}})^{2}d\mu\right)^{1/2}\left(\int_{0}^{1}(t^{\frac{n+1}{2}})^{2}d\mu\right)^{1/2},$

therefore $\mu_{n}^{2}\leq\mu_{n-1}\mu_{n+1}$ and so,
$\left(\frac{\mu_{n}}{\mu_{n+1}}\right)_{n}$ is a decreasing sequence. As
$\frac{\mu_{n}}{\mu_{n+1}}\geq 1$ for every $n\in\mathbb{N},$ then
$\lim_{n}\frac{\mu_{n}}{\mu_{n+1}}$ exists.
(iii) By assumption, there are $a,b\in(0,1)$ such that $\mu([a,b])>0.$ So, we
can find $c_{n}>0$ such that
$\int_{0}^{1}(t^{n}-t^{n+1})d\mu\geq\int_{a}^{b}(t^{n}-t^{n+1})d\mu\geq
c_{n}\mu([a,b])>0,$ and we get $(\mu_{n})_{n}$ is a strictly decreasing
sequence. ∎

For a Borel probability measure on $[0,1)$, the condition $\mu((0,1))>0$ in
Lemma 32 is equivalent to $\mu\neq\delta_{0},$ where $\delta_{0}$ is the Dirac
measure on $[0,1)$.

The _difference sequences_ of a given sequence $(a_{n})_{n\in\mathbb{N}_{0}}$
of real numbers are defined by $\Delta^{0}(a_{n})=a_{n}$,
$\Delta^{1}(a_{n})=a_{n+1}-a_{n}$,
$\Delta^{j}(a_{n})=\Delta^{1}(\Delta^{j-1}(a_{n}))$ for $j>1$. A sequence
$(a_{n})_{n}$ is said to be _completely monotonic_ whenever
$((-1)^{j}\Delta^{j}(a_{n}))_{j\in\mathbb{N}_{0}}$ is a sequence of positive
numbers for every $n\in\mathbb{N}$.

A sequence $(\mu_{n})_{n\in\mathbb{N}_{0}}$ of positive numbers is said to be
_a solution of the Hausdorff moment problem_ if there exists a Borel
probability measure $\mu$ on $[0,1]$ such that
$\mu_{n}=\int_{0}^{1}t^{n}d\mu(t)$ for each $n\in\mathbb{N}_{0}$.

A proof of the following characterization due to Hausdorff can be seen in [28]
Theorem 3.15.

######  Theorem 33 (Hausdorff)

A sequence $(\mu_{n})_{n\in\mathbb{N}_{0}}$ of positive numbers is a sequence
of moments of a Borel probability $\mu$ in $[0,1)$ if and only $\mu_{0}=1$,
$(\mu_{n})_{n\in\mathbb{N}_{0}}$ is completely monotonic and
$\lim_{n}\mu_{n}=0$.

Hausdorff theorem is stated for measures on $[0,1]$ without the condition
$\lim_{n}\mu_{n}=0$. If $\mu$ is a probability on $[0,1]$, the condition
$\mu(\\{1\\})=0$ is easily seen to be equivalent to the condition
$\lim_{n}\mu_{n}=0$, by the Lebesgue dominated convergence theorem. Observe
that $\int_{\\{1\\}}t^{n}d\mu=\mu(\\{1\\})$ for every $n\in\mathbb{N}_{0}$.

###### Example 3

If $0<a<1,$ then $(a^{n})_{n\in\mathbb{N}_{0}}$ is the sequence of moments of
a probability on $[0,1)$, as follows immediately from the Hausdorff theorem
and the equality
$(\Delta^{j}(a^{n}))_{j\in\mathbb{N}_{0}}=((a-1)^{j}a^{n})_{j\in\mathbb{N}_{0}}$.

A function $f\in C^{\infty}([0,\infty))$ is said to be _completely monotonic_
whenever $(-1)^{j}f^{(j)}(x)\geq 0$ whenever $j\in\mathbb{N}_{0}$. The
following theorem is a consequence of the Hausdorff theorem together with [31,
Theorem 11d, p.158].

###### Theorem 34

If $f\in C^{\infty}([0,\infty))$ is a completely monotonic function such that
$f(0)=1$ and $\lim_{x\to\infty}f(x)=0,$ then $(f(n))_{n=0}^{\infty}$ is a
sequence of moment problems for a Borel probability on $[0,1)$.

###### Example 4

Let $\mu_{n}=(n+1)^{-p}$, $p>0$. The sequence $(\mu_{n})_{n\in\mathbb{N}_{0}}$
is a sequence of moments for a Borel probability $\mu$ on $[0,1)$.

###### Lemma 35

Assume that $f_{1},f_{2}\in C^{\infty}[0,\infty)$ and
$(-1)^{j}f_{i}^{(j)}(x)\geq 0$ for all $x\in[0,\infty)$, $j\in\mathbb{N}_{0}$,
$i=1,2$, $f_{1}(0)f_{2}(0)=1$, $f_{1}$ bounded and
$\lim_{x\to\infty}f_{2}(x)=0$. Then $((f_{1}(n)f_{2}(n))_{n=0}^{\infty}$ is a
sequence of moments for a Borel probability on $[0,1)$.

###### Proof.

Under these hypothesis, $\lim_{x\to\infty}f_{1}(x)f_{2}(x)=0$. By applying the
Leibnitz rule, we get

$(-1)^{n}(f_{1}f_{2})^{(n)}(x)=\sum_{j=0}^{n}\left(\begin{array}[]{c}n\\\
j\end{array}\right)(-1)^{j}f_{1}^{(j)}(x)(-1)^{n-j}f_{2}^{(n-j)}(x)\geq 0.$

The result follows now immediately from Theorem 34. ∎

The following example shows that, beyond multiples of the Cesàro operator,
there are more examples satisfying the conditions of Proposition 22.

###### Example 5

Let $\mu_{n}=\frac{n+2}{2(n+1)^{2}}$, $n\in\mathbb{N}_{0}.$ There exists a
probability $\mu$ on $[0,1)$ which is the solution of the Hausdorff moment
problem for $(\mu_{n})_{n\in\mathbb{N}_{0}}$. Indeed, it is enough to observe
that, if one takes $f_{1}(x)=1+\frac{1}{x+1}$ and $f_{2}(x)=\frac{1}{2(1+x)}$,
then $\mu_{n}=f_{1}(n)f_{2}(n)$. The conclusion holds by Lemma 35.

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