Generalized Cesàro operators in weighted Banach spaces of analytic functions with sup-norms

Abstract

An investigation is made of the generalized Cesàro operators $C_t$, for $t\in [0,1]$, when they act on the space $H({{\mathbb {D}}})$ of holomorphic functions on the open unit disc ${{\mathbb {D}}}$, on the Banach space $H^\infty$ of bounded analytic functions and on the weighted Banach spaces $H_v^\infty$ and $H_v^0$ with their sup-norms. Of particular interest are the continuity, compactness, spectrum and point spectrum of $C_t$ as well as their linear dynamics and mean ergodicity.

1 Introduction and preliminaries

The (discrete) generalized Cesàro operators $C_t$, for $t\in [0,1]$, were first investigated by Rhaly [25, 26]. The action of $C_t$ from the sequence space $\omega := {{\mathbb {C}}}^{{{\mathbb {N}}}_0}$ into itself, with ${{\mathbb {N}}}_0:=\{0,1,2,\ldots \}$, is given by

$$\begin{aligned} C_tx:=\left( \frac{t^nx_0+t^{n-1}x_1+\cdots +x_n}{n+1}\right) _{n\in {{\mathbb {N}}}_0},\quad x=(x_n)_{n\in {{\mathbb {N}}}_0}\in \omega . \end{aligned}$$

(1.1)

For $t=0$ and with $\varphi :=(\frac{1}{n+1})_{n\in {{\mathbb {N}}}_0}$ note that $C_0$ is the diagonal operator

$$\begin{aligned} D_\varphi x:= \left( \frac{x_n}{n+1}\right) _{n\in {{\mathbb {N}}}_0}, \quad x=(x_n)_{n\in {{\mathbb {N}}}_0}\in \omega , \end{aligned}$$

(1.2)

and, for $t=1$, that $C_1$ is the classical Cesàro averaging operator

$$\begin{aligned} C_1x:=\left( \frac{x_0+x_1+\cdots +x_n}{n+1}\right) _{n\in {{\mathbb {N}}}_0},\quad x=(x_n)_{n\in {{\mathbb {N}}}_0}\in \omega . \end{aligned}$$

(1.3)

The behaviour of $C_t$ on various sequence spaces has been investigated by many authors. We refer the reader to [25,26,27], to the recent papers [28, 30, 31] and to the introduction of the papers [5, 13] and the references therein. The operator $C_1$ was thoroughly investigated on weighted Banach spaces in [2]; see also [12]. Certain variants of the Cesàro operator $C_1$ are considered in [9, 16].

Our aim is to investigate the operators $C_t$, for $t\in [0,1]$, when they are suitably interpreted to act on the space $H({{\mathbb {D}}})$ of holomorphic functions on the open unit disc ${{\mathbb {D}}}:=\{z\in {{\mathbb {C}}}\,:\ |z|<1\}$, on the Banach space $H^\infty$ of bounded analytic functions and on the weighted Banach spaces $H_v^\infty$ and $H_v^0$ with their sup-norms. The space $H({{\mathbb {D}}})$ is equipped with the topology $\tau _c$ of uniform convergence on the compact subsets of ${{\mathbb {D}}}$. According to [21, §27.3(3)] the space $H({{\mathbb {D}}})$ is a Fréchet–Montel space. A family of norms generating $\tau _c$ is given, for each $0<r<1$, by

$$\begin{aligned} q_r(f):=\sup _{|z|\le r}|f(z)|,\quad f\in H({{\mathbb {D}}}). \end{aligned}$$

(1.4)

A weight v is a continuous, non-increasing function $v:[0,1)\rightarrow (0,\infty )$. We extend v to ${{\mathbb {D}}}$ by setting $v(z):=v(|z|)$, for $z\in {{\mathbb {D}}}$. Note that $v(z)\le v(0)$ for all $z\in {{\mathbb {D}}}$. Given a weight v on [0, 1), we define the corresponding weighted Banach spaces of analytic functions on ${{\mathbb {D}}}$ by

$$\begin{aligned} H_v^\infty :=\{f\in H({{\mathbb {D}}}):\, \Vert f\Vert _{\infty ,v}:=\sup _{z\in {{\mathbb {D}}}}|f(z)|v(z)<\infty \}, \end{aligned}$$

and

$$\begin{aligned} H^0_v:=\{f\in H({{\mathbb {D}}}):\, \lim _{|z|\rightarrow 1^-}|f(z)|v(z)=0\}, \end{aligned}$$

both endowed with the norm $\Vert \cdot \Vert _{\infty ,v}$. Since $\Vert f\Vert _{\infty ,v}\le v(0)\Vert f\Vert _\infty$ whenever $f\in H^\infty$, it is clear that $H^\infty \subseteq H^\infty _v$ with a continuous inclusion. If $v(z)=1$ for all $z\in {{\mathbb {D}}}$, then $H^\infty _v$ coincides with the space $H^\infty$ of all bounded analytic functions on ${{\mathbb {D}}}$ with the sup-norm $\Vert \cdot \Vert _\infty$ and $H^0_v$ reduces to $\{0\}$. Moreover, $H^\infty _v\subseteq H({{\mathbb {D}}})$ continuously. Indeed, fix $0<r<1$. Then $\frac{1}{v(0)}\le \frac{1}{v(z)}\le \frac{1}{v(r)}$ for $|z|\le r$ and so (1.4) implies that

$$\begin{aligned} q_r(f)=\sup _{|z|\le r}\frac{v(z)|f(z)|}{v(z)}\le \frac{1}{v(r)}\sup _{|z|\le r}v(z)|f(z)|\le \frac{1}{v(r)}\Vert f\Vert _{\infty , v},\quad f\in H^\infty _v. \end{aligned}$$

We refer the reader to [10] for a recent survey of such types of weighted Banach spaces and operators between them.

Whenever necessary we will identify a function $f\in H({{\mathbb {D}}})$ with its sequence of Taylor coefficients ${\hat{f}}:=({\hat{f}}(n))_{n\in {{\mathbb {N}}}_0}$ (i.e., ${\hat{f}}(n):=\frac{f^{(n)}(0)}{n!}$, for $n\in {{\mathbb {N}}}_0$), so that $f(z)=\sum _{n=0}^\infty {\hat{f}}(n)z^n$, for $z\in {{\mathbb {D}}}$. The linear map $\Phi :H({{\mathbb {D}}})\rightarrow \omega$ is defined by

$$\begin{aligned} \Phi \left( f=\sum _{n=0}^\infty {\hat{f}}(n)z^n \right) :={\hat{f}},\quad f\in H({{\mathbb {D}}}). \end{aligned}$$

It is injective (clearly) and continuous. Indeed, for each $m\in {{\mathbb {N}}}_0$,

$$\begin{aligned} r_m(x):=\max _{0\le j\le m}|x_j|,\quad x=(x_j)_{j\in {{\mathbb {N}}}_0}\in \omega , \end{aligned}$$

is a continuous seminorm in $\omega$. Fix $0<r<1$, in which case

$$\begin{aligned} r_m(\Phi (f))&=\max _{0\le j\le m}|{\hat{f}}(j)|=\max _{0\le j\le m}\left| \frac{1}{2\pi i}\int _{|z|=r}\frac{f(z)}{z^{j+1}}\, dz\right| \le \max _{0\le j\le m}\sup _{|z|=r}\frac{|f(z)|}{|z|^j}\\ {}&=\max _{0\le j\le m}\frac{1}{r^j}q_r(f)= \frac{1}{r^m}q_r(f), \end{aligned}$$

for each $f\in H({{\mathbb {D}}})$ because $\frac{1}{r^j}\le \frac{1}{r^m}$ for all $0\le j\le m$. Of course, the increasing sequence of seminorms $\{r_m\,\ m\in {{\mathbb {N}}}_0\}$ generates the topology of $\omega$.

We first provide an integral representation of the generalized Cesàro operators $C_t$ defined on $H({{\mathbb {D}}})$, for $t\in [0,1)$. So, fix $t\in [0,1)$ and define $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ by $C_tf(0):=f(0)$ and

$$\begin{aligned} C_tf(z):=\frac{1}{z}\int _0^z\frac{f(\xi )}{1-t\xi }\,d\xi ,\ z\in {{\mathbb {D}}}\setminus \{0\}, \end{aligned}$$

(1.5)

for every $f\in H({{\mathbb {D}}})$. It turns out that $C_t$ is continuous on $H({{\mathbb {D}}})$; see Proposition 2.1. Moreover, the discrete Cesàro operator $C_t:\omega \rightarrow \omega$, when restricted to the subspace $\Phi (H({{\mathbb {D}}}))\subseteq \omega$ is transferred to $H({{\mathbb {D}}})$ as follows. For a fixed $f\in H({{\mathbb {D}}})$ we have $f(\xi )=\sum _{n=0}^\infty a_n \xi ^n$, for $\xi \in {{\mathbb {D}}}$, with ${\hat{f}}=(a_n)_{n\in {{\mathbb {N}}}_0}$ its sequence of Taylor coefficients. Since $\frac{1}{1-t\xi }=\sum _{n=0}^\infty t^n\xi ^n$, for $\xi \in {{\mathbb {D}}}$, we can form the Cauchy product of the two series, thereby obtaining

$$\begin{aligned} \frac{f(\xi )}{1-t\xi }=\sum _{n=0}^\infty (\sum _{k=0}^{n}t^{n-k}a_k)\xi ^n, \quad \xi \in {{\mathbb {D}}}. \end{aligned}$$

Then (1.5) yields

$$\begin{aligned} zC_tf(z)=\int _0^z\sum _{n=0}^\infty (\sum _{k=0}^{n}t^{n-k}a_k)\xi ^n\,d\xi =\sum _{n=0}^\infty \left( \frac{t^na_0+t^{n-1}a_1+\cdots +a_n}{n+1}\right) z^{n+1},\ z\in {{\mathbb {D}}}. \end{aligned}$$

The interchange of the infinite sum and the integral is permissible by uniform convergence of the series. This shows that $C_tf\in H({{\mathbb {D}}})$ also has the series representation

$$\begin{aligned} C_tf(z)&=\sum _{n=0}^\infty \left( \frac{t^na_0+t^{n-1}a_1+\cdots +a_n}{n+1}\right) z^{n}\nonumber \\&=\sum _{n=0}^\infty \left( \frac{t^n{\hat{f}}(0)+t^{n-1}{\hat{f}}(1)+\cdots +{\hat{f}}(n)}{n+1}\right) z^{n}=\sum _{n=0}^\infty (C_t^\omega ({\hat{f}}))_nz^n, \end{aligned}$$

(1.6)

where the coefficients of the series are precisely as in (1.1). For the sake of clarity we will denote the discrete generalized Cesàro operator $C_t:\omega \rightarrow \omega$ by $C_t^\omega$ and reserve the notation $C_t$ for the operator (1.5) acting in $H({{\mathbb {D}}})$. Note that $C_0^\omega =D_\varphi$ (see (1.2)). Moreover, $C_0$ is given by $C_0f(z)=\frac{1}{z}\int _0^z f(\xi )\,d\xi$ for $z\not =0$ and $C_0f(0)=f(0)$, which is the classical Hardy operator in $H({{\mathbb {D}}})$.

The main results for $C_t$ when acting in the Fréchet space $H({{\mathbb {D}}})$ occur in Proposition 2.1 (continuity), Proposition 3.3 (non-compactness), Proposition 3.7 (spectra) and Proposition 3.8 (linear dynamics and mean ergodicity). For the analogous information concerning $C_t$ when acting in the weighted Banach spaces $H^\infty _v$ and $H^0_v$ see Proposition 2.4 and Corollary 2.5 (continuity), Proposition 2.7 (compactness), Proposition 2.8 (spectra) and Proposition 3.2 (linear dynamics and mean ergodicity).

We end this section by recalling a few definitions and some notation concerning locally convex spaces and operators between them. For further details about functional analysis and operator theory relevant to this paper see, for example, [15, 18, 20,21,22, 29].

Given locally convex Haudorff spaces X, Y (briefly, lcHs) we denote by ${{\mathcal {L}}}(X,Y)$ the space of all continuous linear operators from X into Y. If $X=Y$, then we simply write ${{\mathcal {L}}}(X)$ for ${{\mathcal {L}}}(X,X)$. Equipped with the topology of pointwise convergence on X (i.e., the strong operator topology) the lcHs ${{\mathcal {L}}}(X)$ is denoted by ${{\mathcal {L}}}_s(X)$. Equipped with the topology $\tau _b$ of uniform convergence on the bounded subsets of X the lcHs ${{\mathcal {L}}}(X)$ is denoted by ${{\mathcal {L}}}_b(X)$.

Let X be a lcHs space. The identity operator on X is denoted by I. The transpose operator of $T\in {{\mathcal {L}}}(X)$ is denoted by $T'$; it acts from the topological dual space $X':={{\mathcal {L}}}(X,{{\mathbb {C}}})$ of X into itself. Denote by $X'_\sigma$ (resp., by $X'_\beta$) the topological dual $X'$ equipped with the weak* topology $\sigma (X',X)$ (resp., with the strong topology $\beta (X',X)$); see [21, §21.2] for the definition. It is known that $T'\in {{\mathcal {L}}}(X'_\sigma )$ and $T'\in {{\mathcal {L}}}(X_\beta ')$, [22, p. 134]. The bi-transpose operator $(T')'$ of T is simply denoted by $T''$ and belongs to ${{\mathcal {L}}}((X'_\beta )'_\beta )$.

A linear map $T:X\rightarrow Y$, with X, Y lcHs’, is called compact if there exists a neighbourhood ${{\mathcal {U}}}$ of 0 in X such that $T({{\mathcal {U}}})$ is a relatively compact set in Y. It is routine to show that necessarily $T\in {{\mathcal {L}}}(X,Y)$. We recall the following well known result; see [20, Proposition 17.1.1], [22, §42.1(1)].

Lemma 1.1

Let X be a lcHs. The compact operators are a 2-sided ideal in ${{\mathcal {L}}}(X)$.

Given a lcHs X and $T\in {{\mathcal {L}}}(X)$, the resolvent set $\rho (T;X)$ of T consists of all $\lambda \in {{\mathbb {C}}}$ such that $R(\lambda ,T):=(\lambda I-T)^{-1}$ exists in ${{\mathcal {L}}}(X)$. The set $\sigma (T;X):={{\mathbb {C}}}{\setminus } \rho (T;X)$ is called the spectrum of T. The point spectrum $\sigma _{pt}(T;X)$ of T consists of all $\lambda \in {{\mathbb {C}}}$ (also called an eigenvalue of T) such that $(\lambda I-T)$ is not injective. Some authors (eg. [29]) prefer the subset $\rho ^*(T;X)$ of $\rho (T;X)$ consisting of all $\lambda \in {{\mathbb {C}}}$ for which there exists $\delta >0$ such that the open disc $B(\lambda ,\delta ):=\{z\in {{\mathbb {C}}}:\, |z-\lambda |<\delta \}\subseteq \rho (T;X)$ and $\{R(\mu ,T):\, \mu \in B(\lambda ,\delta )\}$ is an equicontinuous subset of ${{\mathcal {L}}}(X)$. Define $\sigma ^*(T;X):={{\mathbb {C}}}{\setminus } \rho ^*(T;X)$, which is a closed set with $\sigma (T;X)\subseteq \sigma ^*(T;X)$. For the spectral theory of compact operators in lcHs’ we refer to [15, 18], for linear dynamics to [6, 17] and for mean ergodic operators to [23], for example.

2 Continuity, compactness and spectrum of $C_t$

In this section we establish, for $t\in [0,1)$, the continuity of $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ as well as the continuity of $C_t$ from $H^\infty$ (resp., $H^\infty _v$) into $H^\infty$ (resp., $H^\infty _v$). The same is true for $C_t:H^0_v\rightarrow H^0_v$ whenever $\lim _{r\rightarrow 1^-}v(r)=0$. It is also shown that the bi-transpose $C_t''$ of $C_t\in {{\mathcal {L}}}(H^0_v)$ is the generalized Cesàro operator $C_t\in {{\mathcal {L}}}(H^\infty _v)$, provided that $\lim _{r\rightarrow 1^-}v(r)=0$. For such weights v it also turns out that both $C_t\in {{\mathcal {L}}}(H^0_v)$ and $C_t\in {{\mathcal {L}}}(H^\infty _v)$ are compact operators (cf. Proposition 2.7); their spectrum is identified in Proposition 2.8. Of particular interest are the standard weights $v_\gamma (z):=(1-|z|)^\gamma$, for $\gamma >0$ and $z\in {{\mathbb {D}}}$.

Proposition 2.1

For every $t\in [0,1)$ the operator $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ is continuous. Moreover, the set $\{C_t:\, t\in [0,1)\}$ is equicontinuous in ${{\mathcal {L}}}(H({{\mathbb {D}}}))$.

Proof

Fix $f\in H({{\mathbb {D}}})$. Taking into account that $C_tf(0)=f(0)$, for all $t\in [0,1)$ and, for each $r\in (0,1)$, that $\sup _{|z|\le r}|C_tf(z)|=\sup _{|z|=r}|C_tf(z)|$, the formula (1.5) implies, for each $z\in {{\mathbb {D}}}{\setminus }\{0\}$, that

$$\begin{aligned} |C_tf(z)|&=\frac{1}{|z|}\left| \int _0^z\frac{f(\xi )}{1-t\xi }d\xi \right| \le \frac{1}{|z|}|z|\max _{\xi \in [0,z]}\frac{|f(\xi )|}{|1-t\xi |}\\ {}&\le \frac{1}{1-|z|}\max _{|\xi |\le |z|}|f(\xi )|=\frac{1}{1-|z|}\max _{|\xi |=|z|}|f(\xi )|, \end{aligned}$$

because $|1-t\xi |\ge 1-t|\xi |\ge 1-|\xi |\ge 1-|z|$, for all $|\xi |\le |z|$. It follows from the previous inequality, for each $r\in (0,1)$, that

$$\begin{aligned} q_r(C_tf)=\sup _{|z|\le r}|C_tf(z)|\le \frac{1}{1-r}\sup _{|\xi |\le r}|f(\xi )|=\frac{1}{1-r}q_r(f);. \end{aligned}$$

see (1.4). This implies the result. $\square$

The following example will prove to be useful in the sequel.

Example 2.2

Consider the constant function $f_1(z):=1$, for every $z\in {{\mathbb {D}}}$, in which case $C_tf_1(0)=f_1(0)=1$ for every $t\in [0,1]$. For $t=0$, it was noted in Sect. 1 that $C_0$ is

the Hardy operator. In particular, $C_0f_1(z)=1$, for every $z\in {{\mathbb {D}}}$. For $t\in (0,1]$, note that $C_tf_1(0)=1$ and

$$\begin{aligned} C_tf_1(z)=\frac{1}{z}\int _0^z \frac{d\xi }{1-t\xi }=-\frac{1}{tz}\log (1-tz),\quad z\in {{\mathbb {D}}}{\setminus }\{0\}. \end{aligned}$$

For $t=1$ this shows, in particular, that $C_1(H^\infty )\not \subset H^\infty$, which is well known. For an investigation of the operator $C_1$ acting in $H^\infty$ we refer to [14].

Concerning $t\in (0,1)$, recall the Taylor series expansion

$$\begin{aligned} -\log (1-z)=z\sum _{n=0}^\infty \frac{z^n}{n+1},\quad z\in {{\mathbb {D}}}, \end{aligned}$$

from which it follows that

$$\begin{aligned} -\frac{\log (1-tz)}{tz}=\sum _{n=0}^\infty \frac{t^n}{n+1}z^n, \quad z\in {{\mathbb {D}}}{\setminus }\{0\}, \end{aligned}$$

with the series having radius of convergence $\frac{1}{t}>1$. The claim is that $\Vert C_tf_1\Vert _\infty =\sup _{|z|<1}|C_tf_1(z)|=-\frac{\log (1-t)}{t}$. Indeed, $C_tf_1$ is clearly holomorhic in $B(0,\frac{1}{t}):=\{\xi \in {{\mathbb {C}}}:\, |\xi |<\frac{1}{t}\}$ hence, continuous in $B(0,\frac{1}{t})$, and satisfies $C_tf_1(1)=-\frac{\log (1-t)}{t}$ with $\lim _{r\rightarrow 1^-}C_tf_1(r)=C_tf_1(1)$. On the other hand, for every $z\in {{\mathbb {D}}}{\setminus }\{0\}$ and $t\in (0,1)$ we have that

$$\begin{aligned} |C_tf_1(z)|=\left| -\frac{\log (1-tz)}{tz}\right| \le \sum _{n=0}^\infty \frac{t^n}{n+1}|z|^n\le \sum _{n=0}^\infty \frac{t^n}{n+1}=-\frac{\log (1-t)}{t}. \end{aligned}$$

This completes the proof of the claim. Observe that $\Vert C_tf_1\Vert _\infty >1$. Indeed, define $\gamma (t)=-\log (1-t)-t$, for $t\in [0,1)$. Then $\gamma (0)=0$, $\lim _{t\rightarrow 1^-}\gamma (t)=\infty$ and $\gamma '(t)=\frac{1}{1-t}-1=\frac{t}{1-t}$, for $t\in [0,1)$. Since $\gamma '(t)>0$, for $t\in (0,1)$, it follows that $\gamma$ is strictly increasing and so $\gamma (t)>0$ for all $t\in (0,1)$. This implies that $\Vert C_tf_1\Vert _\infty =-\frac{\log (1-t)}{t}>1$ for every $t\in (0,1)$. On the other hand, for $t\in (0,1)$, the inequality $\sum _{n=0}^\infty t^n/(n+1)<\sum _{n=0}^\infty t^n$ implies that $-\frac{\log (1-t)}{t}<\frac{1}{1-t}$. So, we have shown that $\Vert C_0f_1\Vert _\infty =1$ and

$$\begin{aligned} 1<\Vert C_tf_1\Vert _\infty <\frac{1}{1-t},\quad t\in (0,1). \end{aligned}$$

We now turn to the action of $C_t$ in various Banach spaces. For $t=1$ it was noted above that $C_1$ fails to act in $H^\infty$.

Proposition 2.3

For $t\in [0,1)$ the operator $C_t:H^\infty \rightarrow H^\infty$ is continuous. Moreover, $\Vert C_0\Vert _{H^\infty \rightarrow H^\infty }=1$ and

$$\begin{aligned} \Vert C_t\Vert _{H^\infty \rightarrow H^\infty }=-\frac{\log (1-t)}{t},\quad t\in (0,1). \end{aligned}$$

Proof

Let $f\in H^\infty$ be fixed. Then

$$\begin{aligned} |C_0f(z)|=\left| \frac{1}{z}\int _0^zf(\xi )d\xi \right| \le \max _{|\xi |\le |z|} |f(\xi )|\le \Vert f\Vert _\infty . \end{aligned}$$

This implies that $\Vert C_0\Vert _{H^\infty \rightarrow H^\infty }\le 1$. On the other hand, $C_0f_1=f_1$ and so we can conclude that $\Vert C_0\Vert _{H^\infty \rightarrow H^\infty }=1$.

Now let $t\in (0,1)$. Then, for the parametrization $\xi :=sz$, for $s\in (0,1)$, it follows from $|1-stz|\ge 1-|stz|\ge 1-st$ that

$$\begin{aligned} |C_tf(z)|&=\left| \frac{1}{z}\int _0^z\frac{f(\xi )}{1-t\xi }d\xi \right| =\left| \int _0^1\frac{f(sz)}{1-stz}ds\right| \le \max _{|\xi |\le |z|}|f(\xi )|\int _0^1\frac{ds}{1-st|z|}\\&\le \Vert f\Vert _\infty \int _0^1\frac{ds}{1-st}=-\frac{\log (1-t)}{t}\Vert f\Vert _\infty . \end{aligned}$$

So, $C_t\in {{\mathcal {L}}}(H^\infty )$ with $\Vert C_t\Vert _{H^\infty \rightarrow H^\infty }\le -\frac{\log (1-t)}{t}$. But, $\Vert C_tf_1\Vert _\infty = -\frac{\log (1-t)}{t}$. Accordingly, $\Vert C_t\Vert _{H^\infty \rightarrow H^\infty }= -\frac{\log (1-t)}{t}$. $\square$

Proposition 2.4

Let v be a weight function on [0, 1). For each $t\in [0,1)$ the operator $C_t:H_v^\infty \rightarrow H_v^\infty$ is continuous. Moreover, $\Vert C_0\Vert _{H^\infty _v\rightarrow H_v^\infty }=1$ and

$$\begin{aligned} 1\le \Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}\le -\frac{\log (1-t)}{t},\quad t\in (0,1). \end{aligned}$$

Proof

Recall that $C_tf(0):=f(0)$ for each $f\in H({{\mathbb {D}}})$ and $t\in [0,1]$. Fix $t\in (0,1)$. Given $f\in H^\infty _v$ and $z\in {{\mathbb {D}}}{\setminus }\{0\}$, observe that

$$\begin{aligned} v(z)|C_tf(z)|&=\frac{v(z)}{|z|}\left| \int _0^z\frac{f(\xi )}{1-t\xi }d\xi \right| =v(z)\left| \int _0^1\frac{f(sz)}{1-stz}ds\right| \\&\le v(z)\int _0^1\frac{|f(sz)|}{|1-stz|}ds\le \int _0^1\frac{v(sz)|f(sz)|}{|1-stz|}ds\\ {}&\le \Vert f\Vert _{\infty ,v}\int _0^1\frac{ds}{|1-stz|}\le \Vert f\Vert _{\infty ,v}\int _0^1\frac{ds}{1-st|z|}\\&=-\frac{\log (1-t|z|)}{t|z|}\Vert f\Vert _{\infty ,v}, \end{aligned}$$

where we used that $v(sz)=v(s|z|)\ge v(|z|)=v(z)$, for $s\in (0,1)$, as v is non-increasing on (0, 1) and that $|1-stz|\ge 1-st|z|$, for $s\in (0,1)$. According to the calculations in Example 2.2 we can conclude that

$$\begin{aligned} \Vert C_tf\Vert _{\infty , v}=\sup _{z\in {{\mathbb {D}}}}|C_tf(z)|v(z)\le \Vert f\Vert _{\infty ,v}\sup _{z\in {{\mathbb {D}}}}\left[ -\frac{\log (1-t|z|)}{t|z|}\right] =-\frac{\log (1-t)}{t}\Vert f\Vert _{\infty ,v}. \end{aligned}$$

This implies that $C_t\in {{\mathcal {L}}}(H^\infty _v)$ and $\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}\le -\frac{\log (1-t)}{t}$.

For $t=0$ observe that

$$\begin{aligned} |C_0f(z)|\le \int _0^1|f(sz)|ds\le \max _{|\xi |\le |z|}|f(\xi )|=\frac{1}{v(z)}\max _{|\xi |=|z|}|f(\xi )|v(\xi )\le \frac{1}{v(z)}\Vert f\Vert _{\infty ,v}, \end{aligned}$$

as $v(\xi )=v(z)$ whenever $|\xi |=|z|$ with $\xi \in {{\mathbb {D}}}$. This shows that $\Vert C_0\Vert _{H^\infty _v\rightarrow H_v^\infty }\le 1$. Since $C_0f_1=f_1$, it follows that actually $\Vert C_0\Vert _{H^\infty _v\rightarrow H_v^\infty }=1$.

It remains to show that $\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}\ge 1$ for $t\in (0,1)$. To this end, fix $t\in (0,1)$ and consider the function $g_0(z):=\frac{1}{1-tz}=\sum _{n=0}^\infty t^nz^n$, for $z\in {{\mathbb {D}}}$. Then $\Vert g_0\Vert _\infty =\frac{1}{1-t}$ and so $g_0\in H^\infty \subseteq H^\infty _v$. Moreover, for every $z\in {{\mathbb {D}}}{\setminus }\{0\}$, it is the case that

$$\begin{aligned} C_tg_0(z)=\frac{1}{z}\int _0^z\frac{d\xi }{(1-t\xi )^2}=\frac{1}{z} \left[ \frac{1}{t(1-t\xi )}\right] _0^z=\frac{1}{tz}\left[ \frac{1}{1-tz}-1\right] =\frac{1}{1-tz}=g_0(z). \end{aligned}$$

It follows that $\Vert g_0\Vert _{\infty ,v}=\Vert C_tg_0\Vert _{\infty ,v}\le \Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}\Vert g_0\Vert _{\infty ,v}$ which implies that $\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}\ge 1$. $\square$

Corollary 2.5

Let v be a weight function on [0, 1) satisfying $\lim _{r\rightarrow 1^-}v(r)=0$. For each $t\in [0,1)$ the operator $C_t:H^0_v\rightarrow H^0_v$ is continuous and satisfies $\Vert C_t\Vert _{H^0_v\rightarrow H^0_v}=\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}$.

Proof

By Proposition 2.4 and the fact that $H^0_v$ is a closed subspace of $H^\infty _v$, to obtain the result it suffices to establish that $C_t(H_v^0)\subseteq H^0_v$. To this effect, observe that $H^\infty \subseteq H^0_v$ and that $H^\infty$ is dense in $H^0_v$, as the space of polynomials is dense in $H^0_v$; see Section 1 of [11] and also [7]. Proposition 2.3 implies that $C_t(H^\infty )\subseteq H^\infty \subseteq H^0_v$. Since $C_t$ acts continuously on $H^\infty _v$, it follows that

$$\begin{aligned} C_t(H^0_v)=C_t(\overline{H^\infty })\subseteq \overline{C_t(H^\infty )}\subseteq H^0_v. \end{aligned}$$

Moreover, $\lim _{r\rightarrow 1^-}v(r)=0$ implies that $H^\infty _v$ is canonically isometric to the bidual of $H^0_v$, [8, Example 2.1], and that the bi-transpose $C_t'':H^\infty _v\rightarrow H_v^\infty$ of $C_t:H^0_v\rightarrow H^0_v$ coincides with $C_t:H^\infty _v\rightarrow H^\infty _v$ (see Lemma 2.6 below), from which the identity $\Vert C_t\Vert _{H^0_v\rightarrow H^0_v}=\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}$ follows. $\square$

Lemma 2.6

Let v be a weight function on [0, 1) satisfying $\lim _{r\rightarrow 1^-}v(r)=0$. For each $t\in [0,1)$, the bi-transpose $C''_t:H^\infty _v\rightarrow H^\infty _v$ of $C_t:H^0_v\rightarrow H_v^0$ coincides with $C_t:H^\infty _v\rightarrow H^\infty _v$.

Proof

By Proposition 2.3 and Corollary 2.5, together with the fact that $H^\infty _v$ is canonically isometric to the bidual of $H^0_v$, both of the operators $C_t''$ and $C_t$ act continuously on $H^\infty _v$.

To show that the bi-transpose $C_t'':H^\infty _v\rightarrow H^\infty _v$ of $C_t:H^0_v\rightarrow H^0_v$ coincides with $C_t:H^\infty _v\rightarrow H^\infty _v$ we proceed via several steps.

First step Given $f\in H({{\mathbb {D}}})$, its Taylor polynomials $p_k(z)=\sum _{j=0}^k{\hat{f}}(j)z^j$, $z\in {{\mathbb {D}}}$, for $k\in {{\mathbb {N}}}_0$, converge to f uniformly on compact subsets of ${{\mathbb {D}}}$. That is, $p_k\rightarrow f$ in $(H({{\mathbb {D}}}),\tau _c)$ as $k\rightarrow \infty$. Accordingly, the averages of $(p_k)_{k\in {{\mathbb {N}}}_0}$, that is, the Cesàro means $f_n(z):=\frac{1}{n+1}\sum _{j=0}^np_j(z)$, for $z\in {{\mathbb {D}}}$ and $n\in {{\mathbb {N}}}_0$, also converge to f in $(H({{\mathbb {D}}}),\tau _c)$ as $n\rightarrow \infty$.

Second step Lemma 1.1 in [7] implies, for every $f\in H^\infty _v$ and $n\in {{\mathbb {N}}}_0$, that $\Vert f_n\Vert _{\infty ,v}\le \Vert f\Vert _{\infty ,v}$, where $f_n$ is the n-th Cesàro mean of f, as defined in the First step. Denote by $U_v$ the closed unit ball of $(H^\infty _v,\Vert \cdot \Vert _{\infty ,v})$. Then, for any given $f\in U_v$, its sequence of Cesàro means satisfies $(f_n)_{n\in {{\mathbb {N}}}_0}\subseteq U_v$ and $f_n\rightarrow f$ in $(H({{\mathbb {D}}}),\tau _c)$ as $n\rightarrow \infty$.

Third step With the topology of uniform convergence on the compact subsets of $U_v$ denoted by $\tau _c$, let $X:=\{F\in (H_v^\infty )':\ F|_{U_v}\ \text { is } \tau _c-\text { continuous }\}$ be endowed with the norm $\Vert F\Vert :=\sup \{|F(f)|:\ f\in U_v\}$. Then [8, Theorem 1.1(a)] ensures that $(X,\Vert \cdot \Vert )$ is a Banach space and that the evaluation map $\Psi :H^\infty _v\rightarrow X'$ defined by $(\Psi (f))(F):=\langle f,F\rangle$, for $F\in X$ and $f\in H^\infty _v$, is an isometric isomorphism onto $X'$ (where $X'$ is the dual Banach space of $(X,\Vert \cdot \Vert )$). Moreover, by [8, Theorem 1.1(b) and Example 2.1] the restriction map $R:X\rightarrow (H^0_v)'$ given by $F\mapsto F|_{H_v^0}$, is also a surjective isometric isomorphism. Therefore, the spaces $H^\infty _v$ and $(H^0_v)''$ are isometrically isomorphic, that is, X and $(H^0_v)'$ are isometrically isomorphic and hence, also $H^\infty _v$ and $(H^0_v)''$ are isometrically isomorphic.

It is easy to see, since the Banach space X above is the predual of $H^\infty _v$, that the evaluation map $\delta _z\in X$, for every $z\in {{\mathbb {D}}}$, where $\delta _z:f\mapsto f(z)$, for $f\in H^\infty _v$, satisfies $|\langle f,\delta _z\rangle |\le \Vert f\Vert _{\infty ,v}/v(z)$. In particular, the linear span L of the set $\{\delta _z:\ z\in {{\mathbb {D}}}\}$ separates the points of $H^\infty _v=X'$ and hence, L is dense in X. Therefore, the pointwise convergence topology $\tau _p$ on $H^\infty _v$ is Hausdorff and coarser than the $w^*$-topology $\sigma (H^\infty _v, X)$.

Fourth step The closed unit ball $U_v$ of $H^\infty _v$ is a $\tau _c$-compact set by Montel’s theorem, as it is $\tau _c$-bounded and closed. On the other hand, $U_v$ is also $\sigma (H^\infty _v,X)$-compact by the Alaoglu-Bourbaki theorem. Since $\tau _p|_{U_v}$ is coarser than $\tau _c|_{U_v}$ and Hausdorff, we can conclude that $\tau _p|_{U_v}=\tau _c|_{U_v}$. In the same way, it follows that $\tau _p|_{U_v}=\sigma (H^\infty _v,X)|_{U_v}$. Accordingly, $\tau _p|_{U_v}=\tau _c|_{U_v}=\sigma (H^\infty _v,X)|_{U_v}$.

We are now ready to prove that $(C_t)''=C_t$. To show this, it suffices to establish that $(C_t)''f=C_tf$ for every $f\in U_v$.

So, fix $f\in U_v$. With $(f_n)_{n\in {{\mathbb {N}}}_0}$ as in the First step it follows from there that $f_n\rightarrow f$ in $(H({{\mathbb {D}}}),\tau _c)$ as $n\rightarrow \infty$ and, by the Second step, that $(f_n)_{n\in {{\mathbb {N}}}_0}\subseteq U_v$. This implies that $C_tf_n\rightarrow C_tf$ in $(H({{\mathbb {D}}}),\tau _c)$ as $n\rightarrow \infty$. Since $C_t\in {{\mathcal {L}}}(H^\infty _v)$ and $f\in U_v$, it is clear that $C_tf\in H^\infty _v$. On the other hand, by the Fourth step the sequence $(f_n)_{n\in {{\mathbb {N}}}_0}$ also converges to f in $(H^\infty _v,\sigma (H^\infty _v,X))=(H^\infty _v,\sigma (H^\infty _v, (H^0_v)'))$. Since $(C_t)'':((H^0_v)'',\sigma ((H^0_v)'',(H^0_v)'))\rightarrow ((H^0_v)'',\sigma ((H^0_v)'',(H^0_v)')$ is continuous, [20, §8.6], that is, $(C_t)'':(H^\infty _v,\sigma (H^\infty _v, X))\rightarrow (H^\infty _v,\sigma (H^\infty _v, X))$ is continuous, it follows that $(C_t)''f_n\rightarrow (C_t)''f$ in $(H^\infty _v,\sigma (H^\infty _v, X))$ as $n\rightarrow \infty$. Now, $(f_n)_{n\in {{\mathbb {N}}}_0}\subset H^\infty \subseteq H^0_v$, as each $f_n$ is a polynomial, and $(C_t)''f_n=C_tf_n$ for every $n\in {{\mathbb {N}}}_0$. Moreover, the sequence $C_tf_n\rightarrow (C_t)''f$ in $(H({{\mathbb {D}}}),\tau _p)$ as $n\rightarrow \infty$. Thus, $(C_t)''f=C_tf$ as desired. $\square$

Proposition 2.7

Let v be a weight function satisfying $\lim _{r\rightarrow 1^-}v(r)=0$. For each $t\in [0,1)$, both of the operators $C_t:H^\infty _v\rightarrow H^\infty _v$ and $C_t\rightarrow H^0_v\rightarrow H^0_v$ are compact.

Proof

Fix $t\in [0,1)$. Since $H^0_v$ is a closed subspace of $H_v^\infty$ and $C_t(H^0_v)\subseteq H_v^0$ (cf. Corollary 2.5), it suffices to show that $C_t:H^\infty _v\rightarrow H_v^\infty$ is compact. First we establish the following Claim:

• (*) Let the sequence $(f_n)_{n\in {{\mathbb {N}}}}\subset H^\infty _v$ satisfy $\Vert f_n\Vert _{\infty ,v}\le 1$ for every $n\in {{\mathbb {N}}}$ and $f_n\rightarrow 0$ in $(H({{\mathbb {D}}}),\tau _{c})$ for $n\rightarrow \infty$. Then $C_tf_n\rightarrow 0$ in $H^\infty _v$.

To prove the Claim, let $(f_n)_{n\in {{\mathbb {N}}}}\subset H^\infty _v$ be a sequence as in (*). Fix $\varepsilon >0$ and select $\delta \in (0,\beta )$, where $\beta :=\min \{1,\frac{\varepsilon (1-t)}{2},\frac{\varepsilon (1-t)}{2v(0)}\}$. Since $\{\xi \in {{\mathbb {C}}}\ \ |\xi |\le (1-\delta )\}$ is a compact subset of ${{\mathbb {D}}}$, there exists $n_0\in {{\mathbb {N}}}$ such that

$$\begin{aligned} \max _{|\xi |\le 1-\delta }|f_n(\xi )|<\delta , \quad n\ge n_0. \end{aligned}$$

Recall that $C_tf_n(0)=f_n(0)$ for every $n\in {{\mathbb {N}}}$. For $z\in {{\mathbb {D}}}{\setminus }\{0\}$ we have seen previously that

$$\begin{aligned} v(z)|C_tf_n(z)|=v(z)\left| \int _0^1\frac{f_n(sz)}{1-stz}ds\right| \le v(z)\int _0^{1-\delta }\frac{|f_n(sz)|}{|1-stz|}ds+v(z)\int _{1-\delta }^1\frac{|f_n(sz)|}{|1-stz|}ds. \end{aligned}$$

Denote the first (resp., second) summand in the right-side of the previous inequality by $(A_n)$ (resp., by $(B_n)$). Using the facts that $|1-stz|\ge 1-st |z|\ge \max \{1-s,1-t,1-|z|\}$, for all $s,t\in [0,1)$ and $z\in {{\mathbb {D}}}$, and that v is non-increasing on [0, 1) it follows, for every $n\ge n_0$, that $\int _0^{1-\delta }|f_n(sz)|\,ds\le (1-\delta )\max _{|\xi |\le (1-\delta )}|f_n(\xi )|$ (as $|sz|\le (1-\delta )$ for all $s\in [0,1-\delta ]$) and hence, that

$$\begin{aligned} (A_n)\le \frac{v(0)(1-\delta )}{1-t}\max _{|\xi |\le 1-\delta }|f_n(\xi )|<\frac{\varepsilon }{2}. \end{aligned}$$

On the other hand, for every $n\ge n_0$, we have (as $\Vert f_n\Vert _{\infty ,v}=\sup _{\xi \in {{\mathbb {D}}}}v(\xi )|f_n(\xi )|\le 1$) that

$$\begin{aligned} (B_n)=\int _{1-\delta }^1 \frac{v(z)}{v(sz)}\frac{v(sz)|f_n(sz)|}{|1-stz|}\,ds\le \int _{1-\delta }^1\frac{\Vert f_n\Vert _{\infty ,v}}{1-t}\,ds\le \frac{\delta }{1-t}<\frac{\varepsilon }{2}. \end{aligned}$$

It follows that $\Vert C_tf_n\Vert _{\infty ,v}<\varepsilon$ for every $n\ge n_0$. That is, $C_tf_n\rightarrow 0$ in $H^\infty _v$ for $n\rightarrow \infty$ and so (*) is proved.

The compactness of $C_t\in {{\mathcal {L}}}(H^\infty _v)$ can be deduced from (*) as follows. Let $(f_n)_{n\in {{\mathbb {N}}}}\subset H^\infty _v$ be any bounded sequence. There is no loss of generality in assuming that $\Vert f_n\Vert _{\infty ,v}\le 1$ for all $n\in {{\mathbb {N}}}$. To establish the compactness of $C_t\in {{\mathcal {L}}}(H^\infty _v)$ we need to show that $(C_tf_n)_{n\in {{\mathbb {N}}}}$ has a convergent subsequence in $H^\infty _v$.

Since $H^\infty _v\subseteq H({{\mathbb {D}}})$ continuously, the sequence $(f_n)_{n\in {{\mathbb {N}}}}$ is also bounded in the Fréchet–Montel space $H({{\mathbb {D}}})$. Hence, there is a subsequence $g_j:=f_{n_j}$, for $j\in {{\mathbb {N}}}$, of $(f_n)_{n\in {{\mathbb {N}}}}$ and $f\in H({{\mathbb {D}}})$ such that $g_j\rightarrow f$ in $H({{\mathbb {D}}})$ with respect to $\tau _c$. In particular, $g_j\rightarrow f$ pointwise on ${{\mathbb {D}}}$. Since $v(z)|g_j(z)|=v(z)|f_{n_j}(z)|\le 1$ for all $z\in {{\mathbb {D}}}$ and $j\in {{\mathbb {N}}}$, letting $j\rightarrow \infty$ it follows that $v(z)|f(z)|\le 1$ for all $z\in {{\mathbb {D}}}$, that is, $f\in H^\infty _v$ with $\Vert f\Vert _{\infty ,v}\le 1$. Let $h_j:=\frac{1}{2}(g_j-f)$, for $j\in {{\mathbb {N}}}$. Then $\Vert h_j\Vert _{\infty , v}\le 1$, for $j\in {{\mathbb {N}}}$, and $h_j\rightarrow 0$ in $H({{\mathbb {D}}})$ with respect to $\tau _c$. Condition (*) implies that $C_th_j\rightarrow 0$ in $H^\infty _v$ from which it follows that $C_tf_{n_j}=C_tg_j=C_t(g_j-f)+C_tf=2C_th_j+C_tf\rightarrow C_tf$ in $H^\infty _v$, as desired. $\square$

Proposition 2.8

Let v be a weight function on [0, 1) satisfying $\lim _{r\rightarrow 1^-}v(r)=0$. For each $t\in [0,1)$ the spectra of $C_t\in {{\mathcal {L}}}(H^\infty _v)$ and of $C_t\in {{\mathcal {L}}}(H^0_v)$ are given by

$$\begin{aligned} \sigma _{pt}(C_t;H^\infty _v)=\sigma _{pt}(C_t;H^0_v)=\left\{ \frac{1}{m+1}\,: m\in {{\mathbb {N}}}_0\right\} ,\end{aligned}$$

(2.1)

and

$$\begin{aligned} \sigma (C_t;H^\infty _v)=\sigma (C_t;H^0_v)=\left\{ \frac{1}{m+1}\,: m\in {{\mathbb {N}}}_0\right\} \cup \{0\}.\end{aligned}$$

(2.2)

Proof

Let $t\in [0,1)$ be fixed. By [13, Lemma 3.6] we know that the point spectrum of the operator $C_t^\omega \in {{\mathcal {L}}}(\omega )$ is given by $\sigma _{pt}(C_t^\omega ;\omega )=\{\frac{1}{m+1}:\, m\in {{\mathbb {N}}}_0\}$ and, for each $m\in {{\mathbb {N}}}_0$, that the corresponding eigenspace $\textrm{Ker}(\frac{1}{m+1}I-C_t^\omega )$ is 1-dimensional and is generated by an eigenvector $x^{[m]}=(x_n^{[m]})_{n\in {{\mathbb {N}}}_0}\in \ell ^1$. Since $H^0_v\subseteq H^\infty _v\subseteq H({{\mathbb {D}}})$ with continuous inclusions and $\Phi :H({{\mathbb {D}}})\rightarrow \omega$ (cf. Sect. 1) is a continuous embedding, this implies that $\sigma _{pt}(C_t;H^0_v)\subseteq \sigma _{pt}(C_t;H^\infty _v)\subseteq \{\frac{1}{m+1}\,:\, m\in {{\mathbb {N}}}_0\}$. Indeed, let $f\in H({{\mathbb {D}}}){\setminus }\{0\}$ and $\lambda \in {{\mathbb {C}}}$ satisfy $C_tf=\lambda f$. Then $\lambda f(z)=\sum _{n=0}^\infty \widehat{(\lambda f)}(n)z^n=\sum _{n=0}^\infty \lambda {\hat{f}}(n)z^n$ and, by (1.6), we have that $(C_tf)(z)=\sum _{n=0}^\infty (C_t^\omega {\hat{f}})_nz^n$. It follows that $C_t^\omega {\hat{f}}=\lambda {\hat{f}}$ in $\omega$ with ${\hat{f}}\not =0$ and so $\lambda \in \sigma _{pt}(C_t^\omega ;\omega )=\{\frac{1}{m+1}\,\ m\in {{\mathbb {N}}}_0\}$.

To conclude the proof, it remains to show that $\{\frac{1}{m+1}\,:\, m\in {{\mathbb {N}}}_0\}\subseteq \sigma _{pt}(C_t;H^0_v)$. To establish this recall, for each $m\in {{\mathbb {N}}}_0$, that the eigenvector $x^{[m]}\in \ell ^1$ and hence, the function $g_m(z):=\sum _{n=0}^\infty (x^{[m]})_nz^n$ belongs to $H^0_v$ because $0\le v(z)|g_m(z)|\le v(z)\Vert x^{[m]}\Vert _{\ell ^1}$ for $z\in {{\mathbb {D}}}$ and $\lim _{r\rightarrow 1^-}v(r)=0$. Moreover, according to (1.5) and (1.6) we have, for each $z\in {{\mathbb {D}}}$, that

$$\begin{aligned} C_tg_m(z)=\sum _{n=0}^{\infty }(C_t^\omega x^{[m]})_nz^n=\sum _{n=0}^{\infty }(\frac{1}{m+1} x^{[m]})_nz^n=\frac{1}{m+1}\sum _{n=0}^\infty (x^{[m]})_nz^n =\frac{1}{m+1}g_m(z). \end{aligned}$$

Thus $g_m$ is an eigenvector of $C_t\in {{\mathcal {L}}}(H^0_v)$ corresponding to the eigenvalue $\frac{1}{m+1}$.

The validity of $\sigma (C_t;H^0_v)=\sigma (C_t;H^\infty _v)=\{\frac{1}{m+1}\,:\, m\in {{\mathbb {N}}}_0\}\cup \{0\}$ follows from the fact that $C_t$ is a compact operator on both spaces. $\square$

We now investigate the norm of $C_t$ on $H^\infty _v$ for the standard weights $v_\gamma (z):=(1-|z|)^\gamma$, for $\gamma >0$ and $z\in {{\mathbb {D}}}$, which satisfy $\lim _{r\rightarrow 1^-}v_\gamma (r)=0$.

Proposition 2.9

Let $t\in (0,1)$ and $\gamma >0$.

• (i) The operator norm $\Vert C_t\Vert _{H^\infty _{v_\gamma }\rightarrow H^\infty _{v_\gamma }}=1$, for every $\gamma \ge 1$.

• (ii) For each $\gamma \in (0,1)$, the inequality $\Vert C_t\Vert _{H^\infty _{v_\gamma }\rightarrow H^\infty _{v_\gamma }}\le \min \{-\frac{\log (1-t)}{t},\frac{1}{\gamma }\}$ is valid.

Proof

We adapt the arguments given for the Cesàro operator $C_1$ in the proof of [2, Theorem 2.3].

Let $\gamma >0$ and $t\in (0,1)$ be fixed. For $f\in H^\infty _{v_\gamma }$ with $\Vert f\Vert _{\infty ,v_\gamma }=1$ we have

$$\begin{aligned} |C_tf(z)|&=\frac{1}{|z|}\left| \int _0^1\frac{f(sz)}{1-stz}ds\right| \le \int _0^1\frac{|f(sz)|}{1-st|z|}ds\\&\le \int _0^1\frac{|f(sz)|}{1-s|z|}ds\le \int _0^1\frac{ds}{(1-s|z|)^{\gamma +1}}=\frac{1}{(1-|z|)^\gamma }\frac{1-(1-|z|)^\gamma }{\gamma |z|}, \end{aligned}$$

as $z\in {{\mathbb {D}}}$ implies that $1-st|z|\ge 1-s|z|$, for $s\in (0,1)$. Accordingly,

$$\begin{aligned} v_\gamma (z)|C_tf(z)|=(1-|z|)^\gamma |C_tf(z)|\le \frac{1-(1-|z|)^\gamma }{\gamma |z|},\quad z\not =0, \end{aligned}$$

and hence,

$$\begin{aligned} \Vert C_tf\Vert _{\infty ,v_\gamma }\le \frac{1}{\gamma } \sup _{z\in {{\mathbb {D}}}}\frac{1-(1-|z|)^\gamma }{|z|}. \end{aligned}$$

Define $\phi (s):=\frac{1-(1-s)^\gamma }{s}$ for $s\in (0,1]$ and $\phi (0)=\gamma$, in which case $\phi$ is continuous. So, the previous inequality yields $\Vert C_tf\Vert _{\infty ,v_\gamma }\le \frac{M_\gamma }{\gamma }$, for all $\Vert f\Vert _{\infty ,v_\gamma }\le 1$, that is, $\Vert C_t\Vert _{H^\infty _{v_\gamma }\rightarrow H^\infty _{v_\gamma }}\le \frac{M_\gamma }{\gamma }$, where $M_\gamma :=\sup _{s\in [0,1]}\phi (s)$. Proposition 2.4 yields that $1\le \Vert C_t\Vert _{H^\infty _{v_\gamma }\rightarrow H^\infty _{v_\gamma }}\le -\frac{\log (1-t)}{t}$ for $t\in (0,1)$. On page 101 of [2] it is shown that $\frac{M_\gamma }{\gamma }\le 1$ whenever $\gamma \ge 1$ and that $M_\gamma \le 1$ for all $\gamma \in (0,1)$. The proof of both parts (i) and (ii) follows immediately. $\square$

Remark 2.10

For each $\gamma >0$ let $v_\gamma (z)=(1-|z|)^\gamma$, for $z\in {{\mathbb {D}}}$. Proposition 2.9 implies that $\sup _{0\le t<1}\Vert C_t\Vert _{H^\infty _{v_\gamma }\rightarrow H^\infty _{v_\gamma }}<\infty$. Moreover, if $\gamma \ge 1$, then $\Vert C_t^n\Vert _{H^\infty _{v_\gamma }\rightarrow H^\infty _{v_\gamma }}=1$ for every $n\in {{\mathbb {N}}}$; see case (i) in the proof of [2, Theorem 2.3] together with the fact that $1\in \sigma _{pt}(C_t, H^\infty _{v_\gamma })$ by Proposition 2.8.

Let $n\in {{\mathbb {N}}}$ be fixed. Consider the weight $v(z)=(\log \frac{e}{1-|z|})^{-n}$, for $z\in {{\mathbb {D}}}$, which satisfies $v(0)=1$ and $\lim _{|z|\rightarrow 1^-}v(z)=0$.

The function $f(z):=[\log (1-z)]^n\in H({{\mathbb {D}}})$ belongs to $H^\infty _v$. Indeed, for each $z\in {{\mathbb {D}}}$, we have that

$$\begin{aligned} |\log (1-z)|=\left| -\sum _{n=1}^\infty \frac{z^n}{n}\right| \le \sum _{n=1}^\infty \frac{|z|^n}{n}=-\log (1-|z|) \end{aligned}$$

and hence, that $|f(z)|=|\log (1-z)|^n\le (-\log (1-|z|))^n$. Since v is given by $v(z)=(1-\log (1-|z|))^{-n}$ and $\lim _{|z|\rightarrow 1^-}\frac{-\log (1-|z|)}{1-\log (1-|z|)}=1$, it follows that $\Vert f\Vert _{\infty ,v}<\infty$ and so $f\in H^\infty _v$. On the other hand,

$$\begin{aligned} C_1f(z)=\frac{1}{z}\int _0^z\frac{(\log (1-\xi )^n)}{1-\xi }d\xi =-\frac{1}{(n+1)z}(\log (1-z))^{n+1},\quad z\in {{\mathbb {D}}}. \end{aligned}$$

Accordingly, $C_1f\not \in H^\infty _v$ since

$$\begin{aligned} \lim _{s\rightarrow s^-}v(s)|(C_1)f(s)|&=\frac{1}{n+1}\lim _{s\rightarrow 1^-}\left| \frac{(\log (1-s))^{n+1}}{s(1-\log (1-s))^n}\right| \\&=\frac{1}{n+1}\lim _{s\rightarrow 1^-}\left| \left( \frac{\log (1-s)}{1-\log (1-s)}\right) ^n\frac{\log (1-s)}{s}\right| =\infty . \end{aligned}$$

This implies that the Cesàro operator $C_1$ is not well-defined on $H^\infty _v$, that is, $C_1(H^\infty _v)\not \subseteq H^\infty _v$. But, by Proposition 2.4 the generalized Cesàro operator $C_t\in {{\mathcal {L}}}(H^\infty _v)$ for every $t\in [0,1)$. At this point, the following question arises: Is $\sup _{t\in [0,1)}\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}<\infty$ for this particular v? Our next two results show that the answer is negative for certain weights v, which includes $v(z)=\left( \log \frac{e}{1-|z|}\right) ^{-n}$ for $z\in {{\mathbb {D}}}$.

Proposition 2.11

Let v be a weight function on [0, 1) such that $\sup _{t\in [0,1)}\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}<\infty$. Then $C_1\in {{\mathcal {L}}}(H^\infty _v)$.

Proof

Proposition 2.1 implies that $\{C_t:\, t\in [0,1)\}$ is equicontinuous in ${{\mathcal {L}}}(H({{\mathbb {D}}}))$. The claim is that $\lim _{t\rightarrow 1^-}C_tf(z)=C_1f(z)$, for every $f\in H({{\mathbb {D}}})$ and $z\in {{\mathbb {D}}}$.

To prove this claim fix $f\in H({{\mathbb {D}}})$ and $z\in {{\mathbb {D}}}{\setminus }\{0\}$. Recall, for $t\in [0,1)$, that

$$\begin{aligned} C_tf(z)=\frac{1}{z}\int _0^z\frac{f(\xi )}{1-t\xi }d\xi =\int _0^1\frac{f(sz)}{1-stz}ds \end{aligned}$$

and

$$\begin{aligned} C_1f(z)=\frac{1}{z}\int _0^z\frac{f(\xi )}{1-\xi }d\xi =\int _0^1\frac{f(sz)}{1-sz}ds. \end{aligned}$$

Moreover, for each $z\in {{\mathbb {D}}}{\setminus }\{0\}$, we have (as $|1-stz|\ge (1-|z|)$) that

$$\begin{aligned} \left| \frac{f(sz)}{1-st z}\right| \le \frac{|f(sz)|}{1-|z|}\le \frac{1}{1-|z|}\max _{|\xi |\le |z|}|f(\xi )|, \quad s\in [0,1], \end{aligned}$$

and that $\lim _{t\rightarrow 1^-}\frac{f(sz)}{1-st z}=\frac{f(sz)}{1-s z}$ for every $s\in [0,1]$. So, we can apply the dominated convergence theorem to conclude that $\lim _{t\rightarrow 1^-}C_tf(z)=C_1f(z)$ for $z\in {{\mathbb {D}}}{\setminus }\{0\}$. For $z=0$ we have $C_tf(0)=f(0)=C_1f(0)$ for each $f\in H({{\mathbb {D}}})$ and $t\in [0,1)$. So, for each $f\in H({{\mathbb {D}}})$, we can conclude that $C_tf\rightarrow C_1f$ pointwise on ${{\mathbb {D}}}$ for $t\rightarrow 1^-$. The claim is thereby established.

We now show that $C_tf\rightarrow C_1f$ in $H({{\mathbb {D}}})$ as $t\rightarrow 1^-$ for every $f\in H^\infty _v$. The assumption $\sup _{t\in [0,1)}\Vert C_t\Vert _{H^\infty _v\rightarrow H_v^\infty }<\infty$ implies that there exists $M>0$ satisfying $\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}\le M$ for every $t\in [0,1)$. Therefore,

$$\begin{aligned} \sup _{z\in {{\mathbb {D}}}}|C_tf(z)|v(z)\le M\Vert f\Vert _{\infty ,v}, \quad f\in H^\infty _v,\ t\in [0,1). \end{aligned}$$

(2.3)

Fix $f\in H^\infty _v$. Then $\{C_tf:\, t\in [0,1)\}$ is a bounded set in $H({{\mathbb {D}}})$. Indeed, given $r\in (0,1)$ and $t\in [0,1)$ we have (as $v(r)\le v(z)$ for all $|z|\le r$) that

$$\begin{aligned} q_r(C_tf)=\sup _{|z|\le r}|C_tf(z)|=\max _{|z|=r}|C_tf(z)|\le \frac{M}{v(r)}\Vert f\Vert _{\infty ,v}. \end{aligned}$$

So, the set $\{C_tf:\, t\in [0,1)\}$ is bounded in the Fréchet–Montel space $H({{\mathbb {D}}})$ and hence, it is relatively compact in $H({{\mathbb {D}}})$. Since $C_tf\rightarrow C_1f$ pointwise on ${{\mathbb {D}}}$ for $t\rightarrow 1^-$, it follows that $C_tf\rightarrow C_1f$ with respect to $\tau _{c}$, that is, in the Fréchet space $H({{\mathbb {D}}})$, for $t\rightarrow 1^-$. In particular, $C_1f\in H({{\mathbb {D}}})$.

Since $H^\infty _v\subseteq H({{\mathbb {D}}})$ and $C_th\rightarrow C_1h$ pointwise on ${{\mathbb {D}}}$ as $t\rightarrow 1^-$, for every $h\in H({{\mathbb {D}}})$, letting $t\rightarrow 1^-$ in (2.3) it follows that

$$\begin{aligned} |C_1f(z)|v(z)\le M\Vert f\Vert _{\infty ,v},\quad z\in {{\mathbb {D}}}, \end{aligned}$$

that is, $\Vert C_1f\Vert _{\infty ,v}\le M\Vert f\Vert _{\infty ,v}$. But, $f\in H^\infty _v$ is arbitrary and so $C_1\in {{\mathcal {L}}}(H^\infty _v)$. $\square$

Proposition 2.12

For each $n\in {{\mathbb {N}}}$, let $v(z)=(\log (\frac{e}{1-|z|}))^{-n}$ for $z\in {{\mathbb {D}}}$. Then $\sup _{t\in [0,1)}\Vert C_t\Vert _{H^\infty _v\rightarrow H^\infty _v}=\infty$.

Proof

Apply Proposition 2.11 and the discussion prior it. $\square$

3 Linear dynamics and mean ergodicity of $C_t$

The aim of this section is to investigate the mean ergodicity and the linear dynamics of the operators $C_t$, for $t\in [0,1)$, acting on $H({{\mathbb {D}}})$, $H^\infty _v$ and $H^0_v$

An operator $T\in {{\mathcal {L}}}(X)$, with X a lcHs, is called power bounded if $\{T^n:\ n\in {{\mathbb {N}}}_0\}$ is an equicontinuous subset of ${{\mathcal {L}}}(X)$. For a Banach space X, this means that $\sup _{n\in {{\mathbb {N}}}_0}\Vert T^n\Vert _{X\rightarrow X}<\infty$. Given $T\in {{\mathcal {L}}}(X)$, the averages

$$\begin{aligned} T_{[n]}:=\frac{1}{n}\sum _{m=1}^nT^m,\quad n\in {{\mathbb {N}}}, \end{aligned}$$

are usually called the Cesàro means of T. The operator T is said to be mean ergodic (resp., uniformly mean ergodic) if $(T_{[n]})_{n\in {{\mathbb {N}}}}$ is a convergent sequence in ${{\mathcal {L}}}_s(X)$ (resp., in ${{\mathcal {L}}}_b(X)$). It is routine to check that $\frac{T^n}{n}=T_{[n]}-\frac{n-1}{n}T_{[n-1]}$, for $n\ge 2$, and hence, $\tau _s$-$\lim _{n\rightarrow \infty }\frac{T^n}{n}=0$ whenever T is mean ergodic. Every power bounded operator on a Fréchet–Montel space X is necessarily uniformly mean ergodic, [1, Proposition 2.8]. Concerning the linear dynamics of $T\in {{\mathcal {L}}}(X)$, with X a lcHs, the operator T is called supercyclic if, for some $z\in X$, the projective orbit $\{\lambda T^nz\,:\ \lambda \in {{\mathbb {C}}},\ n\in {{\mathbb {N}}}_0\}$ is dense in X. Since the closure of the linear span of a projective orbit is separable, if such a supercyclic operator $T\in {{\mathcal {L}}}(X)$ exists, then X is necessarily separable.

Observe that the space $H^\infty _v$ is never separable, [24, Theorem 1.1]. Therefore, every operator $T\in {{\mathcal {L}}}(H^\infty _v)$ is clearly not supercyclic. However, the spaces $H({{\mathbb {D}}})$, [21, Theorem 27.2.5], and $H^0_v$, [24, Theorem 1.1], for every weight v are always separable. Hence, the problem of supercyclicity for non-zero operators $T\in {{\mathcal {L}}}(H({{\mathbb {D}}}))$ and $T\in {{\mathcal {L}}}(H^0_v)$ arises.

The following result, [5, Theorem 6.4], is stated here for Banach spaces.

Theorem 3.1

Let X be a Banach space and let $T\in {{\mathcal {L}}}(X)$ be a compact operator such that $1\in \sigma (T;X)$ with $\sigma (T;X){\setminus } \{1\}\subseteq \overline{B(0,\delta )}$ for some $\delta \in (0,1)$ and satisfying $\textrm{Ker}(I-T)\cap \textrm{Im} (I-T)=\{0\}$. Then T is power bounded and uniformly mean ergodic.

A consequence of the previous theorem is the following result.

Proposition 3.2

Let v be a weight function on [0, 1) satisfying $\lim _{r\rightarrow 1^-}v(r)=0$. For each $t\in [0,1)$ both of the operators $C_t\in {{\mathcal {L}}}(H^\infty _v)$ and $C_t\in {{\mathcal {L}}}(H^0_v)$ are power bounded, uniformly mean ergodic and fail to be supercyclic.

Proof

Fix $t\in [0,1)$. It was already noted that $C_t\in {{\mathcal {L}}}(H^\infty _v)$ cannot be supercyclic. The operator $C_t$ is a compact operator on both $H^\infty _v$ and on $H^0_v$ (cf. Proposition 2.7). Therefore, the compact transpose operators $C_t'\in {{\mathcal {L}}}((H^\infty _v)')$ and $C'_t\in {{\mathcal {L}}}((H^0_v)')$ have the same non-zero eigenvalues as $C_t$ (see, e.g., [15, Theorem 9.10-2(2)]). In view of Proposition 2.8 it follows that $\sigma _{pt}(C_t';(H^\infty _v)')=\sigma _{pt}(C_t';(H^0_v)')=\{\frac{1}{m+1}\,:\, m\in {{\mathbb {N}}}_0\}$. We can apply [6, Proposition 1.26] to conclude that $C_t$ is not supercyclic on $H^0_v$.

By Proposition 2.8 and its proof (as $x^{[0]}=(t^n)_{n\in {{\mathbb {N}}}_0}$) we have that $\textrm{Ker}(I-C_t)=\textrm{span}\{g_0\}$, with $g_0(z)=\sum _{n=0}^\infty t^nz^n$, for $z\in {{\mathbb {D}}}$. On the other hand, $\textrm{Im}(I-C_t)$ is a closed subspace of $H^\infty _v$ (resp., of $H^0_v$), as $C_t$ is compact in $H^\infty _v$ (resp., in $H^0_v$)), and $\textrm{Im}(I-C_t)\subseteq \{g\in H^\infty _v\,:\, g(0)=0\}$ (resp., $\subseteq \{g\in H^0_v\,:\, g(0)=0\}$), because $C_tf(0)=f(0)$ for each $f\in H^\infty _v$ (resp., each $f\in H^0_v$). Moreover, [15, Theorem 9.10.1] implies that $\mathrm{codim\,Im}(I-C_t)=\mathrm{dim\,Ker} (I-C_t)=1$. Accordingly, both $\textrm{Im}(I-C_t)$ and $\{g\in H^\infty _v\,:\, g(0)=0\}=\textrm{Ker}(\delta _0)$ are hyperplanes, where $\delta _0\in (H^\infty _v)'$ is the linear evaluation functional $f\mapsto f(0)$, for $f\in H^\infty _v$. It follows that necessarily $\textrm{Im}(I-C_t)=\{g\in H^\infty _v:\, g(0)=0\}$.

Let $h\in \textrm{Im}(I-C_t)\cap \textrm{Ker}(I-C_t)$. Then $h(0)=0$ and there exists $\lambda \in {{\mathbb {C}}}$ such that $h=\lambda g_0$. This yields that $0=h(0)=\lambda g_0(0)=\lambda$. Hence, $h=0$. So, $\textrm{Im}(I-C_t)\cap \textrm{Ker}(I-C_t)=\{0\}$.

Proposition 2.8 implies that $1\in \sigma (C_t; H^\infty _v)=\sigma (C_t; H^0_v)=\{\frac{1}{m+1}\,;\, m\in {{\mathbb {N}}}_0\}\cup \{0\}$. Consequently, for $\delta =\frac{1}{2}$, all the assumptions of Theorem 3.1 are satisfied. So, we can conclude that $C_t$ is power bounded and uniformly mean ergodic on both $H^\infty _v$ and on $H^0_v$. $\square$

In contrast to the compactness of $C_t$ acting in the Banach spaces $H^\infty _v$ and $H^0_v$ (cf. Proposition 2.7) the situation for the Fréchet space $H({{\mathbb {D}}})$ is different.

Proposition 3.3

For each $t\in [0,1)$ the operator $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ is an isomorphism and, hence, it is not compact.

Proof

Fix $t\in [0,1)$. Consider the operator $T_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$, for $f\in H({{\mathbb {D}}})$, given by

$$\begin{aligned} T_tf(z):= (1-tz)(zf(z))'=(1-tz)(f(z)+zf'(z)),\quad z\in {{\mathbb {D}}}. \end{aligned}$$

Then $T_t$ is clearly well-defined. Moreover, its graph is closed. Indeed, for a given sequence $(f_n)_{n\in {{\mathbb {N}}}}\subset H({{\mathbb {D}}})$, suppose that $f_n\rightarrow f$ in $H({{\mathbb {D}}})$ and $T_tf_n\rightarrow g$ in $H({{\mathbb {D}}})$. Since multiplication operators (by elements from $H({{\mathbb {D}}})$) and the differentiation operator are continuous on $H({{\mathbb {D}}})$ and the evaluation functionals at points of ${{\mathbb {D}}}$ belong to $H({{\mathbb {D}}})'$, it follows that $f'_n\rightarrow f'$ in $H({{\mathbb {D}}})$ and hence, $T_tf_n=(1-tz)(f_n+zf'_n)\rightarrow (1-tz)(f+zf')=T_tf$ in $H({{\mathbb {D}}})$. Accordingly, $g=T_tf$. Since $H({{\mathbb {D}}})$ is a Fréchet space, the closed graph theorem, [20, Corollary 5.4.3], implies that $T_t\in {{\mathcal {L}}}(H({{\mathbb {D}}}))$.

Finally, it is routine to verify that $C_t\circ T_t=T_t\circ C_t=I$. So, the inverse operator $C_t^{-1}=T_t\in {{\mathcal {L}}}(H({{\mathbb {D}}}))$ exists and hence, $C_t$ is a bi-continuous isomorphism of $H({{\mathbb {D}}})$ onto itself. In particular, $C_t$ cannot be compact. $\square$

Let $\Lambda :=\{\frac{1}{n+1}:\, n\in {{\mathbb {N}}}_0\}$ and $\Lambda _0:=\Lambda \cup \{0\}$. We recall from [4, Lemma 2.7] the following lemma, which is an extension of a result of Rhoades [27].

Lemma 3.4

For every $\mu \in {{\mathbb {C}}}{\setminus } \Lambda _0$ there exist $\delta =\delta _\mu >0$ and constants $d_\delta , D_\delta >0$ such that $\overline{B(\mu ,\delta )}\cap \Lambda _0=\emptyset$ and

$$\begin{aligned} \frac{d_\delta }{n^{\alpha (\nu )}}\le \prod _{k=1}^n\left| 1-\frac{1}{k\nu }\right| \le \frac{D_\delta }{n^{\alpha (\nu )}},\quad \forall n\in {{\mathbb {N}}},\ \nu \in B(\mu ,\delta ), \end{aligned}$$

(3.1)

where $\alpha (\nu ):=\textrm{Re}(\frac{1}{\nu })$.

Remark 3.5

As a direct application of Lemma 3.4 we obtain, for every $\mu \in {{\mathbb {C}}}{\setminus } \Lambda _0$, that there exist $\delta >0$ and $d_\delta , D_\delta >0$ such that $\overline{B(\mu , \delta )}\cap \Lambda _0=\emptyset$ and, for every $\nu \in B(\mu ,\delta )$ and $n\in {{\mathbb {N}}}_0$, we have that

$$\begin{aligned} d_\delta D_\delta ^{-1}\left( \frac{n-h}{n+1}\right) ^{\alpha (\nu )}\le \prod _{j=n-h+1}^{n+1}\left| 1-\frac{1}{j\nu }\right| \le D_\delta d_\delta ^{-1}\left( \frac{n-h}{n+1}\right) ^{\alpha (\nu )}, \end{aligned}$$

(3.2)

for all $h\in \{1,\ldots , n-1\}$, where $\alpha (\nu )=\textrm{Re}(\frac{1}{\nu })$.

For each $k\in {{\mathbb {N}}}$ with $k\ge 2$ define $r_k:=(1-\frac{1}{k})$. Define the norms $\Vert \cdot \Vert _k$ and $|||\cdot |||_k$ on $H({{\mathbb {D}}})$ by

$$\begin{aligned} \Vert f\Vert _k:=\sum _{n=0}^\infty |{\hat{f}}(n)|r_k^n,\quad f=\sum _{n=0}^\infty {\hat{f}}(n)z^n, \end{aligned}$$

and

$$\begin{aligned} |||f|||_k:=\sup _{n\in {{\mathbb {N}}}_0} |{\hat{f}}(n)|r_k^n\quad f=\sum _{n=0}^\infty {\hat{f}}(n)z^n. \end{aligned}$$

Lemma 3.6

Each of the sequences $\{\Vert \cdot \Vert _k\}_{k\ge 2}$ and $\{|||\cdot |||_k\}_{k\ge 2}$ is a fundamental system of norms for $(H({{\mathbb {D}}}),\tau _c)$.

Proof

Given $r\in (0,1)$ choose any $k\ge 2$ such that $0<r<(1-\frac{1}{k})$. Then, for every $f\in H({{\mathbb {D}}})$, we have

$$\begin{aligned} q_r(f)=\sup _{|z|=r}\left| \sum _{n=0}^\infty {\hat{f}}(n)z^n\right| \le \sum _{n=0}^\infty |{\hat{f}}(n)|r^n\le \sum _{n=0}^\infty |{\hat{f}}(n)|\left( 1-\frac{1}{k}\right) ^n=\Vert f\Vert _k. \end{aligned}$$

On the other hand, given $k\ge 2$, let $r_k:=(1-\frac{1}{k})<(1-\frac{1}{k+1}):=r_{k+1}$. By the Cauchy inequalities, for $n\in {{\mathbb {N}}}_0$, we have

$$\begin{aligned} |{\hat{f}}(n)|\le \frac{1}{r_{k+1}^n}\max _{|z|=r_{k+1}}|f(z)|= \frac{1}{r_{k+1}^n} q_{r_{k+1}}(f),\quad f\in H({{\mathbb {D}}}), \end{aligned}$$

and hence,

$$\begin{aligned} \Vert f\Vert _{r_k}=\sum _{n=0}^\infty |{\hat{f}}(n)|r_k^n\le q_{r_{k+1}}(f)\sum _{n=0}^\infty \left( \frac{r_k}{r_{k+1}}\right) ^n= cq_{r_{k+1}}(f),\quad f\in H({{\mathbb {D}}}), \end{aligned}$$

with $c=\frac{1}{1-\frac{r_k}{r_{k+1}}}=k^2>0$ as $\frac{r_k}{r_{k+1}}<1$, which is independent of f.

So, the systems $\{q_r\}_{r\in (0,1)}$ and $\{\Vert \cdot \Vert _k\}_{k\ge 2}$ are equivalent on $H({{\mathbb {D}}})$.

Observe, for every $k\ge 2$, that

$$\begin{aligned} |||f|||_k=\sup _{n\in {{\mathbb {N}}}_0}|{\hat{f}}(n)|r^n_k\le \sum _{n=0}^\infty |{\hat{f}}(n)|r_k^n=\Vert f\Vert _k,\quad f\in H({{\mathbb {D}}}), \end{aligned}$$

and that

$$\begin{aligned} \Vert f\Vert _k&=\sum _{n=0}^\infty |{\hat{f}}(n)|r_k^n=\sum _{n=0}^\infty |{\hat{f}}(n)|\left( \frac{r_k}{r_{k+1}}\right) ^nr^n_{k+1}\\&\le \sup _{n\in {{\mathbb {N}}}_0}|{\hat{f}}(n)|r^n_{k+1}\sum _{n=0}^\infty \left( \frac{r_k}{r_{k+1}}\right) ^n= k^2|||f|||_{k+1}, \end{aligned}$$

for $f\in H({{\mathbb {D}}})$, where $\sum _{n=0}^\infty \left( \frac{r_k}{r_{k+1}}\right) ^n=k^2$. Therefore, the systems $\{\Vert \cdot \Vert _k\}_{k\ge 2}$ and $\{|||\cdot |||_k\}_{k\ge 2}$ are equivalent. $\square$

Proposition 3.7

For each $t\in [0,1)$ the spectra of the operator $C_t\in {{\mathcal {L}}}(H({{\mathbb {D}}}))$ are given by

$$\begin{aligned} \sigma _{pt}(C_t;H({{\mathbb {D}}}))=\sigma (C_t;H({{\mathbb {D}}}))=\Lambda \end{aligned}$$

(3.3)

and

$$\begin{aligned} \sigma ^*(C_t;H({{\mathbb {D}}}))=\Lambda _0. \end{aligned}$$

(3.4)

Proof

Let $t\in [0,1)$ be fixed. For any weight function v on [0, 1) satisfying $\lim _{r\rightarrow 1^-}v(r)=0$, we have $H^\infty _v\subseteq H({{\mathbb {D}}})$ continuously and $\Phi :H({{\mathbb {D}}})\rightarrow \omega$ is a continuous imbedding. Accordingly, $\sigma _{pt}(C_t;H_v^\infty )\subseteq \sigma _{pt}(C_t;H({{\mathbb {D}}}))\subseteq \Lambda$; see the proof of Proposition 2.8. Since $\sigma _{pt}(C_t;H^\infty _v)=\Lambda$ (cf. Proposition 2.8) and $\sigma _{pt}(C_t^\omega ;\omega )=\Lambda$ [5, Theorem 3.7], it follows that $\sigma _{pt}(C_t;H({{\mathbb {D}}}))=\Lambda$. Moreover, in view of Proposition 2.8 above and Theorem 3.7 in [5], the eigenspace corresponding to each eigenvalue $\frac{1}{n+1}\in \Lambda$ is 1-dimensional. By Proposition 3.3, the operator $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ is a bi-continuous isomorphism and so $0\not \in \sigma (C_t; H({{\mathbb {D}}}))$.

The claim is that ${{\mathbb {C}}}{\setminus } \Lambda _0\subseteq \rho (C_t;H({{\mathbb {D}}}))$. To establish this claim, fix $\nu \in {{\mathbb {C}}}{\setminus }\Lambda _0$. Given $g(z)=\sum _{n=0}^\infty c_nz^n\in H({{\mathbb {D}}})$, consider the identity

$$\begin{aligned} (C_t-\nu I)f(z)=g(z),\quad z\in {{\mathbb {D}}}, \end{aligned}$$

(3.5)

where $f(z)=\sum _{n=0}^\infty a_n z^n\in H({{\mathbb {D}}})$ is to be determined. It follows from (1.6) that $C_tf(z)=\sum _{n=0}^\infty (\frac{t^na_0+t^{n-1}a_1+\cdots +a_n}{n+1})z^n$ from which the identity $(C_t-\nu I)f(z)=\sum _{n=0}^\infty (\frac{t^na_0+t^{n-1}a_1+\cdots +a_n}{n+1}-\nu a_n)z^n$ is clear. So, (3.5) is satisfied if and only if

$$\begin{aligned} \sum _{n=0}^\infty \left( \frac{t^na_0+t^{n-1}a_1+\cdots +a_n}{n+1}-\nu a_n \right) z^n=\sum _{n=0}^\infty c_nz^n, \quad z\in {{\mathbb {D}}}, \end{aligned}$$

that is, if and only if

$$\begin{aligned} \frac{t^na_0+t^{n-1}a_1+\cdots +a_n}{n+1}-\nu a_n=c_n,\quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

In view of this we can argue, as in the proof of [5, Lemma 3.6], to show that if a function $f\in H({{\mathbb {D}}})$ exists which satisfies the identity (3.5), then the Taylor coefficients $(a_n)_{n\in {{\mathbb {N}}}_0}$ of f must verify the following equalities

$$\begin{aligned} a_0&=\frac{c_0}{1-\nu }\nonumber \\ a_n&=\frac{c_n}{(\frac{1}{n+1}-\nu )}+\sum _{h=1}^n(-1)^h\frac{\nu ^{h-1}t^hc_{n-h}}{(n+1) \prod _{j=n-h+1}^{n+1}(\frac{1}{j}-\nu )}\nonumber \\&=: A_n+B_n, \quad n\ge 1. \end{aligned}$$

(3.6)

Observe, for each $n\ge 1$ and $h\in \{1,\ldots ,n\}$, that

$$\begin{aligned} (-1)^h\prod _{j=n-h+1}^{n+1}\left( \frac{1}{j}-\nu \right) =-\prod _{j=n-h+1}^{n+1}\left( \nu -\frac{1}{j}\right) =-\nu ^{h+1}\prod _{j=n-h+1}^{n+1}\left( 1-\frac{1}{j\nu }\right) \end{aligned}$$

and so

$$\begin{aligned} B_n=-\sum _{h=1}^n\frac{\nu ^{h-1}t^hc_{n-h}}{\nu ^{h+1}(n+1) \prod _{j=n-h+1}^{n+1}(1-\frac{1}{j\nu })}=-\frac{1}{\nu ^2} \sum _{h=1}^n\frac{t^hc_{n-h}}{(n+1)\prod _{j=n-h+1}^{n+1}(1-\frac{1}{j\nu })}. \end{aligned}$$

Accordingly, to verify the claim we need to prove that the power series $\sum _{n=0}^\infty a_nz^n$ is convergent in ${{\mathbb {D}}}$, with $(a_n)_{n\in {{\mathbb {N}}}_0}$ defined according to (3.6). First, observe that the series $g(z)=\sum _{n=0}^\infty c_nz^n$ is convergent in ${{\mathbb {D}}}$ and satisfies

$$\begin{aligned} \limsup _{n\rightarrow \infty }\root n \of {|c_n|}=\limsup _{n\rightarrow \infty } \root n \of {\frac{|c_n|}{|\frac{1}{n+1}-\nu |}}=\limsup _{n\rightarrow \infty }\root n \of {|A_n|}. \end{aligned}$$

Therefore, the series $\sum _{n=1}^\infty A_nz^n$ has the same radius of convergence as the series $\sum _{n=0}^\infty c_nz^n$ and hence, it converges in $H({{\mathbb {D}}})$. Accordingly, $f_1(z):=\sum _{n=1}^\infty A_nz^n$, for $z\in {{\mathbb {D}}}$, belongs to $H({{\mathbb {D}}})$. On the other hand, the series

$$\begin{aligned} \sum _{n=1}^\infty B_nz^n&=-\frac{1}{\nu ^2}\sum _{n=1}^\infty \sum _{h=1}^n\frac{t^h c_{n-h}}{(n+1)\prod _{j=n-h+1}^{n+1}(1-\frac{1}{j\nu })}\\&=-\frac{1}{\nu ^2}\sum _{h=1}^\infty t^hz^h\sum _{n=h}^\infty \frac{c_{n-h}z^{n-h}}{(n+1)\prod _{j=n-h+1}^{n+1}(1-\frac{1}{j\nu })},\quad z\in {{\mathbb {D}}}. \end{aligned}$$

To establish the convergence of the series $\sum _{n=1}^\infty B_nz^n$ in $H({{\mathbb {D}}})$, fix $z\in {{\mathbb {D}}}{\setminus }\{0\}$ and $r\in (|z|,1)$. Recall, for every $n\in {{\mathbb {N}}}_0$, that the Taylor coefficients of g satisfy (as $\frac{1}{r}>1$)

$$\begin{aligned} |c_n|=\left| \frac{g^{(n)}(0)}{n!}\right| =\left| \frac{1}{2\pi i}\int _{|\xi |=r}\frac{g(\xi )}{\xi ^{n+1}}\,d\xi \right| \le \frac{1}{r^{n}}\max _{|\xi |=r}|g(\xi )|\le \frac{C}{r^{n+1}} \end{aligned}$$

where $C:=\max _{|\xi |=r}|g(\xi )|$. Therefore, setting $\alpha :=\alpha (\nu )=\textrm{Re}(\frac{1}{\nu })$ and $d:=d_\delta$ and $D:=D_\delta$ for a suitable $\delta >0$ (cf. Remark 3.5), we obtain via (3.1) and (3.2) that

$$\begin{aligned}&\sum _{h=1}^\infty t^h|z|^h\sum _{n=h}^\infty \frac{|c_{n-h}|\,|z|^{n-h}}{(n+1)\prod _{j=n-h+1}^{n+1}|1-\frac{1}{j\nu }|}\\&\quad \le C\sum _{h=1}^\infty t^h|z|^{h-1}\left( \frac{|z|}{r}d^{-1}(h+1)^{-\alpha -1}+\sum _{n=h+1}^\infty \left( \frac{|z|}{r}\right) ^{n-h+1}Dd^{-1}\left( \frac{n+1}{n-h}\right) ^\alpha \right) \\&\quad =Cd^{-1}\sum _{h=1}^\infty t^h(h+1)^{-\alpha -1}|z|^{h}+ CDd^{-1}\sum _{h=1}^\infty t^h|z|^{h-1}\sum _{n=h+1}^\infty \left( \frac{|z|}{r}\right) ^{n-h+1}\left( \frac{n+1}{n-h}\right) ^\alpha \\&\quad \le Cd^{-1}\left( \sum _{h=1}^\infty t^h(h+1)^{-\alpha -1}|z|^{h}+D\sum _{h=1}^\infty t^h |z|^{h-1}\max \{1, (2+h)^\alpha \}\sum _{n=h+1}^\infty \left( \frac{|z|}{r}\right) ^{n-h+1}\right) , \end{aligned}$$

which is finite after observing that if $\alpha \le 0$, then $\left( \frac{n+1}{n-h}\right) ^\alpha =\left( \frac{n-h}{n+1}\right) ^{-\alpha }\le 1$ for every $h\in {{\mathbb {N}}}$ and every $n\ge h+1$, whereas if $\alpha >0$, then $(\frac{n+1}{n-h})^\alpha =(1+\frac{h+1}{n-h})^\alpha \le (2+h)^\alpha$. This implies that the series $\sum _{n=1}^\infty B^nz^n$ converges in $H({{\mathbb {D}}})$. Accordingly, $f_2(z):=\sum _{n=1}^\infty B_nz^n$, for $z\in {{\mathbb {D}}}$, belongs to $H({{\mathbb {D}}})$.

Set $f(z):=\frac{c_0}{1-\nu }+f_1(z)+f_2(z)$, for $z\in {{\mathbb {D}}}$. Then $f\in H({{\mathbb {D}}})$. Moreover, the arguments above imply that f satisfies (3.5). The identities (3.6) imply that f is the unique solution of (3.5). Accordingly, the inverse operator $(C_t-\nu I)^{-1}:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ exists. In particular, $(C_t-\nu I)^{-1}\in {{\mathcal {L}}}(H({{\mathbb {D}}}))$ as it is the inverse of a continuous linear operator on a Fréchet space.

Since $\nu \in {{\mathbb {C}}}{\setminus }\Lambda _0$ is arbitrary and $0\in \rho (C_t;H({{\mathbb {D}}}))$, we can conclude that $\sigma (C_t;H({{\mathbb {D}}}))=\Lambda$.

It remains to show that $\sigma ^*(C_t;H({{\mathbb {D}}}))=\Lambda _0$. To establish this, fix $\mu \in {{\mathbb {C}}}{\setminus }\Lambda _0$ and observe, by Lemma 3.4, that there exist $\delta >0$ and constants $d_\delta , D_\delta >0$ such that $\overline{B(\mu ,\delta )}\cap \Lambda _0=\emptyset$ and the inequalities (3.1) and (3.2) are satisfied. We will show that $B(\mu ,\delta )\subset \rho (C_t;H({{\mathbb {D}}}))$ and that the set $\{(C_t-\nu I)^{-1}:\, \nu \in B(\mu ,\delta )\}$ is equicontinuous in ${{\mathcal {L}}}(H({{\mathbb {D}}}))$. To see this, first observe that the function $\nu \in \overline{B(\mu ,\delta )} \mapsto \textrm{Re}(\frac{1}{\nu })\in {{\mathbb {R}}}$ is continuous and hence, $\alpha _0:=\max _{\nu \in \overline{B(\mu ,\delta )} }\{\textrm{Re}(\frac{1}{\nu })\}$ exists. For the sake of simplicity of notation set $d:=d_\delta$ and $D:=D_\delta$.

Let $\nu \in B(\mu ,r)$, where $r:=\frac{1}{2}d(\Lambda _0, \overline{B(\mu ,\delta )})>0$ has the property that $|\nu -\frac{1}{j}|>r$ for all $j\in {{\mathbb {N}}}$. It was proved above, for any fixed $g(z)=\sum _{n=0}^\infty c_nz^n\in H({{\mathbb {D}}})$, that

$$\begin{aligned} (C_t-\nu I)^{-1}g(z)=\frac{c_0}{1-\nu }+\sum _{n=1}^\infty \left( \frac{c_n}{\frac{1}{n+1}-\nu }-\frac{1}{\nu ^2} \sum _{h=1}^n\frac{(-1)^ht^hc_{n-h}}{(n+1)\prod _{j=n-h+1}^{n+1}(1-\frac{1}{j\nu })}\right) z^n, \end{aligned}$$

for each $z\in {{\mathbb {D}}}$. So, for $k\ge 2$ fixed, consider the norm $\Vert \cdot \Vert _k$ in $H({{\mathbb {D}}})$. Then we have, via (3.6), that

$$\begin{aligned}&\Vert (C_t-\nu I)^{-1}g\Vert _k\\&\quad \le \frac{|c_0|}{|1-\nu |}+\sum _{n=1}^\infty \left| \frac{c_n}{\frac{1}{n+1}-\nu }- \frac{1}{\nu ^2}\sum _{h=1}^n\frac{(-1)^ht^hc_{n-h}}{(n+1) \prod _{j=n-h+1}^{n+1}(1-\frac{1}{j\nu })}\right| \left( 1-\frac{1}{k}\right) ^n\\&\quad \le \left( \frac{1}{r}\sum _{n=0}^\infty |c_n|\left( 1-\frac{1}{k}\right) ^n\right) +\frac{1}{|\nu |^2}\sum _{n=1}^\infty \sum _{h=1}^n\frac{t^h|c_{n-h}|}{(n+1) \prod _{j=n-h+1}^{n+1}|1-\frac{1}{j\nu }|}\left( 1-\frac{1}{k}\right) ^n\\&\quad =\frac{1}{r}\Vert g\Vert _k+\frac{1}{|\nu |^2}\sum _{h=1}^\infty t^h\left( 1-\frac{1}{k}\right) ^h\sum _{n=h}^\infty \frac{|c_{n-h}|}{(n+1) \prod _{j=n-h+1}^{n+1}|1-\frac{1}{j\nu }|}\left( 1-\frac{1}{k}\right) ^{n-h}. \end{aligned}$$

Moreover, (3.1) and (3.2) with $\alpha (\nu )=\textrm{Re}(\frac{1}{\nu })\le \alpha _0$ imply, for each $h\in {{\mathbb {N}}}$, that

$$\begin{aligned}&\sum _{n=h}^\infty \frac{|c_{n-h}|}{(n+1)\prod _{j=n-h+1}^{n+1}|1-\frac{1}{j\nu }|}\left( 1-\frac{1}{k}\right) ^{n-h}=\sum _{l=0}^\infty \frac{|c_{l}|}{(l+h+1)\prod _{j=l+1}^{l+h+1}|1-\frac{1}{j\nu }|}\left( 1-\frac{1}{k}\right) ^{l}\\&\quad = \frac{|c_0|}{(h+1)\prod _{j=1}^{h+1}|1-\frac{1}{j\nu }|}+\sum _{l=1}^\infty \frac{|c_{l}|}{(l+h+1)\prod _{j=l+1}^{l+h+1}|1-\frac{1}{j\nu }|}\left( 1-\frac{1}{k}\right) ^{l}\\&\quad \le d^{-1}|c_0|(h+1)^{\alpha (\nu ) -1}+d^{-1}D \sum _{l=1}^\infty \frac{|c_l|}{l+h+1}\left( \frac{l+h+1}{l}\right) ^{\alpha (\nu )}\left( 1-\frac{1}{k}\right) ^l\\&\quad \le d^{-1}|c_0|(h+1)^{\alpha _0 -1}+d^{-1}D \sum _{l=1}^\infty \frac{|c_l|}{l+h+1}\left( \frac{l+h+1}{l}\right) ^{\alpha _0}\left( 1-\frac{1}{k}\right) ^l\\&\quad \le \max \{d^{-1},d^{-1}D\}(2+h)^{\alpha _0}\sum _{l=0}^\infty |c_l|\left( 1-\frac{1}{k}\right) ^l=K(2+h)^{\alpha _0}\Vert g\Vert _k, \end{aligned}$$

with $K:=\max \{d^{-1},d^{-1}D\}$, and hence, since $|\nu |>r$ for all $\nu \in B(\mu ,\delta )$, that

$$\begin{aligned}&\frac{1}{|\nu |^2}\sum _{h=1}^\infty t^h\left( 1-\frac{1}{k}\right) ^h\sum _{n=h}^\infty \frac{|c_{n-h}|}{(n+1) \prod _{j=n-h+1}^{n+1}|1-\frac{1}{j\nu }|}\left( 1-\frac{1}{k}\right) ^{n-h}\\&\quad \le \frac{K}{r^2}\Vert g\Vert _k\sum _{h=1}^\infty t^h\left( 1-\frac{1}{k}\right) ^h(2+h)^{\alpha _0}=K'\Vert g\Vert _k, \end{aligned}$$

with $K'=\frac{K}{r^2}\sum _{h=1}^\infty t^h\left( 1-\frac{1}{k}\right) ^h(2+h)^{\alpha _0}<\infty$, by the ratio test, for instance.

We have established, for every $\nu \in B(\mu ,\delta )$, that

$$\begin{aligned} \Vert (C_t-\nu I)^{-1}g\Vert _k\le (\frac{1}{r}+K')\Vert g\Vert _k. \end{aligned}$$

Since $g\in H({{\mathbb {D}}})$ and $k\ge 2$ are arbitrary, this shows that the set $\{(C_t-\nu I)^{-1}:\, \nu \in B(\mu ,\delta )\}$ is equicontinuous. Hence, $\sigma ^*(C_t;H({{\mathbb {D}}}))=\Lambda _0$.$\square$

Proposition 3.8

For each $t\in [0,1)$ the operator $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ is power bounded, uniformly mean ergodic but, it fails to be supercyclic. Moreover, $(I-C_t)(H({{\mathbb {D}}}))$ is the closed subspace of $H({{\mathbb {D}}})$ given by

$$\begin{aligned} (I-C_t)(H({{\mathbb {D}}}))=\{g\in H({{\mathbb {D}}})\, :\, g(0)=0\} \end{aligned}$$

(3.7)

and we have the decomposition

$$\begin{aligned} H({{\mathbb {D}}})=\textrm{Ker}(I-C_t)\oplus (I-C_t)(H({{\mathbb {D}}})). \end{aligned}$$

(3.8)

Proof

Fix $t\in [0,1)$. We first prove that $C_t$ is power bounded. Once this is established, $C_t$ is necessarily uniform mean ergodic because $H({{\mathbb {D}}})$ is a Fréchet- Montel space (see [1, Proposition 2.8]).

Given $k\ge 2$ we have, for every $f\in H({{\mathbb {D}}})$ and with $r_k:=(1-\frac{1}{k})$, that

$$\begin{aligned} |||C_tf|||_k&=\sup _{n\in {{\mathbb {N}}}_0}\left| \frac{1}{n+1}\sum _{j=0}^n t^{n-j}{\hat{f}}(j)\right| r_k\le \sup _{n\in {{\mathbb {N}}}_0}\frac{1}{n+1}\sum _{j=0}^n |{\hat{f}}(j)|r_k^n\\&\le \sup _{n\in {{\mathbb {N}}}_0}\frac{1}{n+1}\sum _{j=0}^n |{\hat{f}}(j)|r_k^j\le \sup _{j\in {{\mathbb {N}}}_0}|{\hat{f}}(j)|r_k^j =|||f|||_k, \end{aligned}$$

because $r_k^n\le r_k^j$ for all $j\in \{0,1,\ldots ,n\}$. It follows, for every $n\in {{\mathbb {N}}}$, that

$$\begin{aligned} |||C_t^nf|||_k\le |||f|||_k,\quad f\in H({{\mathbb {D}}}). \end{aligned}$$

Since $k\ge 2$ is arbitrary, the operator $C_t\in {{\mathcal {L}}}(H({{\mathbb {D}}}))$ is indeed power bounded.

To establish that $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ is not supercyclic, note that the continuous embedding $\Phi :H({{\mathbb {D}}})\rightarrow \omega$ has dense range. The operator $C_t^\omega \in {{\mathcal {L}}}(\omega )$ satisfies $\Phi \circ C_t=C_t^\omega \circ \Phi$ as an identity in ${{\mathcal {L}}}(H({{\mathbb {D}}}),\omega )$, which implies if $C_t:H({{\mathbb {D}}})\rightarrow H({{\mathbb {D}}})$ is supercyclic, then also $C_t^\omega :\omega \rightarrow \omega$ must be supercyclic as $\Phi \circ C_t^n=\Phi \circ C_t\circ C_t^{n-1}=C_t^\omega \circ \Phi \circ C_t^{n-1}=\cdots =(C_t^\omega )^n\circ \Phi$, for all $n\in {{\mathbb {N}}}$, and $\Phi (H({{\mathbb {D}}}))$ is dense in $\omega$. A contradition with [5, Theorem 6.1].

To establish (3.7) note that $(I-C_t)(H({{\mathbb {D}}}))\subseteq \{g\in H({{\mathbb {D}}})\,:\ g(0)=0\}$ because $C_tf(0)=f(0)$ for every $f\in H({{\mathbb {D}}})$. To show the reverse inclusion, let $g\in H({{\mathbb {D}}})$ satisfy $g(0)=0$. Then $h(z):=zg'(z)+g(z)$, for $z\in {{\mathbb {D}}}$, is holomorphic and $h(0)=0$. Accordingly, also $z\mapsto \frac{h(z)}{z}$, for $z\in {{\mathbb {D}}}{\setminus }\{0\}$, and taking the value $h'(0)$ at $z=0$ is holomorphic in ${{\mathbb {D}}}$. Define $f\in H({{\mathbb {D}}})$ by

$$\begin{aligned} f(z):=\frac{1}{tz-1}\int _0^z(1-t\xi )\frac{h(\xi )}{\xi }\,d\xi ,\quad z\in {{\mathbb {D}}}, \end{aligned}$$

and note that $f(0)=0$. Direct calculation reveals that

$$\begin{aligned} \frac{f(z)}{1-tz}-(zf(z))'=h(z)=(zg(z))',\quad z\in {{\mathbb {D}}}, \end{aligned}$$

from which it follows that

$$\begin{aligned} \int _0^z\frac{f(\xi )}{1-t\xi }\,d\xi -zf(z)=zg(z),\quad z\in {{\mathbb {D}}}. \end{aligned}$$

Since $f(0)=0$, we can conclude that

$$\begin{aligned} \frac{1}{z}\int _0^z\frac{f(\xi )}{1-t\xi }\,d\xi -f(z)=g(z),\quad z\in {{\mathbb {D}}}, \end{aligned}$$

that is, $(C_t-I)f=g$ and so $g\in (I-C_t)(H({{\mathbb {D}}}))$. Hence, (3.7) is valid.

To show the validity of (3.8) it suffices to repeat the argument given in the proof of Proposition 3.2. $\square$