On the continuous Cesàro operator in certain function spaces

Abstract

Various properties of the (continuous) Cesàro operator $\mathsf {C}$, acting on Banach and Fréchet spaces of continuous functions and $L^p$-spaces, are investigated. For instance, the spectrum and point spectrum of $\mathsf {C}$ are completely determined and a study of certain dynamics of $\mathsf {C}$ is undertaken (eg. hyper- and supercyclicity, chaotic behaviour). In addition, the mean (and uniform mean) ergodic nature of $\mathsf {C}$ acting in the various spaces is identified.

Appendix

Here we collect a few relevant results concerning the spectrum and mean ergodic properties of continuous linear operators defined on certain classes of Fréchet spaces.

Lemma 5.1

Let $X$ be a Fréchet space and $S\in {\mathcal {L}}(X)$. Suppose that $X=\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j, Q_{j,j+1})$, with $X_j$ a Banach space (having norm $\Vert \ \Vert _j$) and linking maps $Q_{j,j+1}\in {\mathcal {L}}(X_{j+1},X_j)$ which are surjective for all $j\in {\mathbb {N}}$, and suppose, for each $j\in {\mathbb {N}}$, that there exists $S_j\in {\mathcal {L}}(X_j)$ satisfying

$$\begin{aligned} S_jQ_j=Q_jS, \end{aligned}$$

(5.1)

where $Q_j\in {\mathcal {L}}(X,X_j)$, $j\in {\mathbb {N}}$, denotes the canonical projection of $X$ onto $X_j$ (i.e., $Q_{j,j+1}\circ Q_{j+1}=Q_j$). Then

$$\begin{aligned} \sigma (S)\subseteq \cup _{j=1}^\infty \sigma (S_j)\subseteq \sigma (S)\cup \cup _{j=1}^\infty \sigma _{pt}(S_j). \end{aligned}$$

(5.2)

Moreover,

$$\begin{aligned} \sigma _{pt}(S)\subseteq \cup _{j=1}^\infty \sigma _{pt}(S_j). \end{aligned}$$

(5.3)

Proof

It follows from (5.1) that

$$\begin{aligned} (\lambda I_j-S_j)Q_j=Q_j(\lambda I-S) \end{aligned}$$

(5.4)

for all $j\in {\mathbb {N}}$ and $\lambda \in {\mathbb {C}}$, where $I_j$ denotes the identity map on $X_j$.

Fix any $\lambda \in \cap _{j=1}^\infty \rho (S_j)$. If $(\lambda I-S)x=0$ for some $x\in X$, then by (5.4) we have $(\lambda I_j-S_j)Q_jx=Q_j(\lambda I-S)x=0$ for all $j\in {\mathbb {N}}$. It follows that $Q_jx=0$ for all $j\in {\mathbb {N}}$. This implies that $x=0$ as $x\in X= \mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j, Q_{j,j+1})$. The proof of the surjectivity of $(\lambda I-S)$ follows as in the last part of the proof (cf. p. 154) of Theorem 4.1 of [6] via (5.4) and the fact that $(\lambda I_j-S_j)$ is bijective for all $j\in {\mathbb {N}}$. As $X$ is a Fréchet space, we can conclude that $(\lambda I-S)^{-1}\in {\mathcal {L}}(X)$ and so $\lambda \in \rho (S)$. This establishes that $\sigma (S)\subseteq \cup _{j=1}^\infty \sigma (S_j)$.

To verify the second containment in (5.2) we first observe that if $\mu \in \rho (S)$, then $(\mu I- S)$ is invertible in ${\mathcal {L}}(X)$ and hence, $(\mu I_j- S_j)\in {\mathcal {L}}(X_j)$ is surjective for all $j\in {\mathbb {N}}$; this follows routinely from (5.4) and the fact that each operator $Q_j$, for $j\in {\mathbb {N}}$, is surjective. Suppose that $\nu \in \rho (S){\setminus } \cap _{j=1}^\infty \rho (S_j)$. Then $\nu \not \in \rho (S_{j_0})$ for some $j_0\in {\mathbb {N}}$, i.e., $(\nu I_{j_0}-S_{j_0})$ is not invertible in ${\mathcal {L}}(X_{j_0})$. Since $(\nu I_{j_0}-S_{j_0})$ is surjective, it follows that $\nu \in \sigma _{pt}(S_{j_0})$.

Now, let $\lambda \in \cup _{j=1}^\infty \sigma (S_j)$. If $\lambda \in \sigma (S)$, then there is nothing to prove. If $\lambda \not \in \sigma (S)$, then $\lambda \in \rho (S)$. From the previous paragraph $\lambda \in \sigma _{pt}(S_{j_0})$ for some $j_0\in {\mathbb {N}}$, i.e., $\lambda \in \cup _{j=1}^\infty \sigma _{pt}(S_j)$. This establishes the second containment in (5.2). Thereby (5.2) has been proved.

To verify (5.3) let $\lambda \in (\cup _{j=1}^\infty \sigma _{pt}(S_j))^c$, in which case $(\lambda _jI_j-S_j)$ is injective for each $j\in {\mathbb {N}}$. Suppose that $x\in X$ satisfies $(\lambda I- S)x=0$ in which case (5.4) implies that $(\lambda I_j-S_j)Q_jx=0$ for every $j\in {\mathbb {N}}$. Hence, $Q_jx=0$ for every $j\in {\mathbb {N}}$ and so $x=0$. This shows that $(\lambda I-S)$ is injective and so $\lambda \not \in \sigma _{pt}(S)$, i.e., $\lambda \in (\sigma _{pt}(S))^c$. Thereby (5.3) is established. $\square$

A Fréchet space $X$ is always a projective limit of continuous linear operators $R_{j}:\ X_{j+1}\rightarrow X_j$, for $j\in {\mathbb {N}}$, with each $X_j$ a Banach space. If $X_j$ and $R_j$ can be chosen such that each $R_j$ is surjective and $X$ is isomorphic to the projective limit $\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j,R_j)$, then $X$ is called a quojection [9, Section 5]. Banach spaces and countable products of Banach spaces are quojections. Moscatelli [27] gave the first examples of quojections which are not isomorphic to countable products of Banach spaces. Concrete examples of quojection Fréchet spaces are $\omega ={\mathbb {C}}^{\mathbb {N}}$, the spaces $L^p_{loc}(\Omega )$, for $1\le p\le \infty$, and $C^{(m)}(\Omega )$ for $m\in {\mathbb {N}}_0$, with $\Omega \subseteq {\mathbb {R}}^N$ any open set, all of which are isomorphic to countable products of Banach spaces. We refer the reader to the survey paper [26] for further information. Under the assumptions of Lemma 5.1 the Fréchet space $X$ there is necessarily a quojection. The same is true in Lemmas 5.2 and 5.4 to follow.

Lemma 5.2

Let $X$ be a Fréchet space and $\{S_n\}_{n=1}^\infty \subseteq {\mathcal {L}}(X)$. Suppose that $X=\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j, Q_{j,j+1})$, with $X_j$ a Banach space (having norm $\Vert \ \Vert _j$) and linking maps $Q_{j,j+1}\in {\mathcal {L}}(X_{j+1},X_j)$ which are surjective for all $j\in {\mathbb {N}}$, and suppose, for each $j,\ n\in {\mathbb {N}}$, that there exists $S_n^{(j)}\in {\mathcal {L}}(X_j)$ satisfying

$$\begin{aligned} S_n^{(j)}Q_j=Q_jS_n, \end{aligned}$$

(5.5)

where $Q_j\in {\mathcal {L}}(X,X_j)$, $j\in {\mathbb {N}}$, denotes the canonical projection of $X$ onto $X_j$ (i.e., $Q_{j,j+1}\circ Q_{j+1}=Q_j$). Then the following statements are equivalent.

• (i) The limit $\tau _b$-$\lim _{n\rightarrow \infty }S_n=:S$ exists in ${\mathcal {L}}_b(X)$.

• (ii) For each $j\in {\mathbb {N}}$, the limit $\tau _b$-$\lim _{n\rightarrow \infty }S_n^{(j)}=:S^{(j)}$ exists in ${\mathcal {L}}_b(X_j)$.

In this case, the operators $S\in {\mathcal {L}}(X)$ and $S^{(j)}\in {\mathcal {L}}(X_j)$, for $j\in {\mathbb {N}}$, satisfy

$$\begin{aligned} Sx=(S^{(j)}x_j)_{j}, \quad x=(x_j)_{j}\in X. \end{aligned}$$

(5.6)

Proof

For each $j\in {\mathbb {N}}$, define $q_j(x):=\Vert Q_jx\Vert _j$ for $x\in X$. Then $\{q_j\}_{j=1}^\infty \subseteq \Gamma _X$ is a fundamental sequence generating the lc-topology of $X$ (as $X=\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j, Q_{j,j+1})$).

(i) $\Rightarrow$ (ii). The existence in ${\mathcal {L}}_b(X)$ of the stated limit $S\in {\mathcal {L}}(X)$ ensures the existence (in the norm of $X_j$) of

$$\begin{aligned} \lim _{n\rightarrow \infty }S_n^{(j)}Q_jx=\lim _{n\rightarrow \infty }Q_jS_nx=Q_jSx, \end{aligned}$$

(5.7)

for all $j\in {\mathbb {N}}$ and $x\in X$, via the continuity of $Q_j$ and (5.5). In fact, the weaker requirement that $S_n\rightarrow S$ in ${\mathcal {L}}_s(X)$ suffices for this.

Fix $j\in {\mathbb {N}}$. Define $S^{(j)}$ on $X_j=Q_j(X)$ by $S^{(j)}(Q_jx):=Q_jSx$, for $x\in X$. Then $S^{(j)}\in {\mathcal {L}}(X_j)$. Indeed, $S^{(j)}$ is well defined because if $Q_jx=Q_jx^{\prime }$ for some $x,\,x^{\prime }\in X$, then $Q_j(x-x^{\prime })=0$ and so, via (5.5), $0=S_n^{(j)}Q_j(x-x^{\prime })=Q_j S_n(x-x^{\prime })$ for all $n\in {\mathbb {N}}$. Passing to the limit for $n\rightarrow \infty$, it follows that $0=Q_jS(x-x^{\prime })$, i.e., $Q_jSx=Q_j Sx^{\prime }$. Clearly, $S^{(j)}$ is linear as both $Q_j$ and $S$ are linear. Finally, since $S^{(j)}u=\lim _{n\rightarrow \infty }S_n^{(j)}u$, for each $u\in X_j$ [cf. (5.7)] and $\{S_n^{(j)}\}_{n=1}^\infty \subseteq {\mathcal {L}}(X_j)$ with $X_j$ a Banach space, it follows from the Uniform Boundedness Principle that $S^{(j)}$ is continuous and hence, $S_n^{(j)}\rightarrow S^{(j)}$ in ${\mathcal {L}}_s(X_j)$ for $n\rightarrow \infty$. It is routine [via (5.5)] to check that $S^{(j)}Q_j=Q_jS$.

As noted above, $X$ is necessarily a quojection and so there exists $B\in {\mathcal {B}}(X)$ such that $\mathcal {U}_j\subseteq Q_j(B)$ [13, Proposition 1], where $\mathcal {U}_j$ is the closed unit ball of $X_j$. So, by (5.5) we have

$$\begin{aligned} \sup _{u\in \mathcal {U}_j}\left\| \left( S_n^{(j)}-S^{(j)}\right) u\right\| _j&\le \sup _{x\in B}\left\| \left( S_n^{(j)}-S^{(j)}\right) Q_j{x}\right\| _j\\&=\sup _{x\in B}\Vert (Q_j(S_n-S)x\Vert _j=\sup _{x\in B}{q}_j((S_n-S)x) \end{aligned}$$

for all $n\in {\mathbb {N}}$. Since $\sup _{x\in B}{q}_j((S_n-S)x)\rightarrow 0$ for $n\rightarrow \infty$ (by assumption), it follows that $\sup _{u\in \mathcal {U}_j}\Vert (S_n^{(j)}-S^{(j)})u\Vert _j\rightarrow 0$ for $n\rightarrow \infty$, i.e., $\tau _b$-$\lim _{n\rightarrow \infty }S_n^{(j)}=S^{(j)}$. Since $j\in {\mathbb {N}}$ is arbitrary, the proof is complete.

(ii) $\Rightarrow$ (i). Fix $x=(x_j)_{j}\in X=\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j,Q_{j,j+1})$ and set $Sx:=(S^{(j)}x_j)_{j}$. Then $Sx\in X$. Indeed, $Q_jx=x_j$ for all $j\in {\mathbb {N}}$ and so, via (5.5), we have

$$\begin{aligned} Q_{j,j+1}S_{j+1}x_{j+1}&=\lim _{n\rightarrow \infty }Q_{j,j+1}S_n^{(j+1)}Q_{j+1}x =\lim _{n\rightarrow \infty }Q_{j,j+1}Q_{j+1}S_nx\\&=\lim _{n\rightarrow \infty }Q_{j}S_nx=\lim _{n\rightarrow \infty }S_n^{(j)}Q_jx=S^{(j)}x_j, \end{aligned}$$

for all $j\in {\mathbb {N}}$, i.e., $Sx\in X$. Clearly, the linearity of the $S^{(j)}$’s imply the linearity of the map $S:x\mapsto Sx$, for $x\in X$. Moreover, the continuity of $S$ is a consequence of $X=\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j,Q_{j,j+1})$. Next, fix $j\in {\mathbb {N}}$ and $B\in {\mathcal {B}}(X)$. Again via (5.5) we have

$$\begin{aligned} \sup _{x\in B}q_j((S_n-S)x)&=\sup _{x\in B}\Vert (Q_j(S_n-S)x\Vert _j=\sup _{x\in B}\left\| \left( S_n^{(j)}-S^{(j)}\right) Q_jx\right\| _j\\&= \sup _{u\in Q_j(B)}\left\| \left( S_n^{(j)}-S^{(j)}\right) u\right\| _j \end{aligned}$$

for all $n\in {\mathbb {N}}$. Since $Q_j(B)\in {\mathcal {B}}(X_j)$, it follows from the assumption (ii) that $\sup _{u\in Q_j(B)}\Vert (S_n^{(j)}-S^{(j)})u\Vert _j \rightarrow 0$ for $n\rightarrow \infty$. Accordingly, for each $j\in {\mathbb {N}}$ and each $B\in {\mathcal {B}}(X)$ we have $\lim _{n\rightarrow \infty }\sup _{x\in B}q_j((S_n-S)x)= 0$, i.e., (i) holds. $\square$

Remark 5.3

A careful examination of the proof of Lemma 5.2 shows that the equivalence (i) $\Leftrightarrow$ (ii) remains valid if $\tau _b$ is replaced with $\tau _s$.

Lemma 5.4

Let $X=\mathrm{proj\,}_{j\in {\mathbb {N}}}(X_j, Q_{j.j+1})$ be a Fréchet space and operators $S\in {\mathcal {L}}(X)$ and $S_j\in {\mathcal {L}}(X_j)$, for $j\in {\mathbb {N}}$, be given which satisfy the assumptions of Lemma 5.1 (with $Q_j\in {\mathcal {L}}(X,X_j)$, $j\in {\mathbb {N}}$, denoting the canonical projection of $X$ onto $X_j$ and $\Vert \ \Vert _j$ being the norm in the Banach space $X_j$).

• (i) $S\in {\mathcal {L}}(X)$ is power bounded if and only if each $S_j\in {\mathcal {L}}(X_j)$, $j\in {\mathbb {N}}$, is power bounded.

• (ii) $S\in {\mathcal {L}}(X)$ is uniformly mean ergodic if and only if each $S_j\in {\mathcal {L}}(X_j)$, $j\in {\mathbb {N}}$, is uniformly mean ergodic.

• (iii) $S\in {\mathcal {L}}(X)$ is mean ergodic if and only if each $S_j\in {\mathcal {L}}(X_j)$, $j\in {\mathbb {N}}$, is mean ergodic.

Proof

Let $\{q_j\}_{j=1}^\infty \subseteq \Gamma _X$ be the fundamental sequence of seminorms generating the lc-topology of $X$ as given in the proof of Lemma 5.2.

• (i) Suppose that each $S_j\in {\mathcal {L}}(X_j)$, $j\in {\mathbb {N}}$, is power bounded, i.e., there exists $M_j>0$ such that

  $$\begin{aligned} \left\| S_j^nu\right\| _j\le M_j\Vert u\Vert _j,\quad u\in X_j,\ n\in {\mathbb {N}}. \end{aligned}$$

  It follows from (5.1) that $S^n_jQ_j=Q_jS^n$ for all $j,\ n\in {\mathbb {N}}$. Fix $j\in {\mathbb {N}}$. Then, for each $n\in {\mathbb {N}}$ and $x\in X$ we have

  $$\begin{aligned} q_j(S^nx)=\left\| Q_jS^n x\right\| _j=\left\| S_j^nQ_j x\right\| _j\le M_j\Vert Q_jx\Vert _j=M_jq_j(x). \end{aligned}$$

  Since $\{q_j\}_{j=1}^\infty$ generate the lc-topology of the Fréchet space $X$, it follows that $\{S^n\}_{n=1}^\infty \subseteq {\mathcal {L}}(X)$ is equicontinuous, i.e., $S$ is power bounded. Conversely, suppose that $S$ is power bounded. Fix $j\in {\mathbb {N}}$ and let $\mathcal {U}_j$ be the closed unit ball of $X_j$. Since $X$ is a quojection, there exists $B\in {\mathcal {B}}(X)$ with   $\mathcal {U}_{j} \subseteq Q_j(B)$. Moreover, the power boundedness of $S$ implies that $C:=\cup _{n\in {\mathbb {N}}}S^n(B)\in {\mathcal {B}}(X)$ and hence, there exists $M>0$ such that $q_j(z)\le M$ for every $z\in C$. Let $u\in \mathcal {U}_j$. Then $u=Q_jx$ for some $x\in B$ and so

  $$\begin{aligned} \left\| S_j^nu\right\| _j=\left\| S_j^nQ_jx\right\| _j=\left\| Q_jS^nx\right\| _j=q_j(S^nx)\le M, \end{aligned}$$

  for every $n\in {\mathbb {N}}$. This implies that the operator norms satisfy $\Vert S_j^n\Vert _{op}\le M$, for $n\in {\mathbb {N}}$. Accordingly, $S_j\in {\mathcal {L}}(X_j)$ is power bounded.

• (ii) For each $n\in {\mathbb {N}}$ define $\tilde{S}_n:=S_{[n]}\in {\mathcal {L}}(X)$ and $\tilde{S}_n^{(j)}:=(S_j)_{[n]}\in {\mathcal {L}}(X_j)$, for $j\in {\mathbb {N}}$. It follows from (5.1) that $\tilde{S}_n^{(j)}Q_j=Q_j\tilde{S}_n$, for $j,\ n\in {\mathbb {N}}$. Accordingly, we can apply Lemma 5.2 (with $\tilde{S}_n$ in place of $S_n$ and $\tilde{S}_n^{(j)}$ in place of $S_n^{(j)}$) to conclude that $S$ is uniformly mean ergodic if and only if each $S_j$, $j\in {\mathbb {N}}$, is uniformly mean ergodic.

• (iii) Apply the same argument as in part (ii) but now apply Lemma 5.2 with $\tau _s$ in place of $\tau _b$; see Remark 5.3.

$\square$