Spectral properties of generalized Cesàro operators in sequence spaces

Abstract

The generalized Cesàro operators $C_t$, for $t\in [0,1]$, were first investigated in the 1980s. They act continuously in many classical Banach sequence spaces contained in ${{\mathbb {C}}}^{{{\mathbb {N}}}_0}$, such as $\ell ^p$, $c_0$, c, $bv_0$, bv and, as recently shown in Curbera et al. (J Math Anal Appl 507:31, 2022) [26], also in the discrete Cesàro spaces ces(p) and their (isomorphic) dual spaces $d_p$. In most cases $C_t$ ($t\not =1$) is compact and its spectra and point spectrum, together with the corresponding eigenspaces, are known. We study these properties of $C_t$, as well as their linear dynamics and mean ergodicity, when they act in certain non-normable sequence spaces contained in ${{\mathbb {C}}}^{{{\mathbb {N}}}_0}$. Besides ${{\mathbb {C}}}^{{{\mathbb {N}}}_0}$ itself, the Fréchet spaces considered are $\ell (p+)$, $ces(p+)$ and $d(p+)$, for $1\le p<\infty$, as well as the (LB)-spaces $\ell (p-)$, $ces(p-)$ and $d(p-)$, for $1<p\le \infty$.

1 Introduction

The (discrete) generalized Cesàro operators $C_t$, for $t\in [0,1]$, were first investigated by Rhaly, [52]. The action of $C_t$ from $\omega :={{\mathbb {C}}}^{{{\mathbb {N}}}_0}$ into itself (with ${{\mathbb {N}}}_0:=\{0,1,2,\ldots \}$) is given by

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

(1.1)

For $t=0$ 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)

where $\varphi :=\left( \frac{1}{n+1}\right) _{n\in {{\mathbb {N}}}_0}$, and for $t=1$ that $C_1$ is the classical Cesàro averaging operator

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

(1.3)

The spectra of $C_1$ have been investigated in various Banach sequence spaces. For instance, we mention $\ell ^p$ ($1<p<\infty$), [22, 23, 30, 40], $c_0$ [1, 40, 51], c [40], $\ell ^\infty$ [50, 51], the Bachelis spaces $N^p$ ($1<p<\infty$) [25], bv and $bv_0$ [47, 48], weighted $\ell ^p$ spaces [7, 10], the discrete Cesàro spaces ces(p) (for $p\in \{0\}\cup (1,\infty )$), [24], and their dual spaces $d_s$ ($1<s<\infty$), [19]. For the class of generalized Cesàro operators $C_t$, for $t\in (0,1)$, a study of their spectra and compactness properties (in $\ell ^2$) go back to Rhaly, [52, 53]. A similar investigation occurs for $\ell ^p$ ($1<p<\infty$) in [58] and for c and $c_0$ in [55, 59]. The paper [55] also treats $C_t$ when it acts on $bv_0$, bv, c, $\ell ^1$, $\ell ^\infty$ and the Hahn sequence space h. In the recent paper [26] the setting for considering the operators $C_t$ is a large class of Banach lattices in $\omega$, which includes all rearrangement invariant sequence spaces (over ${{\mathbb {N}}}_0$ for counting measure), and many others.

Our aim is to study the compactness, the spectra and the dynamics of the generalized Cesàro operators $C_t$, for $t\in [0,1)$, when they act in certain classical, non-normable sequence spaces $X\subseteq \omega$. Besides $\omega$ itself, the Fréchet spaces considered are $\ell (p+)$, $ces(p+)$ and $d(p+)$, for $1\le p<\infty$, as well as the (LB)-spaces $\ell (p-)$, $ces(p-)$ and $d(p-)$, for $1<p\le \infty$.

In Sect. 2 we formulate various preliminaries that will be needed in the sequel concerning particular properties of the spaces X that we consider, as well as linear operators between such spaces. We also collect some general results required to determine the spectra of operators T acting in the spaces X and the compactness of their dual operator $T'$ acting in the strong dual space $X'_\beta$ of X.

Section 3 is devoted to a detailed study of the operators $C_t$, for $t\in [0,1)$, when they act in $\omega$. These operators are never compact (c.f. Proposition 3.2) and their spectrum is completely described in Theorem 3.7 where, in particular, it is established that the set of all eigenvalues of $C_t$ is independent of t and equals $\Lambda :=\{\frac{1}{n+1}:\, n\in {{\mathbb {N}}}_0\}$. The 1-dimensional eigenspace corresponding to $\frac{1}{n+1}$, for each $n\in {{\mathbb {N}}}_0$, is identified in Lemma 3.4.

The situation for the other mentioned spaces $X\subseteq \omega$, which is rather different, is treated in Sects. 4 and 5. The operator $C_t$, for $t\in [0,1)$, is always compact in these spaces; see Theorem 4.5(i) for the case of Fréchet spaces and Theorem 5.3(i) for the case of (LB)-spaces. The spectra of $C_t$ are fully determined in Theorems 4.5(ii) and 5.3(ii), and the 1-dimensional eigenspace corresponding to each eigenvalue of $C_t$ is identified in Theorems 4.5(iii) and 5.3(iii). We note, for all cases of X and $t\in [0,1)$, that the set of all eigenvalues of $C_t$ is again $\Lambda$. The main tool is a factorization result stating that $C_t=D_\varphi R_t$, where $D_\varphi :X\rightarrow X$ is a compact (diagonal) operator in X and $R_t:X\rightarrow X$ is a continuous linear operator; see Propositions 4.4(iii) and 5.2(iii).

For the definition of a mean ergodic operator and the notion of a supercyclic operator we refer to Sect. 6, where the relevant operators under consideration are $C_t$ acting in the spaces X, for each $t\in [0,1)$. It is necessary to determine some abstract results for linear operators in general lcHs’ (c.f., Theorems 6.2 and 6.4), which are then applied to $C_t$ to show that it is both power bounded and uniformly mean ergodic in all spaces $X\not =\omega$; see Theorem 6.6. The same is true for $C_t$ acting in $\omega$; see Theorem 6.1. In this section we also investigate the properties of the dual operators $C'_t$ acting in $X'_\beta$, which are given by (6.6) and (6.8). The operators $C_t'$ are compact and their spectra are identified in Proposition 6.7, where it is also shown that the set of all eigenvalues of $C'_t$ is $\Lambda$. Moreover, for each $n\in {{\mathbb {N}}}_0$, the eigenvector in $X'_\beta$ spanning the 1-dimensional eigenspace corresponding to $\frac{1}{n+1}\in \Lambda$ is also determined. A consequence of $C'_t$ having a rich supply of eigenvalues is that each operator $C_t:X\rightarrow X$, for $t\in [0,1)$, fails to be supercyclic. Moreover, it is established in Proposition 6.8 that $C'_t:X'_\beta \rightarrow X'_\beta$ is power bounded, uniformly mean ergodic but, not supercyclic. It should be noted that the main results in this section are also new for $C_t$ acting in the Banach spaces $\ell ^p$, ces(p) and $d_p$.

2 Preliminaries

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 $\tau _s$ on X (i.e., the strong operator topology) the lcHs ${{\mathcal {L}}}(X)$ is denoted by ${{\mathcal {L}}}_s(X)$ and for the topology $\tau _b$ of uniform convergence on bounded sets the lcHs ${{\mathcal {L}}}(X)$ is denoted by ${{\mathcal {L}}}_b(X)$. Denote by ${{\mathcal {B}}}(X)$ the collection of all bounded subsets of X and by $\Gamma _X$ a system of continuous seminorms determing the topology of X. The identity operator on X is denoted by I. The dual operator of $T\in {{\mathcal {L}}}(X)$ is denoted by $T'$; it acts in the topological dual space $X':={{\mathcal {L}}}(X,{{\mathbb {C}}})$ of X. Denote by $X'_\sigma$ (resp., by $X'_\beta$) the space $X'$ with the weak* topology $\sigma (X',X)$ (resp., with the strong topology $\beta (X',X)$); see [37, Sect. 21.2] for the definition. It is known that $T'\in {{\mathcal {L}}}(X'_\sigma )$ and $T'\in {{\mathcal {L}}}(X_\beta ')$, [38, p. 134]. For the general theory of functional analysis and operator theory relevant to this paper see, for example, [27, 33, 36, 44, 49, 56].

Lemma 2.1

Let X be a lcHs and $T\in {{\mathcal {L}}}(X)$ be an isomorphism of X onto itself. Then $T'$ is an isomorphism of $X'_\beta$ onto itself. If, in addition, X is complete and barrelled, then T is an isomorphism of X onto itself if, and only if, $T'$ is an isomorphism of $X'_\beta$ onto itself.

Proof

If T is an isomorphism of X onto itself, then $T^{-1}\in {{\mathcal {L}}}(X)$ exists with $TT^{-1}=T^{-1}T=I$. It was already noted that $T', (T^{-1})'\in {{\mathcal {L}}}(X'_\beta )$ and clearly $(T^{-1})'T'=T'(T^{-1})'=I$. Thus, $(T')^{-1}$ exists in ${{\mathcal {L}}}(X'_\beta )$ and $(T')^{-1}=(T^{-1})'$; that is, $T'$ is an isomorphism of $X_\beta '$ onto itself.

Suppose that X is also complete and barrelled and that $T'\in {{\mathcal {L}}}(X'_\beta )$ is an isomorphism of $X'_\beta$ onto itself. As proved above, $T''$ is necessarily an isomorphism of $X''_\beta$ onto itself. By the proof of Lemma 3 in [6] it follows that T is an isomorphism of X onto itself. This completes the proof. $\square$

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. An eigenvalue $\lambda$ of T is called simple if . Some authors (e.g. [56]) 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 $\sigma (T;X)\subseteq \sigma ^*(T;X)$. If X is a Banach space, then $\sigma (T;X)= \sigma ^*(T;X)$. For the spectral theory of compact operators in lcHs’ we refer to [27, 33], for example.

Corollary 2.2

Let X be a complete, barrelled lcHs and $T\in {{\mathcal {L}}}(X)$. Then

$$\begin{aligned} \rho (T;X)=\rho (T';X'_\beta )\ \text{ and } \ \sigma (T;X)=\sigma (T';X'_\beta ). \end{aligned}$$

(2.1)

Moreover,

$$\begin{aligned} \sigma ^*(T'; X'_\beta )\subseteq \sigma ^*(T;X). \end{aligned}$$

(2.2)

Proof

The identities in (2.1) are an immediate consequence of Lemma 2.1.

Fix $\lambda \in \rho ^*(T; X)$. Then there exists $\delta >0$ such that $B(\lambda , \delta )\subseteq \rho (T;X)$ and $\{R(\mu ;T)\,:\, \mu \in B(\lambda ,\delta )\}\subseteq {{\mathcal {L}}}(X)$ is equicontinuous. For each $\mu \in B(\lambda ,\delta )$ it follows from the proof of Lemma 2.1 that $R(\mu , T)'=((\mu I-T)^{-1})'=(\mu I-T')^{-1}=R(\mu , T')$. Then [38, Sect. 39.3(6), p.138] implies that $\{R(\mu , T')\,:\, \mu \in B(\lambda ,\delta )\}\subseteq {{\mathcal {L}}}(X'_\beta )$ is equicontinuous, that is, $\lambda \in \rho ^*(T';X'_\beta )$. So, we have established that $\rho ^*(T;X)\subseteq \rho ^*(T';X'_\beta )$; taking complements yields (2.2). $\square$

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 $T\in {{\mathcal {L}}}(X,Y)$. For the following result see [38, Sect. 42.1(1)] or [36, Proposition 17.1.1].

Lemma 2.3

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

To establish the continuity of $C_t$, for $t\in [0,1]$, in the Fréchet spaces considered in this paper we will need the following result, [14, Lemma 25].

Lemma 2.4

Let $X=\cap _{n=1}^\infty X_n$ and $Y=\cap _{m=1}^\infty Y_m$ be two Fréchet spaces which resp. are the intersection of the sequence of Banach spaces $(X_n,\Vert \cdot \Vert _n)$, for $n\in {{\mathbb {N}}}$, and of the sequence of Banach spaces $(Y_m,|||\cdot |||_m)$, for $m\in {{\mathbb {N}}}$, satisfying $X_{n+1}\subset X_n$ with $\Vert x\Vert _n\le \Vert x\Vert _{n+1}$ for each $n\in {{\mathbb {N}}}$ and $x\in X_{n+1}$ and $Y_{m+1}\subset Y_m$ with $|||y|||_m\le |||y|||_{m+1}$ for each $m\in {{\mathbb {N}}}$ and $y\in Y_{m+1}$. Suppose that X is dense in $X_n$ for each $n\in {{\mathbb {N}}}$. Then a linear operator $T:X\rightarrow Y$ is continuous if, and only if, for each $m\in {{\mathbb {N}}}$ there exists $n\in {{\mathbb {N}}}$ such that the operator T has a unique continuous extension $T_{n,m}:X_n\rightarrow Y_m$.

The following result, based on [8, Lemma 2.1], will be needed to determine the spectra of $C_t$, for $t\in [0,1]$, in the Fréchet spaces considered in this paper.

Lemma 2.5

Let $X=\cap _{n=1}^\infty X_n$ be a Fréchet space which is the intersection of a sequence of Banach spaces $(X_n,\Vert \cdot \Vert _n)$, for $n\in {{\mathbb {N}}}$, satisfying $X_{n+1}\subset X_n$ with $\Vert x\Vert _n\le \Vert x\Vert _{n+1}$ for each $n\in {{\mathbb {N}}}$ and $x\in X_{n+1}$. Let $T\in {{\mathcal {L}}}(X)$ satisfy the following condition:

(A) For each $n\in {{\mathbb {N}}}$ there exists $T_n\in {{\mathcal {L}}}(X_n)$ such that the restriction of $T_n$ to X (resp. of $T_n$ to $X_{n+1}$) coincides with T (resp. with $T_{n+1}$).

Then the following properties are satisfied.

• (i) $\sigma (T;X)\subseteq \cup _{n=1}^\infty \sigma (T_n;X_n)$ and $\sigma _{pt}(T;X)\subseteq \cap _{n=1}^\infty \sigma _{pt}(T_n;X_n)$.

• (ii) If $\cup _{n=1}^\infty \sigma (T_n;X_n)\subseteq \overline{\sigma (T;X)}$, then $\sigma ^*(T;X)=\overline{\sigma (T;X)}$.

• (iii) If for each $\lambda \in \cap _{n=1}^\infty \sigma _{pt}(T_n;X_n)$ and $m\in {{\mathbb {N}}}$, then $\sigma _{pt}(T;X)=\cap _{n=1}^\infty \sigma _{pt}(T_n;X_n)$.

Proof

In view of [8, Lemma 2.1] it remains to show the validity of the inclusion $\sigma _{pt}(T;X)\subseteq \cap _{n=1}^\infty \sigma _{pt}(T_n,X_n)$ in the statement (i) and the identity in (iii).

The inclusion $\sigma _{pt}(T;X)\subseteq \cap _{n=1}^\infty \sigma _{pt}(T_n;X_n)$ is clear. Indeed, if $(\lambda I-T)x=0$ for some $x\in X{\setminus }\{0\}$ and $\lambda \in {{\mathbb {C}}}$, then in view of $X\subseteq X_n$ and $T_n|_X=T$, for $n\in {{\mathbb {N}}}$, (see condition (A)), we have that $x\in X_n{{\setminus }}\{0\}$ and $(\lambda I-T_n)x=0$ for every $n\in {{\mathbb {N}}}$. Hence, $\lambda \in \cap _{n=1}^\infty \sigma _{pt}(T_n;X_n)$.

To establish the validity of (iii), fix $\lambda \in \cap _{n=1}^\infty \sigma _{pt}(T_n;X_n)$. Then, for each $n\in {{\mathbb {N}}}$, there exists $x_n\in X_n{\setminus }\{0\}$ such that $(\lambda I-T_n)x_n=0$. Since $x_{n+1}\in X_{n+1}\subseteq X_n$, for $n\in {{\mathbb {N}}}$, condition (A) implies that also $(\lambda I-T_n)x_{n+1}=0$ in $X_n$ for each $n\in {{\mathbb {N}}}$. So, for each $n\in {{\mathbb {N}}}$, we have that $x_{n+1}=\mu _n x_n$ for some $\mu _n\in {{\mathbb {C}}}{\setminus }\{0\}$. Therefore, $x_n=(\prod _{j=1}^{n-1}\mu _j)x_1$, with $\prod _{j=1}^{n-1}\mu _j\not =0$. Accordingly, $x_1\in X_n$ for each $n\in {{\mathbb {N}}}$ and hence, $x_1\in X$. On the other hand, applying again condition (A), we can conclude that $(\lambda I-T)x_1=(\lambda I-T_1)x_1=0$, i.e., $\lambda \in \sigma _{pt}(T;X)$. $\square$

Fréchet spaces X which satisfy the assumptions of Lemma 2.5 are often called countably normed Fréchet spaces; for the general theory of such spaces see [29], for example.

A Hausdorff locally convex space $(X,\tau )$ is called an (LB)-space if there is a sequence $(X_k)_{k\in {{\mathbb {N}}}}$ of Banach spaces satisfying $X_k\subseteq X_{k+1}$ continuously for $k\in {{\mathbb {N}}}$, $X=\cup _{k=1}^\infty X_k$ and $\tau$ is the finest locally convex topology on X such that the natural inclusion $X_k\subset X$ is continuous for each $k\in {{\mathbb {N}}}$, [44, pp. 290–291]. In this case we write $X= \mathrm{ind\,}_k X_k$. If, in addition, X is a regular (LB)-space, [36, p. 83], then a set $B\subset X$ is bounded if and only if there exists $m\in {{\mathbb {N}}}$ such that $B\subset X_m$ and B is bounded in the Banach space $X_m$. Complete (LB)-spaces are regular, [37, Sect. 19.5(5)]. All of the (LB)-spaces of sequences considered in this note will be regular because of the following result, [44, Proposition 25.19(2)].

Lemma 2.6

Let $X=\mathrm{ind\,}_k X_k$ be an (LB)-space with an increasing union of reflexive Banach spaces $X=\cup _{k=1}^\infty X_k$ such that each inclusion $X_k\subseteq X_{k+1}$, for $k\in {{\mathbb {N}}}$, is continuous. Then X is complete and hence, also regular.

An (LB)-space $X=\mathrm{ind\,}_k X_k$ is said to be boundedly retractive if for every $B\in {{\mathcal {B}}}(X)$ there exists $k\in {{\mathbb {N}}}$ such that B is contained and bounded in $X_k$, and X and $X_k$ induce the same topology on B. The (LB)-space X is said to be sequentially retractive if for every null sequence in X there exists $k\in {{\mathbb {N}}}$ such that the sequence is contained and converges to zero in $X_k$. Finally, the (LB)-space X is said to be compactly regular if for every compact subset C of X there exists $k\in {{\mathbb {N}}}$ such that C is compact in $X_k$. Each of these three notions implies the completeness of X, [57, Corollary 2.8]. Neus [46] proved that all these notions are equivalent even for inductive limits of normed spaces.

In the setting of boundedly retractive (LB)-spaces, the following general statement on the compactness of certain dual operators is valid.

Proposition 2.7

Let X be a lcHs, $Y=\mathrm{ind\,}_k Y_k$ be a boundedly retractive (LB)-space and $T\in {{\mathcal {L}}}(X,Y)$ be compact. Then $T'\in {{\mathcal {L}}}(Y'_\beta , X'_\beta )$ is compact.

Proof

The compactness of T implies that there exists a closed, absolutely convex neighbourhood ${{\mathcal {U}}}$ of 0 in X such that $T({{\mathcal {U}}})$ is a relatively compact set in Y. So, the closure $B:=\overline{T({{\mathcal {U}}})}\in {{\mathcal {B}}}(Y)$ of $T({{\mathcal {U}}})$ is a compact set in Y. But, Y is a boundedly retractive (LB)-space. Accordingly, there exists $k\in {{\mathbb {N}}}$ such that B is contained and bounded in $Y_k$, and Y and $Y_k$ induce the same topology on B. Therefore, B is also a compact set in $Y_k$ and $T(X)\subseteq Y_k$. Accordingly, the operator T acts compactly from X into $Y_k$. Denote by $T_1$ the operator T when interpreted to be acting from X into $Y_k$ and by $i_k$ the continuous inclusion of $Y_k$ into Y. So, $T_1\in {{\mathcal {L}}}(X,Y_k)$ is compact and $T=i_kT_1$. Denote by p the continuous seminorm on X corresponding to ${{\mathcal {U}}}$ and let $X_p$ denote the normed quotient space $\left( \frac{X}{\textrm{Ker}p}, p\right)$. Then there exists a unique continuous linear operator S from $X_p$ into $Y_k$ such that $SQ=T_1$, where Q denotes the canonical quotient map from X into $X_p$ and hence, is an open map. Since $T_1\in {{\mathcal {L}}}(X,Y_k)$ is compact and $Q\in {{\mathcal {L}}}(X,X_p)$ is open, the operator $S\in {{\mathcal {L}}}(X_p, Y_k)$ is necessarily compact. By Schauder’s theorem, [38, Sect. 42(7), p. 202], it follows that $S'\in {{\mathcal {L}}}(Y'_k, X_p')$ is compact. So, $T'_1=Q'S'\in {{\mathcal {L}}}(Y'_k,X'_\beta )$ is compact and hence, $T'=T'_1i_k'\in {{\mathcal {L}}}(Y'_\beta , X'_\beta )$ is compact (cf. Proposition 17.1.1 in [36]). This completes the proof. $\square$

A Fréchet space X is said to be quasinormable if for every neighbourhhod ${{\mathcal {U}}}$ of 0 in X there exists a neighbourhhod ${{\mathcal {V}}}$ of 0 in X so that, for every $\varepsilon > 0$, there exists $B\in {{\mathcal {B}}}(X)$ satisfying ${{\mathcal {V}}}\subseteq B+\varepsilon {{\mathcal {U}}}$. Thus, every Fréchet-Schwartz space is quasinormable [44, Remark, p. 313]. The strong dual $X'_\beta$ of a quasinormable Fréchet space X is necessarily a boundedly retractive (LB)-space [18, Theorem]. Thus, the strong dual of any Fréchet-Schwartz space (briefly, (DFS)-space) is a boundedly retractive (LB)-space.

Corollary 2.8

Let X and Y be two Fréchet spaces and $T\in {{\mathcal {L}}}(X,Y)$. If $T''\in {{\mathcal {L}}}(X''_\beta , Y''_\beta )$ is compact, then T is compact.

If, in addition, X is quasinormable and T is compact, then $T''\in {{\mathcal {L}}}(X''_\beta , Y''_\beta )$ is compact.

Proof

Suppose that $T''\in {{\mathcal {L}}}(X''_\beta , Y''_\beta )$ is compact. Since X, Y are Fréchet spaces, they are isomorphic to their respective natural image in $X''_\beta$, $Y''_\beta$ (in which they are closed subspaces). Moreover, the restriction of $T''$ to X coincides with T and takes its values in $Y\subseteq Y''_\beta$. Then the compactness of T follows from that of $T''$.

Suppose that X is quasinormable and that $T\in {{\mathcal {L}}}(X,Y)$ is compact. Since X is quasinormable, its strong dual $X'_\beta$ is a boundedly retractive (LB)-space. Moreover, Y being a Fréchet space implies that $T':Y'_\beta \rightarrow X'_\beta$ is compact, [27, Corollary 9.6.3]. It follows from Proposition 2.7, with $Y'_\beta$ in place of X and $X'_\beta$ in place of $Y=\mathrm{ind\,}_k Y_k$ and $T'$ in place of T, that $T''\in {{\mathcal {L}}}(X''_\beta , Y''_\beta )$ is compact. $\square$

To identify the spectrum of $C_t$ acting in the (LB)-spaces arising in this paper we will require the following two results; the first one, i.e. Lemma 2.9, is a direct consequence of Grothendieck’s factorization theorem (see e.g. [44, Theorem 24.33]), and the second one, i.e. Lemma 2.10, is proved in [11, Lemma 5.2].

Lemma 2.9

Let $X=\mathrm{ind\,}_n X_n$ and $Y=\mathrm{ind\,}_mY_m$ be two (LB)-spaces with increasing unions of Banach spaces $X=\cup _{n=1}^\infty X_n$ and $Y=\cup _{m=1}^\infty Y_m$. Let $T:X\rightarrow Y$ be a linear map. Then T is continuous (i.e., $T\in {{\mathcal {L}}}(X,Y)$) if and only if for each $n\in {{\mathbb {N}}}$ there exists $m\in {{\mathbb {N}}}$ such that $T(X_n)\subseteq Y_m$ and the restriction $T:X_n\rightarrow Y_m$ is continuous.

Lemma 2.10

Let $X=\mathrm{ind\,}_k X_k$ be a Hausdorff inductive limit of a sequence of Banach spaces $(X_k,\Vert \cdot \Vert _k)$. Let $T\in {{\mathcal {L}}}(X)$ satisfy the following condition:

(A$\prime$) For each $k\in {{\mathbb {N}}}$ the restriction $T_k$ of T to $X_k$ maps $X_k$ into itself and $T_k\in {{\mathcal {L}}}(X_k)$.

Then the following properties are satisfied.

• (i) $\sigma _{pt}(T;X)=\cup _{k=1}^\infty \sigma _{pt}(T_k;X_k)$.

• (ii) If $\cup _{k=m}^\infty \sigma (T_k;X_k)\subseteq \overline{\sigma (T;X)}$ for some $m\in {{\mathbb {N}}}$, then $\sigma ^*(T;X)=\overline{\sigma (T;X)}$.

• (iii) $\sigma (T;X)\subseteq \cap _{m\in {{\mathbb {N}}}}\left( \cup _{n=m}^\infty \sigma (T_n;X_n)\right)$.

Another useful fact for our study is the following result.

Lemma 2.11

Let $T\in {{\mathcal {L}}}(\omega )$. Let X be a Fréchet space or an (LB)-space continuously included in $\omega$. If $T(X)\subseteq X$, then $T\in {{\mathcal {L}}}(X)$.

Proof

The result follows from the closed graph theorem, [44, Theorem 24.31], after recalling that X is ultrabornological, [44, Remark 24.15(c) & Proposition 24.16] and has a web, [44, Corollary 24.29 & Remark 24.36]. So, it is enough to show that the graph of T in X is closed. To do this, we assume that a net $(x_\alpha )_\alpha \subseteq X$ satisfies $x_\alpha \rightarrow x$ and $T(x_\alpha )\rightarrow y$ in X. Since the inclusion $X\subseteq \omega$ is continuous, $x_\alpha \rightarrow x$ in $\omega$ and hence, $T(x_\alpha )\rightarrow T(x)$ in $\omega$. On the other hand, by the continuity of the inclusion $X\subseteq \omega$ also $T(x_\alpha )\rightarrow y$ in $\omega$. Then $T(x)=y$. So, (x, y) belongs to the graph of T. This shows that the graph of T is closed. $\square$

For X a barrelled lcHs, every bounded subset of ${{\mathcal {L}}}_s(X)$ is equicontinuous, [44, Proposition 23.27]. It is known that every Fréchet space is barrelled, [44, Remark, p. 296], and that every (LB)-space is barrelled, [44, Proposition 24.16].

The operator norm of a Banach space operator $T\in {{\mathcal {L}}}(X,Y)$ will be denoted by $\Vert T\Vert _{X\rightarrow Y}$. The Banach spaces $\ell ^p=\ell ^p({{\mathbb {N}}}_0)$, for $1\le p<\infty$, with their standard norm $\Vert \cdot \Vert _p$ are classical. For $1<p<\infty$ these spaces are reflexive. The spectra of $C_t$ acting in such spaces are given in the following result; see [58] for $1<p<\infty$ and also [55, Sect. 8] for $1\le p<\infty$. Recall from Sect. 1 that

$$\begin{aligned} \Lambda :=\left\{ \frac{1}{n+1}:\, n\in {{\mathbb {N}}}_0\right\} . \end{aligned}$$

Proposition 2.12

For each $t\in [0,1)$ the operator $C_t\in {{\mathcal {L}}}(\ell ^p)$, for $1\le p<\infty$, is a compact operator satisfying

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

and

$$\begin{aligned} \left( \sum _{n=0}^\infty \left( \frac{t^n}{n+1}\right) ^p\right) ^{1/p}\le \Vert C_t\Vert _{\ell ^p\rightarrow \ell ^p}\le \left( \frac{1}{t}\log \left( \frac{1}{1-t}\right) \right) ^{1/p},\ 1<p<\infty ,\ t\in (0,1), \end{aligned}$$

with $\Vert C_0\Vert _{\ell ^p\rightarrow \ell ^p}=1$. Moreover,

$$\begin{aligned} \sigma _{pt}(C_t;\ell ^p)=\Lambda \ \hbox { and }\ \sigma (C_t; \ell ^p)=\Lambda \cup \{0\}. \end{aligned}$$

(2.3)

Concerning the classical Cesàro operator $C_1$ (c.f. (1.3)) in ${{\mathcal {L}}}(\ell ^p)$ we have the following result.

Proposition 2.13

Let $1<p<\infty$.

• (i) The operator $C_1\in {{\mathcal {L}}}(\ell ^p)$ with $\Vert C_1\Vert _{\ell ^p\rightarrow \ell ^p}=p'$, where $\frac{1}{p}+\frac{1}{p'}=1$.

• (ii) The spectra of $C_1$ are given by

  $$\begin{aligned} \sigma _{pt}(C_1;\ell ^p)=\emptyset \ \text{ and } \ \sigma (C_1;\ell ^p)=\left\{ z\in {{\mathbb {C}}}:\, \left| z-\frac{p'}{2}\right| \le \frac{p'}{2}\right\} . \end{aligned}$$

Moreover, the range $(C_1-zI)(\ell ^p)$ is not dense in $\ell ^p$ whenever $|z-\frac{p'}{2}|<\frac{p'}{2}$.

For part (i) we refer to [34, Theorem 326] and for part (ii) see [30, 40, 54] and the references therein. In particular, $C_1$ is a not a compact operator.

G. Bennett thoroughly investigated the discrete Cesàro spaces

$$\begin{aligned} ces(p):=\{x\in \omega :\, C_1|x|\in \ell ^p\},\quad 1<p<\infty , \end{aligned}$$

where $|x|:=(|x_n|)_{n\in {{\mathbb {N}}}_0}$, which satisfy $\ell ^p\subseteq ces(p)$ continuously and are reflexive Banach spaces relative to the norm

$$\begin{aligned} \Vert x\Vert _{ces(p)}:=\Vert C_1|x|\Vert _p,\quad x\in ces(p); \end{aligned}$$

(2.4)

see, for example, [17], as well as [15, 24, 31, 41] and the references therein. The following result, [26, Proposition 5.6] describes the spectra of $C_t$ acting in ces(p).

Proposition 2.14

Let $t\in [0,1)$ and $1<p<\infty$. The operator $C_t\in {{\mathcal {L}}}(ces(p))$ is compact and satisfies

$$\begin{aligned} \Vert C_t\Vert _{ces(p)\rightarrow ces(p)}\le \min \left\{ \frac{1}{1-t},\frac{p}{p-1}\right\} . \end{aligned}$$

Moreover,

$$\begin{aligned} \sigma _{pt}(C_t;ces(p))=\Lambda \ \hbox { and }\ \sigma (C_t; ces(p))=\Lambda \cup \{0\}. \end{aligned}$$

(2.5)

The situation for $C_1\in {{\mathcal {L}}}(ces(p))$ is quite different. Indeed, $\Vert C_1\Vert _{ces(p)\rightarrow ces(p)}=p'$ and the spectra are given by

$$\begin{aligned} \sigma _{pt}(C_1;ces(p))=\emptyset \ \text{ and } \ \sigma (C_1; ces(p))=\left\{ z\in {{\mathbb {C}}}:\, \left| z-\frac{p'}{2}\right| \le \frac{p'}{2}\right\} \end{aligned}$$

for each $1<p<\infty$; see Theorem 5.1 and its proof in [24]. In particular, $C_1$ is not a compact operator.

The dual Banach spaces $(ces(p))'$, for $1<p<\infty$, are rather complicated, [35]. A more transparent isomorphic identification of $(ces(p))'$ is given in Corollary 12.17 of [17]. It is shown there that

$$\begin{aligned} d_p:=\left\{ x\in \ell ^\infty :\, {\hat{x}}:=\left( \sup _{k\ge n}|x_k|\right) _{n\in {{\mathbb {N}}}_0}\in \ell ^p\right\} ,\quad 1<p<\infty , \end{aligned}$$

is a Banach space for the norm

$$\begin{aligned} \Vert x\Vert _{d_p}:=\Vert {\hat{x}}\Vert _p,\quad x\in d_p, \end{aligned}$$

(2.6)

which is isomorphic to $(ces(p'))'$, where $p'$ is the conjugate exponent of p. The sequence ${\hat{x}}$ is called the least decreasing majorant of x. The duality is the natural one given by

$$\begin{aligned} \langle w,x\rangle :=\sum _{n=0}^\infty w_nx_n,\quad w\in ces(p'),\ x\in d_p. \end{aligned}$$

In particular, $d_p$ is reflexive for each $1<p<\infty$. Since $|x|\le |{\hat{x}}|$, it is clear that $\Vert x\Vert _p\le \Vert {\hat{x}}\Vert _p=\Vert x\Vert _{d_p}$, for $x\in d_p$, that is, $d_p\subseteq \ell ^p$ continuously. So, for all $1<p<\infty$, we have $d_p\subseteq \ell ^p\subseteq ces(p)$ with continuous inclusions. The following result is Theorem 6.9 of [26].

Proposition 2.15

Let $t\in [0,1)$ and $1<p<\infty$. The operator $C_t\in {{\mathcal {L}}}(d_p)$ is compact and satisfies

$$\begin{aligned} \Vert C_t\Vert _{d_p\rightarrow d_p}\le (1-t)^{-1-(1/p)}. \end{aligned}$$

Moreover,

$$\begin{aligned} \sigma _{pt}(C_t;d_p)=\Lambda \ \hbox { and }\ \sigma (C_t; d_p)=\Lambda \cup \{0\}. \end{aligned}$$

(2.7)

Concerning the operator $C_1\in {{\mathcal {L}}}(d_p)$, $1<p<\infty$, it is known that $\Vert C_1\Vert _{d_p\rightarrow d_p}=p'$ and that its spectra are given by

$$\begin{aligned} \sigma _{pt}(C_1;d_p)=\emptyset \ \text{ and } \ \sigma (C_1; d_p)=\left\{ z\in {{\mathbb {C}}}:\, \left| z-\frac{p'}{2}\right| \le \frac{p'}{2}\right\} ; \end{aligned}$$

see Proposition 3.2 and Corollary 3.5 in [19].

3 The operators $C_t$ acting in $\omega$

Given an element $x=(x_n)_{n\in {{\mathbb {N}}}_0}\in \omega$ we write $x\ge 0$ if $x=|x|=(|x_n|)_{n\in {{\mathbb {N}}}_0}$. By $x\le z$ it is meant that $(z-x)\ge 0$. The sequence space $\omega$ is a non-normable Fréchet space for the Hausdorff locally convex topology of coordinatewise convergence, which is determined by the increasing sequence of seminorms

$$\begin{aligned} r_n(x):=\max _{0\le j\le n}|x_j|,\quad x\in \omega , \end{aligned}$$

(3.1)

for each $n\in {{\mathbb {N}}}_0$. Observe that $r_n(x)=r_n(|x|)\le r_n(|y|)=r_n(y)$ whenever $x,y\in \omega$ satisfy $|x|\le |y|$. Let $e_n:=(\delta _{nj})_{j\in {{\mathbb {N}}}_0}$ for each $n\in {{\mathbb {N}}}_0$ and set ${{\mathcal {E}}}:=\{e_n:\, n\in {{\mathbb {N}}}_0\}$. It is clear from (1.1) that each $C_t:\omega \rightarrow \omega$ is a linear map which is represented by a lower triangular matrix with respect to the unconditional basis ${{\mathcal {E}}}$ of $\omega$. Namely,

$$\begin{aligned} C_t\simeq \left( \begin{array}{ccccc} 1 &{}\quad 0 &{}\quad 0 &{} \quad 0 &{}\quad \cdots \\ t/2 &{}\quad 1/2 &{}\quad 0 &{}\quad 0 &{}\quad \cdots \\ t^2/3 &{}\quad t/3 &{}\quad 1/3 &{}\quad 0 &{}\quad \cdots \\ t^3/4 &{}\quad t^2/4 &{}\quad t/4 &{}\quad 1/4 &{}\quad \cdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{} \\ \end{array}\right) \end{aligned}$$

(3.2)

with main diagonal the positive, decreasing sequence given by

$$\begin{aligned} \varphi :=\left( \frac{1}{n+1}\right) _{n\in {{\mathbb {N}}}_0}\in c_0. \end{aligned}$$

(3.3)

The following properties of $C_t$ are recorded in [26, Lemma 2.1], except for part (iv).

Lemma 3.1

Let $t\in [0,1)$.

• (i) Each $C_t$ is a positive operator on $\omega$,i.e., $C_tx\ge 0$ whenever $x\ge 0$.

• (ii) Let $0\le r\le s\le 1$. Then

  $$\begin{aligned} 0\le |C_rx|\le C_r|x|\le C_s|x|,\quad x\in \omega . \end{aligned}$$

• (iii) For each $t\in [0,1)$ the identities

  $$\begin{aligned} C_te_n=\sum _{k=0}^\infty \frac{t^k}{k+n+1}e_{k+n}\in \ell ^1, \quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

  and

  $$\begin{aligned} C_t(e_n-te_{n+1})=\frac{1}{n+1}e_n,\quad n\in {{\mathbb {N}}}_0, \end{aligned}$$

  are valid.

• (iv) For each $1<q<\infty$ we have $d_q\subseteq \ell ^q\subseteq ces(q)\subseteq \omega$ with continuous inclusions.

Proof

(iv) In view of the discussion after (2.6) it remains to establish that $ces(q)\subseteq \omega$ continuously. Fix $x\in ces(q)$. Given $n\in {{\mathbb {N}}}_0$ observe that

$$\begin{aligned} |x_k|\le (n+1)\frac{|x_0|+|x_1|+\cdots +|x_n|}{n+1}\le (n+1)\Vert C_1|x|\Vert _q=(n+1)\Vert x\Vert _{ces(q)}, \ 0\le k\le n. \end{aligned}$$

It follows from (3.1) that $r_n(x)\le (n+1)\Vert x\Vert _{ces(q)}$. Since $n\in {{\mathbb {N}}}_0$ is arbitrary, we can conclude that $ces(q)\subseteq \omega$ continuously. $\square$

The classical Cesàro operator $C_1:\omega \rightarrow \omega$ is a bicontinuous topological isomorphism (and hence, is not a compact operator) with spectra given by

$$\begin{aligned} \sigma (C_1;\omega )=\sigma _{pt}(C_1;\omega )=\Lambda \ \hbox { and }\ \sigma ^*(C_1;\omega )=\Lambda \cup \{0\}; \end{aligned}$$

see [8, p. 285 and Proposition 4.4]. So, we will only consider the case $t\in [0,1)$.

Let $t\in [0,1)$ and fix $n\in {{\mathbb {N}}}_0$. According to (1.1) and (3.1), for each $x\in \omega$, it is the case that

$$\begin{aligned} r_n(C_tx)=\max _{0\le k\le n}\left| \frac{1}{k+1}\sum _{i=0}^{k-1}t^{k-i}x_i\right| \le \max _{0\le k\le n}\frac{1}{k+1}\sum _{i=0}^{k-1}|x_i|\le r_n(x). \end{aligned}$$

(3.4)

This implies that $C_t\in {{\mathcal {L}}}(\omega )$ and that the family of operators $\{C_t:\, t\in [0,1)\}$ is an equicontinuous subset of ${{\mathcal {L}}}(\omega )$.

Proposition 3.2

For each $t\in [0,1)$ the operator $C_t\in {{\mathcal {L}}}(\omega )$ is a bicontinuous isomorphism of $\omega$ onto itself with inverse operator $(C_t)^{-1}:\omega \rightarrow \omega$ given by

$$\begin{aligned} (C_t)^{-1}y=((n+1)y_n-nty_{n-1})_{n\in {{\mathbb {N}}}_0},\quad y\in \omega \ (\hbox {with}\ y_{-1}:=0). \end{aligned}$$

(3.5)

In particular, $C_t$ is not a compact operator.

Proof

Fix $t\in [0,1)$. Let $x\in \omega$ satisfy $C_tx=0$. Considering the coordinate 0 of $C_tx=0$ yields $x_0=0$; see (1.1). The equation for coordinate 1 of $C_tx=0$ is $\frac{tx_0+x_1}{2}=0$ (cf. (1.1)) which yields $x_1=0$. Proceed inductively for successive coordinates reveals that $x_n=0$ for all $n\in {{\mathbb {N}}}_0$. Hence, $C_t$ is injective.

Given $y\in \omega$ let $x\in \omega$ be the element on the right-side of (3.5). Direct calculation shows that $C_tx=y$. Accordingly, $C_t$ is surjective.

By the open mapping theorem for Fréchet spaces (cf. Corollary 24.29 and Theorem 24.30 in [44]) the operator $C_t$ is a bicontinuous isomorphism.

Since $C_t$ is a bicontinuous isomorphism of $\omega$, which is an infinite dimensional Fréchet space, $C_t$ cannot be a compact operator. $\square$

To determine the spectrum of $C_t\in {{\mathcal {L}}}(\omega )$ requires some preparation. Define

$$\begin{aligned} {{\mathcal {S}}}:=\left\{ x\in \omega :\, \beta (x):=\lim _{n\rightarrow \infty }\frac{|x_{n+1}|}{|x_n|}<1\right\} , \end{aligned}$$

(3.6)

with the understanding that there exists $N\in {{\mathbb {N}}}_0$ such that $x_n\not =0$ for $n\ge N$ and the limit $\beta (x)$ exists. Analogously to $d_p$, for $1<p<\infty$, define

$$\begin{aligned} d_1:=\left\{ x\in \ell ^\infty :\, {\hat{x}}:=\left( \sup _{k\ge n}|x_k|\right) _{n\in {{\mathbb {N}}}_0}\in \ell ^1\right\} ; \end{aligned}$$

(3.7)

see [17, 19, 24, 31] and the references therein. Then $d_1$ is a Banach lattice for the norm $\Vert x\Vert _{d_1}:=\Vert {\hat{x}}\Vert _1$ and the coordinatewise order. Since $0\le |x|\le {\hat{x}}$, for $x\in \ell ^\infty$, it is clear that $\Vert x\Vert _1\le \Vert x\Vert _{d_1}$ for $x\in d_1$, that is, $d_1\subseteq \ell ^1$ with a continuous inclusion. Clearly, $d_1\subseteq d_p$, for all $1<p<\infty$, and $d_1\subseteq \ell ^1$ implies that $d_1\subseteq \ell ^p$, for all $1<p<\infty$. Moreover, $\ell ^p\subseteq ces(p)$ (cf. Section 2) and so also $d_1\subseteq ces(p)$, for $1<p<\infty$. All inclusions are continuous. In view of Lemma 3.1(iv) it is clear that $d_1\subseteq \omega$ and $\ell ^1\subseteq \omega$ continuously. It is known that ${\mathcal {S}}\subseteq d_1$, [26, Lemma 3.3].

Remark 3.3

Proposition 2.15 is also valid for $p=1$; see [26, Theorem 6.9].

The following result, [26, Lemma 3.6], will be required.

Lemma 3.4

Let $t\in [0,1)$ and $\varphi$ be as in (3.3). For each $m\in {{\mathbb {N}}}$ define $x^{[m]}\in \omega$ by

$$\begin{aligned} x^{[m]}:=\alpha _m \left( 0,\ldots ,0,1, \frac{(m+1)!}{m!\,1!}t,\frac{(m+2)!}{m!\,2!}t^2,\frac{(m+3)!}{m!\,3!}t^3,\ldots \right) , \end{aligned}$$

(3.8)

with $\alpha _m\in {{\mathbb {C}}}{\setminus }\{0\}$ arbitrary, where 1 is in position m. For $m=0$ define $x^{[0]}:=\alpha _0(t^n)_{n\in {{\mathbb {N}}}_0}$ with $\alpha _0\in {{\mathbb {C}}}{\setminus }\{0\}$ arbitrary.

• (i) For each $m\in {{\mathbb {N}}}_0$, the vector $x^{[m]}$ is the unique solution in $\omega$ of the equation $C_tx=\varphi _m x=\frac{1}{m+1}x$ whose m-th coordinate is $\alpha _m$.

• (ii) The vector $x^{[m]}\in d_1\subseteq \omega$, for each $m\in {{\mathbb {N}}}_0$.

Remark 3.5

Let $t\in [0,1)$ and X be any Banach space in $\{\ell ^1,d_1\}\cup \{\ell ^p, ces(p), d_p:\, 1<p<\infty \}$. For each $\nu \in \sigma _{pt}(C_t;X)=\Lambda$, it is the case that $\textrm{dim}\,\textrm{Ker}(\nu I-C_t)=1$. Indeed, $d_1\subseteq X$; see the discussion prior to Remark 3.3. Given $\nu \in \Lambda$ there exists $m\in {{\mathbb {N}}}_0$ such that $\nu =\varphi _m$. According to Lemma 3.4 the 1-dimensional eigenspace corresponding to $\nu \in \sigma _{pt}(C_t;\omega )$ is spanned by $x^{[m]}$ with $x^{[m]}\in d_1$. The claim is thereby proved.

The next lemma places a restriction on where $\sigma (C_t;\omega )$ can be located in ${{\mathbb {C}}}$.

Lemma 3.6

Let $t\in [0,1)$. For each $\nu \in {{\mathbb {C}}}{\setminus }\Lambda$ the operator $C_t-\nu I$ is a bicontinuous isomorphism of $\omega$ onto itself. In particular, $\sigma (C_t;\omega )\subseteq \Lambda$.

Proof

Fix $\nu \not \in \Lambda$. Let $(C_t-\nu I)x=0$ for $x\in \omega$. It follows from (3.2), by equating the coordinate 0 of $C_tx=\nu x$, that $x_0=\nu x_0$ and hence, as $\nu \not =1$, that $x_0=0$. Equating the coordinate 1 of $C_tx=\nu x$ yields $\frac{tx_0+x_1}{2}=\nu x_1$. Since $x_0=0$ and $\nu \not =\frac{1}{2}$, it follows that $x_1=0$. Considering coordinate 2 gives $\frac{t^2x_0+tx_1+x_2}{3}=\nu x_2$. Then $x_0=x_1=0$ and $\nu \not =\frac{1}{3}$ imply $x_2=0$. Proceed inductively to conclude that $x=0$, that is, $C_t-\nu I$ is injective.

To verify the surjectivity of $C_t-\nu I$ fix $y\in \omega$. It is required to show that there exists $x\in \omega$ satisfying $(C_t-\nu I)x=y$. Equating coordinate 0 gives $x_0-\nu x_0=y_0$, that is, $x_0=y_0/(1-\nu )$. Considering coordinate 1 yields $\frac{tx_0}{2}+(\frac{1}{2}-\nu )x_1=y_1$. Substituting for $x_0$ gives $(\frac{1}{2}-\nu )x_1=y_1-\frac{t}{2(1-\nu )}y_0$, that is,

$$\begin{aligned} x_1=\frac{y_1}{\left( \frac{1}{2}-\nu \right) }-\frac{ty_0}{2\left( \frac{1}{2}-\nu \right) (1-\nu )}. \end{aligned}$$

Next, an examination of coordinate 2 yields $\frac{t^2}{3}x_0+\frac{t}{3}x_1+(\frac{1}{3}-\nu )x_2=y_2$. Substituting for $x_0$ and $x_1$ we can conclude that

$$\begin{aligned} x_2=\frac{y_2}{\left( \frac{1}{3}-\nu \right) }-\frac{ty_1}{3\left( \frac{1}{3}-\nu \right) \left( \frac{1}{2}-\nu \right) }+\frac{\nu t^2y_0}{3\left( \frac{1}{3}-\nu \right) \left( \frac{1}{2}-\nu \right) (1-\nu )}. \end{aligned}$$

Continuing inductively yields

$$\begin{aligned} x_n&=\frac{y_n}{\left( \frac{1}{n+1}-\nu \right) }-\frac{ty_{n-1}}{(n+1)\left( \frac{1}{n+1}-\nu \right) \left( \frac{1}{n}-\nu \right) }+\nonumber \\&\quad +\frac{\nu t^2y_{n-2}}{(n+1)\left( \frac{1}{n+1}-\nu \right) \left( \frac{1}{n}-\nu \right) \left( \frac{1}{n-1}-\nu \right) }\nonumber \\&\quad -\frac{\nu ^2 t^3y_{n-3}}{(n+1)\left( \frac{1}{n+1}-\nu \right) \left( \frac{1}{n}-\nu \right) \left( \frac{1}{n-1}-\nu \right) \left( \frac{1}{n-2}-\nu \right) }+ \cdots \nonumber \\&\quad +(-1)^n \frac{\nu ^{n-1} t^ny_{0}}{(n+1)\left( \frac{1}{n+1}-\nu \right) \left( \frac{1}{n}-\nu \right) \ldots \left( \frac{1}{2}-\nu \right) (1-\nu )}. \end{aligned}$$

(3.9)

Then $x\in \omega$ satisfies $(C_t-\nu I)x=y$. Hence, $C_t-\nu I$ is surjective. $\square$

Combining the previous results yields the main result of this section.

Theorem 3.7

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

$$\begin{aligned} \sigma (C_t;\omega )=\sigma _{pt}(C_t;\omega )=\Lambda , \end{aligned}$$

with each eigenvalue being simple, and

$$\begin{aligned} \sigma ^*(C_t;\omega )=\Lambda \cup \{0\}. \end{aligned}$$

The 1-dimensional eigenspace corresponding to the eigenvalue $1/(m+1)\in \Lambda$ is spanned by $x^{[m]}$ (cf. (3.8)), for each $m\in {{\mathbb {N}}}_0$.

Proof

It is clear from Lemma 3.4 that $\Lambda \subseteq \sigma _{pt}(C_t;\omega )$ and that each point $1/(m+1)\in \Lambda$ is a simple eigenvalue of $C_t$, whose corresponding eigenspace is spanned by $x^{[m]}$, for each $m\in {{\mathbb {N}}}_0$. Since $\sigma (C_t;\omega )\subseteq \Lambda$ (cf. Lemma 3.6) and $\sigma _{pt}(C_t;\omega )\subseteq \sigma (C_t;\omega )$, we can conclude that $\sigma (C_t;\omega )=\sigma _{pt}(C_t;\omega )=\Lambda$. The containment $\sigma (C_t;\omega )\subseteq \sigma ^*(C_t;\omega )$ and the fact that $\sigma ^*(C_t;\omega )$ is a closed set imply that $0\in \sigma ^*(C_t;\omega )$.

It remains to show that every $\nu \not \in (\Lambda \cup \{0\})$ belongs to $\rho ^*(C_t;\omega )$. So, fix $\nu \not \in (\Lambda \cup \{0\})$. Select $\delta >0$ such that the distance $\epsilon$ of $B(\nu ,\delta )$ to the compact set $\Lambda \cup \{0\}$ is strictly positive. It follows from $0\le t<1$ and the identity (3.9) which is coordinate n of $(C_t-\nu I)^{-1}y$, for each $y\in \omega$, that for any given $k\in {{\mathbb {N}}}_0$ there exists $M_k>0$ such that

$$\begin{aligned} r_k((C_t-\mu I)^{-1}y)\le \frac{M_k}{\epsilon ^{k+1}}\left( \max _{0\le j< k}|\nu |^j\right) r_{k}(y),\quad \mu \in B(\nu ,\delta ), \end{aligned}$$

where $r_k$ is the seminorm (3.1), with k in place of n. This implies that $\{(C_t-\mu I)^{-1}\,:\, \mu \in B(\nu ,\delta )\}$ is a bounded set in ${{\mathcal {L}}}_s(\omega )$ and hence, by the barrelledness of $\omega$, it is an equicontinuous subset of ${{\mathcal {L}}}(\omega )$. Accordingly, $\nu \in \rho ^*(C_t;\omega )$. $\square$

4 $C_t$ acting in the Fréchet spaces $\ell (p+)$, $d(p+)$ and $ces(p+)$

Given $1\le p<\infty$, consider any strictly decreasing sequence $\{p_k\}_{k\in {{\mathbb {N}}}}\subseteq (p,\infty )$ which satisfies $p_k\downarrow p$. Then $X_k:=\ell ^{p_k}$ satisfies $X_{k+1}\subseteq X_k$ with $\Vert x\Vert _{\ell ^{p_k}}\le \Vert x\Vert _{\ell ^{p_{k+1}}}$ for each $k\in {{\mathbb {N}}}$ and $x\in X_{k+1}$. Moreover, $X=\cap _{k=1}^\infty X_k$ (i.e., $\ell (p+):=\cap _{k=1}^\infty \ell ^{p_k}$) is a Fréchet space of the type given in Lemma 2.5 whose topology is generated by the increasing sequence of norms $u_k$, for $k\in {{\mathbb {N}}}$, given by

$$\begin{aligned} u_k:x\mapsto \Vert x\Vert _{\ell ^{p_k}}, \quad x\in \ell (p+). \end{aligned}$$

(4.1)

That is, $u_k\le u_{k+1}$ for $k\in {{\mathbb {N}}}$. Moreover, $p_k>p$ implies that the natural inclusion map $\ell (p+)\hookrightarrow \ell ^{p_k}$ is continuous for each $k\in {{\mathbb {N}}}$. Clearly the Banach space $\ell ^p\subseteq \ell (p+)$ continuously and also $\ell (p+)\subseteq \omega$ continuously, as $\ell ^q\subseteq \omega$ continuously, for every $1\le q<\infty$ (cf. Lemma 3.1(iv)). The space $\ell (p+)$ is independent of the choice of $\{p_k\}_{k\in {{\mathbb {N}}}}$.

Changing the Banach spaces, now let $X_k:=ces(p_k)$, in which case again $X_{k+1}\subseteq X_k$ with $\Vert x\Vert _{ces(p_k)}\le \Vert x\Vert _{ces(p_{k+1})}$ for each $k\in {{\mathbb {N}}}$ and $x\in X_{k+1}$; see [13, Proposition 3.2(iii)]. Then $X=\cap _{k=1}^\infty X_k$ (i.e., $ces(p+):=\cap _{k=1}^\infty ces(p_k)$) is a Fréchet space of the type given in Lemma 2.5 whose topology is generated by the increasing sequence of norms $v_k$, for $k\in {{\mathbb {N}}}$, given by

$$\begin{aligned} v_k:x\mapsto \Vert x\Vert _{ces(p_k)}, \quad x\in ces(p+). \end{aligned}$$

(4.2)

That is, $v_k\le v_{k+1}$ for $k\in {{\mathbb {N}}}$. Again $ces(p)\subseteq ces(p+)$ (if $p>1$) and $ces(p+)\subseteq \omega$ with both inclusions continuous, where we again use Lemma 3.1(iv). The Fréchet spaces $ces(p+)$, for $1\le p<\infty$, have been intensively studied in [9, 14].

Finally, consider the family of Banach spaces $X_k:=d_{p_k}$, in which case $X_{k+1}\subseteq X_k$ with $\Vert x\Vert _{d_{p_k}}\le \Vert x\Vert _{d_{p_{k+1}}}$ for each $k\in {{\mathbb {N}}}$ and $x\in X_{k+1}$; see [19, Proposition 5.1(iii)]. So, $X=\cap _{k=1}^\infty X_k$ (i.e., $d(p+):=\cap _{k=1}^\infty d_{p_k}$) is a Fréchet space of the type given in Lemma 2.5 whose topology is generated by the increasing sequence of norms $w_k$, for $k\in {{\mathbb {N}}}$, given by

$$\begin{aligned} w_k:x\mapsto \Vert x\Vert _{d_{p_k}}, \quad x\in d(p+). \end{aligned}$$

(4.3)

That is, $w_k\le w_{k+1}$ for $k\in {{\mathbb {N}}}$. With continuous inclusions we have $d_p\subseteq d(p+)\subseteq \omega$; see [20, Sect. 4] or, argue as for $\ell ^p$ and $\ell (p+)$.

It is known that the canonical vectors ${\mathcal {E}}$ belong to $\ell (p+)$, $d(p+)$ and $ces(p+)$, for $1\le p<\infty$, and form an unconditional basis in each of these spaces; see [20, Proposition 3.1], [20, Lemma 4.1] and [9, Proposition 3.5(i)], respectively.

In this section we consider the compactness and determine the spectra of $C_t$ when they act in the Fréchet spaces $\ell (p+)$, $d(p+)$ and $ces(p+)$, for $1\le p<\infty$. The decreasing sequence $\{p_k\}_{k\in {{\mathbb {N}}}}$ always has the properties listed above. Crucial for the proofs is the existence of a particular factorization available for $C_t$ (cf. Proposition 4.4).

The decreasing sequence $\varphi$ given in (3.3) satisfies $\Vert \varphi \Vert _\infty =1$. Define the linear map $D_\varphi :\omega \rightarrow \omega$ by

$$\begin{aligned} D_\varphi x:=(\varphi _0x_0,\varphi _1 x_1,\varphi _2 x_2,\ldots )=\left( \frac{x_n}{n+1}\right) _{n\in {{\mathbb {N}}}_0},\quad x\in \omega . \end{aligned}$$

(4.4)

The diagonal (multiplication) operator $D_\varphi \in {{\mathcal {L}}}(\omega )$ since, for each $n\in {{\mathbb {N}}}_0$,

$$\begin{aligned} r_n(D_\varphi x)\le r_n(x),\quad x\in \omega , \end{aligned}$$

where $r_n$ is the seminorm (3.1). Define the right-shift operator $S:\omega \rightarrow \omega$ by

$$\begin{aligned} Sx:=(0, x_0,x_1,\ldots ),\quad x\in \omega . \end{aligned}$$

(4.5)

For each $n\in {{\mathbb {N}}}$ note that $r_n(Sx)=\max _{0\le k<n}|x_k|\le r_n(x)$ and for $n=0$ that $r_0(Sx)=0\le r_0(x)$ for each $x\in \omega$. So, for every $n\in {{\mathbb {N}}}_0$, the operator S satisfies

$$\begin{aligned} r_n(Sx)\le r_n(x),\quad x\in \omega , \end{aligned}$$

(4.6)

which implies that $S\in {{\mathcal {L}}}(\omega )$. The following result is Lemma 2.2 in [26].

Lemma 4.1

For each $t\in [0,1)$ we have the representation

$$\begin{aligned} C_t=\sum _{n=0}^\infty t^nD_\varphi S^n \end{aligned}$$

with the series being convergent in ${{\mathcal {L}}}_s(\omega )$. Equivalently,

$$\begin{aligned} C_tx=\sum _{n=0}^\infty t^nD_\varphi S^n x,\quad x\in \omega , \end{aligned}$$

with the series being convergent in $\omega$.

Fix $t\in [0,1)$ and $x\in \omega$. For each $n\in {{\mathbb {N}}}_0$ it follows from (4.6) that

$$\begin{aligned} r_n\left( \sum _{k=0}^\infty t^kS^kx\right) \le \sum _{k=0}^\infty r_n(t^kS^kx)\le \frac{1}{1-t}r_n(x). \end{aligned}$$

Accordingly, the series

$$\begin{aligned} R_t:=\sum _{n=0}^\infty t^nS^n, \quad t\in [0,1), \end{aligned}$$

(4.7)

is absolutely convergent in the quasicomplete lcHs ${{\mathcal {L}}}_s(\omega )$. In particular, $R_t\in {{\mathcal {L}}}(\omega )$. Combining this with Lemma 4.1 and the fact that $D_\varphi \in {{\mathcal {L}}}(\omega )$ yields the following factorization of $C_t$.

Proposition 4.2

For each $t\in [0,1)$ the operators $D_\varphi$, $R_t$, $C_t$ belong to ${{\mathcal {L}}}(\omega )$ and

$$\begin{aligned} C_t=D_\varphi R_t=\sum _{n=0}^\infty t^n D_\varphi S^n, \end{aligned}$$

(4.8)

with the series being absolutely convergent in ${{\mathcal {L}}}_s(\omega )$.

Our aim is to to extend Proposition 4.2 to ${{\mathcal {L}}}(X)$ with $X\in \{\ell (p+),ces(p+),d(p+):\, 1\le p<\infty \}$, to show that $D_\varphi \in {{\mathcal {L}}}(X)$ is compact and then to apply Lemma 2.3 to conclude that $C_t\in {{\mathcal {L}}}(X)$ is compact.

Proposition 4.3

Let X be any Fréchet space in $\{\ell (p+),ces(p+),d(p+)\,:\, 1\le p<\infty \}$. Then $D_\varphi$ maps X into X and $D_\varphi \in {{\mathcal {L}}}(X)$ is compact.

Proof

Recall that $\varphi \in c_0$ with $\Vert \varphi \Vert _\infty =1$. We consider each of the three possible cases for X. It was shown above that $D_\varphi \in {{\mathcal {L}}}(\omega )$ and that $X\subseteq \omega$ continuously.

(a) Suppose that $X=\ell (p+)$ for some $1\le p<\infty$. Clearly, $D_\varphi (X_k)\subseteq X_k$ for each $k\in {{\mathbb {N}}}$ and so $D_\varphi \in {{\mathcal {L}}}(X)$; see Lemma 2.11. In the notation of [14] it is clear from (4.4) that $D_\varphi$ is precisely the multiplication operator $M_\varphi$ defined there. Such a multiplication operator is compact if and only if $\varphi \in \ell (\infty -)=\cup _{s>1}\ell ^s$, [14, Proposition 17], which is surely the case as $\varphi \in \ell ^2$, for example. So, $D_\varphi \in {{\mathcal {L}}}(\ell (p+))$ is a compact operator.

(b) Suppose that $X=ces(p+)$ for some $1\le p<\infty$. It follows from (2.4) that $D_\varphi (X_k)\subseteq X_k$ for each $k\in {{\mathbb {N}}}$ and so $D_\varphi :X\rightarrow X$. Lemma 2.11 yields that $M_\varphi =D_\varphi \in {{\mathcal {L}}}(ces(p+))$. Moreover, if $\varphi \in d(\infty -)=\cup _{s>1} d_s$, then $M_\varphi$ is also compact, [14, Proposition 10]. But, $\varphi$ is a positive decreasing sequence and so $\varphi ={\hat{\varphi }}$. Accordingly, by choosing $s=2$ say, we see that

$$\begin{aligned} \Vert \varphi \Vert _{d_2}:=\Vert {\hat{\varphi }}\Vert _2=\Vert \varphi \Vert _2<\infty . \end{aligned}$$

Hence, $\varphi \in d_2\subseteq d(\infty -)$ and so $D_\varphi =M_\varphi \in {{\mathcal {L}}}(ces(p+))$ is indeed compact.

(c) Suppose $X=d(p+)$ for some $1\le p<\infty$. Since $|D_\varphi x|= D_\varphi |x|\le |x|$, for $x\in \ell ^\infty$, it is clear that $\widehat{D_\varphi x}\le {\hat{x}}$. Then (2.6) implies that $D_\varphi (X_k)\subseteq X_k$ for all $k\in {{\mathbb {N}}}$ and so $D_\varphi :X\rightarrow X$. Again Lemma 2.11 yields that $D_\varphi \in {{\mathcal {L}}}(d(p+))$. Note that the operator $M_{d(p+)}^\varphi$ in [21] is precisely $D_\varphi :d(p+)\rightarrow d(p+)$. It was verified in (b) above that $\varphi \in d(\infty -)$ which, together with $D_\varphi \in {{\mathcal {L}}}(d(p+))$, implies that $D_\varphi$ is compact, [21, Theorem 4.13(i)]. $\square$

Proposition 4.4

Let $t\in [0,1)$, and X be any Fréchet space in $\{\ell (p+),ces(p+),d(p+):\, 1\le p<\infty \}$.

• (i) The generalized Cesàro operator $C_t$ maps X into itself and $C_t\in {{\mathcal {L}}}(X)$.

• (ii) The right-shift operator S given by (4.5) maps X into itself and belongs to ${{\mathcal {L}}}(X)$.

• (iii) The operator $R_t$ given by (4.7) maps X into itself and belongs to ${{\mathcal {L}}}(X)$, with the series $\sum _{n=0}^\infty t^nS^n$ being absolutely convergent in ${{\mathcal {L}}}_s(X)$. Moreover,

  $$\begin{aligned} C_t=D_\varphi R_t=\sum _{n=0}^\infty t^nD_\varphi S^n. \end{aligned}$$

Proof

(i) Again we consider the three possible cases for X. Fix $t\in [0,1)$. According to Proposition 3.2 the operator $C_t\in {{\mathcal {L}}}(\omega )$.

(a) Suppose that $X=\ell (p+)$ for some $1\le p<\infty$. Proposition 2.12 implies that $C_t(X_k)\subseteq X_k$ for all $k\in {{\mathbb {N}}}$, with $X_k=\ell ^{p_k}$, and so $C_t(X)\subseteq X$. In view of Lemma 2.11, with $T:=C_t$, it follows that $C_t\in {{\mathcal {L}}}(\ell (p+))$.

(b) Suppose that $X=ces(p+)$ for some $1\le p<\infty$. Proposition 2.14 shows that $C_t(X_k)\subseteq X_k$ for all $k\in {{\mathbb {N}}}$, with $X_k=ces(p_k)$, and so $C_t(X)\subseteq X$. Again, for $T:=C_t$, Lemma 2.11 implies that $C_t\in {{\mathcal {L}}}(ces(p+))$.

(c) Suppose that $X=d(p+)$ for some $1\le p<\infty$. Proposition 2.15 shows that $C_t(X_k)\subseteq X_k$ for all $k\in {{\mathbb {N}}}$, with $X_k=d_{p_k}$, and so $C_t(X)\subseteq X$. Yet again, for $T:=C_t$, Lemma 2.11 implies that $C_t\in {{\mathcal {L}}}(d(p+))$.

(ii) Again we check the three separate cases for X. Prior to Lemma 4.1 it was shown that $S\in {{\mathcal {L}}}(\omega )$.

(a) Suppose that $X=\ell (p+)$ for some $1\le p<\infty$. Using the fact that the Banach space right-shift operator $S:\ell ^{p_k}\rightarrow \ell ^{p_k}$ is an isometry, for every $k\in {{\mathbb {N}}}$, we see that $S(X)\subseteq X$. It follows that $S\in {{\mathcal {L}}}(\ell (p+))$; see Lemma 2.11 for $T:=S\in {{\mathcal {L}}}(\omega )$.

(b) Suppose that $X=ces(p+)$ for some $1\le p<\infty$. It is known, for each $k\in {{\mathbb {N}}}$, that $S\in {{\mathcal {L}}}(ces(p_k))$ and $\Vert S\Vert _{ces(p_k)\rightarrow ces(p_k)}\le 1$, [26, Lemma 5.4]. Accordingly, $S(X)\subseteq X$ and so Lemma 2.11, for $T:=S\in {{\mathcal {L}}}(\omega )$, implies that $S\in {{\mathcal {L}}}(ces(p+))$.

(c) Suppose that $X=d(p+)$ for some $1\le p<\infty$. Fix $k\in {{\mathbb {N}}}$. It is known that $S\in {{\mathcal {L}}}(d_{p_k})$ and

$$\begin{aligned} \Vert S^m\Vert _{d_{p_k}\rightarrow d_{p_k}}=(m+1)^{1/p_k},\quad m\in {{\mathbb {N}}}_0, \end{aligned}$$

(4.9)

[26, Lemma 6.2]. For $m=1$ we can conclude that $S(d_{p_k})\subseteq d_{p_k}$ for $k\in {{\mathbb {N}}}$, that is, $S(X)\subseteq X$. So, in view of Lemma 2.11, for $T:=S\in {{\mathcal {L}}}(\omega )$, it follows that $S\in {{\mathcal {L}}}(d(p+))$.

(iii) (a) Suppose that $X=\ell (p+)$ for some $1\le p<\infty$. Fix $k\in {{\mathbb {N}}}$ and $x\in \ell (p+)\subseteq \ell ^{p_k}$. It follows from S being an isometry in $\ell ^{p_k}$ that $u_k(S^n x)= u_k(x)$ for all $n\in {{\mathbb {N}}}_0$ and hence, that

$$\begin{aligned} \sum _{n=0}^\infty u_k(t^n S^n x)=\sum _{n=0}^\infty t^n u_k(S^n x)\le \frac{1}{1-t}u_k(x)<\infty . \end{aligned}$$

Accordingly, the series $\sum _{n=0}^\infty t^n S^n x$ is absolutely convergent in the Fréchet space $\ell (p+)$ for each $x\in \ell (p+)$. By part (ii) the sequence $\{\sum _{n=0}^m t^n S^n \}_{m\in {{\mathbb {N}}}_0}\subseteq {{\mathcal {L}}}(\ell (p+))$ and so, by the Banach-Steinhaus theorem (as $\ell (p+)$ is barrelled), the series $\sum _{n=0}^\infty t^n S^n$ is absolutely convergent in ${{\mathcal {L}}}_s(\ell (p+))$; its sum is denoted by $R_t\in {{\mathcal {L}}}(\ell (p+))$.

It has been established that each of the operators $C_t$, $D_\varphi$, $R_t$ belongs to ${{\mathcal {L}}}(\ell (p+))$. The identities $C_t=D_\varphi R_t=\sum _{n=0}^\infty t^nD_\varphi S^n$ are valid in ${{\mathcal {L}}}(\ell (p+))$ because they are valid in ${{\mathcal {L}}}(\omega )$; see Lemma 4.1 and both (4.7) and (4.8).

(b) Suppose $X=ces(p+)$ for some $1\le p<\infty$. Fix $k\in {{\mathbb {N}}}$ and $x\in ces(p+)\subseteq ces(p_k)$. Using $\Vert S^n\Vert _{ces(p_k)\rightarrow ces(p_k)}\le 1$, for all $n\in {{\mathbb {N}}}_0$ (see the proof of part (ii)(b)), we can argue as in (a) to conclude that

$$\begin{aligned} \sum _{n=0}^\infty v_k(t^nS^n x)\le \frac{1}{1-t}v_k(x)<\infty . \end{aligned}$$

Hence, the series $\sum _{n=0}^\infty t^n S^n x$ is absolutely convergent in $ces(p+)$ for each $x\in ces(p+)$. Then argue as in (a) to deduce that the series $R_t:=\sum _{n=0}^\infty t^n S^n$ is absolutely convergent in ${{\mathcal {L}}}_s(ces(p+))$, with $R_t\in {{\mathcal {L}}}(ces(p+))$, and that the identities $C_t=D_\varphi R_t=\sum _{n=0}^\infty t^nD_\varphi S^n$ are valid in ${{\mathcal {L}}}(ces(p+))$.

(c) Let $X=d(p+)$ for some $1\le p<\infty$. Fix $k\in {{\mathbb {N}}}$ and $x\in d(p+)\subseteq d_{p_k}$. It follows from (4.9) that

$$\begin{aligned} w_k(S^mx)=\Vert S^m x\Vert _{d_{p_k}}\le \Vert S^m\Vert _{d_{p_k}\rightarrow d_{p_k}} \Vert x\Vert _{d_{p_k}}=(m+1)^{1/p_k}w_k(x),\quad m\in {{\mathbb {N}}}_0, \end{aligned}$$

and hence, since $0\le t<1$, that

$$\begin{aligned} \sum _{n=0}^\infty w_k(t^nS^n x)\le \left( \sum _{n=0}^\infty t^n (n+1)^{1/p_k}\right) w_k(x)<\infty . \end{aligned}$$

Now argue as in (a) to conclude that the series $R_t:=\sum _{n=0}^\infty t^n S^n$ is absolutely convergent in ${{\mathcal {L}}}_s(d(p+))$, with $R_t\in {{\mathcal {L}}}(d(p+))$, and that the identities $C_t=D_\varphi R_t=\sum _{n=0}^\infty t^nD_\varphi S^n$ are valid in ${{\mathcal {L}}}(d(p+))$. $\square$

We come to the main result of this section, which should be compared with Proposition 3.2 and Theorem 3.7.

Theorem 4.5

Let $t\in [0,1)$ and X be any Fréchet space in $\{\ell (p+), ces(p+),d(p+):\, 1\le p<\infty \}$.

• (i) The generalized Cesàro operator $C_t\in {{\mathcal {L}}}(X)$ is compact.

• (ii) The spectra of $C_t$ are given by

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

  (4.10)

  and

  $$\begin{aligned} \sigma ^*(C_t;X)=\sigma (C_t;X)=\Lambda \cup \{0\}. \end{aligned}$$

  (4.11)

• (iii) For each $\lambda \in \sigma _{pt}(C_t;X)$ the subspace $(\lambda I-C_t)(X)$ is closed in X with $\textrm{codim}\, (\lambda I- C_t)(X)=1$. Moreover, the 1-dimensional eigenspace $\textrm{Ker}(\frac{1}{m+1}I-C_t)=\textrm{span} (x^{[m]})$, for each $m\in {{\mathbb {N}}}_0$, with $x^{[m]}\in d_1\subseteq X$ given by (3.8).

Proof

(i) Since $D_\varphi \in {{\mathcal {L}}}(X)$ is compact (cf. Proposition 4.3) and $R_t\in {{\mathcal {L}}}(X)$ (cf. Proposition 4.4(iii)), the compactness of $C_t$ follows from the factorization $C_t=D_\varphi R_t$ (cf. Proposition 4.4(iii)) and Lemma 2.3.

(ii) Since $X\subseteq \omega$, we can conclude from Theorem 3.7 that

$$\begin{aligned} \sigma _{pt}(C_t;X)\subseteq \sigma _{pt}(C_t;\omega )=\Lambda . \end{aligned}$$

(4.12)

Fix $1\le p<\infty$. Then $d_1\subseteq \ell ^1\subseteq \ell ^p\subseteq \ell (p+)$. Since $\ell ^p\subseteq ces(p)\subseteq ces(p+)$ (cf. (1) on p. 2 of [24]), it follows that also $d_1\subseteq ces(p+)$. Moreover, $d_1\subseteq d_p\subseteq d(p+)$. So, $d_1\subseteq X$. Given $\nu \in \Lambda$ there exists $m\in {{\mathbb {N}}}_0$ such that $\nu =\varphi _m$. According to Lemma 3.4 the 1-dimensional eigenspace corresponding to $\nu \in \sigma _{pt}(C_t;\omega )$ is spanned by $x^{[m]}$ with $x^{[m]}\in d_1$. Since $d_1\subseteq X$, it follows that $\nu \in \sigma _{pt}(C_t;X)$. So, it has been established that $\Lambda \subseteq \sigma _{pt}(C_t;X)$. Combined with (4.12) we can conclude that (4.10) is valid.

The spectrum of a compact operator in a lcHs is necessarily a compact subset of ${{\mathbb {C}}}$ (see [27, Theorem 9.10.2], [33, Theorem 4 & Proposition 6]) and it is either a finite set or a countable sequence of non-zero eigenvalues with limit point 0. It follows from part (i) and (4.10) that

$$\begin{aligned} \sigma (C_t;X)=\Lambda \cup \{0\}. \end{aligned}$$

(4.13)

The discussion in the first three paragraphs of this section, with the notation from there, shows that $X=\cap _{k=1}^\infty X_k$ is a Fréchet space of the type given in Lemma 2.5. Setting there $T:=C_t\in {{\mathcal {L}}}(X)$ and $T_n:=C_t\in {{\mathcal {L}}}(X_n)$ for $n\in {{\mathbb {N}}}$ (see Propositions 2.12, 2.14 and 2.15), it is clear that condition (A) is satisfied. Moreover, $\sigma (T_n;X_n)=\Lambda \cup \{0\}$ for every $n\in {{\mathbb {N}}}$ (cf. (2.3), (2.5) and (2.7) with $p_n$ in place of p) and so, via (4.13), we have that

$$\begin{aligned} \cup _{n=1}^\infty \sigma (T_n;X_n)=\Lambda \cup \{0\}=\sigma (T;X)=\sigma (C_t;X). \end{aligned}$$

In particular, $\cup _{n=1}^\infty \sigma (T_n;X_n)\subseteq \overline{\sigma (T;X)}$ and so we can conclude from Lemma 2.5 that (4.11) is valid.

(iii) First observe that $(\nu I-C_t)=\nu (I- \nu ^{-1}C_t)$, for $\nu \in {{\mathbb {C}}}{\setminus } \{0\}$, with $\nu ^{-1}C_t$ being a compact operator by part (i). So, by [27, Theorem 9.10.1(i)], the subspace $(\nu I-C_t)(X)$ is closed in X with for every $\nu \in \sigma _{pt}(C_t;X)$. But, for $\nu \in \sigma _{pt}(C_t;X)$, as observed in the proof of part (ii), where it was also established that $\textrm{Ker}(\frac{1}{m+1}I -C_t)=\textrm{span}(x^{[m]})$, for each $m\in {{\mathbb {N}}}_0$. $\square$

Remark 4.6

(i) The identity (4.10), established in the proof of part (ii) of Theorem 4.5, can also be deduced from Lemma 2.5(ii).

(ii) Let $t\in [0,1)$ and X be any Fréchet space in $\{\ell (p+), ces(p+), d(p+)\,:\, 1\le p<\infty \}$. Since $X\subseteq \omega$ and $C_t\in {{\mathcal {L}}}(\omega )$ is injective (cf. Lemma 3.6), also $C_t\in {{\mathcal {L}}}(X)$ is injective. Moreover, as $C_t\in {{\mathcal {L}}}(X)$ is compact (cf. Theorem 4.5(i)) it cannot be surjective, otherwise it would be an isomorphism thereby implying that $0\in \rho (C_t;X)$, which is not the case (see (4.11)). Recall that ${\mathcal {E}}$ is a basis for X and, by Lemma 3.1(iii), that the range $C_t(X)$ is a proper, dense subspace of X. Hence, 0 belongs to the continuous spectrum of $C_t$. This is in contrast to the situation of $\omega$, where $0\in \rho (C_t;\omega )$; see Theorem 3.7.

(iii) Concerning the case when $t=1$, it is known that $\sigma _{pt}(C_1;\ell (p+))=\emptyset$ and

$$\begin{aligned} \sigma (C_1;\ell (p+)){=}\left\{ z\in {{\mathbb {C}}}\,:\, \left| z-\frac{p'}{2}\right| <\frac{p'}{2}\right\} \cup \{0\}\ \text{ and } \ \sigma ^*(C_1;\ell (p+)){=}\overline{\sigma (C_1;\ell (p+))}, \nonumber \\ \end{aligned}$$

(4.14)

for every $1<p<\infty$, [8, Theorem 2.2]. For $p=1$, again $\sigma _{pt}(C_1;\ell (1+))=\emptyset$ whereas

$$\begin{aligned} \sigma (C_1;\ell (1+))=\left\{ z\in {{\mathbb {C}}}\,:\, \textrm{Re}\,z>0\right\} \cup \{0\}\ \text{ and } \ \sigma ^*(C_1;\ell (1+))=\overline{\sigma (C_1;\ell (1+))},\nonumber \\ \end{aligned}$$

(4.15)

[8, Theorem 2.4]. For the Fréchet space $ces(p+)$, both (4.14) and (4.15) are also valid (with $ces(p+)$, resp. with $ces(1+)$, in place of $\ell (p+)$, resp. in place of $\ell (1+)$), as well as $\sigma _{pt}(C_1;ces(p+))=\emptyset$ for all $1\le p<\infty$, [14, Theorem 3]. For the Fréchet space $d(p+)$, both (4.14) and (4.15) are again valid with $d(p+)$ (resp. with $d(1+)$), in place of $\ell (p+)$ (resp. of $\ell (1+)$), as well as $\sigma _{pt}(C_1;d(p+))=\emptyset$ for all $1\le p<\infty$, [21, Theorem 3.2].

5 $C_t$ acting in the (LB)-spaces $\ell (p-)$, $d(p-)$ and $ces(p-)$

Given $1<p\le \infty$, consider any strictly increasing sequence $\{p_k\}_{k\in {{\mathbb {N}}}}\subseteq (1,p)$ which satisfies $p_k\uparrow p$. The Banach spaces $X_k:=\ell ^{p_k}$ satisfy $X_k\subset X_{k+1}$ with a continuous inclusion, for each $k\in {{\mathbb {N}}}$, and $X=\cup _{k=1}^\infty X_k$ is an (LB)-space, necessarily regular by Lemma 2.6. The (LB)-space X is denoted by $\ell (p-)=\mathrm{ind\,}_k\ell ^{p_k}$. If we set $X_k:=ces(p_k)$, then again $X_k\subset X_{k+1}$ for $k\in {{\mathbb {N}}}$ (see the discussion prior to Proposition 3.3 in [13]) with a continuous inclusion. The (LB)-space $X:=\cup _{k=1}^\infty X_k$, necessarily regular by Lemma 2.6, is denoted by $ces(p-):=\mathrm{ind\,}_k ces(p_k)$. Finally, the Banach spaces $X_k:=d_{p_k}$ satisfy $X_k\subset X_{k+1}$ with a continuous inclusion, for $k\in {{\mathbb {N}}}$ (see Propositions 2.7(ii) and 5.1(iii) in [19]). The (LB)-space $X:=\cup _{k=1}^\infty X_k$, necessarily regular by Lemma 2.6, is denoted by $d(p-):=\mathrm{ind\,}_k d_{p_k}$. The discussion after (3.7) shows that $d_1$ is continuously included in each space in $\{\ell ^p, ces(p), d_p:\,1<p<\infty \}$, from which it follows that $d_1\subseteq X$ continuously, for each $X\in \{\ell (p-), ces(p-),d(p-):\, 1<p\le \infty \}$. Indeed, by the definition of the inductive limit topology, $\ell ^p\subseteq \ell (p-)$ and $d_p\subseteq d(p-)$ and $ces(p)\subseteq ces(p-)$ with all inclusions continuous. In all of these (LB)-spaces the canonical vectors ${\mathcal {E}}$ form a Schauder basis. Indeed, concerning $\ell (p-)$ recall that ${{\mathcal {E}}}$ is a basis for each Banach space $\ell ^{p_k}$ and the natural inclusion $\ell ^{p_k}\subseteq \ell (p-)$ is continuous for each $k\in {{\mathbb {N}}}$. It follows that ${{\mathcal {E}}}$ is a Schauder basis for $\ell (p-)$. For the (LB)-spaces $ces(p-)$, resp. $d(p-)$, see [12, Proposition 2.1], resp. [20, Theorem 4.6]. It follows from [44, Proposition 24.7] together with Lemma 3.1(iv) that $X\subseteq \omega$ continuously. For further properties of the (LB)-spaces $\ell (p-)$, $ces(p-)$ and $d(p-)$, and operators acting in them, we refer to [12, 20, 21], for example, and the references therein.

For each of the three cases above it is clear that the diagonal (multiplication) operator $D_\varphi \in {{\mathcal {L}}}(\omega )$ as defined in (4.4) satisfies $D_\varphi (X_k)\subseteq X_k$ for all $k\in {{\mathbb {N}}}$ (cf. proof of Proposition 4.3) and so $D_\varphi (X)\subseteq X$. By Lemma 2.11 it follows that $D_\varphi \in {{\mathcal {L}}}(X)$. Actually, $D_\varphi \in {{\mathcal {L}}}(X)$ is a compact operator. For the case $X=\ell (p-)$, since $\varphi \in \ell ^2\subseteq \ell (\infty -)$, Proposition 4.5 of [12] implies that $D_\varphi \in {{\mathcal {L}}}(\ell (p-))$ is compact. Suppose now that $X:=ces(p-)$. By Proposition 4.2 of [12] it follows that $D_\varphi \in {{\mathcal {L}}}(ces(p-))$ is compact provided that ${\hat{\varphi }}\in \ell ^t$ for some $t>q$ (with $\frac{1}{p}+\frac{1}{q}=1$). But, it is clear from (3.3) that ${\hat{\varphi }}=\varphi \in \cap _{s>1}\ell ^s$ and so $D_\varphi$ is a compact operator in $ces(p-)$. Consider now when $X:=d(p-)$. Since ${\hat{\varphi }}\in \ell ^2$ and ${\hat{\varphi }}=\varphi$, it follows that $\varphi \in d_2\subseteq d(\infty -)$ and so Proposition 4.13(ii) of [21] implies that $D_\varphi$ is a compact operator in $d(p-)$. So, we have established the following result.

Proposition 5.1

Let X be any (LB)-space in $\{\ell (p-), ces(p-), d(p-)\,:\, 1<p\le \infty \}$. Then $D_\varphi$ maps X into itself and $D_\varphi \in {{\mathcal {L}}}(X)$ is a compact operator.

The following result will also be required.

Proposition 5.2

Let $t\in [0,1)$ and X be any (LB)-space in $\{\ell (p-), ces(p-), d(p-)\,:\, 1<p\le \infty \}$.

• (i) The right-shift operator S given by (4.5) maps X into X and belongs to ${{\mathcal {L}}}(X)$.

• (ii) The generalized Cesàro operator $C_t$ maps X into X and satisfies $C_t\in {{\mathcal {L}}}(X)$.

• (iii) The operator $R_t$ given by (4.7) maps X into X and belongs to ${{\mathcal {L}}}(X)$, with the series $\sum _{n=0}^\infty t^nS^n$ being convergent in ${{\mathcal {L}}}_s(X)$. Moreover,

  $$\begin{aligned} C_t=D_\varphi R_t=\sum _{n=0}^\infty t^n D_\varphi S^n. \end{aligned}$$

  (5.1)

Proof

(i) It was observed in the proof of Proposition 4.4(ii) that $S\in {{\mathcal {L}}}(\omega )$ as well as $S(\ell ^{p_k})\subseteq \ell ^{p_k}$ and $S(ces(p_k))\subseteq ces(p_k)$ and $S(d_{p_k})\subseteq d_{p_k}$, for each $k\in {{\mathbb {N}}}$, from which it is clear that $S(X)\subseteq X$. By Lemma 2.11 it follows that $S\in {{\mathcal {L}}}(X)$.

(ii) In each of the three cases $\ell (p-)$, $ces(p-)$, $d(p-)$ for X it is clear that $C_t:\omega \rightarrow \omega$ (cf. (1.1)) satisfies $C_t(X_k)\subseteq X_k$ for all $k\in {{\mathbb {N}}}$ (see the proof of Proposition 4.4(i)) and hence, $C_t(X)\subseteq X$. Since $C_t\in {{\mathcal {L}}}(\omega )$, via Proposition 3.2, again by Lemma 2.11 we can conclude that $C_t\in {{\mathcal {L}}}(X)$.

(iii) According to part (i) the sequence $\{\sum _{n=0}^kt^nS^n\}_{k\in {{\mathbb {N}}}_0}\subseteq {{\mathcal {L}}}(X)$.

Claim. $\{\sum _{n=0}^kt^nS^n\,:\, k\in {{\mathbb {N}}}_0\}$ is an equicontinuous subset of ${{\mathcal {L}}}(X)$.

Suppose first that $X=\ell (p-)$ or $X=ces(p-)$. Since X is barrelled, to establish the Claim it suffices to show, for each $x\in X$, that

$$\begin{aligned} B(x):=\left\{ \sum _{n=0}^kt^nS^nx\,:\, k\in {{\mathbb {N}}}_0\right\} \end{aligned}$$

is a bounded subset of $X=\mathrm{ind\,}_r X_r$. Since X is a regular (LB)-space, the set B(x) will be bounded if there exists $m\in {{\mathbb {N}}}$ such that $B(x)\subseteq X_m$ and B(x) is bounded in the Banach space $X_m$. But, $x\in X=\cup _{r=1}^\infty X_r$ and so there exists $m\in {{\mathbb {N}}}$ such that $x\in X_m$. Since $S^n\in {{\mathcal {L}}}(X_m)$ for all $n\in {{\mathbb {N}}}_0$, it is clear that $B(x)\subseteq X_m$. Moreover, in the proof of Proposition 4.4(ii) it was noted that $\Vert S\Vert _{X_m\rightarrow X_m}\le 1$ and hence, $\Vert S^n\Vert _{X_m\rightarrow X_m}\le 1$ for all $n\in {{\mathbb {N}}}_0$. Accordingly,

$$\begin{aligned} \Vert \sum _{n=0}^k t^n S^n x\Vert _{X_m}\le \sum _{n=0}^\infty t^n \Vert S^n x\Vert _{X_m}\le \sum _{n=0}^\infty t^n \Vert S^n\Vert _{X_m\rightarrow X_m}\Vert x\Vert _{X_m}\le \frac{\Vert x\Vert _{X_m}}{(1-t)},\quad k\in {{\mathbb {N}}}_0, \end{aligned}$$

which implies that B(x) is a bounded set in $X_m$. In the event that $X=d(p-)$, an analogous argument applies except that now $X_m=d_{p_m}$ and so $\Vert S^n\Vert _{d_{p_m}\rightarrow d_{p_m}}=(n+1)^{1/p_m}$ for $n\in {{\mathbb {N}}}_0$; see (4.9). In this case the previous inequality becomes

$$\begin{aligned} \Vert \sum _{n=0}^k t^n S^n x\Vert _{d_{p_m}}\le \left( \sum _{n=0}^\infty t^n (n+1)^{1/p_m}\right) \Vert x\Vert _{d_{p_m}},\quad k\in {{\mathbb {N}}}_0, \end{aligned}$$

which implies that B(x) is a bounded set in $d_{p_m}$ as $\sum _{n=0}^\infty t^n (n+1)^{1/p_m}<\infty$. The proof of the Claim is thereby complete.

In view of the Claim, to show that the series $\sum _{n=0}^\infty t^n S^n$ converges in ${{\mathcal {L}}}_s(X)$ it suffices to show that the limit

$$\begin{aligned} R_tx:=\lim _{k\rightarrow \infty }\sum _{n=0}^k t^n S^n x=\sum _{n=0}^\infty t^n S^n x \end{aligned}$$

(5.2)

exists in X for all $x\in X$ in some dense subset of X. Since ${{\mathcal {E}}}$ is a Schauder basis for X, its linear span $\textrm{span}\, {{\mathcal {E}}}$ is a dense subspace of X and so it suffices to show that the limit in (5.2) exists for each $x\in {{\mathcal {E}}}$. Let $x:=e_r=(0,\ldots , 0,1,0, \ldots )$, for any fixed $r\in {{\mathbb {N}}}_0$, where 1 is in position r. Then $S^ne_r=e_{r+n}$ for all $n\in {{\mathbb {N}}}_0$. Fix $k\in {{\mathbb {N}}}_0$. It follows that

$$\begin{aligned} \sum _{n=0}^kt^nS^ne_r=\sum _{n=0}^kt^n e_{r+n}=(0,\ldots , 1, t, t^2,\ldots , t^k, 0,0,\ldots ), \end{aligned}$$

(5.3)

where 1 is in position r and $t^k$ is in position $r+k$. Observe that $\Vert e_j\Vert _{\ell ^{p_1}}=1$ for $j\in {{\mathbb {N}}}_0$. Direct calculation via (2.6) shows that $\Vert e_j\Vert _{d_{p_1}}=(j+1)^{1/p_1}$, for $j\in {{\mathbb {N}}}_0$, and by Lemma 4.7 in [17], there exists $K>0$ such that $\Vert e_j\Vert _{ces(p_1)}\le K$ for all $j\in {{\mathbb {N}}}_0$. It follows that $\sum _{j=r}^\infty t^j \Vert e_j\Vert _{\ell ^{p_1}}=\frac{t^r}{(1-t)}\le \frac{1}{(1-t)}$, that $\sum _{j=r}^\infty t^j \Vert e_j\Vert _{ces(p_1)}\le \frac{Kt^r}{(1-t)}\le \frac{K}{(1-t)}$ and that $\sum _{j=r}^\infty t^j \Vert e_j\Vert _{d_{p_1}}\le \sum _{j=r}^\infty t^j(j+1)^{1/p_1}<\infty$. Accordingly, the series

$$\begin{aligned} y^{[r]}:=\sum _{j=r}^\infty t^je_j=(0,\ldots ,0, 1,t, t^2, \ldots ), \end{aligned}$$

(5.4)

with 1 in position r, is absolutely convergent in the Banach space $X_1$ belonging to $\{\ell ^{p_1}, ces(p_1), d_{p_1}\}$ and defines an element of $X_1$, that is, $y^{[r]}\in X_1$. Since the inclusion $X_1\subseteq X$ is continuous, the series (5.4) is also convergent to $y^{[r]}$ in X. For any $k>r$ we have

$$\begin{aligned} \Vert y^{[r]}-\sum _{n=0}^kt^n S^n e_r\Vert _{X_1}=\Vert \sum _{j=r+k+1}^\infty t^j e_j \Vert _{X_1}\rightarrow 0, \quad k\rightarrow \infty , \end{aligned}$$

being the tail of the absolutely convergent series (5.2). So, the sequence in (5.3) converges to $y^{[r]}$ in $X_1$ for $k\rightarrow \infty$ and hence, also to $y^{[r]}$ in X. Since $r\in {{\mathbb {N}}}_0$ is arbitrary, we have proved that the limit in (5.2) exists in X for each $x\in \textrm{span}\, {{\mathcal {E}}}$ and hence, by the Claim, it exists for every $x\in X$. Accordingly, the limit operator $R_t=\lim _{k\rightarrow \infty }\sum _{n=0}^k t^n S^n$ exists in ${{\mathcal {L}}}_s(X)$. Since $D_\varphi , R_t, C_t\in {{\mathcal {L}}}(X)$ and $X\subseteq \omega$ continuously, the equality $C_t=D_\varphi R_t=\sum _{n=0}^\infty t^n D_\varphi S^n$ follows from Proposition 4.2. $\square$

The main result of this section is as follows.

Theorem 5.3

Let $t\in [0,1)$ and X be any (LB)-space in $\{\ell (p-), ces(p-),d(p-):\, 1<p\le \infty \}$.

• (i) The generalized Cesàro operator $C_t\in {{\mathcal {L}}}(X)$ is compact.

• (ii) The spectra of $C_t$ are given by

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

  (5.5)

  and

  $$\begin{aligned} \sigma ^*(C_t;X)=\sigma (C_t;X)=\Lambda \cup \{0\}. \end{aligned}$$

  (5.6)

• (iii) For each $\lambda \in \sigma _{pt}(C_t;X)$ the subspace $(\lambda I-C_t)(X)$ is closed in X with $\textrm{codim}\, (\lambda I- C_t)(X)=1$. Moreover, the 1-dimensional eigenspace $\textrm{Ker}(\frac{1}{m+1}I-C_t)=\textrm{span}(x^{[m]})$, for each $m\in {{\mathbb {N}}}_0$, with $x^{[m]}\in d_1\subseteq X$ given by (3.8).

Proof

(i) Since $D_\varphi \in {{\mathcal {L}}}(X)$ is compact (cf. Proposition 5.1) and $R_t\in {{\mathcal {L}}}(X)$ (cf. Proposition 5.2(iii)), the compactness of $C_t\in {{\mathcal {L}}}(X)$ follows from the factorization in (5.1) and Lemma 2.3.

(ii) The (LB)-space $X=\mathrm{ind\,}_k X_k$ is an inductive limit of the type in Lemma 2.10. Moreover, $T:=C_t\in {{\mathcal {L}}}(X)$ has the property, for each $k\in {{\mathbb {N}}}$, that the restriction $T_k$ of T to the Banach space $X_k$ maps $X_k$ into itself and satisfies $T_k\in {{\mathcal {L}}}(X_k)$. That is, T satisfies condition (A$\prime$) of Lemma 2.10. Then, by Lemma 2.10(i) it follows that $\sigma _{pt}(C_t;X)=\cup _{k=1}^\infty \sigma _{pt}(T_k; X_k)=\Lambda$ (cf. Propositions 2.12, 2.14 and 2.15). Since $C_t\in {{\mathcal {L}}}(X)$ is compact by part (i), the analogous argument used to prove (4.13), now with (4.10) replaced by (5.5), can be used to show that

$$\begin{aligned} \sigma (C_t;X)=\Lambda \cup \{0\}. \end{aligned}$$

(5.7)

Moreover, $\sigma (T_k;X_k)=\sigma (C_t;X_k)=\Lambda \cup \{0\}$ for every $k\in {{\mathbb {N}}}$ and so, for $m=1$ say, we note (via (5.7)) that

$$\begin{aligned} \cup _{k=m}^\infty \sigma (T_k; X_k)=\Lambda \cup \{0\}\subseteq \overline{\sigma (T;X)}. \end{aligned}$$

We can conclude again from Lemma 2.10(ii) that $\sigma ^*(C_t;X)=\overline{\sigma (C_t;X)}$. Combined with (5.7) this yields (5.6).

(iii) The analogous argument used to prove part (iii) of Theorem 4.5 also applies to establish the given statement. Again, since $d_1\subseteq X$ (see the introduction to Sect. 5), it follows that $\textrm{Ker}(\frac{1}{m+1}I-C_t)=\textrm{span}(x^{[m]})$, for each $m\in {{\mathbb {N}}}_0$. $\square$

Remark 5.4

(i) An examination of the arguments given in Remark 4.6 shows that, when suitably adapted, they also apply here to conclude that $C_t(X)$ is a proper, dense subspace of X. That is, 0 belongs to the continuous spectrum of $C_t$.

(ii) Concerning $t=1$, it is known that $\sigma _{pt}(C_1;ces(p-))=\emptyset$, [12, Proposition 3.1] with

$$\begin{aligned} \{0\}\cup \left\{ z\in {{\mathbb {C}}}:\, \left| z-\frac{p'}{2}\right| <\frac{p'}{2}\right\} \subseteq \sigma (C_1; ces(p-))\subseteq \left\{ z\in {{\mathbb {C}}}:\, \left| z-\frac{p'}{2}\right| \le \frac{p'}{2}\right\} \nonumber \\ \end{aligned}$$

(5.8)

and

$$\begin{aligned} \sigma ^*(C_1;ces(p-))=\left\{ z\in {{\mathbb {C}}}:\, \left| z-\frac{p'}{2}\right| \le \frac{p'}{2}\right\} =\overline{\sigma (C_1;ces(p-))},\ 1<p\le \infty ,\qquad \quad \end{aligned}$$

(5.9)

[12, Propositions 3.2 and 3.3].

For the (LB)-space $d(p-)$, both (5.8) and (5.9) are also valid (with $d(p-)$ in place of $ces(p-)$) as well as $\sigma _{pt}(C_1; d(p-))=\emptyset$, for all $1<p\le \infty$; see Theorem 3.6 in [21].

The spectrum of $C_1$ acting in $\ell (p-)$ is covered by the next result.

Recall that the space $\ell (p'+)$ is the strong dual of $\ell (p-)$, [20, Proposition 3.4(i)], and that the dual operator $C_1'\in {{\mathcal {L}}}(\ell (p'+))$ of $C_1\in {{\mathcal {L}}}(\ell (p-))$ is given by

$$\begin{aligned} C_1'x=\left( \sum _{i=n}^\infty \frac{x_i}{i+1}\right) _{n\in {{\mathbb {N}}}_0},\quad x=(x_n)_{n\in {{\mathbb {N}}}_0}\in \ell (p'+), \end{aligned}$$

see, for instance, [40, p. 123].

Proposition 5.5

Let $p\in (1,\infty ]$ and let $p'\in [1,\infty )$ satisfy $\frac{1}{p}+\frac{1}{p'}=1$.

• (i) $\sigma _{pt}(C_1;\ell (p-))=\emptyset$ and $\{z\in {{\mathbb {C}}}\,:\, |z-\frac{p'}{2}|<\frac{p'}{2}\}\subseteq \sigma _{pt}(C_1';\ell (p'+))$.

• (ii) $\{0\}\cup \{z\in {{\mathbb {C}}}\,:\, |z-\frac{p'}{2}|<\frac{p'}{2}\}\subseteq \sigma (C_1;\ell (p-))\subseteq \{z\in {{\mathbb {C}}}\,:\, |z-\frac{p'}{2}|\le \frac{p'}{2}\}$.

• (iii) $\sigma ^*(C_1;\ell (p-))=\{z\in {{\mathbb {C}}}\,:\, |z-\frac{p'}{2}|\le \frac{p'}{2}\}=\overline{\sigma (C_1; \ell (p-))}$.

Proof

(i) The first part of (i) follows from Lemma 2.10(i), the definition $\ell (p-)=\cup _{k=1}^\infty \ell ^{p_k}$ with $1<p_k\uparrow p$, and the fact that $\sigma _{pt}(C_1;\ell ^q)=\emptyset$ for every $1<q<\infty$; see Proposition 2.13(ii).

To establish the second part, fix $z\in {{\mathbb {C}}}$ with $|z-\frac{p'}{2}|<\frac{p'}{2}$. Since $1<p_k\uparrow p$, it follows that $p'_k\downarrow p'$ and hence, the open disk $B(\frac{p'}{2},\frac{p'}{2})\subseteq B(\frac{p'_k}{2},\frac{p'_k}{2})$ for every $k\in {{\mathbb {N}}}$. Accordingly, $|z-\frac{p'_k}{2}|<\frac{p_k'}{2}$ for all $k\in {{\mathbb {N}}}$. So, by [40, Theorem 1(b)], for each $k\ge 1$ there exists $x_k\in \ell ^{p'_k}{{\setminus }}\{0\}$ such that $C'_1x_k=z x_k$ with $x_k=(x_{k,i})_{i\in {{\mathbb {N}}}_0}$ satisfying $x_{k,i+1}=x_{k,0}\prod _{h=0}^i(1-\frac{1}{z (h+1)})$ for all $i\in {{\mathbb {N}}}_0$ (see (1) on p. 125 of [40]) for some $x_{k,0}\in {{\mathbb {C}}}{{\setminus }}\{0\}$. Setting $x_{k,0}:=1$ for each $k\in {{\mathbb {N}}}$, it follows that $x_k=x_{1}=:x$ for all $k\in {{\mathbb {N}}}$ and hence, $x\in \cap _{k\in {{\mathbb {N}}}}\ell ^{p'_k}=(\ell (p-))'=\ell (p'+)$. On the other hand, it is clear that $C_1'x=z x$. This shows the second part of (i).

(ii) To establish the second containment in (ii) we note that an analogous proof as that given for Proposition 3.2 in [12] also applies here. The use of Theorem 3.1 and Lemma 3.1(ii) there needs to be replaced, respectively, with the fact that $\sigma (C_1;\ell ^q)=\{z\in {{\mathbb {C}}}\,:\, |z-\frac{q'}{2}|\le \frac{q'}{2}\}$ for $1<q<\infty$ (cf. Proposition 2.13(ii)) and Lemma 2.10(iii).

Concerning the first containment in (ii), observe that $C_1$ is not surjective on $\ell (p-)$. Indeed, the element $y:=(\frac{1-(-1)^{n+1}}{2(n+1)})_{n\in {{\mathbb {N}}}_0}$ belongs to $\ell ^{p_1}$ with $\ell ^{p_1}\subseteq \ell (p-)$ and so $y\in \ell (p-)$. On the other hand, $x:=C_1^{-1}y=((-1)^n)_{n\in {{\mathbb {N}}}_0}$ belongs to $\omega$ but, $x\not \in \ell ^{p_k}$ for every $k\in {{\mathbb {N}}}$ implies that $x\not \in \ell (p-)=\cup _{k=1}^\infty \ell ^{p_k}$. Since x is the unique element in $\omega$ satisfying $y=C_1x$ (as $C_1\in {{\mathcal {L}}}(\omega )$ is a bicontinuous isomorphism), it follows that y is not in the range of $C_1\in {{\mathcal {L}}}(\ell (p-))$ for every $1<p\le \infty$. In particular, $0\in \sigma (C_1;\ell (p-))$.

Fix $\lambda \in {{\mathbb {C}}}{\setminus }\{0\}$. If $\lambda \in \rho (C_1;\ell (p-))$, then $(\lambda I-C_1)(\ell (p-))=\ell (p-)$. Since $\ell (p-)$ is dense in $\ell ^p$, it follows (with the bar denoting the closure in $\ell ^p$) that

$$\begin{aligned} \ell ^p=\overline{\ell ^p}=\overline{(\lambda I -C_1)(\ell (p-))}\subseteq \overline{(\lambda I- C_1)(\ell ^p)}\subseteq \ell ^p. \end{aligned}$$

By Proposition 2.13 we can conclude that $|\lambda -\frac{p'}{2}|\ge \frac{p'}{2}$. Accordingly, $|\lambda -\frac{p'}{2}|<\frac{p'}{2}$ implies that $\lambda \in \sigma (C_1;\ell (p-))$.

(iii) An analogous argument used for the proof of Propostion 3.3 in [12] also applies here. One only needs to replace the use of Proposition 3.2 and Theorem 3.1 there by part (ii) above and Proposition 2.13, respectively. $\square$

6 Dynamics of the generalized Cesàro operators $C_t$

The aim of this section is to investigate the mean ergodicity and linear dynamics of the operator $C_t$, for $t\in [0,1]$, in $\omega$, in the Fréchet spaces $\{\ell (p+), ces(p+), d(p+):\, 1\le p<\infty \}$ and in the (LB)-spaces $\{\ell (p-), ces(p-), d(p-):\, 1<p\le \infty \}$. For the Banach spaces $\ell ^1$, $d_1$ and $\ell ^p$, ces(p), $d_p$, for $1<p<\infty$, these results are also new. We also study the compactness, spectra and linear dynamics of the dual operators $C_t'$.

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

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

(6.1)

is 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 follows from (6.1) that

$$\begin{aligned} \frac{T^n}{n}= T_{[n]}- \frac{n-1}{n}T_{[n-1]}, \end{aligned}$$

for $n\ge 2$. Hence, necessarily $\frac{T^n}{n}\rightarrow 0$ in ${{\mathcal {L}}}_s(X)$ (resp., in ${{\mathcal {L}}}_b(X)$) as $n\rightarrow \infty$, whenever T is mean ergodic (resp., uniformly mean ergodic). A relevant text is [39].

Concerning the dynamics of a continuous linear operator T defined on a separable lcHs X, recall that T is said to be hypercyclic if there exists $x\in X$ whose orbit $\{T^nx:\, n\in {{\mathbb {N}}}_0\}$ is dense in X. If, for some $x\in X$, the projective orbit $\{\lambda T^nx:\, \lambda \in {{\mathbb {C}}},\ n\in {{\mathbb {N}}}_0\}$ is dense in X, then T is called supercyclic. Clearly, any hypercyclic operator is also supercyclic. As general references, we refer to [16, 32].

We begin with a study of the dynamics of generalized Cesàro operators acting in $\omega$. For this, we will require, for each fixed $n\in {{\mathbb {N}}}_0$, the combinatorial identity

$$\begin{aligned} \sum _{k=n-i}^n(-1)^{(n-i)-k}\left( {\begin{array}{c}n+1\\ k+1\end{array}}\right) =\left( {\begin{array}{c}n\\ i\end{array}}\right) ,\quad i=0,\ldots , n. \end{aligned}$$

(6.2)

For the proof we proceed by induction on $i=0,\ldots ,n$. For $i=0$ observe that

$$\begin{aligned} \sum _{k=n}^n (-1)^{n-k}\left( {\begin{array}{c}n+1\\ k+1\end{array}}\right) =(-1)^0\left( {\begin{array}{c}n+1\\ n+1\end{array}}\right) =1=\left( {\begin{array}{c}n\\ 0\end{array}}\right) . \end{aligned}$$

Assume that (6.2) is valid for some $0\le i<n$. For $i+1$ it follows that

$$\begin{aligned} \sum _{k=n-(i+1)}^n&(-1)^{(n-i-1)-k}\left( {\begin{array}{c}n+1\\ k+1\end{array}}\right) =(-1)^0\left( {\begin{array}{c}n+1\\ n-i\end{array}}\right) +(-1)^{-1}\sum _{k=n-i}^n (-1)^{(n-i)-k}\left( {\begin{array}{c}n+1\\ k+1\end{array}}\right) \\&=\left( {\begin{array}{c}n+1\\ n-i\end{array}}\right) -\left( {\begin{array}{c}n\\ i\end{array}}\right) =\frac{(n+1)!}{(n-i)!(i+1)!}-\frac{n!}{i!(n-i)!}\\&=\frac{n!}{i!(n-i)!}\left[ \frac{n+1}{i+1}-1\right] =\frac{n!}{(i+1)!(n-i-1)!}=\left( {\begin{array}{c}n\\ i+1\end{array}}\right) . \end{aligned}$$

Since this is identity (6.2) for $i+1$, the proof is complete.

Theorem 6.1

Let $t\in [0,1)$ and $x^{[0]}:=\alpha _0(t^n)_{n\in {{\mathbb {N}}}_0}$ with $\alpha _0\in {{\mathbb {C}}}{\setminus }\{0\}$; see (3.8).

• (i) The generalized Cesàro operator $C_t\in {{\mathcal {L}}}(\omega )$ is power bounded and uniformly mean ergodic.

• (ii) $\textrm{Ker}(I-C_t)=\textrm{span}\,\{x^{[0]}\}$ and the range

  $$\begin{aligned} (I-C_t)(\omega )=\{x\in \omega :\, x_0=0\}=\overline{\textrm{span}\,\{e_r:\, r\in {{\mathbb {N}}}\}} \end{aligned}$$

  (6.3)

  of $(I-C_t)$ is closed in $\omega$.

• (iii) The operator $C_t$ is not supercyclic in $\omega$.

Proof

(i) That $C_t$ is power bounded follows from the barrelledness of $\omega$ and $r_n(C_tx)\le r_n(x)$, for $x\in \omega$ and $n\in {{\mathbb {N}}}_0$ (cf. (3.4)), which implies, for every $x\in \omega$, that

$$\begin{aligned} r_n(C^m_tx)\le r_n(x), \quad m, n\in {{\mathbb {N}}}_0. \end{aligned}$$

Since $\omega$ is Montel, $C_t$ is uniformly mean ergodic, [3, Proposition 2.8].

(ii) By part (i) and [5, Theorem 3.5] we can conclude that $(I-C_t)(\omega )$ is closed in $\omega$ and that

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

(6.4)

Moreover, Lemma 3.4(i) yields that $\textrm{Ker}(I-C_t)=\textrm{span}\{x^{[0]}\}$. Since $(C_tx)_0=x_0$ for each $x\in \omega$ (cf. (1.1)), we have $(I-C_t)(\omega )\subseteq \{x\in \omega :\, x_0=0\}=\overline{\textrm{span}\,\{e_r:\, r\in {{\mathbb {N}}}\}}$. In order to establish (6.3), it remains to show that $e_r\in (I-C_t)(\omega )$ for each $r\ge 1$. Observe, via Lemma 3.1(iii), that

$$\begin{aligned} (I-C_t)(e_n-te_{n+1})=(e_n-te_{n+1})-\frac{1}{n+1}e_n=\frac{n}{n+1}e_n-te_{n+1},\quad n\in {{\mathbb {N}}}_0.\qquad \quad \end{aligned}$$

(6.5)

Arguing by induction and using (6.5) we can conclude that $e_r\in (I-C_t)(\omega )$ for each $r\ge 1$. Indeed, if $n=0$, then (6.5) yields $(I-C_t)(e_0-te_1)=-te_1$ and hence, $e_1\in (I-C_t)(\omega )$. Suppose that $e_n\in (I-C_t)(\omega )$. Then (6.5) implies that $\frac{n}{n+1}e_n-te_{n+1}= (I-C_t)(e_n-te_{n+1})\in (I-C_t)(\omega )$. Since $e_n\in (I-C_t)(\omega )$, by the induction hypothesis, it follows that $e_{n+1}\in (I-C_t)(\omega )$. This completes the proof of (6.3).

(iii) To verify that $C_t\in {{\mathcal {L}}}(\omega )$ is not supercyclic we proceed as follows. It follows from (6.4), by a duality argument, that $(\omega )'_\beta = \textrm{Ker}(I-C'_t)\oplus (I-C'_t)((\omega )'_\beta )$ and that , where $C'_t\in {{\mathcal {L}}}((\omega )'_\beta )$ is the dual operator of $C_t$. Accordingly, $1\in \sigma _{pt}(C_t';(\omega )'_\beta )$. On the other hand, a direct calculation shows that the dual operator $C'_t\in {{\mathcal {L}}}((\omega )'_\beta )$ is given by the transpose matrix of (3.2), that is,

$$\begin{aligned} C'_tz=\left( \sum _{k=i}^\infty \frac{t^{k-i}}{k+1}z_k\right) _{i\in {{\mathbb {N}}}_0},\quad z=(z_k)_{k\in {{\mathbb {N}}}_0}\in (\omega )'_\beta . \end{aligned}$$

(6.6)

Recall that $(\omega )'_\beta$ consists of vectors $z=(z_n)_{n\in {{\mathbb {N}}}_0}\in {{\mathbb {C}}}^{{{\mathbb {N}}}_0}$ with only finitely many non-zero coordinates. Define

$$\begin{aligned} z^{[n]}:=\sum _{i=0}^n(-1)^i\left( {\begin{array}{c}n\\ i\end{array}}\right) t^ie_{n-i}\in (\omega )'_\beta {\setminus }\{0\}, \quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

It is shown below that

$$\begin{aligned} C'_tz^{[n]}=\frac{1}{n+1}z^{[n]}, \quad n\in {{\mathbb {N}}}_0. \end{aligned}$$

(6.7)

This reveals that $\Lambda =\{\frac{1}{n+1}:\, n\in {{\mathbb {N}}}_0\}\subseteq \sigma _{pt}(C_t'; (\omega )_\beta ')$. Since $\sigma (C_t; \omega )=\sigma _{pt}(C_t;\omega )=\Lambda$ (cf. Theorem 3.7), it follows via (2.1) in Corollary 2.2 that also $\sigma _{pt}(C_t'; (\omega )'_\beta )\subseteq \sigma (C_t'; (\omega )'_\beta )=\Lambda$. So,

$$\begin{aligned} \sigma _{pt}(C_t'; (\omega )'_\beta )=\sigma (C_t'; (\omega )'_\beta )=\Lambda . \end{aligned}$$

In particular, $C'_t$ has a plenty of eigenvalues which implies that $C_t$ cannot be supercyclic, [16, Proposition 1.26].

It remains to establish (6.7). Note, for $n\in {{\mathbb {N}}}_0$ fixed, that $(z^{[n]})_i=0$ if $i>n$ and $(z^{[n]})_{n-i}=(-1)^i\left( {\begin{array}{c}n\\ i\end{array}}\right) t^i$ for $i=0,\ldots ,n$. In particular, $z^{[n]}\in (\omega )'_\beta {\setminus }\{0\}$. For $i>n$ it is clear that

$$\begin{aligned} (C_t'z^{[n]})_i=\sum _{k=i}^\infty \frac{t^{k-i}}{k+1}(z^{[n]})_k=0=\frac{1}{n+1}\cdot 0=\frac{1}{n+1}(z^{[n]})_i. \end{aligned}$$

To verify that $(C'_tz^{[n]})_{n-i}=\frac{1}{n+1}(z^{[n]})_{n-i}$ for $i=0,\ldots , n$ observe that

$$\begin{aligned} (C'_tz^{[n]})_{n-i}&=\sum _{k=n-i}^\infty \frac{t^{k-(n-i)}}{k+1}(z^{[n]})_{k}=\sum _{k=n-i}^n\frac{t^{k-(n-i)}}{k+1}(z^{[n]})_{k}\\&=\sum _{k=n-i}^n\frac{t^{k-(n-i)}}{k+1}(z^{[n]})_{n-(n-k)}=\sum _{k=n-i}^n\frac{t^{k-(n-i)}}{k+1}(-1)^{n-k}\left( {\begin{array}{c}n\\ n-k\end{array}}\right) t^{n-k}\\&=\sum _{k=n-i}^n\frac{t^{i}}{k+1}(-1)^{n-k}\frac{n!}{(n-k)!\,k!}\cdot \frac{n+1}{n+1}\\&=\frac{t^i(-1)^i}{n+1}\sum _{k=n-i}^n(-1)^{(n-i)-k}\left( {\begin{array}{c}n+1\\ k+1\end{array}}\right) =\frac{(-1)^i}{n+1}t^i\left( {\begin{array}{c}n\\ i\end{array}}\right) , \end{aligned}$$

where the last equality follows from (6.2). But, as noted above, $(-1)^i\left( {\begin{array}{c}n\\ i\end{array}}\right) t^i=(z^{[n]})_{n-i}$ and so $(C_t'z^{[n]})_{n-i}=\frac{1}{n+1}(z^{[n]})_{n-i}$ for $i=0,\ldots , n$. The identity (6.7) is thereby established and the proof is complete. $\square$

We now turn to the dynamics of generalized Cesàro operators $C_t$ acting in the other sequence spaces considered in this paper, for which we first need to establish some general results on bounded linear operators acting in lcHs’. Recall that a linear operator $T:X\rightarrow Y$, with X, Y lcHs’, is said to be bounded if there exists a neighbourhood ${{\mathcal {U}}}$ of $0\in X$ such that $T({{\mathcal {U}}})$ is a bounded subset of Y. It is routine to verify that necessarily $T\in {{\mathcal {L}}}(X,Y)$. A lcHs X is called locally complete if, for each closed, absolutely convex subset $B\in {{\mathcal {B}}}(X)$, the space $X_B:=\textrm{span}\, (B)$ equipped with the Minkowski functional $\Vert \cdot \Vert _B$, [44, p. 47], is a Banach space, whose closed unit ball is B. Such a set B is also called a Banach disc, [36, Sect. 8.3].

Theorem 6.2

Let X be a locally complete lcHs and $T\in {{\mathcal {L}}}(X)$ be a bounded operator satisfying $\sigma (T;X)\subseteq \overline{B(0,\delta )}$ for some $\delta \in (0,1)$. Then $T^n\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$ as $n\rightarrow \infty$. In particular, T is both power bounded and uniformly mean ergodic.

Proof

Since T is a bounded operator, there exists a closed, absolutely convex neighbourhood ${{\mathcal {U}}}$ of $0\in X$ such that $T({{\mathcal {U}}})\in {{\mathcal {B}}}(X)$. So, we can select a closed, absolutely convex subset $B\in {{\mathcal {B}}}(X)$ such that $T({{\mathcal {U}}})\subseteq B$. By the assumptions, $(X_B, \Vert \cdot \Vert _B)$ is a Banach space. Since $T({{\mathcal {U}}})\subseteq B$, the map $S:X\rightarrow X_B$ defined by $Sx:=Tx$ for $x\in X$, is well defined and it is clearly continuous. Let $j:X_B\rightarrow X$ denote the canonical inclusion of $X_B$ into X, i.e., $j(x):=x$ for $x\in X_B$. Then $j\in {{\mathcal {L}}}(X_B, X)$ and $T=jS\in {{\mathcal {L}}}(X)$. On the other hand $Sj\in {{\mathcal {L}}}(X_B)$. So, by [33, Proposition 5, p. 199] we have that

$$\begin{aligned} \sigma (jS;X){\setminus }\{0\}=\sigma (Sj;X_B){\setminus }\{0\}. \end{aligned}$$

Accordingly, $\sigma (Sj;X_B)=\sigma (T;X)\subseteq \overline{B(0,\delta )}$. This implies that the spectral radius r(Sj) of Sj satisfies $r(Sj)\le \delta <1$. Since $r(Sj)=\lim _{n\rightarrow \infty }\left( \Vert (Sj)^n\Vert _{X_B\rightarrow X_B}\right) ^{1/n}$, it follows via standard arguments that $(Sj)^n\rightarrow 0$ in ${{\mathcal {L}}}_b(X_B)$ as $n\rightarrow \infty$. The claim is that this implies $T^n\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$ as $n\rightarrow \infty$. To establish the claim, fix any $C\in {{\mathcal {B}}}(X)$ and any absolutely convex neighbourhood ${{\mathcal {V}}}$ of $0\in X$. Then there exist $\lambda >0$ such that $C\subseteq \lambda {{\mathcal {U}}}$ and $\mu >0$ such that $B\subseteq \mu {{\mathcal {V}}}$. Since B is the unit closed ball of $X_B$ and $(Sj)^n\rightarrow 0$ in ${{\mathcal {L}}}_b(X_B)$, there exists $n_0\in {{\mathbb {N}}}$ such that $(Sj)^n(B)\subseteq \frac{1}{\lambda \mu }B$ for all $n\ge n_0$. So, for each $n> n_0$, it follows that

$$\begin{aligned} T^n(C)&\subseteq \lambda T^n({{\mathcal {U}}})= \lambda T^{n-1}T({{\mathcal {U}}})\subseteq \lambda T^{n-1}(B)=\lambda T^{n-1}(j(B))=\lambda (jS)^{n-1}(j(B))\\&=\lambda j(Sj)^{n-2}S(j(B))=\lambda j[(Sj)^{n-1}(B)]\subseteq \lambda j\left( \left( \frac{1}{\lambda \mu }\right) B\right) =\left( \frac{1}{\mu }\right) j(B)\\&=\left( \frac{1}{\mu }\right) B\subseteq {{\mathcal {V}}}. \end{aligned}$$

This means, with $W(C,{{\mathcal {V}}}):=\{R\in {{\mathcal {L}}}(X):\, R(C)\subseteq {{\mathcal {V}}}\}$, that $T^n\in W(C,{{\mathcal {V}}})$ for each $n>n_0$. Since $C\in {{\mathcal {B}}}(X)$ and ${{\mathcal {V}}}$ are arbitrary and the sets $W(C,{{\mathcal {V}}})$ form a basis of neighbourhoods for 0 in ${{\mathcal {L}}}_b(X)$, the claim is proved, i.e., $T^n\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$ as $n\rightarrow \infty$. It follows that T is power bounded (clearly) and that $T_{[n]}\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$ as $n\rightarrow \infty$ (i.e., T is uniformly mean ergodic). Indeed, let q be any $\tau _b$-continuous seminorm. Then (6.1) implies that $q(T_{[n]})\le \frac{1}{n}\sum _{m=1}^nq(T^m)$ for $n\in {{\mathbb {N}}}$. Since $q(T^n)\rightarrow 0$ in $[0,\infty )$, also its arithmetic means $\frac{1}{n}\sum _{m=1}^nq(T^m)\rightarrow 0$ for $n\rightarrow \infty$, that is, $\lim _{n\rightarrow \infty }q(T_{[n]})=0$. So, we can conclude that $T_{[n]}\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$ for $n\rightarrow \infty$. $\square$

Theorem 6.2 permits us to formulate and prove the following general criterion for power boundedness and uniform mean ergodicity. To state it, recall that a lcHs X is said to be ultrabornological if it is an inductive limit of Banach spaces, [36, Sect. 13.1], [44, p. 283]. For instance, Fréchet spaces, [36, Corollary 13.1.4], and (LB)-spaces are ultrabornological. A lcHs X is called a webbed space if a web can be defined on X. For the definition of a web and the properties of webbed spaces we refer to [36, Sect. 5.2] and [38, Ch. 2.4]. Recall from Sect. 2 that Fréchet spaces and (LB)-spaces are webbed spaces. Moreover, sequentially closed subspaces and quotients of webbed spaces are webbed spaces, [36, Theorem 5.3.1].

For what follows we require the next result concerning algebraic sums in ultrabornological lcHs’ which can be found in [38, Sect. 35.5(4), p. 66].

Proposition 6.3

Let X be an ultrabornological lcHs such that $X=X_1\oplus X_2$ algebraically with both $X_1, X_2\subseteq X$ webbed spaces for the topology induced by X. Then $X_1$ and $X_2$ are closed subspaces of X and $X=X_1\oplus X_2$ topologically, i.e., the canonical projections $P_i:X\rightarrow X_i$ are continuous for $i=1,2$.

In general compact operators need not be mean ergodic. Just consider $T=\alpha I$ with $|\alpha |>1$ in a finite dimensional space.

Theorem 6.4

Let X be a locally complete, webbed and ultrabornological lcHs. 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 $\textrm{Ker}(I-T)\cap (I-T)(X)=\{0\}$. Then T is both power bounded and uniformly mean ergodic.

Proof

Since $T\in {{\mathcal {L}}}(X)$ is a compact operator, the following properties hold true: (a) $(I-T)(X)$ is closed in X, (b) (1 is necessarily an eigenvalue of T as it is an isolated point of $\sigma (T;X)$ and T is compact), and (c) $\textrm{codim}\, (I-T)(X)=\textrm{dim}\, \textrm{Ker}(I-T)<\infty$, see, e.g., [27, Theorem 9.10.1]. Since $\textrm{Ker}(I-T)\cap (I-T)(X)=\{0\}$ by assumption, it follows that $X=\textrm{Ker}(I-T)\oplus (I-T)(X)$ algebraically. Moreover, $(I-T)(X)$ and $\textrm{Ker}(I-T)$ are closed complemented subspaces of X and hence, are webbed spaces, [36, Theorem 5.3.1]. So, we can apply Proposition 6.3 to conclude that $X=\textrm{Ker}(I-T)\oplus (I-T)(X)$ holds topologically.

Set $Y:=(I-T)(X)$ and $S:=T|_Y$. It is routine to verify that $S(Y)\subseteq Y$ and $S:Y\rightarrow Y$ is a compact operator. So, $\sigma (S;Y){{\setminus }}\{0\} =\sigma _{pt}(S;Y)\subseteq \sigma _{pt}(T;X)\subseteq \sigma (T;X)$. But, $1\not \in \sigma (S;Y)$. Otherwise, there exists $y\in Y{{\setminus }}\{0\}$ such that $Sy=y$, i.e., $Ty=y$ or, equivalently, $(I-T)y=0$. Thus, $y\in Y\cap \textrm{Ker}(I-T)=(I-T)(X)\cap \textrm{Ker}(I-T)=\{0\}$ and hence, $y=0$; a contradiction. Hence, $\sigma (S;Y) \subseteq \sigma (T;X){\setminus }\{1\}\subseteq \overline{B(0,\delta )}$ with $\delta \in (0,1)$. Since S is compact, it is also bounded and hence, we can apply Theorem 6.2 to conclude that $S^n\rightarrow 0$ in ${{\mathcal {L}}}_b(Y)$ as $n\rightarrow \infty$, after noting that the closed subspace Y of X is locally complete.

Denote by $P:X\rightarrow X$ the continuous projection onto $\textrm{Ker}(I-T)$ along $(I-T)(X)=Y$, i.e., for each $z\in X$ there exist unique elements $x\in \textrm{Ker}(I-T)$ and $y\in Y$ such that $z=x+y$ and so $Pz:=x$. The claim is that $T^n\rightarrow P$ in ${{\mathcal {L}}}_b(X)$ as $n\rightarrow \infty$. To establish this fix $B\in {{\mathcal {B}}}(X)$ and a neighbourhood ${{\mathcal {U}}}$ of $0\in X$. As $(I-P)\in {{\mathcal {L}}}(X)$, we have that $(I-P)(B)\in {{\mathcal {B}}}(Y)$. Taking into account that $S^n\rightarrow 0$ in ${{\mathcal {L}}}_b(Y)$ as $n\rightarrow \infty$, there exists $n_0\in {{\mathbb {N}}}$ such that $S^n((I-P)(B))\subseteq {{\mathcal {U}}}\cap Y$ for every $n\ge n_0$. On the other hand, for each $z\in X$ we have that $Pz\in \textrm{Ker}(I-T)$, i.e., $TPz=Pz$, and hence, $T^n(Pz)=Pz$ for each $n\in {{\mathbb {N}}}$. Accordingly, as $S=T$ on $(I-P)(X)=(I-T)(X)=Y$ we get, for each $z\in B$ and $n\ge n_0$, that

$$\begin{aligned} T^nz-Pz&=T^n(Pz+(z-Pz))-Pz=T^n(z-Pz)=T^n((I-P)z)\\&=S^n((I-P)z)\in S^n((I-P)(B))\subseteq {{\mathcal {U}}}\cap Y, \end{aligned}$$

where we used the fact that $(I-P)z\in Y$. Since $z\in B$ is arbitrary, this implies that $T^n-P\in W(B,{{\mathcal {U}}}):=\{R\in {{\mathcal {L}}}(X):\, R(B)\subseteq {{\mathcal {U}}}\}$ for each $n\ge n_0$. So, by the arbitrariness of B and ${{\mathcal {U}}}$, the claim is proved. $\square$

Remark 6.5

(i) Let X be a sequentially complete lcHs and $T\in {{\mathcal {L}}}(X)$. If $\frac{T^n}{n}\rightarrow 0$ in ${{\mathcal {L}}}_s(X)$ as $n\rightarrow \infty$, then $\sigma (T;X)\subseteq \overline{B(0,1)}$, [2, Proposition 5.1 & Remark 5.3]; see also [28, Proposition 4.4]. In particular, if T is power bounded, then $\sigma (T;X)\subseteq \overline{B(0,1)}$. In view of this fact, Theorem 6.4 can be seen as a sort of converse result (observe that every sequentially complete lcHs is locally complete, [44, Corollary 23.14]).

(ii) Theorem 6.2 should also be compared with [6, Theorem 10] in which it is proved, for $T\in {{\mathcal {L}}}(X)$ with X a prequojection Fréchet space, that $T^n\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$ as $n\rightarrow \infty$ if, and only if, $\sigma (T;X)\subseteq B(0,1)$ and $\frac{T^n}{n}\rightarrow 0$ in ${{\mathcal {L}}}_b(X)$. Since $\sigma (C_t;\omega )\not \subseteq B(0,1)$ (as $1\in \sigma (C_t;\omega )$ but $1\not \in {B(0,1)}$) and $\omega$ is a prequojection Fréchet space, for each $t\in [0,1)$, it follows that $(C_t)^n\not \rightarrow 0$ in ${{\mathcal {L}}}_b(\omega )$ for $n\rightarrow \infty$.

Combining Theorem 6.4 with the results in the preceding sections we get the following result.

Theorem 6.6

Let $t\in [0,1)$. Let X belong to any one of the sets: $\{d_p,\ell ^p:\, 1\le p<\infty \}\cup \{ces(p):\, 1<p<\infty \}$ or $\{\ell (p+), ces(p+), d(p+):\, 1\le p<\infty \}$ or $\{\ell (p-), ces(p-), d(p-):\, 1< p\le \infty \}$. Then $C_t\in {{\mathcal {L}}}(X)$ is power bounded and uniformly mean ergodic, but not supercyclic.

Proof

From the results of the preceding sections recall that $C_t\in {{\mathcal {L}}}(X)$ is a compact operator on X and $\sigma (C_t;X)=\Lambda \cup \{0\}$. Hence, $\sigma (C_t;X){\setminus }\{1\}\subseteq \overline{B(0,1/2)}$. Moreover, $(I-C_t)(X)$ is also closed in X. Since $x^{[0]}\in d_1\subseteq X$, we can adapt the arguments in the proof of Theorem 6.1 to argue that $(I-C_t)(X)=\{x\in X:\, x_0=0\}=\overline{\textrm{span}\{e_r:\, r\in {{\mathbb {N}}}\}}$ and $\textrm{Ker}(I-C_t)=\textrm{span}\, \{x^{[0]}\}$. Hence, $\textrm{Ker}(I-C_t)\cap (I-C_t)(X)=\{0\}$. So, all the assumptions of Theorem 6.4 (for $\delta =\frac{1}{2}$ and $T:=C_t$) are satisfied. Then we can conclude that $C_t$ is power bounded and uniformly mean ergodic.

To show that $C_t:X\rightarrow X$ is not supercyclic we proceed as follows. Since $C_t\in {{\mathcal {L}}}(X)$ is compact, the operators $C_t:X\rightarrow X$ and $C'_t:X'_\beta \rightarrow X'_\beta$ have the same non-zero eigenvalues, [27, Theorem 9.10.2(2)]. Hence, $\sigma _{pt}(C'_t;X'_\beta )=\sigma _{pt}(C_t;X)=\Lambda$. According to [16, Proposition 1.26] it follows that the operator $C_t:X\rightarrow X$ cannot be supercyclic. $\square$

A first consequence of the results collected above is the following one concerning the dual operators $C'_t$. First we recall the relevant dual spaces involved. Namely, for $p, p'$ satisfying $\frac{1}{p}+\frac{1}{p'}=1$ we have (see Proposition 3.4(i), Proposition 4.3 and Remark 4.4 in [20], respectively):

$\ell (p-)\simeq (\ell (p'+))'_\beta$ and $(\ell (p-))'_\beta \simeq \ell (p'+)$, for $1<p\le \infty$;

$d(p-)\simeq (ces(p'+))'_\beta$ and $(ces(p-))'_\beta \simeq d(p'+)$, for $1< p\le \infty$;

$ces(p-)\simeq (d(p'+))'_\beta$ and $ces(p'+)\simeq (d(p-))'_\beta$, for $1<p\le \infty$.

Proposition 6.7

Let $t\in [0,1)$ and X belong to any one of the sets: $\{d_p,\ell ^p:\, 1\le p<\infty \}\cup \{ces(p):\, 1<p<\infty \}$ or $\{\ell (p+), ces(p+), d(p+):\, 1\le p<\infty \}$ or $\{\ell (p-), ces(p-), d(p-):\, 1<p\le \infty \}$.

• (i) The dual operator $C'_t\in {{\mathcal {L}}}(X'_\beta )$ of $C_t\in {{\mathcal {L}}}(X)$ is compact and is given by

  $$\begin{aligned} C'_ty=\left( \sum _{k=i}^\infty \frac{t^{k-i}}{k+1}y_k\right) _{i\in {{\mathbb {N}}}_0},\quad y=(y_k)_{k\in {{\mathbb {N}}}_0}\in X'_\beta . \end{aligned}$$

  (6.8)

• (ii) The point spectrum of $C'_t\in {{\mathcal {L}}}(X'_\beta )$ is given by

  $$\begin{aligned} \sigma _{pt}(C'_t;X'_\beta )=\sigma _{pt}(C_t;X)=\Lambda . \end{aligned}$$

  (6.9)

Each eigenvalue $\frac{1}{n+1}$, for $n\in {{\mathbb {N}}}_0$, is simple and its corresponding eigenspace is spanned by

$$\begin{aligned} y^{[n]}=\sum _{i=0}^n(-1)^i \left( {\begin{array}{c}n\\ i\end{array}}\right) t^ie_{n-i}\in X'_\beta {\setminus }\{0\}, \ n\in {{\mathbb {N}}}_0. \end{aligned}$$

Moreover,

$$\begin{aligned} \sigma ^*(C_t';X'_\beta )=\sigma (C'_t; X'_\beta )=\Lambda \cup \{0\}. \end{aligned}$$

Proof

(i) Recall that ${{\mathcal {E}}}$ is an unconditional basis in $\ell (p+), ces(p+), d(p+)$, for $1\le p<\infty$ (cf. Section 4) and an unconditional basis in $\ell (p-)$, $ces(p-)$, $d(p-)$, for $1<p\le \infty$ (cf. Section 5). Moreover, ${{\mathcal {E}}}$ is also an unconditional basis in the dual Banach spaces $(\ell ^p)'=\ell ^{p'}$ for $1<p<\infty$, in the dual Banach spaces $(ces(p))'\simeq d_{p'}$ for $1<p<\infty$, [19], and in the dual Banach spaces $(d_p)'\simeq ces(p')$ for $1<p<\infty$ (cf. [17, 24]), as well as in $(d_1)'\simeq ces(0)$, [25, Sect. 6]. In view of the description of $X'_\beta$ (for X non-normable) given prior to this Proposition it follows, for all $X\not =\ell ^1$, that the linear space $\text{ span }({{\mathcal {E}}})=(\omega )'$ is dense in $X'_\beta$. The continuity of $C'_t:X'_\beta \rightarrow X'_\beta$ then implies that (6.6) can be extended to an inequality for every $y\in X'_\beta$, that is, (6.8) is valid.

For $X=\ell ^1$, the linear space $\text{ span }({{\mathcal {E}}})=(\omega )'$ is not dense in $X'_\beta =\ell ^\infty$. So, in this case we argue as follows. Define $Ty:=\left( \sum _{k=i}^\infty \frac{t^{k-i}}{k+1}y_k\right) _{i\in {{\mathbb {N}}}_0}$ for $y\in \ell ^\infty$, in which case $T\in {{\mathcal {L}}}(\ell ^\infty )$. Indeed, for $y\in \ell ^\infty$, note that

$$\begin{aligned} \Vert Ty\Vert _\infty&=\sup _{i\in {{\mathbb {N}}}_0} \left| \sum _{k=i}^\infty \frac{t^{k-i}}{k+1}y_k\right| \le \sup _{i\in {{\mathbb {N}}}_0}\sum _{k=i}^\infty \frac{t^{k-i}}{k+1}|y_k|\le \Vert y\Vert _\infty \sup _{i\in {{\mathbb {N}}}_0}\sum _{k=i}^\infty \frac{t^{k-i}}{k+1}\\&\le \Vert y\Vert _\infty \sup _{i\in {{\mathbb {N}}}_0} \sum _{k=i}^\infty t^{k-i}=\Vert y\Vert _\infty \sum _{j=0}^\infty t^j=\frac{1}{1-t}\Vert y\Vert _\infty \ \ (\text{ as } 0\le t<1). \end{aligned}$$

Accordingly, $\Vert T\Vert _{\ell ^\infty \rightarrow \ell ^\infty }\le \frac{1}{1-t}$, that is, $T\in {{\mathcal {L}}}(\ell ^\infty )$. For each $x\in \ell ^1$ and $y\in \ell ^\infty$, a direct calculation yields

$$\begin{aligned} \langle C_t x, y\rangle =\langle x, Ty\rangle , \end{aligned}$$

which implies that $T=C'_t$.

For any Fréchet space $X\in \{\ell (p+), ces(p+), d(p+)\,:\, 1\le p<\infty \}$ and any Banach space $X\in \{\ell ^1, d_1\}\cup \{\ell ^p, ces(p), d_p\,:\, 1<p<\infty \}$ the operator $C_t\in {{\mathcal {L}}}(X)$ is compact (cf. Propositions 2.12, 2.14, 2.15 and Remark 3.3 and Theorem 4.5(i)). Accordingly, the dual operator $C'_t\in {{\mathcal {L}}}(X'_\beta )$ of $C_t\in {{\mathcal {L}}}(X)$ is compact, [27, Corollary 9.6.3].

For any (LB)-space $X\in \{\ell (p-), ces(p-), d(p-)\,:\, 1<p\le \infty \}$ the operator $C_t\in {{\mathcal {L}}}(X)$ is also compact (cf. Theorem 5.3(i)). So, the compactness of $C_t'\in {{\mathcal {L}}}(X'_\beta )$ follows from Proposition 2.7, after observing that X is a boundedly retractive (LB)-space. Indeed, $X=\ell (p-)$, for $1<p\le \infty$, is a boundedly retractive (LB)-space, as it is the strong dual of the quasinormable Fréchet space $\ell (p'+)$, [45, p. 12]. On the other hand, $X\in \{ces(p-), d(p-):\, 1<p\le \infty \}$ is a boundedly retractive (LB)-space, as it is a (DFS)-space, [20, Proposition 2.5(ii) & Lemma 4.2(i)].

(ii) It was shown in the proof of Theorem 6.1 that each vector $z^{[n]}\in (\omega )'_\beta {\setminus }\{0\}\subseteq X'_\beta$ satisfies $C'_tz^{[n]}=\frac{1}{n+1}z^{[n]}$, for every $n\in {{\mathbb {N}}}_0$. Accordingly.

$$\begin{aligned} \Lambda \subseteq \sigma _{pt}(C'_t;X'_\beta ). \end{aligned}$$

(6.10)

Moreover, $0\not \in \ \sigma _{pt}(C'_t;X'_\beta )$ as $C'_t$ is injective. To verify this let $z\in X'_\beta$ satisfy $C'_tz=0$. By considering the individual coordinates in (6.8) it follows that

$$\begin{aligned} \frac{1}{i+1}z_i=(C_t'z)_i-t(C'_tz)_{i+1}, \quad i\in {{\mathbb {N}}}_0, \end{aligned}$$

that is, $z=0$ and so indeed $0\not \in \ \sigma _{pt}(C'_t;X'_\beta )$. The compactness of $C'_t\in {{\mathcal {L}}}(X'_\beta )$ then implies that

$$\begin{aligned} \sigma (C'_t; X'_\beta )=\{0\}\cup \sigma _{pt}(C'_t;X'_\beta ) \ \text{ and } \ 0\not \in \ \sigma _{pt}(C'_t;X'_\beta ). \end{aligned}$$

(6.11)

It follows from (2.1) in Corollary 2.2 (with $T:=C_t$), from (6.11) and from the fact that $\sigma _{pt}(C_t;X)=\Lambda$, that (6.9) is valid.

Parts (1) and (2) of [27, Proposition 9.10.2] imply that each eigenvalue of $C'_t$ is simple, as this is the case for $C_t$; see Propositions 2.12, 2.14, 2.15 and Remark 3.3 and Theorems 4.5, 5.3, which also include the identities

$$\begin{aligned} \sigma ^*(C_t;X)=\sigma (C_t;X)=\Lambda \cup \{0\}. \end{aligned}$$

(6.12)

Setting $T:=C_t$ it follows from (2.2) in Corollary 2.2, together with (6.12), that

$$\begin{aligned} \sigma ^*(C'_t;X'_\beta )\subseteq \sigma ^*(C_t;X)=\Lambda \cup \{0\}. \end{aligned}$$

From general theory (cf. Section 2) we also have that

$$\begin{aligned} \sigma (C_t';X'_\beta )\subseteq \sigma ^*(C'_t;X'_\beta ). \end{aligned}$$

Since (6.9) and (6.11) imply that $\sigma (C_t';X'_\beta )=\Lambda \cup \{0\}$, we can conclude that

$$\begin{aligned} \Lambda \cup \{0\}=\sigma (C_t';X'_\beta )\subseteq \sigma ^*(C_t';X'_\beta )\subseteq \Lambda \cup \{0\}. \end{aligned}$$

This, together with (6.12), yields $\sigma ^*(C_t';X'_\beta )=\sigma ^*(C_t;X)= \Lambda \cup \{0\}$. $\square$

A consequence of Theorem 6.6 is the next result.

Proposition 6.8

Let $t\in [0,1)$. Let X belong to any one of the sets: $\{d_p, \ell ^p\,:\, 1\le p<\infty \}\cup \{ces(p)\,:\, 1<p<\infty \}$ or $\{\ell (p+), ces(p+), d(p+):\, 1\le p<\infty \}$ or $\{\ell (p-), ces(p-), d(p-)\,:\, 1<p\le \infty \}$. Then $C_t'\in {{\mathcal {L}}}(X_\beta ')$ is power bounded and uniformly mean ergodic, but not supercyclic.

Proof

By Theorem 6.6 the operator $C_t\in {{\mathcal {L}}}(X)$ is power bounded. Since $(C_t')^n=(C_t^n)'$, for every $n\in {{\mathbb {N}}}_0$, it follows from [38, Sect. 39.3(6)] that also $C'_t\in {{\mathcal {L}}}(X'_\beta )$ is power bounded. The operator $C_t\in {{\mathcal {L}}}(X)$ is also uniformly mean ergodic in X, again by Theorem 6.6. Since X is barrelled (hence, quasi-barrelled), Lemma 2.1 in [4] implies that $C'_t$ is uniformly mean ergodic in $X'_\beta$. If $X\not \in \{\ell ^1, d_1\}$, then $X'_\beta$ is reflexive with $(X'_\beta )'_\beta =X$ (cf. the proof of Proposition 6.7) and hence, $(C'_t)'=C_t$. It follows from (6.9) that $C''_t=C_t$ has plenty of eigenvalues so that $C_t'\in {{\mathcal {L}}}(X'_\beta )$ cannot be supercyclic [16, Proposition 1.26]. Finally, suppose that $X\in \{\ell ^1, d_1\}$. Since $C_t$ is compact with $\sigma _{pt}(C_t; X)=\Lambda$ (cf. Proposition 2.12 and Remark 3.3), it follows that $\sigma _{pt}(C'_t;X'_\beta )=\sigma _{pt}(C_t; X)=\Lambda$; see [27, Proposition 9.10.2(2)]. Schauder’s theorem implies that $C'_t\in {{\mathcal {L}}}(X'_\beta )$ is also compact and hence, again by Proposition 9.10.2(2) of [27], now applied to $C'_t$, we can conclude that $\sigma _{pt}(C''_t;X''_\beta )=\sigma _{pt}(C'_t;X'_\beta )=\Lambda$. So, $C''_t\in {{\mathcal {L}}}(X''_\beta )$ has plenty of eigenvalues which implies that $C'_t$ is not supercyclic. $\square$

Remark 6.9

The dynamics of $C_1\in {{\mathcal {L}}}(X)$, with $X\not \in \{\ell ^1,d_1\}$ belonging to one of the sets in Theorem 6.6, is quite different. Consider first the Banach space case. For $1<p<\infty$, the operator $C_1\in {{\mathcal {L}}}(\ell ^p)$ is neither power bounded nor mean ergodic, [5, Proposition 4.2]. Since $\{z\in {{\mathbb {C}}}\,:\, |z-\frac{p'}{2}|<\frac{p'}{2}\}\subseteq \sigma _{pt}(C_1'; \ell ^{p'})$ with $\frac{1}{p}+\frac{1}{p'}=1$, [40, Theorem 1(b)], $C_1\in {{\mathcal {L}}}(\ell ^p)$ cannot be supercyclic, [16, Proposition 1.26]. Similarly, $C_1\in {{\mathcal {L}}}(ces(p))$, for $1<p<\infty$, is not mean ergodic, not power bounded and not supercyclic, [13, Proposition 3.7(ii)]. Also, $C_1\in {{\mathcal {L}}}(d_p)$ is not mean ergodic and not supercyclic, [19, Propositions 3.10 & 3.11]. Since power bounded operators in reflexive Banach spaces are necessarily mean ergodic, [43], $C_1$ cannot be power bounded in $d_p$. Turning to Fréchet spaces, for $1\le p<\infty$ the operator $C_1\in {{\mathcal {L}}}(\ell (p+))$ is not mean ergodic, not power bounded and not supercyclic, [8, Theorems 2.3 & 2.5], as is the case for $C_1\in {{\mathcal {L}}}(ces(p+))$, [14, Proposition 5], and for $C_1\in {{\mathcal {L}}}(d(p+))$, [21, Proposition 3.5]. For (LB)-spaces, with $1< p\le \infty$, the operator $C_1\in {{\mathcal {L}}}(ces(p-))$ is not mean ergodic, not power bounded and not supercyclic, [12, Propositions 3.4 & 3.5], as is the case for $C_1\in {{\mathcal {L}}}(d(p-))$, [21, Proposition 3.8]. Finally, the dynamics of $C_1\in {{\mathcal {L}}}(\omega )$ is the same as for $C_t\in {{\mathcal {L}}}(\omega )$, with $t\in [0,1)$; see Theorem 6.1 above and [8, Proposition 4.3].

The dynamics of $C_1$ acting in $\ell (p-)$ is covered by our final result.

Proposition 6.10

Let $p\in (1,\infty ]$. The Cesàro operator $C_1\in {{\mathcal {L}}}(\ell (p-))$ is not mean ergodic, not power bounded and not supercyclic.

Proof

In view of Proposition 5.5(i) the proof follows in a similar way to that of [8, Theorem 2.3]. For the sake of completeness, we indicate the details.

By the discussion prior to Proposition 6.7 we know that $(\ell (p-))'_\beta \simeq \ell (p'+)$. Proposition 5.5(i) implies that $\frac{1+p'}{2}>1$ belongs to $\sigma _{pt}(C_1';\ell (p'+))$, where $\frac{1}{p}+\frac{1}{p'}=1$. So, there exists a non-zero vector $u\in \ell (p'+)$ satisfying $C_1'(u) =\frac{1+p'}{2}u$. Choose any $x\in \ell (p-)$ such that $\langle x,u\rangle \not =0$. Then

$$\begin{aligned} \left\langle \frac{1}{n}(C_1)^n(x),u\right\rangle =\left\langle x, \frac{1}{n}(C'_1)^n(u)\right\rangle =\frac{1}{n}\left( \frac{1+p'}{2}\right) ^n\langle x,u\rangle ,\quad n\in {{\mathbb {N}}}. \end{aligned}$$

This means that the sequence $\{\frac{1}{n}(C_1)^n(x)\}_{n\in {{\mathbb {N}}}}\subseteq \ell (p-)$ cannot be bounded in $\ell (p-)$. Accordingly, $C_1$ is not mean ergodic and not power bounded.

Applying again Proposition 5.5(i), we see that $C'_1$ has a plenty of eigenvalues. So, $C_1$ cannot be supercyclic, [16, Proposition 1.26]. $\square$