\begin{document}

\author{Ersin Kızgut}
\address{Universitat Politècnica de València \\ Instituto Universitario de Matemática Pura y Aplicada (IUMPA)\\ E-46071 Valencia, Spain}
\email{erkiz@upv.es}

\title{The Cesàro operator on smooth sequence spaces of finite type}

\date{\today}

\keywords{Cesàro operator, smooth sequence spaces of finite type, generalized power series spaces, spectrum, Fréchet space}
\subjclass[2010]{47A10, 47B37, 46A45, 46A04}

\thanks{This article has been accepted in Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Math. RACSAM}

\maketitle

\begin{abstract}
	The discrete Cesàro operator $\mathsf{C}$ is investigated in the class of smooth sequence spaces $\lambda_0(A)$ of finite type. This class contains properly the power series spaces of finite type. Of main interest is its spectrum, which is distinctly different in the cases when $\lambda_0(A)$ is nuclear and when it is not. The nuclearity of $\lambda_0(A)$ is characterized via certain properties of the spectrum of $\mathsf{C}$. Moreover, $\mathsf{C}$ is always power bounded and uniformly mean ergodic on $\lambda_0(A)$.
\end{abstract}

	
\section{Introduction}
	
The discrete Cesàro operator $\ces$ defined on $\cnn$ is defined by
\[
	\ces x:=\left(x_1, \frac{x_1+x_2}{2}, \frac{x_1+x_2+x_3}{3}, \dots, \frac{x_1+\dots +x_i}{i}, \dots \right), \quad x=(x_i)_{i \in \nn}.
\]
The Cesàro operator $\ces$ has been investigated on many Banach sequence spaces. We refer the reader to the introduction of \cite{ABRwc0}. The behaviour of $\ces$ when acting on the Fréchet spaces $\cnn$, $\ell_{p+}=\bigcap_{q>p}\ell_q,\, 1 \leq p<\infty$, and on the power series space $\Lambda_0(\alpha)$ of finite type was studied in \cite{ABR13, ABR17, ABR18}. In this paper we extend the results of \cite{ABR18} to the strictly more general setting of the smooth sequence spaces $\kotz$ of finite type. These spaces were introduced by Terzioğlu \cite{Ter69, Ter73, Ter75, Ram79}. We also refer to Kocatepe \cite{Koc85, Koc88, Koc89}. Some of our proofs are inspired by \cite{ABR18}, but new ingredients are needed in our setting. We describe precisely our context. Let $A=(a_n)_n$, where $a_n=(a_n(i))_i$. $A$ is called a \textit{Köthe matrix} if the following conditions are satisfied:
\begin{enumerate}[label=(K\arabic*)]
	\item $0 \leq a_n(i) \leq a_{n+1}(i)$, for all $i,n \in \nn$.
	\item For all $i \in \nn$, there exists $n \in \nn$ such that $a_n(i)>0$.
\end{enumerate} 
The \textit{Köthe echelon space of order} 0 associated to $A$ is defined by
\[
\kotz=\{x \in \cnn:\limi a_n(i) \, x_i=0, \, \forall n \in \nn\},
\]
which is a Fréchet space when equipped with the increasing system of seminorms
\[
	p_n(x):=\supi  a_n(i)|x_i|, \quad x \in \kotz,\, n \in \nn.
\]
Then $\kotz=\bigcap_{n \in \nn} c_0(a_n)$, with $c_0(a_n)$ the usual Banach space. We consider the projective limit topology in $\kotz$, that is, $\kotz=\proj_{n \in \nn} c_0(a_n)$. For further reading in the theory of Köthe echelon spaces $\lambda_p(A),\, 1 \leq p \leq \infty$ or $p=0$, see \cite[Section 27]{Vog97}.

The space $\kotz$ is said to be a \textit{smooth sequence space of finite type} (or a $G_1$-space) \cite[Section 3]{Ter69} if $A$ satisfies
\begin{enumerate}[label=(G1-\arabic*)]
	\item $0 < a_n(i+1) \leq a_n(i)$, for all $n \in \nn$ and $i \in \nn$.
	\item For all $n \in \nn$ there exist $m>n$ and $C>0$ such that $a_n(i) \leq C {a_m(i)}^2$, for all $i \in \nn$.
\end{enumerate}
Condition (G1-2) is actually equivalent to say that $\kotz$ is a \textit{Fréchet algebra} with respect to pointwise multiplication. Our first result is probably known. We include it for the sake of completeness and to explain the assumption~(\ref{sequences vanish}) below.
\begin{proposition}\label{G1 Schwartz}
	Let $A$ satisfy (G1-1). Then, either $\kotz$ is normable (and necessarily isomorphic to $c_0$) or $\limi a_n(i)=0$ for all $n \in \nn$.
\end{proposition}

\begin{proof}
	Since $A$ satisfies (G1-1), there exists $L_n \geq 0$ such that $\limi a_n(i) = L_n$, for all $n \in \nn$, and $L_n \leq a_n(i) \leq a_n(1)$, for all $i,n \in \nn$. Then we have two possibilities
	\begin{enumerate}[label=\normalfont(\arabic*)]
		\item There exists $n_0 \in \nn$ such that $L_{n_0} > 0$,
		\item For all $n \in \nn$, $L_n=0$.
	\end{enumerate}
	In case (1), for all $n \geq n_0$ and for all $i \in \nn$, $0 < L_{n_0} \leq a_{n_0}(i) \leq a_n(i)$. So $0 <L_{n_0} \leq L_n$ for $n \geq n_0$. Therefore $0 < L_{n_0} \leq a_n(i) \leq a_n(1)$, for all $i \in \nn$, for all $n \geq n_0$. The latter implies
	\[
		a_n(i)|x_i| \leq a_n(1) |x_i| = \frac{a_n(1)}{L_{n_0}} L_{n_0} |x_i| \leq \frac{a_n(1)}{L_{n_0}} a_{n_0}(i)|x_i|.
	\]
	Taking supremum on both sides for $i \in \nn$, we obtain
	\[
		p_n(x) \leq \frac{a_n(1)}{L_{n_0}} p_{n_0}(x).
	\]
	Hence $\kotz$ is normable since $p_{n_0}$ is the norm defining the topology of $\kotz$.  
\end{proof}
In the light of Proposition~\ref{G1 Schwartz}, given the $G_1$-space $\kotz$, we shall assume 
\begin{equation}\label{sequences vanish}
\limi a_n(i)=0, \quad \forall n \in \nn ,
\end{equation}
since otherwise the situation is totally clarified in \cite{ABRwc0}. Therefore, in our case, every $G_1$-space $\kotz$ is a Schwartz space, hence a Montel space \cite[3.4]{Ter69}.

A power series space $\Lambda_0(\alpha)=\{x \in \cnn | \limi \exp(-\alpha_i/n)|x_i|=0\}$ of finite type for $\alpha_i \igoes \infty$ is a $G_1$-space (see \cite[Section 29]{Vog97} for power series spaces of finite type). But the converse is false, in general as shown in Example~\ref{g1_non_pss} below. Before that, let us remind: A Fréchet space $E$ with a fundamental system $(p_n(\cdot))_n$ of seminorms is said to have the property (\ul{DN}) \cite[pp. 368]{Vog97} if there exists $k \in \nn$ so that for every $n \in \nn$ there exist $m \in \nn$, $0 < \tau <1$ and $C>0$ with
\[
p_n(x) \leq C \, p_k(x)^{1-\tau}\,p_m(x)^\tau, \quad \forall x \in E. \tag{\ul{DN}}\label{weak-DN}
\]

\begin{example}\label{g1_non_pss}
\normalfont
The space $X:=\{x \in \cnn:\limi a_n(i)x_i=0,\, \forall n \in \nn\}$, where $(a_n(i))_{i,n}:=\{\exp(-n e^{i/n}) \}$, is a nuclear $G_1$-space which is not isomorphic to a power series space of finite type.
\end{example}

\begin{proof}
\begin{enumerate}[wide, labelwidth=!, labelindent=0pt, label=(\textbf{\roman*})]
\item \textit{$X$ is a Köthe echelon space}: For all $i,n \in \nn$ satisfying $i \geq n(n+1)\log(1+\frac{1}{n})$, we have
		\begin{align*}
			& &\frac{i}{n(n+1)}  =  i \left(\frac{1}{n}-\frac{1}{n+1}\right) \geq & \log\left(1+\frac{1}{n}\right) = \log(n+1)-\log(n) \\
			&\Rightarrow &\log(n)+\frac{i}{n}  \geq & \, \frac{i}{n+1}+\log(n+1) \\
			&\Rightarrow & n e^{\frac{i}{n}}  \geq & (n+1) e^{\frac{i}{n+1}}\\
			 &\Rightarrow & \frac{1}{ne^{\frac{i}{n}}}  \leq  & \frac{1}{(n+1)e^{\frac{i}{n+1}}}.
		\end{align*}
Hence we have $a_n(i) \leq a_{n+1}(i)$, for all $i \geq n(n+1)\log(1+\frac{1}{n})$. Since only a finite number of indices for $i$ remain for each $n$, one can inductively choose $\{b_k\}$ each of which greater than 1, such that $a_k(i) \leq b_k(i) a_{k+1}(i)$ is satisfied for all $i \in \nn$. Observing $X \simeq \lambda_0(C)$ with the choice $c_n:=(b_n a_n)$ we deduce that $X$ is a Köthe echelon space of order 0.
		
\item \textit{$X$ satisfies (G1-1)}: Obviously for all $i \in \nn$, one has $\exp(-n e^{\frac{i+1}{n}}) < \exp(-n e^{\frac{i}{n}})$, so $a_n(i+1) < a_n (i)$, for all $i \in \nn$ and (G1-1) is satisfied.
		
\item \textit{$X$ satisfies (G1-2)}: Given $n$, choose $m=2n$. Then, one has $a_m(i)^2=\exp(-2me^{i/m})=\exp(-4ne^{i/{2n}})$. Thus $a_n(i)=\exp(-ne^{i/n}) \leq a_m(i)^2$ if and only if $-ne^{i/n} \leq -4n e^{i/{2n}}$ if and only if $i/n \geq \log(4)+(1/{2n})$ if and only if $i \geq n(2 \log (4))$. Since $1 < \log(4) < 2$, the latter is certainly satisfied for all $i \geq 4n$. So $a_n(i) \leq a_m(i)^2$, for all $i \geq 4n$. Because only a finite number of indices for $i$ remain, and $a_k(i)>0$ for all $i,k \in \nn$, there is $M>0$ such that $a_n(i) \leq Ma_m(i)^2$, for all $i \in \nn$.

\item \textit{$X$ is nuclear}: Observe first $\frac{\log(i)}{i} \igoes 0$. Given $n$, find $i_0 \in \nn$ such that $\frac{\log (i)}{i} \leq \frac{1}{n}$, for all $i>i_0$. Then
		\[
			\log (i)<\frac{i}{n} \quad \Rightarrow \quad i < e^{i/n} < n e^{i/n} \quad \Rightarrow \quad -n e^{i/n} < -i,
		\]
		for all $i>i_0$. Thus, $a_n(i)=\exp(-ne^{i/n})<e^{-i}$. But then, since $X$ is a $G_1$-space, for any $n \in \nn$, one may find $m>n$ and $C>0$ such that
		\[
			\sum_{i=i_0}^\infty \frac{a_n(i)}{a_m(i)} \leq \sum_{i=i_0}^\infty \frac{C\, a_m(i)^2}{a_m(i)} = \sum_{i=i_0}^\infty C\, a_m(i) \leq \sum_{i=i_0}^\infty e^{-i} < \infty.
		\]
		Hence the Grothendieck-Pietsch criterion for nuclearity (see e.g. \cite[Theorem 28.15]{Vog97}) is satisfied, so $X$ is a nuclear $G_1$-space. 
		
\item \textit{$X$ is not isomorphic to a power series space of finite type}: By \cite[Example 5-(5)]{Bra88}, this space fails the property \eqref{weak-DN}. However, by \cite[Lemma 29.12]{Vog97}, every power series space has property \eqref{weak-DN}. Therefore, $X$ cannot be isomorphic to a power series space of finite type. 
\end{enumerate}
 \end{proof}
More examples can be seen in Lemma~\ref{equivalences on Sn} and Remark~\ref{more G1}.

\section{Continuity and compactness of $\ces$ on $\kotz$}
An operator $T$ on a Fréchet space $X$ into itself is called bounded (resp. compact) if there exists a neighborhood $U$ of the origin of $X$ such that $TU$ is a bounded (resp. relatively compact) set in $X$. The following result is well-known (see e.g. \cite[Lemma 25]{ABR18-2}).

\begin{lemma}\label{proj lim cts}
	Let $E=\proj_m E_m$ and $F=\proj_n F_n$ be Fréchet spaces such that $E$ (resp. $F$) is the intersection of the sequence of Banach spaces $E_m$ (resp. $F_n$), $E$ is dense in $E_m$ and $E_{m+1} \subset E_m$ with continuous inclusion for each $m$ (resp. $F$ is dense in $F_n$ and $F_{n+1}\subset F_n$ with continuous inclusion for each $n$). Let $T\colon E \to F$ be a linear operator. Then
	\begin{enumerate}[label=\normalfont(\arabic*)]
		\item $T$ is continuous if and only if for each $n$ there is $m$ such that $T$ has a unique continuous linear extension $T_{m,n}\colon E_m \to F_n$.
		\item Assume $T$ is continuous. Then $T$ is bounded if and only if there is $m$ such that for each $n$, $T$ has a unique continuous linear extension $T_{m,n}\colon E_m \to F_n$.
	\end{enumerate}
\end{lemma}

\begin{proposition}
	Let $\kotz$ be a Köthe echelon space of order 0. Then, $\ces\colon\kotz \to \kotz$ is continuous if and only if for all $n \in \nn$ there exists $m>n$ such that
	
	\begin{equation}\label{continuity criterion}
		\left\{\frac{a_n(i)}{i} \sum_{j=1}^i \frac{1}{a_m(j)} \right\}_{i \in \nn} \in \ell_\infty.
	\end{equation}
\end{proposition}

\begin{proof}
	Follows from Lemma~\ref{proj lim cts}, and \cite[Proposition 2.2(i)]{ABRwc0}.  
\end{proof}

\begin{corollary}
	Let $A$ be a Köthe matrix satisfying the condition (G1-1). Then, $\ces\colon\kotz \to \kotz$ is continuous.
\end{corollary}

\begin{proof}
	Since $a_m(i+1) \leq a_m(i)$, for all $i \in \nn$, we find $m>n$ such that
	\[
		\frac{a_n(i)}{i} \sum_{j=1}^i \frac{1}{a_m(j)} \leq \frac{a_n(i)}{i} \frac{i}{a_m(i)} = \frac{a_n(i)}{a_m(i)} \leq 1, \quad \forall i \in \nn.
	\]
	Hence $\ces$ is continuous on $\kotz$. 
\end{proof}
The following proposition is a direct consequence of \cite[Proposition 2.2.(ii)]{ABRwc0} and Lemma~\ref{proj lim cts}.
\begin{proposition}
	Let $A$ be a Köthe matrix satisfying condition (G1-1). Then the Cesàro operator $\ces\colon\kotz \to \kotz$ is compact if and only there exists $m \in \nn$ such that for all $n$
	\begin{equation}\label{compactness criterion}
		\left\{\frac{a_n(i)}{i}\sum_{j=1}^i \frac{1}{a_m(j)}\right\}_{i \in \nn} \in c_0
	\end{equation}
	is satisfied.
\end{proposition}

Let $D\colon\cnn \to \cnn$ be the formal operator of differentiation $D(x_1, x_2, x_3, \dots) = (x_2, 2x_3, 3x_4, \dots)$, $x=(x_i)_i$.
\begin{proposition}
	 Let $A$ be a Köthe matrix satisfying condition (G1-1). Then, the following statements are equivalent:
	 \begin{enumerate}[label=\normalfont(\arabic*)]
	 	\item $D\colon\kotz \to \kotz$ is continuous.
	 	\item For all $n \in \nn$, there exist $m>n$ and $M>0$ such that
	 	\begin{equation}
	 		ia_n(i) \leq M a_m(i+1), \quad \forall i \in \nn.
	 	\end{equation}
	 \end{enumerate}
\end{proposition}

\begin{proof}
	(1) $\Rightarrow$ (2) Assume that $D$ is continuous on $\kotz$. Then for all $n \in \nn$, there exist $m>n$ and $M > 0$ such that
	\[
		\supi  a_n(i)|(Dx)_i| \leq M \supi  a_m(i) |x_i|, \quad \forall x \in \kotz.
	\]
	Letting $x=(e_j)$, the canonical basis in $\cnn$ for $j\geq 2$ we obtain
	\[
		(j-1)a_n(j-1) \leq M a_m(j), \quad \forall j \geq 2,
	\]
	which is precisely (2).
	
	(2) $\Rightarrow$ (1) Given $n$, pick $m$ and $M>0$ as in condition (2) and for $x \in \kotz$ we have
	\begin{align*}
		a_n(i)|(Dx)_i|= i a_n(i)|x_{i+1}| \leq & M a_m(i+1)|x_{i+1}| \\
										  \leq & M p_m(x),
	\end{align*}
	which implies that $D$ is continuous on $\kotz$. 
\end{proof}

\begin{example}
\normalfont
	Consider the nuclear $G_1$-space $X$ constructed in Example~\ref{g1_non_pss}. Picking $m=2n$ yields
	\begin{align*}
		\frac{ia_n(i)}{a_m(i+1)} = \frac{i\exp(-ne^{i/n})}{\exp(-2ne^{(i+1)/2n})} & = \exp(\log(i)+2ne^{1/2n}e^{i/2n}-ne^{i/n})\\
		& = \exp\left(\frac{\log(i)}{ne^{i/n}}+2e^{1/2n}e^{-i/2n}-1\right)ne^{i/n} \igoes 0.
	\end{align*}
	Hence $D\colon X \to X$ is continuous.
\end{example}

A Köthe echelon space $\kotz$ of order 0 is called \textit{regular} \cite{Dra70} if
\[
	\frac{a_n(i)}{a_{n+1}(i)} \geq \frac{a_n(i+1)}{a_{n+1}(i+1)}, \quad \forall i,n \in \nn.
\]

\begin{proposition}\label{nuclearity_criterion1}
	Let $\kotz$ be a $G_1$-space. Then the following statements are equivalent:
	\begin{enumerate}[label=\normalfont(\arabic*)]
		\item $\kotz$ is nuclear.
		\item For all $n \in \nn$, there exists $m>n$ such that
		\begin{equation}\label{nuclearity with i}
			\supi  \frac{ia_n(i)}{a_m(i)} < \infty.
		\end{equation}
		\item Given $\alpha \in \rn$, for all $n \in \nn$, there exists $m>n$ such that
		\begin{equation}\label{nuclearity with alpha}
			\supi  \frac{i^\alpha a_n(i)}{a_m(i)} < \infty.
		\end{equation}
	\end{enumerate}
\end{proposition}

\begin{proof}
	(1) $\Rightarrow$ (2) Let $\kotz$ be a nuclear $G_1$-space. Without loss of generality assume that $a_n(i)=1$ if $n \geq i$ and also $a_n(i) \leq a_{n+1}(i)^2$, for all $i,n$. Define $b_n(i):=1$ if $i<n$ and $b_n(i):=\prod_{j=n}^i a_j(i)$ if $i\geq n$. By \cite[Theorem 2]{Rob78} $\lambda_0(B)$ is regular and $\kotz=\lambda_0(B)$ both algebraically and topologically, that is, one has
	\begin{equation}\label{A equals B}
		\forall n \quad \exists m,\, C>0:\quad a_n(i) \leq Cb_m(i), \quad \forall i \in \nn, 
	\end{equation}
	and
	\begin{equation}\label{B equals A}
		\forall n \quad \exists m,\,D>0:\quad b_n(i) \leq Da_m(i),\quad \forall i \in \nn.
	\end{equation}
	Since $\lambda_0(B)$ is nuclear, fix $n \in \nn$, and pick $m>n$ such that $\sumi \frac{b_n(i)}{b_m(i)}<\infty$ by Grothendieck-Pietsch criterion. By regularity, $(\frac{b_n(i)}{b_m(i)})_i$ is decreasing. Then, by \cite[Theorem 3.3.1]{Kno56} for all $n \in \nn$ we can find $m>n$ such that $\limi \frac{ib_n(i)}{b_m(i)} = 0$, which implies
\begin{equation}\label{B satisfies}
	\supi  \frac{ib_n(i)}{b_m(i)} < \infty.
\end{equation}
	Now consider $a_n(i)$ for a fixed $n \in \nn$. By~(\ref{A equals B}) and~(\ref{B satisfies}), there exist $\tilde{m}$ and $M>0$ such that
\[
a_n(i) \leq C b_m(i) \leq CM\frac{b_{\tilde{m}}(i)}{i}, \quad \forall i \in \nn,
\]
Then by~(\ref{B equals A}) find $\tilde{n}>\tilde{m}$ such that
\[
\frac{b_{\tilde{m}}(i)}{i} \leq D \frac{a_{\tilde{n}}(i)}{i}, \quad \forall i \in \nn.
\]
Combining the arguments, one has
\[
	\supi  \frac{ia_n(i)}{a_{\tilde{n}}(i)} \leq CMD <\infty.
\]
	
	(2) $\Rightarrow$ (3) If $\alpha \leq 1$, it is trivial. Let $\alpha>1$. For any $n \in \nn$ we find $m_1 > n$ and $M_1>0$ such that $i a_n(i) \leq M_1 a_{m_1}(i)$, for all $i \in \nn$. Then one may find $m_2 > m_1$ and $M_2>0$ such that $i^2 a_n(i) \leq M_1M_2 a_{m_2}(i)$, for all $i \in \nn$. Continuing up to $k=\lfloor\alpha\rfloor+1$ one has $i^k a_n(i) \leq M_1 \dots M_k a_{m_{k}}(i)$, for all $i \in \nn$ so we obtain (3).
	
	(3) $\Rightarrow$ (1) Take $\alpha=2$. Given $n \in \nn$ apply (3) to find $m>n$ and $M>0$ with $i^2a_n(i) \leq Ma_m(i)$, for all $i \in \nn$. Then $\frac{a_n(i)}{a_m(i)}\leq \frac{M}{i^2}$, for all $i \in \nn$ and $(\frac{a_n	}{a_m}) \in \ell_1$. The space $\kotz$ is nuclear by the Grothendieck-Pietsch criterion. 
\end{proof}

\section{Spectrum of $\ces$ on $\kotz$}
For a locally convex Hausdorff space $X$ and $T \in \clo(X)$, the \textit{resolvent set} $\rho(X;T)$ of $T$ consists of all $\lambda \in \cn$ such that $(\lambda I-T)^{-1}$ exists in $\clo(X)$. The set $\sigma(T;X):=\cn\setminus\rho(T;X)$ is called the \textit{spectrum} of $T$ on $X$. The \textit{point spectrum} $\sigma_{pt}(T;X)$ of $T$ on $X$ consists of all $\lambda \in \cn$ such that $(\lambda I-T)$ is not injective. We denote $\Sigma:=\{\frac{1}{k}:k \in \nn\}$ and $\Sigma_0:=\Sigma \cup \{0\}$.

The inverse Cesàro operator $\ices\colon\cnn \to \cnn$ is given by the explicit formula
	\begin{equation}
		y=(y_i)_{i \in \nn} \mapsto \ices(y) = (iy_i-(i-1)y_{i-1})_{i \in \nn}, \quad y_0:=0.
	\end{equation}

\begin{proposition}\label{invertibility}
	Let $A$ be a Köthe matrix satisfying (G1-1). Then, the following statements are equivalent:
	 \begin{enumerate}[label=\normalfont(\arabic*)]
	 	\item $0 \notin \sigma(\ces;\kotz)$.
	 	\item For any $n \in \nn$, there exists $m>n$ such that
	\[
		\supi  \frac{ia_n(i)}{a_m(i)} < \infty.
	\]
\end{enumerate}
\end{proposition}

\begin{proof}
		(2) $\Rightarrow$ (1) For given $n \in \nn$ pick $m>n$ as in \eqref{nuclearity with i}. Then we have
	\begin{align*}
		a_n(i)|\ices(y)| \leq & a_n(i)|iy_i-(i-1)y_{i-1}|
					 \leq  ia_n(i)|y_i| + (i-1)a_n(i)|y_{i-1}| \\
					 \leq & ia_n(i)|y_i|+(i-1)a_n(i-1)|y_{i-1}| \\
					 =	  & \frac{ia_n(i)}{a_m(i)}a_m(i)|y_i|+\frac{(i-1)a_n(i-1)}{a_m(i-1)}a_m(i-1)|y_{i-1}|,
	\end{align*}
	for some $M>0$. Then
	\[
		\supi  a_n(i) |\ices y| \leq  M a_m(i)|y_i|+Ma_m(i-1)|y_{i-1}| \leq  2M\supi a_m(i)|y_i|.
	\]
	Therefore, $\ices$ is continuous.
	
	(1) $\Rightarrow$ (2) Let $\ices:\kotz \to \kotz$ be continuous. Then for all $n \in \nn$, there exists $m > n$ and $M>0$ with
	\[
		\supi  a_n(i)|iy_i-(i-1)y_{i-1}| \leq M \supi  a_m(i) |y_i|.
	\]
	Let $y=(e_j), j\geq 2$ to get $ja_n(j) \leq M a_m(j)$, for all $j \geq 2$, and hence
	\[
		\supi  \frac{ia_n(i)}{a_m(i)} \leq M < \infty. 
	\]
This completes the proof.
 
\end{proof}

\begin{lemma}\textup{\cite[Proposition 2.6]{ABRwc0}}\label{1/s in ptsp}
	The following statements are equivalent for the Cesàro operator $\ces$ defined on a Köthe echelon space of order zero $\kotz$ and $s \in \nn$.
	\begin{enumerate}[label=\normalfont(\arabic*)]
		\item $\frac{1}{s} \in \ptsp$.
		\item $\limi i^{s-1}a_n(i)=0$, for all $n\in \nn$.
	\end{enumerate}
\end{lemma}

\begin{proposition}\label{G11 supfinite}
	Let $A$ be a Köthe matrix satisfying the condition (G1-1). Then $\Sigma= \ptsp$ if for all $n \in \nn$, there exist $m>n$ and $M>0$ such that
	\[
		\supi  \frac{ia_n(i)}{a_m(i)} < \infty.
	\]
\end{proposition}

\begin{proof}
It is clear that $\ptsp \subseteq \sigma_{pt}(\ces;\cnn)=\Sigma$ \cite[Lemma 2.5(i)]{ABRwc0}. For the other inclusion, we prove that $i^sa_n(i)\igoes0$, for all $s,n \in \nn$ by induction on $s$.  Observe that $0 < ia_n(i) \leq M a_m(i) \igoes 0$ and hence $ia_n(i) \igoes 0$, for all $n \in \nn$. Let now $i^sa_n(i) \igoes 0$ for some $s$, and consider
\[
	i^{s+1}a_n(i)=i^s i a_n(i) \leq M i^sa_m(i) \igoes 0,
\]
by the induction hypothesis. That implies $i^{s+1}a_n(i) \igoes 0$, for all $s \in \nn$ and for all $n \in \nn$, which means $(i^s)_i \in \kotz$, for all $s \in \nn$, and hence by Lemma~\ref{1/s in ptsp}, $\frac{1}{s} \in \ptsp$ for all $s\in \nn$. So $\Sigma \subseteq \ptsp$. 
\end{proof}

\begin{theorem}\label{spectrum main theorem}
	For a $G_1$-space $\kotz$, the following statements are equivalent:
	\begin{enumerate}[label=\normalfont(\arabic*)]
		\item $0 \notin \spec$.
		\item $\frac{1}{2} \in \ptsp$.
		\item There exists $s \in \nn$, $s>1$ such that $\frac{1}{s} \in \ptsp$.
		\item $\Sigma=\ptsp$.
	\end{enumerate}
\end{theorem}

\begin{proof}
	(1) $\Rightarrow$ (2) By Proposition~\ref{invertibility}, for each $n \in \nn$ find $m>n$ and $M>0$ such that $ia_n(i)\leq Ma_m(i)\igoes0$. Hence $ia_n(i) \igoes 0$, for all $n \in \nn$, which implies $\frac{1}{2} \in \ptsp$ by Lemma~\ref{1/s in ptsp}.
	
	(2) $\Rightarrow$ (1) Fix $n \in \nn$. By (G1-2) find $m>n$ and $M>0$ such that $a_n(i) \leq M a_m(i)^2$, for all $i \in \nn$. Then, by Lemma~\ref{1/s in ptsp}
	\[
		\frac{ia_n(i)}{a_m(i)} \leq M i a_m(i) \igoes 0.
	\]
	Thus (1) follows by Propositions~\ref{nuclearity_criterion1} and~\ref{invertibility}.
	
	(2) $\Leftrightarrow$ (3) Clear.
	
	(3) $\Rightarrow$ (4) By Lemma~\ref{1/s in ptsp}, $\frac{1}{s} \in \ptsp$ if and only if $(i^{s-1}) \in \kotz$. Again by Lemma~\ref{1/s in ptsp}, $1 \in \ptsp$ since $(1,1,\dots) \in \kotz$. Assume there is $s \in \nn$ with $(i^s)_i \in \kotz$. Then $(i)_i \in \kotz$. We claim $\limi i^{2^k}a_n(i)=0$, for all $k,n \in \nn$. We prove it by induction on $k$ and all $n\in \nn$. For $k=0$, we have $\lim_i ia_n(i)=0$, for all $n \in \nn$. Suppose that $i^{2^r}a_n(i) \igoes 0$ for some $r \in \nn$, fix $n \in \nn$. By (G1-2) select $m>n$ and $M>0$ such that $a_n(i) \leq M a_m(i)^2$, for all $i \in \nn$. By the induction hypothesis $i^{2^r}a_m(i) \igoes 0$. Notice that $(i^{2^r}a_m(i))^2=i^{2^{r+1}}a_m(i)^2 \igoes 0$. Then
	\[
		0 < i^{2^{r+1}}a_n(i) \leq M i^{2^{r+1}} a_m(i)^2 \igoes 0.
	\]	
	Clearly $\lim_i i^{2^k}a_n(i)=0$, for all $k,n \in \nn$ implies $\limi i^t a_n(i)$, for all $t \in \nn$. Hence $\frac{1}{t} \in \ptsp$, for all $t \in \nn$.
	
	(4) $\Rightarrow$ (3) Trivial. 
	\end{proof}
	The following examples show that the assumption that $\kotz$ is a $G_1$-space in Theorem~\ref{spectrum main theorem} cannot be removed. We start with the following remark.
\begin{remark}\label{the phi for nong1}
\normalfont The function $\varphi(x)=x^\alpha e^{-x}$, $x \in (0,\infty)$, $\alpha>0$, is strictly decreasing on $(\alpha,\infty)$, has a local and global maximum at $x=\alpha$ and $\lim_{x \to 0^+}\varphi(x)=0$. In particular, if $0<\alpha<1$, then $\varphi$ is strictly decreasing on $[1,\infty)$.
\end{remark}

\begin{example}
\normalfont
\begin{enumerate}[wide, labelwidth=!, labelindent=0pt, label=(\textbf{\roman*})]
\setlength\itemsep{0.5em}	

\item Fix $0<\alpha<1$, take a strictly increasing sequence $(\alpha_n)_n \subset (0,\infty)$ tending to $\alpha$. Define the Köthe matrix $A=(a_n)_n$, $a_n(i)=i^{\alpha_n}e^{-i}$, $i,n \in \nn$. The Köthe echelon space $\kotz$ of order zero satisfies the condition (G1-1) and~(\ref{sequences vanish}) by Remark~\ref{the phi for nong1}. However, it is not a $G_1$-space. We first show that $0 \notin \spec$. Indeed, taking $n=1$, we get for each $m>1$,
\[
\supi  \frac{ia_1(i)}{a_m(i)}=\supi  i^{1-\alpha_m+\alpha_1}=\infty,
\]
since $1-\alpha_m+\alpha_1>0$. The condition follows from Proposition~\ref{invertibility}. On the other hand, for each $s,n\in \nn$, we have $0<i^{s-1}a_n(i)=i^{s+\alpha_n-1}e^{-i}\leq i^se^{-i}$, which tends to 0 as $i \to \infty$. Therefore $(i^{s-1})_i\in \kotz$ and $\frac{1}{s} \in \ptsp$ for each $s \in \nn$. This shows that condition (4) does not imply condition (1) in Theorem~\ref{spectrum main theorem} in general.

\item Fix $s \geq 1$, $s \in \nn$ and define the Köthe matrix $A=(a_n)_n$ by
\[
	a_n(i)=\frac{1}{i^{s-\frac{1}{2}+\frac{1}{n+1}}}, \quad i,n \in \nn.
\]
The Köthe echelon space $\kotz$ of order zero satisfies (G1-1) and~(\ref{sequences vanish}) but it is not a $G_1$-space. In this case $\ptsp=\{1,\frac{1}{2},\dots,\frac{1}{s}\}$, and condition (3) in Theorem~\ref{spectrum main theorem} does not imply condition (4). This is easy to show since $(i^{s-1})_i \in \kotz$, hence $(i^{m-1})_i \in \kotz$ for $m=1,2,\dots,s$ but $(i^s)_i\notin \kotz$, because $i^sa_1(i)=1$ for each $i \in \nn$. Thus $\frac{1}{s+1} \notin \ptsp$, which implies $\frac{1}{m} \notin \ptsp$ for each $m>s$, by Lemma~\ref{1/s in ptsp}. 
\end{enumerate}
\end{example}

The next lemma is well-known.

\begin{lemma}\label{spec_inclusion}
	Let $E$ be a Fréchet space, and let $T\colon E \to E$ be a continuous linear operator with the dual $T'\colon E' \to E'$. Then $\sigma_{pt}(T';E') \subset \sigma(T;E)$.
\end{lemma}

\begin{proof}
	Let $\lambda \in \sigma_{pt}(T';E')$. Then there exists a nonzero $u\in E'$ such that $T'u=\lambda u$. For $x \in E$,
	\[
	\langle (\lambda I-T)x,u \rangle =\langle x,(\lambda I-T')u \rangle = \langle x, \lambda u-T'u \rangle=0.
	\]
	If $(\lambda I-T)E$ is dense in $E$, then $u(z)=0$, for all $z \in E$. Contradiction. Therefore $(\lambda I-T)E$ is not dense in $E$, so $\lambda I-T$ is not surjective, and hence $\lambda \in \sigma(T;E)$. 
\end{proof}

\begin{proposition}\label{spectrum is sigma}
	Let $\kotz$ be a nuclear $G_1$-space. Then, $\Sigma = \spec$.
\end{proposition}

\begin{proof}
	The dual of the Cesàro operator $\ces'\colon \kotz' \to \kotz'$ is given by the formula
	\begin{equation}
		\ces'y = \left( \sum_{j=i}^\infty \frac{y_j}{j} \right)_{i \in \nn},
	\end{equation}
	for $y=(y_i)_{i \in \nn} \in \kotz' = \bigcup_{n=1}^\infty \ell_1(\frac{1}{a_n}) =  \{x \in \cnn| \, \exists n \in \nn: \sumi  \frac{|x_i|}{a_n(i)} < \infty\}$.
\begin{enumerate}[wide, labelwidth=!, labelindent=0pt, label=(\textbf{\roman*})]
\setlength\itemsep{0.5em}
\item $0 \notin \spec$: Follows by Proposition~\ref{invertibility}.

\item $\Sigma \subset \sigma_{pt}(\ces';\kotz')$: Let $\lambda \in \cn\setminus\{0\}$. Then $\ces'y=\lambda y$ for $y \in \kotz'$ if and only if $\lambda y_i = \sum_{j=i}^\infty \frac{y_j}{j}$, for all $i \in \nn$. The latter yields $\lambda(y_i-y_{i+1})=\frac{y_i}{i}$, and hence 
		  \begin{equation}\label{recurrence_relation}
		  y_{i+1} =  \left(1-\frac{1}{\lambda i}\right) y_i \quad \Rightarrow \quad
		  y_i     =  y_1 \prod_{j=1}^{i-1} \left(1-\frac{1}{\lambda j}\right), \, \, i \in \nn. 
		  \end{equation}
Now let $\lambda \in \Sigma$, that is, $\lambda=\frac{1}{s}$, for some $s \in \nn$. Let us define $y \in \kotz'$ by
		  \begin{equation}
		  	y_i = y_1 \prod_{j=1}^{i-1} \left(1-\frac{s}{j}\right),
		  \end{equation}
for $1 < i \leq s$ and $y_i=0$ for $i>s$ so that $y$ satisfies~(\ref{recurrence_relation}). Therefore $y$ belongs to the space of finitely supported sequences $c_{00}$, thus clearly to the space $\kotz'$, and satisfies $\ces'y = \lambda y$. Hence $ \lambda \in \sigma_{pt}(\ces';\kotz')$.
		  
\item $\Sigma \subseteq \spec$: follows by Lemma~\ref{spec_inclusion}.
		   
\item $\spec \subseteq \Sigma$: We show $(\ces-\lambda I)^{-1}$ is continuous on $\kotz$ for every $\lambda \in \cn \setminus \Sigma_0$. The $i$-th row of the matrix $(\ces-\lambda I)^{-1}$ is given by \cite{Rea85}:
		   \[
		   	\begin{cases}
		   	\displaystyle	\frac{-1}{i \lambda^2 \prod_{k=j}^i (1-\frac{1}{k\lambda})} & \text{if } 1 \leq j <i 
\\
		   	\displaystyle	\frac{1}{\frac{1}{i}-\lambda} & \text{if } i=j 
\\
		   		0 & \text{otherwise}.
		   	\end{cases}
		   \]
 For $D_\lambda=(d_{ij})_{i,j}$ and $E_\lambda=(e_{ij})_{i,j}$, one may formulate $(\ces-\lambda I)^{-1}=D_\lambda-\frac{1}{\lambda^2}E_\lambda$, where
		  \begin{multicols}{2}
 \noindent
		   \[
		   	d_{ij}= \begin{cases}
		   			\displaystyle \frac{1}{\frac{1}{i}-\lambda} & \text{if } i=j 
 \\
		   			0 & \text{otherwise}.
		   	\end{cases}	
		   \]
\noindent
		   \[
		   	e_{ij}= \begin{cases}
		   	\displaystyle	\frac{1}{i\prod_{k=j}^i \left(1-\frac{1}{k\lambda}\right)} & \text{if } 2 \leq j <i 
 \\
		   		0 & \text{otherwise}.
		   	\end{cases}
		   \]
\end{multicols}
By assumption, $\lambda \notin \Sigma_0$. Then $D_\lambda \in \clo(\kotz)$. $E_\lambda\colon \cnn \to \cnn$ acts continuously on $\kotz$ if and only if given $\lambda \in \cn\setminus \Sigma_0$ for every $n \in \nn$ there exists $m>n$ such that there exists a unique continuous extension $E^{nm}_\lambda\colon c_0(a_m) \to c_0(a_n)$. $E^{nm}_\lambda$ is continuous on $c_0(a_m)$ if and only if $\tilde{E}^{nm}_\lambda= \phi_n \circ E^{nm}_\lambda \circ \phi_m^{-1}$ is continuous on $c_0$, where $\phi_n(x)=(a_n(i)x_i)_i$, and $\tilde{E}^{nm}_\lambda=(\tilde{e}^{nm}_{ij})_{i,j}$. To prove that, we need to verify for each $n \in \nn$ \cite[Theorem 4.51-C]{Tay58}:
		   		\begin{enumerate}[label=\normalfont(\arabic*)]
		   			\item $\limi \tilde{e}^{nm}_{ij}=0$, for each $j \in \nn$.
		   			\item $\supi  \sumj  |\tilde{e}^{nm}_{ij}|<\infty$.
		   		\end{enumerate}
		   
Given $\alpha:=\operatorname{Re}(\frac{1}{\lambda}) \in \rn$ select $m>n$ by the nuclearity criterion~(\ref{nuclearity with alpha}) such that
		   \[
		   	\limi \frac{i^\alpha a_n(i)}{a_m(i)} = 0.
		   \]
For part (1), let $j \in \nn$ be fixed. Use the estimate in \cite[Lemma 7]{Rea85} to see
		   \[
		   	|\tilde{e}^{nm}_{ij}| \leq C \frac{a_n(i)}{a_m(j)} \frac{1}{i^{1-\alpha}j^\alpha} = \frac{C}{a_m(j)j^\alpha}i^{\alpha-1}a_n(i) \leq \frac{C'}{a_m(j)j^\alpha}\frac{a_m(i)}{i} < \varepsilon, \quad \forall i>i_0, 
		   \] 
 for any given $\varepsilon>0$ and for some $i_0 \in \nn$ where $C,C'>0$. For part (2), consider
		   \[
		   	\sumj  |\tilde{e}^{nm}_{ij}|=\sum_{j=1}^{i-1} \frac{a_n(i)}{a_m(j)}|e_{ij}| \leq C \sum_{j=1}^{i-1} \frac{a_n(i)}{a_m(j)} \frac{1}{i^{1-\alpha}j^\alpha} = C \frac{1}{i}\sum_{j=1}^{i-1} \frac{a_n(i) i^\alpha}{a_m(j)j^\alpha},
		   \]
for some $C>0$. Now let $\alpha<1$. Observing $0<\frac{a_n(i)}{a_m(i)} \leq 1$, for all $i \in \nn$ and for all $m>n$, 
		   \[
		   	\sumj  |\tilde{e}^{nm}_{ij}| \leq \frac{C}{i^{1-\alpha}} \sum_{j=1}^{i-1} \frac{1}{j^\alpha} \leq C \max \left(1, \frac{1}{1-\alpha}\right) < \infty,
		   \]
where the last inequality is due to the proof of \cite[Corollary 3.6]{ABRwc0}. If $\alpha\geq 1$, then
		   \[
		   	\sumj  |\tilde{e}^{nm}_{ij}| \leq C \frac{i^\alpha}{i}a_n(i) \sum_{j=1}^{i-1}\frac{1}{a_m(j)j^\alpha} \leq C \frac{i^\alpha}{i}a_n(i)(i-1)\frac{1}{a_m(i)} \leq C \frac{i^\alpha a_n(i)}{a_m(i)} <\infty,
		   \]
Therefore, we deduce that $E_\lambda$ is continuous on $\kotz$ which implies that $(\ces-\lambda I)^{-1}$ is continuous for every $\lambda \in \cn \setminus \Sigma_0$. 
\end{enumerate}
\end{proof}

\begin{corollary}
	Let $\ces\colon \kotz \to \kotz$, where $\kotz$ is a nuclear $G_1$-space. Then $\ces$ is neither compact nor weakly compact.
\end{corollary}

\begin{proof}
	Since $\kotz$ is a $G_1$-space with $a_n(i) \igoes 0$, for all $n \in \nn$, $\kotz$ is a Schwartz space. In particular, it is a Montel space. So there is no distinction between compactness and weak compactness in $\kotz$. Now let $\ces$ be compact. Then, by \cite[Theorem 9.10.2(4)]{Edw65}, $\spec$ is necessarily a compact set. However, that contradicts Proposition~\ref{spectrum is sigma}. 
\end{proof}

For the Köthe matrix $A=(a_n)_n$, define
\begin{equation}\label{sn definition}
	S_n(A):=\left\{s \in \rn: \sumi  \frac{1}{i^s a_n(i)} < \infty \right\}, \quad n \in \nn.
\end{equation}
It follows from~(\ref{sn definition}) and (K1) that $S_n(A) \subseteq S_{n+1}(A)$, for all $n \in \nn$. If $S_{n_0}(A) \neq \varnothing$ for some $n_0 \in \nn$, then $S_n(A) \neq \varnothing$ for all $n \geq n_0$. However, as shown in Example~\ref{Sn counterexample}, $S_{n_0}(A) \neq \varnothing$ does not always imply that $S_n(A) \neq \varnothing$ for all $n<n_0$.

\begin{example}\label{Sn counterexample}
\normalfont	Let $a_n(i)=\exp(-e^{\alpha_i/n})$, where $\alpha_i=2\log(\log(i+2))$. Assume $S_1(A) \neq \varnothing$, then there exists $s \in \rn$ and $M>0$ such that $\frac{\exp(e^{\alpha_i})}{i^s} \leq M$, for all $i \in \nn$. Then,
	\begin{align*}
		e^{2\log(\log(i+2))}  & \leq \log(M)+ s \log(i+2) \\
		(\log(i+2))^2 & \leq \log(M)+ s \log(i+2) \\
		\underbrace{\log(i+2)}_{\igoes\infty} & \leq \underbrace{\frac{\log(M)}{\log(i+2)}}_{\igoes0} +\, s,
	\end{align*}
	for all $i \in \nn$. This is obviously a contradiction, and hence $S_1(A)=\varnothing$. However, for $s=2+\varepsilon$,
	\[
		\frac{1}{i^s a_2(i)}=\frac{\exp(e^{\alpha_i/2})}{i^{2+\varepsilon}} = \frac{\exp(\log(i+2))}{i^{2+\varepsilon}} = \frac{i+2}{i^{2+\varepsilon}} \simeq \frac{1}{i^{1+\varepsilon}} \in \ell_1,
	\]
	for any given $\varepsilon>0$. Then,
	\[
		\sumi  \frac{1}{i^{2+\varepsilon}a_2(i)} < \infty,
	\]
	which implies $S_2(A)\neq \varnothing$. 
\end{example}

For some $n_0 \in \nn$, let $S_{n_0}\neq \varnothing$. We define $s_0(n):=\inf S_n(A)$. We have $s_0(n+1) \leq s_0(n)$, for all $n\geq n_0$. Moreover, since $a_n(i) \downarrow 0$, there is $i_0 \in \nn$ such that
\[
	\sum_{i=i_0}^\infty \frac{1}{i^t a_n(i)} \geq \sum_{i=i_0}^\infty \frac{1}{i^t},
\]
for all $t \in S_n(A)$. This also shows that, if $A$ is a (G1-1) Köthe matrix with $S_n(A) \neq \varnothing$, then $s_0(n) \geq 1$.

\begin{lemma}\label{Sn reduces to limit}
	For a fixed $n \in \nn$ and a Köthe matrix $A$, the following statements are equivalent:
	\begin{enumerate}[label=\normalfont(\arabic*)]
	\setlength\itemsep{0.5em}
		\item $S_n(A) \neq \varnothing$.
		\item There exists $t>1$ such that $[t,\infty) \subset S_n(A)$.
		\item There exists $t>1$ such that for all $\displaystyle s \in [t,\infty), \,\limi \frac{1}{i^s a_n(i)}=0$.
		\item There exists $t>1$ such that for all $\displaystyle s \in [t,\infty),\, \limi \frac{1}{i^s a_n(i)}<\infty$.
	\end{enumerate}
\end{lemma} 

\begin{proof}
	(1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) is trivial.
	
	(4) $\Rightarrow$ (1) Fix a real $t>1$ such that for all $s \geq t$ we have $\supi \frac{1}{i^sa_n(i)}<\infty$. Then, if $s>t+2$ it follows that
	\[
		\sumi \frac{1}{i^sa_n(i)} \leq \sumi \frac{1}{i^{t+2}a_n(i)} = \sumi \frac{1}{i^2 \, i^t a_n(i)} \leq M \sumi  \frac{1}{i^2} < \infty,
	\]
	since there exists $M>0$ such that $\frac{1}{i^ta_n(i)} \leq M$, for all $i \in \nn$. This implies $(t+2,\infty)\subset S_n(A)$. 
\end{proof}

\begin{lemma}\label{equivalences on Sn}
	Let $A$ be a Köthe matrix with $a_n(i):=\exp(-f(\alpha_i/n))$, where $\alpha=(\alpha_i)_i \uparrow \infty$, and let $f$ be a strictly increasing real function. Then, for a fixed $n \in \nn$, the following statements are equivalent:
	\begin{enumerate}[label=\normalfont(\arabic*)]
	\setlength\itemsep{0.5em}
		\item $S_n(A) \neq \varnothing.$	
		\item There exist $t>1$ and $M>0$ such that $\displaystyle \exp(f(\alpha_i/n)) \leq M i^t.$
		\item $\displaystyle \supi  \frac{f(\alpha_i/n)}{\log(i)} < \infty.$
		\item $\displaystyle \supi  \frac{\alpha_i}{f^{-1}(\log(i))} < \infty.$
	\end{enumerate}
\end{lemma}

\begin{proof}
	(1) $\Leftrightarrow$ (2) Follows by Lemma~\ref{Sn reduces to limit}.

	(2) $\Rightarrow$ (3) Since $f(\alpha_i/n) \leq \log(Mi^t)$. Then for $\tilde{M}:=M^{1/t}$ we have,
	\[
		\frac{f(\alpha_i/n)}{\log(i)} \leq t \frac{\log(\tilde{M}i)}{\log(i)}= t \left(\frac{\log(\tilde{M})}{\log(i)}+1 \right) \igoes t < \infty.
	\]
	(3) $\Rightarrow$ (2) Since there exists $M>0$ such that $f(\alpha_i/n) \leq M \log(i) = \log(i)^M$, for all $i \in \nn$, we have $\exp(f(\alpha_i/n)) \leq C i^M$, for $C>0$ which implies (2).
	
	(3) $\Leftrightarrow$ (4) $\sup_i \frac{\alpha_i}{f^{-1}(\log(i))} < \infty$ if and only if there exists $C \in \nn$, $C>0$ such that $\alpha_i \leq \frac{C}{n} n f^{-1}(\log(i))$, for all $i \in \nn$ if and only if $f(\alpha_i/n) \leq \frac{C}{n}\log(i)$, for all $i \in \nn$ if and only if $\sup_i \frac{f(\alpha_i/n)}{\log(i)}<\infty$.  
\end{proof}

\begin{remark}\label{more G1}
\normalfont
Let $f$ defined in Lemma~\ref{equivalences on Sn} be an odd and logarithmically convex in addition. Then the Köthe echelon space $\kotz$ of order 0 for which $A$ is as in Lemma~\ref{equivalences on Sn} is a $G_1$-space if there exists $k \in \nn$ such that $2f(x) \leq f(kx)$ for all $x \geq 1$. In this case, $\kotz=:L_f(\alpha_i,0)$ is called a \textit{Dragilev space of finite type}.
\end{remark}

\begin{lemma}\label{lemma on Sn}
	Let $\kotz$ be a $G_1$-space. If $S_{n_0}(A) \neq \varnothing$, for some $n_0 \in \nn$, then $\kotz$ is not nuclear.
\end{lemma}
\begin{proof}
If $S_{n_0}(A) \neq \varnothing$, there exists $t \in \rn$ such that 
	\begin{equation}\label{sn assumption}
		\sumi  \frac{1}{i^t a_{n_0}(i)}<\infty.
	\end{equation} 
	Suppose that $\kotz$ is nuclear. Then, by Proposition~\ref{nuclearity_criterion1} we find $m>n_0$ and $M>0$ such that
	$\frac{i^ta_{n_0}(i)}{a_m(i)}<M$. Hence
	\[
		\sumi  \frac{1}{i^ta_{n_0}(i)} \geq \frac{1}{M} \sumi  \frac{1}{a_m(i)} \xrightarrow{} \infty,
	\]
	which contradicts~(\ref{sn assumption}). So $\kotz$ is not nuclear. 
\end{proof}

\begin{lemma}\label{G1 arbitrary}
	Let $\kotz$ be a $G_1$-space. Then, for each $n \in \nn$ and for all $k \in \nn$ there exists $m>n$ and $M>0$ such that
	\[
		\frac{1}{a_m(i)^k} \leq M \frac{1}{a_n(i)}, \quad \forall i\in \nn.
	\]
\end{lemma}

\begin{proof}
	Let us prove the assertion for $k=2^s$ by induction over $s$. First observe that (G1-1) implies that there exists $i_0 \in \nn$ such that for any $n \in \nn$, $\frac{1}{a_n(i)}>1$, for all $i \geq i_0$. For $s=1$, for each $n \in \nn$ we find $m>n$ and $M>0$ with
	\[
		\frac{1}{a_m(i)^2} \leq M \frac{1}{a_n(i)}, \quad \forall i \in \nn,
	\]
	by condition (G1-2). Assume that $\frac{1}{a_m(i)^{2^s}} \leq M \frac{1}{a_n(i)}$, for all $i \in \nn$. For some $\tilde{m}>m$, it follows by (G1-2) and the induction hypothesis that
	\[
		\frac{1}{a_{\tilde{m}}(i)^{2^{s+1}}} = \left(\frac{1}{a_{\tilde{m}}(i)^{2^s}}\right)^2 \leq \tilde{M} \frac{1}{a_m(i)^{2^s}} \leq \overline{M} \frac{1}{a_n(i)}, \quad \forall i \geq i_0, 
	\]
	where $\tilde{M}, \, \overline{M}>0$. 
\end{proof}

\begin{lemma}\label{infimum of SnA}
	Let $A$ be a $G_1$-Köthe matrix. If there exists $n_0 \in \nn$ with $S_{n_0}(A) \neq \varnothing$, then $s_0(A):=\inf_n\{\inf S_n(A)\}=1$.
\end{lemma}

\begin{proof}
	Let $n_0 \in \nn$ with $S_{n_0}(A) \neq \varnothing$. Due to the remarks prior to Lemma~\ref{Sn reduces to limit} we only need to show that $s_0(n) \leq 1$. For some $t \in \rn$, $\sum_{i \in \nn} \frac{1}{i^t a_{n_0}(i)}<\infty$. Then there exists $i_0 \in \nn$ such that $\frac{1}{a_{n_0}(i)}<i^t$, for all $i \geq i_0$. Fix $\varepsilon>0$ and take $n_1 \in \nn$ with $\frac{t}{n_1}<\varepsilon$. By Lemma~\ref{G1 arbitrary}, we find $m>n_0$ and $M>0$ such that $\frac{1}{a_{m}(i)^{n_1}} \leq M \frac{1}{a_{n_0}(i)}$, for all $i \geq i_0$. Hence we have
	\[
		\frac{1}{a_m(i)} \leq M^{1/n_1} \frac{1}{a_{n_0}(i)^{1/n_1}} \leq \tilde{M} i^{t/n_1} < \tilde{M} i^\varepsilon, \quad \forall i \geq i_0.
	\]
	Now choose $0<\tilde{\varepsilon}<\varepsilon$ with $\frac{1}{a_m(i)} \leq \tilde{M}i^{\tilde{\varepsilon}}$. Therefore
	\[
		\sum_{i=i_0}^\infty \frac{1}{i^{1+\varepsilon}a_m(i)} \leq \tilde{M} \sum_{i=i_0}^\infty \frac{1}{i^{1+\varepsilon-\tilde{\varepsilon}}}< \infty,
	\]
	which means $s_0(m) \leq 1+\varepsilon$ implying that $s_0(A)=1$. 
\end{proof}

For each $r \geq 1$, define the open disk $D(r):=\{\lambda \in \cn:|\lambda-\frac{1}{2r}|<\frac{1}{2r}\}$. It is routine to establish that if $|\lambda-\frac{1}{2s_0(n)}|<\frac{1}{2s_0(n)}$, then $\operatorname{Re}(\frac{1}{\lambda}) > s_0(n)$.

\begin{proposition}
	Let $\kotz$ be a $G_1$-space with $S_{n_0}(A) \neq \varnothing,$ for some $n_0 \in \nn$. Then
	\begin{equation}\label{nonnuclear spectrum}
		D(1) \cup \{1\} = \bigcup_{n \in \nn} D(s_0(n)) \cup \Sigma \subseteq \spec.
	\end{equation}
\end{proposition}

\begin{proof}
	Lemma~\ref{infimum of SnA} yields $s_0(A)=1$. We get $\Sigma \subseteq \spec$ directly by Proposition~\ref{spectrum is sigma} part (iii) since the proof is independent of nuclearity. Fix $n \in \nn$ and let $\lambda \in \cn \setminus \Sigma$ satisfy $|\lambda-\frac{1}{2s_0(n)}|<\frac{1}{2s_0(n)}$ so that $\lambda \in D(s_0(n))\setminus \Sigma$. For any $y_1 \in \cn\setminus \{0\}$ define $y \in \cnn\setminus \{0\}$ by $y_{i+1}:=y_1 \prod_{j=1}^i (1-\frac{1}{j\lambda})$ for $i \in \nn$. Then $\ces'_ny=\lambda y$, where $\ces'_n\colon \ell_1(1/a_n)\to \ell_1(1/a_n)$ is the dual of the Cesàro operator $\ces_n\colon c_0(a_n) \to c_0(a_n)$. Moreover,
	\[
	\sumi  \frac{|y_i|}{a_n(i)} = \frac{|y_1|}{a_n(1)}+|y_1|\sum_{i=2}^\infty \prod_{j=1}^i \left|1-\frac{1}{j\lambda}\right|\frac{1}{a_n(i)} \leq \frac{|y_1|}{a_n(1)}+C|y_1| \sum_{i=2}^\infty\frac{1}{i^\alpha a_n(i)},
	\]
	where $\alpha=\operatorname{Re}(\frac{1}{\lambda}),\, C>0$, and the inequality is due to the estimate in \cite[Lemma 7]{Rea85}. But since $\alpha>s_0(n) \geq 1$, we have $\alpha \in S_n(A)$ and hence $y \in \kotz'$. For $z \in \kotz \subseteq c_0(a_n)$ the vector $(\ces-\lambda I)z$ with $\ces \in \clo(\kotz)$ belongs to $c_0(a_n)$. Since $y \in c_0(a_n)'\subseteq \kotz'$ it follows that $\langle (\ces-\lambda I)z,y \rangle=\langle (\ces_n-\lambda I)z,y \rangle=\langle z, (\ces'_n-\lambda I)y \rangle=0$. Thus $\langle u,y \rangle=0$ for every $u \in \overline{\operatorname{Im}(\ces-\lambda I)} \subseteq \kotz$ with $y \in \kotz'\setminus \{0\}$. Then $\ces-\lambda I \in \clo(\kotz)$ is not surjective and hence $\lambda \in \spec$. This implies $D(s_0(n)) \setminus \Sigma \subseteq \spec$ and so $\Sigma \cup D(s_0(n)) \subseteq \spec$. Since $n \in \nn$ was arbitary and $\lim_n \frac{1}{2s_0(n)}=\frac{1}{2}$, by Lemma~\ref{infimum of SnA} we establish~(\ref{nonnuclear spectrum}). 
\end{proof}

\section{Iterates of $\ces$ and mean ergodicity}
Let $X$ be a Fréchet space topologized by the increasing system of seminorms $(p_n(\cdot))_{n \in \nn}$. Then, the strong operator topology $\tau_s$ in $\clo(X)$ is determined by the seminorms
\[
p_{n,x}(S):= p_n(Sx), \quad \forall x \in X, \, \forall n \in \nn,
\]
where $S \in \clo(X)$. In this case we write $\clo_s(X)$. Let $\mathcal{B}(X)$ be the family of bounded subsets of $X$. Then, the uniform topology $\tau_b$ in $\clo(X)$ is defined by the family of seminorms
\[
p_{n,B}(S):=\sup_{x \in B} p_n(Sx), \forall n \in \nn, \, \forall B \in \mathcal{B}(X),
\]
where $S \in \clo(X)$. In this case we write $\clo_b(X)$. $\tau_b$ is called the topology of uniform convergence on bounded subsets of $X$.

An operator $T \in \clo(X)$, where $X$ is a Fréchet space, is \textit{power bounded} if $(T^k)_{k=1}^\infty$ is an equicontinuous subset of $\clo(X)$. Given $T \in \clo(X)$, the averages
\[
	T_{[k]}:=\frac{1}{k} \sum_{j=1}^k T^j, \quad k \in \nn,
\]
are called the \textit{Cesàro means} of $T$. The operator $T$ is said to be \textit{mean ergodic} (resp., \textit{uniformly mean ergodic}) if $(T_{[k]})_k$ is a convergent sequence in $\clo_s(X)$ (resp., in $\clo_b(X)$). A Fréchet space $X$ is called mean ergodic if every power bounded operator on $X$ is mean ergodic.
\begin{proposition}\label{mean ergodic}
	Let $\kotz$ be a $G_1$-space. Then $\ces\colon \kotz \to \kotz$ is power bounded and uniformly mean ergodic. In particular, 
\begin{equation}
	\kotz=\ker(I-\ces) \oplus \overline{(I-\ces)(\kotz)}. 
\end{equation}
	Moreover, $\ker(I-\ces)=\operatorname{span}\{\boldsymbol{1}\}$ and 
\begin{equation}	
	\overline{(I-\ces)(\kotz)}=\{x \in \kotz:x_1=0\}=\overline{\operatorname{span}(e_i)_{i \geq 2}}.
\end{equation}
\end{proposition}

\begin{proof}
	Let $\ces^k$ denote the $k$-th iterate of $\ces$. Pick an arbitary $x \in \kotz$. For $k=1$, one has $p_n(\ces x) \leq p_n(x)$ for any $n \in \nn$. Suppose that $p_n(\ces^r x) \leq p_n(x)$ for some $r \in \nn$. Then,
	\[
		p_n(\ces^{r+1}x) = p_n(\ces(\ces^rx)) \leq p_n(\ces^rx) \leq p_n(x).
	\]
	This implies $(\ces^k)_{k=1}^\infty$ is equicontinuous, hence $\ces$ is power bounded on $\kotz$. But since $\kotz$ is a $G_1$-space with $a_n(i) \igoes 0$, for all $n \in \nn$, $\kotz$ is a Schwartz space. In particular, it is a Montel space and hence a mean ergodic space \cite[Proposition 2.13]{ABR09}. Hence $\ces$ is uniformly mean ergodic. The facts that $\ker(I-\ces)=\operatorname{span}\{\boldsymbol{1}\}$ and $\overline{(I-\ces)(\kotz)}=\{x \in \kotz:x_1=0\}=\overline{\operatorname{span}(e_i)_{i \geq 2}}$ follow by applying the same arguments used in \cite[Proposition 4.1]{ABR13}. 
\end{proof}

\begin{proposition}
	Let $\kotz$ be a nuclear $G_1$-space. Then $(I-\ces)^m(\kotz)$ is a closed subspace of $\kotz$ for each $m \in \nn$.
\end{proposition}

\begin{proof}
	Using Proposition~\ref{mean ergodic}, we proceed as in the proof of the analogous result when $\ces$ acts on the power series space of finite type, $\Lambda_0(\alpha)$ \cite[Proposition 3.4]{ABR18}. Consider first $m=1$. By Proposition~\ref{mean ergodic} we have $\overline{(I-\ces)(\kotz)}=\{x \in \kotz:x_1=0\}$. Set $X_1(A):=\{x \in \kotz:x_1=0\} \subseteq \kotz$. Clearly $(I-\ces)(\kotz)\subseteq X_1(A)$. We claim that $(I-\ces)(X_1(A))=(I-\ces)(\kotz)$. One inclusion is obvious. To establish the other, observe that
	\begin{equation}\label{(I-C)x}
		(I-\ces)x=\left(0,x_2-\frac{x_1+x_2}{2},x_3-\frac{x_1+x_2+x_3}{3}, \dots\right), \, x=(x_i)_i \in \kotz,
	\end{equation}
	and in particular
	\begin{equation}\label{(I-C)y}
		(I-\ces)y=\left(0, \frac{y_2}{2}, y_3-\frac{y_2+y_3}{3}, y_4-\frac{y_2+y_3+y_4}{4}, \dots \right), \, y=(y_i)_i \in X_1(A).
	\end{equation}
	Fix $x \in \kotz$. It follows from~(\ref{(I-C)x}) that
	\begin{equation}\label{xj}
		x_j-\frac{1}{j}\sum_{k=1}^jx_k=\frac{1}{j}\left((j-1)x_j-\sum_{k=1}^{j-1}x_k\right), \quad j \geq 2,
	\end{equation}
	is the $j$-th coordinate of the vector $(I-\ces)x$. Set $y_i:=x_i-x_1$ for all $i \in \nn$ and observe that $y \in X_1(A)$ because $(0,1,1,1,\dots) \in \kotz$. Now, by~(\ref{(I-C)y}), the $j$-th coordinate of $(I-\ces)y$ is given by~(\ref{xj}) for $j \geq 2$. So $(I-\ces)x=(I-\ces)y \in (I-\ces)(X_1(A))$.
	
	To show that $(I-\ces)(\kotz)$ is closed in $\kotz$ is suffices to prove that the continuous linear operator $(I-\ces)|_{X_1(A)}\colon X_1(A) \to X_1(A)$ is bijective. It is clearly injective by~(\ref{(I-C)y}), so we prove it is surjective. Observe $X_1(A)=\bigcap_{n \in \nn} X^{(n)}$ with $X^{(n)}:=\{x \in c_0(a_n):x_1=0\}$ is a closed subspace of $c_0(a_n)$ for all $n \in \nn$. Set $\tilde{a}_n(i):=a_n(i+1)$, and $\tilde{A}=(\tilde{a}_n)_n$, for all $i, \, n \in \nn$. Then $X_1(A)$ is topologically isomorphic to $\lambda_0(\tilde{A}):=\bigcap_{n \in \nn} c_0(\tilde{a}_n)$ via the left shift operator $S\colon X_1(A) \to \lambda_0(\tilde{A})$ given by $S(x)=(x_2,x_3,\dots)$ for $x=(x_1,x_2,\dots) \in X_1(A)$. We claim that the operator $T\colon =S \circ (I-\ces)|_{X_1(A)} \circ S^{-1} \in \clo(\lambda_0(\tilde{A}))$ is bijective with $T^{-1} \in \clo(\lambda_0(\tilde{A}))$.
	
	To verify the claim, first we observe that the operator $T\colon \cnn \to \cnn$ is bijective, and its inverse $R:=T^{-1}\colon \cnn \to \cnn$ is determined by direct calculation as a lower triangular matrix $R=r_{ij}$ with entries
	\[
		r_{ij}=
		\begin{cases}
			\displaystyle \frac{1}{j} & \text{ if } 1 \leq j < i
 \\
			\displaystyle \frac{i+1}{i} & \text{ if } i=j
 \\
			0 & \text{ if } j>i.
		\end{cases}
	\]
	To show that $R$ is also the inverse of $T$ on $\lambda_0(\tilde{A})$, we need to verify $R \in \clo(\lambda_0(\tilde{A}))$. This is equivalent to show that for each $n \in \nn$ there exists $m>n$ such that $\phi_n \circ R \circ \phi^{-1}_m \in \clo(c_0)$, where the operator $\phi_k\colon c_0(\tilde{a}_k) \to c_0$ is given by $\phi_k(x)=(\tilde{a}_k(i)x_i)_i$ for $x \in c_0(\tilde{a}_k)$ is a surjective isometry. Whenever $n \in \nn$ and $m>n$, the lower triangular matrix corresponding to $\phi_n \circ R \circ \phi^{-1}_m$ is given by $D_{mn}:=\left(\frac{a_n(i+1)}{a_m(j+1)}r_{ij}\right)_{i,j}$. For each fixed $j \in \nn$, $D_{mn}$ satisfies
	\[
		\limi \frac{a_n(i+1)}{a_m(j+1)}r_{ij}=\frac{1}{ja_m(j+1)}\limi a_n(i+1)=0.
	\]
	Moreover, by (G1-1) and the nuclearity criterion~(\ref{nuclearity with i}),
	\begin{align*}
		\sumj  \frac{a_n(i+1)}{a_m(j+1)}r_{ij} & =\sum_{j=1}^{i-1} \frac{a_n(i+1)}{ja_m(j+1)}+\frac{i+1}{i}\frac{a_n(i+1)}{a_m(i+1)} \\
		& \leq \frac{a_n(i+1)}{a_m(i+1)} \sum_{j=1}^{i-1}\frac{1}{j}+\frac{i+1}{i}\frac{a_n(i+1)}{a_m(i+1)} \\
		& = \frac{a_n(i+1)}{a_m(i+1)}\left(\sum_{j=1}^{i-1}\frac{1}{j}+\frac{i+1}{i}\right) \\
		& \leq (i+1) \frac{a_n(i+1)}{a_m(i+1)} < \infty,
	\end{align*}
	Thus, by \cite[Theorem 4.51-C]{Tay58}, $\phi_n \circ R \circ \phi^{-1}_m \in \clo(c_0)$, as desired. Hence $(I-\ces)(\kotz)$ is closed in $\kotz$. Since $(I-\ces)(\kotz)$ is closed, it follows from Proposition~\ref{mean ergodic} that $\kotz=\ker(I-\ces)\oplus(I-\ces)(\kotz)$. Hence, by \cite[Remark 3.6]{ABR13}, $(I-\ces)^m(\kotz)$ is closed for all $m \in \nn$.   
\end{proof}

A Fréchet space operator $T \in \clo(X)$, where $X$ is separable, is called \textit{hypercyclic} if there exists $x \in X$ such that the orbit $\{T^kx:k \in \nn_0\}$ is dense in $X$. If, for some $z \in X$, the projective orbit $\{\lambda T^kz:\lambda \in \cn, k \in \nn_0\}$ is dense in $X$, then $T$ is called \textit{supercyclic}. Clearly, if $T$ is hypercyclic then $T$ is supercyclic.

\begin{proposition}
	Let $A$ be a Köthe matrix. Then $\ces \in \clo(\kotz)$ is not supercyclic, and hence not hypercyclic either.
\end{proposition}

\begin{proof}
	Follows from \cite[Proposition 4.13]{ABRwc0}. 
\end{proof}

\end{document}