# Are Ritt operators mean ergodic?

In the following, $T$ is a bounded operator on a Banach space $X$.

• $T$ is called "power bounded" if $\sup_{n\in \mathbb N}\|T^n\|<\infty$;
• $T$ is called "mean ergodic" if the Cesàro sums $\frac{1}{n}\sum_{k=1}^n T^k$ converge strongly as $n\to \infty$;
• $T$ is called a "Ritt operator" if its spectrum is contained in the unit disk and $\sup_{\{|\lambda|>1\}}\|(\lambda-1)R(\lambda,T)\|<\infty$.

Now, the following is known:

• In a reflexive space every power bounded operator is mean ergodic (Lorch 1939).
• This not true any more if we drop the assumption of reflexivity (counterexample in a large class of non-reflexive Banach spaces by Fonf-Lin-Wojtaszczyk 2001).
• Furthermore, it is known that Ritt property is strictly stronger than power boundedness (Nagy-Zemanek and Lyubich, both 1999).

So in particular in reflexive spaces each Ritt operator is mean ergodic.

Is it known whether Ritt property implies mean ergodicity in general (i.e., possibly non-reflexive) spaces?

I wouldn't bother imposing some geometric conditions on the Banach space, or even restricting to sequence spaces.

• could you please clarify the notation $R(\lambda,T)$? Thanks. Commented Apr 7, 2016 at 18:25
• @user89334 $R(\lambda,T):=(\lambda-T)^{-1}$ Commented Apr 8, 2016 at 12:36

There exists a Ritt operator $T$ on $\ell^\infty$ which is not mean ergodic.
Indeed, let $T$ be the "multiplication operator" given by $Tx = \big((1-\frac{1}{n})x_n\big)_{n \in \mathbb{N}}$ for each $x = (x_n)_{n \in \mathbb{N}} \in \ell^\infty$.
The spectrum of $T$ is given by $\sigma(T) = \{1-\frac{1}{n}: \; n \in \mathbb{N}\} \cup \{1\}$, and for every $\lambda$ in the resolvent set of $T$ we have $$\|R(\lambda,T)\| = \frac{1}{\operatorname{dist}(\lambda, \sigma(T))}$$ (it is easy to check that this is true for every multiplication operator). In particular, $\|R(\lambda,T)\| \le \frac{1}{\operatorname{dist}(\lambda, [0,1])}$ whenever $|\lambda| > 1$. This implies that $T$ is a Ritt operator.
On the other hand, $T$ is not mean ergodic. To see this, note that $1$ is not an eigenvalue of $T$, but an eigenvalue of its dual operator $T'$ (for instance, if $\mathcal{U}$ is a free ultra filter on $\mathbb{N}$, then the functional $\varphi \in (\ell^\infty)'$ given by $\langle \varphi, x\rangle = \lim_{n \to \mathcal{U}}x_n$ for each $x \in \ell^\infty$ is an eigenvector of $T'$ for the eigenvalue $1$). Hence, the fixed space of $T$ does not separate the fixed space of $T'$, so it follows from the mean ergodic theorem that $T$ is not mean ergodic.