A convergence problem in Banach spaces related to ergodic theory Suppose $X$ is a Banach space, $T\in B(X)$, satisfied the following condition.


*

*$\sup \Big\lVert\frac{1}{n}\sum \limits_{i=0}^{n-1}T^{i}\Big\rVert<\infty$

*$\frac{1}{n}\lVert T\rVert^{n}\rightarrow0$, as $n \rightarrow\infty$


For $x \in X$, take $x_{n}=\frac{1}{n} \sum\limits_{i=0}^{n-1}T^{i}x$. If there is a subquence $\{x_{n_{k}}\}$ which has a weak limit  $x^{*}$(in the weak topology), prove that $x_{n} $ is convergent to $ x^{*}$, in the norm topology, and $Tx^{*}=x^{*}$
This can been seen as a generalization of the von Neumann Ergodic Theorem for Banach spaces.
Any advice and discussions will be appreciated.
 A: For me it is a bit easier to work with the Cesàro averaging operators 
$$
S_n = \frac{1}{n} \sum_{i=0}^{n-1} T^i
$$
than with the sequences $x_n$. Since $S_nx = x_n$, the translation is straightforward.
Observe that $(1-T)S_n = S_n(1-T) = \frac{1}{n}(1-T^{n})$.
I assume that the condition in the second bullet point $\frac{1}{n} \lVert T\rVert^n \to 0$ (which is equivalent to $\Vert T\rVert \leq 1$) is a typo for the weaker condition $\frac{1}{n} \lVert T^n\rVert \to 0$ saying that $\lVert T^n\rVert$ grows slower than linearly. We have
$$\tag{$\ast$}
\lVert S_n(1-T)\rVert = \lVert(1-T)S_n\rVert \leq \frac{1}{n}(1+\lVert T^n\rVert) \xrightarrow{n\to\infty} 0.
$$
We will also make use of the identity
$$
\tag{${\ast\ast}$}
1-S_n = \frac{1}{n} \sum_{i=0}^{n-1} (1-T^i) 
= (1-T) \frac{1}{n}\sum_{i=0}^{n-1} i \cdot S_i.
$$

Now we're ready for the proof. Assume that $x \in X$ and that $n_{k}$ is an increasing sequence such that $S_{n_k} x \to x^\ast$ weakly. We want to show that
$$
\lVert x^\ast - S_nx\rVert \xrightarrow{n\to\infty} 0.
$$


*

*Since bounded operators are weak-weak continuous, we have $TS_{n_k}x \to Tx^\ast$ weakly, and on the other hand for $\varphi \in X^\ast$ we have
$$
\lvert \varphi((1-T)x^\ast)\rvert 
= \lim_{k\to\infty} \lvert\varphi ((1-T)S_{n_k}x)\rvert\leq \lVert\varphi\rVert\lVert x \rVert\lim_{k\to\infty}\lVert(1-T)S_{n_k}\rVert = 0
$$
where the first equality follows from weak convergence and last one follows from $(\ast)$.
Thus $\varphi(x^\ast - Tx^\ast) = 0$ for all $\varphi \in X^\ast$ and since $X^\ast$ separates the points of $X$ we conclude that $x^\ast = Tx^\ast$. We also have $S_{n}x^\ast = x^\ast$.

*Let $y = x- x^\ast$ and note that
$$\tag{$\ast\ast\ast$}
S_ny = S_n x - x^\ast,
$$
so that weak convergence $S_{n_k}x \to x^\ast$ implies weak convergence $S_{n_k}y \to 0$.
With $(\ast\ast)$ we get
$$
y - S_{n_k}y = (1-T)\sum_{i=0}^{n_k-1}i\cdot S_iy,
$$
so $y = x-x^\ast$ is in the weak closure of the range of $(1-T)$.

*For convex sets the weak closure and the norm closure coincide, so $y = x- x^\ast$ is in the norm closure of the range of $(1-T)$.
Let $\varepsilon \gt 0$. 
There is $z = (1-T)w$ such that $\lVert y-z\rVert \lt \varepsilon$. Again with $(\ast)$ we conclude that $\lVert S_n z\rVert \to 0$.
Finally, the hypothesis $C = 
\sup_{n \in \mathbb{N}} \lVert S_n \rVert \lt \infty$ (which we haven't used so far) shows that for $n$ large enough we have 
$$
\lVert S_n y \rVert \leq \lVert S_n(y-z)\rVert + \lVert S_n z\rVert \leq (C+1)\varepsilon. 
$$
Recalling $(\ast\ast\ast)$ this gives
$$
\lVert S_n x - x^\ast\rVert \xrightarrow{n\to\infty} 0,
$$
as we wanted.
