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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.

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What are your thoughts so far? Is everything a complete mystery or are there some parts of this exercise that are clear? Do you see how this generalizes the usual von Neumann ergodic theorem? –  t.b. Jun 1 '12 at 8:15
Some hints: using the second bullet point and the fact that a bounded operator is weak-weak continuous, prove first that $Tx^\ast = x^\ast$. Then prove that $x-x^\ast$ is in the weak closure of the range of $1-T$ using that $x - x_n = (1-T)\sum_{i=0}^{n-1} x_i$. Since the weak closure of a convex set is equal to its norm closure, $x-x^\ast$ is in the norm-closure of the range of $1-T$. But for $z$ in the range of $1-T$ the sequence $z_n = \frac{1}{n} \sum_{i=1}^{n-1} T^i z$ norm-converges to zero by the first bullet point. –  t.b. Jun 2 '12 at 8:46
You seem to prefer to appreciate advice and discussions rather silently... –  t.b. Jun 14 '12 at 1:26
@t.b. If I'm honest, this is why I fear that your thoroughness in answering terse/vague/homeworky questions is not always a good use of your time, or even something good for the OP –  user16299 Jun 18 '12 at 7:52
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1 Answer

up vote 5 down vote accepted

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 $$ \lVert x^\ast - S_nx\rVert \xrightarrow{n\to\infty} 0. $$

  1. 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 $S_{n}x^\ast = x^\ast$.

  2. 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 $(1-T)$.

  3. 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.

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