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Let $\lbrace x_n\rbrace\in l_2$ be a sequence weakly converges to $x$.

I want to prove that there is a subsequence $\lbrace x_{n_k}\rbrace $ such that the Cesàro means $\frac{1}{k}(x_{n_1}+x_{n_2}+\ldots x_{n_k})$ converges to $x$ in $l_2$.

Isn't it true that $x_n$ weakly converges to $x$ implies that there is a subsequence $x_{n_k}$ which converges to $x$?

Then, I think that if the sequence converges to $x$ in $l_2$ then the Cesàro mean of the sequence also converges to $x$ in $l_2$.

Isn't it?

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No. Say $x_n$ is the sequence which has a $1$ in the $n$th place and $0$s everywhere else. Then $x_n$ converges weakly to $0$, but since $\|x_n \|_2=1$, no subsequence can converge in $l_2$. –  Chris Eagle May 30 '11 at 14:25
    
Then, how can I make use the weak convergence? –  Forthepiece May 30 '11 at 14:28
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up vote 5 down vote accepted

First of all, the following are equivalent

  1. $x_{n} \to x$ in norm, that is $\|x_n - x\| \to 0$.
  2. $x_{n} \to x$ weakly and $\|x_n\| \to \|x\|$.

That 2. implies 1. is a nice exercise in applying the polarization identity.

A standard example of a weakly convergent sequence that doesn't contain a norm-convergent sequence is an orthonormal system, as can be seen from Bessel's inequality, for example.

However, the result you're asking about is true. Slightly more generally we have:

Theorem (Banach-Saks) Every bounded sequence has a subsequence such that its Cesàro means converge.

The proof is not hard: Let $\|x_n\| \leq C/2$ for all $n$. Since the closed ball of radius $C/2$ in the span of the $(x_n)$ is compact metrizable in the weak topology, we may assume that the sequence converges weakly in the first place. By translating the sequence, we may even assume that it converges weakly to zero, and the sequence will certainly be bounded by $C$.

Now choose the sub-sequence $y_k = x_{n_k}$ inductively, using weak convergence to zero:

  • Choose $y_1 = x_1$.
  • Assume that $y_1, \ldots, y_k$ are already chosen. Since $|\langle y_i, x_n\rangle| \to 0$ as $n \to \infty$ we may choose $y_{n+1}$ such that $|\langle y_{n+1}, y_i\rangle| \leq 1/(n+1)$ for all $i = 1,\ldots,n$.

Estimate $$\left\Vert \frac{1}{n} \sum_{k=1}^{n} y_{k} \right\Vert^{2}$$ and show that it converges to zero. If you lose yourself in the estimates here, don't worry, strengthen them by Nate's suggestion which makes the argument a little easier.

Finally, the result you're asking about follows from the Banach-Saks theorem by applying the uniform boundedness principle to see that a weakly convergent sequence is bounded.

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As Chris Eagle has pointed out, the argument you suggest doesn't work, since in general no subsequence converges in norm.

Here's a hint. Let's assume for simplicity that our weak limit is $x=0$. Then the square norm of the $m$th Cesaro mean is $$\bigg\lVert \frac{1}{m} \sum_{k=1}^m x_{n_k} \bigg\rVert^2 = \frac{1}{m^2} \sum_{k=1}^m ||x_{n_k}||^2 + \frac{2}{m^2} \sum_{1 \le j < k \le m} \langle x_{n_j}, x_{n_k} \rangle.$$

Since a weakly convergent sequence is bounded in norm (do you know why?), the first term goes to 0. For the second term, see if you can choose your subsequence so as to make this small. For instance, perhaps you could choose it so as to guarantee that for each $k$, $$\max_{1 \le j < k} |\langle x_{n_j}, x_{n_k} \rangle| \le 2^{-k}.$$

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