Problem. Convergence. Banach space. Weak topology

Question

Let E be a Banach space and let $(X_n)$ be a sequence such that $X_n \rightharpoonup x$ in the weak topology σ(E,E'). Set:

$S_n=\frac{1}{n}\sum_{k=1}^n(-1)^kX_k$

Does $Sn \rightharpoonup x $ in the weak topology σ(E,E')?

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I was thinking that I should write something like this:

• let $T \in E'$:

  $T(S_n)=\frac{1}{n}\sum_{k=1}^n(-1)^kT(X_k)$

• $|T(S_n)-T(x)| \le \frac{1}{n}\sum_{k \le n}(-1)^k |T(X_n)-T(x)| \le \frac{1}{n}\sum_{k \le n}|T(X_n)-T(x)|$

• $ \forall \epsilon >0: \exists N$ such that $\forall n>N$:

  $|T(X_n)-T(x)| < \epsilon$

• $|T(S_n)-T(x)| \le \frac{1}{n}\sum_{k \le N}|T(X_n)-T(x)|+\epsilon$

• for n sufficiently large:

  $|T(S_n)-T(x)|\le2\epsilon$

=> $Sn \rightharpoonup x $ in the weak topology σ(E,E')

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Is this right? Can you tell me where I am wrong, please?

Answer 1

$S_n$ converge weakly to $0$:

Let $y\in E'$, $\epsilon > 0$ and $n_0$ the rank where $\forall k > n_0, |\langle x_k-x, y \rangle |\leq \epsilon$

Then

$$\langle S_n, y \rangle = \langle \frac{n_0}{n} S_{n_0} + \frac{1}{n} \sum_{k=n_0+1}^n (-1)^k x_k , y \rangle$$

$$ = \frac{1}{n} \langle n_0s_{n_0}, y \rangle + \frac{1}{n}\sum_{k=n_0+1}^n (-1)^k \langle (x_k-x) , y \rangle + \frac{1}{n} \langle \sum_{k=n_0+1}^n (-1)^k x , y \rangle $$

Hence

$$|\langle S_n, y \rangle| \leq \frac{1}{n} C + \epsilon + \frac{1}{n} \langle x,y\rangle \to \epsilon$$

And this is true for every $\epsilon$, so it converge to $0$ for each $y\in E'$