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Let $H$ Hilbert Space. Show that if $x_n\rightharpoonup x$ then there exists a subsequence $\{x_{nk}\}$ of $\{x_{n}\}$ such that the sequence $\lim_{m\rightarrow \infty } \frac{1}{m}\sum_{k=1}^{m}x_{nk}=x$. One suggestion please.

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  • $\begingroup$ @Raff: as far as I can tell, Mazur's Lemma will give you convex combinations, but not necessarily averages. $\endgroup$ – Martin Argerami Nov 27 '14 at 21:21
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This is the Banach-Saks theorem. Its proof goes as follows. Let me denote the inner product by $(\cdot,\cdot)$.

We construct the subsequence inductively. Denote $n_1:=1$. Assume $n_1\dots n_{k}$ are chosen already. Then choose $n_{k+1}$ such that $$ |( x_{n_i}-x, \ x_{n_{k+1}}-x )| \le \frac1k \quad \forall i=1\dots k. $$ This is possible since by weak convergence $( x_{n_i}-x, \ x_m-x ) \to 0$ for $m\to\infty$.

Since $(x_n)$ converges weakly, it is bounded, say $\|x_n-x \|\le M$ for all $n$. Let us compute the norm of $\sum_{i=1}^k (x_{n_i}-x)$: $$ \begin{split} \| \sum_{i=1}^k (x_{n_i}-x)\|^2 & \le 2 \sum_{i=1}^k \sum_{j=1}^{i-1} |(x_{n_i}-x,x_{n_j}-x )| + \sum_{i=1}^k\|x_{n_i}-x\|^2\\ &\le 2 \sum_{i=1}^k\sum_{j=1}^{i-1}\frac1{i-1} + \sum_{i=1}^k M^2 \le k(2+M^2). \end{split}$$ This implies $$ \|\frac1k \sum_{i=1}^k (x_{n_i}-x)\|^2 \le \frac{2+M^2}k, $$ and it follows $$ \frac1k \sum_{i=1}^k (x_{n_i}-x) \to 0, $$ or equivalently $$ \frac1k \sum_{i=1}^k x_{n_i} \to x $$ for $k\to\infty$.

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  • $\begingroup$ Could you clarify what happens starting from: $\| \sum_{i=1}^k (x_{n_i}-x) \|^2$. Like what happens with the indices, do you use triangle ineq. or something? $\endgroup$ – mavavilj Feb 17 '19 at 21:52

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