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Suppose a bounded sequence $(x_n)$ converges to $x$ in the Cesaro sense (i.e., $\frac{1}{n}(x_1 + x_2 + \dots + x_n)\rightarrow x$) in a separable Hilbert space $H$. How to prove that some subsequence $(x_{n_k})$ converges weakly to $x$?

flag	asked Jul 19 2010 at 14:24 Kestutis Cesnavicius 1,198●5●12

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Reminds me of Banach-Saks theorem, which goes the other way. – Gjergji Zaimi Jul 19 2010 at 15:07

Indeed, looks like the exercise might've been intended the other way. For the record: the proof of Banach-Saks can be found in books.google.com/… – Kestutis Cesnavicius Jul 19 2010 at 16:53

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Whatever the corrected question is supposed to be, it almost certainly is not research level. – Zen Harper Jul 19 2010 at 18:13

falagar · Accepted Answer · 2010-07-19 14:58:22Z UTC

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If we take $x_n = (-1)^n x$ then $x_n$ converges to $0$ in Cesaro sence. But no subsequence of $x_n$ converges weakly to $0$. $x_n$ is also a bounded sequence. Hence your statements seems wrong.

answered Jul 19 2010 at 14:58

falagar
1,703●4●23

Thanks. It was an exercise that I couldn't solve, so I assumed it should hold. – Kestutis Cesnavicius Jul 19 2010 at 15:13

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@Kestutis Cesnavicius: A good habit to be in is that if you cannot prove a statement, try to break it. Either you succeed in falsifying the statement, or you gain more understanding of why the statement should be true. – Theo Johnson-Freyd Jul 19 2010 at 17:57

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Kestutis, may be it was asked (or they wanted to ask) to prove that some subsequence weakly converges, but not necessary to $x$? Which is not very deep, too, but it is true at least. – Fedor Petrov Jul 19 2010 at 18:06

Cesaro convergence implies weak convergence of a subsequence

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