# Cesaro convergence implies weak convergence of a subsequence

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 $x$?

• Reminds me of Banach-Saks theorem, which goes the other way. – Gjergji Zaimi Jul 19 '10 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 '10 at 16:53
• Whatever the corrected question is supposed to be, it almost certainly is not research level. – Zen Harper Jul 19 '10 at 18:13

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.
• 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 '10 at 18:06