Cesaro convergence implies weak convergence of a subsequence - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T00:31:12Zhttp://mathoverflow.net/feeds/question/32502http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/32502/cesaro-convergence-implies-weak-convergence-of-a-subsequenceCesaro convergence implies weak convergence of a subsequenceKestutis Cesnavicius2010-07-19T14:24:47Z2010-07-19T14:58:22Z
<p>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 $x$?</p>
http://mathoverflow.net/questions/32502/cesaro-convergence-implies-weak-convergence-of-a-subsequence/32508#32508Answer by falagar for Cesaro convergence implies weak convergence of a subsequencefalagar2010-07-19T14:58:22Z2010-07-19T14:58:22Z<p>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.</p>