Cesaro convergence implies weak convergence of a subsequence - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:31:12Z http://mathoverflow.net/feeds/question/32502 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32502/cesaro-convergence-implies-weak-convergence-of-a-subsequence Cesaro convergence implies weak convergence of a subsequence Kestutis Cesnavicius 2010-07-19T14:24:47Z 2010-07-19T14:58:22Z <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#32508 Answer by falagar for Cesaro convergence implies weak convergence of a subsequence falagar 2010-07-19T14:58:22Z 2010-07-19T14:58:22Z <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>