Weak versus strong convergence - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:46:07Z http://mathoverflow.net/feeds/question/100378 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100378/weak-versus-strong-convergence Weak versus strong convergence dcs24 2012-06-22T18:06:24Z 2012-06-23T01:41:15Z <p>This is my first time posting.</p> <p>I am well aware that an $L^2$ weakly converging sequence is not convergent in the corresponding strong topology. However, my question is as follows, do the sequence of norms corresponding to a weakly convergent sequence converge? </p> <p>Take for instance the sine function on (0,1), specifically $\sin(x/\varepsilon)$, this weakly converges to zero, and the norms converge to the mean of $|\sin^2|$.</p> <p>So despite no strong convergence, do the norms still converge to something else?</p> <p>Many thanks for you help and time in advance,</p> <p>Daniel</p> http://mathoverflow.net/questions/100378/weak-versus-strong-convergence/100380#100380 Answer by mohanravi for Weak versus strong convergence mohanravi 2012-06-22T18:21:48Z 2012-06-22T18:21:48Z <p>No, of course not. Take two different sequences converging weakly to zero and interleave them.</p> http://mathoverflow.net/questions/100378/weak-versus-strong-convergence/100408#100408 Answer by Andreas Blass for Weak versus strong convergence Andreas Blass 2012-06-22T22:22:54Z 2012-06-22T22:22:54Z <p>Any bounded sequence $\langle s_n\rangle$ of non-negative reals is the sequence of norms of a weakly convergent sequence in $L^2$, for example the sequence $\langle s_n e_n\rangle$, where $\langle e_n\rangle$ is your favorite orthonormal basis for $L^2$.</p>