Question

This is my first time posting.

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?

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 $|\sin^2|$.

So despite no strong convergence, do the norms still converge to something else?

Many thanks for you help and time in advance,

Daniel

Answer 1

No, of course not. Take two different sequences converging weakly to zero and interleave them.

Answer 2

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 $L^2$.