Take the 2-minute tour ×

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,


share|improve this question
Also, I know that the sequence of norms are bounded in $\mathbb R$, so contain a convergent subsequence. I just wonder if the whole sequence converges? –  dcs24 Jun 22 '12 at 18:17

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer
Thank you. Whilst trying to be as abstract as possible, I dropped to many restrictions from the problem in mind. I will bear this counter example in mind. –  dcs24 Jun 22 '12 at 18:27

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 $L^2$.

share|improve this answer
I like this. I would vote it up, but I'm not reputable enough yet! –  dcs24 Jun 23 '12 at 15:35

Your Answer


