Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

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
add comment

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
add comment

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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.