# Inner products of weakly convergent sequences

I have a weakly convergent sequence in $L^2(U)$ (for $U$ some bounded open domain with smooth boundary), $u_k\rightharpoonup u$.

I want to show that there is a sequence $v_k\rightharpoonup v$, such that

$$\langle u_k,v_k \rangle \nrightarrow \langle u, v \rangle.$$

My idea is to make a sequence $v_k\rightharpoonup 0$, such that $\langle u_k,v_k \rangle=1$

but I can't figure out how to do it.

Thanks!

• If $u_{k}$ is an arbitrary weakly convergent sequence, then you cannot in general guarantee the existence of $v_{k}$. For instance, if $u_{k}$ converges in norm to $u$, then for any weakly convergent sequence $v_{k}$, $\left<u_{k}, v_{k}\right> \to \left<u, v\right>$. – user14717 Nov 10 '14 at 4:20
• To add to Nick's comment, this works precisely if $\|u-u_k\|\not\to 0$; in this case, $\|u\|<\limsup \|u_k\|$, so you can take $v_k=u_k$. – user138530 Nov 10 '14 at 4:48
• how can I see that if $\|u-u_k\|\to 0$ then $\left<u_{k}, v_{k}\right> \to \left<u, v\right>$ ? – Porufes Nov 10 '14 at 5:20

## 1 Answer

There are two cases:

1. the sequence $(u_n)_{n\geqslant 1}$ converges in norm to $u$. In this case, we have $\langle u_k,v_k\rangle\to \langle u,v\rangle$ for each sequence $(v_k)_{k\geqslant 1}$ converging weakly to $v$. This can be seen from the triangle inequality and the fact that $(v_k)_{k\geqslant 1}$ is bounded.

2. The sequence $(u_n)_{n\geqslant 1}$ does not converges in norm to $u$. Expanding $\langle u-u_k\rangle^2$, we can see that the sequence $(\lVert u_k\rVert)_{k\geqslant 1}$ does not converge to $\lVert u\rVert$, therefore we can choose $u_k:=v_k$.