Inner products of weakly convergent sequences

Question

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!

Answer 1

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 $u_k:=v_k$.