L2 convergence, subsequence

Question

I'm sai. I have a question about $L^{2}$ convergence.

Let $U \subset \mathbb{R}^{d}$ be open.

Suppose $u_{n} \in C_{0}^{\infty}(U), n\in \mathbb{N} $ satisfies the following assertion:

For any $v \in C_{0}^{\infty}(U)$, $u_{n}v\to 0\,\,{\rm in}\,\,L^{2}(U;dx)$. ($dx$ is Lebesgue measure on $U$)

Does subsequence $(u_{n_{k}})_{k=1}^{\infty}$ of $(u_{n})_{n=1}^{\infty}$ such that $u_{n_{k}}\to 0 \,\,dx{\rm -a.e.} $ exist?

I think subsequence like this exists. To prove the existence, I only need to show that $u_{n}\to 0$ in $L^{2}(U;dx)$.

Since \begin{eqnarray*} \|u_{n}\|_{L^{2}}&=&\|u_{n}-u_{n}v+u_{n}v\|_{L^{2}}\\ &\leq& \|u_{n}-u_{n}v\|_{L^{2}}+\|u_{n}v\|_{L^{2}}\\ &\leq& \|u_{n}\|_{L^{2}}^{1/2}\|1-v\|^{1/2}_{L^{2}}+\|u_{n}v\|_{L^{2}}\\ \end{eqnarray*}

, I only need to show that $v \in C_{0}^{\infty}$ such that $\|1-v\|_{L^{2}}=0$ exists.

But I think $v\in C_{0}^{\infty}$ as described above does not exists. what would be a good way to do it? Thanks.

Answer 1

Let $U=(0,1)$. Take $u_n(t) = 1_{(0,1)} (t) \sin (2 \pi n t)$. Then for any $v\in L^2(0,1)$ (which is contained in $L^1(0,1)$) we have $\langle u_n, v \rangle \to 0$, but since $\|u_n\| = {1 \over \sqrt{2} } $ no subsequence of $u_n$ can converge to zero.