Weak convergence-exercice

Question

Let $\Omega$ be an open set in $\mathbb{R}^n$ and let $(u_n)$ be a bounded sequence in $H^1_0(\Omega).$

Who's the theorem say that we can extract a subsequence denoted $u_{n}$ as $u_n$ weakly converge to $u$ in $H^1_0(\Omega)$?
Why if $f_n(x)$ converge strongly to $f(x)$ in $(L^{\infty})$ and $u_n$ weakly converge to $u$ in $H^1_0(\Omega)$ then $f_n u_n$ converge weakly to $fu$ in $L^2(\Omega)$? Thank's.

As a consequence of Banach-Alaoglu theorem any bounded sequence have a weak* converging subsequence. But as $ H^1_0(\Omega) $ is Hilbert space, hence weak and weak* topologies coincide.

Davide Giraudo · Answer 1 · 2013-04-21 20:52:57Z

I don't know whether there is a name for this result. The best is to remember why it is true. The statement reminds Bolzano-Weierstass theorem. You know that the Hilbert space $H^1_0(\Omega)$ is separable, so let $(e_k,k\geqslant 1)$ be a Hilbert basis for this space. For each $k$, the sequence $(\langle u_n,e_k\rangle,k\geqslant 1)$ is bounded, so we can extract a convergent susbsequence. By a Cantor's diagonal argument, we can choose a subsequence $(u_{n_j},j\geqslant 1)$ such that $(\langle u_{n_j},e_k\rangle, j\geqslant 1)$ is convergent for all $k\geqslant 1$. Calling $F$ the closure of the vector space spanned by the $u_j$ and using a decomposition $H^1_0(\Omega)=F\oplus^\perp F^\perp$ wget the wanted $u$.
Recall that a weakly convergent sequence is bounded, then use that for each $\phi\in L^2(\Omega)$, $$\left|\int (f_nu_n-fu)\phi dx\right|\leqslant \int |f_n-f||u_n||\phi|+\left|\int \phi f(u_n-u)\right|.$$

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