# Weak convergence-exercice

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

1. Who's the theorem say that we can extract a subsequence denoted $u_{n}$ as $u_n$ weakly converge to $u$ in $H^1_0(\Omega)$?

2. 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. – smiley06 Apr 21 '13 at 13:13

1. 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 $u$.
2. Recall that a weakly convergent sequence is bounded, then use that for each $\phi\in L^2(\Omega)$, $$\left|\int (f_nu_n-fu)\phi dx\right|\leqslant \int |f_n-f||u_n||\phi|+\left|\int \phi f(u_n-u)\right|.$$
The answer to your first question is Alaoglu. His theorem states that the unit ball is weak-$\star$ compact. Since $H_0^1 (\Omega)$ is a Hilbert its isomorphic to its dual (Riestz representation theorem) and so weak and weak-$\star$ topologies coincide. Thus the unit ball in $H_0^1 (\Omega)$ is weak compact. The same applies to any other ball. So if a sequence is bounded then it is contained in a weak compact, and so it has a weakly convergent subsequence.