1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – smiley06 Apr 21 '13 at 13:13
1
$\begingroup$
  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 $\phi\in L^2(\Omega)$, $$\left|\int (f_nu_n-fu)\phi dx\right|\leqslant \int |f_n-f||u_n||\phi|+\left|\int \phi f(u_n-u)\right|.$$

$\endgroup$
1
$\begingroup$

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.

Hope this helps. Regards, D

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.