3
$\begingroup$

I'm really surprised I can't seem to find this statement anywhere, even though it seems to follow from compactness theorems, I'm therefore wondering whether I made a mistake in my proof below.

Claim: Let $\Omega$ be a sufficiently nice bounded domain. Let $u_n \in H^1(\Omega)$ with $u_n \rightharpoonup u$ in $L^2$, where $\rightharpoonup$ denotes weak convergence. Assume further that $\| \nabla u_n\|_{L^2} \leq C$ for some $C > 0$ independent of $n$. Then $u_n \rightharpoonup u$ in $H^1$.

Proof: Since $u_n$ is bounded in $H^1$, it must have a $H^1$-weakly convergent subsequence to some $\tilde u \in H^1$ by the sequential Banach-Alaoglu theorem. Since this subsequence is also $L^2$-weakly convergent and weak limits are unique, it is true that $u = \tilde u$ and in particular $u \in H^1$. Now assume that the sequence $u_n$ does not converge weakly to $u$. Then there must be some subsequence that is completely outside of a neighborhood $U$ of $u$. By the same compactness argument, this subsequence again has a $H^1$-weakly convergent subsequence which must converge to $u$, and hence be in $U$ infinitely often, a contradiction.

$\endgroup$

1 Answer 1

1
$\begingroup$

The result is true and your argument is correct. A slightly better phrasing might be to show that every subsequence of $u_n$ has a further subsequence that converges weakly to $u$ in $H^1$. It is then a standard topological result that this implies that $u_n \rightharpoonup u$. In fact, the last part of your argument would essentially prove the general topological result.

Perhaps the reason that you cannot find this exact result anywhere is that it is a not too difficult instance of the well known criterion that says that a relatively compact sequence with a unique limit point must converge to that limit point.

$\endgroup$
2
  • $\begingroup$ thanks! Since strong $L^2$ convergence implies weak $L^2$ convergence, this in particular implies that the $H^1$-unit ball is $L^2$ closed? Moreover, by the Rellich-Kondrachev theorem, a similar argument should give that the $u_n \rightarrow u$ in $L^2$ (or, as a corollary: sequential weak $H^1$ convergence implies strong $L^2$ convergence). I guess I'm just surprised I haven't come across any of these results, but only versions that deal with subsequences and not the whole sequence. $\endgroup$ Jul 25, 2019 at 10:51
  • $\begingroup$ All the results mentioned in your comment are also true, yes. I feel like I must have seen a result like this written in a book but I can't think which one without actually finding and checking a bunch of them. $\endgroup$ Jul 25, 2019 at 11:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .