Sobolev space $H^1_0$ weakly closed

Question

Let $\{u_n\}$ be a sequence of functions in $H^1_0(\Omega)$ such that $u_n$ converges weakly to $u\in H^1(\Omega)$.

Then, can I conclude that $u$ is in fact in $H^1_0$?

Yes, because $H_0^1(\Omega)$ is a convex set which is norm-closed. — Svetoslav, Sep 24 '15 at 14:10
@Svetoslav I just looked it up and you are referring to Mazur's Theorem I suppose? — mononono, Sep 24 '15 at 14:19

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

Yes, because $H_0^1(\Omega)$ is a convex set which is norm-closed. -- Svetoslav

edited Apr 13 '17 at 12:20

Community

answered Sep 26 '15 at 04:20

1 Answers1