# Is $f$ weakly lower semicontinuous?

Given a compact subset $$\Omega$$ of $$\mathbb{R}^N$$, I wonder if $$f(u)=\int_\Omega (1-|u|^2)^2\ dx$$ is weakly lower semicomtinuous (w.l.s.c) on $$H^1(\Omega)$$, meaning that if $$\lbrace u_n\rbrace$$ tends to $$u$$ weakly, then $$f(u)\leq \liminf f(u_n)$$. I know that the norm $$\Vert u \Vert _{H^1}= \Vert u \Vert_{L^2} + \Vert \nabla u \Vert_ {L^2}$$ is w.l.s.c., and so $$\int_\Omega |u|^2\ dx$$ is w.l.s.c.

What can be said about $$f$$? Thanks in advance.

I’m proving that $$f$$ is weakly continuous. I hope that I’m not wrong.
Assume $$u_n \rightarrow u$$ weakly. By Reillich, $$u_n \rightarrow u$$ strongly wrt the $$L^2$$ norm. So we have a subsequence $$u_{p_n}$$ that converges ae to $$u$$.
Then by the dominated convergence theorem, $$f(u_{p_n}) \rightarrow f(u)$$. Since the same statement can be made for any subsequence of $$u_n$$ instead, it follows that $$f$$ is continuous wrt the weak topology.
• Thanks for your answer. Im trying now to deal with fatou's lemma. I need to check that $\liminf (1-|u_n|^2)^2 = (1-|u|^2)^2$ – Senna Oct 24 '19 at 18:44
• If $H^1$ embeds compactly into L^4 then this works, so $N<4$ – daw Oct 24 '19 at 19:09
• $1$ is not a dominating function. For $u \equiv 3$, you have $(1-u^2)^2 = 64$. For $N \ge 4$, Fatou seems to be the way to go. – gerw Oct 25 '19 at 6:02