Given a compact subset $\Omega$ of $\mathbb{R}^N$, I wonder if $$F(u)=\int_\Omega f(u)\ dx =\int_\Omega (1-|u|^2)^2\ dx$$ is weakly lower semicontinuous (w.l.s.c) on $H^1(\Omega)$, meaning that if $\lbrace u_n\rbrace$ tends to $u$ weakly, then $F(u)\leq \liminf F(u_n)$. This is my reasoning: Since $u_n$ weakly converges to $u$ un $H^1$, also converges weakly in $L^2$ and so, (up to a subsequence) one has $u_{\sigma (n)}\rightarrow u$ a.e. Now using this and continuity of $x\mapsto (1-|x|^2)^2$, $f( u_{\sigma (n)})\rightarrow f(u)$ and Fatou's lemma says $$ F(u)=\int f(u)\ dx \leq \liminf \int f( u_{\sigma (n)} )\ dx= F( u_{\sigma (n)}) $$ My question is how can I obtain the same inequality with the original sequence? And, is Rellich theorem useful here? Any answer is welcome! Thank you.
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$\begingroup$ That is not a problem since you can start out by fixing a subsequence on which you converge to the $\liminf$ and then run the whole argument for that sequence. (Another comment is that weak convergence in $L^2$ of course doesn't imply pointwise convergence on a subsequence, but I guess it's fine for weak convergence in $H^1$.) $\endgroup$– Christian RemlingOct 25, 2019 at 17:54
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$\begingroup$ Thank you Christian. $\endgroup$– R. N. MarleyOct 25, 2019 at 17:57
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$\begingroup$ So my reasoning is correct, isnt it? $\endgroup$– R. N. MarleyOct 25, 2019 at 17:58
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$\begingroup$ I think it's fine after the adjustments I suggested, but I didn't really think about it carefully, so can't say anything for sure. $\endgroup$– Christian RemlingOct 25, 2019 at 18:01
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$\begingroup$ You are indeed using Rellich Theorem when stating that you have convergence a.e. of a subsequence (or at least I do not see any other way to prove this here). $\endgroup$– Ayman MoussaOct 25, 2019 at 19:51
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