TITLE: Is this functional weakly continuous?
QUESTION [5 upvotes]: Take a $C^1$ function $G \colon \mathbb{R}\to \mathbb{R}$ and define a functional 
$$\mathcal{G}(u)=\int_0^1G(u(t))\, dt, \quad u \in H^1(0, 1).$$
We then have $\mathcal{G}\in C^1\big(H^1(0, 1)\to \mathbb{R}\big)$. Now, I would like to apply Weierstrass's theorem to this functional, and so I need to show that it is weakly lower semicontinuous. 

Question 1 Is it true?


Some course notes I'm reading act as if $\mathcal{G}$ were weakly continuous, because they claim the differential 
$$\mathcal{G}' \colon H^1(0, 1) \to \big[ H^1(0, 1) \big] ' $$
is weak-strong continuous. (This trivially implies the claim). To show that, they first compute 
$$\langle \mathcal{G}'(u), v \rangle = \int_0^1 G'(u)v\, dt,$$
which is clear to me, and then factor the mapping 
$$u \in H^1 \mapsto \mathcal{G}'(u) \in \big[ H^1 \big]'$$
as 
$$u \in H^1 \mapsto u \in L^\infty \mapsto G'\circ u \in L^\infty \mapsto \mathcal{G}'(u) \in \big[ H^1 \big]';$$
then, since the first embedding is compact (so they say) and the other arrows are continuous, the whole mapping is weak-strong continuous. 

Question 2 This reasoning seems wrong to me, because the embedding
  $H^1(0, 1) \hookrightarrow L^\infty(0, 1)$ is not compact. Am I wrong?

REPLY [6 votes]: The embedding $H^1(0,1) \hookrightarrow L^\infty(0,1)$ is indeed compact.  This follows from general Sobolev embedding theorems, but in this special case it makes a nice exercise in using the Arzela-Ascoli theorem.  Leave a comment if you want hints.