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?