# Is this functional weakly continuous?

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?

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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.
With a clear mind it is easy! Take a bounded sequence $u_n \in H^1(0,1)$. Then $u_n$ is equibounded and equi-Hölder continuous and so it has a uniformly convergent subsequence. In fact $$u_n(x)-u_n(y)=\int_y^x u_n'(s)\, ds$$ so that $$\lvert u_n(x) -u_n(y)\rvert \le \lVert u_n \rVert_{H^1}\lvert x-y\rvert^{\frac{1}{2}}$$ which proves equicontinuity. Now observe that $H^1$ is embedded in $L^\infty$ and so a $H^1$-bounded sequence is $L^\infty$-bounded also. This proves equiboundedness. –  Giuseppe Negro May 31 '11 at 17:56