Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
Thank you very much. Now I can finish reading those notes without worrying! Later on I will think at your problem also. Ascoli-Arzelà theorem is not directly applicable but I have some workarounds in mind. I'll let you know. –  Giuseppe Negro May 30 '11 at 7:04
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.