MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering whether the set $\lbrace f\in H_0^1(\Omega)|\|f\|_{L^\infty(\Omega)}\leq 1\rbrace$ is compact in $H_0^1(\Omega)$ or not. Here $\Omega$ is a convex domain in $\mathbb{R}^3$ with Lipschitz boundary.


share|cite|improve this question
up vote 18 down vote accepted

No. As a general rule, in order to obtain compactness in some norm, one needs control of a higher regularity than what is associated to that norm, in order to shut down an "escape to frequency infinity". For instance, $H^1_0$ has one degree of regularity, so one needs to control a norm involving more than one derivative in order to obtain compactness. (But in many cases one only needs an epsilon more regularity; for instance, Arzela-Ascoli tells us that equicontinuity, which can be thought of as an infinitesimal amount of regularity, is enough (together with some additional hypotheses) to obtain compactness in the uniform norm, which has zero degrees of regularity.)

In this specific case, a concrete counterexample can be obtained by considering the unit cube $\Omega = [0,1]^3$ and the sequence $f_n := \frac{1}{n} \sin(2\pi n x) \sin(2\pi y) \sin(2\pi z)$, which goes to zero in $L^\infty$ but stays away from zero in $H^1_0$, and so cannot have any $H^1_0$ convergent subsequence.

share|cite|improve this answer

Your Answer


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.