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.
Thanks.
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. Thanks. 


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, ArzelaAscoli 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. 

