How does Rellich–Kondrachov not lead to a contradiction?

by stats_model   Last Updated June 23, 2018 05:20 AM

I'm currently working through Lawrence Evan's PDE's textbook, and in it, he states the Rellich-Kondrachov theorem:

Assume $U$ is a bounded open set of $\mathbb R^n$, and $\partial U$ is $C^1$. Suppose $1 \leq p < n$. Then $$W^{1,p}(U) \subset \subset L^p(U)$$ for each $1 \leq q < p^*$ where $p^*$ is the Sobolev conjugate of $p$.

I am currently failing to see how this does not contradict the fact that a linear space is compact if and only if it is finite dimensional. In particular the argument I have in mind is that, if $\overline{W^{1,p}(U)}$ is compact, then so is the unit ball in $W^{1,p}(U)$, being a closed subset of a compact set. But, $W^{1,p}(U)$ is clearly not finite dimensional.

Where does the above argument go wrong? Thanks!

Related Questions

Definitions of fractional Sobolev Spaces

Updated July 03, 2018 08:20 AM