TITLE: Compactness result PDEs
QUESTION [5 upvotes]: There is an argument that I see is used often in Evans PDE book, that I don't really get.  We take a bounded sequence, say $(u_m) \in W^{1,q}(\Omega)$.  By some functional analysis results, we know that since this is a reflexive Banach space, there is a subsequence that converges weakly to some element, say $u_{m_j} \rightharpoonup u$ .  But now, there is the claim that this subsequence converges strongly in $L^q$.  Why is this?  I'm assuming that it has to do with the Rellich compactness result, $W^{1,p} \subset \subset L^q$ for all $1\leq q< p^{*}=\frac{np}{n-p}$.    How does this argument work then?

REPLY [2 votes]: Compact linear operators on Banach spaces map weakly convergent sequences to strongly convergent sequences; such operators are also called completely continuous. The embedding $W^{1,q} \to L^q$ is such an operator.