Compactness result PDEs

Question

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?

Answer 1

Let's begin with a weakly convergent sequence: $u_j \rightharpoonup u$ in $W^{1,p}$, $1<p<\infty$. A weakly convergent sequence is bounded. By the Rellich-Kondrachov theorem, it has a subsequence $u_{j_k}$ that converges strongly in $L^p$. How do we conclude that the entire sequence $u_j$ converges strongly in $L^p$?

Note that we also have $u_j \rightharpoonup u$ in $L^p$. Therefore, the limit of $(u_{j_k})$ is nothing but $u$, because strong convergence implies weak, and weak limits are unique (the weak topology is Hausdorff). It remains to invoke an easy fact from general topology: if $u,u_j$ are elements of a Hausdorff space and every subsequence of $(u_j)$ has a subsubsequence that converges to $u$, then $u_j\to u$. (Proof is by contradiction, and is left as an exercise).

Answer 2

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.