Weakly convergent sequence and compensated compactness

Question

This question is about a claim made in the proof of theorem 2.1.1 in the book Hyperbolic Conservation Laws and the Compensated Compactness Method by Yunguang Lu. (For simplicity I will only write done a special case.) The general idea is that weak convergence + compactness of derivatives gives some form of (weak) continuity of a quadratic form.

Assume that $u^k\in L^2(U)$ converges weakly to $0$ and the sequence $\partial_{x_j} u^k$ is contained in a compact subset of $H^{-1}(U).$ Then for a cutoff function $\phi\in C_c^\infty(U), w^k = \phi u^k,$ we have $$ \partial_{x_j} w^k =\phi \partial_{x_j}u^k + u^k\partial_{x_j}\phi. $$ The book (page 11) says that: the first term $\phi \partial_{x_j}u^k $ is contained in a compact subset of $H^{-1} (U);$ therefore $\partial_{x_j} w^k$ is contained in a compact subset of $H^{-1} (U).$ So we can extract a strongly convergent subsequence, and so on.

It's obvious that $\phi \partial_{x_j}u^k $ is contained in a compact set by assumption. But how does this imply $\partial_{x_j} w^k$ is also contained in a compact set? After all, we do not really know at this point if $u^k\partial_{x_j}\phi$ is contained in a compact set.

Answer 1

Let $U \subset \mathbf{R}^n$ be a bounded open domain. Then by the Rellich--Kondrachov theorem the embedding $W_0^{1,2}(U) \to L^2(U)$ is a compact map. Therefore the map $L^2(U) \to W^{-1,2}(U) = (W_0^{1,2})^*(U)$ that sends $u \in L^2(U)$ to the functional $v \in W_0^{1,2}(U) \mapsto \int_U uv$ is compact also.

Let $(u_k \mid k \in \mathbf{N})$ be a sequence of functions in $L^2(U)$ as in the question, converging weakly in $L^2(U)$ to some $u \in L^2(U)$. Given $\phi \in C_c^\infty(U)$, the sequence $u_k \partial_j \phi$ is weakly convergent in the same sense. Therefore $\{ u_k \partial_j \phi \mid k \in \mathbf{N} \}$ is contained in a compact subset of $W^{-1,2}(U)$, and so is $\{ \partial_j w_k = \phi \partial_j u_k + u_k \partial_j \phi \mid k \in \mathbf{N} \}$.