Inequalities for $L^2$ norms of gradients of functions that weakly converge in a Sobolev space

Question

Let $\Sigma$ be a $k$-dimensional compact manifold with boundary. Suppose that $W^{1,k}(\Sigma) \subset L^2(\Sigma)$ is compact and that $\{\phi_j \}$ is a sequence that converges weakly in $W^{1,k}(\Sigma)$, strongly in $L^2(\sum)$, an pointwise a.e. to a map $\psi: \Sigma \to \mathbb{R}$.

Why do we have the following? $$\int_{\Sigma}|\nabla \psi|^2dv \leq \liminf _{j \to \infty}\int_{\Sigma}|\nabla \phi_j|^2dv $$

I believe that is possible show $L^k(\Sigma) \subset L^2(\Sigma)$, but I am not sure.

Answer 1

I think it's implicitly assumed that $k\ge 2$; otherwise the gradient need not be in $L^2$.

To begin with, all the three limits of $\{\phi_j\}$ are the same. Indeed, let $\phi$ be its weak limit in $W^{1,k}$. Since the embedding into $L^2$ is compact (and compact operators map weakly convergent sequences to strongly convergent ones), we have $\phi_j\to \phi$ strongly in $L^2$. Consequently, there is a subsequence $\phi_{j_l}$ converging to $\phi$ a.e. This implies $\phi=\psi$ a.e.

Since $\Sigma$ is compact, it has finite volume. An application of Hölder's inequality shows the $L^p$ spaces are nested. In particular, $L^k\subset L^2$ and $W^{1,k} \subset W^{1,2}$.

Linear operators preserve weak convergence. Indeed, if $x_j\to x$ weakly in a space $X$, and $T:X\to Y$ is a bounded linear operator, then for every $z\in Y^*$ we have $z(T(x_j))\to z(T(x))$ because $z\circ T\in X^*$.

In particular, the gradient operator, being linear and bounded from $W^{1,k}$ to $L^2$, preserves weak convergence. The norm is lower semicontinuous under weak convergence. This completes the proof.