1

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.

A.D.
  • 508
  • 1
    Thats a tough first question lol – Trajan Dec 07 '15 at 17:55
  • you get the liminf via the weak convergence, see here http://math.stackexchange.com/questions/631110/lim-inf-with-norm-and-weak-convergence – Zimkovic Dec 07 '15 at 17:58
  • 1
    You are correct, but I want understand why $|\nabla \phi_j|$ converge weakly for $|\nabla \psi|$ in $L^2$. – A.D. Dec 07 '15 at 18:12

1 Answers1

2

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.

  • I already find the solution for above problem. Your proof is quite similar the my proof. Thanks Sally . – A.D. Feb 24 '16 at 13:51