Weakly convergence in $W^{1,q}, 1<q<\infty$

Question

Let $(x_n)$ be weakly convergent against $x$ in the Sobolev space $W^{1,q},1<q<\infty$.

Now I have to show, that $(\dot{x_n})$ converges weakly against $\dot{x}$ in $L^q$.

(With the point I name the derivation.)

How can I show that, please?

Answer 1

The Sobolev inequality tells you that you can embed certain Sobolev spaces in others; this gives you a bounded linear operator from one space to the other.

Now you only need to note that a bounded linear operator from one Banach space to another also maps weakly convergent sequences to weakly convergent sequences.

Answer 2

1. The sequence $\{\dot{x_n}\}$ is bounded in $L^q$.
2. Let $\varphi$ be a test function. We have $$\int_\Omega \dot{x_n}\varphi dt=-\int_{\Omega}x_n\dot{\varphi}dt,$$ so if $x_{k'}\to u$ and $\dot{x_{k'}}\to v$ weakly in $L^q$, then $v=\dot u$.
3. As $W^{1,q}$ is separable and reflexive, the unit ball is weakly compact, so we just need to check that if $x_{k'}\to y$ in $W^{1,q}$ then $y=x$.