I was wondering under which conditions the (weak) differential operator $D: W^{k,p}(\Omega)\rightarrow L^p(\Omega)$, $u \mapsto Du$ from a Sobolev space into the underlying $L^p$-space (on some open subset $\Omega\subset\mathbb{R}^n$) is compact.
My "idea": For the classical differentiable case, i.e. $D:C^m(\Omega)\rightarrow C(\Omega)$, one can use Arzela-Ascoli to prove that $D$ is compact iff $m\geq 2$. And for $\gamma:=k-\frac{n}{p}>m+\alpha$, the Sobolev embedding theorem tells us that $W^{k,p}(\Omega)\subset C^{m,\alpha}(\Omega)$. So for sufficiently big $p$ and $k$, one would have that $D:W^{k,p}(\Omega)\rightarrow C(\Omega)$ is compact… however, this compactness is only given with respect to the topology of uniform convergence, not w.r.t. the $L^p$-norm.
Best Answer
If $\Omega\subset\mathbb{R}^n$ has finite measure, uniform convergence implies $L^p$-convergence: Assume that $(f_n)_{n\in\mathbb{N}}\subset C(\Omega)$ converges against $f\in C(\Omega)$ w.r.t $\|\cdot\|_\infty$, then:
$$\left(\int\limits_\Omega |f(x)-f_n(x)|^p dx\right)^{1/p} \leq \|f-f_n\|_\infty\cdot \mu(\Omega)^{1/p}$$