Is the space $W^{2,p}(\mathbf{U})$, compactly embedded to $W^{1,\infty}(\mathbf{U})$, where $\mathbf{U}$ is the unit disk.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ By Rellich-Kondrachov yes if $p>2$. $\endgroup$– Pietro MajerMay 4, 2019 at 11:29
-
$\begingroup$ And no for $p\le2$: you can do the same construction here mathoverflow.net/questions/81034/… $\endgroup$– Pietro MajerMay 4, 2019 at 11:59
-
$\begingroup$ For a non-zero smooth function $f$ with compact support in $U$ consider $f_\epsilon(x):={\epsilon}f({x\over\epsilon})$. It converges to zero uniformly for $\epsilon\to0$, and it has bounded $W^{2,2}$-norm (because $\| D^2 f_\epsilon\|_{2}=\|D^2 f\|_{2}$ and $\|f_\epsilon\|_{1,2}=o(1)$, so it converges to $0$ also weakly $W^{2,2}$. But it has constant non-zero $\|\nabla f_\epsilon\|_{\infty}=\|\nabla f\|_{\infty}$ so no subsequence can converge in $W^{1,\infty}$ norm . $\endgroup$– Pietro MajerMay 4, 2019 at 14:41
Add a comment
|