Is the embedding $W^{1,\infty}(\Omega) \hookrightarrow L^{2}(\Omega)$ compact? All I could find is the Rellich Kondrachov Theorem but it only gives that $W^{1,\infty}(\Omega) \hookrightarrow L^{\infty}(\Omega)$ is compact. My guess is we could use the fact that $L^{\infty}(\Omega) \hookrightarrow L^{2}(\Omega)$ is continuous and deduce the result?
Asked
Active
Viewed 38 times
0
-
1Yes you are right. – timur Jan 26 '22 at 22:50
-
1Exactly. This uses a very common and useful argument: If in a chain of continuous embeddings at least one is compact, then the embedding from the first to last space in this chain is also compact – Claudio Moneo Jan 26 '22 at 22:54