$L^2(U)$ compact embedded in $H^{-1}(U)$?

Question

Let $U$ be an open subset of $R^d$. We already knew that $L^2(U)$ is a subset of $H^{-1}(U)$. Question: is this a compact embedding?

score 3 · Accepted Answer · edited Mar 08 '24 at 14:16

3

Yes, this embedding is compact. This can be seen, since this is the adjoint of the embedding of $H_0^1(U)$ in $L^2(U)$.

edited Mar 08 '24 at 14:16

daw

answered Mar 25 '13 at 12:39

gerw

thank you. Can you please add some more detail or some references for the fact that the duality preserve the compact embedding property? – Hong DM Mar 25 '13 at 12:45
See http://math.stackexchange.com/questions/41432/easy-proof-adjointcompact-compact. – gerw Mar 25 '13 at 12:49
thank you. your answer is very helpful – Hong DM Mar 25 '13 at 12:54

1 Answers1