Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H¹_0$?

Question

Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H_0^1$?

In this article https://drive.google.com/open?id=0B0MDtuRPAebZQ3NvRFVQekJ4VEpTMldiRWxmbU9GLVM4MDNF the authors claim, in the proof of Proposition 4, that $L^3(\Omega)$ is continuously embedded into $H^{-1}(\Omega)$, the dual of $H_0^1(\Omega)$, being $\Omega \in \mathbb{R}^3$ a bounded open set.

The injections $H_0^1 \subset L^2 \subset H^{-1}$ are well known, but I am not able to prove the claim for $L^3$.

Any references, hints or suggestions will be the most appreciated.

Answer 1

By Sobolev, $H^1_0 \subset L^{2^*}=L^6$, because $\frac1{2^*}=\frac12-\frac13=\frac16$.

So $L^3\subset L^{6/5} = (L^6)^* \subset (H^1_0)^* = H^{-1}$.

This shows that the lowest exponent that you can embed is $6/5$, and $3$ is much more than needed.

I'm assuming $|\Omega|<\infty$.

Answer 2

The embedding $L^3\subset H^{-1}$ follows from the embeddings $L^3\subset L^2$ and $L^2\subset H^{-1}$ by composition of the embeddings.