$W^{1,p}(\Omega) \hookrightarrow L^p(\Omega)$

functional-analysissobolev-spaces

Let $\Omega \subset \mathbb{R}^n$ be open and bounded. I want to show that there is a compact embedding of the Sobolev space into $L^p$.
$$W^{1,p}(\Omega) \hookrightarrow L^p(\Omega)$$

I wanted to check if my reasoning is correct for the $p=n$ case.

I claim that, since the $L^p$ spaces are nested continuously, applying this fact to the function and its first order weak derivatives, we should have a continuous embedding
$$W^{1,n}(\Omega) \hookrightarrow W^{1,q}(\Omega), $$
for any $q \in [1,n)$. Then, we can apply Rellich-Kondrachov to get a compact embedding.

Is this the correct idea? I was wondering if, in particular, my claim $W^{1,n}(\Omega) \hookrightarrow W^{1,q}(\Omega)$ is correct?

Thank you

Best Answer

Yes, the reasoning is correct, but you need $\Omega$ to be a Lipschitz domain, otherwise there are counterexamples where we fail even the embedding. Assuming a Lipschitz boundary means you assume "some regularity" of the boundary.

Moreover, note how important is that $\Omega$ is bounded for these inclusions, otherwise all the "nested" spaces are no more nested and the reasoning fails: indeed give a look here where $\Omega$ is not bounded to see that your Sobolev embedding fails. Hence, your claim $W^{1,n}(\Omega) \hookrightarrow W^{1,q}(\Omega)$ is correct since $q \le n$, and actually note that another interesting question is if we can gain some integrability conditions by adding some "differentiability condition": see my answer here as well.

Let me add some references in cases you wanted to go deeper with some embeddings result: “Giovanni Leoni - A first course in Sobolev Spaces” “Adams - Sobolev Spaces”.

Other two referenes, if you are interested in Sobolev Spaces from the point of view of PDEs then:

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