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From the Rellich-Kondrachov theorem we know that $H^m(\Omega)\hookrightarrow_c L^2(\Omega)$ when $\Omega$ is bounded of class $C^1$ and $m\geq 1$ is an integer. Also this is not true if $\Omega:=\mathbb{R}$. See for example Chapter 9 of Bresis - Functional analysis, Sobolev spaces and partial differential equations, 2010, or Chapter 6 of Adams, Fournier - Sobolev spaces, 2003 for details.

My question is do we have the same compact embedding in the cases:

  1. $H^m(\Omega)$ and $L^2(\mathbb{R})$, where $\Omega \subset \mathbb{R}$,

  2. $H^m(\Omega_1)$ and $L^2(\Omega_2)$, where $\Omega_1 \subset \mathbb{R}$, $\Omega_2 \subset \mathbb{R}$ and $\Omega_1 \subset \Omega_2$?

The reason I am asking is that I work on a problem where I would like to use the Aubin-Lions theorem to show some compact embedding. Reminder of the Aubin-Lions theorem: Let $X_1 \hookrightarrow_c X_2 \hookrightarrow X_3$ be three Banach spaces, where $X_1$ and $X_3$ are reflexive and as usual $\hookrightarrow$ and $\hookrightarrow_c$ stand for the continuoous and compact embeddings.. Let $a \in (0,1)$. Then $$L^2(0,T;X_1)\cap W^{a,2}(0,T;X_3) \hookrightarrow_c L^2(0,T;X_2).$$

Additional questions:

$\bullet$ I was wondering could $X_1, X_2, X_3$ be given on a three different spaces e.g. $X_1(\Omega_1), X_2(\Omega_2), X_3(\mathbb{R})$ where $\Omega_1 \subset \mathbb{R}$, $\Omega_2 \subset \mathbb{R}$ and $\Omega_1 \subset \Omega_2$? Or more precisely in the case I am interested in could I use $X_1:=H^m(\Omega_1)$, $X_2:=L^2(\Omega_2)$ (this is why I am interested in the Rellich-Kondrachov theorem) and $X_3:=H^{-m}(\mathbb{R})$? I noticed that in the Aubin-Lions theorem it is nowhere said that the spaces $X_1, X_2, X_3$ need to be on the same set (whether it is $\Omega$ or $\mathbb{R}$).

$\bullet$ And also I am not sure could we have compact embeddings on $\mathbb{R}$ or it always it needs to be on some subset of $\mathbb{R}$?

I usually avoid dealing with compact embedding theorems so I don't know much about them and I hope that I didn't asked the obvious. Help with this would be great and I definitely need it. Thanks in advance.

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    $\begingroup$ You get compactness by composing the the (compact) embedding of $H^m(\Omega_1)$ into $L^2(\Omega_1)$ with the isometric embedding of the latter into $L^2(\Omega_2)$. $\endgroup$
    – Giorgio Metafune
    Jan 12, 2021 at 11:40
  • $\begingroup$ @GiorgioMetafune, thank you for the comment. That should answer the case 2. of the first question. Is the same working for the case 1. of the first question, i.e. do I have a compact embedding when the sets are $H^m(\Omega)$ and $L^2(\mathbb{R})$? I am never sure about compact embeddings when I have $\mathbb{R}$ involved. $\endgroup$
    – Mark
    Jan 12, 2021 at 12:03
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    $\begingroup$ Yes, the same. Boundedness Is used only for $\Omega_1$. $\endgroup$
    – Giorgio Metafune
    Jan 12, 2021 at 12:30
  • $\begingroup$ @GiorgioMetafune, Thank you for the follow up. Concerning two additional questions: That means (I think) that I can use the Aubin-Lions theorem if $X_1 (\Omega_1)$ and the rest could be even $X_2(\mathbb{R})$, $X_3 (\mathbb{R})$. Also (I think) then I could have compact embeddings such as $X \hookrightarrow_c Y$ as long as $X$ is defined on some bounded set. (I hope that my conclusions are not wrong) $\endgroup$
    – Mark
    Jan 12, 2021 at 13:46
  • $\begingroup$ Well I do not know the setting precisely, but if you compose bounded linear emebdeddings (or operators) and one is compact, then the whole is compact. I used only that; if this is the case in your setting, there is no problem. $\endgroup$
    – Giorgio Metafune
    Jan 12, 2021 at 14:03

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