Background: I work on a PDE problem where I have some approximating sequence of measure-valued functions and I need to compactly embed it into some negative Sobolev space $W^{-m,q}$ on the bounded interval in $\mathbb{R}$. I am mostly interested in the spaces where $q=2$. I found only one such embedding in the one theorem from the paper:

Evans - Weak convergence methods for nonlinear partial differential equations, 1990.

Theorem 6 (Compactness for measures, page 7): Assume the sequence $\{\mu_k\}_{k=1}^{\infty}$ is bounded in $\mathcal{M}(U)$, $U \subset \mathbb{R}^n$. Then $\{\mu_k\}_{k=1}^{\infty}$ is precompact in $W^{-1,q}(U)$ for each $1 \leq q <1^*$.

Here $\mathcal{M}(U)$ represents space of signed Radon measures on $U$ with finite mass, $U \subset \mathbb{R}^n$ is an open, bounded, smooth subset of $\mathbb{R}^n, n \geq 2$ and $1^*=\frac{n}{n-1}$ represents a Sobolev conjugate.

The identical theorem (Lemma 2.55, page 38) is given in the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, with a difference that instead of $1 \leq q <1^*$, in there is written $1 \leq q <\frac{n}{n-1}$ (here it isn't written explicitly that $n\geq 2$).

My question: does the Theorem 6 works in one dimension ($n=1$)? That is do we have a compact embedding of space $\mathcal{M}(U)$ into the space $W^{-1,q}(U)$, where $U \subset \mathbb{R}$?

And additionaly:

  • I assume that if we have compact embedding into $W^{-1,q}(U)$, then we have it also in the $W^{-m,q}(U), m\geq 1$?
  • Are there any other measure spaces (e.g. space of finite positive measures $\mathcal{M}_+$, space of probability measures with finite first moment $Pr_1$, etc.) that are compactly embedded into some negative Sobolev spaces $W^{-m,q}(U)$?

I think that if we use definition of the Sobolev conjugate: $\frac{1}{p^*}=\frac{1}{p}-\frac{1}{n}$, we get for $p=1,n=1$ the $\frac{1}{1^*}=\frac{1}{1}-\frac{1}{1}\Rightarrow 1^*=\infty$. So we would have that theorem 6 (maybe) works for every $1 \leq q < \infty$ (and then for $q=2$ also)? If we use $p^*=\frac{np}{n-p}$ we would have for $n=1,$ $p^*=\frac{p}{1-p}$ and here we could not take $p=1$ and get $p^*$.

I usually do not deal with the measure-valued and negative Sobolev spaces, so I don't know much about them. Help with this would be great and I definitely need it. And any additional reference besides the two mentioned above would be nice. Thanks in advance.

  • $\begingroup$ yes, the statement is written in dimensions $n=1$ too, see my answer below. As a general rule of thumbs things are easier in lower dimensions, and clearly here the statement is distinguished because the threshold for Morrey's inequality $W^{1,p}\subset C^\alpha$ in dimension 1 is $p=1$, so in this particular case the statement of Theorem 6 holds verbatim for all $p\geq 1$. $\endgroup$ Commented May 10, 2020 at 14:19
  • $\begingroup$ as should be clear from my answer below, the key is to first obtain a continuous embedding $W^{1,p}\subset C^{\alpha}$, which is then automatically transferred to a compact embedding via Ascoli-Arzelà. And in dimension 1 this works for all $p\geq 1$. $\endgroup$ Commented May 10, 2020 at 14:21

1 Answer 1


Here is a partial answer, which has to do with dual compact embeddings: If the embedding between (resonable) Banach spaces $X\subset\subset Y$ is compact then the dual embedding is compact too, $Y^*\subset\subset X^*$.

This is useful here since the space of Radon measures is the dual of continuous bounded functions, $\mathcal M(U)=(C_b(U))^*$. Now for $p>n$ we have that $W^{1,p}$ is continuously embedded into some Hölder space $C^\alpha$ (for some $\alpha\equiv \alpha(n,p)$). By the Arzelà-Ascoli theorem this shows that the embedding $$ W^{1,p}(U)\subset\subset C_b(\bar U) $$ is compact too. As a consequence we have that the embedding $$ \mathcal M(U)\subset\subset W^{-1,q}(U) $$ provided that $q=p'$ is such that $p>n$, ie for all $q<1^*=\frac{n}{n-1}$ (this is exactly why the cirical $1^*$ exponent appears in your Theorem 6).

As for the second part of the question: since the embedding $W^{m,p}\subset W^{1,p}$ is trivially continuous for $m>1$, the reversed embedding $W^{-1,q}\subset W^{-m,q}$ is continuous. Then the composition of "compact$\circ$continuous = compact" $\mathcal M\subset\subset W^{-1,q}\subset W^{-m,q}$ also gives compactness for $$ \mathcal M(U)\subset\subset W^{-m,q}(U). $$

  • $\begingroup$ Thank you for the answer and the comments. It is nice to be sure that Theorem 6 works in one dimension also. In the case I am interested in $W^{1,2}$ is continuously embedded into $C^{0,\alpha}, 0<\alpha\leq \frac{1}{2}$. By Arzela-Ascoli it is compactly embedded in $C_b$ and by duality argument $\mathcal{M}$ is compactly embedded in $W^{-1,2}$. I wasn't sure in the Theorem 6 is it true that $1^* = \infty$? If it is true, then $\mathcal{M}$ is compactly embedded in $W^{-1,q},q<\infty$ based on your answer? And thanks again. $\endgroup$
    – Mark
    Commented May 11, 2020 at 9:12
  • $\begingroup$ Yes, formally you can tink of $1^*=\infty$, and indeed $\mathcal M\subset\subset W^{-1,q}$ for all $q<\infty$ in dimension 1. $\endgroup$ Commented May 11, 2020 at 10:06
  • $\begingroup$ Thank you for the follow up. I think that this will help me a lot. $\endgroup$
    – Mark
    Commented May 11, 2020 at 14:34
  • $\begingroup$ I am happy with the answer, but I always leave a week or two before I accept the answer. I take the time to try to use the answer I got on the problem I am trying to solve. And also I would like to see if there is an answer to the second part of my question (about other measure spaces). If not, this answer is quite nice and it will be accepted definitely. Thanks again. $\endgroup$
    – Mark
    Commented May 11, 2020 at 14:52
  • $\begingroup$ Oh sure, sorry about that, for some reason I thought you were a new user here and that you didn't know about our MO system. I'm sorry. And I'm not chasing for points either, it just annoys me when there are nice answers given to a quiestion (I'm not saying mine is!) but users just leave them hanging out there. I'll remove my comment ASAP, now I feel terrible. $\endgroup$ Commented May 11, 2020 at 14:54

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