# Compact embedding of space of signed Radon measures into Sobolev space $W^{-1,q}$ from Evans paper; Does it work in one space dimension?

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 $$U \subset \mathbb{R}$$?

• 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.

• 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 $p\geq 1$. – leo monsaingeon May 10 at 14:19
• 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 $p\geq 1$. – leo monsaingeon May 10 at 14:21

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 $$\mathcal M(U)\subset\subset W^{-m,q}(U).$$
• 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. – Mark May 11 at 9:12
• Yes, formally you can tink of $1^*=\infty$, and indeed $\mathcal M\subset\subset W^{-1,q}$ for all $q<\infty$ in dimension 1. – leo monsaingeon May 11 at 10:06