# Compact, continuous embeddings of $H^s := W^{s,2} \leftrightarrow C^{(\alpha)}$

The sobolev-space $H^s([-\pi,\pi])$ can be embedded into $C^{(\alpha)}([-\pi,\pi])$ (space of $\alpha$-Hölder-continuous functions) and vice-versa.
My question is for which exponents $s, \alpha$ can we reach those embeddings and for which exponents $s, \alpha$ are these embeddings compact?

$H^s \subset C^{(\alpha)}$ continuous for $s > \alpha + [?]$
$H^s \subset C^{(\alpha)}$ compact for $s > \alpha + [?]$
$C^{(\alpha)} \subset H^s$ continuous for $s < \alpha - [?]$
$C^{(\alpha)} \subset H^s$ compact for $s < \alpha - [?]$

What I have found so far:
$C^{(\alpha)} \subset H^s$ continuous for $1 > \alpha > s \geq 0$ (Bernstein's theorem)

-

Sobolev spaces normally have integrability exponent $s\ge 1$, just as Lebesgue spaces do. Indeed, otherwise they would not qualify as normed spaces.

Also, in much Sobolev space literature $H^s$ stands for $W^{s,2}$, not $W^{1,s}$. This makes sense if we think that $H$ stands for Hilbert. For this reason, I write $W^{1,s}$ below.

1. $W^{1,1}$ continuously embeds into $C^0$ (by the Fundamental Theorem of Calculus), but not into any $C^\alpha$ with $\alpha>0$. Indeed, it contains $|x|^\epsilon$ for every $\epsilon>0$.
2. For $s> 1$, $W^{1,s}$ continuously embeds into $C^\alpha$ with $\alpha=1-1/s$, by Morrey's inequality. The constant $1-1/s$ cannot be improved. The embedding is not compact. To verify both of the above, consider a sequence of triangles with base $n^{-2}$ and height $n^{2/s-2}$. Putting them next to each other on the real line, you get a sequence of functions which is bounded in $W^{1,s}$ and has no convergent subsequence in $C^{\alpha}$ with $\alpha=1-1/s$.
3. When $0<\alpha<\beta <1$, the space $C^\beta$ compactly embeds into $C^\alpha$. See Is there a reference for compact imbedding of Hölder space? This and item 2 imply that $W^{1,s}$ compactly embeds into $C^\alpha$ when $0<\alpha<1-1/s$.

I think you are mistaken about the embedding of $C^\alpha$ into $W^{1,s}$. For any $\alpha\in (0,1)$, there exist $C^\alpha$ functions that are not of bounded variation, and therefore do not belong to any Sobolev space. You can build them from a chain of triangles similar to item 2 above: if the total height of triangles is infinite, the function does not have bounded variation.

The space of Lipschitz functions ($\alpha=1$) is the same as $W^{1,\infty}$. See relation between $W^{1,\infty}$ and $C^{0,1}$. This embedding is not compact because it's an invertible operator.

As you can see, going from Hölder to Sobolev scale is not as easy as the other way around: you get nothing when $\alpha<1$, and then suddenly get everything when $\alpha=1$.

-
A very good explanation, in my opinion. A tiny quibble: various other bits of information make the origin of the "H" for the $L^2$ version of "W" spaces hazy... Nevertheless, the mnemonic that it is "Hilbert (-space)" is very good. :) – paul garrett Aug 6 '13 at 22:29
Sorry, I assumed the common notation $W^{1,p}$ would refer to those. I meant the $W^{s,2}$-spaces as you correctly noticed. Thank you neverteless for the clarifications. As I have found the necessary constants for $W^{s,2}$ already (Sobolev's theorem, Berstein's theorem and a generalisation of Sobolev's theorem), I marked the question as solved. Do you know, whether the embedding $H^s \subset C^{(\alpha)}$ for the "actual" $H^s$ and $0 < \alpha < s-\frac{1}{2}$ is continuous or compact? – AlexR Aug 7 '13 at 9:45
@AlexR: It is compact. – timur Aug 10 '13 at 0:38
@timur Thank you very much. – AlexR Aug 15 '13 at 12:48