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.
- $W^{1,1}$ 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 $\epsilon>0$.
- 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 $\alpha=1-1/s$.
- 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 $\alpha=1$.
