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.
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 $\alpha=1$.