What is the relationship between Hölder spaces and differentiability?

Question

I'm porting this question over from MSE as it did not get any responses other than one comment on there.

Let $C^{k,\alpha}$ be a Hölder space where $0 \leq \alpha \leq 1$. I have seen various sources informally describe these spaces as functions having at least "$k + \alpha$" derivatives. How does some sense of fractional differentiability follow from the definition of the Hölder norm?

Answer 1

Let $D\subset\mathbb{R}^n$ be a bounded domain. The fractional Sobolev space $W^{s,p}(D)$, $0<s<1$, $1<p<\infty$ is defined as the space of all $f\in L^p(D)$ such that $$ [f]_{s,p}=\left(\int_D\int_D\frac{|f(x)-f(y)|^p}{|x-y|^{n+sp}}\, dxdy\right)^{1/p}<\infty. $$ The space is equipped with the norm $\Vert f\Vert_{s,p}=\Vert f\Vert_p+[f]_{s,p}$.

It is easy to prove (by a straightforward estimate of the integral that $C^{0,\alpha}(D)\subset W^{1-\frac{1}{p},p}(D)$ whenever $1<p<\infty$ and $1-\frac{1}{p}<\alpha\leq 1$. In that sense H"older functions belong to some fractional Sobolev spaces.

You can find some basic facts in

Di Nezza, Eleonora; Palatucci, Giampiero; Valdinoci, Enrico, Hitchhiker’s guide to the fractional Sobolev spaces, Bull. Sci. Math. 136, No. 5, 521-573 (2012). ZBL1252.46023.

Answer 2

Theorem 14 in https://eudml.org/doc/167995 says that if $f\in C^\alpha$ and $\beta<\alpha$ then $D^\beta f$ exists and we have that $D^\beta f\in C^{\alpha-\beta}$, where $D^\beta$ is the fractional derivative.

For example $D^\beta x^\alpha=\operatorname{Const} \cdot x^{\alpha-\beta}$.

See:

https://math.stackexchange.com/questions/2685689/does-taking-a-fractional-derivative-remove-a-fractional-amount-of-holder-regular