# Does taking a fractional derivative remove a fractional amount of Holder regularity?

We define the space $C^{n+\alpha}$ as functions who are $n$ times differentiable whose $n$th derivative is $\alpha$ Holder. That is, each time we take a derivative we remove one number of regularity.

When functions have no more regularity to give to take a derivative we get to negative $C$ spaces. I.e. it can be shown that a function that is $C^{-.5}$ is the weak/distributional derivative of a $C^{.5}$ function.

How far can this be generalized? Can this be generalized to fractional derivatives? Meaning if we take a $.5$ derivative of a $C^{.9}$ function do we get a $C^{.4}$ function? Recall that we define $D^\alpha f(x)=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dx}\int_0^x\frac{f(t)}{(x-t)^\alpha}dt$.

This is true for $f(x)=x^{\alpha}$, which is $C^{\alpha}$. Fractional derivatives give $cx^{\alpha-\beta}$ which is $C^{\alpha-\beta}$.

This is true. The main point is that the integral operator $$f\mapsto \int_0^x\frac{f(t)}{(x-t)^\alpha}dt$$ adds $(1-\alpha)$ to the Hölder exponent, after which $d/dx$ removes $1$. This property of fractional integral was proved by Hardy and Littlewood: see Theorem 14 in Some properties of fractional integrals. I. Mathematische Zeitschrift (1928), Volume: 27, page 565-606. The theorem is on page 587. Note that your $\alpha$ is their $1-\alpha$.