If $f\in C^{\alpha}([a,b])$, $f\geq0$ and $0<p<1$, is it true $f^p\in C^{\beta}([a,b])$ for some $0<\beta<1$?

Question

Let $f\in C^{\alpha}([a,b])$ and $f\geq0$, where $C^{\alpha}$ denotes the $\alpha$-Hölder space.

If $p\geq1$, then $f^p\in C^{\alpha}([a,b])$, by the mean value theorem applied to $g(x)=x^p$. Indeed, $f^p(x)-f^p(y)=g(f(x))-g(f(y))=p\,\xi_{x,y}^{p-1}(f(x)-f(y))$, with $\xi_{x,y}\in [f(x),f(y)]$, and since $f$ is bounded, say by $M>0$, then $\xi_{x,y}\leq M$, so $|f^p(x)-f^p(y)|\leq C|f(x)-f(y)|$.

What about $0<p<1$? In such a case, $\xi_{x,y}^{p-1}$ may not be bounded, as it can go to $0$. Thus, do we have $f^p\in C^{\beta}([a,b])$ for some $0<\beta<1$?

Answer 1

Let us start from the following well-known inequality (here $p \in (0,1)$): $$ | x^p - y^p| \leq |x-y|^p\qquad \forall x,y \geq 0. $$ If $f$ is a non-negative $\alpha$-Holder function and $p\in (0,1)$, we get $$ |f(x)^p - f(y)^p|\leq |f(x) - f(y)|^p \leq C |x-y|^{\alpha p}, $$ hence $f\in C^{\alpha p}$.