Square root of a continuous function

Question

Is it square root of a real $\alpha-$Holder continuous function $f$ defined on $[0,1]$ a $\alpha/2$ Holder continuous, provided $\sqrt{f(x)}$ it exists and is continuous, i.e. whether $|f(x)-f(y)|\le C|x-y|^\alpha$ implies that $|\sqrt{f(x)}-\sqrt{f(y)}|\le C'|x-y|^{\alpha/2}$.

Answer 1

$$ \begin{aligned} \left|\sqrt{f(x)}-\sqrt{f(y)}\right|&=\frac{\bigl|f(x)-f(y)\bigr|}{\sqrt{f(x)}+\sqrt{f(y)}} \\ &\le\min\left(\sqrt{f(x)}+\sqrt{f(y)},\frac{C|x-y|^\alpha}{\sqrt{f(x)}+\sqrt{f(y)}}\right) \\ &\le\sqrt C\,|x-y|^{\alpha/2}, \end{aligned} $$ as desired. (The inequality here follows because $\min(u,c/u)\le\sqrt c$ for all positive real $c$ and $u$.)