Holder continuity of the derivate of $|x|^\alpha$ for $\alpha>1$ Suppose $U=B(0,1)$ is an open ball in $\mathbb R^m$ and $\alpha>1$. My question is if $|x|^\alpha\in C^{1,\gamma}(U)$ for some $\gamma\in (0,1]$. I know that $|x|^\alpha\in C^1(U)$ and
$$
f_i(x):=\frac{\partial}{\partial x_i}|x|^\alpha=\alpha|x|^{\alpha-2}x_i, \mbox{ if }x\neq 0.
$$
If $\alpha\geq 2$, it is easy because we obtain $|x|^\alpha\in C^2(U)\subset C^{1,1}(U)$, with $\gamma=1$ in this case.
We can suppose $1<\alpha<2$. If $|x|^\alpha\in C^{1,\gamma}(U)$ then using $x_n=\frac{e_i}n$ and $y_n=-\frac{e_i}n$ we obtain
$$
\frac{|f_i(x_n)-f_i(y_n)|}{|x_n-y_n|^\gamma}=\frac{\alpha}{2^{\gamma-1}}n^{\gamma-(\alpha-1)}\overset{n\rightarrow+\infty}\longrightarrow+\infty,
$$
if $\gamma>\alpha-1$. If this $\gamma$ exist we must have $\gamma\leq\alpha-1$. I guess that $|x|^\alpha\in C^{1,\alpha-1}(U)$.
I try proof that exist $C>0$ such that
$$
|f_i(x)-f_i(y)|\leq C|x-y|^{\alpha-1},
$$
but I only succeeded in the case $m=1$.
 A: First, let me prove this for $m=1$. The question is, whether the function
$$
g(x):= sign(x)|x|^\gamma
$$
is in $C^\gamma([-1,1])$ for $\gamma\in (0,1)$. Let $x,y$ be such that $0<x<y$. Then
$$
g(y)-g(x) = \int_x^y \gamma t^{\gamma-1}dt 
$$
and by Hoelder's inequality
$$
|g(y)-g(x)| \le  \int_x^y \gamma t^{\gamma-1}dt \le \gamma (x-y)^\gamma \cdot \left(\int_x^y |t|^{(\gamma-1)\cdot \frac{\gamma}{1-\gamma}} dt\right)^{\frac{1-\gamma}\gamma}
=  \gamma (x-y)^\gamma ( y^{1-\gamma}-x^{1-\gamma})^{\frac{1-\gamma}\gamma}
\le \gamma (x-y)^\gamma .
$$
In case $0<x<y$, we have $|x-y| \ge |x|,|y|$, which implies
$$
\frac{g(y)-g(x)}{|y-x|^\gamma} =\frac{|y|^\gamma+ |x|^\gamma}{|y-x|^\gamma} \le2.
$$
This shows $g\in C^\gamma([-1,1])$.
Let now $m>1$. Set $\gamma:=\alpha-1$. Then with your functions $f_i$ we have for $0<|x|\le |y|$
$$
f_i(x)- f_i(y) = \gamma( |x|^\gamma ( \frac{x_i}{|x|} -  \frac{y_i}{|y|})+ (|x|^{\gamma-1}  -  |y|^{\gamma-1}) \frac{y_i}{|y|}).
$$
The second term can be reduced to the case $m=1$. The first term is unproblematic
if $|x|\le |x-y|$. Now suppose $|x|>|x-y|$, which implies $|y|\le 2|x|$.
Then
$$
\left|\frac{x_i}{|x|} -  \frac{y_i}{|y|} \right|=\left| \frac{x_i(|y|-|x|)-|y|(x_i-y_i)}{|x|\cdot |y|}\right|\le\frac{(|x|+|y|)|x-y|}{|x|\cdot |y|}
\le 2\frac{|x-y|}{|x|} \le 2|x-y|^\gamma |x|^{-\gamma}.
$$
This enables to prove $f_i\in C^\gamma(\overline{B_1(0)})$ and $f\in C^\alpha(\overline{B_1(0)})$ .
