Recently I had to make use of the fact that the map $x\mapsto |x|^s_2$ is globally $s$-Hölder continuous $\left(|\cdot|_2\text{ is the Euclidean distance}\right)$, which can be reduced to the $1$-dimensional problem by letting $a:=|x|_2$, $b:=|y|_2$ and using the triangle inequality one gets
$||x|^s_2-|y|_2^s|=|a^s-b^s|\leq |a-b|^s=||x|_2-|y|_2|^s\leq |x-y|^s_2$.
After that I wondered what happens in the similar situation where we consider the function
$f:\mathbb{R}^n\rightarrow \mathbb{R},x\mapsto |x|^s_2x$.
It is clearly of class $C^1$ and naively it should behave like $x^{s+1}$ in the $1$-$d$ case which is of class $C^{1,s}$. But so far I was not able to come up with a proof. I was able to reduce the problem to showing that for any given $1\leq i,j\leq n$ the function
\begin{gather} \label{1} x\mapsto x_ix_j|x|^{s-2}_2 \text{is globally Hölder continuous on the set }U:=\{x\in \mathbb{R}^n| x_i>0\text{ and }x_j>0\}.\tag{1} \end{gather}
It is then of course enough to consider the cases $i=1=j$ and $i=1,j=2$ (in case it makes sense to distinguish these situations). Also $U$ is convex, so I tried to use the mean value theorem but to no avail.
Anyone has an idea how to show the Hölder continuity of (1)?