Is this vector-valued map Hölder-continuous? Pick $0<q<1$ and consider the map from $\mathbb{R}^n$ to $\mathbb{R}^n$ that sends $x$ to $|x|^{q-1}x$. Is this map Hölder-continuous (I guess with exponent $\leq q$)? In dimension one, I can exploit the homogeneity of the inequality defined by Hölder-continuity; but how can I proceed in higher dimension?
 A: Here is a  proof of a more general result. Fix $q\in (0,1]$. Let $Y $ be a normed vector space, and let $f:\mathbb R^n\to Y$ be a map such that 


*

*The restriction of $f$ to the unit sphere of $\mathbb R^n$ is $q$-Hölder continuous.

*$f(tx)=t^q f(x)$ for all $t\ge 0$ and all $x\in \mathbb R^n$.  


Claim:   $f$ is $q$-Hölder continuous. 
Step I. Let $C$ be such that 
$$\|f(x)-f(y)\|\le C|x-y|^q \tag{*} $$ for all unit vectors $x,y$. Property 2 implies that the same inequality   holds whenever $|x|=|y|$. Thus, 

(A)  $f$ satisfies (*) on every sphere centered at the origin 

Step II. Let $x$ be a unit vector. The restriction of $f$ to the  ray $\{tx: t\ge 0\}$    is given by $t^q f(x)$. This is $q$-Hölder continuous with the constant $\|f(x)\|$. Since $ \|f\|$ is bounded on the unit sphere by some $M$, it follows that  

(B) $f$ satisfies $ \|f(x)-f(y)\|\le M|x-y|^q  $  on every halfline emanating from the origin.

Step III. The combination of two highlighted statement implies that $f$ is $q$-Hölder continuous on $\mathbb R^n$. Proof:  given two points $x$ and $y$ with $|x|\le |y|$, let $z$ be the radial projection of $y$ onto the sphere with radius $|x|$ centered at the origin. Note that $|z-x|\le |y-x|$ (projection is a contraction). By (A), $$\|f(z)-f(x)\|\le C |z-x |^q\le C |y-x |^q$$
By (B), 
$$\|f(z)-f(y)\|\le M |z-y |^q\le M |y-x |^q$$
The triangle inequality completes the proof:
$$\|f(x)-f(y)\|\le (C+M)  |y-x |^q$$
A: Let $0<q\leq1$. The map $x \mapsto \|x\|^q$ on $\mathbb{R}^n$ is $q$-Hölder continuous. Assume the similar result for $x \mapsto |x|^q$ on $\mathbb{R}_{\geq 0}$. Then write the given norm-power map as a composition of $x \mapsto \|x\|$, which is Lipschitz, and $\|x\| \mapsto \|x\|^\alpha$ to conclude. Save this fact.
Let $0<\alpha\leq1$. The map $x \mapsto \|x\|^{\alpha-1}x$ on $\mathbb{R}^n$ is $\alpha$-Hölder continuous. The proof technique comes from a paper by Cazenave, Fang and Han, where it is shown that the complex map $z \mapsto |z|^{\alpha-2}z^2$ is $\alpha$-Hölder. You'd think that with the $z^2$ factor, the proof method wouldn't apply, but it goes through.
Applying the previous lemma and the triangle inequality,
\begin{align}
\left\| \|x\|^{\alpha-1}x - \|y\|^{\alpha-1}y \right\|
 &= \left\| \|x\|^{\alpha-1}x - \|x\|^\alpha\frac{y}{\|y\|} + \|x\|^\alpha\frac{y}{\|y\|} - \|y\|^{\alpha-1}y \right\| \\
 &\leq \|x\|^\alpha \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|} \right\| + \left| \|x\|^\alpha-\|y\|^\alpha \right| \\
 &\leq \|x\|^\alpha \left\|\frac{x}{\|x\|} - \frac{y}{\|x\|} + \frac{y}{\|x\|} - \frac{y}{\|y\|} \right\| + \|x-y\|^\alpha \\
 &= \|x\|^{\alpha-1} \left\| x - y + \frac{y}{\|y\|}(\|y\|-\|x\|) \right\| + \|x-y\|^\alpha \\
 &\leq 2\|x\|^{\alpha-1}\|x-y\| + \|x-y\|^\alpha.
\end{align}
Without loss of generality, assume $0 < \|y\| \leq \|x\|$ and proceed by cases. If $\|x-y\| \leq \|x\|$, then $\alpha-1\leq0$ implies
\begin{equation}
\|x\|^{\alpha-1} \leq \|x-y\|^{\alpha-1}.
\end{equation}
Multiplying through by $\|x-y\|$ yields the result in this case. Now assume $\|x-y\| \geq \|x\|$ and note
\begin{equation}
\|x-y\| \leq \|x\| + \|y\| \leq 2\|x\|
\end{equation}
by the assumption $0 < \|y\| \leq \|x\|$. Multiplying through by $\|x\|^{\alpha-1}$ yields
\begin{equation}
\|x\|^{\alpha-1}\|x-y\| \leq 2\|x\|^\alpha \leq 2\|x-y\|^\alpha.
\end{equation}
Taking the larger of the constants in the two cases gives
\begin{equation}
\left\| \|x\|^{\alpha-1}x - \|y\|^{\alpha-1}y \right\| \leq 5\|x-y\|^\alpha.
\end{equation}
