Prove that $||a|^\alpha a- |b|^\alpha b|\leq C(|a|^\alpha+|b|^\alpha)|a-b|$ for some $C>0$ Let $a,b\in \mathbb{C}$ and $\alpha\geq 0$. Is this following inequality true?
$||a|^\alpha a- |b|^\alpha b|\leq C(|a|^\alpha+|b|^\alpha)|a-b|$ for some constant $C>0$ independent of $a$, $b$ and $\alpha$.
I have spent many hours to figure it out yet.
For the real case, let's set $f(x)=|x|^\alpha x$. Then by MVT, the result follows easily.... But in the case of complex, $f$ is even not differentiable...
 A: The inequality is true, at least if you allow the constant $C$ to depend on $\alpha$.
In order to prove this, define $f:\mathbb{R}_+\to \mathbb{R}_+$ by $f(x)=x^{\alpha}$. Furthermore assume without loss of generality that $|a|\geq|b|>0$. By the triangle inequality
\begin{align*}
||a|^{\alpha}a-|b|^{\alpha}b|&\leq ||a|^{\alpha}a-|b|^{\alpha}a|+||b|^{\alpha}a-|b|^{\alpha}b|\\
&\leq||a|^{\alpha}-|b|^{\alpha}|\cdot|a|+|b|^{\alpha}|a-b|.\end{align*}
By the mean value theorem there exists a $\xi\in [|b|,|a|]$ such that $f'(\xi)(|a|-|b|)=\alpha\cdot \xi^{\alpha-1}(|a|-|b|)=|a|^{\alpha}-|b|^{\alpha}$. Plugging this into the expression above and using the fact that $\xi\leq|a|$ as well as the inverse triangle inequality (i.e. $||a|-|b||\leq |a-b|$) , we get
\begin{align*}
||a|^{\alpha}a-|b|^{\alpha}b|&\leq \alpha \xi^{\alpha-1}\cdot||a|-|b||\cdot |a|+|b|^{\alpha}|a-b|\\
&\leq \alpha |a|^{\alpha-1}\cdot|a-b|\cdot |a|+|b|^{\alpha}|a-b|\\
&\leq (1+\alpha) (|a|^{\alpha}+|b|^{\alpha})|a-b|.
\end{align*}
