Show u^2|u|^(p-3) is Holder continuous (derivative of power nonlinearity) I am trying to show that the function $f:\mathbb{C}\rightarrow \mathbb{C}$ given by $f(u)=u^2|u|^{p-3}$ is (uniformly) Holder continuous for $p\in(1,2)$ (so $p-3\in(-2,-1)$, this ensures that $f$ is continuous but not differentiable at $0$). I think it should/want it to be Holder continuous of degree $p-1$, like $|u|^{p-1}$, but having tried many approaches now the closest I have got is that
$$|f(u)-f(v)|\lesssim_p |u-v|^{p-1}+|v|^{p-1}|u|^{-2}(|u|+|v|)|u-v|$$
and the coefficient in front of the second difference, $|v|^{p-1}|u|^{-2}(|u|+|v|)$, blows up as $u$, $v$ get small.
Does anyone know how to prove (or disprove!) the Holder continuity?
(context: this term appears in the derivative of the nonlinearity for the nonlinear Schrodinger equation)
 A: Let $f(z) = z^2|z|^a$ for $a \in (-2,-1)$. We prove that $f$ is uniformly
$(a+2)$-Hölder continuous in $\mathbb C$, i.e. there exists $C > 0$ such that
$$|f(z)-f(y)| \leq C|z-y|^{a+2} \quad \forall z,y \in \mathbb C \quad (\star)$$
Without loss of generality, $z$ and $y$ are non-zero. Observe that if we have proved ($\star$) for any $z \in \mathbb C^*$ and $y = 1$, then one would get
$$|z^2|z|^a - y^2|y|^a| = |y|^{a+2} \left| \left( \frac{z}{y}\right)^2 \left| \frac{z}{y}\right|^a -1 \right| \leq C |y|^{a+2} \left| \frac{z}{y}-1 \right|^{a+2} = C\left| z-y \right|^{a+2} \quad \forall z, y \in \mathbb C^*$$
Hence, without loss of generality, we will assume that $y = 1$.
Let $0 < \varepsilon < 1/2$ and consider $U = B_{\epsilon}(1) \subset \mathbb C$ the ball of radius $\varepsilon$ around $1$.
On $U$, $f(x+iy)$ is real-differentiable, so for $z \in U$, the mean value theorem applies :
$$|f(z) - 1| \leq \sup_{v \in U}||Df(v)|| \cdot |z-1|$$
Moreover,
$$|z-1| = |z-1|^{a+2}|z-1|^{-a-1} \leq \varepsilon^{-a-1} |z-1|^{a+2} \quad \forall z \in U$$
If $z \in U^c \setminus \{0\}$, one writes
$$z^2|z|^a - 1 = (|z|^{a+2}-1)\left( \frac{z^2}{|z|^2} -1\right) + (|z|^{a+2}-1) + \left( \frac{z^2}{|z|^2} -1\right)$$
so that
$$|z^2|z|^a - 1| \leq 3||z|^{a+2}-1| + 2$$
Letting $C = 2(\min_{z \in U^c}||z|^{a+2}-1|)^{-1} \neq 0$, one gets
$$|z^2|z|^a - 1| \leq (C+3)||z|^{a+2}-1|$$
and one can use the fact that $x \mapsto |x|^{a+2}$ is uniformly $(a+2)$-Hölder continuous in $\mathbb R$ (see here for example) to conclude that
$$||z|^{a+2}-1| \leq C ||z|-1|^{a+2} \leq C|z-1|^{a+2}$$
