$F(u)=u|u|^a$ is locally Lipschitz In my PDE class we have solved the nonlinear Schrödinger equation $$i\partial_t u+\Delta u =F(u), u(t=0)=u_0$$ where $F(u)=u|u|^a,a>0$ using a fixed point argument (under certain conditions which is not important for my question).
This involves showing the following inequality for $u,v \in \mathbb{C}$ $$|F(u)-F(v)|\leq C_a |u-v|(|u|^a+|v|^a)$$ for some $C_a$ depending on $a$. I have some problem seeing this. One can factor as follows $$|F(u)-F(v)|=|(u-v)(|u|^a+|v|^a)+v|u|^a-u|v|^a|$$ and thus I wonder if $$|v|u|^a-u|v|^a|\leq C_a |u-v|(|u|^a+|v|^a)$$ and it's here I got stuck. It's easy to see that $$|v|u|^a-u|v|^a|\leq (|u|+|v|)(|u|^a+|v|^a)$$ but how can I get the factor $|u-v|$?
Related question: Is the map $u\mapsto |u|^2 u$ globally or locally Lipschitz continuous in the $H_0^1$ norm?
 A: Here is what I could get for $a$ an even integer to show that:
$$\big\lvert v\lvert u\rvert^a - u\lvert v\rvert^a\big\rvert \leq C\lvert u-v\rvert(\lvert u\rvert^a + \lvert v \rvert^a)$$
First notice that $\lvert u \rvert^a = uu^{\frac a2-1}\bar u^{\frac a2}$ (with $\bar u$ the complex conjugate):
$$ \big\lvert v\lvert u\rvert^a - u\lvert v\rvert^a\big\rvert = \lvert uv\rvert\cdot\lvert u^{\frac a2-1}\bar u^{\frac a2} - v^{\frac a2-1}\bar v^{\frac a2} \rvert=\lvert uv\rvert\cdot\big\lvert u\lvert u\rvert^{a-1} - v\lvert v\rvert^{a-1}\big\rvert$$
and repetitively using that $$\big\lvert u\lvert u\rvert^{\alpha} - v\lvert v\rvert^{\alpha}\big\rvert \leq \lvert u-v\rvert(\lvert u \rvert^\alpha + \lvert v \rvert^\alpha) + \big\lvert v\lvert u \rvert^\alpha - u\lvert v \rvert^\alpha\big\rvert $$
you get a bound:
$$
\big\lvert v\lvert u\rvert^a - u\lvert v\rvert^a\big\rvert \leq C\lvert u-v\rvert(\lvert u\rvert^a + \lvert v \rvert^a) + \lvert uv\rvert^\frac a2\cdot\lvert u-v\rvert.
$$
Conclude by using Young's inequality on the $uv$ term:
$$\lvert u\rvert^\frac a2 \lvert v\rvert^\frac a2 \leq \frac 12(\lvert u \rvert^a + \lvert v \rvert^a).$$
I could not find how to do it for more general $a$.
