I am reading a proof for the existence of a solution to the Local Cauchy problem of the non-linear Schrodinger equation
$$
i\partial\_t u+\Delta u +\epsilon u |u|^{2} = 0 \\
u(x,0)=u\_0(x)
$$


The structure of the proof is due to J. Ginibre and G. Velo (*On a class of nonlinear Schrödinger equations. I. The Cauchy problem, general case*). Note that details of the proof are NOT necessary for understanding what I am asking here.


Denote, by $L\_{T}^p L\_x^q$ the spacetime banach space $L([0,T]; L^q (\mathbb R^2))$. The norm is given by
$$
\left(\int\_0^T \|u(\cdot, t)\|\_{L^q (\mathbb R^2)}^p dt\right)^{1/p}.
$$


In the proof it seems to be implied that for all functions $u, v$ with sufficient regularity (say, in $C\_c^\infty(\mathbb R^d \times \mathbb R)$), we have for some $C>0$ independent of $u,v$,
$$
\|\nabla(u|u|^2)-\nabla(v|v|^2)\|\_{L^1\_TL^2\_x}\\
\leq C(\||\nabla(u-v)| (|u|^2+|v|^2)\|\_{L^1\_TL^2\_x} \\
+ \| |u-v|(|u|+|v|)(|\nabla u|+|\nabla v|) \|\_{L^1\_TL^2\_x}).
$$


I do not understand why this inequality holds. **How to prove this inequality?**


I proved a similar inequality regarding $u$ rather than $\nabla u$ by showing that for complex numbers $u,v$, $|u|u|^2-v|v|^2|\leq C |u-v|(|u^2|+|v^2|)$, but it seems to be really hard to extend this to $\nabla(u|u|^2)$.



An inequality of spacetime Banach space for non-linear Schrodinger equation
It is sufficient to write
$$u^2\bar{u}-v^2\bar{v}=(u-v)u\bar{u}+(u-v)v\bar{u}+(\bar{u}-\bar{v})v^2$$
and then differentiate each term.


