# An inequality of spacetime Banach space for non-linear Schrodinger equation

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 $$\nabla(u|u|^2)$$.

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.