Let $u,v \in B := \{ u \in C([0,T]; L^1(\mathbb{R}^N) \cap L^\infty(\mathbb{R}^N)) \ : \nabla^\alpha u(t) \in L^1 \cap L^\infty \ \text{for all} \ |\alpha| \leq 2, \ \text{and} \ \ ||u||_{B} := ^{\ \ \ \text{sup}}_{0<t<T} (||u(t)||_{L^1 \cap L^\infty} + t^{1/2}||\nabla u(t)||_{L^1 \cap L^\infty} + t||\nabla^2 u(t)||_{L^1 \cap L^\infty} \leq R ) \}$
where $\alpha$ is a multi-index, $R > 0$. $||\cdot||_{L^1 \cap L^\infty} = ||\cdot||_{L^1} + ||\cdot||_{L^\infty}$
I would like to find a bound on the norm $|| \nabla(u^3 - v^3) ||_{L^1 \cap L^\infty}$, in terms of $|| \nabla(u - v) ||_{L^1 \cap L^\infty}$. That is, I would like to say $|| \nabla(u^3 - v^3) ||_{L^1 \cap L^\infty} \leq K|| \nabla(u - v) ||_{L^1 \cap L^\infty}$, for some constant $K$.
I was previously able to find a bound on $||(u^3 - v^3) ||_{L^1 \cap L^\infty}$ in terms of $||(u - v) ||_{L^1 \cap L^\infty}$. Using the fact that $f(a) - f(b) = \int^{1}_{0} \frac{d}{d\theta} f(\theta a + (1- \theta)b) \text{d}\theta$ and setting $f(x) = x^3$, I found that $u^3 - v^3 = \int^{1}_{0} \frac{d}{d\theta} f(\theta u^3 + (1- \theta)v^3) \text{d}\theta \leq \int^{1}_{0} 3||u| + |v||^2 (u-v) \text{d}\theta \leq 3(2R)^2(u-v)$ since $u,v \in B$ .
However, this is presumably of no use here, unless I can somehow bound $||\nabla u||_{L^1 \cap L^\infty}$ by $|| u||_{L^1 \cap L^\infty}$? I'm tempted to think that there is at least a similar technique that can be used here though. Any ideas?