Kato's inequality Let u be a smooth function defined in a Riemannian manifold $(M,g)$. The well known Kato's inequality states
$$|∇|∇u||^2≤|∇^2u|^2$$
where $∇^2$ represents the Hessian operator of $M$.
I would like ask if someone can shows me a short proof for this inequality or giving me a reference within I can find the proof.
Thank you for helping me
 A: Where $\nabla u \ne 0,$ this is simple calculus: The product rule tells us
$$\nabla |\nabla u| = \frac{\nabla (g(\nabla u,\nabla u))}{2 |\nabla u|}=\frac{g(\nabla^2u,\nabla u)}{|\nabla u|}.$$
Taking the squared norm and estimating the numerator with Cauchy-Schwarz gives us $$|\nabla|\nabla u||^2 \le \frac{|\nabla^2 u|^2 |\nabla u|^2}{|\nabla u|^2} = |\nabla^2 u|^2$$ as desired.
In general, we have to interpret things weakly: for an arbitrary smooth function $u,$ $|\nabla u|$ is only Lipschitz, so we must interpret $\nabla |\nabla u|$ as a weak/distributional derivative. Thankfully, we at least know (thanks to Rademacher) that this distribution is in fact in $L^\infty_\rm{loc};$ so both sides of the proposed distributional inequality $$|\nabla|\nabla u||^2\le |\nabla^2u|^2$$ are in $L^\infty_\rm{loc}$ and thus it suffices to prove the inequality almost everywhere. 
At points where $\nabla u \ne 0,$ the calculation I did at the start of my answer suffices. 
At a point where $|\nabla u| = 0$ and $|\nabla u|$ is classically differentiable, we know $|\nabla u|$ is at a local minimum and thus $\nabla |\nabla u| = 0,$ so the inequality holds since the RHS is manifestly non-negative. 
The set of points where $\nabla u=0$ and $|\nabla u|$ is not classically differentiable has measure zero, so we're done.
