Let $u$ be a smooth function defined in a Riemannian manifold $(M, g)$. The well known Kato’s inequality states

kato’inequality application in Hilbert space

Theorem 1. $$
|\nabla| \nabla u |^{2} \leq\left|\nabla^{2} u\right|^{2}
$$


where $\nabla^{2}$ represents the Hessian operator of $M$.

Proof .

Where $\nabla u \neq 0$, this is simple calculus: The product rule tells us
$$
\nabla|\nabla u|=\frac{\nabla(g(\nabla u, \nabla u))}{2|\nabla u|}=\frac{g\left(\nabla^{2} u, \nabla u\right)}{|\nabla u|}
$$
Taking the squared norm and estimating the numerator with Cauchy-Schwarz gives us
$$
|\nabla| \nabla u |^{2} \leq \frac{\left|\nabla^{2} u\right|^{2}|\nabla u|^{2}}{|\nabla u|^{2}}=\left|\nabla^{2} u\right|^{2}
$$
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 $L_{\mathrm{loc}}^{\infty}$; so both sides of the proposed distributional inequality
$$
|\nabla| \nabla u |^{2} \leq\left|\nabla^{2} u\right|^{2}
$$
are in $L_{\mathrm{loc}}^{\infty}$ and thus it suffices to prove the inequality almost everywhere.
At points where $\nabla u \neq 0$, 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

.

\begin{remark}

The fact that a function is classically differentiable almost everywhere does not imply that this a.e. derivative is its distributional derivative (e.g. in the case of a step functino), and it’s usually the distributional derivative that we actually care about. Here, the full version of Rademacher (including the result $C^{0,1}=W^{1, \infty}$ ) tells us that the distributional and a.e. derivatives are one and the same; so after this point we don’t have to worry about it

\end{remark}

An application of Kato inequality is in the area of PDE

three is a striking properties of Schrödinger semigroups on $L_{1}\left(\mathbb{R}^{d}\right)$ (holomorphy and closedness of $-\Delta+V$, the test functions are a core) with the same elegant argument Kato
gave, but extending the results to possibly non-symmetric elliptic operators. In the second part,
we consider the Dirichlet problem
$(-\Delta+V) u=0,\left.\quad u\right|{\partial \Omega}=\varphi, \quad u \in C(\bar{\Omega})$ where $V \in L{\infty}(\Omega, \mathbb{R})$ and $\Omega$ is a bounded Wiener regular set.

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