# Kato's Inequality with Laplacian

Let $u \in C^2(\mathbb R^n)$ be complex-valued. I am trying to prove the inequality

$$\Delta |u| \geq \Re\left(\frac{\bar u}{u} \Delta u\right),$$

in the distributional sense. I tried calculating both sides explicitly in terms of the real and imaginary parts of $u$, but the expressions I get don't seem to be directly comparable. Is there a simple proof?

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This inequality is mentioned in Exercise 8 of Chapter 7 of Lieb-Loss Analysis. Have you tried looking there? – Giuseppe Negro Dec 29 '12 at 23:15
please post the proof for those who don't have the book – Koushik Dec 30 '12 at 0:46
I feel that the "right" form of this inequality is $\Delta \sup_\alpha u_\alpha\ge \sup_\alpha \Delta u_\alpha$, for general families of functions. (Some mild technical assumptions will be needed, I expect). Then the result in this post would be obtained from $|u|=\sup_{\alpha\in[0,2\pi]} \Re e^{i\alpha }u$. Informal reasoning: Laplacian is akin to a convexity measure, and taking supremum produces a "more convex" function that the original ones. A limit formula like $\Delta u(x) = \frac{n+2}{2n}\,\lim_{r\to 0}\frac{1}{r^{n+2}} \int_{|y-x|\le r} (u(y)-u(x))\,dy$ might help. – user53153 Dec 30 '12 at 3:03

There is almost certainly a beautiful proof that generalizes this inequality to elliptic operators and makes use of some nice, general properties, but here is an elementary proof using direct calculation: Set $|u| = (u \bar{u})^{1/2}$, and calculate componentwise:

\begin{align*} \Delta |u| & = \sum_i \frac{1}{2} |u|^{-1} D^2_{x_i}|u| - \frac{1}{4} |u|^{-3} [D_{x_i} |u|]^2 \\ & = \sum_i \frac{1}{2} |u|^{-1} (\bar{u} u_{x_i x_i} + 2 u_{x_i} \bar{u}_{x_i} + u \bar{u}_{x_i x_i}) - \frac{1}{4} |u|^{-3} (u^2 \bar{u}_{x_i}^2 + 2|u|^2 u_{x_i} \bar{u}_{x_i} + \bar{u}^2 u_{x_i}^2 ) \\ & = \frac{1}{2} |u|^{-1} (\bar{u} \Delta u + u \Delta \bar{u}) - \frac{1}{4} |u|^{-3} \sum_i (u \bar{u}_{x_i} - u_{x_i} \bar{u})^2 \end{align*}

Now we claim that each term $u \bar{u}_{x_i} - u_{x_i} \bar{u}$ is purely imaginary. One can check this easily (write $u = f + ig$ with $f,g$ purely real). Therefore the square of such a term is nonpositive, so in the above inequality we are subtracting a nonpositive term. Thus we have

\begin{align*} \Delta |u| & \geq \frac{1}{2} |u|^{-1} (\bar{u} \Delta u + u \Delta \bar{u}) \\ & = \Re\left[ \frac{\bar{u}}{|u|} \Delta u \right]. \end{align*}

Edit: This differs from the inequality you stated above, but as far as I can tell, this must be the correct version of the inequality, because if $u$ is purely real, then a direct calculation of $\Delta |u|$ yields

$$\Delta |u| = \frac{u}{|u|} \Delta u$$

whereas in the inequality you stated, were it true, then for $u$ purely real, we would have $\Delta |u| \geq \Delta u$, which is false if $u(x) < 0$ but $\Delta u(x) > 0$.

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I have actually made a mistake here. Let me attempt to fix it. – Christopher A. Wong Dec 30 '12 at 17:06