Kato’s inequality

Lemma 1 Suppose that ${u\in H^1(\mathbb{R}^n)}$ satisfies ${-\Delta u\geq f}$ in the weak sense for some for some ${f \in L^{1}\left(\mathbb{R}^n\right)}$ . Then ${\Delta u_{-} \geqslant f \cdot \chi_{\{u<0\}}}$ in the weak sense. Here ${u_-(x)=\max\{0,-u(x)\}}$ .

Proof: For any ${\varepsilon>0}$ define ${u_{\varepsilon}=\sqrt{u^2+\varepsilon^2}}$ . Then it’s easy to see that ${u_{\varepsilon}}$ and ${\frac{u}{u_{\varepsilon}}}$ are in ${H_{lo c}^{1}\left(\mathbb{R}^n\right)}$ and that

$\displaystyle u_{\varepsilon} \rightarrow|u| \quad \text { in } H_{loc}^{1}\left(\mathbb{R}^n\right) \text { as } \varepsilon \rightarrow 0$

Now, let ${\varphi}$ be a non-negative function in ${C_0^{1}}$ . A straightforward computation shows that

$\displaystyle \nabla u_{\varepsilon} \cdot \nabla \varphi \leqslant \nabla u \cdot \nabla\left(\frac{u}{u_{\varepsilon}} \varphi\right) \text { a.e. }$

Set ${\varphi_{\varepsilon}=\frac{1}{2}\left(1-\frac{u}{u_\varepsilon}\right) \varphi}$ . It follows from the above inequality that

$\displaystyle \int \frac{1}{2} \nabla\left(u_{\varepsilon}-u\right) \cdot \nabla \varphi \leqslant-\int \nabla u \cdot \nabla \varphi_{\varepsilon}$

Note that ${\varphi_{\varepsilon}}$ is a non-negative function in ${H_0^{1} \cap L^{\infty}}$ . It follows from ${-\Delta u \geqslant f}$ that

$\displaystyle \int \nabla u \cdot \nabla \varphi_{\varepsilon} \geqslant \int f \cdot \varphi_{\varepsilon}$

One combines the last two inequalities to get

$\displaystyle \int \frac{1}{2} \nabla\left(u_{\varepsilon}-u\right) \cdot \nabla \varphi \leq-\int f \cdot \varphi_{\varepsilon}.$

Letting ${\varepsilon \rightarrow 0}$ , using the fact that ${0 \leq \varphi_{\varepsilon} \leq \varphi}$ and ${\varphi_{\varepsilon} \rightarrow \chi_{\{u<0\}} \varphi}$ . It implies that

$\displaystyle \int \nabla u_{-} \cdot \nabla \varphi \leqslant-\int f \chi_{\{u<0\}} \varphi$

This is equivalent to saying that ${\Delta u_{-} \geqslant f \cdot \chi_{\{u<0\}}}$ .

$\Box$

Lemma 2 Suppose that ${u\in H^1}$ satisfies ${-\Delta u \leqslant f}$ then

$\displaystyle -\Delta u_{+} \leqslant f \cdot \chi_{\{u>0\}}.$

In other words, if ${u}$ is a sub-solution, then ${u_+}$ is also a sub-solution.

Proof: The assumption is equivalent to ${-\Delta(-u) \geqslant-f}$ . Using the previous lemma, one obtains that

$\displaystyle \Delta(-u)_{-} \geqslant-f \cdot \chi_{\{-u<0\}}.$

Since ${(-u)_{-}=u_{+}}$ , then ${-\Delta u_{+} \leqslant f \cdot \chi_{\{u>0\}}}$

$\Box$

Now if ${-\Delta u=f}$ , then combine the previous two lemmas, we get the following Kato’s inequality

$\displaystyle -\Delta |u|\leq f\cdot \text{sgn} (u).$

Remark: S. Agmon, On positivity and decay of solutions of second order elliptic equations on Riemannian manifolds, in Methods of functional analysis and theory of elliptic equations (Naples, 1982), 19-52, Liguori, Naples, 1983.

lim Practice= Perfect