# How to show that $p-$Laplacian operator is monotone?

Define $$\langle -\Delta_p u, v \rangle_{(W^{1,p})', W^{1,p}} = \int_{\Omega}|\nabla u |^{p-2}\nabla u \nabla v.$$

How do I show that this operator is monotone? I get $$\langle-\Delta_p u_1 - \Delta_p u_2, u_1-u_2\rangle = \int(\nabla u_1 - \nabla u_2)(|\nabla u_1|^{p-2}\nabla u_1 - |\nabla u_2|^{p-2}\nabla u_2)$$ Now adding and subtracting $|\nabla u_1|^{p-2}\nabla u_2$ in the brackets still doesn't help with one of the terms..

(Recall an operator is monotone if $\langle Tx - Ty, x-y\rangle \geq 0.$)

• Hint: If you expand out the values under the integral you should be able to use something of the form $a^{2} + b^{2} \geq 2ab$, by noting that $\nabla u_{1} \nabla u_{2} = \nabla u_{2} \nabla u_{1}$. I think that this is enough, but just had a quick look before leaving – Keeran Brabazon Sep 20 '13 at 10:54
• @aere Out of interest why are you denoting as -$\Delta_{p}$ and not $\Delta_{p}$? – user100431 Dec 21 '14 at 21:13

We have, following Keeran's idea from the comments - for $p \ge 2$: $\def\sp#1{\left\langle#1\right\rangle}\def\abs#1{\left|#1\right|}\def\Lp{\Delta_p}\def\np#1{\abs{\nabla #1}^{p-2}}$ \begin{align*} \sp{-\Lp u_1 + \Lp u_2, u_1 - u_2} &= \int_\Omega \bigl(\np{u_1}\nabla u_1 - \np{u_2}\nabla u_2\bigr)(\nabla u_1 - \nabla u_2)\\ &= \int_\Omega \abs{\nabla u_1}^p - \bigl(\np{u_1}+\np{u_2}\bigr)\nabla u_1\nabla u_2 + \abs{\nabla u_2}^p\\ &\ge \int_\Omega \abs{\nabla u_1}^p + \abs{\nabla u_2}^p - \frac 12\bigl(\np{u_1} + \np{u_2}\bigr)\bigl(\abs{\nabla u_1}^2 + \abs{\nabla u_2}^2\bigr)\\ \def\nn#1#2{\abs{\nabla #1}^{#2}} &= \frac 12 \int_\Omega \nn{u_1}p + \nn{u_2}p - \np{u_1}\nn{u_2}2 - \np{u_2}\nn{u_1}2\\ &= \frac 12 \int_\Omega \bigl(\np{u_1} - \np{u_2}\bigr)\bigl(\nn{u_1}2 - \nn{u_2}2\bigr)\\ &\ge 0. \end{align*} For the final inequality we used that $x\mapsto x^2$ and $x \mapsto x^{p-2}$ are monotonically increasing on $\mathbb R$.

Let $x,y\in \mathbb{R}^N$ and note that $$\tag{1}|x|^{p-2}x-|y|^{p-2}y=\int_0^1\frac{d}{dt}\left(|y+t(x-y)|^{p-2}(y+t(x-y))\right)dt$$

By calculating the derivative, we conclude from $(1)$ that $$\tag{2}|x|^{p-2}x-|y|^{p-2}y=(y-x)\int_0^1|y+t(y-x)|^{p-2}dt+\\+ (p-2)\int_0^1|y+t(y-x)^{p-4}|\langle y+t(y-x),x-y\rangle(y+t(y-x))dt$$

It follows from $(2)$ that $$\tag{3}\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle =|x-y|^2\int_0^1|y+t(y-x)|^{p-2}dt+\\ +(p-2)\int_0^1|y+t(y-x)|^{p-4}\langle y+t(y-x),x-y\rangle^2dt$$

If $p\geq 2$, then we conclude the positivity from $(3)$. If $p<2$ we note that

$$\int_0^1|y+t(y-x)|^{p-4}\langle y+t(y-x),x-y\rangle^2dt\leq|x-y|^2\int_0^1|y+y(y-x)|^{p-2}$$

which again implies the positivity. I would like to remark that in fact we have a more Strong inequality (which is used, for example, in showing that $-\Delta_p$ is $S_+$)

$$\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle \geq \left\{ \begin{array}{cc} c_p|x-y|^p &\mbox{ if p\geq 2} \\ c_p\frac{|x-y|^2}{(|x|+|y|)^{2-p}} &\mbox{if p<2} \end{array} \right.$$

where $c_p>0$ is a constant. For this strongly inequality see the notes of Ireneo Peral in the Appendix.

Although the answers given by other ones are good enough, I would like to give a very simple proof the inequality below: $$\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle\geq 0, \quad 1<p<\infty.$$ which can induce the desired result and only takes advantage of Young's inequality. By a direct computation, $$\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle=|x|^{p}+|y|^p-|x|^{p-2}x\cdot y -|y|^{p-2}y\cdot x,$$ By Young's inequality, $$||x|^{p-2}x\cdot y|\leq |x|^{p-1}|y|\leq \frac{|x|^p}{p'}+\frac{|y|^p}{p},$$ where $p'=\frac{p}{p-1}$. Similarly, $||y|^{p-2}y\cdot x|\leq \frac{|y|^p}{p'}+\frac{|x|^p}{p}$. Hence, $$-|x|^{p-2}x\cdot y -|y|^{p-2}y\cdot x\geq -|x|^p-|y|^p.$$ Hence, $$\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle\geq 0.$$ Of course, this inequality fails to be optimal but good enough to show that p-Laplacian is monotone.

• Interesting calculation. I think you should explicitly write that $$\langle |x|^{p-2}x-|y|^{p-2}y,x-y\rangle \ge 1/p|x|^p+1/p'|y|^p.$$ – Tomás Jun 10 '17 at 11:55