5
$\begingroup$

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 $\langle Tx - Ty, x-y\rangle \geq 0.$)

$\endgroup$
  • 3
    $\begingroup$ 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 $\endgroup$ – Keeran Brabazon Sep 20 '13 at 10:54
  • $\begingroup$ @aere Out of interest why are you denoting as -$\Delta_{p}$ and not $\Delta_{p}$? $\endgroup$ – user100431 Dec 21 '14 at 21:13
3
$\begingroup$

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 $\mathbb R$.

$\endgroup$
  • $\begingroup$ @martini Out of interest why are you denoting as -$\Delta_{p}$ and not $\Delta_{p}$? $\endgroup$ – user100431 Dec 21 '14 at 21:19
6
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.