## An inequality useful for the p-Laplacian

Fix $p>1$. Let $u,v\in{\mathbb R}^n$, then

$\displaystyle (u-v)\cdot(|u|^{p-2}u-|v|^{p-2}v)\geq 0.$

Proof: Unfold and rewrite what we want to show as:

$\displaystyle (|u|^{p-2}+|v|^{p-2})(u\cdot v)\leq |u|^{p}+|v|^{p}.$

By Cauchy-Schwarz the left hand-side is less than

$\displaystyle |u|^{p-1}|v|+|v|^{p-1}|u|,$

which can be written as

$\displaystyle |u|^{p}+|v|^{p}+(|u|-|v|)(|v|^{p-1}-|u|^{p-1}).$

But, since $p>1$,

$\displaystyle (|u|-|v|)(|v|^{p-1}-|u|^{p-1})\leq 0.$

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### 2 Responses to An inequality useful for the p-Laplacian

1. Leonid says:

A more general version: if $\Phi$ is a convex function on ${\mathbb R}^n$ and $F : {\mathbb R}^n\to{\mathbb R}^n$ is its gradient, then
$(u-v)\cdot (F(u)-F(v)) \ge 0$,
i.e., $F$ is monotone. With $F(u)=|u|^p$ you get the above. $p=1$ works too, if the gradient at 0 is defined as some vector in the subdifferential of $F$.

Indeed, the restriction of $\Phi$ to the line through $u$ and $v$ is a convex function of one real variable. Therefore, its derivative is nondecreasing, and this is exactly what the inequality says.

[PPC: I’ve edited the LaTeX, thanks. You forgot to write ‘latex’ after the dollar sign]

2. Leonid says:

I was vaguely hoping for the TeX markup to magically transform into formulas.