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.

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