Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The function $\mathbb{R}^n\to \mathbb{R}$, $x\mapsto |x|^p$ (where $p>1$) is convex and thus the inequality $$|y|^p-|x|^p\ge p(y-x)\cdot x |x|^{p-2}$$ is valid. In some lecture notes of Peter Lindqvist, it is remarked that this inequality can be strengthened to $$|y|^p-|x|^p\ge p(y-x)\cdot x |x|^{p-2} + C(p) |y-x|^p$$ (of course $C(p)>0$) at least for $p>2$. Does anyone know a proof of this inequality?

share|improve this question
    
This is a two-dimensional problem. You may assume $x=(1,0)$, $y=(1+\alpha,\beta)$. Then $|y|^p-|x|^p=((1+\alpha)^2+\beta^2)^{p/2}-1$ and $p(y-x) \cdot x|x|^{p-2}=p\alpha$. –  Christian Blatter May 19 '11 at 16:03

1 Answer 1

(I'm expanding my comment.)

This is a two-dimensional problem. One may assume $x=(1,0)$, $y=(1+\alpha, \beta)$ with $\sqrt{\alpha^2+\beta^2}=:r$. Then $$|y|^p-|x|^p=((1+\alpha)^2 +\beta^2)^{p/2}-1\ ,\qquad p(y-x)\cdot x |x|^{p-2}=p\alpha\ .$$ It follows that we have to prove an inequality of the form $$(1+2\alpha+r^2)^{p/2}\geq 1 + p\alpha + Cr^p \qquad\qquad (1)$$ for a suitable $C$, and we may assume $p\geq2$.

Putting $r:=0$ in (1) the statement reads $(1+2\alpha)^{p/2}\geq 1 + p\alpha$, and this is true for $p\geq2$ by Bernoulli's inequality. Now the derivative of the left side of (1) with respect to $r$ is $${p\over 2}(1+2\alpha +r^2)^{p/2 -1}\ 2r\geq {p\over 2}r^{p-2}\ 2r=pr^{p-1}\ ,$$ and the derivative of the right side of (1) with respect to $r$ is $Cp r^{p-1}$. So if $C=1$ the left side of (1) grows faster with $r$ than the right side. It follows that (1) is true with $C=1$.

share|improve this answer
    
This argument does not quite work when $\alpha<-1/2$. I think this can be fixed somehow though. –  Florian May 26 '11 at 11:56
    
For $x:=(1,0)$, $y:=(-1,0)$, $p>2$ the inequality becomes $0\ge -2p + 2^p C$, which is not true for $C=1$. Thus a smaller $C$ (which has to depend on $p$) must be chosen. –  Florian May 26 '11 at 12:30
    
I solved the problem: The function $(r,\alpha)\mapsto r^{-p}((1+2\alpha+r^2)^{p/2}-1-p\alpha)$ for $\alpha\in(-r,r),r>0$ is positive and has the limits $\infty$ for $r\to 0$ and $1$ for $r\to\infty$ uniformly in $\alpha$ and is therefore bounded from below by a positive constant. –  Florian May 27 '11 at 9:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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