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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?

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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

(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 $C=1$.

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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

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