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?
A convexity inequality
4
$\begingroup$
real-analysis
-
0This 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 $p(y-x) \cdot x|x|^{p-2}=p\alpha$. – 2011-05-19