Letting $\epsilon\geq0$ be small, $a,b\in \mathbb{R}^N$ and $\ell >1$. Is the following inquality $$\tag{E}\label{E} \Big((\sqrt{\epsilon^2+|a|^2})^{\ell-2}a- (\sqrt{\epsilon^2+|b|^2})^{\ell-2}b\Big)(a-b)\geq 0, $$ true?

If $\epsilon=0$, \eqref{E} becomes $$(|a|^{\ell-2}a- |b|^{\ell-2}b\Big)(a-b)\geq 0, $$ which is true by using the following inequalities: \begin{align} \left(|a|^{\ell-2}a-|b|^{\ell-2}b\right)\cdot (a-b)\geq 2^{-\ell}|a-b|^2,&\quad \ell\geq 2,\\ \left(|a|^{\ell-2}a-|b|^{\ell-2}b\right)\cdot (a-b)\geq (\ell-1)|a-b|^2\left(|a|+|b|\right)^{\ell-2} ,&\quad 1<\ell \leq 2. \end{align} $a\cdot b$ denotes the usual inner product in $\mathbb{R}^N$.

Can any one know if \eqref{E} is true or not?

Here my try for the answer: Letting $\epsilon$ small. Denoting by $f$ the function defined as $$f(\mathbf{x})=\frac{1}{\ell}(\epsilon^2+|\mathbf{x}|^2)^{\ell/2}.$$ $f$ is convexe in $\mathbf{x}$. Then, by the convexity proprity we have $$(f'(\mathbf{x})-f'(\mathbf{y}))\cdot (\mathbf{x}-\mathbf{y})\geq c\|\mathbf{x}-\mathbf{y}\|^2.$$ Then \eqref{E} holds. I don't know there is a mistake or not.

  • $\begingroup$ Is the product taken to be the ordinary inner product? $\endgroup$ – I was suspended for talking Apr 25 '18 at 19:16
  • $\begingroup$ Yes, $a.b$ denotes the usual inner product in $\mathbb{R}^N$. $\endgroup$ – Achaire Apr 25 '18 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.