2
$\begingroup$

Let $a$ be a vector in $R^m$. Under which condition the following inequality is always true: $$ \sqrt m \|a\|_{\infty}\leq \|a\|_2^2-\frac{\left(\sum_{i=1}^ma_i\right)^2}{m}. $$

  • 0
    Why do you ask? Almost never for $m=1$.2012-03-23
  • 1
    I don't see where the question is coming from either. But for $m=2$ there is already a significant set where it holds, namely $$\{(x,-x):\ x\geq1\}$$2012-03-23
  • 0
    Thank you. But I am lookimg for some proprrty (dilation, translatiom...) of the vrctor $a$ that inequality would be true.2012-03-23
  • 0
    Nick, the above question was asked by curiosity.. Maybe you were just wondering?2012-03-24

1 Answers 1

0

There is little hope to find a full answer, in particular this is no such answer, but rather a discussion in the hope it might trig some idea to someone else...

Let us choose $x=(1,1,\ldots,1)$ and consider the Cauchy-Schwarz inequality, $$\left|\sum x_na_n\right|^2\leq m \sum |a_n|^2$$
This tells us not only that $$0\le \sum|a_n|^2 -\frac{\left|\sum a_n\right|^2}{m}$$ but also that we have equality when $a=\lambda x$, hence in order to have an inequality, like the one in the OP, we need to spot some condition like ``$a$ must be outside of some skew-cone neighbourhood (to be found/defined) of the line $\lambda x$, $\lambda\in\mathbb{R}$''.

If we look at the case $\mathbb{R}^2$. Consider $a=(x,y)$ where $x, then $$ \sqrt{2}\|a\|_\infty\leq \|a\|_2^2-\frac{(x+y)^2}{2} $$ simplifies to $$2\sqrt{2}|y|\leq (y-x)^2$$ and then $$x\leq y - 2^{3/4}\sqrt{|y|}.$$

After this one might wish to see some repetition for the $\mathbb{R}^3$ case - but unfortunately that is more a more delicate problem, ending up with $$3^{3/2}\max(|x|,|y|,|z|)\leq (x-y)^2 + (x-z)^2+(y-z)^2$$