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We know from basic linear algebra that $\forall x \neq 0, \frac{||x||_2}{||x||_{\infty}} \leq \sqrt{n}$ (where $n$ is the dimension).We also know that the equality occurs if and only if all coordinates are equal.

When, on the contrary, all coordinates are $0$ except one, then $\frac{||x||_2}{||x||_{\infty}} = 1$.

It appears that the more distant the coordinates are, the smaller this ratio.

I am looking for an (in)equality linking $\frac{||x||_2}{||x||_{\infty}}$ with $\sigma(x)$ the standard deviation of the $x_i$, or another measure of how distant the coordinates are.

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  • $\begingroup$ What is $n$ here? $\endgroup$ – Botond May 21 at 15:09
  • $\begingroup$ The dimension. (I edited the question.) $\endgroup$ – sjulliot May 21 at 15:31
  • $\begingroup$ How do you choose a random vector from $\Bbb{R}^n$? $\endgroup$ – Hume2 May 22 at 9:26
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"It appears that the more distant the coordinates are, the smaller this ratio."

That is not really correct, you can try an almost identical case when all coordinates equal and non-zero except one that is zero. You will get $\frac{||x||_2}{||x||_{\infty}} = \sqrt{n-1}$.

In terms of connection with standard deviation, let $\mu=\frac{\sum\limits_ix_i}{n}$ than $\sigma^2=\frac{||x||_2^2}{n}-\mu^2$. So you can have for example: $\frac{||x||_2^2}{||x||_{\infty}^2}=n\left(\frac{\sigma^2}{||x||_{\infty}^2}-\frac{\mu^2}{||x||_{\infty}^2}\right)$, but $\mu^2\leq \frac{||x||_2^2}{n} $ (https://en.wikipedia.org/wiki/Generalized_mean) so:

$\frac{||x||_2^2}{||x||_{\infty}^2}\geq -\frac{||x||_2^2}{||x||_{\infty}^2}+n\frac{\sigma^2}{||x||_{\infty}^2}$

$\frac{||x||_2^2}{||x||_{\infty}^2}\geq n\frac{\sigma^2}{2||x||_{\infty}^2}$

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  • $\begingroup$ What is $\delta$ here ? Is it $\sigma$ ? Besides, shouldn't it be $||x||_2^2 = n(\sigma^2 + \mu^2)$ ? $\endgroup$ – sjulliot May 27 at 7:46

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