Norm 2 against norm inf

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.

• What is $n$ here? – Botond May 21 at 15:09
• The dimension. (I edited the question.) – sjulliot May 21 at 15:31
• How do you choose a random vector from $\Bbb{R}^n$? – Hume2 May 22 at 9:26

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