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How could you show that:

$$\|x\|_\infty \le \|x\|_2 \le \sqrt{n} \|x\|_\infty. $$

I was able to show the left hand side but got stuck showing the right hand side. What would be the best way to approach it?

For the LHS: $$\|x\|_\infty = \max\limits_{j}|x_j| \le \sqrt{\sum_i {x_i^2}} = \|x\|_2 $$.

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  • $\begingroup$ Isn't the infinity norm the maximum absolute value of the entries of the vector? There seems to be some confusion surrounding your last line. Besides, the right hand side is answered below. $\endgroup$
    – Sarvesh Ravichandran Iyer
    Commented Feb 22, 2018 at 23:02
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    $\begingroup$ Note that your formula for $\|x\|_2$ on the last line was incorrectly written. You can fix this by editing the question. (One of us could fix it, of course, but I think it's better to have corrections of that kind made by the original author of the question.) $\endgroup$
    – David K
    Commented Feb 22, 2018 at 23:14

1 Answer 1

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For the RHS (since, for all $i$, $|x_i|\leq \sup_j |x_j|:=\|x\|_{\infty} \ \Rightarrow \ x_i^2 \leq \|x\|_{\infty}^2$) : \begin{align*} \|x\|_2=\sqrt{\sum_i x_i^2}& \leq \sqrt{\sum_i \|x\|_{\infty}^2} \\ & =\sqrt{n \|x\|_{\infty}^2}=\sqrt{n} \|x\|_{\infty} \end{align*}

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  • $\begingroup$ @John Rolding It is related with the $n$! In the sum, you replace each $x_i^2$ by $\|x\|_{\infty}^2$... : maybe it will be more easy to see it with the following notation : $\|x\|_2=\sqrt{x_1^2+\cdots +x_n^2}\leq\sqrt{\underbrace{\|x\|_{\infty}^2+\cdots +\|x\|_{\infty}^2}_{n \text{ times } \|x\|_{\infty}^2}} = \sqrt{n\|x\|_{\infty}^2 } $ $\endgroup$
    – Netchaiev
    Commented Feb 22, 2018 at 23:37

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