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: $$\|x_j\|_\infty = \max|x_j| \le \sqrt{\sum_i {x_i^2}} = \|x\|_2 $$.
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*}