Proving $\|x\|_{\infty}\leq\|x\|_2\leq\sqrt{n}\|x\|_{\infty}$

Question

I am trying to prove the inequality $\|x\|_{\infty}\leq\|x\|_2\leq\sqrt{n}\|x\|_{\infty}.$ I have made an attempt below, but I am unsure of my proof for the second inequality (during the third line where I introduce the inequality sign).

\begin{align} \|x\|_{\infty}&=\max_{1\leq j\leq n} |x_j| \\ &=|x_p| \ \ \ \ \text{for some} \ 1\leq p\leq n \\ &=\left(|x_p|^2\right)^{1/2} \\ &\leq\left(|x_1|^2+|x_2|^2+..+|x_n|^2\right)^{1/2} \\ &=\|x\|_2 \\ \therefore \|x\|_{\infty}&\leq\|x\|_2\\ \\ \|x\|_2&=\left(|x_1|^2+|x_2|^2+..+|x_n|^2\right)^{1/2} \\ \|x\|_2^2&=|x_1|^2+|x_2|^2+..+|x_n|^2 \\ &\leq n\max_{1\leq j\leq n} |x_j|^2 \\ &=n\|x\|_{\infty}^2 \\ \therefore \|x\|_2&\leq\sqrt{n}\|x\|_{\infty} \end{align}

Would it be better notation to write $\left(\max_{1\leq j\leq n} |x_j|\right)^2$, as opposed to $\max_{1\leq j\leq n} |x_j|^2$?

Answer 1

Since $x \mapsto x^2$ is increasing for $x\geq 0$ you have, for $x,y\geq 0$, that $x\leq y \implies x^2\leq y^2$.

Also, $\max(|x_1|,...,|x_n|)^2=|x_p|^2 $ for some $p$. Now since for every $j$ you have $|x_j|\leq |x_p|$ then for every $j$ you also have $|x_j|^2\leq |x_p|^2$, so that $\max(|x_1|^2,...,|x_n|^2)=|x_p|^2 $.

Thus $\max(|x_1|^2,...,|x_n|^2)=\max(|x_1|,...,|x_n|)^2$.

To conclude, since since for every $j$ you have $|x_j|\leq |x_p|$, then $|x_1|^2+...+|x_n|^2\leq |x_p|^2+ ...+|x_p|^2=n\max(|x_1|,...,|x_n|)^2$.

Your notation suggests two different interpretations for $\max$ which result yielding the same value, so it shouldn't be a problem.