Let $A\in \mathbb{C}^{m,n}$ be a matrix. Show that that $\|A\|_\infty \le \sqrt{n}\|A\|_2$. I'm wondering, however, if this statement is actually true at all. Could it be that the $\|\cdot\|_\infty$ and $\|\cdot\|_2$ were misplaced?
On a different note, we know that $\|Ax\|_\infty \le \|Ax\|_2$, which immediately implies that $\|Ax\|_\infty \le \sqrt{n}\|Ax\|_2$, so that $\|A\|_\infty \le \sqrt{n}\|A\|_2$. Is this correct?
If $y\in\mathbb R^m$ then $\|y\|_\infty \le \sqrt{y_1^2+\ldots+y_m^2}$ and therefore, $\|Ax\|_\infty \le \|Ax\|_2$ for all $x\in\mathbb R^n$.
Moreover, $\|Ax\|_2\le \|A\|_2\|x\|_2\le \sqrt{n}\|A\|_2\|x\|_\infty$. Hence, if $x\ne 0$, $$\frac{\|Ax\|_\infty}{\|x\|_\infty}\le \sqrt{n}\|A\|_2.$$ Taking the $\sup_{x\ne 0}$ on the LHS yields the result.