Prove $\frac{1}{\sqrt{n}}\|A\|_{\infty} \leq\|A\|_{2} \leq \sqrt{m}\|A\|_{\infty} $
Solution 1:
If $\|x\|_2 = 1$, then
$$\|Ax\|_2^2 = \sum_{i = 1}^m\left\lvert \sum_{j = 1}^n A_{ij}x_j\right\rvert^2 \le \sum_{i = 1}^m\left(\sum_{j = 1}^n |A_{ij}|^2\right) \le m\max_{1\le i \le m} \left(\sum_{j = 1}^n |A_{ij}|\right)^2 \le m \|A\|_\infty^2$$ Taking square roots, then taking the supremum over all $x$ with $\|x\|_2 = 1$, we obtain your right bound.