For $A \in \mathbb{R}^{m \times n}$m need to prove the following bounds:
\begin{align*} ||A|| &\leq \sqrt{m} \max_{i \in \{1, 2, \dots, m\}} \big( \sum_{j=1}^n A_{ij}^2 \big)^{1/2}\\ ||A|| &\geq \frac{1}{\sqrt{mn}} \sum_{i=1}^m | \sum_{j=1}^nA_{ij}| \end{align*}
where $||A||$ is the operator norm $||A||= max \{ ||Ax||_{l_2}: ||x||_{l_2}=1\}$
Attempt: For the lower bound: \begin{align*} ||A|| &= \max_{||x||=1} \{ \langle x, A^TAx \rangle \} \end{align*} cannot proceed beyond this
For the upper bound:
\begin{align*} ||Ax|| &= \sqrt{\sum_{i=1}^m\big(\sum_{j=1}^n a_{ij}x_j \big)^2} \\ &\leq \sqrt{\sum_{i=1}^m ( \sum_{j=1}^na_{ij}^2) ( \sum_{k=1}^nx_k^2 )}\\ &\leq \sqrt{m}\max_{i \in \{1, 2, \dots, m\}} \big( \sum_{j=1}^n A_{ij}^2 \big)^{1/2} \end{align*}