Prove $||A||_{2,\infty} = \max_{1\le i \le m}\sqrt{\sum_{j=1}^{n} |a_{ij}|^{2}}$ Let $||.||_{2, \infty}$ denote an induced matrix norm from $L_{2}$ to $L_{\infty}$ and $A \in \mathbb{R}^{m \times n}$. I have to prove that
$$||A||_{2,\infty} = \max_{1\le i \le m}\sqrt{\displaystyle \sum_{j=1}^{n} |a_{ij}|^{2}}$$ by showing both $||A||_{2,\infty} \le \max_{1\le i \le m}\sqrt{\displaystyle \sum_{j=1}^{n}  |a_{ij}|^{2}}$ and $||A||_{2,\infty} \ge \max_{1\le i \le m}\sqrt{\displaystyle \sum_{j=1}^{n} |a_{ij}|^{2}}$.
I am completely stuck in the first part so far, below is what I have
$$
\begin{align*}
||Ax||_{\infty} &\le ||Ax||_{2} \\
&= ||\sum_{j=1}^{n} a_{j} x_{j}||_{2}\\
&\le  \sum_{j=1}^{n} |x_{j}| ||a_j||_{2}\\
&= \sum_{j=1}^{n} |x_{j}| \sqrt{\sum_{i=1}^{m}|a_{ij}|^{2}}\\
&\le ||x||_{\infty}\sum_{j=1}^{n} \sqrt{\sum_{i=1}^{m}|a_{ij}|^{2}}\\
&\le ||x||_{2} \sum_{j=1}^{n}\sqrt{\sum_{i=1}^{m}|a_{ij}|^{2}}\\
\end{align*}
$$
It isn't quite close to $||A||_{2,\infty} \le \max_{1\le i \le m}\sqrt{\displaystyle \sum_{j=1}^{n}  |a_{ij}|^{2}}$. Any help will be appreciated.
 A: You used $\|Ax\|_\infty \le\|Ax\|_2$ at the beginning. Though mathematically correct, it also means that at the end you can only estimates $\|A\|_{2, 2}$ instead of $\|A\|_{2,\infty}$.
So don't do that. Instead,
\begin{align*}
\|Ax\|_\infty &= \max _{1\le i\le m} \left| \sum_{j=1}^n A_{ij} x_j\right| \\
&\le \max _{1\le i\le m} \sqrt{\sum_{j=1}^n A_{ij}^2} \sqrt{\sum_{j=1}^n |x_j|^2} \\
&= \left(\max _{1\le i\le m} \sqrt{\sum_{j=1}^n A_{ij}^2}\right) \ \|x\|_2
\end{align*}
where we use Cauchy-Schwarz. This already shows that
$$\|A\|_{2, \infty} \le  \max _{1\le i\le m} \sqrt{\sum_{j=1}^n A_{ij}^2}$$
and the equality follows essentially from the equality case of Cauchy-Schwarz: for any $i_0$, let $x = (A_{i_01}, A_{i_02} , \cdots, A_{i_0n})$. Then
\begin{align}
\|Ax\|_{\infty} &= \max _{1\le i\le m} \left| \sum_{j=1}^n A_{ij} x_j\right|  \\
&\ge \left| \sum_{j=1}^n A_{i_0j} x_j\right|\\
&= \sum_{j=1}^n A_{i_0j}^2\\
&= \sqrt{\sum_{j=1}^n A_{i_0j}^2}\sqrt{\sum_{j=1}^n A_{i_0j}^2} \\
&= \sqrt{\sum_{j=1}^n A_{i_0j}^2} \ \|x\|_2. 
\end{align}
Since this holds for all $i_0$, pick such $i_0$ so that
$$\sqrt{\sum_{j=1}^n A_{i_0j}^2} = \max_{1\le i\le m} \sqrt{\sum_{j=1}^n A_{ij}^2}$$
Then one has
$$ \|A\|_{2, \infty} \|x\|_2 \ge \|Ax\|_{\infty} \ge \max_{1\le i\le m} \sqrt{\sum_{j=1}^n A_{ij}^2} \|x\|_2,$$
which implies
$$\|A\|_{2, \infty}  \ge\max_{1\le i\le m} \sqrt{\sum_{j=1}^n A_{ij}^2}.$$
A: Using Cauchy's inequality on the rows $\mathbf{a}_i$ of $A$, $$\|A\mathbf{x}\|_\infty=\max(|\mathbf{a}_1\cdot\mathbf{x}|,\ldots,|\mathbf{a}_m\cdot\mathbf{x}|)\le\max_i\|\mathbf{a}_i\|_2\|\mathbf{x}\|_2$$ as required. All that remains is to find one vector where equality holds, for example using $\mathbf{x}=\mathbf{a}_k$, the row that has the maximum 2-norm.

Note: $\max_i\|\mathbf{a}_i\|_2=\max_{1\le i\le m}\sqrt{\sum_{j=1}^n|a_{ij}|^2}$.
