Equivalence induced matrix norms! The equivalence of vector norms on finite dimensional spaces immediately implies that all induced matrix norms are equivalent. Theorem 1, here, gives some of the popular induced norms. I am interested in the $\|A\|_{2,1}$ and $\|A\|_{2,2}$ norm, where
\begin{equation*}
\|A\|_{2,1}=\sup\{\|A\mathbf{x}\|_1:\|\mathbf{x}\|_2=1\}
\end{equation*}
and
\begin{equation*}
\|A\|_{2,2}=\sup\{\|A\mathbf{x}\|_2:\|\mathbf{x}\|_2=1\}
\end{equation*}
According to my analysis so far; $\|A\|_{2,1}$ upper-bounds $\|A\|_{2,2}$. Can someone verify?
For a matrix $A\in\mathbb{R}^{n\times m}$
$\|A\|_{2,1}=\max_{u\in\{1,-1\}^n}\|A^Tu\|_2\leq \sqrt{\sum_{i=1}^m(\sum_{j=1}^n|a_{ij}|)^2}$. 
And 
$\|A\|_{2,2}=\lambda_{\max}(A^TA)=\sigma_{\max}(A)\leq\|A\|_F= \sqrt{\sum_{i=1}^m\sum_{j=1}^n|a_{ij}|^2}$, 
$\implies$
\begin{equation}
\|A\|_{2,2}^2\leq{\sum_{i=1}^m\sum_{j=1}^n|a_{ij}|^2}\leq{\sum_{i=1}^m\left(\sum_{j=1}^n|a_{ij}|\right)^2}
\end{equation}.
 A: Your inequalities all go in the same direction, so you cannot possibly prove inequalities both ways. 
You have the easy inequalities $$\|x\|_2\leq\|x\|_1\leq\sqrt n\,\|x\|_2.$$ So you immediately get (from the definition, not the characterization you quote)
$$\tag{1}\bbox[5px,border:2px solid green]{\|A\|_{2,2}\leq\|A\|_{2,1}\leq\sqrt n\|A\|_{2,2}.}
$$
These inequalities are sharp. For instance, if $A=E_{11}$, then 
$$
\|Ax\|_1=\|(x_1,0,\ldots,0)\|_1=|x_1|=\|x\|_2.
$$
So $\|A\|_{2,2}=\|A\|_{1,1}$. And if $A=I_m$, then $\|Ax\|_1=\|x\|_1$, so
$$
\|A\|_{2,1}=\max\{\|x\|_1:\ \|x\|_2=1\}=\sqrt n,
$$
while $\|A\|_{2,2}=1$. So $\|A\|_{2,1}=\sqrt n\|A\|_{2,2}$.
Edit: here is a short proof of the inequalities $(1)$, in case they are not obvious.
You have, for any nonzero $x$. 
$$
\frac{\|Ax\|_2}{\|x\|_2}\leq\frac{\|Ax\|_1}{\|x\|_2}\leq\|A\|_{2,1}.
$$
Now you take supremum (forget about the term on the middle) and you obtain
$$
\|A\|_{2,2}\leq\|A\|_{1,1}.
$$
Similarly, start with 
$$
\frac{\|Ax\|_1}{\|x\|_2}\leq\sqrt{n}\,\frac{\|Ax\|_2}{\|x\|_2}\leq\sqrt{n}\,\|A\|_{2,2}.
$$
Now forget the middle term and take supremum, to obtain
$$
\|A\|_{2,1}\leq\sqrt{n}\,\|A\|_{2,2}.
$$
