Which entry-wise $L_{p,q}$ matrix norms satisfy $\|\mathbf{A x} \|_l \le \|\mathbf{A}\|_{p,q} \|\mathbf{x} \|_l$? The following submultiplicative inequality holds when, e.g. $p=q=l=2$.
$$\|\mathbf{A x} \|_l  \le \|\mathbf{A}\|_{p,q} \|\mathbf{x} \|_l.$$
Here, $\mathbf{A}$ is a real square matrix, $\mathbf{x}$ is a real vector and $\|\mathbf{A}\|_{p,q}$ is the "entry-wise" $L_{p,q}$ matrix norm, not the induced norm.
Is there a general expression for $p, q$ and $l$ when the above inequality is true?
 A: Here is at least a generalisation of your quoted result.
Let $c^i$ be the $i$th column of $A$, and let $y_i = \|c^i\|_l$, and $x'_i=|x_i|$. Then $\|y\|_k = \|A\|_{l,k}$, $\|x'\|_l=\|x\|_l$,and $Ax = \sum_i c^i x_i$, so
$$ \|Ax\|_l =\left\|\sum_i c^i x_i \right\|_l\le \sum_i |x_i|\|c^i\|_l = \langle x',y\rangle\le \|y\|_k \|x'\|_l = \|A\|_{l,k} \|x\|_l,$$
Where $k$ is the Holder conjugate of $l$, $\frac1l + \frac1k=1$. Also, using Minkowski's inequality like here, you should be able to show that $\|A\|_{p,q} \le \|A\|_{q,p}$ whenever $ q\ge p$. In addition, the $p$ norms are ordered in the sense that
$$ p\ge q \implies \|x\|_p \le \|x\|_q$$
So we have the following. For any $l$, $\|Ax\|_l \le \|A\|_{p,q}\|x\|_l$ for any $p,q$ in the rectangle $$1\le p\le l,   1\le q \le k$$
For $1\le l\le 2$, $k\geq2\ge l$ so by Minkowski we also have all the exponents in the 'mirror' rectangle
$$1\le p\le k,   1\le q \le l$$
I haven't thought about if this is optimal.

I'll leave this less useful calculation below. It proceeds by splitting into rows, which is not quite the definition of the $L^{l,k}$ norm.
Let $r^i$ be the $i$th row of $A$ i.e. $r^i_j = a_{ij}$. Then $(Ax)_i = r^i \cdot x$,
$$ \|Ax\|^l_l = \sum_i |r^i\cdot x|^l \leq \sum_i \|r^i\|^l_{k}\|x\|^l_l,$$
so that
$$\|Ax\|_l \leq \|A^{\top}\|_{{k,l}} \|x\|_l,$$
where again $k$ is the Holder conjugate.
