Recall that for a given norm $\|\cdot\|$ on $\mathbb{R}^n$, the dual norm is defined as a function $\|\cdot\|_*: \mathbb{R}^n \rightarrow \mathbb{R}$ with:
$\|y\|_* = \max \limits_x \{x^Ty: \|x\|\le1 \}$
Or equivalently:
$\|y\|_* = \max \limits_x \{\frac{x^Ty}{\|x\|}: \|x\| \neq 0 \}$
It trivially follows form the definition that:
$\frac{x^Ty}{\|x\|} \le \max \limits_x \frac{x^Ty}{\|x\|} = \|y\|_*$
for $\|x\| \neq 0$. Then we obtain something like a generalized version of Cauchy-Schwarz:
$x^Ty \le \|x\| \| y\|_*$
From this we can derive that $\|\cdot\|_{\infty}$ is the dual norm of $\|\cdot\|_1$ (and vice versa).
Let's talk about matrix norms now which is a particular case (correct me if I am wrong) of an operator norm. So really just to remind you, given two normed vector spaces $U$ and $V$ over (in my case) $\mathbb{R}$, a continuos linear operator $A: U \rightarrow V$ is defined as:
$\|Au\|_V \le C\|u\|_U \forall u \: \in U$
Then the norm of such an operator is given as:
$\|A\| = \min \limits_{C \ge 0} \{\|Au\|_V \le C\|u\|_U \forall u \: \in U \}$
where $\|\cdot\|_U$ and $\|\cdot\|_V$ are norms associated with vector spaces $U$ and $V$, respectively.
Dual norm is a special of an operator norm. I believe that alternatively we can write:
$\|A\| = \max \limits_{u \neq 0} \frac{\|Au\|_V}{\|u\|_U}$
So I would think that kind of the same inequality will hold:
$\|Au\|_V \le \|u\|_U \|A\|$
I would like to use this inequality to show that matrix L1 norm is a dual of matrix infinity norm (and vice versa). So I shall choose the norm of $U$ as L1 norm but what I should choose as a norm of $V$? And surely in this case $u$ are matrices - should we have something like $u \in \mathbb{R}^{n \times n}$?