What is the relation between $\|AB\|_1$ and $\|A\|_1 \|B\|_1$?

Question

For a matrix $A \in \mathbb{R}^{k \times m}$ let

$$\|A\|_1 = \sum\limits_{i=1}^k\sum\limits_{j=1}^m|A_{ij}|$$

be the sum of the absolute values of entries of $A$.

1. Given two matrices $A \in \mathbb{R}^{k \times m}$ and $B \in \mathbb{R}^{m \times n}$, what is the relation between $\|AB\|_1$ and $\|A\|_1 \|B\|_1$?

2. It is known that for the Frobenius norm we have that $$\|AB\|_F \leq \|A\|_F \|B\|_F$$ Does the same inequality also hold for the $\|\cdot\|_1$ norm?

My guess is that there is some $c$ depending on $k,m,n$ such that $\|AB\|_1 \leq c\|A\|_1 \|B\|_1$.

Answer 1

We indeed have $\|AB\|_1 \leq \|A\|_1 \|B\|_1$. In particular, we have a sort of "Hölder's inequality" with $\|A\|_{\infty} = \max_{i,j} |A_{ij}|$: noting that $\|A\|_\infty \leq \|A\|_1$, we have $$ \|AB\|_1 \leq \|A\|_1 \cdot \|B\|_\infty \leq \|A\|_1 \cdot \|B\|_1. $$ In fact, $c = 1$ is the lowest number for which we have $\|AB\|_1 \leq c\|A\|_1 \|B\|_1$ for all $k,m,n$, as can be demonstrated by taking $A$ and $B$ to each be the matrix with a $1$ in the upper-left corner and zeros in all other positions.