The max-entry norm is not submultiplicative

Idea. $\|\cdot\|_{\max}$ is the sup-norm on $\mathbb{R}^{n^2}$, so it is automatically a norm; but multiplication of matrices adds up $n$ products, which can exceed the product of the maxima.

Step 1: it is a norm.
- Definiteness: $\max|a_{ij}|=0$ iff all entries vanish.
- Homogeneity: $\max|\lambda a_{ij}|=|\lambda|\max|a_{ij}|$.
- Triangle inequality: $|a_{ij}+b_{ij}|\le|a_{ij}|+|b_{ij}|\le\|A\|_{\max}+\|B\|_{\max}$ for every $(i,j)$.

Step 2: the counterexample. Let $J$ be the matrix whose entries are all equal to $1$. Then $\|J\|_{\max}=1$ and
$$(J^2)_{ij}=\sum_{k=1}^n1\cdot1=n,\quad\text{so}\quad J^2=nJ,\quad\|J^2\|_{\max}=n.$$
Since $n\ge2$, $\|J^2\|_{\max}=n>1=\|J\|_{\max}^2$.

Fix. The rescaled norm $n\|A\|_{\max}$ is submultiplicative: $n|(AB)_{ij}|\le n\sum_k|a_{ik}||b_{kj}|\le n\cdot n\|A\|_{\max}\|B\|_{\max}$, i.e. $\|AB\|'\le\|A\|'\|B\|'$ for $\|A\|'=n\|A\|_{\max}$.

Remark. For $n=1$ the statement fails (the norm is $|a|$, which is multiplicative). Being submultiplicative is exactly what is needed to write $\|A^k\|\le\|A\|^k$ and to use geometric series.