Let $f,g:\mathbb{R}^{n\times n}\to \mathbb{R}$ be given by
$$\begin{aligned} f(A) &= \max_{i=1,...,n}i\cdot \sum_{j=1}^{n}|a_{ij}|,\\ g(A) &= n\cdot \max_{i,j=1,...,n}|a_{ij}| \end{aligned}$$
Is g submultiplicative? Are $f,g$ compatible with $\| \cdot \|_1$, given by $\|x\|_1 = \sum_{i=1}^{n}|x_i|$?
How does one go about showing whether these are true or false? Every counterexample I try out seems to fail, even though I know that $g$ is just a variant of the so-called maximum-norm (which is not submultiplicative), still, that doesn't seem to help much.
Thank you very much in advance, I would very much appreciate help with this.
In fact, $g$ is submultiplicative. For $a,b \in \Bbb R^n$, note that $$ |a^Tb| \leq \|a\|_\infty \cdot \|b\|_1 \leq n \cdot \|a\|_\infty \cdot \|b\|_\infty. $$ Now, if $a_i$ is the $i$th row of $A$ and $b_j$ is the $j$th column of $B$, we have $$ n \cdot |(AB)_{i,j}| = n \cdot |a_i^T b_j| \leq (n \cdot \|a_i\|_\infty) \cdot (n \cdot \|b_j\|_\infty) \leq g(A)\cdot g(B). $$ Conclude that we indeed have $g(AB) \leq g(A)g(B)$.
As for compatibility: it should be easy to find a counterexample for $g$. I suspect that $f$ will be compatible; it is probably useful to note that $f(A) = \|DA\|_\infty$. Here, $\|\cdot \|_\infty$ denotes the operator norm induced by the $\infty$-norm, which is to say that $\|M\|_\infty$ is the maximal absolute row-sum. $D$ is the diagonal matrix matrix $D = \operatorname{diag}(1,2,\dots,n)$.