For a matrix $A \in \mathbb{R}^{n\times n}$ (if it matters, is invertible) and a vector $x \in \mathbb{R}^{n}$, do we have in general that $\Vert Ax \Vert_{\infty} \leq \Vert A\Vert_{\infty}\Vert x\Vert_{\infty}$? From this page I think the answer is no for entry-wise matrix norm and yes for the induced matrix norm.
I came upon a paper that suggests this inequality holds (without specifying whether the matrix norm used is the induced norm or the entry-wise norm) to prove a lemma but skipped some steps. The vector $x$ here is in the interior of the norm ball $\mathcal{B} = \{v \in \mathbb{R}^n \,\vert \, \Vert v \Vert_{\infty} \leq 1\}$.
$$A=\begin{bmatrix} 1 & 1 \\ 1 &1 \end{bmatrix}$$ And $x= \begin{bmatrix} 1 \ 1 \end{bmatrix}$$
Show that this is not true for the entrywise matrix norm.
The definition of the operator norm is $\| A \|_\infty$ is the smallest number such that for all $x$ you have $$\Vert Ax \Vert_{\infty} \leq \Vert A\Vert_{\infty}\Vert x\Vert_{\infty}$$ so yes, the inequality holds for the operator norm. When written in this inequality form, it usually means the operator norm.