$\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ is always bounded Given the matrix $A= (a_{i,j}) \in M_{n,n}$
$||A||=\sup\limits_{x\in X}\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ where  $|| $ . $|| _n$ is $ R^N$ norm
$\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ is always bounded:
$||Ax||^2_{n}=(\sum\limits_{i,j=0}^n a_{i,j} x_j)^2\le\sum\limits_{i,j=0}^{n} a^2_{i,j} x^2_j
=(\sum\limits_{i,j=0}^n a^2_{i,j})(\sum\limits_{j=0}^n x^2_j)= M ||x||^2_n $
where $M=(\sum\limits_{i,j=0}^n a^2_{i,j})=||A||^2_n$
Is a correct proof?
Maybe:
$||Ax||^2_{n}=\sum\limits_{i}^n(\sum\limits_{j}^n a_{i,j} x_j)^2 \le\sum\limits_{i}^n(\sum\limits_{j}^n a^2_{i,j} x^2_j)
=\sum\limits_{i}^n(\sum\limits_{j}^n a^2_{i,j})(\sum\limits_{j=0}^n x^2_j)= M ||x||^2_n $

We have
$$
\left\{ \frac{\|\mathcal Ax\|_{n}}{\|x\|_n} : x \in \mathbb R^n \right\}
=
\left\{ \|\mathcal Ax\|_{n} : x \in \mathbb R^n, \|x\|_n=1\right\}
$$
The set on the right is compact because $\mathcal A$ and norm are continuous and the unit sphere is compact. Hence, the set on the left is bounded.
