$\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ 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 $

I propose you here a way to show it regardless of the norm used:
Note that for every $x \neq 0$, holds $$\|A\frac{x}{\|x\|} \| = \frac{\|A\|}{\|x\|}$$ and thus
$$\|A\|:= \sup \left\{\frac{\|Ax\|}{\|x\|}: x \in \mathbb{R}^n\right\} = \sup \left\{\|A\frac{x}{\|x\|} \|: x \in \mathbb{R}^n\right\}= \sup \left\{\|Ax\|: \|x\|\leq 1, x \in \mathbb{R}^n\right\}.   $$
Now suppose by contradiction that $\|A\| $ is not bounded, then for ever $m \in \mathbb{N}$ there is some $x_m $ such that $\|x_m\| \leq 1$ and $\|Ax\| \geq m$. Doing this for every $m$ we get a sequence $\{x_m\}$ contained in the closed open unit ball $B = \{x : \|x\| \leq 1\}$. In particular $B$ is compact (use Riesz's Lemma and the fact $\mathbb{R}^n$ is of finite dimension). Hence we may extract a converging subsequence $x_{m_k}$ converging to some $x^*$ such that $\|x^*\| = 1$. Finally by continuity of the norm we get $$\|Ax^*\| = \lim_{m \to \infty} \|Ax_m\| \geq \lim_{m \to \infty} m = \infty.$$
This is a contradiction to the fact that $Ax^* \in \mathbb{R}^n$ (which is true since $A:\mathbb{R}^n \to \mathbb{R}^n$ and $x^* \in \mathbb{R}^n$).
Edit: As said by Ihf you may also directly conclude by saying that the function $x \mapsto \|Ax\|$ is continuous (composition of continuous applications) on the compact $B$ and thus reaches its maximum.
