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

Since you are using double sums, I highly advise you to use two summation signs. For example, the $i$-th component of $Ax$ equals $(Ax)_i = \displaystyle \sum_{j=0}^n a_{ij}x_j$, so the value of $||Ax||^2$ is $\displaystyle\sum_{i=0}^n(Ax)_i^2$ which equals
$$\sum_{i=0}^n \left(\sum_{j=0}^n a_{ij}x_j\right)^2$$
To prove what you need, I advise you use the Cauchy-Schwarz inequality, since each element of the vector $Ax$ is actually an inner product of one row of $A$ with the vector $x$.
Alternatively, you can prove that $||Ax||$ is bounded on $S=\{x \mid \lVert x\rVert = 1\}$ (because it is a continuous function on a compact set) and, since $$\frac{||Ax||}{||x||} = \left|\left|A\left(\frac{x}{||x||}\right)\right|\right|,$$ each value of $\frac{||Ax||}{||x||}$ is equal to some value of $Ay$ for $y\in S$, so it is bounded.
