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

Question

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 $

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

Thank you, I understood your proofs.

But, why mine is not correct? I used Cauchy–Schwarz inequality

$||Ax||^2_{n}=(\sum\limits_{i,j=1}^n a_{i,j} x_j)^2 = \sum\limits_{i=1}^n (\sum\limits_{j=1}^n a_{i,j} x_j)^2 \le \sum\limits_{i=1}^n (\sum\limits_{j=1}^n a^2_{i,j})(\sum\limits_{j=1}^n x^2_j)= M ||x||^2_n $

now it's better?