Difficulty in understanding a step in a definition in the book Walter Rudin In the book Principles of mathematical analysis by Walter Rudin,He writes:
"For $ A\in L(\Bbb R^n,\Bbb R^m)$, define the norm $||A||$ of $A$ to be sup of all numbers $|Ax|$, where $x$ ranges over all vectors in $\Bbb R^n$ with $|x|\le1$.
Observe that the inequality
$$ |Ax|\le||A|||x| $$
holds for all $ x\in \Bbb R^n$. Also, if $\lambda$ is such that $|Ax|\le \lambda|x| $ for all $ x\in \Bbb R^n $, then $||A|| \le \lambda $."
Here I am not able to understand the last line. where he says $||A|| \le \lambda$.
Since $ |Ax|\le||A|||x| $ and $|Ax|\le \lambda|x|$ both hold for all $ x\in \Bbb R^n $, then why necessarily $||A|| \le \lambda$ ?
 A: By definition $||A||$ is the supremum of all $|Ax|$ where $x$ ranges over the unit ball. Now assume that $|Ax| \leq \lambda|x|$ for all $x \in \mathbb{R}^n$. First observation is that $\lambda \geq 0$ since $|Ae_1| \geq 0$ and $|e_1| = 1$ (here $e_1$ is some vector with norm $1$). 
Let $x \in \mathbb{R}^n$ with $|x| \leq 1$. Then by definition $|Ax| \leq \lambda |x| \leq \lambda$ where the last relation holds because $|x| \leq 1$ and $\lambda \geq 0$. Hence $||A||$ is the supremum of a set where all the elements are less than $\lambda$, hence $||A|| \leq \lambda$.
A: This comes from an equivalent definition of the Norm $\| A \|$:
$$ \| A \| := \text{inf} \{\lambda \geq 0 | \forall x \in \mathbb{R}^n: |Ax| \leq \lambda |x| \} $$
A: If $|Ax|\leq \lambda|x|$ for all $x\in\mathbb{R}^n$, then in particular it holds for all $|x|\leq 1$. Then
$$\|A\|=\sup_{|x|\leq 1}|Ax|\leq\sup_{|x|\leq 1}\lambda|x|\leq\lambda.$$
A: I think the inequality $|Ax| \leq \| A \| |x|$ really doesn't imply what you need. Here is another suggestion of how to see it: from the definition of $\|A\|,$ which says
$$\|A\| = \sup_{|x| \leq 1} |Ax|,$$
it is clear that if $|Ax| \leq \lambda |x|$ for every $x \in \mathbb{R}^n,$ then the set of real numbers $\{|Ax|: |x| \leq 1\}$ is a subset of the set $\{ |Ax| : |Ax| \leq \lambda\}:$
$$\{|Ax|: |x| \leq 1\} \subset \{ |Ax| : |Ax| \leq \lambda\}$$
Now if you believe the standard fact that $\sup A \leq \sup B$ whenever $A \subset B,$ then it is clear that
$$\sup \{|Ax|: |x| \leq 1\} \leq \sup \{ |Ax| : |Ax| \leq \lambda\}.$$
But plainly $\sup \{ |Ax| : |Ax| \leq \lambda\} \leq \lambda,$ so that
$$\|A\| = \sup_{|x| \leq 1} |Ax| = \sup \{|Ax|: |x| \leq 1\} \leq \lambda,$$
as needed.
