induced matrix norm proof I'm not sure how to approach the following question
prove the induced matrix norm satisfies 
$$||A||=\max_{||x||=1}||Ax||$$
I'm assuming that I can use the multiplicativity property and say 
$||Ax||<=||A|| ||x||$ and if $||x||=1$, then $||A|| ||x||= ||A|| $
Thanks in advance.
P. S. $||x=1||$ is under max. I don't know how to make it small.
 A: A norm $|\cdot|$ on the vector space ${\mathbb R}^n=:V$ is a function $|\cdot|:\>V\to{\mathbb R}_{\geq0}$ satisfying certain axioms, e.g., $|\lambda x|=|\lambda|\>|x|$. The standard example is the euclidean norm
$$|x|:=\sqrt{x_1^2+\ldots +x_n^2}\ ,$$
but there are many others.
Assume that such a norm on $V$ has been fixed  once and for all. 
We shall need the following fact: The hyperplane $H_i:\>x_i=1$ does not contain the origin, hence the function $x\mapsto|x|$ (which is continuous) assumes a positive minimum $m_i$ on $H_i$. Let $\min_{1\leq i\leq n} m_i=:m>0$. Then it is easy to see that
$$|x_i|\leq {|x|\over m}\qquad(1\leq i\leq n)\ .\tag{1}$$
Denote by ${\cal L}(V)$ the space of all linear transformations $A:\>V\to V$. The given norm on $V$ "induces" a norm on ${\cal L}(V)$, as described in the following. Inducing means that this norm can be defined without providing additional data, as e.g., some extraneous coefficient matrix.
Let an $A\in{\cal L}(V)$ be given. Then for any $x\in V$ one has, using $(1)$,
$$|Ax|=\left|\sum_{i=1}^n x_i\>Ae_i\right|\leq \sum_{i=1}^n|x_i|\>|Ae_i|\leq C|x|$$
for a certain constant $C>0$. Linearity then implies
$$|Ax-Ay|\leq C|x-y|\ ,$$
i.e. the transformation $A$ is Lipschitz continuous on $V$ with Lipschitz constant $C$. Now the norm $\|A\|$  of $A$ is the "optimal Lipschitz constant" for $A$. A worst case analysis shows that necessarily
$$\|A\|=\max_{|x|=1}|Ax|\ ,\tag{2}$$
because on the "unit sphere" (with respect to $|\cdot|$) there is a point $\xi$ where $|Ax|$ assumes its maximum, and for this point one has $$|A\xi-0|=|A\xi|=\max_{|x|=1}|Ax|=\max_{|x|=1}|Ax|\>|\xi-0 |\ ;$$ hence the "optimal Lipschitz constant" has at least the value given by $(2)$, and it is easy to see that this value is admissible.
