Verify that the function $|| \space ||_0 : (X \to X ) \to \mathbb R$ is a norm Verify that the function $\| \space \|_0 : (X \to X ) \to \mathbb R$  (where
$X$ is a unitary $n$-dimensional vector space, and  $(X \to X )$  is the set of all linear transformations from $X$ to $X$) defined as $$\|A\|_0=\sup_{\|x\| \le1}\|Ax\| \qquad A\in (X \to X )$$
is a norm on $ (X \to X )$, and then show that $$\|A\|_0=\sup_{\|x\| \le1 , \\ \|y\| \le1}|(Ax,y)|$$
Can somebody tell me is my proof correct?

*

*$\|A\|_0=\sup_{\|x\| \le1}\|Ax\|\ge0$


*$\|A\|_0=\sup_{\|x\| \le1}\|Ax\|=0  \Rightarrow \|Ax\|=0  $ for $\|x\| \le1  \Rightarrow A=0$


*$\| \lambda A\|_0=\sup_{\|x\| \le1}|\lambda| \|Ax\|=|\lambda| \|A\|_0$


*$\|A+B\|_0=\sup_{\|x\| \le1}\|Ax+Bx\| \le\sup_{\|x\| \le1}(\left\|Ax\|+\|Bx\|\right)  \le\sup_{\|x\| \le1}\|Ax\|+\sup_{\|x\| \le1}\|Bx\|=\|A\|_0+\|B\|_0$
I am a little uncertain for the 2. part and I don't know how to show that the equality $\|A\|_0=\sup_{\|x\| \le1 , \\ \|y\| \le1}|(Ax,y)|$     holds . I would appreciate some help
 A: Your proof is in the right direction but it only has calculations and lacks any explanations. For instance, for 1., you should add something like "let $x\in X$ be arbitrary" at the start, and perhaps something like  "because $\| \cdot \|$ is a norm, and therefore $\|A x\|\ge 0$" at the end.
For 2, to conclude, you need to know what the meaning of $=$ is, i.e.  $A\in(X\to X) =  0 $ means $Ax = 0$ for all $x\in X$. And this is what you need to prove. Its not exactly the same as $Ax=0$ for each $x$ such that $\|x\|\le 1$ (this is what you have, after you use the corresponding property for $\|\cdot\|$). To finish, if $x\in X$ with $x\neq 0$, then $Ax = A(\|x\| \frac x{\|x\|}) = \|x\| A(\frac x{\|x\|}) =0$, where I first used linearity, then what you proved for 2. This completes the proof of 2. Try to add explanations for 3 and 4 as well.
For the identity, one direction follows from a named result (Cauchy–Schwarz); for the other, one way would be to start from
$$\sup_{\substack{\|x\| \le 1\\ \|y\|\le 1}} (Ax, y) \ge \sup_{\substack{\|x\|\le 1 \\ Ax\neq 0}} \left (Ax , \frac{Ax}{\|Ax\|}\right) $$
A: Sketch of part 2:
In $(Ax, y)$, put $y = \frac{Ax}{\left\| Ax \right\|}$ (don't forget to treat the case $\| Ax \| = 0$). Notice that
$$
|(Ax, y)| = \frac{1}{\|Ax \|} | (Ax, Ax) | = \frac{\| Ax \|^2}{\|Ax \|} = \left\| Ax \right\|.
$$
What does this say about $\sup_{\| x \| \leq 1 \\ \| y \| \leq 1}(Ax, y)$? For the other direction you may want to look up Cauchy-Schwartz.
