First question: I know that the following definitions of norm of the operator are equivalent:
$$\begin{align*} a&= \inf\{ k\;\colon\; \lVert Av\rVert\leq k\lVert v\rVert \text{ for all }v\in V\},\\ \\ b&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\},\\ \\ c&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \},\\\\ d&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$
I found somewhere one more definition (let's call it e ): $ e = \sup \{ \frac{ ||Av||}{||v||}: v \in V , ||v|| \leq 1 \} $ . I want to prove that this definition is also equivalent, so can someone just check this method:
I want to prove that $ a \leq e $ and $ e \leq d$. The second inequality follows because supremum on the subset is smaller, so we just have to prove the first inequality.
$a = \inf \{ k: ||Av|| \leq k||v||, v \in V \} $, so it follows that $||Av|| \leq a ||v|| $ for all $v \in V, ||v|| \leq 1$ and $a$ is the least constant for which this holds.
$ \sup_{v \in V, ||v|| \leq 1} \frac{||Av||}{||v||} = e $, so $||Av|| \leq e||v|| $ for all $ v \in V, ||v|| \leq 1 $. Now it follows that $a \leq e$ and all norms are same now. Is this correct?
Second question: If linear operator does not reach its norm on the set $\{ v \in V: ||v|| \leq 1\} $ does it mean that it doesn't reach nowhere and conversely? I think that if there is $v \in V $ such that $||A|| = \frac{||Av|| }{||v|| } $, then we can take $ y= \frac{v}{||v|| } $, $||y||=1$, and $||A||= ||Ay|| $. Is this reasoning correct? Thank you in advance.