[Math] Equivalent Definitions of the Operator Norm

Question

How do you prove that these four definitions of the operator norm are equivalent?
$$\begin{align*}
\lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\
&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\
&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\
&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}.
\end{align*}$$

Answer 1

Let $$\begin{align*} I &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ S_1&=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ S_2&=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ S_3&=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$ Notice that $S_2 \le S_1$ and as $\|Av\| /\|v\| = \| A(v / \|v\|)\|$ we have $S_3 \le S_2$. Now if $\|v\|\le 1$ we have $\|Av\| \le \|Av\| /\|v\|$. Then $S_1 \le S_3$ and $$ S_1=S_2=S_3.$$ Now note that $$ \|Av\| \le S_3 \|v\| \quad \forall v \in V.$$ Then $I \le S_3$ and by definition of $\sup$ we have $$ I \ge \|Av_n\| /\|v_n\| \ge S_3 - 1/n \quad \forall n.$$ Then $S_3 = I$.