Equivalent Definitions of the Operator Norm

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

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.

Attribution
Source : Link , Question Author : KiaSure , Answer Author : Mike Pierce

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