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Let $A$ be a bounded linear operator on a complex Hilbert space $V$.

It is well known that $$\lVert A\rVert=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}.$$

I want to understand why $$\begin{align*} I &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ &=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$

Proof: Now note that by definition of $\|A\|$ we have $$ \|Av\| \le \|A\| \|v\| \quad \forall v \in V.$$ Then $I \le \|A\|$.

On the other hand by definition of $\sup$ we have $$ I \ge \|Av_n\| /\|v_n\| \ge \|A\| - 1/n \quad \forall n.$$ Then $\|A\| = I$.

I don't understand why $$I \ge \|Av_n\| /\|v_n\|?$$

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$I+\frac 1 n> c_n$ for some $c_n$ with $\|Av\| \leq c_n\|v\|$ for all $v$. Hence $I+\frac 1 n> \frac {\|Av\|} {\|v\|}$. Since this is true for all $v \neq 0$ we get $I+\frac 1 n>\sup \{ \frac {{\|Av\|}} {{\|v\|}}: v \neq 0\}$. Now let $n \to \infty$.

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