# norms of Hilbert space operators

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 $$I \ge \|Av_n\| /\|v_n\|?$$

$$I+\frac 1 n> c_n$$ 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 $$n \to \infty$$.