A question about norm of the bounded linear operator

Question

How to prove that $||T||=\sup \{||Tx||:||x||\le 1\}=\inf \{k:k\ge0\; \text{ and } ||Tx||\le k||x||, \forall x\}$

what I know is $||T||=\sup\{\frac{||Tx||}{||x||}: x\ne0\}=\sup\{ ||Tx||:||x||=1\}$

Answer 1

If $0<\|x\| \le 1$, then $\|Tx\| = \|T\hat x\|\cdot\|x\| \le \|T\|\cdot \|x\| \le \|T\|$, where $\hat x = x\cdot\|x\|^{-1}$. Hence $$\sup\{\|Tx\| : \|x\| \le 1\} \le \|T\|.$$ Since $\|T\| = \sup\{\|Tx\|:\|x\| = 1\}$ and $\{\|Tx\|:\|x\| = 1\}\subset\{\|Tx\| : \|x\| \le 1\},$ we also have $$\|T\| \le \sup\{\|Tx\| : \|x\| \le 1\},$$ and the equality follows by combining inequalities.

Let $k\ge0$ be such that $\|Tx\| \le k\|x\|$ for each $x$. In particular this holds for all unit vectors $x$, so $\|T\| \le k$. Hence $$\|T\|\le \inf\{k : k\ge 0,\ \|Tx\|\le k\|x\|,\ \text{for all $x$}\}.$$ For each $x$ we have $\|Tx\| \le \|T\|\cdot \|x\|$, hence $$\inf\{k : k\ge 0,\ \|Tx\|\le k\|x\|,\ \text{for all $x$}\} \le \|T\|.$$ We get the desired equality by combining inequalities again.