$||T||=\sup\{||Tx||:||x||=1\}$, This is known as operator norm.
$\beta =\sup\{||Tx||:||x||\lt 1\}$. I want to prove $||T||= \beta.$
Now let us take $x$ s.t $||x|| \lt1$, $||T(\frac{x}{||x||})|| \le||T||$.
It follows $\beta \le ||T||.$ But I found difficult in true reverse inequality. I have seen similar results in this site, but It was defined as the supremum of $\{||Tx||:||x|| \le1\}$. In my question it is strict inequality. Can somone help with this?
If $\|x\| = 1$, let $x_n = (1-\frac{1}{n})x$ so that $\|x_n\|< 1$. Then since you assume that $T$ is a continuous operator you have that $\|Tx_n\| \to \|Tx\|$. Since $\beta \geq \|Tx_n\|$ for each $n$, you then get that $\beta \geq \lim_{n \to \infty} \|Tx_n\| = \|Tx\|$. Since $x$ was an arbitrary element of norm $1$ this result tells you that $\beta \geq \|T\|$.