formula for the norm of a normal operator

Question

In Rudin's Functional analysis, he does a theorem which shows that for a normal operator

$\Vert T\Vert=\sup\left\{|\langle Tx,x\rangle|\colon \Vert x \Vert \leq 1\right\}$. Why can't $\Vert x \Vert \leq 1$ be sharpened to $||x||=1$ or is he simply accounting for the zero space?

Answer 1

You are right that one can replace $\|x\|\le 1$ by $\|x\|=1$ at least when $X\neq \lbrace 0 \rbrace$. In the latter case it would be a matter of the definition of $\sup \emptyset$ (in $\mathbb R_+$ the least upper bound of $\emptyset$ is $0$).