# Operator Norm ( Confusion)

I am reading a book about operator theory and it states the following;

If $X$ and $Y$ are two normed spaces and $T:X\rightarrow Y$ is an operator, define it's norm by; $$\|T(x)|| = \sup \{ \| T(x) \|: \text{ with }\|x\| \le 1\} = \sup\{ \| T(x) \|: \text{ with } \|x\| = 1\}$$

My Question is: How the statement of $\|x\| \le 1$ changed to $\|x\| = 1$ ?

We know $T$ is linear. Let $x$ be any vector with norm less than or equal to 1. Then we have $||Tx||=||x||*||\frac{Tx}{||x||}||\leq||\frac{Tx}{||x||}||=||T(\frac{x}{||x||})||$.
What does this mean? Well, it means that if $||x||\leq 1$, there is some other point vector, given by $x'=x/||x||$, with norm 1, such that $||Tx||\leq||Tx'||$. Thus, if we're computing the sup over all $x$ such that $||x||\leq 1$, we can just forget about the points with norm less than one, since we know the sup must occur on the boundary.
Because if $\;0<\lVert x\rVert\le 1$, then $\Biggl\lVert\dfrac{x}{\lVert x\rVert}\Biggr\rVert=1$, and because $\lVert\lambda x\rVert=\lvert \lambda\rvert\lVert x\rVert$.