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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 $\|x\| = 1$ ?

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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.

The intuition is: linear operators generally take smaller vectors to smaller vectors, and larger vectors to larger vectors (in the sense made precise above). So if you're calculating the sup on the disk, only worry about the larger vectors.

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  • $\begingroup$ Thanks for the answer and detailed explaination. $\endgroup$ – user247511 Aug 20 '15 at 21:43
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Because if $\;0<\lVert x\rVert\le 1$, then $\Biggl\lVert\dfrac{x}{\lVert x\rVert}\Biggr\rVert=1$, and because $\lVert\lambda x\rVert=\lvert \lambda\rvert\lVert x\rVert$.

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