- #1

Kreizhn

- 743

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I've been looking over some results from functional analysis, and have a question. It seems that often times in functional analysis, when we want to show something is true, it often suffices to show it holds for the unit ball. That is, if X is a Banach space, then define [itex] b_1(X) = \left\{ x \in X : \| x \| = 1 \right\} [/itex].

Examples of this is when calculating the operator norm of a bounded operator T, we have

[tex] \| T \| = \sup_{x \in X} \frac{ \| Tx \| }{\| x \|} = \sup_{x \in b_1(X)} \| Tx \| [/tex]

Another example is that of compact operators. One definition is that a linear operator K is compact if it maps bounded sets to relatively compact sets (sometimes called precompact sets). Equivalently, another definition is that K takes the unit ball [itex] b_1(X) [/itex] to a relatively compact set.

I'm wondering, without explicitly showing that this is true in all of these cases, why does this work? Is it because we can map elements of X to [itex] b_1(X) [/itex] by

[tex] x \mapsto \frac x{\| x \|} [/tex] ?

This map doesn't seem invertible, so I don't think it's an isomorphism between X and [itex] b_1(X) [/itex] but it might nonetheless allow us to things like cast closed subsets of X as closed subsets of [itex] b_1(X) [/itex] or something along those lines.