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Compact Operators and the Unit Ball

  1. Dec 10, 2009 #1
    Hi all,

    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.
     
  2. jcsd
  3. Dec 10, 2009 #2

    dextercioby

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    The only vector of zero norm in a Banach space is the zero vector. That's why the zero vector is not interesting for the study of linear operators and their properties like boundedness and continuity. Take out out zero and you have the bijection.
     
  4. Dec 10, 2009 #3
    There's nothing special about the unit ball; the fact is that when you're talking about linear operators, scaling issues are more or less irrelevant, and so "the unit ball" is a convenient proxy for "any bounded set".
     
  5. Dec 10, 2009 #4
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