Compact Operators and the Unit Ball

In summary, in functional analysis, it is often sufficient to show that something holds for the unit ball in order to prove its validity for the entire space. This is because the unit ball acts as a convenient proxy for any bounded set in a Banach space. This can be seen in examples such as calculating the operator norm and defining compact operators. The map x \mapsto \frac x{\| x \|} allows us to map elements of X to b_1(X), making it useful for showing properties of subsets of X. Additionally, the zero vector is not of interest in the study of linear operators, and dilation is a homeomorphism in a topological vector space.
  • #1
Kreizhn
743
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.
 
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  • #2
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.
 
  • #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".
 
  • #4

Related to Compact Operators and the Unit Ball

What is a compact operator?

A compact operator is a linear transformation between two topological vector spaces that maps bounded sets to relatively compact sets. In other words, it maps bounded sets to sets that are "smaller" in some sense.

What is the unit ball?

The unit ball in a normed vector space is the set of all points whose distance from the origin is less than or equal to 1. In other words, it is the set of all points that have a norm (length) of 1.

What is the relationship between compact operators and the unit ball?

Compact operators are often studied in relation to the unit ball because they have a special property: they map the unit ball to a relatively compact set. This makes them useful in certain areas of functional analysis and applied mathematics.

How are compact operators and the unit ball used in functional analysis?

Compact operators are a fundamental tool in functional analysis, as they allow us to extend theorems and results from finite-dimensional spaces to infinite-dimensional spaces. The unit ball is often used as a starting point for studying the behavior of compact operators.

Can compact operators and the unit ball be generalized to other mathematical structures?

Yes, the concepts of compactness and unit balls can be generalized to other mathematical structures beyond normed vector spaces. For example, compact operators can be defined for general Banach spaces, and the unit ball can be defined for any metric space.

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