Compact operators on normed spaces

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    Compact Operators
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The discussion centers on the characterization of compact linear operators T: X → Y in normed spaces, specifically the equivalence of T being compact and the closure of the image of the closed unit ball \overline{B} being compact in Y. The user demonstrates that if A is a bounded set, then the image under T of a scaled version of A is contained within the compact closure of T applied to \overline{B}. Furthermore, they illustrate that for any bounded sequence x_n, a convergent subsequence can be derived from T(x_n), confirming the compactness of T.

PREREQUISITES
  • Understanding of compact operators in functional analysis
  • Familiarity with normed spaces and bounded sets
  • Knowledge of sequences and convergence in metric spaces
  • Proficiency in linear transformations and their properties
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  • Study the properties of compact operators in Banach spaces
  • Explore the relationship between bounded sets and compactness in normed spaces
  • Learn about the Arzelà-Ascoli theorem and its implications for compactness
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Mathematicians, particularly those specializing in functional analysis, graduate students studying operator theory, and researchers exploring the properties of linear operators in normed spaces.

AxiomOfChoice
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Is it easy to show that [itex]T: X \to Y[/itex] is a compact linear operator -- i.e., that the closure of the image under [itex]T[/itex] of every bounded set in [itex]X[/itex] is compact in [itex]Y[/itex] -- if and only if the image of the closed unit ball [itex]\overline B = \{x\in X: \|x\|\leq 1\}[/itex] has compact closure in [itex]Y[/itex]? One direction is (of course) easy, since [itex]\overline B[/itex] is bounded, but I'm having trouble with the other one. Can someone help please? Thanks!
 
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Hi AxiomOfChoice! :smile:

Let A be a bounded set, then there exists an [itex]\alpha>0[/itex] such that

[tex]\alpha A\subseteq \overline{B}[/tex]

(with of course [itex]\alpha A:=\{\alpha a~\vert~a\in A\}[/itex]).

Then

[tex]\alpha T(A)=T(\alpha A)\subseteq T(\overline{B})[/tex]

which means that the closure of [itex]\alpha T(A)[/itex] is compact, which implies that T(A) is compact.

I'll leave it to you to work out the details.
 
The definition way might be even easier. Take a bounded sequence x_n. We have to show that T(x_n) has a convergent subsequence. Take a subsequence x_j with x_j non-zero. Look at T(x_j/||x_j||). This has a convergent subsequence. Then T(x_j) = ||x_j|| T(x_j/||x_j||). The ||x_j|| have a convergent subsequence in R by boundedness, hence T(x_n) has a convergent subsequence.
 

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