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Compact operators on normed spaces

  1. Aug 3, 2011 #1
    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!
  2. jcsd
  3. Aug 3, 2011 #2
    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]).


    [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.
  4. Aug 3, 2011 #3
    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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