Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Compact operators on normed spaces
  1. Compactness of a space (Replies: 3)

  2. Normed space (Replies: 5)

Loading...