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Homework Help: Double Orthogonal Closed Subspace Inner Product => Hilbert

  1. May 13, 2015 #1
    Let [itex]X[/itex] be an Inner Product Space. If for every closed subspace [itex]M[/itex], [itex]M^{\perp \perp} = M[/itex], then [itex]X[/itex] is a Hilbert Space (It's complete).
    Hint: Use the following map: [itex]T : X \longrightarrow \overset{\sim}{X}: T(y)=(x,y)=f(x)[/itex] where [itex](x,y)[/itex] is the inner product of [itex]X[/itex].

    Relevant equations:
    [itex]S^{\perp}[/itex] is always closed to every [itex]S \subset X[/itex]

    Attemp to solution.
    I don't really know how to solve it, most theorems I have read, have Hilbert Space as a hyphotesis.
    The only idea I had was trying to prove that [itex]\overline{T(M^{\perp})} = M^{\circ}[/itex] (Where [itex]M^{\circ}[/itex] is the Null Space of [itex]M[/itex]).

    Edit: How do I do to have latex without jumping lines? I uploaded a pdf with a itext more closely formated to what I tried to write.

    Edit2: I read other posts to find out how to do it, I only had to add an i to the "tex".

    Attached Files:

    Last edited: May 13, 2015
  2. jcsd
  3. May 19, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. May 19, 2015 #3
    I'm thinking about the family of subspaces of [itex] \overset{\sim}{X}[\itex] that are intersection of nullspaces of one dimensional subspaces of [itex]X[\itex], they are all closed. I think that if I can probe that they're all closed subspaces of [itex]X[\itex] I may, somehow, be able to prove the required statement.
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