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Hilbert Space: Closest point property

  1. Dec 6, 2007 #1
    1. The problem statement, all variables and given/known data

    The theorem about the closest point property says:
    If A is a convex, closed subspace of a hilbert space H, then

    [tex] \forall x \in H\,\, \exists y \in A:\,\,\,\, \| x-y\| = \inf_{a\in A}\|x-a\|[/tex]

    I have to show that it is enough to show this theorem for x = 0 only, by using the isometry [itex]T_{x_0}(x) = x_0 + x [/itex].

    3. The attempt at a solution

    So I would have to show that [itex] \exists y \in A [/itex] such that [itex]\|y\| = \inf_{a\in A}\|a\|[/itex], that is if A contains an element with least length, than for any point x in H there is a point in A that is closest to x, than any other in A.
    Then what?
    Any hint is appreciated.
  2. jcsd
  3. Dec 6, 2007 #2
    If A is a subspace, the claim [itex]\exists y[/itex] so that [itex]\|y\| = \textrm{inf}_{a\in A} \|a\|[/itex], is trivial, because such y is the origo. So I suppose the A was not supposed to be assumed to be a subspace. A being convex and closed should be enough. Then.... A remains convex and closed in translations.
  4. Dec 6, 2007 #3
    Yeah that's right. A isn't assumed to be a subspace, just a subset.
    And you are saying that "A remains convex and closed in translations", how can this be used to show what I want. Because [itex] x -a [/itex] need not lie in A, so I can't use the property with least length. Or how would I be able to use it?
  5. Dec 6, 2007 #4
    I needed to stop and think myself... but let's see, I think I got this.

    One hint that I must give you, is that draw pictures! These are kind of tasks where there is too many different things that you can start doing, and it's difficult to guess what you are supposed to start doing without first drawing some kind of picture.

    A set

    -x + A = \{-x+a\;|\; a\in A\}

    seems to can handy. Choose [itex]y'\in -x+A[/itex] so that it has the smallest norm in this set. Then the desired y should be almost found.
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