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Projection Theorem

  1. Jan 17, 2008 #1
    [SOLVED] Projection Theorem

    1. The problem statement, all variables and given/known data
    If M is a closed subspace of a Hilbert space H, let x be any element in H and y in M, then I have to show that

    [tex] \|x-y\| =\inf_{m\in M}\|x-m\| [/tex]

    implies (equivalent to) that

    [tex] x-y\in M^{\perp} [/tex]


    3. The attempt at a solution

    I have shown the implication "<=", ie that [itex] x-y\in M^{\perp} [/itex] implies the 1st statement. And I've been told that the implication (=>) I now want to show is basically in the that proof.

    The proof for "<=" goes like: If [itex] x-y\in M^{\perp} [/itex] then (x-y,z) = 0 for all z in M, so by Pythagoras thm

    [tex] \|x-y+z\|^2 = \|x-y\|^2 +\|z\| \geq \|x-y\|^2 [/tex]

    M subspace => [itex] m = y-z \in M[/itex]. So [itex] \|x-y\| \leq \|x-m\|[/itex] for all m in M. So this gives the implication the other way around.

    But I don't see how I can go back, nor how the 'proof' for "=>" is basically contained in my this proof.
    And how or where should I use that M is closed, I feel like it should have been used to conclude the implication "<=" from knowing [itex] \|x-y\| \leq \|x-m\|[/itex], or is it not needed for "<="?
     
    Last edited: Jan 17, 2008
  2. jcsd
  3. Jan 17, 2008 #2

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    Here's a summary of what you've already shown (assume x, y are fixed vectors, with y in M, and z is an arbitrary vector in M, so that w=z-y is also an arbitrary vector in M):

    [tex]x-y \in M^\perp \Rightarrow (x-y,z)=0 \Rightarrow ||x-y+z||^2 = ||x-y||^2 + ||z||^2 \Rightarrow ||x-y|| \leq ||x-w|| \Rightarrow ||x-y|| = \inf_{m\in M}\|x-m\| [/tex]

    Now to get the other direction, start by seeing how many of those [itex]\Rightarrow[/itex]'s you can turn into [itex]\Leftarrow[/itex]'s (or to get both directions at once, [itex]\Leftrightarrow[/itex]'s). The one that should give you the most difficulty is:

    [tex] ||x-y+z||^2 = ||x-y||^2 + ||z||^2 \Leftarrow||x-y|| \leq ||x-y+z|| [/tex]

    The condition can be written as:

    [tex] ||x-y||^2 \leq ||x-y+z||^2 = ||x-y||^2 + 2 \Re (x-y,z) + ||z||^2 [/tex]

    [tex] \Leftrightarrow 0 \leq ||z||^2 + 2\Re (x-y,z) [/tex]

    Think about what happens when z is really small.
     
    Last edited: Jan 17, 2008
  4. Jan 18, 2008 #3
    So using Cauchy Schwartz, since (x-y,z) must be real, I get

    [tex]0 \leq \|z\|^2 + 2(x-y,z) \leq \|z\|(\|z\| + 2\|x-y\|)[/tex]

    So that [itex] \|z\|^2 + 2(x-y,z)[/itex] gets 'squeezed' to zero, and hence (x-y,z) too? But if z varies it'll affect (x-y,z) too, so that thing can get small too. So how does varying z helps me in this?
     
  5. Jan 18, 2008 #4

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    Sorry, I'm not following your last post. One way to continue would be to write:

    [tex]f(t) = ||tz||^2 + 2 \Re (x-y,tz) [/tex]

    Given the above inequality, we can write:

    [tex] 0 \leq f(t) = t^2||z||^2 + 2 \Re (x-y,z) t [/tex]

    But's it's easy to see that this will be negative at some points unless [itex]\Re(x-y,z)=0[/itex]. There's a trick to get the same thing for the imaginary part. But I have a feeling there's a simpler way to do this.
     
    Last edited: Jan 18, 2008
  6. Jan 20, 2008 #5
    That was my mistake I thought (x-y,z) should be real so I used Cauchy-Schwartz, but never mind that.

    I see that there could be some points where that inequality could fail, but doesn't that need to be proved first? t would have to fulfill

    [tex] (t \|z\|^2 + 2Re(x-y,z))t < 0 [/tex]
    [tex]0< t < -\frac{2Re(x-y,z)}{\|z\|^2}[/tex]

    So that f(t) would be negative. Don't we need to show that such a z in M would exist? Ie any z such that [itex]-2Re(x-y,z)/\|z\|^2>0[/itex] or [itex]Re(x-y,z)<0[/itex].
    Otherwise it is clear how Re(x-y,z) has to be 0.

    But I'm a little clueless on how to show Im(x-y,z) = 0, what is the trick you were talking about?
     
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