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Converging Inner Product Sequence in Hilbert Space

  1. Jun 7, 2008 #1
    The problem statement, all variables and given/known data
    Let H be a Hilbert space. Prove that if [tex] \left\{ x _{n} \right\} [/tex] is a sequence such that lim[tex]_{n\rightarrow\infty}\left\langle x_{n},y\right\rangle[/tex] exists for all [tex]y\in H[/tex], then there exists [tex]x\in H[/tex] such that lim[tex]_{n\rightarrow\infty} \left\langle x_{n},y\right\rangle = \left\langle x,y\right\rangle[/tex].

    The attempt at a solution
    I'm pretty sure the idea is to show that [tex] \left\{ x _{n} \right\} [/tex] is Cauchy and hence convergent (since H is complete). Then [tex] x_{n} \rightarrow x [/tex] for some [tex] x \in H [/tex]. And since [tex] y \rightarrow y [/tex] and also the inner product is continuous, we have [tex] \left\langle x_{n},y\right\rangle \rightarrow \left\langle x,y\right\rangle[/tex].

    I'm not convinced my argument for [tex] \left\{ x _{n} \right\} [/tex] being Cauchy is sound though and would appreciate any feedback:

    [tex] \left\langle x_{m}-x_{n},y\right\rangle = \left\langle x_{m},y\right\rangle - \left\langle x_{n},y\right\rangle \rightarrow 0; m,n\rightarrow\infty[/tex].

    And because this is true for all [tex] y \in H [/tex] we must have [tex] x_{m}-x_{n} \rightarrow 0[/tex] (this is where I think things might fall apart) and therefore [tex] \left\|x_{m}-x_{n}\right\| \rightarrow 0; m,n\rightarrow\infty[/tex]. And so [tex] \left\{ x _{n} \right\} [/tex] is Cauchy.

    Unrelated to the above problem, I'd also really like to know why ( [tex] x_{n} \rightarrow x [/tex] ) is not properly aligned in the text. (This is my first time using Latex.)
     
  2. jcsd
  3. Jun 7, 2008 #2

    Hurkyl

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    I think the key you're looking for is the nondegeneracy axiom of the inner product:

    If [itex]\forall y: \langle x, y \rangle = 0[/itex], then [itex]x = 0[/itex]​
     
  4. Jun 7, 2008 #3
    Thanks Hurkyl.

    That is what I was trying to use, but I hadn't managed to convince myself that I could replace the equality with a limit.
     
  5. Jun 7, 2008 #4

    Hurkyl

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    Oh, by the way, to get properly aligned LaTeX, use [ itex ] ... [ /itex ] tags, instead of [ tex ] ... [ /tex ] tags.

    And I do think you have a point -- I hadn't noticed that problem in my initial skim of your method. I guess to formally state it, you would like to invoke this conjecture (which is a special case of what you're trying to prove):

    If [itex]\forall y: \lim_{n \rightarrow +\infty} \langle x_n, y \rangle = 0[/itex], then [itex]\lim_{n \rightarrow +\infty} x_n = 0[/itex]​

    and you're right to worry about it, since this is something that deserves to be proved. (And I certainly agree that the method you have chosen seems like a reasonable approach to the problem -- this case might be easier to prove directly than the general case!)


    Considering special cases is often helpful; what happens if you let y be one of your x's? I suspect it will help, but I don't yet see how.
     
    Last edited: Jun 7, 2008
  6. Jun 7, 2008 #5
    Again, many thanks!
     
  7. Jun 7, 2008 #6

    Hurkyl

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    My only other thought at the moment is to consider proving the contrapositive -- what if [itex]\langle x_n, x_n \rangle[/itex] didn't converge to zero; can you get any help there?



    By the way, are we considering an arbitrary Hilbert space? Or are we sticking to ones with countable dimension? Or only to finite dimensional spaces?
     
  8. Jun 7, 2008 #7
    Very definitely general...

    Will try out your suggestions though.
     
  9. Jun 8, 2008 #8

    Dick

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    Let x_n=e_n where e_n is an orthonormal basis of the hilbert space. Then <x_n,y> does converge for any y. To zero. But x_n doesn't converge. So don't try to show x_n converges. It might not.
     
  10. Jun 8, 2008 #9
    Good point! Have had a growing suspicion that it would not converge (strongly) otherwise the question would have been "show there exists [itex] x \in H[/itex] such that [itex]x_n \rightarrow x [/itex]" as opposed to asking to show weak convergence only. Grrr... But thanks for the counter example, Dick. Hmmm... back to the drawing board!

    (Oh, and "itex" rocks... shot-a-lot Hurkyl!)
     
  11. Jun 8, 2008 #10

    Hurkyl

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    Oh phooey, I had tricked myself into thinking that wouldn't work, but it clearly does. I was mixing up kinds of convergence in L²(R), grumble grumble....
     
  12. Jun 8, 2008 #11
    Since the given limit exists for all [itex] y \in H [/itex], we have [itex] lim_{x \rightarrow \infty} \langle x_{n},y \rangle [/itex] is a function from H into C (or R). With a little messy manipulation (reversing of the inner product - which I think I can "unreverse" at the end) it is quite easy to show that it (the reversed version) is linear. (I think the above is sesquilinear - but I don't wish to go there!)

    IF (big), I could also show it is bounded, then it would be a bounded linear functional on H and by the Riesz Representation Theorem, there exists an [itex]x[/itex] such that [itex] lim_{x \rightarrow \infty} \langle y,x_{n} \rangle = \langle y,x \rangle [/itex].

    I am not convinced I can show it's bounded though. (It would be easy if [itex] \|x_{n}\| [/itex] were bounded... but I think my bootstraps are just not long enough for me to hoist myself over the final hurdle!! *Frown*)
     
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