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Inner Product Sum Inequality

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let V be a real inner product space, and let v1, v2, ... , vk be a set of orthonormal vectors.
    Prove
    Ʃ (from j=1 to k)|<x,vj><y,vj>| ≤ ||x|| ||y||

    When is there equality?

    2. Relevant equations



    3. The attempt at a solution

    I've tried using the two inequalities given to us in lectures, namely Cauchy-Schwarz Inequality which states

    |<v,w>| ≤ ||v|| ||w||

    But surely, using this inequality, we get Ʃ (from j=1 to k)|<x,vj><y,vj>| ≤ k(||x|| ||v|| ||y|| ||v|| = k( ||x|| ||y||) since the v are orthonormal!

    I understand this is an inequality, and so obviously the inequality above is a better approximation than the one I've just shown, but I'm not sure where to go.

    The other inequality is Bessel's Inequality which states

    ||v||2 ≥ Ʃ|<v, ei>|2 if ei is a set of orthonormal elements.

    Thanks
     
    Last edited: Nov 20, 2012
  2. jcsd
  3. Nov 20, 2012 #2

    micromass

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    The space [itex]\mathbb{R}^k[/itex] is an inner product space for the usual inner product. What does the Cauchy-Schwarz inequality say in this special case [itex]\mathbb{R}^k[/itex]??
     
  4. Nov 21, 2012 #3
    Ok, so according to Wikipedia (I haven't been taught this in lectures), the Cuachy-Schwarz inequality over ℝn is:

    (Ʃ xiyi)2 ≤ Ʃxi2 Ʃyi2

    Do I replace multiplication with inner products? I've tried that, but I must be doing something wrong.


    (Ʃ <x,vj>)2 ≤ Ʃ<x,x> Ʃ<vj,vj> = k||x||2Ʃ<vj,vj> = k||x||2 x k since ||vj||=1 ?
    But then where should I go from here, if here is where I should be?
    Sorry
     
  5. Nov 21, 2012 #4

    micromass

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    The version of Cauchy-Schwarz that is most standard is actually

    [tex]\sum_{k=1}^n |\alpha_k\beta_k|\leq \sqrt{\sum_{k=1}^n |\alpha_k|^2}\sqrt{\sum_{k=1}^n |\beta_k|^2}[/tex]

    Now, apply this on your original problem

    [tex]\sum_{k=1}^n |<x,v_k><y,v_k>|[/tex]
     
  6. Nov 21, 2012 #5
    Thank you so much! I think I have it, although it seems very easy, which always seems suspicious to me in maths. Here goes:

    Ʃ |<x,vj><y,vj>| ≤ √(Ʃ<x,vj>2) √(Ʃ<xy,vj>2)

    Then by Bessel's Inequality

    √Ʃ<x,vj>2√Ʃ<xy,vj>2 ≤ √||x||2 √||y||2

    So Ʃ |<x,vj><y,vj>| ≤ ||x|| ||y|| as required!
     
    Last edited: Nov 21, 2012
  7. Nov 21, 2012 #6

    micromass

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    That's it!!
     
  8. Nov 21, 2012 #7
    Thank you very much! You've made my night!
     
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