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Cauchy Schwarz's proof

  1. Aug 31, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'm trying to follow the demonstration of the Cauchy-Schwarz's inequality proof given in http://mathworld.wolfram.com/SchwarzsInequality.html.
    I am stuck at the last step, namely that [itex]\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow |\langle f , g \rangle |^2 \leq \langle f , f \rangle \langle g , g \rangle[/itex].


    2. Relevant equations

    I don't know.

    3. The attempt at a solution
    [itex]\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle \bar f , g \rangle \langle \bar f , g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle[/itex]. I'm stuck here.
    I know that [itex]||f||=\sqrt {\langle f , f \rangle}[/itex] but I don't even know if I can use this fact. Any tip is appreciated.
     
  2. jcsd
  3. Sep 1, 2012 #2

    fluidistic

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    I've made some progress I think.
    Mathworld didn't specify it explicitely but I think that f and g are real functions.
    So that I reach [itex]\langle \bar g , f \rangle \langle f , \bar g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle \bar f , g \rangle \langle \bar f , g \rangle \leq \langle \bar f , f \rangle \langle \bar g , g \rangle \Rightarrow \langle f , g \rangle ^2 \leq \langle f , f \rangle \langle g , g \rangle[/itex]. So I "almost" reach the proof. I have a missing absolute value though. Any idea why?
     
  4. Sep 2, 2012 #3

    fluidistic

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    Ok I got it wrong, f and g aren't real valued function because lambda (which is complex) is defined by some inner products involving f and g and their complex conjugate only.
    If someone could tell me how to understand the last step I'd be grateful.
     
  5. Sep 2, 2012 #4

    HallsofIvy

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    This is incorrect. [itex]\langle f, g\rangle^2[/itex] is a complex number. You want [itex]\left|\langle f, g\rangle\right|^2[/itex].

    Again, those are wrong. You want [itex]\langle f, \bar f\rangle= |f|^2[/itex] and [itex]\langle g, \bar g\rangle= |g|^2[/itex].
     
  6. Sep 2, 2012 #5

    fluidistic

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    Thanks HallsofIvy!
    Yeah you are right, I realized this in my previous post.


    Ok... How do I prove these, for any inner product? I'm looking at the properties of Hermitian inner product given there: http://mathworld.wolfram.com/HermitianInnerProduct.html but I've no clue how to relate it with the [itex]|.|^2[/itex] (same as norm squared?)
     
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