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Cauchy-Schwartz Inequality for Step Functions

  1. Oct 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Let

    [tex]\phi,\psi : [a,b] \rightarrow \Re [/tex]

    be step functions.

    Prove that

    [tex](\int \phi\psi)^{2} \leq (\int\phi^{2})(\int\psi^{2}) .[/tex]

    Hint: Consider the quadratic function of a real variable t defined by

    [tex]Q(t)=\int(t\phi+\psi)^2 .[/tex]

    3. The attempt at a solution

    I really don't know where to start with this, and the hint only confuses me more! :p

    Any help appreciated, thanks!
     
    Last edited: Oct 17, 2009
  2. jcsd
  3. Oct 17, 2009 #2

    Dick

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    Q(t)>=0, since it's the integral of a nonnegative function (a square). Expand Q(t) out and differentiate with respect to t. Solve Q'(t)=0 for t and put that value of t back into the expression Q(t)>=0 and see what you get.
     
  4. Oct 18, 2009 #3
    Yeah I get a similar thing. So we get a turning point of Q at some value t=-psi/phi, and when you put this back into Q you get

    [tex]\int0 = constant[/tex].

    Am I being really dumb cos I can't seem to get anything like the inequality from this :((((

    Cheers.
     
  5. Oct 18, 2009 #4

    Dick

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    I meant integrate first. I.e.
    [tex]
    t^2 \int \phi^{2} + 2t \int\phi \psi + \int\psi^{2} \geq 0.
    [/tex]

    Now minimize that. The minimum occurs at a value of t that is a ratio of two integrals.
     
  6. Oct 18, 2009 #5
    Ah yeah I got it :-) Thanks!
     
  7. Oct 18, 2009 #6

    Dick

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    Whatever variable phi and psi are functions of. Call it x. So write psi(x) and phi(x).
     
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