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Let F and y both be continuous for simplicity. Knowing that:[tex]

  1. Jun 10, 2010 #1
    Let F and y both be continuous for simplicity. Knowing that:
    [tex] \int_0^x F'(t)y^2(t) dt = F(x) \quad \forall x \geq 0 [/tex]
    can you say that the function [tex] y[/tex] is bounded? Why? I know that [tex] \int_0^x F'(t) dt = F(x) [/tex] but I can't find a suitable inequality to prove rigorously that y is bounded.
     
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  3. Jun 10, 2010 #2

    mathman

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    Re: Inequality

    Taking derivatives of both sides F'(x)y2(x)=F'(x) for all x>0, so |y(x)|=1.
     
  4. Jun 10, 2010 #3
    Re: Inequality

    Unless, [itex]F(x) = C[/itex] in which case the condition you gave us is [itex]0 = C[/itex]. So, if [itex]F(x) \equiv 0[/itex] we can't say anything about the function [itex]y(x)[/itex].
     
  5. Jun 10, 2010 #4
    Re: Inequality

    Thanks, I was missing something very easy.
     
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