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Theorem Reminder

  1. Mar 16, 2013 #1
    Could someone remind me what theorem is this:

    ##let \ \phi: [0,1] \rightarrow \mathbb{R} \ , \ \phi \geqslant 0 \ , \ \phi \in \mathbb{R} \\[20pt]

    if \ \ \int_0^1 \mathrm{\phi(t)} \ \mathrm{d}t = 0 \ \Rightarrow \ \phi \equiv 0##

    Thanks
     
    Last edited: Mar 17, 2013
  2. jcsd
  3. Mar 16, 2013 #2

    jbunniii

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    I assume you mean ##\phi \geq 0##, not ##\phi > 0## (else the conclusion ##\phi = 0## would be impossible). It's false without further assumptions. Let ##\phi## be the indicator function for any finite subset of ##[0,1]## for a counterexample. Do you mean to assume ##\phi## is continuous?
     
  4. Mar 17, 2013 #3

    pwsnafu

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    [strike]He's written ##\phi \equiv 0## so he probably means a.e.[/strike]

    Maybe not. But yeah equality is only up to almost everywhere.
     
  5. Mar 17, 2013 #4

    jbunniii

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    Good point. OP, can you give us more context? Is it a Riemann integral or a Lebesgue integral?
     
  6. Mar 17, 2013 #5
    fixed. ##\phi \geqslant 0, \ \phi## is continuous.
     
  7. Mar 17, 2013 #6
    Riemann integral
     
  8. Mar 17, 2013 #7

    jbunniii

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    OK, I don't know if the theorem has a name, but it's easy to prove. If ##\phi## is not identically zero, take a point ##x## where ##\phi(x) > 0##. Then there is a neighborhood of positive radius around ##x## where ##\phi(y) > \phi(x)/2## for all ##y## in the neighborhood. So the integral is at least ##\phi(x)/2## times the width of the neighborhood.
     
  9. Mar 17, 2013 #8
    Do we get this because of the continuity of ##\phi##?

    did you mean then that the integral is at least ##\epsilon \cdot \frac{\phi(x)}{2}## Hence ##\equiv## to 0?
     
  10. Mar 17, 2013 #9

    jbunniii

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    Yes.
    It's at least ##\epsilon \cdot \frac{\phi(x)}{2}## (assuming ##\epsilon## is the width of the neighborhood described earlier), so the integral is NOT zero.

    To summarize, we proved that if ##\phi## is continuous and ##\phi(x) > 0## for some ##x##, then ##\int \phi > 0##. Equivalently, if ##\int \phi = 0##, then there cannot be any ##x## for which ##\phi(x) > 0##, i.e., ##\phi## must be identically zero.
     
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