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Integral of product of two functions = 0

  1. Nov 13, 2011 #1
    If [itex] f \in L^1(0,1) [/itex] and if [itex] \int_0^1 fg = 0 [/itex] for any continuous [itex]g \colon (0,1) \rightarrow \mathbb{R}[/itex], then is it the case that [itex]f = 0[/itex] almost everywhere in [itex](0,1)[/itex]?

    Thanks.
     
  2. jcsd
  3. Nov 13, 2011 #2
    Consider the case g = f.
     
  4. Nov 13, 2011 #3
    I've got it, thanks.
     
  5. Nov 13, 2011 #4
    Can you post your proof? This is an interesting problem, and I've been unable to prove it so far.
     
  6. Nov 14, 2011 #5

    disregardthat

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    It is not necessarily so that the integral of f*g is 0 if g = f, since f isn't necessarily continuous.
     
  7. Nov 14, 2011 #6
    Oh! Would g(x)=0 also be a counter example? With g(x)=0, f(x)g(x)=0 for all x and thus [itex]\int^1_0 f(x)g(x)=0[/itex].
     
  8. Nov 14, 2011 #7

    disregardthat

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    A counter-example must be an f which is 0 on a set of non-zero measure such that for any continuous function g, the integral of fg from 0 to 1 is 0...

    By the way, is L^1(0,1) the space of riemann-integrable functions, or lebesgue integrable functions?
     
    Last edited: Nov 14, 2011
  9. Nov 14, 2011 #8

    pwsnafu

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    Similarly, g is not necessarily L1. The domain is (0,1) not [0,1].

    Err...

    In order to be a counter example you need to produce a [itex]f \in L^1 (0,1)[/itex] and [itex]g \in C(0,1)[/itex] such that
    1. the integral of fg is 0,
    2. the set of x such that [itex]f(x) \neq 0[/itex] is not equal to zero.
    In measure notation [itex]\mu(f^{-1}(\mathbb{R}\setminus \{0\})) > 0[/itex].

    That would be Lebesgue.

    Tyler, consider g(x) = 1.
     
  10. Nov 14, 2011 #9
    Let \eps>0. Then since f is L-integrable, f^2 also is, so there is a \delt>0 such that if m(E)<\delt, then

    [itex] \int_E f^2 < \eps [/itex]

    Also, since f is L-integrable, it's measurable, so Lusin implies that there exists a measurable F contained in (0,1) such that M((0,1)-F) < \delt and for which the restriction of f to F is continuous. Define the following:

    [itex] u(x) = f|_F(x) [/itex] if [itex] x \in F [/itex], [itex] 0 [/itex] if [itex] x \in F^C [/itex]
    [itex] v(x) = f|_{F^C}(x) [/itex] if [itex] x \in F^C [/itex], [itex] 0 [/itex] if [itex] x \in F [/itex]

    Then:

    [itex]

    \int_0^1 f^2 = \int_0^1 f \cdot (u+v) = \int_0^1 f \cdot u + \int_0^1 f \cdot v = \int_F f \cdot u + \int_{F^C} f^2 = \int_{F^C} f^2 < \eps

    [/itex]

    So [itex] \int_0^1 f^2 = 0 [/itex]. Since f^2 is nonnegative on (0,1), Tschebyshev's inequality implies f^2 = 0 ae on (0,1). So f = 0 ae on (0,1).

    I guess u would actually be piecewise continuous, but I think it can be patched up fairly easily.
     
    Last edited: Nov 14, 2011
  11. Nov 14, 2011 #10

    disregardthat

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    Nice proof resolvent.

    I don't think so :confused:

    The premise is that f is integrable, and that fg has integral = 0 for any continuous function g. A counter-example must satisfy this (you can't just bring about a single continuous g), but also not satisfy that the set of zeroes has measure 0.
     
    Last edited: Nov 14, 2011
  12. Nov 14, 2011 #11

    pwsnafu

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    You're right.

    Edit: No, you are right about g. But not about zeroes of f. The theorem concludes that "f = 0 a.e." That means the set of x where f is not 0 has zero measure.

    So for a counter example you need "not f = 0 a.e." which is saying "the set of f=/=0 has non-zero measure". You can have "f=0" have measure 1/2 and "f=/=0" also have measure half, which isn't zero a.e.
     
    Last edited: Nov 14, 2011
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