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Math Help - Proving a function is identically zero

  1. #1
    Senior Member Pinkk's Avatar
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    Proving a function is identically zero

    If f is a continuous function on [a,b] such that \int_{a}^{b}fg = 0 for all continuous functions g. Then f=0 on [a,b].

    So yeah, I know I have to invoke that U(fg)=L(fg)=0 but after that I'm stumped. Thanks.
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  2. #2
    Member mabruka's Avatar
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    Here is an idea.


    We have \int_a^b f^2=0 where f^2\geq 0, so f^2=0 on [a,b].

    Hence f=0 on [a,b]


    What do you think?
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  3. #3
    Senior Member Pinkk's Avatar
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    I think that makes sense, yeah.
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  4. #4
    Senior Member Pinkk's Avatar
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    And I just want to make sure that proving there cannot be some point x_{0}\in [a,b] where f^{2} > 0. So suppose there is (at least) one point x_{0} such that f^{2}(x_{0}) > 0. Then since f^{2} is continuous on [a,b], f^{2} > 0 on some interval (x_{0} - \delta, x_{0} + \delta), \delta > 0, so f^{2} > 0 on the closed interval [x_{0} -\frac{\delta}{2}, x_{0} + \frac{\delta}{2}]. Clearly \inf \{f^{2}(x): x\in [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]\} > 0 since the function must achieve a minimum at a point on the interval by its continuity on a bounded interval, so for any partition P, L(f^{2}, P) > 0, so \int_{x_{0} - \frac{\delta}{2}}^{x_{0} + \frac{\delta}{2}} f^{2} > 0 and since the function is zero everywhere else, then \int_{a}^{b}f^{2} > 0, which is a contradiction.

    I think this seems a little overboard but I just want to make sure if this is correct or if there is an easier proof that shows f^{2} cannot be positive at any point. Thanks.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    And I just want to make sure that proving there cannot be some point x_{0}\in [a,b] where f^{2} > 0. So suppose there is (at least) one point x_{0} such that f^{2}(x_{0}) > 0. Then since f^{2} is continuous on [a,b], f^{2} > 0 on some interval (x_{0} - \delta, x_{0} + \delta), \delta > 0, so f^{2} > 0 on the closed interval [x_{0} -\frac{\delta}{2}, x_{0} + \frac{\delta}{2}]. Clearly \inf \{f^{2}(x): x\in [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]\} > 0 since the function must achieve a minimum at a point on the interval, so for any partition P, L(f^{2}, P) > 0 so then \int_{x_{0} - \frac{\delta}{2}}^{x_{0} + \frac{\delta}{2}} f^{2} > 0 and since the function is zero everywhere else, then \int_{a}^{b}f^{2} > 0, which is a contradiction.

    I think this seems a little overboard but I just want to make sure if this is correct or if there is an easier proof that shows f^{2} cannot be positive at any point. Thanks.
    That's similar to what I do. You can weaken this to something like if the above is true for any polynomial g(x)
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