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Thread: Proving a function is identically zero

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

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

    So yeah, I know I have to invoke that $\displaystyle U(fg)=L(fg)=0$ 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 $\displaystyle \int_a^b f^2=0$ where $\displaystyle f^2\geq 0$, so $\displaystyle f^2=0$ on $\displaystyle [a,b]$.

    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 $\displaystyle x_{0}\in [a,b]$ where $\displaystyle f^{2} > 0$. So suppose there is (at least) one point $\displaystyle x_{0}$ such that $\displaystyle f^{2}(x_{0}) > 0$. Then since $\displaystyle f^{2}$ is continuous on $\displaystyle [a,b]$, $\displaystyle f^{2} > 0$ on some interval $\displaystyle (x_{0} - \delta, x_{0} + \delta), \delta > 0$, so $\displaystyle f^{2} > 0$ on the closed interval $\displaystyle [x_{0} -\frac{\delta}{2}, x_{0} + \frac{\delta}{2}]$. Clearly $\displaystyle \inf \{f^{2}(x): x\in [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]\} > 0$ since the function must achieve a minimum at a point on the interval by its continuity on a bounded interval, so for any partition $\displaystyle P$, $\displaystyle L(f^{2}, P) > 0$, so $\displaystyle \int_{x_{0} - \frac{\delta}{2}}^{x_{0} + \frac{\delta}{2}} f^{2} > 0$ and since the function is zero everywhere else, then $\displaystyle \int_{a}^{b}f^{2} > 0$, 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 $\displaystyle f^{2}$ 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 $\displaystyle x_{0}\in [a,b]$ where $\displaystyle f^{2} > 0$. So suppose there is (at least) one point $\displaystyle x_{0}$ such that $\displaystyle f^{2}(x_{0}) > 0$. Then since $\displaystyle f^{2}$ is continuous on $\displaystyle [a,b]$, $\displaystyle f^{2} > 0$ on some interval $\displaystyle (x_{0} - \delta, x_{0} + \delta), \delta > 0$, so $\displaystyle f^{2} > 0$ on the closed interval $\displaystyle [x_{0} -\frac{\delta}{2}, x_{0} + \frac{\delta}{2}]$. Clearly $\displaystyle \inf \{f^{2}(x): x\in [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]\} > 0$ since the function must achieve a minimum at a point on the interval, so for any partition $\displaystyle P$, $\displaystyle L(f^{2}, P) > 0$ so then $\displaystyle \int_{x_{0} - \frac{\delta}{2}}^{x_{0} + \frac{\delta}{2}} f^{2} > 0$ and since the function is zero everywhere else, then $\displaystyle \int_{a}^{b}f^{2} > 0$, 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 $\displaystyle f^{2}$ 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 $\displaystyle g(x)$
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