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Math Help - Reimann integral question

  1. #1
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    Reimann integral question

    Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

    Why isn't g(x)= 0 why does it have to be f(x)? Why not both?
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  2. #2
    Super Member girdav's Avatar
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    Re: Reimann integral question

    What we suppose works for all Riemann-integrable function g, and we only know that \int_a^b fg =0.
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  3. #3
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    Re: Reimann integral question

    Quote Originally Posted by CountingPenguins View Post
    Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].
    From the given do you know that \int_a^b {f^2 }  = 0~?

    Is that possible if f(t)\not=0 for some t\in[a,b]~?
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  4. #4
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    Re: Reimann integral question

    f is continuous and g is integrable and the product of the integral of fg over the interval [a,b] is 0. So f(x) must be 0 for all x in [a,b], is all the information I have been given. I will start with the definition of continuous for f and integrable for g and then hope I see something useful after that.
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  5. #5
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    Re: Reimann integral question

    Did you see reply #3. That answers the question.
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Reimann integral question

    Quote Originally Posted by CountingPenguins View Post
    Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

    Why isn't g(x)= 0 why does it have to be f(x)? Why not both?
    I came up with the following solution, I hope it is not so wrong.


    Suppose that f is not 0 for all x\in[a,b], then there exists x_0\in[a,b] such that f(x_0)>0 (If the is no such x_0, there must be x_1\in[a,b] such that f(x_1)<0.In that case we will look at -f).

    f is continuous function therefor exists \delta>0 for which f(x)>\epsilon>0 for all x\in (x_0-\delta,x_0+\delta).

    Now we define h(x)-'staircase'-characteristic function of open interval (x_0-\delta,x_0+\delta).

    h(x)=\left\{\begin{matrix}1 & x\in (x_0-\delta,x_0+\delta) \\ 0 & \text{otherwise} \end{matrix}\righ

    Hence, we will get:

    0=\int_{a}^{b}f(x)h(x)dx=\int_{a}^{x_0-\delta}f(x)h(x)dx+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+\int_{x_0+\delta}^{  b}f(x)h(x)dx=0+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+0>2\epsilon\delta .

    So, we got a contradiction to the fact that if f is continuous on [a,b] and h is staircase function \int_{a}^{b}f(x)h(x)dx=0 \Rightarrow f\equiv 0.

    But we also know that if g is integrable function on [a,b], so for every \epsilon_0>0 exists staircase function h so that

    \int_{a}^{b}|g(x)-h(x)|<\epsilon_0


    Combining these two results we will conclude the needed.
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  7. #7
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    Re: Reimann integral question

    Quote Originally Posted by Also sprach Zarathustra View Post
    I came up with the following solution, I hope it is not so wrong.
    Your solution is correct.
    Note that f \text{ is continuous on }[a,b] so as such that means that both f~\&~f^2 is are integrable on [a,b].

    By the given \int_a^b {f \cdot f}  = \int_a^b {f^2 }  = 0.
    But by your reasoning that must imply that f  \equiv 0.
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