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Thread: A problem of Riemann Integrable

  1. #1
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    Talking A problem of Riemann Integrable

    If $\displaystyle f(x)$ is Riemann Integrable in $\displaystyle [a,b]$,and for any integer $\displaystyle n\geq0$, we have $\displaystyle \int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion:$\displaystyle f(x)=0$ a.e.?
    (a.e. stands for almost everywhere)
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  2. #2
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    Yes. Note that if $\displaystyle \int_{a} ^{b} \ x^nf(x)dx =0$ for all $\displaystyle n \in \mathbb{N}$ then $\displaystyle \int_{a} ^{b} \ p(x)f(x)dx =0$ for any polynomial $\displaystyle p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $\displaystyle C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $\displaystyle I: C^0([a,b]) \rightarrow \mathbb{R}$ $\displaystyle g \mapsto \int_{a} ^{b} \ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\displaystyle \int_{a} ^{b} \ g(x)f(x)dx =0$ for all $\displaystyle g(x) \in C^0([a,b])$. It is a well known result that under this conditions $\displaystyle f \equiv 0$.

    Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.
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    Quote Originally Posted by Xingyuan View Post
    If $\displaystyle f(x)$ is Riemann Integrable in $\displaystyle [a,b]$,and for any integer $\displaystyle n\geq0$, we have $\displaystyle \int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion:$\displaystyle f(x)=0$ a.e.?
    (a.e. stands for almost everywhere)

    Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.

    Tonio
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  4. #4
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    Quote Originally Posted by tonio View Post
    Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.

    Tonio
    No,In the problem,I am not assuming $\displaystyle f(x)\geq0$
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    Quote Originally Posted by Xingyuan View Post
    No,In the problem,I am not assuming $\displaystyle f(x)\geq0$

    I didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.

    Tonio
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    Quote Originally Posted by tonio View Post
    Well you made your question almost trivial allowing n >= 0 ==> taking n = 0 we get that INT{1*f(x)} dx = 0 ==> f(x) = 0 a.e.

    Tonio
    Huh? That doesn't follow.
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  7. #7
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    Quote Originally Posted by tonio View Post
    I didn't say anything about f(x) >= 0. I took n = 0, since you wrote n >= 0, and then x^0 =1 and INT{x^0*f(x)} dx = INT fx dx = 0 ==> f(x) = 0 a.e.

    Tonio
    Consider $\displaystyle \int_{-\pi}^{\pi}\sin(x)\,dx$

    It's only $\displaystyle 0$ at the two endpoints and $\displaystyle x=0$, but the whole integral is $\displaystyle 0$.

    Or, for that matter, consider $\displaystyle \int_{-a}^a f(x)\,dx$ where $\displaystyle f(x)$ is any odd function.
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    Quote Originally Posted by rn443 View Post
    Huh? That doesn't follow.

    Yes it does

    Tonio
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    Quote Originally Posted by tonio View Post
    Yes it does

    Tonio
    Are you assuming f is nonnegative?
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    Quote Originally Posted by rn443 View Post
    Are you assuming f is nonnegative?
    Ok, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.

    Tonio
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    Does the theorem still hold if we instead assume $\displaystyle \int_{-\infty}^\infty x^n f(x) \mathrm dx = 0$ for all n >= 0, provided that $\displaystyle \int_{-\infty}^\infty |x^n f(x)| \mathrm dx$ exists and is finite?
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  12. #12
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    Lets suppose, without loss of generality, that is $\displaystyle a=-\pi$ and $\displaystyle b=\pi$. Since $\displaystyle f(*)$ is Riemann integrable in $\displaystyle [-\pi,\pi]$ is can be expanded in Fourier series ...

    $\displaystyle f(x)= \frac{a_0}{2} + \sum_{k=1}^{\infty} a_{k}\cdot \cos kx + b_{k}\cdot \sin kx$ (1)

    ... where...

    $\displaystyle a_{k}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cdot \cos kx\cdot dx$

    $\displaystyle b_{k}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cdot \sin kx\cdot dx$ (2)

    Now is...

    $\displaystyle \cos kx = \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot (kx)^{2n}}{(2n)!}$

    $\displaystyle \sin kx = \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot (kx)^{2n+1}}{(2n+1)!}$ (3)

    ... and $\displaystyle \forall n\ge 0$ is...

    $\displaystyle \int_{-\pi}^{\pi} x^{n}\cdot f(x)\cdot dx =0$ (4)

    The conclusion is that all the integrals in (2) are null and the same is for the $\displaystyle a_{k}$ and $\displaystyle b_{k}$ ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  13. #13
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    Quote Originally Posted by Jose27 View Post
    Yes. Note that if $\displaystyle \int_{a} ^{b} \ x^nf(x)dx =0$ for all $\displaystyle n \in \mathbb{N}$ then $\displaystyle \int_{a} ^{b} \ p(x)f(x)dx =0$ for any polynomial $\displaystyle p(x)$. By Wierstrass aproximation theorem, this set (polynomials) is dense in $\displaystyle C^0 ([a,b])$ where this last is considered with the uniform norm. The functional $\displaystyle I: C^0([a,b]) \rightarrow \mathbb{R}$ $\displaystyle g \mapsto \int_{a} ^{b} \ g(x)f(x)dx$ is linear and continous and is zero on a dense subset of the domain so it must be zero everywhere. So $\displaystyle \int_{a} ^{b} \ g(x)f(x)dx =0$ for all $\displaystyle g(x) \in C^0([a,b])$. It is a well known result that under this conditions $\displaystyle f \equiv 0$.

    Edit: Actually, we would need f to be continous, but since it's R-I it's continous a.e. so there is no real problem.
    Yes,Riemann Integrable function is continous a.e.
    But Weierstrass' approximation theorem is:
    Every function defined and continuous on the finite interval $\displaystyle [a,b] $can be approximated uniformly on $\displaystyle [a,b]$ by polynomials to any degree of accuracy.
    Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $\displaystyle g(x)$
    If we denote the set $\displaystyle E$ of all discontinuity point,then define:

    $\displaystyle g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$

    then $\displaystyle g(x)$ is a continuous function on$\displaystyle [a,b]$,right?

    then use the conclusion above,we get the conclusion $\displaystyle g(x)\equiv0$.
    so $\displaystyle f(x)=0$a.e.
    Is this proof right?
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  14. #14
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    Quote Originally Posted by Xingyuan View Post
    Yes,Riemann Integrable function is continous a.e.
    But Weierstrass' approximation theorem is:
    Every function defined and continuous on the finite interval $\displaystyle [a,b] $can be approximated uniformly on $\displaystyle [a,b]$ by polynomials to any degree of accuracy.
    Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $\displaystyle g(x)$
    If we denote the set $\displaystyle E$ of all discontinuity point,then define:

    $\displaystyle g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$

    then $\displaystyle g(x)$ is a continuous function on$\displaystyle [a,b]$,right?

    then use the conclusion above,we get the conclusion $\displaystyle g(x)\equiv0$.
    so $\displaystyle f(x)=0$a.e.
    Is this proof right?
    Notice that I'm using Weierstrass to expand the set of functions $\displaystyle g$ such that $\displaystyle \int_{a} ^{b} g(x)f(x)dx =0$, I'm not saying anything about $\displaystyle f$.

    Another thing, the $\displaystyle g$ you defined could be not well defined if f has a jump discontinuity.

    Ok, I thought you gave f(x) >= 0 but you actually didn't, but then the claim is false as redsoxfan showed already.
    This doesn't contradict the claim since $\displaystyle \int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$
    Last edited by Jose27; Oct 15th 2009 at 10:19 AM.
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  15. #15
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Jose27 View Post
    This doesn't contradict the claim since $\displaystyle \int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$
    I was only contradicting the case $\displaystyle n=0$.
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