# Thread: A problem of Riemann Integrable

1. ## A problem of Riemann Integrable

If $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\geq0$, we have $\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ a.e.?
(a.e. stands for almost everywhere)

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

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.

3. Originally Posted by Xingyuan
If $f(x)$ is Riemann Integrable in $[a,b]$,and for any integer $n\geq0$, we have $\int_{a}^{b}x^{n}f(x)dx = 0$. can we get the conclusion: $f(x)=0$ 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

4. Originally Posted by tonio
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 $f(x)\geq0$

5. Originally Posted by Xingyuan
No,In the problem,I am not assuming $f(x)\geq0$

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

6. Originally Posted by tonio
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.

7. Originally Posted by tonio
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 $\int_{-\pi}^{\pi}\sin(x)\,dx$

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

Or, for that matter, consider $\int_{-a}^a f(x)\,dx$ where $f(x)$ is any odd function.

8. Originally Posted by rn443
Huh? That doesn't follow.

Yes it does

Tonio

9. Originally Posted by tonio
Yes it does

Tonio
Are you assuming f is nonnegative?

10. Originally Posted by rn443
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

11. Does the theorem still hold if we instead assume $\int_{-\infty}^\infty x^n f(x) \mathrm dx = 0$ for all n >= 0, provided that $\int_{-\infty}^\infty |x^n f(x)| \mathrm dx$ exists and is finite?

12. Lets suppose, without loss of generality, that is $a=-\pi$ and $b=\pi$. Since $f(*)$ is Riemann integrable in $[-\pi,\pi]$ is can be expanded in Fourier series ...

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

... where...

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

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

Now is...

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

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

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 $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$
If we denote the set $E$ of all discontinuity point,then define:

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

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

then use the conclusion above,we get the conclusion $g(x)\equiv0$.
so $f(x)=0$a.e.
Is this proof right?

14. Originally Posted by Xingyuan
Yes,Riemann Integrable function is continous a.e.
But Weierstrass' approximation theorem is:
Every function defined and continuous on the finite interval $[a,b]$can be approximated uniformly on $[a,b]$ by polynomials to any degree of accuracy.
Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $g(x)$
If we denote the set $E$ of all discontinuity point,then define:

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

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

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

Another thing, the $g$ 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 $\int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$

15. Originally Posted by Jose27
This doesn't contradict the claim since $\int_{-\pi } ^{\pi } x\sin(x) dx=2\int_{0 } ^{ \pi } x\sin(x) dx >0$
I was only contradicting the case $n=0$.

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