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Is it true that : If for $f,g \in C[0,1]$, $k\in \mathbb{N}$ it holds : $$\int_0^1 f(x)x^kdx= \int_0^1g(x)x^kdx$$

Then $$f=g$$

What have I tried :

One can rewrite: $$ \int_0^1 (f(x)-g(x))x^kdx=0$$

f,g are continuous on $C[0,1]$ and from the approximation theorem of Stone-Weierstrass one knows that there is a sequence $\{p_n\}$ of polynomials that converges uniformly to $f(x)$ for $n\rightarrow \infty$ and also one $\{p_t\}$ which converges uniformly to $g(x)$ for $t\rightarrow \infty $

so: $$\int_0^1 \lim_{n\rightarrow \infty} \{p_n\}x^kdx = \int_0^1 \lim_{t\rightarrow \infty} \{p_t\}x^kdx =\lim_{n\rightarrow \infty}\int_0^1\{p_n\}x^kdx=\lim_{t\rightarrow \infty}\int_0^1\{p_t\}x^k dx $$

this doesnt work

How does one go about showing it?

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up vote 2 down vote accepted

This is Application 11.6 in N. L. Carothers - Real Analysis.

The idea is that there is a sequence $(p_n)$ of polynomial functions converging uniformly to $f$, so you get that $p_n\cdot f$ converges uniformly to $f^2$. Then evaluate $\int_0^1 f^2(x)\,dx = \lim_{n\to\infty} \int_0^1 p_n(x)f(x)\,dx$. But all of the integrals involved in the limit on the RHS are $0$, so the RHS equals $0$ and thus $f$ must be the $0$ function.

PS: - this search would probably have dug up something useful for you. It helps knowing what the problem one is studying is called.

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Thank you very much, this is very helpful ! – bakabakabaka Jan 24 '13 at 11:31

Note that $$\int_0^1 h(x)x^k~dx=0 \implies h(x)=0 \forall x\in [0,1]$$Since here $x \geq 0$.

Put $h(x)=f(x)-g(x)$ You get your result.

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Thanks, does this work for any intervall if $C[a,b]$ is given (and the integral boundaries are changed to : $\int_a^b f(x)x^k)$ ? – bakabakabaka Jan 24 '13 at 11:35
@bakabakabaka: If $a > 0 ,\forall k$ then the result holds again if $k$ is even then the result holds for all $[a,b]$ – A.D Jan 24 '13 at 16:24

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