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Let $f$ be a continuous function on $[a,b]$ such that $\int_a^bx^nf(x)dx=0$ for all nonnegative integer $n$. Prove that $f(x)=0$ for all $x\in[a,b]$.

The equation tells us that $\int_a^bp(x)f(x)dx=0$ for all polynomials $p$. I was thinking to choose a nonzero polynomial $p$ such that $p(x)f(x)\ge0$ for all $x\in[a,b]$. However this is apparently impossible if $f$ has infinitely many roots. Then I tried to make $p(x)f(x)$ very large in some interval, and very small in the other intervals. Again this does not seem to work if the roots of $f$ is dense in $[a,b]$.

Edit: Okay, so I now know that the roots of $f$ cannot be dense in $[a,b]$ (otherwise $f$ is constant and we're done). Now $f$ is positive (wlog) in some interval $[c,d]$. I want $p(x)$ to be a polynomial such that $p(x)f(x)$ is very large in $[c,d]$ and very small elsewhere. I'm definitely sure that such a polynomial exists, but don't know how to prove it. Can my idea be finished?

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    This question is a duplicate of http://math.stackexchange.com/questions/16831/a-nonzero-function-in-c0-1-for-which-int-01-fxxn-dx-0-forall-n/2011-11-12

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The polynomials are dense in $C([a,b])$. Thus you can choose a sequence of polynomials $p_n$ such that $p_n(x) \to f(x)$ uniformly on $[a,b]$. Then $0 = \lim_{n \to \infty} \int_a^b p_n(x) f(x) dx = \int_a^b f^2(x) dx.$

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    I've not learned that far, but thanks for taking time to explain. Can you help me continue my idea? See my edits above.2011-11-12