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What can we say about $f$ if $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$?

I want to prove that for $f \in L^1(\mathbb{T})$ s.t $\forall g \in C(\mathbb{T}) \ \int_{\mathbb{T}} fg = 0$ then $\int_{\mathbb{T}} |f| = 0$.

(where $\mathbb{T}$ is the unit circle).

How to show this?

I thought approximating $f$ by some $h$ continuous, i.e $||f-h||_1 \leq \epsilon$, but I don't see how to procceed from here, any hints or full solutions will be appreciated.

Thanks.

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    See http://math.stackexchange.com/questions/17026/what-can-we-say-about-f-if-int-01-fxpxdx-0-for-all-polynomials-p. It follows from the uniqueness of Fourier coefficients.2012-06-02
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    Does C(T) represent the continuous functions?2012-06-02
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    If F was nonzero on a set of positive measure, couldn't you approximate the indicator function for that set by continuous functions?2012-06-02
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    Thanks Jonas, didn't think it was that much easy.2012-06-02
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    For a full solution, maybe not so easy; you have to prove that Fourier coefficients are unique! Fortunately it *is* easy to Google for references on that.2012-06-02

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