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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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    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

2 Answers 2

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$|f(x)| = f(x) \cdot \text{sgn} f(x)$ Consider $h_\epsilon$ - a continuous approximation of $\text{sgn} f$ such that $h(x') \neq \text{sgn} f(x')$ only in $\epsilon$ neighborhoods of points where $f$ changes sign. Then $x_\epsilon = \int_{\mathbb T} fh = 0$ and $\lim \limits_{\epsilon \rightarrow \ 0} x_\epsilon = \int_{\mathbb T} |f| = 0$.

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    My plan is to construct a sequence of functions that converges to $\text{sgn} f$. I now see that my construction is naive (assumes that the set of points where $f$ changes sign is countable), but I still think that such sequence can be constructed and that integration preserves it's limit (the latter is due to continuity of $F(g) = \int fg$, it seems). You're right, though. My proof is poor.2012-06-02
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The link posted by @Jonas has a Fourier-based solution, but my personal inclination is to use Lebesgue differentiation. Namely, suppose $f\ne 0$ in $L^1$ and let $a$ be a Lebesgue point of $f$ with $f(a)>0$ (wlog). There exists $R>0$ such that $\int_{a-r}^{a+r}f(x)\,dx\ge r f(a)$ for all $0. Hence $\int_0^R \int_{a-r}^{a+r}f(x)\,dx\,dr \ge \frac12 R^2 f(a)>0$ On the other hand, Fubini yields $\int_0^R \int_{a-r}^{a+r}f(x)\,dx\,dr = \int_{a-R}^{a+R} \int_{|x-a|}^R f(x)\,dr\,dx = \int_{a-R}^{a+R} (R-|x-a|) f(x)\,dx$ which is zero by assumption, with $g=(R-|x-a|)^+$. Contradiction.