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Let $f$ be a complex valued function of a real variable for which $ \int_a^b \text{Im}(f(t))e^{int}dt=0 \text{, for all } n. $ Is $\text{Im}(f(t))$ identicaly zero on $[a,b]$?

I think the answer is yes, but I don't know why.

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    In that case we might as well assume $f$ is real-valued and replace $\mathrm{Im}(f(t))$ with $f(t)$.2012-09-16

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It is about the $L_2[a,b]$ function space: http://en.wikipedia.org/wiki/L2_space

Probably it is important that $a$ and $b$ are multiples of $\pi$.

The key is that the $\varphi_n:=t\mapsto e^{int}$ functions generate the $L_2$ space: form an orthogonal base in the sence that every $f\in L_2$ can be uniquely written in the (potentially infinite) sum: $\sum_n \alpha_n\cdot\varphi_n$ for some $\alpha_n\in\mathbb C$. Then, using the "scalar product" $\langle \varphi,\psi\rangle := \int_a^b \overline\varphi\cdot\psi$, we basically arrive that your $f$ is orthogonal to all base vector $\varphi_n$.