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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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Certainly not. Consider $f$ any real-valued function. Perhaps you mean $\mathrm{Im}(f(t)e^{int})$ instead? – Alex Becker Sep 16 '12 at 20:52
@AlexBecker Good eyes, I meant $\text{Im}(f(t))$ is identically zero. Thank you. – Nicolas Essis-Breton Sep 16 '12 at 20:55
In that case we might as well assume $f$ is real-valued and replace $\mathrm{Im}(f(t))$ with $f(t)$. – Alex Becker Sep 16 '12 at 20:56
up vote 2 down vote accepted

It is about the $L_2[a,b]$ function 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 $\varphi_n$.

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