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Given the following system of integral equations for an integrable function $f(x)$:

For all integers $k \ge 1$ holds

$\int_{0}^{2\pi} [f(x)]^k e^{(ikx)} dx = 0$.

If $f(x)$ is real-valued and non-negative, is $f(x)=const$ almost everywhere the only solution to the system? Any ideas are very much appreciated.

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    $\begingroup$ For complex-valued functions certainly not: take $f(x)=e^{2ix}$. Is your $f$ real-valued? $\endgroup$
    – fedja
    Aug 25, 2018 at 15:32

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Another simple solution that comes to mind : $f(x) = \cos(x)$, because then $[f(2\pi-x)]^k \sin(k(2\pi-x)) =[\cos(2\pi-x)]^k \sin(2k\pi-kx) = [\cos(x)]^k [-sin(kx)] = -[f(x)]^k\sin(kx)$ and thus $\int_0^{\pi}+\int_{\pi}^{2\pi} = 0$.

Any similarly symmetric function would work, e.g. one that is positive like $f(x) = |\pi-x|$.

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  • $\begingroup$ Sorry I wanted to search for solutions were f is real valued and non negative. $\endgroup$
    – Schmelli
    Aug 25, 2018 at 16:27
  • $\begingroup$ Any similarly symmetric function would work, e.g. one that is positive like $f(x) = |\pi-x|$. $\endgroup$ Aug 25, 2018 at 20:21
  • $\begingroup$ How would be the situation if the functions sin(kx) are replaced by exp(ikx)? Still f is supposed to be real-valued and non-negative. Thanks a lot! $\endgroup$
    – Schmelli
    Aug 26, 2018 at 7:18

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