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Let's say $f(t)$ is a periodic and bounded signal, so it can be represented with Fourier series:

$f(t)= a_0 + \sum_{n=1}^{\infty}a_n\mathbf{cos}(n\omega _0t) + \sum_{n=1}^{\infty}b_n\mathbf{sin}(n\omega _0t)$

If $0 \leq f(t) \leq 1 $, what constraints can be on $a_0$, $a_n$ and $b_n$ ($n \geq 1 $)?

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    Is $f(t)$ continuous? – 2017-01-12
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    Yes. $f(t)$ is continuous – 2017-01-12
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    Tell us about what you tried. – 2017-01-12
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    Well I don't know even what I can try. – 2017-01-12

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By Parseval equation we get $$ \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)=\frac{1}{\pi}\int_{-\pi}^{\pi}\lvert f(x)\rvert^2\, dx\, \leq \frac{1}{\pi}\int_{-\pi}^{\pi}dx\, =2 $$

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    Thanks! I've forgotten the Parseval's theorem. – 2017-01-12