# What constraints can be on the Fourier coefficients of $f(t)$ if $0 \leq f(t) \leq 1$

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 $b_n$ ($n \geq 1$)?

• Is $f(t)$ continuous? – Mark Jan 12 '17 at 4:30
• Yes. $f(t)$ is continuous – regress Jan 12 '17 at 4:35
• Tell us about what you tried. – SchrodingersCat Jan 12 '17 at 4:40
• Well I don't know even what I can try. – regress Jan 12 '17 at 4:45

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