# A problem concerning finite number of Fourier coefficients

Is there a smooth, non-zero $2\pi$-periodic function $f,$ with support of $f$ contained in an interval $[a,b]\subset[0,2\pi],$ such that $b-a<2\pi$ and only finitely many Fourier coefficients of $f,$ $$\hat{f}(n)=\int_0^{2\pi}f(x)e^{inx}dx$$ are non-zero? Explain your answer.

My attempt: The fourier expansion has the form $$\frac{a_0}{2}+\sum_{n=1}^Na_n\sin(nx)+\sum_{n=1}^Nb_n\cos(nx).$$ Probably we could take points where the function has values $0$(there are infinitely many points) to conclude that all Fourier coefficients are $0$?

Also what puzzles me is that how could a periodic function defined on $\mathbb{R}$ only have a bounded support. Can anyone help me please?

• Of course it is intended that the function is extended periodically to the whole of $\mathbb{R}$. – Giuseppe Negro Aug 2 '14 at 13:06

## 1 Answer

HINT. Use the exponential formulation of Fourier series. You are trying to prove that, if $$f(x)=\sum_{-N}^N c_n e^{i n x}$$ vanishes at an infinite number of points, then $f$ is identically zero. You are doing well.

To conclude, try a clever substitution that makes $f$ into (the restriction to the unit circle of) a complex polynomial.

• Still not sure which substitution to use. Could you be more specific? Many thanks. – lovelesswang Aug 2 '14 at 15:03
• Note that $e^{inx}=\left(e^{ix}\right)^n$. What happens if you set $z=e^{ix}$? – Giuseppe Negro Aug 2 '14 at 16:07
• Do I need to choose $2n+1$ different values: $z_1,z_2,\ldots,z_{2n+1}$ and then try to prove the matrix with rows consisting of $n$ powers of $z_i$ is non-singular? – lovelesswang Aug 2 '14 at 16:26
• Noo, there is no need to do that. It is way easier. Did you try the substitution above? What did you get? Remember, a complex polynomial $F(z)$ can vanish only at a finite number of points, otherwise $F\equiv 0$. – Giuseppe Negro Aug 2 '14 at 17:32
• Yes, a polynomial can't have infinitely many zeroes in $\mathbb{C}$. – Shine Aug 3 '14 at 14:06