A funtion and its fourier transformation cannot both be compactly supported unless f=0 Problem : Suppose that $f$ is continuous on $\mathbb{R}$. Show that $f$ and $\hat f$ cannot both be compactly supported unless $f=0$.
Hint : Assume $f$ is supported in [0,1/2]. Expand $f$ in a Fourier series in the interval [-,1], and note that as a result, f is a trigonometric polynomial.
I proved that f is trigonometric polynomial by using hint.
But, I don't know how to prove function's fourier transform cannot compactly supported function. 
Can I get some hints?

Suppose the support of $f$ is contained in $[-1,1],$ and $\hat f (y) = 0$ for $|y|>N \in \mathbb N.$ Applying a standard Fourier series argument on $[-\pi,\pi]$ then shows
$$f(x) = \sum_{-N}^{N}\hat f (n) e^{inx}, x \in [-\pi,\pi].$$
Thus $f$ is a trigonometric polynomial that vanishes on $[1,\pi].$ But a trigonometric polynomial on $[-\pi,\pi]$ that vanishes at an infinite number of points must vanish identically. Thus $f\equiv 0.$
