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?

If $f$ is compactly supported, then $\hat{f}$ is holomorphic. You can see that because the Fourier transform integral extends into the complex plane and is differentiation in the complex variable $s$:
$$
          \hat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-ist}f(t)dt \\
          \hat{f}'(s) = \frac{1}{\sqrt{2\pi}}\int_{-R}^{R}(-it)e^{-ist}f(t)dt.
$$
Holomorphic functions cannot have too many zeros, which would happen if $\hat{f}$ were compactly supported.
