fourier analysis - A proof of the Bernstein inequality

The Fourier transform of $f'(x)$ is $ixihat{f}(xi)$, which has the same support as $hat{f}(xi)$. So we can write $ixihat{f}(xi)$ = $ixihat{f}(xi)phi(xi)$, where $phi(xi)$ is a smooth bump function depending on the support of $hat{f}$, that is equal to one on the support of $hat{f}$. Taking inverse Fourier transforms, we get $f'(x) = f(x) star g(x)$, where $g(x)$ is the inverse Fourier transform of $ixiphi(xi)$. From the definition of convolution, one gets $|f'(x)| leq ||f||_{infty}||g||_1$. Since this holds for any $x$ and $||g||_1$ depends only on the support of $hat{f}$, you get the desired inequality with $C = ||g||_1$.