A friend is looking for a clean proof of the following inequality of Bernstein: If $f: R \to R$ is a bounded function whose Fourier transform has compact support, then $ \f'\_{\infty} \le C \ f \_{\infty} $ where $C$ only depends on the support of the Fourier transform. Any reference would be very much appreciated.
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The Fourier transform of $f'(x)$ is $i\xi\hat{f}(\xi)$, which has the same support as $\hat{f}(\xi)$. So we can write $i\xi\hat{f}(\xi)$ = $i\xi\hat{f}(\xi)\phi(\xi)$, 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 $i\xi\phi(\xi)$. 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 $C = g_1$. 

