# A proof of the Bernstein inequality

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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There is something not quite right with your statement of the inequality. – Yemon Choi Jul 3 '10 at 18:13
By the way, what sources has your friend tried? Does Google and then a search in the default texts (Zygmund, Katznelson, etc.) not do the trick? – Yemon Choi Jul 3 '10 at 18:13
Is one of your $\|f\|_\infty$s really a $\|\hat{f}\|_\infty$? – Robin Chapman Jul 3 '10 at 18:14
There is a proof in Nikolsky, S. M., Approximation of Functions of Several Variables and Imbedding Theorems, Nauka, Moscow, 1977. that seems to be unreadable. I think he has looked up Katznelson. – Keivan Karai Jul 3 '10 at 18:19
@Robin: The first one is $\|f'\|_{\infty}$. For some reason, I cannot get it fixed in the main question. – Keivan Karai Jul 3 '10 at 18:20

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 $C = ||g||_1$.

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very neat. Thanks! – Keivan Karai Jul 3 '10 at 20:30