A proof of the Bernstein inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:00:40Z http://mathoverflow.net/feeds/question/30434 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/30434/a-proof-of-the-bernstein-inequality A proof of the Bernstein inequality Keivan Karai 2010-07-03T18:09:33Z 2010-07-03T18:53:10Z <p>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 <code>$\|f'\|_{\infty} \le C \| f \|_{\infty}$</code> where $C$ only depends on the support of the Fourier transform. Any reference would be very much appreciated.</p> http://mathoverflow.net/questions/30434/a-proof-of-the-bernstein-inequality/30447#30447 Answer by Michael Greenblatt for A proof of the Bernstein inequality Michael Greenblatt 2010-07-03T18:53:10Z 2010-07-03T18:53:10Z <p>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 $C = ||g||_1$.</p>