Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
3  
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
show 2 more comments

1 Answer 1

up vote 11 down vote accepted

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 $C = ||g||_1$.

share|improve this answer
    
very neat. Thanks! –  Keivan Karai Jul 3 '10 at 20:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.