MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ be an integrable function on $\mathbb{R}$ where support($\hat{f}$) $\subseteq$ [$-\gamma, \gamma$] for some $ 0 < \gamma < 1$

Prove that | $f(x) - f(0)$| $ \leq c \gamma$ |x| $\underset{ y \in \mathbb{R}}{sup}(1+|y|)|f(y)|$ for some absolute constant $c$.

share|cite|improve this question
1  
Maybe math.SE is a better place for this question. – Daniel Moskovich May 19 '11 at 13:13

You can successively establish the following:

$$\| f \|_2\le C\sup (1+|y|)|f(y)|,$$

$$\|\hat f\|_2\le C\|f\|_2,$$

$$\|f'\|_\infty\le C\gamma^{3/2}\|\hat f\|_2.$$

Here $\|\cdot\|_p$ denotes the $L^p$-norm. The last inequality uses the assumption about the support of $\hat f$.

share|cite|improve this answer
    
how did you get the first inequality? – jessica May 21 '11 at 20:59
    
$$\int f^2\,dy=\int f^2(1+|y|)^2(1+|y|)^{-2}\,dy$$ $$\le [\sup |f(y)|(1+|y|)]^2\int (1+|y|)^{-2}\,dy.$$ – Michael Renardy May 23 '11 at 0:06

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.