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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 $c$.

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

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 $\hat f$.

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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
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