inequality for an integrable real valued function with a compactly supported fourier transform

Question

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 $c$.

$\begingroup$ Maybe math.SE is a better place for this question. $\endgroup$
– Daniel Moskovich
Commented May 19, 2011 at 13:13 — Daniel Moskovich, Commented May 19, 2011 at 13:13

Michael Renardy · Accepted Answer · 2011-05-19 13:59:36Z

1

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 $\hat f$.

answered May 19, 2011 at 13:47

Michael Renardy

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$\begingroup$ how did you get the first inequality? $\endgroup$
– jessica
Commented May 21, 2011 at 20:59
$\begingroup$ $$\int f^2\,dy=\int f^2(1+|y|)^2(1+|y|)^{-2}\,dy$$ $$\le [\sup |f(y)|(1+|y|)]^2\int (1+|y|)^{-2}\,dy.$$ $\endgroup$
– Michael Renardy
Commented May 23, 2011 at 0:06

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