# Inequality for an integrable real valued function with a compactly supported Fourier transform

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

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

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Write Fourier inversion formula for $f(x)$ and $f(0)$. You get an expression of $f(x)-f(0)$ as an integral on the compact set $[-\gamma, \gamma]$. Then you just have to bound all the terms in the integral (to bound $1-e^{iyx}$, you may want to express it as an integral).
yes this is what i've been trying to do. I got the expression of $f(x) - f(0)$ as an integral, and i used cauchy schwatrz and plancherel, then bounded $||f||_{2} \leq c \underset{y \in \mathbb{R}}{sup}(1+ |y|)(|f(y)|)$, but i cant get the right bound from the exponential - 1. how do i write it as an integral? – jessica May 22 '11 at 18:25
actually, i could solve the integral $e^{iyx} - 1$ but it doesnt help since i cant get a bound for it with |x| in it – jessica May 22 '11 at 18:48
I think you can use $e^{ixy} - 1 = \int_0^x iy e^{ity} \, dt$. – Joel Cohen May 22 '11 at 18:54
The $|x|$ appears when you write $|\int_0^x e^{ity}| \le |x|$. – Joel Cohen May 22 '11 at 18:57