I am looking at the proof that if $f$ is a Scwhartz function, i.e., $f \in \mathcal{S}(\mathbb{R})$ then the Fourier transform of $-2\pi ixf(x)$ is $\frac{d}{d\xi} \hat{f}(\xi)$.
For the proof, let $\epsilon >0$ and consider $$\frac{\hat{f}(\xi +h)-\hat{f}(\xi)}{h}-(\widehat{-2\pi ixf})(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi ix\xi}\Big[\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\Big] dx.$$
I have a question about the step in the attached proof below. In the proof, we make $$\int_{|x|\ge N}\Big|f(x)e^{-2\pi ix\xi}\Big[\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\Big]\Big|dx \le C\epsilon.$$ To show this I think the authors intend to split this into two integrals, so $\int_{|x|\ge N}\Big|f(x)e^{-2\pi ix\xi} 2\pi ix\Big|dx\le 2\pi \epsilon$ and $\int_{|x|\ge N}\Big|f(x)e^{-2\pi ix\xi}\frac{e^{-2\pi ixh}-1}{h}\Big|dx\le C'\epsilon.$ However, I don't know how to bound the second integral using $\int_{|x|\ge N} |f(x)|dx \le \epsilon$, since here we have $\Big|\frac{e^{-2\pi ixh}-1}{h}\Big|\le \frac{2}{|h|}$ and $|h|$ is supposed to be small, hence this part goes unbounded. So how can we deal with this fraction when $|x|\ge N$ to bound the integral as in the proof below? I would greatly appreciate any help.
You have to actually use that $$ |e^{-2πixh}-1+2πixh|\le C\,x^2h^2 $$ and that $$ \int_{-∞}^∞|f(x)|\,|x|^2\,dx $$ is finite for the rapidly falling Schwartz test functions.
\begin{align} |e^{-2iu}-1+2iu|&=|-2i\sin(u)+2iue^{iu}| \\ &\le2\,|\sin(u)-u\cos(u)|+2|u\sin(u)| \\ &\le|\sin(\theta u)u^2|+2|u|\min(1,|u|)\le 3u^2 \end{align} as per Taylor $$0=\sin(0)=\sin(u-u)=\sin(u)-\cos(u)u-\frac12\sin(θu)u^2$$ with $θ\in(0,1)$.
One could argue that the first term is locally of third order, but that does not help the proof.