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The following is the proof from Stein and Shakarchi's Fourier Analysis that given a function $f \in \mathcal{S}(\mathbb{R})$ the Fourier transform of $-2\pi ixf(x)$ is $\frac{d}{d\xi}\hat{f}(\xi)$.

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My question is, in the second paragraph of the proof, how do we find a $h_0$ such that for all $|x|\le N$, and $|h|<h_0$, we have $$\Big|\frac{e^{-2\pi ixh}-1}{h}+2\pi ix\Big| \le \frac{\epsilon}{N}.$$ Clearly, for each $x$ we can find a $h_0$ that satisfies the inequality, but to complete the proof, we need to show that we can find a $h_0$ such that the inequality holds uniformly over $|x|\le N$. However, I don't recall any theorems that guarantee such uniform differentiability. I would greatly appreciate any suggestions or solutions.

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  • $\begingroup$ Do you know the DCT? It's simple to prove using it. $\endgroup$
    – 3-in-441
    Sep 24, 2016 at 0:54
  • $\begingroup$ @3-in-441 This is proved before DCT or measure theory is introduced. Would it not be possible to show this without using DCT? $\endgroup$ Sep 24, 2016 at 0:55
  • $\begingroup$ @3-in-441 By the way, how do we use the DCT to prove uniform differentiability here? $\endgroup$ Sep 24, 2016 at 1:00
  • $\begingroup$ @3-in-441 it is a bad idea to use over-complicated theorems for simple things, as differentiation under an integral. $\endgroup$
    – reuns
    Sep 24, 2016 at 4:02
  • $\begingroup$ @user1952009 Doesn't the proof of differentiation under an integral require at least bounded convergence theorem? At any rate, I learned measure theory before Fourier analysis and was under the impression that's usually how it's done. My apologies to the OP for over complicating. $\endgroup$
    – 3-in-441
    Sep 24, 2016 at 4:09

1 Answer 1

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Observe that $1 - 2\pi i x h\ $ is the sum of the first two terms of the Taylor series for $e^{-2\pi i x h}$. Therefore, $e^{-2\pi i x h} - 1 + 2\pi i x h\ $ is the sum of all except the first two terms: $$e^{-2\pi i x h} - 1 + 2\pi i x h = \sum_{n=2}^{\infty}\frac{(2\pi i x h)^n}{n!}$$ Taking absolute values of both sides and applying the constraint $|x| \leq N$, we obtain $$ |e^{-2\pi i x h} - 1 + 2\pi i x h| \leq \sum_{n=2}^{\infty}\frac{|2\pi x h|^n}{n!} \leq \sum_{n=2}^{\infty}\frac{|2\pi N h|^n}{n!}$$ Divide both sides by $|h| \neq 0$ to get $$\left|\frac{e^{-2\pi i x h} - 1}{h} + 2\pi i x\right| \leq \sum_{n=2}^{\infty}\frac{|2\pi N|^n}{n!}|h|^{n-1} = \sum_{k=1}^{\infty}\frac{|2\pi N|^{k+1}}{(k+1)!}|h|^k$$ The right hand side is a power series with infinite radius of convergence, so certainly it is continuous at the origin $|h| = 0$, where its value is zero. Therefore, as claimed, there is some $h_0$ such that the right hand side will be bounded by $\epsilon/N$ for all $|h| < h_0$.

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  • $\begingroup$ ?? for $h$ small enough : $|e^{-2i \pi xh} - 1+2i\pi xh| < 4\pi^2 x^2h^2 $ that's what you need here, since $f(x)$ is Schwartz, $\int_{-\infty}^\infty |4\pi^2 x^2 f(x)e^{-2i \pi \xi x}|dx$ converges and $|\frac{\hat{f}(\xi+h)-\hat{f(\xi)}}{h}-\widehat{(-2i \pi x f)}(\xi)| = |\int_{-\infty}^\infty f(x)(e^{-2i \pi xh}-1+2i\pi xh)e^{-2i \pi \xi x}dx| \to 0$ as $h \to 0$ $\endgroup$
    – reuns
    Sep 24, 2016 at 4:11

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