2
$\begingroup$

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.

enter image description here

$\endgroup$
8
  • $\begingroup$ come on we answered that on your other question math.stackexchange.com/questions/1938990/… so make more efforts please, read different courses, do some research on the forum $\endgroup$
    – reuns
    Oct 3, 2016 at 9:35
  • $\begingroup$ @user1952009 Oh I just read your comment below about the inequality. But still I think the answer below helps as it shows how to get it. Anyways thanks for your comment :) $\endgroup$ Oct 3, 2016 at 9:38
  • $\begingroup$ Can you write a proof of $f \in C^2 \implies $ what I wrote ? (hint : intermediate value theorem) $\endgroup$
    – reuns
    Oct 3, 2016 at 9:55
  • 1
    $\begingroup$ a proof that if $f \in C^2(\mathbb{R})$ then there exists $c$ such that $|f(y) - f(0)-y f'(0)| < c y^2$ for $|y| < 1$ $\endgroup$
    – reuns
    Oct 3, 2016 at 10:00
  • 1
    $\begingroup$ It is the proof of Taylor's theorem. By the mean value theorem, there is a $c(y) \in (0,y)$ such that $ f'(c(y)) =\frac{f(y)-f(0)}{y}$. And there is a $d(y) \in (0,c(y))$ such that $\frac{f'(c(y))-f'(0)}{c(y)} = f''(d(y))$, hence $f(y) = f(0)+y f'(c(y)) = f(y)+f(0) + y f'(0)+ y c(y) f''(d(y))$ and by definition $|y(c(y)) f''(d(y))| \le | y^2| \sup_{ t \in [-1,1]} |f''(t)|$ for $|y| < 1$ $\endgroup$
    – reuns
    Oct 3, 2016 at 10:25

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Can you tell me where we can get the first bound? $\endgroup$ Oct 2, 2016 at 17:10
  • 2
    $\begingroup$ @takecare $f(y) = e^{-2i \pi y}$ is $C^2$ at $y = 0$ so as $y \to 0$ : $f(y) = f(0) + yf'(0) + \mathcal{O}(y^2)$ which means precisely that there exists a $C$ such that $|f(y) - f(0) - y f'(0)| < C y^2$ for $|y|$ small enough, say for $|y| < 1$. Since $e^{-2i \pi y}$ is $1$ periodic, it means that $|f(y) - f(0) - y f'(0)| < C y^2$ is true for every $y \in \mathbb{R}$. Finish with $y = xh$ $\endgroup$
    – reuns
    Oct 3, 2016 at 9:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .