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This question follow up from a discussion on this post Prove that there exists a constant $c>0$ such that for $R>0$ and $t\in \Bbb R$ we have, $$\left|\int_{|x|<R} \frac{\sin x}{x}e^{-ixt}dx\right|<c$$

This aim to rpove that the Fourier trnasform of $g: x\mapsto \frac{\sin x}{x}$

Given by $$\widehat{g}(t) = \lim_{R\to \infty}\int_{|x|<R} \frac{\sin x}{x}e^{-ixt}dx$$ is bounded. Please I would like to see a prove with elementary tools. I mean a prove that can ignore the Fourier analysis here.

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    $\begingroup$ Why would you want to ignore Fourier analysis in a question about a Fourier transform? $\endgroup$ – Daniel Fischer Dec 1 '17 at 16:53
  • $\begingroup$ @DanielFischer from discussion [here][math.stackexchange.com/questions/2545959/…} it seems to be obvious using plancherel. Although I don't see why? $\endgroup$ – Guy Fsone Dec 1 '17 at 16:59
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    $\begingroup$ Up to constant factors (depending on the normalisation of the Fourier transform), $\frac{\sin x}{x}$ is the Fourier transform of the characteristic function of a bounded interval. By the inversion theorem [for the Fourier transform on $L^2$], the Fourier transform of $\frac{\sin x}{x}$ is [a multiple of] the characteristic function of a bounded interval. $\endgroup$ – Daniel Fischer Dec 1 '17 at 17:05
  • $\begingroup$ @DanielFischer that is really the point. I don't want to use that argument. since we don't how the Fourrier transform on $L^2$ look like. similar can be found here math.stackexchange.com/questions/2535865/… $\endgroup$ – Guy Fsone Dec 1 '17 at 17:08
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    $\begingroup$ The $L^2$ Fourier transform (i) is an isometry on $L^2$ and (ii) agrees with the $L^1$ Fourier transform on $L^2\cap L^1$. That tells us everything we need to know about what it looks like. $\endgroup$ – David C. Ullrich Dec 1 '17 at 21:17
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$$\int_{-R}^R 2\frac{\sin x}{x} e^{-ixt}dx =\int_{-R}^R \int_{-1}^{1}e^{i x u}du e^{-ixt}dx =\int_{-1}^{1}\int_{-R}^R e^{ix(u-t)}dx du = \int_{-1}^1 2\frac{\sin (R(u-t))}{u-t}du=\int_{-1-t}^{1-t} 2\frac{\sin (Rv)}{v}dv=\int_{-R(1+t)}^{R(1-t)} 2\frac{\sin w}{w}dw$$

Thus for $|t| \ne 1$ writing $$\int_{-R(1+t)}^{R(1-t)} (...)= 1_{|t|\le 1}\int_{-R(1+t)}^{R(1-t)}(...)+ 1_{|t|>1}\int_{-R(1+t)}^{R(1-t)}(...)$$ we have, $$ \lim_{R \to \infty} 1_{|t|>1}\int_{-R(1+t)}^{R(1-t)}(...)=0 $$ Since $|t|>1$ implies $t+1$ and $t-1$ are of the same sign.

On the other hand $|t|<1$ we have, $1+t>0$ and $1-t>0$. whence, $$\lim_{R \to \infty} \int_{-R}^R 2\frac{\sin x}{x} e^{-ixt}dx = \lim_{R \to \infty} \int_{-R(1+t)}^{R(1-t)} 2\frac{\sin w}{w}dw \\= 1_{|t|<1}\lim_{R \to \infty} \int_{-R(1+t)}^{R(1-t)} 2\frac{\sin w}{w}dw = 1_{|t|<1}\int_{-\infty}^{\infty} 2\frac{\sin w}{w}dw\\ =2\pi1_{|t|<1}$$

The same method works for showing the Fourier inversion theorem for any $f \in L^1,f' \in L^1$ or $f'$ piecewise $L^1$ as here.

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  • $\begingroup$ How do you obtain the first equality? also do you directly consider $t>0?$ $\endgroup$ – Guy Fsone Dec 1 '17 at 17:22
  • $\begingroup$ By integrating $e^{ixu}$ $\endgroup$ – reuns Dec 1 '17 at 17:23
  • $\begingroup$ Then it should be $e^{-ixu}$ $\endgroup$ – Guy Fsone Dec 1 '17 at 17:24
  • $\begingroup$ What did you use in the two last equalities? $\endgroup$ – Guy Fsone Dec 1 '17 at 17:34
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    $\begingroup$ What do you not understand in David's answer ?.. $\endgroup$ – reuns Dec 1 '17 at 17:40

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