2
$\begingroup$

Fourier transform ($F$) of tempered distribution $\operatorname{sinc}$ is tempered distribution $\operatorname{Rect}$.

Does it mean that tempered distributions $F(\operatorname{sinc})$ and $\operatorname{Rect}$ behave the same as far as integration against Schwartz functions goes?

If so, how the latter (intuitively) proves that the Fourier transform (not in the sense of distributions) of $\operatorname{sinc}$ is indeed $\operatorname{Rect}$?

In other words : how does this prove that Fourier transform of $\operatorname{sinc}$ in the sense of distribution is Fourier transform of $\operatorname{sinc}$ in the sense of functions?

$\endgroup$
0
0
$\begingroup$

To answer your first question: yes, when you say that two distributions are equal you mean that they behave the same when integrated against test functions.

To show that the Fourier transform of $\operatorname{sinc}$ is $\operatorname{rect}$, I would actually go the other way around: take the Fourier transform of $\operatorname{rect}$. It's a very basic one to calculate and will yield $\operatorname{sinc}$. Then note that $\mathcal{F}^2(f) = \tilde{f}$, where $\tilde{f} = f(-x)$. But both $\operatorname{rect}$ and $\operatorname{sinc}$ are symmetric with respect to this reflection, thus you have your desired result.

$\endgroup$
4
  • 1
    $\begingroup$ Thanks, I get this, however how does F(sinc) = Rect in the sense of distributions proves that Fourier transform of sinc is Rect in the sense of functions ? $\endgroup$
    – Elaws
    Dec 3 '20 at 22:09
  • $\begingroup$ Hmm, okay I apologize I didn't quite understand your question. Well by definition for a tempered distribution $T$ then we define $(\mathcal{F}T)(f) = T(\mathcal{F}f)$ for $f$ a test function. So we'd need to use that $rect(f) = sinc(\mathcal{F}f)$ to then derive $\mathcal{F}sinc = rect$ as functions. But unfortunately I'm not quite sure how you define $rect$ and $sinc$ as distributions. . . $\endgroup$
    – Oreomair
    Dec 3 '20 at 22:39
  • $\begingroup$ Aren't they defined as their integration against $f$ over $\mathbb{R}$ ? Rect is indeed the correct spectrum of sinc, so the equality must somehow be true in the sense of functions ? But how to get from distribution equality to associated functions equality ? $\endgroup$
    – Elaws
    Dec 3 '20 at 22:48
  • $\begingroup$ Yes, I believe that works. I thought perhaps you were using the relationship $sinc$ can have to the dirac distribution (en.wikipedia.org/wiki/…). It's actually a very interesting question you ask, I'll continue to think on it. $\endgroup$
    – Oreomair
    Dec 4 '20 at 0:33
0
$\begingroup$
  • Do you mean the Fourier transform of $sinc$ in $L^2$ sense or in pointwise sense ie. the improper Riemann integrals $\int_{-\infty}^\infty sinc(x)e^{-2i\pi \xi x}dx$ ?

  • The Fourier transform of $sinc$ in the sense of distributions (call it $T$) tells us $\lim_{n\to \infty}\int_{-\infty}^\infty sinc(x) e^{-2i\pi \xi x} e^{-\pi x^2/n^2}dx \tag{1}$

    not the same as the improper Riemann integral (well it is the same for $|\xi|\ne 1/2$ but it needs a proof specific to $sinc$)

  • $(1)$ gives the Fourier transform of $sinc$ in $L^2$ sense as $\lim_{n\to \infty} T \ast ne^{-\pi n^2 x^2}$, limit in $L^2$ sense, since $T$ is the distribution $rect$ then it is $\lim_{n\to \infty} rect \ast ne^{-\pi n^2 x^2}=rect$, limit in $L^2$ sense.

$\endgroup$
3
  • $\begingroup$ Thanks. I was talking about $\int_{-\infty}^\infty sinc(x)e^{-2i\pi s x}dx$ (I don't know about the limit you describe, having more engineering rather than mathematical background^^) : How come this expression can somehow be equal to the Fourier transform of its associated distribution ? What is the link between Fourier transform of a function and Fourier transform of its associated distribution ? I'm glad I can have Fourier transform of distribution associated with sinc, but why does this also gives me the true spectrum of function sinc ? $\endgroup$
    – Elaws
    Dec 5 '20 at 0:55
  • $\begingroup$ $sinc$ is not $L^1$ so $\int_{-\infty}^\infty sinc(x)e^{-2i\pi \xi x}dx$ is not a natural thing to look at. Prove that for $|\xi|\ne 1/2$ this improper integral converges and is equal to $$\lim_{n\to \infty}\int_{-\infty}^\infty sinc(x) e^{-2i\pi \xi x} e^{-\pi x^2/n^2}dx$$ The latter is $$=\lim_{n\to \infty}\int_{-\infty}^\infty T(y)n e^{-\pi (y-\xi)^2 n^2}dy$$ by definition of the Fourier transform in the sense of distributions and our knowledge of the Gaussians. So it is $$=\lim_{n\to \infty}\int_{-\infty}^\infty rect(y)n e^{-\pi (y-\xi)^2 n^2}dy=rect(\xi)$$ $\endgroup$
    – reuns
    Dec 5 '20 at 1:21
  • $\begingroup$ What is $T(y)$ in the second line ? How do you go from functions to distributions in this explanation ? If I understand correctly, it has to do with approximating a function of infinite support with function of compact support (by multiplying by a gaussian) ? $\endgroup$
    – Elaws
    Dec 5 '20 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.