# Fourier transform of a tempered distribution : why a link with classical Fourier transform?

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?

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.

• 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 ? Dec 3 '20 at 22:09
• 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. . . Dec 3 '20 at 22:39
• 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 ? Dec 3 '20 at 22:48
• 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. Dec 4 '20 at 0:33
• 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.

• 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 ? Dec 5 '20 at 0:55
• $sinc$ 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 $$=\lim_{n\to \infty}\int_{-\infty}^\infty rect(y)n e^{-\pi (y-\xi)^2 n^2}dy=rect(\xi)$$ Dec 5 '20 at 1:21
• 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) ? Dec 5 '20 at 11:57