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?
 A: 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.
A: *

*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.
