I just finished reading the distribution chapter of Brad Osgood's lecture notes on Fourier Transform : For someone more with an engineering background, he does an amazing job at giving intuition so you feel the mathematical background without entering too much in it (I'd love to understand this more deeply, but I mainly want to apply Fourier Transform without getting lost too much in maths !).
I understand that the framework of distributions allows us to compute Fourier transforms of "functions" we couldn't have before. It gives really nice and easy formulas to work with sines, cosines, dirac delta, Heaviside function, etc... Which we couldn't have with classical Fourier transform.
However, on an intuitive level, I do not understand (I) why having the Fourier transform of a distribution (so some functional whose associated function integrates nicely with a Schwartz function) somehow also gives us the classical Fourier transform of the associated function (how could it be if the classical Fourier transform does not exist like for sine, cosine, etc... ?) ? (II) What integration with some arbitrary class of rapidly decreasing function has to do with it ? Intuitively, it would seem unrelated to me ! How this ideal case of Fourier transform of distributions has to do with classical Fourier transform of functions ? Because above all : I want the frequency spectrum of the functions sine, cosine, heaviside function, etc... not the Fourier transform of some conceptual distribution whose associated function integrates nicely with Schwartz functions.
(III) Is there some theory to prove the equality between the Fourier transform of a distribution and the Fourier transform of its associated function ? So if we can't compute the classical Fourier transform of sine, we can still get it by the Fourier transform of its associated distribution ?
It seems the extract just below is also about this (but I'm not sure) : "If the classical Fourier transform of a function defines a distribution, then that distribution is the Fourier transform of the distribution that the function defines". (IV) But I don't know if it includes the following functions : sines, cosines, Heaviside unit step, dirac, etc... because the "classical Fourier transform" of the latter does not exist... So it can't define a distribution ?
In this extract from a lecture of Brad Osgood (47'13 to 47'33), he explains an equality that has to be understood as an equality of distributions, not of pointwise functions. (V) When and why can the line be blurred ?