# Discussion : Does Fourier transform of a distribution give classical Fourier transform of its associated function ? Why?

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 ?

• You can't really prove an equality. Because the fourier transform of a distribution is indeed a distribution while the fuorier transform of a function is indeed a function. What you can show is how the fourier transform of a distribution whose function is defined in L2 is a distribution whose function is the standard fourier function of that function, this isn't hard to show from the definition of fourier transform of a distribution. Dec 6, 2020 at 1:32
• (I), (III) and (V) seem to be almost the same question. They are all more or less about the identity $T_{\mathcal{F}f} = \mathcal{F}T_f.$ Dec 6, 2020 at 21:00

Every locally integrable function $$f \in L^1_{\text{loc}}$$ induces a distribution $$T_f$$ by integration against the function: $$\langle T_f, \varphi \rangle := \int f(x)\,\varphi(x)\,dx.$$ Moreover, if $$T_f = T_g$$ then $$f=g$$ almost everywhere, so we can identify $$T_f$$ with $$f \in L^1_{\text{loc}}$$.

Not every distribution is given in this way, though. For example, the Dirac $$\delta$$ distribution is not. Those distributions that are given by a function constitute a subspace of the distributions. Because of the identification of $$f$$ with $$T_f$$ we can look at $$L^1_{\text{loc}}$$ as a subspace of the distributions.

In general, distributions are defined as linear functionals over $$C^\infty_c,$$ the space of infinitely differentiable functions with compact support. The space of such distributions is denoted by $$\mathcal{D}'.$$ Many operations on distributions are defined by moving the operation to the test function (with some modification of the operation). The definitions are usually based on how it works if $$u$$ is given by a function. For example, if $$f \in L^1_{\text{loc}}$$ and $$\varphi\in C^\infty_c$$ then, by integration by parts, we have $$\int f'(x)\,\varphi(x)\,dx = - \int f(x)\,\varphi'(x)\,dx$$ and therefore we define the derivative of a distribution $$u$$ by $$\langle u', \varphi \rangle := \langle u, -\varphi' \rangle \quad \forall\varphi\in C^\infty_c.$$

The use of test functions with compact support (the $${}_c$$ part of $$C^\infty_c$$) give the distributions a local behavior. We can say that $$u=v$$ on a region $$\Omega$$ if $$\langle u, \varphi \rangle = \langle v, \varphi \rangle$$ for all $$\varphi \in C^\infty_c$$ with support on $$\Omega.$$ For example, $$\delta=0$$ on $$(0,\infty)$$.

For the Fourier transform we have $$\int \mathcal{F}f(\xi)\,\varphi(\xi)\,d\xi = \int f(\xi)\,\mathcal{F}\varphi(\xi)\,d\xi$$ so we want to define the Fourier transform of a distribution by $$\langle \mathcal{F}u, \varphi \rangle = \langle u, \mathcal{F}\varphi \rangle.$$ There's a problem, however; it's not certain that $$\mathcal{F}\varphi \in C^\infty_c$$ so $$\langle u, \mathcal{F}\varphi \rangle$$ is not defined.

Therefore, when defining the Fourier transform of distributions one instead takes test functions from the Schwartz space $$\mathcal{S},$$ a space of rapidly decreasing functions (also the derivatives decrease rapidly). This space is closed under Fourier transform so the definition above makes sense. Unfortunately, since $$\mathcal{S}\supset C^\infty_c$$ the space of distributions over $$\mathcal{S}$$ is smaller than the space of distributions over $$C^\infty_c.$$ This space of distributions is denoted by $$\mathcal{S}'$$ and the distributions are called tempered.

The classical Fourier transform is not defined for $$\sin, \cos, H, \delta$$ since these are not in $$L^1$$ (or even $$L^2$$). But these functions define tempered distributions and as such they have a Fourier transform. For $$\delta$$ the Fourier transform is $$T_{\mathbf{1}},$$ where $$\mathbf{1}$$ is the function with value $$1$$ everywhere. And since we identify $$T_f$$ with $$f$$ one usually just says that $$\mathcal{F}\delta=\mathbf{1}.$$

• Great answer! Just out of curiosity, what would the Fourier transforms of $\sin$, $\cos$, and $H$ (as tempered distributions) be?
– Unit
Dec 6, 2020 at 22:12
• Since $f(x) = \frac{1}{2\pi} \int \mathcal{F}f(\xi) \, e^{ix\xi}\,d\xi$ and $$\sin x = \frac{1}{2i}(e^{ix}-e^{-ix}) = \frac{1}{2\pi} \int \frac{2\pi}{2i}(\delta(\xi-1)-\delta(\xi+1)) e^{ix\xi} \, d\xi \\ \cos x = \frac{1}{2}(e^{ix}+e^{-ix}) = \frac{1}{2\pi} \int \frac{2\pi}{2}(\delta(\xi-1)+\delta(\xi+1)) e^{ix\xi} \, d\xi$$ we can conclude that $$\mathcal{\sin}(\xi) = -i\pi(\delta(\xi-1)-\delta(\xi+1)) \\ \mathcal{\cos}(\xi) = \pi(\delta(\xi-1)+\delta(\xi+1))$$ Dec 6, 2020 at 23:20
• You can find a calculation of the Fourier transform of the Heaviside step function here. Dec 6, 2020 at 23:23