# How this relates to the usual Fourier transform?

I am used to the Fourier transform defined on $\mathcal{S}(\mathbb{R}^n)$. In that case we define the Schwartz space as the space of all $C^\infty(\mathbb{R}^n)$ functions such that for every multi indexes $\alpha,\beta$ we have

$$\sup_{x\in \mathbb{R}^n}|x^\alpha||D_\beta f(x)| < \infty.$$

We thus define

$$\hat{f}(k)=\int f(x) e^{-ik\cdot x}d^nx.$$

and we end up with one inverse operation, so that we can write

$$f(x)=\dfrac{1}{(2\pi)^n}\int \hat{f}(k)e^{ik\cdot x}d^nk.$$

Finaly, if we consider the topological dual $\mathcal{S}'(\mathbb{R}^n)$ we can define the Fourier transform of a tempered distribution. We simply put for $\zeta\in \mathcal{S}'(\mathbb{R}^n)$

$$\hat{\zeta}[f]=\zeta[\hat{f}].$$

On the other hand, I'm reading one paper that brings something that seems different. It is talking about Minkowski space, so it is $\mathbb{R}^4$ with the semi-Riemannian metric $\eta = \operatorname{diag}(1,-1,-1,-1)$.

Following the terminology of Gel'fand & Shilov (1964), let $K$ be the space of all $C^\infty$ scalar functions on $\mathbb{R}^4$ of compact support, and $Z$ be the space of all slowly increasing analytic functions on $\mathbb{R}^4$. Then the operation of Fourier transformation provides a bijection of $K$ onto $Z$, with $$\phi(x)\mapsto \hat{\phi}(k)=\int\phi(x)e^{ik\cdot x}d^4x.$$ where the dot denotes the Minkowskian inner product. If we now consider $f$ as a continuous linear functional on $K$ whose value at $\phi\in K$ is $\langle f,\phi\rangle$, then $\hat{f}$ is the generalized function on $Z$ defined by $$\langle \hat{f},\hat{\phi}\rangle=(2\pi)^4 \langle f,\phi\rangle.$$

A comment: this function $f$ is continuous and doesn't have compact support.

This seems like a quite different point of view on the Fourier transform. The Schwartz space and tempered distributions doesn't even appear, the Fourier transform is defined as a distribution.

All this makes me quite confused. How all of this relates to the usual Fourier transform based on the Schwartz space? How the two approaches make contact, and why this approach is reasonable?

• anything unclear ? The Schwartz space is easier because it is its own Fourier transform. Unfortunately it isn't adapted to the Laplace transform (which is important in the definition of many special functions) – reuns Oct 24 '17 at 3:37

To take a concrete example look at the Laplace transform sending a Gaussian $\phi_a(x) = e^{-a^2 x^2/2}, a > 0$ to a Gaussian $\Phi_a(s) = \int_{-\infty}^\infty \phi_a(x)e^{-sx}dx = a^{-1} e^{s^2/a^2}$. From this you can define the analytic functional $\delta(s-s_0)$ as $$\langle \delta(s-s_0) ,\Phi_a \rangle = \Phi_a(s_0)$$ Where $\Phi_a(s), s \in \mathbb{C}$ is defined as $$\Phi_a(s) = \int_{-\infty}^\infty \phi_a(x)e^{-sx}dx= \int_{-\infty}^\infty (\int_{-\infty}^\infty \Phi_a(i\omega)e^{i \omega x} d\omega)e^{-sx}dx$$ and extend to the vector space generated by the shifted Gaussians and their closure under some topology, namely the space of Laplace transforms of functions decreasing faster than any exponential (so that the Laplace transform is entire).