Applying Fourier transforms in two variables separately

Question

Let $G(x, t) := (4\pi t)^{-d / 2}e^{-|x|^2 / 4t}$ for all $x \in \mathbb{R}^d$ and $t \in (0, \infty)$. The function $G$ is not integrable on $\mathbb{R}^d \times (0, \infty)$ since $$ \int_0^\infty \int_{\mathbb{R}^d} G(x, t)\,dx\,dt = \int_0^\infty \,dt = \infty, $$ so the Fourier transform of $G$ is not well-defined. However, $G$ is integrable when viewed as functions of the variables $x$ and $t$ separately. In particular, if we let $\mathcal{F}_x$ and $\mathcal{F}_t$ denote the Fourier transforms in $x$ and $t$, respectively, then we can compute $$ \mathcal{F}_x G(\xi, t) = e^{-4\pi^2t|\xi|^2}, $$ and $$ \mathcal{F}_t \mathcal{F}_x G(\xi, \tau) = \int_0^\infty e^{-4\pi^2t|\xi|^2} e^{-2\pi i \tau t}\,dt = \frac{1}{4\pi^2|\xi|^2 + 2\pi i\tau}. $$ But it seems like we have computed the Fourier transform of $G$! What is going on here? I am currently learning (the very basics) of tempered distributions, and it seems like this computation is justified in a certain sense (because $\mathcal{F} = \mathcal{F}_t \mathcal{F}_x$ is in fact true for Schwartz functions). On the other hand, the computation above doesn't even mention distribution theory—is there any other rigorous way of understanding this computation without requiring such machinery?

Answer 1

You can explain the integral in the sense of distribution, even the integral does not converge in the Lebsgue sense.

Let me take a simple example. Formally, We can abuse notation and write $$ \delta(\xi) = \hat{1}(\xi) = \int_{\mathbb{R}^n} e^{-2\pi i\xi\cdot x}\,dx. $$ The above integral does not converge, but you can explain it by applying it to a test function $\phi\in\mathcal{S}(\mathbb{R}^n)$: $$ \langle\int_{\mathbb{R}^n} e^{-2\pi i\xi\cdot x}\,dx, \phi(\xi)\rangle := \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\phi(\xi)e^{-2\pi i\xi\cdot x}\,d\xi\,dx = \int_{\mathbb{R}^n}\hat{\phi}(x)\,dx = \phi(0) = \langle\delta, \phi\rangle. $$ Your problem can be explained in the same way. Such notations are convenient for computation, because they enable us to pretend 'generalized functions' as functions.

Answer 2

1. What is going on here?

Indeed, in the $L^1$ theory of the Fourier transform, you need Fubini-type theorems to exchange the order of Fourier transforms and to write $\mathcal F_{t,x} = \mathcal F_t \mathcal F_x$. Here indeed, $G$ is not integrable and as you can see in your last computation, the Fourier transform is actually still not defined when $\xi = 0$, and the result is not bounded, while it would be continuous if $G$ was $L^1$.

Your computation is rigorous and makes sense without distribution theory but is saying that for all $\xi\neq 0$, $\mathcal F_t \mathcal F_x G = (|2\pi\,\xi|^2 + 2i\pi\,\tau)^{-1}$. It does not tell anything about $\mathcal F_{t,x}G$ and about $\mathcal F_x\mathcal F_tG$ and about what happens when $\xi=0$. To investigate these questions, distribution theory is indeed a good framework.

2. Distribution theory

For tempered distributions, the Fourier transform is defined by $\langle \mathcal F f,\varphi\rangle = \langle f,\mathcal F \varphi\rangle$ for any Schwarz class test function $\varphi$, and coincides with the classical integral definition if $f\in L^1$, so indeed the relation $\mathcal F_{t,x} = \mathcal F_t \mathcal F_x = \mathcal F_x \mathcal F_t$ is valid, as well as your first classical computation of the Fourier transform of a Gaussian with $t>0$.

Warning however, distribution theory is not just a way to justify formal computations. For example $\Delta(1/|x|) = 0$ when $x\neq 0$ in dimension $3$ but $\Delta(1/|x|) = 4\pi\,\delta_0$ in distribution theory. Here, in your last computation, there is a singularity when $\xi = 0$ and so one should use test functions or approximations to justify it.