## ODE Proof of Gaussian Fourier Transform

Consider the Gaussian function $f(x) := e^{-\pi x^{2}}$. It is well-known that, if we define the Fourier transform of a Schwartz function $g: \mathbb{R}^{n} \rightarrow \mathbb{R}$ to be

$\displaystyle \widehat{g}(\xi) := \int_{\mathbb{R}^{n}}g(x)e^{-2\pi i x\cdot\xi}dx, \indent \forall \xi \in \mathbb{R}^{n}$

then $\widehat{f}(\xi) = f(\xi)$. The proof of this result I learned as an undergraduate used Cauchy’s theorem of complex analysis. I did not really understand the proof at the time, as I was not to take complex analysis until the following semester. However, I would like to describe a (well-known) proof of this result which relies on properties of the Fourier transform with respect to differentiation and global uniqueness of solutions to certain ordinary differential equations (ODEs).

Let $f$ be defined as above. Observe that by the Picard-Lindelof theorem, $f$ uniquely satisfies the ordinary differential equation

$\displaystyle (\partial_{x}f)(x) = -2\pi x f(x), \indent \forall x \in \mathbb{R}$

Since $f$ is Schwartz function (i.e. $f \in \mathcal{S}(\mathbb{R})$), we can take its Fourier transform to another Schwartz function $\widehat{f}$. Taking the Fourier transform of both sides of the ODE above, we see that $\widehat{f}$ satisfies

$\displaystyle (2\pi i\xi) \widehat{f}(\xi) = -i(\partial_{x}\widehat{f})(\xi) \Longleftrightarrow -2\pi\xi\widehat{f}(\xi) = (\partial_{x}\widehat{f})(\xi), \indent \forall \xi \in \mathbb{R}$

By uniqueness, we conclude that $\widehat{f}(\xi) = e^{-\pi\xi^{2}}$, for all $\xi \in \mathbb{R}$.

We can use this one-dimensional result to compute the Fourier transform of the $n$-dimensional Gaussian

$\displaystyle f(x) = f(x_{1},\cdots,x_{n}) = e^{-\pi\left|x\right|^{2}} = e^{-\pi(x_{1}^{2}+\cdots+x_{n}^{2})}, \indent \forall x \in \mathbb{R}^{n}$

It follows from the one-dimensional case $f \in \mathcal{S}(\mathbb{R}^{n})$, and therefore the integral defining $\widehat{f}$ makes sense and is absolutely convergent. By Fubini’s theorem,

$\displaystyle \begin{array}{lcl}\widehat{f}(\xi) = \int_{\mathbb{R}^{n}}e^{-\pi\left|x\right|^{2}}e^{-2\pi i \xi \cdot x}dx &=& \left(\int_{\mathbb{R}}e^{-\pi x_{1}^{2}}e^{-2\pi i \xi_{1}x_{1}}dx_{1}\right)\cdots\left(\int_{\mathbb{R}}e^{-\pi x_{n}^{2}}e^{-2\pi i \xi_{n}x_{n}}dx_{n}\right)\\&=& e^{-\pi \xi_{1}^{2}}\cdots e^{-\pi\xi_{n}^{2}}\\&=& e^{-\pi\left|\xi\right|^{2}}\end{array}$

I’ll later upload to the problem solutions section of this site the complex-analytic proof I learned as an undergrad.