Inversion of the Weierstrass transform

Remarks

The kernel $k(u) = \frac{1}{\sqrt{4\pi}} e^{-u^2/4}$ is integrable with $\hat k(\xi) = e^{-\xi^2}$, and $F = k * f$. The convolution theorem on $L^1 * L^2$ gives $\hat F = \hat k \hat f$, and the Plancherel theorem makes the Fourier transform a bijection of $L^2(\R)$ up to the factor $2\pi$, which yields the formula. Formally $W = e^{D^2}$ with $D = d/dx$, and the inverse is $e^{-D^2}$; for example, $W[e^{ax}] = e^{a^2} e^{ax}$ and $e^{-D^2} e^{ax} = e^{-a^2} e^{ax}$. The inversion is unstable: the factor $e^{\xi^2}$ grows very fast, so a small change of $F$ at high frequencies can change $f$ a lot. Not every function in $L^2$ is a Weierstrass transform: $F$ must satisfy $e^{\xi^2} \hat F \in L^2$, and every such $F$ extends to an entire function.

Going deeper

For a student

The inversion is the uniqueness theorem for the backward heat equation at a fixed time, in the class of square-integrable data. On the Fourier side it is division by the Gaussian $e^{-\xi^2}$, which never vanishes; this is why the transform is injective. The same reasoning applies to any convolution with an integrable kernel whose Fourier transform has no zeros. The operator notation $e^{D^2}$ also explains how $W$ acts on polynomials: it maps the Hermite polynomial $H_n(x/2)$ to $x^n$, since $e^{-D^2} x^n = H_n(x/2)$ for the physicists' Hermite polynomials $H_n$.