Weierstrass transform

Remarks

The kernel $\frac{1}{\sqrt{4\pi}} e^{-u^2/4}$ is the density of the normal distribution with mean $0$ and variance $2$. So $Wf(x) = \mathbb{E}[f(x + Y)]$ for a random variable $Y$ with that law. It is the heat kernel $\frac{1}{\sqrt{4\pi t}} e^{-u^2/(4t)}$ at $t = 1$, so $Wf = u(\cdot, 1)$ where $u$ solves the heat equation $u_t = u_{xx}$ with $u(\cdot, 0) = f$. For integrable $f$ and $\hat f(\xi) = \int f(x) e^{-i x \xi}\, dx$, one has $\widehat{Wf}(\xi) = e^{-\xi^2} \hat f(\xi)$. The growth condition can be weakened to $\int \abs{f(y)} e^{-c y^2}\, dy < \infty$ for one $c < \frac14$, which is enough for the integral to converge for every $x$.

Example: $f(x) = x^2$. With $Y$ of mean $0$ and variance $2$, $Wf(x) = \mathbb{E}[(x + Y)^2] = x^2 + 2x\, \mathbb{E}[Y] + \mathbb{E}[Y^2] = x^2 + 2$. Likewise $W[e^{a x}](x) = e^{a x} \mathbb{E}[e^{aY}] = e^{a^2} e^{a x}$ for every real $a$.

Non-example: $f(x) = e^{x^2/4}$ fails the condition for $c < \frac14$, and indeed $\int_\R e^{y^2/4} e^{-(x-y)^2/4}\, dy = e^{-x^2/4} \int_\R e^{x y/2}\, dy$ diverges for every $x$.