$f, F:\Bbb{R}\to \Bbb{R}^+$ such that $$\int_{-\infty}^\infty f(x)=1\\ \int_{-\infty}^\infty F(x)=1\\ 2\pi\int_x^\infty tF(t) dt =f(x)\\ f(x)f(y)f(z) = F(x^2+y^2+z^2)$$
where in the last equation, $x, y, z$ are independent variables (like Cartesian coordinates, maybe).
This comes up in solving Maxwell-Boltzmann’s distribution. And a solution is $$f(x) = Ae^{Bx^2}$$ where $A, B$ are arbitrary real constants.
My question is whether there is a more general solution to the above system of equations.
First, note that the conditions force $f,F$ to be smooth outside zero at least.
Define $g(x)=\ln{f}(\sqrt{x})$. Then there exists a continuous $G$ such that for positive $x,y,z$, $G(x+y+z)=g(x)+g(y)+g(z)$. Therefore, $G(x)=3g(x/3)$, hence $\frac{g(x)+g(y)+g(z)}{3}$ depends only on $x+y+z$.
You can conclude that for any $r \in \mathbb{R}$, the quantity $g_1(r)=g(a)-g(b)$ does not depend on the couple $(a,b)$ of positive real numbers such that $a-b=r$, and $g_1$ is smooth, and $g_1(x)=g(2x)-g(x)=g(3x)-g(2x)$.
We also have $g_1(x)+g_1(y)+g_1(z)$ depending only on $x+y+z$, and by definition $g_1$ is odd. It follows that $g_1$ is linear. Now, $g(x)=g_1(x-1)+g(1)$, so $g$ is affine, Q.E.D.