Solving the functional equation $f(0)^2 f(\sqrt{x^2+y^2+z^2}) = f(x)f(y)f(z)$

Question

The Main Question

I am interested in solving the functional equation

$$f(0)^2 f(\sqrt{x^2 + y^2 + z^2}) = f(x)f(y)f(z)$$ .

where $x,y,z$ are real numbers.

One set of solutions take the form $f(x) = C e^{A x^2}$ (ie. Gaussian) for some constants $C$ and $A$.

My question is: Do all solutions of this functional equation take this form? Can we prove that there are no solutions where $f$ does not take this form?

Extra Context

I am interested in Maxwell's derivation of the Maxwell-Boltzmann distribution, as found here. One can make a physical argument as to why the distribution of velocities in a gas at equilibrium should obey this functional equation. This later paper makes the claim that 'In essence the proof relies on the fact that the only solution of the functional equation $f^2(0) · f (|v|) = f (v_x) · f (v_y ) · f (v_z )$ is the Gaussian $f (v_x) = C · e^{A·v_x^2}$ ' (emphasis mine). In this context, $|v| = \sqrt{v_x^2+v_y^2 + v_z^2}$ is the magnitude of the velocity (ie. the speed) and $v_x, v_y, v_z$ are the components of the velocity in the three directions. While I agree that this is indeed a solution (and the most obvious one at that), I cannot convince myself that it is the only solution.

Answer 1

This is related to Cauchy's functional equation $f(x+y)=f(x)+f(y)$, see here for a nice collection of results.

More concretely, you can do the following: if $f(0)=0$, then $x=y=z$ gives $f\equiv 0$. Therefore, without loss of generality, we may assume $f(0)\neq 0$. Setting $y=z=0$ then gives $f(|x|)=f(x)$, so $f$ is an even function. In particular, it suffices to obtain the restriction $f|_{[0,\infty)}$.

We introduce the auxiliary function $$ \begin{align*}g\colon\mathbb{R}_{\geq 0}&\to\mathbb{R}\\ x&\mapsto f(\sqrt{x})/f(0)\end{align*} $$ so that $f(x)=g(x^2)f(0)$ and $g(0)=1$. From the original functional equation, we then obtain $$ f(0)^3g(x^2+y^2+z^2)=f(0)^3g(x^2)g(y^2)g(z^2) $$ and so with $a=x^2$, $b=y^2$ and $z=0$ we obtain $$ g(a+b)=g(a)g(b) $$ for all $a,b\geq 0$. In particular, from $a=b$ we already obtain $g(a)\geq 0$ for all $a\geq 0$. If there exists $c>0$ such that $g(c)=0$, then the equation gives $g(a)=0$ for all $a\geq x$, and then $g(2a)=g(a)^2$ gives $g(a)=0$ for all $a>0$. Hence we may assume without loss of generality that $g(a)>0$ for all $a\geq 0$.

We can then define $$ \begin{align*}h\colon\mathbb{R}&\to\mathbb{R}\\ x&\mapsto \begin{cases}\log g(x) &\text{if }x\geq 0\\-\log g(-x) &\text{else,}\end{cases}\end{align*} $$ and it is not hard (but cumbersome), to verify that we have $$ h(x+y)=h(x)+h(y) $$ for all $x,y\in\mathbb{R}$. That is, $h$ satisfies the Cauchy equation. There are some pathological solutions to this, but as soon as $h$ satisfies any sort of 'niceness', the only solutions are $h(x)=Ax$ for some real constant $A$. For example, as you seem to be interested in physics, a reasonable hypothesis would be that $f$ is continuous. In this case $h$ is continuous too, which rules out any pathological solutions.

Putting everything back together, we obtain $$ f(x)=f(0)g(x^2)=f(0)e^{h(x^2)}=f(0)e^{Ax^2}, $$ which is precisely the solution you were looking for.