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

Let $f:\mathbb{R}\to\mathbb{R}$ be a Lebesgue-measurable function such that there exists $g:\mathbb{R}_{\geq 0}\to\mathbb{R}$ for which $$f(x)\,f(y)=g\left(\sqrt{x^2+y^2}\right)\tag{*}$$ for all $x,y\in\mathbb{R}$. Then, note that by plugging in $y:=0$, we have $$f(x)\,f(0)=g\big(|x|\big)\text{ for every }x\in\mathbb{R}\,.\tag{#}$$ If $f(0)=0$, then $g\equiv 0$, whence $f(x)\,f(y)=0$ for every $x,y\in\mathbb{R}$. This shows that $f\equiv 0$ (which is not a probability density function).

We now assume that $c:=f(0)$ is nonzero. From (*) and (#), $$f(x)\,f(y)=c\,f\left(\sqrt{x^2+y^2}\right)\text{ for all }x,y\in\mathbb{R}\,.$$ Note that, by (#), $f$ is an even function. Hence, it suffices to solve for $f(x)$ when $x\geq 0$. Define $h:\mathbb{R}_{\geq 0}\to\mathbb{R}$ to be $h(t):=\dfrac{1}{c}\,f(\sqrt{t})$ for all $t\geq 0$. Then, $$h(s)\,h(t)=h(s+t)\text{ for all }s,t\geq 0\,.$$ That is, $$h(t)=h\left(\frac{t}{2}+\frac{t}{2}\right)=\left(h\left(\frac{t}{2}\right)^{\vphantom{a^a}}\right)^2\geq 0\,.$$ If $h(\tau)=0$ has a solution $\tau\geq 0$, then note that $$h(s+\tau)=h(s)\,h(\tau)=0\text{ for all }s\geq 0\,,$$ whence $h(t)=0$ for all $t\geq \tau$.

If $h(t)=0$ for all $t\geq 0$, then $f(x)=0$ for all $x\geq 0$. This contradicts the assumption that $f(0)\neq 0$. If there exists $t\geq 0$ such that $h(t)\neq 0$, then let $$\sigma:=\sup\big\{t\geq 0\,\big|\,h(t)=0\big\}\,.$$ Note that $\sigma\leq \tau$. If $\sigma>0$, then note that $$0=h(2\sigma)=\big(h(\sigma)\big)^2\,,$$ whence $h(\sigma)=0$. That is, $$0=h(\sigma)=\left(h\left(\frac{\sigma}{2}\right)^{\vphantom{a^a}}\right)^2\,,$$ implying $h\left(\dfrac{\sigma}{2}\right)=0$. By the same argument as the previous paragraph, $h(t)=0$ for all $t\geq \dfrac{\sigma}{2}$. This contradicts the definition of $\sigma$. Thus, $\sigma=0$. Now, $$h(0)=h(0+0)=\big(h(0)\big)^2$$ implies that $h(0)=0$ or $h(0)=1$. Since $\sigma=0$, we get $h(0)=1$, leading to a solution $$h(t)=\left\{\begin{array}{ll}1&\text{if }t=0\,,\\0&\text{if }t>0\,.\end{array}\right.$$ That is, $$f(x)=\left\{\begin{array}{ll}c&\text{if }x=0\,,\\0&\text{if }x\neq 0\,.\end{array}\right.$$ (This makes $f$ a measurable function, but this solution is not a probability density function.)

From now on, we assume that $h(t)>0$ for all $t\geq 0$. Define $$\eta(t):=\ln\big(h(t)\big)\text{ for }t\ge 0\,.$$ Then, $$\eta(s+t)=\eta(s)+\eta(t)\text{ for all }s,t\geq 0\,.\tag{@}$$ This is Cauchy's functional equation. Since $f$ is measurable, $h$ is also measurable, and so is $\eta$. Therefore, there exists a constant $a\in\mathbb{R}$ for which $$\eta(t)=at\text{ for each }t\geq 0\,.$$ That is, $$h(t)=\exp(at)\text{ for each }t\geq 0\,,$$ whence $$f(x)=c\,\exp(ax^2)\text{ for all }x\in\mathbb{R}\,.$$ Now, if $f$ is a probability density function, $a=-b$ for some $b>0$, and $c=\sqrt{\dfrac{b}{\pi}}$.

Remark. There are solutions $f$ that are not Lebesgue measurable. All such solutions are given by $$f(x):=c\,\exp\big(\eta(x^2)\big)$$ for all $x\in\mathbb{R}$, where $\eta:\mathbb{R}_{\geq 0}\to\mathbb{R}$ is any solution to (@) which is not Lebesgue measurable. To find such $\eta$, you need the Axiom of Choice.