Functions $f(x)/g(x), g(x)/h(x),h(x)/f(x)$ are constant

Solution 1:

EDIT: I am using continuity of the function $h$ below. (Which is not among the assumptions in the question.) I do not know how to proceed without this assumption. (Unless I missed something, continuity of $h$ at one point of the domain is sufficient for this proof to go through.)

Notice that for any $x,y\in(0,\infty)$ you have $$f(x)g(y)=h(\sqrt{x^2+y^2})=f(y)g(x)$$ which implies $$\frac{f(x)}{g(x)}=\frac{f(y)}{g(y)}.$$ This implies that $f(x)/g(x)$ is a constant function.

As far as $h(x)/f(x)$ is concerned, we get for any $r>0$ and $\varphi\in(0,\frac\pi2)$: $$h(r)=f(r\cos\varphi)g(r\sin\varphi)$$ which implies (for any fixed $r>0$, taking limit $\varphi\to0$) $$h(r)=f(r) \lim\limits_{t\to0^+} g(t),$$ assuming the limit $\lim\limits_{t\to0^+} g(t)$ exists.

To see that this limit exists, just fix some $x>0$ and use that $$g(t)=\frac{h(\sqrt{x^2+t^2})}{f(x)}$$ which implies $$\lim\limits_{t\to0^+} g(t)=\lim\limits_{t\to0^+} \frac{h(\sqrt{x^2+t^2})}{f(x)}=\frac{h(x)}{f(x)}$$ by continuity of the function $h$. This shows that the limit used in the previous argument indeed exists. And it also gives a different argument that $h(x)/f(x)$ is constant.