A functional equation over a circle: $f(r \cos \phi)+f(r\sin \phi)=f(r)$

Question

I am interested in the functional equation $$f(r \cos \phi)+f(r\sin \phi)=f(r),\qquad r\geq 0,\ \ \phi\in[0,\pi/2]\text.$$ Let's assume that $f:[0,\infty)\to\mathbb R$ is monotone. Clearly, $f(x)=ax^2$ is a solution. Are there any other solutions?

Answer 1

The functions $x\rightarrow ax^2$ are the only continuous solutions in fact :

Let $f$ be a non zero solution of your equation.

It means that in cartesian coordinates $f(x)+f(y)=f(\sqrt{x^2+y^2})$ for all $x,y\geq 0$.

Let $g(x)=f(\sqrt x)$. We have for all $x,y\geq 0$ : $$g(x^2)+g(y^2)=g(x^2+y^2)$$

A classical result states that $\phi:\mathbb{R}\rightarrow\mathbb{R}$ continuous such that $\phi(x)+\phi(y)=\phi(x+y)$ is of the form $\phi(x)=ax$.

The same holds if $\phi$ is continuous from $\mathbb{R}_+\rightarrow \mathbb{R}$, $\phi$ with the same property.

Then there exists $a\in\mathbb{R}$ such that $g(x)=ax$.

This way you get $f(x)=g(x^2)=ax^2$.