Solutions of the functional equation $f\big(x+g(y)\big)=f(x)+f\big(g(y)\big)$

Question

I know the solutions of the well-known Cauchy functional equation $$f(x+y)=f(x)+f(y)\text.$$ But what does it change if I have the following form $$f\big(x+g(y)\big)=f(x)+f\big(g(y)\big)\text?$$ What can I say about $g$?

Answer 1

There are lots of possibilities. For example, you might have $g(y) = 1$ for all $y$, in which case $f(x+1) = f(x) + f(1)$ just says that $f(x) = f(1) x + p(x)$ where $p(x)$ is periodic with period $1$ and $p(1) = p(0) = 0$. On the other hand, if $g$ is surjective, the equation reduces to Cauchy's.

EDIT: For any possible nonzero value $v$ of $g$, we have $$f(x + v) = f(x) + f(v)$$ and then $$f(x) = p(x) + \dfrac{f(v)}{v} x$$ where $p(x) = f(x) - \dfrac{f(v)}{v} x$ is periodic with period $v$. Nonlinear cases with two or more incommensurate $v$'s are going to be quite exotic (nonmeasurable, I would expect, just like nonlinear solutions to the Cauchy equation).