Redundancy Conditions for the Functional Equation $f(x+h(x))=f(x)+f(h(x))$

Abstract

Consider the functional equation $f(x+h(x))=f(x)+f(h(x))$, where $h:\mathbb{R}\to \mathbb{R}$ is a given continuous function, $h(0)=0$. It is proved if the set of all zeros of $h$ and of all points where $h(x)=-x$ is not "too much dense", then the continuous and at $x=0$ differentiable solution $f:\mathbb{R}\to \mathbb{R}$ of the functional equation under consideration is $f(x)=xf’(0)$ for all real $x$.