Non-trivial real function such that $|f(x) - f(ax+b)|$ depends only on $a$ and $b$

Question

Let $a,b \in\mathbb{R}$ be constants.

Find a (non-trivial) continuous function $f: \mathbb{R} \to \mathbb{R}$ such that for all $x \in \mathbb{R}$,

\begin{align*} |f(x)-f(ax+b)| \end{align*} depends only upon $a$ and $b$.

Answer 1

Suppose $\vert f(x)-f(ax+b)\vert = g(a,b)$ for all $x,a,b\in \mathbb{R}$. Then for all $x$ we have $$\vert f(x) - f(0) \vert = g(0,0)$$ and so $$f(x) = f(0)\pm g(0,0)$$ In order for $f$ to be continuous, we need $f(x) = f(0)+g(0,0)$ or $f(x) = f(0)-g(0,0)$. In either case, $f$ is constant.