Find function $f(ax+b)+c \le x \le f(x+c)+b$

Question

Find all functions $f: \Bbb R \rightarrow \Bbb R$ such that

$f(ax+b)+c \le x \le f(x+c)+b$

for all $x \in \Bbb R$ and $a,b,c$ real constant.

I've done some trivial idea as $x=0$, $x=-c$, $x=-b/a$ but I got nothing more.

Answer 1

Notice that if $a = 0$ there can't be any function satisfying your hypothesis, since the number $f(b) + c$ would have to be smaller than every real $x$. Suppose $a \neq 0$. Using the first inequality, if we let $y = ax + b$ we get $$ f(y) \le \frac{y - b}{a} - c. $$ Using $y = x + c$ in the second inequality, we obtain $$y - c -b \le f(y). $$ Putting together both inequalities, we get $$ y - b \le \frac{y}{a} - \frac{b}{a}, \ \text{for every}\ y \in \mathbb{R}. $$ $$ (1 - 1/a)y \le b(1 - 1/a), \ \text{for every}\ y \in \mathbb{R}. $$ The last inequality is only possible if $a = 1$. In that case, we can go back to our two inequalities to obtain $$ y - b - c \le f(y) \le y - b - c \ \Longrightarrow \ f(y) = y - b - c.$$ It is easy to check that the function above satisfies the original functional equation.