Specify the monotonicity based on the functional equation $f\bigl(f(x)\bigr)=f(x)-\frac{x}{4}$ I have a strictly monotone function $f:\mathbb{R}\to\mathbb{R}$ for which is true that:
$$f\bigl(f(x)\bigr)=f(x)-\frac{x}{4}\ \forall x\in\mathbb{R}$$
I want to prove that $f$ is strictly increasing, so I need to prove that for:
$$a<b\implies f(a)<f(b)$$
For $x=0$ from the given relationship I get $f(0)=0$ (and $f(x)\ne0 \ \forall x\ne0$).
Now I am trying to find another value of $f$ to compare with $f(0)$, without any luck. Ideas?
 A: You know that $f$ is stricly monotone, so it is either strictly increasing or strictly decreasing. Assume now that it is strictly decreasing. Then if $a < b$ you have $f(a) > f(b)$. This means
$$f(f(a)) - f(f(b)) > \frac{1}{4}(b-a) > 0$$
and thus $f(f(b)) < f(f(a))$. Since $f$ is supposed decreasing, this can be true only if $f(b) > f(a)$. But $f(a) > f(b)$ by assumption, contradiction. Then $f$ is strictly increasing.
A: Since $f$ is strictly monotone, either $f$ is strictly increasing, or $f$ is strictly decreasing.

Suppose $f$ is strictly decreasing.
\begin{align*}
\text{Then}\;\;&f(1) < f(0)&&\text{[since $f$ is strictly decreasing]}\\[4pt]
\implies\;&f(1) < 0&&\text{[since $f(0)=0$]}\\[4pt]
\implies\;&f(f(1)) > f(0)&&\text{[since $f$ is strictly decreasing]}\\[4pt]
\implies\;&f(f(1)) > 0&&\text{[since $f(0)=0$]}\\[4pt]
\implies\;&f(1) - {\small{\frac{1}{4}}} > 0&&\text{[by the functional equation]}\\[4pt]
\implies\;&f(1) > 0\\[4pt]
\end{align*}
contradiction.

It follows that $f$ is strictly increasing.
