$f: \mathbb{R} \to \mathbb{R}$ satisfying the following property $|f(x) - f(y)| \geq |x-y| $ for all $x,y \in \mathbb{R}$

Question

Let $f$ be a continuous function, $f: \mathbb{R} \to \mathbb{R}$ satisfying the following property $|f(x) - f(y)| \geq |x-y| $ for all $x,y \in \mathbb{R}$. Can we conclude that the function $f$ is monotone on $\mathbb{R}$?

Attempt :

$\frac{|f(x)-f(y)|}{|x-y|} \geq 1 \Rightarrow |f'(x)| \geq 1 for x \neq y$ I am not able to continue beyond this.

Answer 1

Suppose that it is not monotone. Then there exists a triple $x_1 < x_2 < x_3$ such that, for example, $f(x_1) < f(x_2) > f(x_3)$. Suppose that in this case we have $f(x_1) \ge f(x_3)$. Then by continuity of $f$ we have $y \in [x_2,x_3]$ such that $f(y) = f(x_1)$. Then

$$0 = |f(x_1) - f(y)| \ge |x_1-y| \ge |x_1-x_2| > 0$$

Contradiction.