Prove $f$ is surjective if $\lvert f(x)-f(y) \rvert \geq k \lvert x-y \rvert$ for all $x,y\in\mathbb{R}$

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a continuous function and satisfies the "reverse-Lipschitz" criterion. That is, for all $x,y\in\mathbb{R}$ and some $k> 0$, we have \begin{equation*} \lvert f(x)-f(y) \rvert \geq k \lvert x-y \rvert \end{equation*} How do we go about showing that such a function is a surjection?

Breaking $\mathbb{R}$ into postive and negative values has provided me with inequalities \begin{equation} kx-\lvert f(0) \rvert \leq \lvert f(x) \rvert \text{ if $x \geq 0$} \end{equation} \begin{equation} -kx-\lvert f(0) \rvert \leq \lvert f(x) \rvert \text{ if $x \leq 0$} \end{equation} Which both show $\lvert f(x) \rvert$ is unbounded, and if I could show in general that one set of values has no upper bound and the other no lower bound, then I could finish instantly by applying the Intermediate Value Theorem I believe.

Alternatively, I've considered trying to show that the image of $f$ is both open and closed and thus must be all of $\mathbb{R}$ but have made as equally little progress there.

• WLOG k=1. WLOG x>y – Jacob Wakem Mar 22 '17 at 2:15

You are almost done. If $f$ is unbounded above for both the positive and negative $x,$ then (by the intermediate value theorem whereof you speak), for any $M> f(0),$ there exist an $x_1 < 0$ and an $x_2 > 0,$ such that $f(x_1) = f(x_2) = M,$ so $|f(x_1) - f(x_2)| = 0 < k|x_1 - x_2|.$
The function is strictly (WLOG) increasing; for if it weren't then we would have $f(x)-f(y)=0$ and $x-y$ nonzero (by intermediate value theorem). From this it is obvious the range is unbounded in both directions.