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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.

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  • $\begingroup$ WLOG k=1. WLOG x>y $\endgroup$ – Jacob Wakem Mar 22 '17 at 2:15
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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 $|f(x_1) - f(x_2)| = 0 < k|x_1 - x_2|.$

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  • $\begingroup$ Aha, so if both are bounded "in the same direction" above or below so to speak, we'd arrive at a contradiction. So then we pin down the final result by The Intermediate Value Theorem. Give me a moment to work with this and I'll select this as the answer. Cheers. $\endgroup$ – zzz Mar 22 '17 at 2:22
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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.

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  • $\begingroup$ This is also a strong answer because in the end it illustrates that the real important part is that the map is injective. And since it's injective and continuous, it's strictly increasing or decreasing. Thanks for this. $\endgroup$ – zzz Mar 22 '17 at 3:09

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