# $|f(x)-f(y)|\geq k|x-y|$.Then $f$ is bijective and its inverse is continuous.

My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:

$$|f(x)-f(y)|\geq k|x-y|$$

Then $f$ is bijective and its inverse is continuous.

Well, there's a Theorem , Invariance of domain, that says

"If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".

But I'm not knowing how to proceed...need a clue...Thanks for attention!!!

-
Invariance of domain in one dimension is not very hard to prove, it's essentially the statement that a strictly monotonic function on an interval has a continuous inverse which is what Brian proves in his answer. The case $n \geq 2$ is a completely different story... – t.b. Sep 16 '12 at 3:32

Proving that $f$ is injective is easy: if $f(x)=f(y)$, then $|f(x)-f(y)|=0$, and therefore ... ?
Once you’ve shown that $f$ is injective, consider $f(0)$ and $f(1)$; they can’t be equal, so either $f(0)<f(1)$, or $f(0)>f(1)$. Show that in the first case $f$ must be strictly monotonically increasing, and in the second case it must be strictly monotonically decreasing; you’ll need the fact that $f$ is continuous.
After you’ve shown that $f$ is strictly monotonic, you can use the fact that $|f(x)-f(y)|\ge k|x-y|$ for all $x,y\in\Bbb R$ to show that the range of $f$ is unbounded in both directions. Then you can use continuity again to show that $f$ is surjective and hence a bijection.
At this point it should be easy to show that if $I$ is an open interval in $\Bbb R$, then $f[I]$ is also an open interval, which immediately implies that $f^{-1}$ is continuous.