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Let $f:\mathbb{R}^n \to \mathbb{R}^n$ be continuous and let there exist $\alpha > 0$ such that $||f(\mathbf{x}) - f(\mathbf{y})|| \geq \alpha || \mathbf{x} - \mathbf{y}||$ for all $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. Prove that $f$ is one-one, onto and that $f^{-1}$ is continuous.

One-one is trivial. It is onto-ness that I can't show.

Write $S = f(\mathbb{R}^n)$. Using sequential continuity, it is possible to show that $S$ is closed. If I could show $S$ is open, I would be done, but I can't.

Also, writing $g(\mathbf{x}) := \dfrac{f(\mathbf{x})}{\alpha}$, the condition can be converted to that of proper expansive map, $||g(\mathbf{x}) - g(\mathbf{y})|| \geq || \mathbf{x} - \mathbf{y}||$. But since $\mathbb{R}^n$ is not compact, I cannot use the result here.

Any help is appreciated!

EDIT: As commented below, the Invariance of Domain theorem seems to work in this case, but that result does not use the expansive-type condition provided here (except for showing the injectivity), and so it appears that an easier proof would be possible.

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    $\begingroup$ This might be too big of a hammer, but see en.wikipedia.org/wiki/Invariance_of_domain $\endgroup$
    – Lukas Geyer
    Feb 18, 2019 at 15:18
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    $\begingroup$ But maybe there is an easier proof using the expansive condition ? $\endgroup$
    – RandomStudent
    Feb 18, 2019 at 15:52
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    $\begingroup$ Yes, that is what I meant, it solves your problem, but there might be a more direct and easier way here. $\endgroup$
    – Lukas Geyer
    Feb 18, 2019 at 15:58
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    $\begingroup$ It is unlikely you can get away with using just general topology and metric spaces theory since surjectivity fails for expanding maps of some reasonably nice locally compact planar sets (which locally are Sierpinsky carpets) or even totally disconnected subsets of the real line. One has to use, in addition, some real analysis. What book are you using in your class? $\endgroup$
    – Moishe Kohan
    Feb 27, 2019 at 18:18
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    $\begingroup$ The simplest example to think about is the map $z\mapsto 2z$ of the set of natural numbers. It is expanding but not surjective. $\endgroup$
    – Moishe Kohan
    Feb 27, 2019 at 18:44

1 Answer 1

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Actually the Invariance of Domain theorem solves your problem, but not the way you think : indeed, if you apply it with $U = \mathbb{R}^n$, you only get that $f$ is a homeomorphism on its image, and you don't get that its image is $\mathbb{R}^n$.

Here is the way you can use it correctly, and your expansion condition is really necessary :

For $r > 0$, denote by $B(a,r)$ the open ball of center $a \in \mathbb{R}^n$ and radius $r$. For all $r > 0$, the image $f(B(0,r))$ is open, by the Invariance of Domain theorem. So $f(B(0,r)) \cap B(f(0), \alpha r)$ is open in $B(f(0), \alpha r)$.

But $f(\overline{B(0,r)}) \cap B(f(0), \alpha r) = f(B(0,r)) \cap B(f(0), \alpha r)$ : indeed, if $||x|| = r$, then $||f(x)-f(0)|| \geq \alpha r$, so $f(x) \notin B(f(0), \alpha r)$. So you get that $f(B(0,r)) \cap B(f(0), \alpha r)$ is also closed in $B(f(0), \alpha r)$.

By connectivity, $f(B(0,r)) \cap B(f(0), \alpha r)$ is open and closed in $B(f(0), \alpha r)$, and not empty (because it contains $f(0)$), so it is equal to $B(f(0), \alpha r)$. In other words you have, for all $r > 0$, $$ B(f(0), \alpha r) \subset f(B(0,r)) $$

Obviously this implies that $f$ is surjective.

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    $\begingroup$ Probably this can be simplified. Namely, it is sufficient to prove that $f(\mathbb R^n)$ is open and closed at the same time. The Invariance of Domain shows that it is open. And in the OP it is mentioned that $f(\mathbb R^n)$ is closed. Indeed, if $f(x_n)\to y$ as $n\to \infty$ then by the expansion condition the sequence $x_n$ is Cauchy, hence by continuity $y=f(\lim_{n\to\infty} x_n) \in f(\mathbb R^n)$. $\endgroup$
    – Skeeve
    Feb 28, 2019 at 7:43
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    $\begingroup$ Yes, this is what I had in mind when I said that showing that $f(\mathbb{R}^n)$ is open is sufficient (that it is also closed, and thus equal to $\mathbb{R}^n)$. $\endgroup$
    – RandomStudent
    Feb 28, 2019 at 8:21
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    $\begingroup$ Yes, sorry, I didn't see that you mentioned that $f(\mathbb{R}^n)$ is closed. Then I agree with you, but in both cases, you use the expansion condition and not only the Invariance of Domain. $\endgroup$
    – TheSilverDoe
    Feb 28, 2019 at 9:51
  • $\begingroup$ The difficult part of the problem is to prove the Invariance of Domain Theorem without invoking any algebraic topology but using the expansion property. $\endgroup$
    – Moishe Kohan
    Mar 1, 2019 at 18:01

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