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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.

RandomStudent
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    This might be too big of a hammer, but see https://en.wikipedia.org/wiki/Invariance_of_domain – Lukas Geyer Feb 18 '19 at 15:18
  • Isn't it exactly what I need? With $U = \mathbb{R}^n$ ? But unfortunately "The proof uses tools of algebraic topology, notably the Brouwer fixed point theorem." – RandomStudent Feb 18 '19 at 15:42
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    But maybe there is an easier proof using the expansive condition ? – RandomStudent Feb 18 '19 at 15:52
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    Yes, that is what I meant, it solves your problem, but there might be a more direct and easier way here. – Lukas Geyer Feb 18 '19 at 15:58
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    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? – Moishe Kohan Feb 27 '19 at 18:18
  • No idea what you're referring to, but seems interesting, so will read up on this stuff. As for our class, we're using Real Analysis by Carothers. Probably this was mistakenly present in the provided problemset, but atleast I'm convinced at this point that my elementary attempts had no chance of success. – RandomStudent Feb 27 '19 at 18:32
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    The simplest example to think about is the map $z\mapsto 2z$ of the set of natural numbers. It is expanding but not surjective. – Moishe Kohan Feb 27 '19 at 18:44
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    I actually spent some time thinking about compactification of $\mathbb R^n$, and maybe a continuous extension of $f$ giving results. It did not work out for me, but I thought I'd put the idea forward anyway. It did not work because I could not show that the extended $f$ is continuous with respect to the metric on the compactification on $\mathbb R^n$. – Sarvesh Ravichandran Iyer Feb 28 '19 at 16:08
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    You can use a proof from Terry Tao's blog here: https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/. The point is that since in your case $f^{-1}$ is Lipschitz, it sends sets of measure zero to sets of measure zero, hence, there is no need to use polynomial approximation and, accordingly, no need to appeal to Brouwer's fixed point theorem. The rest of the proof is pure analysis. Let me know if you want to see details. It is more instructive to work it out yourself, which is the whole point of the assigned exercise. – Moishe Kohan Mar 01 '19 at 18:23
  • It is easy to see that $f(\infty)=\infty$. So, one might think of $f$ as $S^2 \to S^2$, which is compact! We just need the right metric with which $f$ is still expansive. – Behnam Esmayli Nov 15 '19 at 13:42

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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.

TheSilverDoe
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    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 $y=f(\lim_{n\to\infty} x_n) \in f(\mathbb R^n)$. – Skeeve Feb 28 '19 at 07:43
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    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 $\mathbb{R}^n)$. – RandomStudent Feb 28 '19 at 08:21
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    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. – TheSilverDoe Feb 28 '19 at 09:51
  • The difficult part of the problem is to prove the Invariance of Domain Theorem without invoking any algebraic topology but using the expansion property. – Moishe Kohan Mar 01 '19 at 18:01