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In my real analysis class my professor gave us the problem of proving that if $F: \mathbb{R}^{m} \rightarrow \mathbb{R}^{m}$ is continuous and satisfies $\| F(x) - F(y)\| \geq \lambda \| x - y \|$ then $F$ is a bijection with continuous inverse. ($∥⋅∥$ is the Eucliden norm and $\lambda$ is some positive real number.)

The problem of injectivity is easy enough since if $x \neq y$ then $\|F(x) - F(y)\| \geq \lambda \|x-y\| > 0$.

Also given that F is a continuous bijection then the continuity of the inverse $g$ is also obvious since fixing $x = F(u)$ and $y = F(v)$ we have that $\|g(x) - g(y)\| = \|u-v\| \leq \frac{1}{\lambda}\|x-y\|$ so g is Lipschitz and therefore continuous.

My question is, how exactly is one supposed to prove surjectivity? It seems easy enough by intermediate value theorem if we restrict $F:\mathbb{R} \rightarrow \mathbb{R}$. But I can't seem to figure it out more generally. any hints would be much appreciated!

2 Answers2

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This is a partial answer.

Call $X=F(\mathbb{R}^m)$. Then the inverse map $g: X \longrightarrow \mathbb{R}^m$ is Lipschitz continuous. I show that this implies that $X$ is a closed subset of $\mathbb{R}^m$.

Let $x_n \subset X$ be a convergent sequence with $x_n \to x$. It is enough to show that $x \in X$. Now, $g(x_n)$ is a Cauchy sequence in $\mathbb{R}^m$, so it converges to some $y$. Applying $F$ we get $x_n \to F(y) $ hence $x = F(y) \in X$.

If one is able to show that $X$ is open as well (maybe using that $F$ is an embedding?), by connectedness argument we have $X = \mathbb{R}^m$.

Crostul
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  • Well, invariance of domain is one way to see that $X$ is open. – Jose27 Jan 20 '15 at 00:28
  • I suspect we could deduce surjectivity from the continuity method (by considering $F_t(x)=tF(x)+(1-t)x$), or something similar, but I haven't been able to make it work. – Jose27 Jan 20 '15 at 01:14
  • I've been wracking my brain for a while and can't see quite how one would go about proving that $X$ is open. I'd prefer not to use invariance of domain if possible, since my professor only glazed over it without proof in class. –  Jan 21 '15 at 01:56
  • $\mathbb{R}^m$ is connected. This is equivalent on saying that any non-empty subset which is both open and closed must be $\mathbb{R}^m$ itself. See http://en.wikipedia.org/wiki/Connected_space – Crostul Jan 25 '15 at 10:58
  • @EricSilva did you figure out a proof of $X$ being open, without using invariance of domain in the end? – Aryaman Jal Apr 28 '18 at 07:34
  • Why is $g:X\to \mathbb{R}^m $ Lipschitz continuous? – hallo007 Jul 01 '22 at 00:13
  • @hallo97 this is an ovious consequence of our hypothesis. $$|F(x) - F(y)| \ge \lambda |x-y|$$ is the same as $$|g(x) - g(y)| \le \lambda |x-y|$$ – Crostul Jul 01 '22 at 16:49
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here's an idea with $m=2.$

First take $\lambda=1.$ We can rescale to reduce to this case.

Now take the axes, these must map to curves going off to infinity. So we have a curly X in the image.

Let's take a point $(a,b)$ and try to find the point mapping to it. It must be in one part of the $X.$

Let's assume WLOG the part is the image of the upper right quadrant.

If we go far enough along the axes to points $(x,0)$ and $(0,y)$ the line between must map to a curve giving a curly triangle than contains $(a,b)$. This follows from the main hypothesis since the mapped to point must be further from the origin.

Now consider the rays going from the origin to the line between $(x,0)$ and $(0,y).$ We can parametrize these by distance from $(x,0).$ At some point their images must transit from being below $(a,b)$ to above it. So take the supremum of points which give curves that lie below. $(a,b)$ must lie on the image of that ray.

Mark Joshi
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