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A problem from my differential geometry class:

Suppose $f:\mathbb{R}^2 \to \mathbb{R}^2$ is a $C^1$ mapping, and for every $x\in \mathbb{R}^2$

$$\| Df(x) - \text{Id} \| < 10^{-10}.$$

Prove or disprove: $f$ must be a bijection.

My intuition tells me that there should be a counterexample. $f$ linear will not work, because either it is an isomorphism, or it has range of dimension $\leq 1$, in which case $\|Df - \text{Id} \| = \|f - \text{Id}\|$ will be too big.

One idea I had is a function which on the complex plane would be expressed as $z\mapsto z^{\alpha}$, where $\alpha - 1$ is positive and very close to zero. This would certainly not be a bijection, but should not move anything on the unit circle too much. The derivative of such a function would be the matrix representing the complex number $\alpha (a+bi)^{\alpha -1}$. It seems like this matrix would be sort of tricky to come up with. $\alpha(a+bi)^{\alpha -1} :=\alpha e^{(\alpha-1) \log{(a+bi)}}$, and how can I simplify this?

I know this function is not complex-differentiable at zero, but perhaps it is differentiable when viewed from $\mathbb{R}^2 \to \mathbb{R}^2$?

Any hints or ideas? No need to feed me the answer. Thanks

Eric Auld
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2 Answers2

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Hint: Let $g(x) = f(x) - x$. If $f(x) = f(y)$, $x - y = \int \ldots$

Robert Israel
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  • The Mean Value Inequality might be better than a nebulous integral. But I was hoping people would let Eric figure it out rather than giving it away. – Ted Shifrin Sep 13 '13 at 04:16
  • Well, that's why I made it nebulous. – Robert Israel Sep 13 '13 at 05:36
  • I've heard of cloud computing, but cloud integrals are new to me. – copper.hat Sep 13 '13 at 06:22
  • Hmmm...I'm not seeing it completely. $Dg = Df - \text{Id}$, so we can bound $|g(y) - g(x)| = |x-y|$ above by using mean value inequality, but what good is bounding $|x-y|$ above? Also, I'm not seeing what $x-y$ has to do with any integral. (Here $x,y\in \mathbb{R}^2$, right?) – Eric Auld Sep 13 '13 at 15:37
  • If $0 \le d \le c d$ where $c < 1$, what can you conclude about $d$? – Robert Israel Sep 14 '13 at 01:44
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This is basically the same as Robert's solution but with solving $y=f(x)$ as the motivation.

Consider looking for solutions to $y=f(x)$ by reformulating as a fixed-point problem. In this case we could try $y-x+x-f(x) = 0$, which can be written as $y+x-f(x) = x$. So, we look for fixed points of $\phi_y(x) = y+x-f(x)$.

We see that $\|\phi_y(x_1)-\phi_y(x_2) \| \le \sup_\xi \|I-\frac{\partial f(\xi)}{\partial x} \| \|x_1-x_2\|$, and so $\phi_y$ is a contraction.

copper.hat
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  • Where you write $\frac{\partial f (\xi)}{\partial x}$, do you mean $Df(\xi)$? (Just different notation, I think.) So this argument would seem to imply that if $|Df - \text{Id}| \leq \eta <1$ on $\mathbb{R}^2$, then $f$ must be a bijection. This is not what I expected! – Eric Auld Sep 13 '13 at 15:50
  • Yes, $Df(\xi) = \frac{\partial f(\xi) }{\partial x}$. And yes, $\eta<1$ is sufficient. – copper.hat Sep 13 '13 at 16:17
  • Can the $\sup$ in your comment be taken over $\xi$ on the line between $x_1$ and $x_2$, or must it be over a larger set? – Eric Auld Sep 13 '13 at 17:50
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    If you want to show that $f$ is bijective for all $x$, then the $\sup$ is over all $x$. – copper.hat Sep 13 '13 at 17:59