2011-09-09 23:09:04 8 Comments

(This question was posed to me by a colleague; I was unable to answer it, so am posing it here instead.)

Let $f: {\bf R}^n \to {\bf R}^n$ be an everywhere differentiable map, and suppose that at each point $x_0 \in {\bf R}^n$, the derivative $Df(x_0)$ is nonsingular (i.e. has non-zero determinant). Does it follow that $f$ is locally injective, i.e. for every $x_0 \in {\bf R}^n$ is there a neighbourhood $U$ of $x_0$ on which $f$ is injective?

If $f$ is *continuously* differentiable, then the claim is immediate from the inverse function theorem. But if one relaxes continuous differentiability to everywhere differentiability, the situation seems to be much more subtle:

- In one dimension, the answer is "Yes"; this is the contrapositive of Rolle's theorem, which works in the everywhere differentiable category. (The claim is of course false in weaker categories such as the Lipschitz (and hence almost everywhere differentiable) category, as one can see from a sawtooth function.)
- The Brouwer fixed point theorem gives local surjectivity, and degree theory gives local injectivity if $\det Df(x_0)$ never changes sign. (This gives another proof in the case when $f$ is continuously differentiable, since $\det Df$ is then continuous.)
- On the other hand, if one could find an everywhere differentiable map $f: B \to B$ on a ball $B$ that was equal to the identity near the boundary of $B$, whose derivative was always non-singular, but for which $f$ was not injective, then one could paste infinitely many rescaled copies of this function $f$ together to produce a counterexample. The degree theory argument shows that such a map does not exist in the orientation-preserving case, but maybe there is some exotic way to avoid the degree obstruction in the everywhere differentiable category?

It seems to me that a counterexample, if one exists, should look something like a Weierstrass function (i.e. a lacunary trigonometric series), as one needs rather dramatic failure of continuity of the derivative to eliminate the degree obstruction. To try to prove the answer is yes, one thought I had was to try to use Henstock-Kurzweil integration (which is well suited to the everywhere differentiable category) and combine it somehow with degree theory, but this integral seems rather unpleasant to use in higher dimensions.

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## 4 comments

## @Bruce Blackadar 2013-12-04 05:13:41

From this result one can obtain a differentiable but nonsmooth version of the Implicit Function Theorem by the usual argument.

There are two interesting closely related questions apparently still remaining:

Is there a corresponding version of the continuous Implicit Function Theorem not requiring continuous differentiability in the variables being solved for?

In the original inductive proof of the Implicit Function Theorem, continuous differentiability was needed to insure a decreasing chain of locally nonvanishing minors for the Jacobian determinant. Does there always exist such a chain if simple differentiability with nonzero Jacobian is assumed?

A positive answer to 2 gives a positive answer to 1 and an inductive proof of the differentiable nonsmooth Inverse Function Theorem.

For a more careful description of these problems, see the exposition of the Implicit Function Theorem in my real analysis manuscript on my website http://wolfweb.unr.edu/homepage/bruceb/ .

## @András Bátkai 2013-12-04 07:28:10

This great question should be posted as a separate question. Here it is somehow lost and probably does not receive proper attention.

## @S. Carnahan 2013-12-04 11:29:23

I agree. If you decide to write a separate question, it may help to add a link to this one (and perhaps write a comment on the question at the top, pointing to your question).

## @Bruce Blackadar 2013-12-04 22:48:41

Thanks, I did that; see mathoverflow.net/questions/150856/…

## @Julien Melleray 2011-09-12 19:21:43

For another proof that your question has a positive answer, you may look at the folowing paper by Jean Saint Raymond: http://www.math.jussieu.fr/~raymond/preprints/inversion.dvi

It seems that he was not aware of the reference given by David Preiss above - his proof is in English so at least it should avoid you having to learn Russian just for that...

## @Terry Tao 2011-09-13 00:10:39

Thanks! It is a nice argument; I decided to write it up (so that I could properly understand it) at terrytao.wordpress.com/2011/09/12/…

## @Julien Melleray 2011-09-13 14:13:17

You're more than welcome. I'm amazed how quickly you read the paper and wrote about it...

## @David Preiss 2011-09-10 21:25:32

The usual reference to the proof is A. V. Cernavskii in "Finite-to-one open mappings of manifolds", Mat. Sb. (N.S.), 65(107) (1964), 357–369 and "Addendum to the paper "Finite-to-one open mappings of manifolds"", Mat. Sb. (N.S.), 66(108) (1965), 471–472. If I remember it correctly, he does not state it explicitly, but it follows from what is there.

## @Gil Kalai 2011-09-10 21:49:19

Welcome to MO, David!

## @Terry Tao 2011-09-11 00:30:07

Thanks David! (Now I just need to get a physical copy of the paper and learn some Russian...)

## @Terry Tao 2011-09-11 00:44:28

(Ah, there is a translated copy at American Mathematical Society Translations, Series 2. Vol. 100: Fourteen papers on logic, geometry, topology and algebra. American Mathematical Society, Providence, R.I., 1972. iv+316 pp. Still need to get to the library, though. )

## @Tyler Helmuth 2011-09-10 05:15:02

From http://arxiv.org/abs/1011.1288 by Ivar Ekeland:

## @Terry Tao 2011-09-10 16:06:07

Thanks for the interesting reference! Reading it, though, it seems that this paper is focused on the infinite dimensional case, and only establishes local surjectivity rather than local injectivity (it seems oriented towards the task of proving existence (but not uniqueness) for PDE. It also has an additional hypothesis that $Df(x_0)^{-1}$ is locally bounded.