Does the inverse function theorem hold for everywhere differentiable maps?

Question

(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:

1. 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.)
2. 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.)
3. 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.

Answer 1

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.

Answer 2

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...

Answer 3

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

I present an inverse function theorem for differentiable maps between Frechet spaces which contains the classical theorem of Nash and Moser as a particular case. In contrast to the latter, the proof does not rely on the Newton iteration procedure, but on Lebesgue's dominated convergence theorem and Ekeland's variational principle. As a consequence, the assumptions are substantially weakened: the map F to be inverted is not required to be C^2, or even C^1, or even Frechet-differentiable.

Answer 4

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:

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

2. 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/ .

Answer 5

S. Radulescu and M. Radulescu gave a proof of the inverse function theorem for everywhere differentiable maps in 1989. The title of the paper, published in J. Math. Anal. Appl., is “Local inversion theorems without assuming continuous differentiability”.