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From the result discussed in Does the inverse function theorem hold for everywhere differentiable maps? (which I'll call the differentiable nonsmooth Inverse Function Theorem) 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/ .

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If I remember correctly, strict differentiability at one point $x_0$ (and of course invertibility of $df(x_0)$) is sufficient for the implicit function theorem to hold. This means something like $$ df(x_0)(h) = lim_{t\to 0, x\to x_0}\frac1t f(x+th). $$ I am traveling and I do not have access to the books now. This is in:

  • MR0724435 Ver Eecke, Paul Fondements du calcul différentiel. (French) [Foundations of differential calculus] Mathématiques. [Mathematics] Presses Universitaires de France, Paris, 1983. 345 pp. ISBN: 2-13-038180-4 (Reviewer: William Eames) 26-01 (46G05 58-01 58C20)

  • MR0817719 (87e:58001) Ver Eecke, Paul(F-PCRD) Applications du calcul différentiel. (French) [Applications of differential calculus] Mathématiques. [Mathematics] Presses Universitaires de France, Paris, 1985. 397 pp. ISBN: 2-13-038961-9 58-01 (26E15)

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  • $\begingroup$ English treatments can also be found in Schechter's Handbook of Analysis and its Foundations. See sections 25.12 and 25.13. (What is described as "strict" differentiability above, Schechter chooses to call "strong" differentiability.) $\endgroup$ Dec 6, 2013 at 9:05
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    $\begingroup$ Thanks for the answer and references, but this is not really what I'm looking for. Strict differentiability is close enough to continuous differentiability to see that the usual proofs can be adapted to this case. The questions I asked seem to be more subtle, along the lines of the question asked by Tao (and answered) in mathoverflow.net/questions/75049/… . $\endgroup$ Dec 6, 2013 at 18:39
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The answer to the first question is YES. There were two proofs of this result, one by Jean Saint Raymond (Mathematika, 2002), the other by L. Hurwicz and M. K. Richter (Discussion paper no. 279, Dept. of Economics, University of Minnesota, 1994).

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