# Is there a differentiable but nonsmooth version of the continuous Implicit Function Theorem?

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/ .

• I know this is an old question, but I am slightly confused by it - could the OP (or someone else) clarify what is meant by the "continuous implicit function theorem" here, in contrast to the "differentiable nonsmooth inverse function theorem"? I think the latter is meant to be the theorem stated in the paper by Jean Saint Raymond mentioned in one of the answers. Commented Feb 5 at 9:38

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: