Is a differentiable (but not $C^1$) function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ with invertible derivative everywhere an open map?

Question

Is a differentiable (but not $C^1$) function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ with invertible derivative everywhere an open map? I know that if we assume the function is $C^1$, then this is a consequence of the inverse function theorem (or a step on the way to proving the inverse function theorem). I've convinced myself no counter-example exists if $n = 1$, but I haven't been able to prove it or come up with a counter-example in the case $n = 2$. I've spent more time trying to come up with a counter-example, mostly trying variants involving $x^2 \sin(1/x)$.

Answer 1

The result holds in every dimension.

For the sake of simplicity, we assume that $f(0)=0$ and $f’(0)=Id$.

There is a continuous function (with vector values) $\epsilon$ such that for each $x$, $f(x)=x+|x|\epsilon(x)$.

Let $R >0$ be such that if $|x| \leq R$, $|\epsilon(x)| \leq 0.5$.

Let $0<r<R$, let $z$ be such that $|z| < r/4$.

It is easy to prove that for any $x$ with $|x|=r$, $|f(x)-z| > |f(0)-z|$.

So consider $g(x)=f(x)-z$ and $h(x)=|g(x)|$ for $|x| \leq r$.

Assume that $g$ does not vanish, then $h$ is differentiable, without critical points, and does not reach its minimum on the border of the disc, which is impossible.

So $g$ does vanish, ie, for all $z$ with $|z|$ small enough, there is some $y$ with $|y| \leq 5|z|$ sich that $f(y)=z$.

In other words: if $V$ is a neighborhood of $0$, so is $f(V)$, which ends the proof.