I'd like an example of a function $f:(a,b)\to\mathbb R$ and a point $c\in(a,b)$ such that:
- $f$ is invertible.
- $f$ is continuous at $c$.
- $f^{-1}$ is discontinuous at $f(c)$.
Motivation: There is a calculus book that states the following.
Let $f$ be an invertible function defined on an interval $I$. If $f$ is differentiable at $c\in I$ and $f'(c)\neq 0$, then $f^{-1}$ is differentiable at $f(c)$.
In the proof, the continuity of $f^{-1}$ at $f(c)$ is essential. Usually, the said essential fact is an hypothesis (if the domain is not an interval) or it is implied by the hypothesis that $f$ is continuous in a neighborhood of $c$ (if the domain is an interval). But in the said book, both hipothesis are missing and the fact is justified as follows:
As $f$ is differentiable at $c$, $f$ is continuous at $c$. Therefore, $f^{-1}$ is continuous at $f(c)$.
I suspect continuity at $c$ does not imply continuity of the inverse at $f(c)$ due to the following facts:
- It seems it is not a common result in analysis books.
- In the usual proofs that the inverse of a continuos map (on an interval or on a compact set) is continous, in order to prove that the inverse is continuous at a given point, we need the continuity of $f$ in the whole domain.
- In more recent editions of the said book, the statement was modified (now, it is supposed that $f$ is differentiable in a neighborhood of $c$, which implies what is needed).
However, I do not have a counterexample.
Best Answer
Here is an example of such a function. It comes from the book Les contre-exemples en mathématiques by Bertrand Hauchecorne (more precisely, section 8.22, page 150, in the second edition).
Consider $g\colon \Bbb R_+ \to \Bbb R_+$ defined by $$ g(x) = \begin{cases} \frac{n}{2} & \text{if} \quad x=n \quad \text{is an even integer},\\ \frac{1}{n+2} & \text{if} \quad x=n \quad \text{is an odd integer},\\ \frac{1}{2(n-1)} & \text{if} \quad x=\frac{1}{n} \quad \text{with} \quad n \geqslant 2 \quad \text{integer},\\ x & \text{in any other case}. \end{cases} $$ Now, consider $f\colon \Bbb R \to \Bbb R$ defined as $$ f(x) = \begin{cases} g(x) & \text{if} \quad x \geqslant 0,\\ -g(-x) & \text{if} \quad x < 0. \end{cases} $$ One shows with a bit of effort that $f$ is a bijection.
Note that $|f(x)| \leqslant |x|$ for all $x$, so that $f$ is continuous at $0$ with $f(0)=0$. However, for all $n \geqslant 1$, $f(2n-1) = \frac{1}{2n+1}$ shows that $f^{-1}\left(\frac{1}{2n+1}\right) = 2n-1 \underset{n \to \infty}{\longrightarrow} \infty$, so that $f^{-1}$ is not continuous at $0$.