Example of function that is differentiable at $0$, and has inverse function that is not continuous at $0$?

Question

Is there any example of a function $f(x)$ differentiable at $x=0$, with an inverse function that is not continuous when $x=0$? Any help where to start, or maybe even if someone has the example of such a function would be greatly appreciated.

Answer 1

For $n\in\mathbb N$, let $A_n$ be sets with the following properties:

• $A_n$ is uncountable,
• $0\notin A_n$,
• $0\in \overline{A_1}$
• $A_n\cap A_k=\emptyset$ for $k\ne n$.
• $(-1,-\frac1{n^2}]\cup[-\frac1{n^2},1)\subseteq \bigcup_{k< n}A_k$

One can write such a partition down explicitly (see below), but I'm afraid the formalism might hide theses four essential properties.

For each $n$, let $$f_n\colon \left(-\tfrac1n,-\tfrac1{n+1}\right]\cup\left[\tfrac1{n+1},\tfrac1n\right)\to A_n$$ be any bijection and let $$f(x)=\begin{cases}0&\text{if }x=0\\f_n(x)&\text{if }\frac1{n+1}\le |x|<\frac1n\end{cases}$$ Then $f\colon(-1,1)\to(-1,1)$ is a bijection. Let $g$ be its inverse.

If $\frac1{n+1}\le |x|<\frac1n$, we have $f(x)\in A_n$, hence $|f(x)|<\frac1{n^2}\le (1+\frac1n)^2|x|^2\le 4|x|^2$, hence $f'(0)=0$.

But $g$ is not continuous: Of course $g(0)=0$. But if $\epsilon>0$, because $0$ is a limit point of $A_1$, there are $x$ with $|x|\ge \frac12$ and $f(x)<\epsilon$. Hence there are $x$ with $|x|<\epsilon$ and $|g(x)|\ge \frac12$.

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Here's an explicit description of $A_n$:

$$A_1=\left\{x\in\mathbb (-1,1)\mid x\ne 0, \left\lceil \tfrac1{|x|}\right\rceil\le4\text{ or even}\right\}$$ $$A_n=\left\{x\in\mathbb (-1,1)\mid x\ne 0, n^2<\left\lceil \tfrac1{|x|}\right\rceil\le(n+1)^2\text{ and odd}\right\}\quad\text{if }n\ge2$$

Answer 2

Unless there are more conditions that you're not sharing, then $f(x)=e^x$ is such a function. Its inverse function $g(x)=\ln x$ isn't even defined at $x=0$, much less continuous there.

Answer 3

Let $f(x)$ be the pdf of a standard lognormal distribution and $x=a$ be its mode. The domain of $f$ is $(0,\infty)$, but let us also define $f(0)=0$. Then the right hand derivative of $f$ at zero is zero. Now, let \begin{align} Q &= f^{-1}(\mathbb{Q}),\\ Q^c &= f^{-1}(\mathbb{R}\setminus\mathbb{Q}) = \mathbb{R}^+\setminus Q,\\ g(x) &=\begin{cases} f(x) &\textrm{ if } x\in [0,a]\cap Q,\\ -f(x) &\textrm{ if } x\in [0,a]\cap Q^c,\\ f(x) &\textrm{ if } x\in (a,\infty)\cap Q^c,\\ -f(x) &\textrm{ if } x\in (a,\infty)\cap Q.\\ \end{cases} \end{align} Then $g(0)=g'(0)=0$ and $g$ has an inverse, but $g^{-1}$ is not continuous at $0$ because $\lim_{x\rightarrow\infty}g(x)=0=g(0)$.

Answer 4

WLOG we will asssume $f(0) = 0$ and $f'(0) = 0$.

Recall that if the function $f$ is continuous in a neighborhood of $0$ then it is either monotonic (in which case it has a continuous inverse) or non monotonic (in which case it fails to be injective). So our $f$ needs to be continuous (and differentiable) in $0$ but not in any neighborhood of $0$.

We want $f^{-1}$ not to be continuous in $f(0) = 0$, meaning that there must exist $\varepsilon>0$ such that for any $\delta >0$ there is a $x$, with $|x| < \delta$, for which $|f^{-1}(x)|\geq \varepsilon$.

Translating this into a requirement on $f$ yields the following. There exists $\varepsilon > 0$ such that, for each $\delta > 0$ we can find a $|x|\geq \varepsilon$ that gives $|f(x)| < \delta$. In other words, we want $f(x)$ to attain arbitrarly small values, for 'large enough' values of $x$, which is not in contradiction with the requirement on $f$ we have in our hypotheses. The hardest part of the requirement is that of obtaining a function which has the above mentioned property and yet is injective.

We will construct a bijective function $f: [-2,2] \to [-2,2]$ by first creating a function $g$ that satisfies the requirements for $0\leq x \leq 2$ and then get $f$ by simmetry.

We will start by letting, for $0\leq x \leq 1$, $$g_1(x) = p(x)= x^2,$$ which has the required properties save for the fact that its inverse is continuous in $0$, and force our modification of $g_1$ to lie within the parabola and the cubic function $$c(x) = x^3.$$ We create the first 'hole' in $g_1$ at $2^{-1}$ by letting $$g_1\left(2^{-1}\right) = c\left(2^{-1}\right) = 2^{-3}.$$ This action frees $2^{-2}$ that later on will be used as image of a larger value of $x$.

In order to preserve injectivity we now have to modify the image of $2^{-3/2}$. Again we remap this point using $c$, which yields $$g_1\left(2^{-3/2}\right) = c\left(2^{-3/2}\right)=c\circ p^{-1} \circ c\left(2^{-1}\right) = 2^{-9/2}. $$

This process must be repeated indefinitely by letting, for $n=0,1,2\dots$, $$g_1\left(t^{(n)}\left(2^{-1}\right)\right)=c\left(t^{(n)}\left(2^{-1}\right)\right),$$ where $$t(x) = p^{-1} \circ c(x),$$ and $$t^{(n)}(x) = \underbrace{t\circ t \circ \cdots \circ t}_{n} (x)$$ for $n>0$, and $t^{(0)}(x) = x$.

Now, as mentioned earlier, we have room to let $g(x) = 2^{-2}$ for some $1<x \leq 2$. Fixing $g\left(2-2^{-2}\right) = 2^{-2}$ yields the scenario depicted below, where $$r : r(x) = 2-x.$$

Now we can go back to the neighborhoods of $0$ and start another iteration at $3^{-1}$. Using the sequence of primes $p_k$, $k=1,2,\dots$ prevents any 'interference' between different iterations. Hence we have, for $0\leq x \leq 1$, $$g_1(x)= \begin{cases} c(x) & \left(x=t^{(n)}\left(p_k^{-1}\right), \ \mbox{for some} \ k>0 \ \mbox{and some} \ n>0\right)\\ p(x) & (\mbox{otherwise}). \end{cases} $$ Note that $g_1$ is injective (what is its range?).

Finally me make the domain compact and connected by means of a reflection with respect to line $r$. As a result we have $$ g(x) = \begin{cases} g_1(x) & (0\leq x \leq 1)\\ p_k^{-2} & \left(x = 2-p_{k}^{-2},\mbox{for some}\ k>0\right)\\ 2-g_1^{-1}(2-x) & (\mbox{otherwise}). \end{cases} $$ The result after a few iterations on $k$ is sketched below.

Now the required $f$ is $$ f(x) = \begin{cases} g(x) & (0\leq x \leq 2)\\ -g(-x) & (-2\leq x <0). \end{cases} $$

A final note. The reader can easily verify that $f(x)$ is differentiable almost everywhere, i.e. at all points $-2 < x <2$ such that $f(x) = \pm p(x)$.