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.
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For $n\in\mathbb N$, let $A_n$ be sets with the following properties:
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 $|g(x)|\ge \frac12$. Here's an explicit description of $A_n$: $$A_1=\left\{x\in\mathbb (-1,1)\mid x\ne 0, \left\lceil \tfrac1{|x|}\right\rceil\le4\text{ or even}\right\}$$ $$A_n=\left\{x\in\mathbb (-1,1)\mid x\ne 0, n^2<\left\lceil \tfrac1{|x|}\right\rceil\le(n+1)^2\text{ and odd}\right\}\quad\text{if }n\ge2$$ |
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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. |
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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 $\lim_{x\rightarrow\infty}g(x)=0=g(0)$. |
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I happened to see a paper on the arxiv last night that gives an example answering the OP's question: see Example 2 here. (Note that here $f(0)= 0 $, as the OP probably intended.) I confess that I haven't looked at the construction in detail, but superficially it looks similar to Hagen's answer...which in fact was posted to the internet first! |
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