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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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So you wnat $f,g$ such that $f(0)=g(0)=0$, $f(g(x))=g(f(x))=x$, $f$ differentiable at $0$, $g$ not continuous at $0$? –  Hagen von Eitzen Dec 10 '12 at 23:21
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3 Answers

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 $|g(x)|\ge \frac12$.


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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I think you're mixing up the domain and codomain here. The inverse of $y = e^x$ is $x = \ln y$. This is perfectly defined and differentiable when $x=0$, i.e. when $y=1$. –  Nate Eldredge Dec 10 '12 at 23:15
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@NateEldredge Umm, I would say the function is $f(x) = e^x$ and the inverse is $f(x) = \ln x$, both of which are functions of $x$. This agrees with the answer. I think the question is not clear and the OP should clarify. –  Graphth Dec 10 '12 at 23:20
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@NateEldredge: $x=\ln y$ isn't the inverse of $y=e^x$. It's precisely the same curve! For the inverse, we'd need $y=\ln x$ or $x=e^y$. –  Cameron Buie Dec 11 '12 at 5:47
    
@Graphth: Well, my point was that saying something about the inverse function "when $x=0$" could be a statement about a value in its domain or in its codomain. And the latter makes more sense in the question: we know what $f(x)$ does near $x=0$. There's no reason this should have anything to do with what $u = f^{-1}(v)$ does near $v=0$ (choosing completely different letters to avoid further confusion), but it could plausibly be related to what happens near $u=0$. –  Nate Eldredge Dec 11 '12 at 14:38
    
Fair enough, @Nate. I see where you're coming from. I'm still not entirely clear on what the OP intended, here. –  Cameron Buie Dec 11 '12 at 15:48
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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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