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Does there exist a differentiable bijection $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f'(x) = f^{-1}(x)$ ?

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  • $\begingroup$ Motivated by your question, I asked the following question, in case you might be interested. $\endgroup$
    – Dejan Govc
    Commented Jan 15, 2013 at 20:23

1 Answer 1

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No such bijection exists.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable bijection. Then $f$ is increasing or decreasing. If $f$ is increasing, then $f'(x)\ge0$ for all $x\in\mathbb{R}$, while if $f$ is decreasing then $f'(x)\le0$ for all $x\in\mathbb{R}$. If $f'=f^{-1}$, then $f^{-1}(x)\ge0$ or $f^{-1}(x)\le0$ for all $x\in\mathbb{R}$, contradicting the fact that $f$, and hence $f^ {-1}$, is a bijection.

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  • $\begingroup$ Possible counterpoint? What about $f(x)=x^{\phi}*\phi^{1-\phi}$ where $\phi$ is the Golden Ratio. (I stole this solution from mathoverflow.net/questions/34052/function-satisfying-f-1-f/…) $\endgroup$
    – Platatat
    Commented Apr 22, 2014 at 4:58
  • $\begingroup$ $f$ is a bijection satisfying $f'=f^{-1}$ on $(0,\infty)$. When considered as a bijection fron $\mathbb{R}$ to itself, it has to be defined as $-f(|x|)$ for $x<0$, and then $f'(x)>0$ for all $x$. $\endgroup$ Commented Apr 23, 2014 at 9:42

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