17
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Motivated by your question, I asked the following question, in case you might be interested. $\endgroup$ – Dejan Govc Jan 15 '13 at 20:23
20
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Apr 22 '14 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 $x$. $\endgroup$ – Julián Aguirre Apr 23 '14 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.