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I've found a nice problem concerning analytic functions. Here it is:

Let $f: (0, \infty) \rightarrow \mathbb{R}$ be a function differentiable on $(0, \infty)$ and such that $f^{-1} = f'$. Prove that $f$ is analytic on $(0, \infty)$.

I'm not sure if it's relevant, but I know that $f$ cannot be a bijection :)

Could you help me?

Andrew
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  • is $f$ infinitely differentiable? this is essentially what the question appears to be asking – DanZimm Apr 23 '13 at 16:43
  • @DanZimm: Analyticity is stronger than infinite differentiability. – Jonas Meyer Apr 23 '13 at 16:49
  • Just to confirm - is $f^{-1}$ actually the inverse function s.t. $f\circ f^{-1} \equiv \iota$? (Then we assume $f$ injective.) If so, does $f^{-1} = f'$ require that the domains of both sides are the same (in which case the image of $f$ is $(0,\infty)$ as well), or that they agree on their domains (so that $f'(-1)$ is ignored if $f(x)=-1$ for some $x$), or what? – not all wrong Apr 23 '13 at 17:01
  • @JonasMeyer ah right right, my bad – DanZimm Apr 23 '13 at 17:05
  • Yes, $f^{-1}$ is the inverse function of $f$. I think we can assume that they agree on their domains. Actually, I've rewritten the whole problem. – Andrew Apr 23 '13 at 17:07
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    See http://math.stackexchange.com/questions/279517/does-there-exist-f0-infty-to0-infty-such-that-f-f-1 and http://mathoverflow.net/questions/34052/function-satisfying-f-1-f/ for the same setup, though they ask different questions. – not all wrong Apr 23 '13 at 23:28

1 Answers1

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The answer is affirmative, as proven in this MathOverflow post. The answerer, José Hdz. Stgo., assumes $f:[0,\infty) \to [0, \infty)$ and $f(0) = 0$, but these additional hypotheses don't affect the argument regarding analyticity.

MathematicsStudent1122
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