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Find the functions $f:\Bbb R_*^+ \to \Bbb R_*^+$ such that:

$$f(x)f \left(\frac{1}{x}\right)=1$$

  • 0
    What is $R_\ast^+$?2012-10-24
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    I presume it's better known, and less ambiguously written, as $\Bbb R_{> 0}$.2012-10-24
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    Am I missing something? $f(x) = x$.2012-10-24
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    Where's your proof that no other $f$ does the job?2012-10-24
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    None. $f(x) = 1$ also works. As does $f(x) = \frac{1}{x}$. I guess the question is to characterize all such functions...2012-10-24
  • 0
    What about $f(x) = 1+(x-1)\chi_{\Bbb Q}$? No continuity conditions...2012-10-24
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    How about $f(x)$ being $x^c$ and $-x^c$ ?2012-10-24

2 Answers 2

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In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe2111.pdf.

The general solution is $f(x)=\pm e^{C\left(x,\frac{1}{x}\right)}$ , where $C(u,v)$ is any antisymmetric function.

5

For any function $\tilde{f}:[1,\infty)\rightarrow \mathbb{R}_+^*$ such that $f(1)=1$, there is a unique extension $f:(0,\infty)\rightarrow \mathbb{R}_+^*$ defined for $x<1$ by $f(x)=1/\tilde{f}(1/x)$.