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

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

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What is $R_\ast^+$? – Asaf Karagila Oct 24 '12 at 15:05
I presume it's better known, and less ambiguously written, as $\Bbb R_{> 0}$. – Lord_Farin Oct 24 '12 at 15:06
Am I missing something? $f(x) = x$. – copper.hat Oct 24 '12 at 15:07
Where's your proof that no other $f$ does the job? – Lord_Farin Oct 24 '12 at 15:08
None. $f(x) = 1$ also works. As does $f(x) = \frac{1}{x}$. I guess the question is to characterize all such functions... – copper.hat Oct 24 '12 at 15:08
up vote 1 down vote accepted

In fact this functional equation belongs to the form of

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

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+1 for paul. I deserve a point for my comment imho :) – mick Oct 29 '12 at 22:33

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 $f(x)=1/\tilde{f}(1/x)$.

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+1 very nice... – copper.hat Oct 24 '12 at 15:47

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