TITLE: Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x\in \Bbb R^+$ the following is valid: $f\bigg(\frac{1}{f(x)}\bigg)=\frac{1}{x}$
QUESTION [9 upvotes]: Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x\in \Bbb R^+$ the following is valid:


$$f\bigg(\frac{1}{f(x)}\bigg)=\frac{1}{x}$$


I tried to substitute $\frac{1}{x}$ for $x$ and compare the equations:
$$f\bigg(\frac{1}{f(\frac{1}{x})}\bigg)=x $$
From this I found one solution $f(x)=x.$

REPLY [5 votes]: Using the hint of @Somos we can substitute
$$g(x)=-\log{(f(\exp{(x)}))}$$
so that the equation becomes
$$g(g(x))=x$$
So we just need $g(x)$ to be an involution (of which there are infinitely many). Choosing any involution $g(x)$ defined over $\mathbb{R}$ gives a solution
$$f(x)=\exp{(-g(\log{(x)}))}$$
Some solutions include
$$(g(x),f(x))\in\{(x,1/x),(-x,x)\}$$
I believe these are the only continuous solutions but there should be infinitely many discontinuous ones.