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

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 $$f(x)=x.$$

• Let $g(x):=\log(f(\exp(x))$. Then $x=-g(-g(x)).$ Jul 25, 2020 at 20:01

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.