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Find all function $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ satisfying $f(xf(y))=yf(x)$ for all $x,y\in\mathbb{R}^+$.

Let $P(x,y)$ denote $f(xf(y))=yf(x)$.

Similarly from Functional equation : $f(xf(y))=yf(x)$ ,

we will get $f(1)=1$ and $f(f(x))=x$, $\forall x\in \mathbb{R^+}$. Then $f$ is bijective.

Moreover, consider $P(x,f(y))$: we obtain $f(xy)=f(x)f(y)$, $\forall x,y\in\mathbb{R}^+$

From this, it is not hard to show that

$f(x^p)=(f(x))^p$, $\forall x\in\mathbb{R}^+,$ for every rational number $p$.

It seems all solution to the problem are only $f(x)=x$ and $f(x)=\frac{1}{x}$.

How to prove it or are there other solutions?

Note: If we can show $f(x^p)=(f(x))^p$ for irrational $p$ then we will get all solution are $f(x)=x$ and $f(x)=\frac{1}{x}$.

Thank in advances.

Jean Marie
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Bless
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  • Under the assumption of continuity, $$f(xy)=f(x)f(y)$$ means that there exists some $a \in \Bbb{R}$ such that $f(x)=x^a$ – Crostul Feb 11 '18 at 17:23

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Bad news: there are weird solutions involving the axiom of choice.

Take $B$ a basis for $\mathbb{R}$ over $\mathbb{Q}$. Then $f$ is determined uniquely by its values on points of the form $e^b$, for $b \in B$.

Now, all we have left to take care of is the condition that $ f\circ f = id$. For this, pair the points of $B$ and define $f$ such that $f(e^b)=e^{b'}$, where $b,b'$ are paired.

idok
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