Find $f$ such that $f(af(b))=bf(a)$

Asked May 28 '23 at 13:28

Active May 28 '23 at 13:28

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Find $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that $$\forall a,b \in \mathbb{R}^+, f(af(b))=bf(a)$$

$f(x)=x$ is a trivial solution, but I'm not sure if this is the only one.

I can see that $f(f(x))=x$ and therefore $f$ is bijective, but not anything else.

asked May 28 '23 at 13:28

Saturday

2

I believe $f(x)=\frac 1x$ may be another solution... – abiessu May 28 '23 at 13:33
I see $f(f(x))=xf(1)$ so, if $f(1)=0$ , it's not bijective – lulu May 28 '23 at 13:35
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@lulu, the range is $\mathbb{R}^{+}.$ – aqualubix May 28 '23 at 13:36
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Ah, missed that. But still, there's the $f(1)$ – lulu May 28 '23 at 13:36
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This seems like it's a duplicate of this: https://math.stackexchange.com/q/2646128/1104384 – Bruno B May 28 '23 at 13:42

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