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Let $f$ be a continuous and bijective function such as $f(\frac{a}{b}) = \frac{f(a)}{f(b}$.
What are the possible functions for $f$?

My guess is that it can only be a polynomial of the shape $f : x \mapsto x^n$.
I have already found a few properties:

  • $\forall n,m \in \mathbb{Z}, f(x^n) \cdot f(x^m) = f(x^{n+m})$
  • $\forall n \in \mathbb{Z}, f^n(\frac{a}{b}) = \frac{f^n(a)}{f^n(b)}$
  • $f(1) = 1$
  • $f(0) = 0 \text{ if } f \neq x \mapsto 1$

Any ideas on how to prove that it is only polynomials $X^n$ that works ?

Ottavio
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Adrien Chaffangeon
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  • If you restrict $f$ to be defined on $(0,\infty)$ then $x \mapsto x^\alpha$ with $\alpha \in \mathbb R$ also works. On what your $f$ is defined? – mathcounterexamples.net Dec 07 '20 at 12:58
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    https://math.stackexchange.com/questions/43964/if-fxy-fxfy-then-show-that-fx-xt-for-some-t may help. – Tan Dec 07 '20 at 13:16

1 Answers1

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If $f$ is defined on $(0,\infty)$, take $g(x) := \ln(f(e^x))$. Then since $f$ is continuous, $g$ is continuous $$ g(a-b) = \ln(f(e^a/e^b)) = \ln(f(e^a)/f(e^b)) = g(a) - g(b) $$ Since $f(1) = 1$, then $g(0) = 0$, so $g(-b) = -g(b)$ and $$ g(a+b) = g(a)+g(b) $$ The problem then reduces to Cauchy's functional equation. , and so $g$ is linear, i.e. $$ f(x) = e^{g(\ln(x))} = e^{a\ln(x)} = x^a $$ for all $x>0$.

If $f$ is defined on $\mathbb R$, then for $x<0$, $f(x) = f(|x|/(-1)) = f(x)/f(-1) = |x|^a / f(-1)$. Since $f$ is bijective, $f(-1)<0$.

Since $f(-1) = f(-1/1) = f(1/-1) = f(-1) = 1/f(-1)$ we obtain $f(-1) = -1$. Finally $$ \boxed{f(x) = |x|^{a-1} x} $$ for some $a>0$.

LL 3.14
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