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Is there any way of constructing a non-strictly monotonic bijective function $f:\Bbb R\to(0,+\infty)$ satisfying: $$f(x+y)=f(x)f(y), f(0)=1, f(1)=a>0\space$$

(without a Hammel basis for $\Bbb R$ over $\Bbb Q$)?

This question, without the condition that $f$ is not strictly monotonic, has already been asked many times, but I couldn't think of any discontinuous bijection from $\Bbb R$ to $(0,+\infty)$ with the properties above. I know that strict monotonicity implies $f(x)=a^x,\space\forall x\in\Bbb R$. One idea was to take some dense additive subgroup $G\subset\Bbb R$ and define $f(x)=a^x,\space\forall x\in G$, but then, as we require injectivity and $f>0$, the problem arises with $f(\Bbb R\setminus G)$. I found a related answer where it is proven that $f$ is either identically $0$ or $f>0\space\forall x\in\Bbb R$, but I couldn't use that answer to construct a function I'm looking for because we haven't learned about a Hammel basis in real analysis lectures yet. I also eliminated $f(x)=\alpha x,\alpha\in\Bbb R$ after realizing I couldn't fix one $\alpha$.

Is there any more elementary method I'm failing to see?

Thank you in advance!

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You'll have hard time to build such a solution based on elementary methods. See this answer for more details.

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  • $\begingroup$ Thank you very much for the link and the feedback! $\endgroup$ – Invisible Oct 22 at 17:40

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