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Consider the field $\mathbb{R}(X)$ together with the total ordering given by: $a > b$ $\Leftrightarrow$ $a(x) - b(x)$ is eventually positive.

Say that a map $f: \mathbb{R}(X) \to \mathbb{R}(X)$ is a contraction if there exists a constant $C \in \mathbb{R}(X)$ with $0 \leq C < 1$ such that for all $a, b \in \mathbb{R}(X)$, we have $$ |f(a) - f(b)| \leq C \cdot |a - b| $$ While $|a|:=\max \{a, -a\}$.

Must every contraction $f$ on $\mathbb{R}(X)$ possess a unique fixed point?

Striving for a counterexample, I'm particularly worried about the case $C = 1 - \frac{1}{X}$. However, I haven't been able to come up with anything specific.

Tina
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Smiley1000
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1 Answers1

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The map $$f:a\mapsto 1/x-a''/x $$ is a counterexample. Note that the derivative is a well-defined map on $\mathbb R(X)$.

The map $f$ is a contraction with arbitrarily small $C\in(0,1)\subset\mathbb R$ as in your definition, since differentiating and dividing by $x$ both make rational functions grow strictly slower at infinity. Moreover, a fixed point of $f$ must be a solution of the ODE $$y''=1-xy,$$ which does not seem to have rational solutions.

Lorenzo Pompili
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  • I don't see how $C$ can be "arbitrarily small". For example, consider $C = \frac{1}{X^3}$, $a = 0$ and $b = \frac{1}{2} X^2$. Then we have $f(a) = \frac{1}{X}$ and $f(b) = 0$. Hence we have $|a - b| = \frac{1}{2} X^2$, $C \cdot |a - b| = \frac{1}{2 X}$ and $|f(a) - f(b)| = \frac{1}{X}$. This shows that $|f(a) - f(b)| > C \cdot |a - b|$, contradicting the claim. – Smiley1000 Oct 04 '24 at 14:58
  • I believe that something like $C = \frac{1}{X^2}$ should work, though. – Smiley1000 Oct 04 '24 at 15:20
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    Right, sorry, I overlooked that $C$ is not necessarily a real number. I meant that it is a contraction for arbitrary $C\in (0,1)$. In any case, my counterexample works, and I agree that it is a contraction with $C=1/X^2$. And it looks likely that it is possible to generalize what I wrote to obtain a counterexample for any $C>0$. – Lorenzo Pompili Oct 04 '24 at 16:35