Abstract
Given an involution $f$ on $(0,\infty ),$ we prove that the set $\mathcal {C} (f) : = $ {$\lambda > 0 : \lambda f $ is an an involution} is a closed \multiplicative subgroup of $(0, \infty)$ and therefore $\mathcal{C}(F)$ is $\{1\}$, $% \intervalo$ or $\lambda ^{\mathbb{Z}}=\{\lambda ^{n}:n\in \mathbb{Z}\}$ for some $\lambda >0$, $\lambda \neq 1$. Moreover, we provide examples of involutions possessing each one of the above types as the set $\mathcal {C} (f) $ and prove that the unique involutions $f$ such that $\mathcal {C} (f) = (0,\infty)$ are $f(x)=\frac{c}{x},$ $c>0$.
Citation
A. Linero Bas. J. S. Cánovas. G. Soler López. "On Closed Subgroups Associated with Involutions." Real Anal. Exchange 33 (2) 395 - 404, 2007/2008.
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