Denote $e_A$ the identity map from $A$ to itself. Questions such like solving $f$ in the functional equation $f\circ f=e_\mathbb{R}$ or $f\circ f=e_{\mathbb{R}\setminus\{a_1,\ldots,a_n\}}$ with or without limitation of the continuousness of $f$ have already been discussed in this question. Given that there are some solutions like $f:x\mapsto1/x$ with domain $\mathbb R\setminus\{0\}$, it is natural to try to extend the domain to $\mathbb{R}\cup \{\infty\}$. To find the solutions with better properties, it is necessary to add a restriction of the solution. As it is hard to analysis of the nature of $f$ at infinity, the following question arises since $\mathbb{R}\cup\{\infty\}$ and $S^1$ are same to some extent.

Let $S^1\subset\mathbb R^2$ be the unit circle centered at the origin, can we find all continuous solutions to the equation $f\circ f=e_{S^1}$ with the restriction that there exists a real-analytic function $g$ s.t. $(\cos g(x),\sin g(x))=f((\cos x,\sin x))$ for all $x\in\mathbb R$?

It is not hard to see that there exists a constant integer $n$ such that $g(x)-2\pi nx$ have a period $2\pi$. Plus, because $f$ is a continuous bijection, $n$ has to be $\pm1$. So the question boils down to finding a function with the following conditions: $g\in C^\omega(\mathbb R)$, $g(x)=\pm x+\sum_{n=-\infty}^\infty c_ne^{inx}$ and a functional equation with respect to $g$. I'm still struggling with the functional equation.

  • 2
    $\begingroup$ It seems that in the end you view $S^1$ as subset of $\Bbb R^2$ and as $\Bbb R/(2\pi\Bbb Z)$ at the same time. Or do you want identify $\Bbb R^2$ with $\Bbb C$ and meant $f(e^{ix})$ instead of $f(x\bmod 2\pi)$? Or explicitly $f(\cos x,\sin x)$? Are you in fact rather looking for real-analytic $f$ that is $2\pi$-periodic and $f(f(x))=x+2k\pi$? $\endgroup$ – Hagen von Eitzen Jun 23 at 9:11
  • $\begingroup$ This is only a partial answer: since $id=f^2: S^1 \to S^1$, then applying the homology functor $id_*=(f_*)^2 : H_*(S^1) \to H_*(S^1)$. This implies that $f$ has degree $1$. Since the degree of maps is a complete invariant on the circle, this implies that $f$ is homotopic to the identity map. The problem then reduces to finding functions homotopic to the identity that satisfy the desired condition. $\endgroup$ – Lance Jun 23 at 10:52
  • 1
    $\begingroup$ @Hagen Sorry for the self-contradition and the unclear of my question. In fact I only want to find a "good" real-analytic function mapping $\Bbb R\cup\{\infty\}$ to itself and I tried to extend the definition of real-analytic to the infinity point. If it is still unclear, feel free to close this question as I have not yet find a good expression to it. $\endgroup$ – Kemono Chen Jun 23 at 10:57
  • $\begingroup$ Your question makes sense. I think it would be more clear if you required $f$ to be a smooth map of smooth manifolds. Btw, in response to my earlier comment, $f$ could also be homotopic to the antipodal map. $\endgroup$ – Lance Jun 23 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.