Consider the equation
$$f(x)=f\left(\frac{1}{x}\right)$$
Where we want $f$ to be real-meromorphic.
Are all solutions $f$ of the form
$$f(x) = g\left(\frac{x}{1+x^2}\right)$$
Where $g$ is a continuous function ?
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3I don't think so; $f(x)=(\log|x|)^2$ ? Take an arbitrary even function and then replace the main variable $x$ with $\log|x|$. – Arash Dec 16 '15 at 12:35
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But $log |x|$ is not meromorphic near the real line. It has a singularity at $x= 0$. – mick Dec 16 '15 at 20:18
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Related : http://math.stackexchange.com/questions/1579723/solve-gx-frac1x-xy-frac1xy?noredirect=1#comment3216666_1579723 – mick Dec 17 '15 at 20:48
1 Answers
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Note that $h\colon \Bbb R\setminus\{0\}\to \Bbb R\setminus (-2,2)$, $x\mapsto x+\frac1x$ is continuous and almost injective, in the sense that $h(x)=h(y)$ implies $y=x$ or $y=\frac1y$. Also, $h$ is never $0$. As a consequence, $f$ can be written as $f(x)=g(\frac 1{h(x)})$ for some $g\colon [-\frac12,\frac12]\setminus\{0\}$.
The function $g$ is also continuous. This is because if we restrict $h$ to $\Bbb R\setminus(-1,1)$ it becomes bijective with continuous inverse.
However, note that $g$ need only be continuous as function defined on $[-\frac12,\frac12]\setminus\{0\}$. Specifically, it is not guaranteed that a continuous extension to $[-\frac12,\frac12]$ exists. This exists only if $\lim_{x\to 0}g(x)$ exists, i.e., if $\lim_{x\to+\infty}f(x)=\lim_{x\to-\infty}f(x)$. If you demand that $f(x)=f(\frac1x)$ should also hold for $x=0$ in the sense that $\lim_{|x|\to\infty} f(x)=f(0)$, then this is the case.
Hagen von Eitzen
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Im confused about your answer. What exactly did you prove ? Or are you just telling me the facts without proof ? You seem to split Up the cases but im not sure in what. Im not sure how limits relate. Notice that i do not consider $log |x|$ An answer ( see comments ). – mick Dec 16 '15 at 20:25
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Im not so smart sorry. But Maybe others will not get it Neither. Thanks for the efforts so far. – mick Dec 16 '15 at 20:28