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Let $R(m,n)$ be defined on all the integers such that $R(m,0)=m, R(0,n)=n, R(m,n)=R(n,m)$ and $R(R(m,n),p)=R(m,R(n,p))$ for all integers $p$. Thus $R$ satisfies the Abel associativity equation. Let $f$ be defined for all integers, and $f(0)=0$. Then the Forsyth/Abel functional equation is

$$f(m)+f(n)=f(R(m,n))$$

The problem is to prove that $f(n)=n\theta$ where $\theta$ is a constant. It is the fundamental step in Forsyth's proof that a meromorphic function which satisfies an algebraic addition theorem and is not rational is necessarily periodic. (Theory of Functions, Chapter 18). A special case was treated in question 288554.

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2 Answers 2

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This is wrong. Let $f$ be any permutation of the integers such that $f(0)=0$, and $R(m,n) = f^{-1}(f(m)+f(n))$.

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  • $\begingroup$ If one writes $f(n)=ng(n)$ (because $f(0)=0$) the equation becomes $mg(m)+ng(n)=R(m,n)g(R(m,n))$. If $f$ is a permutation of the integers then $ng(n)$ also is which suggests that $g$ is a unit, i.e., the $\theta$ in the proposed solution $\endgroup$
    – mark
    Commented Feb 13, 2018 at 19:09
  • $\begingroup$ @mark your $g(n)$ is not obliged to be integer $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 13, 2018 at 21:18
  • $\begingroup$ It is if $ng(n)$ is a permutation of the integers $\endgroup$
    – mark
    Commented Feb 13, 2018 at 21:19
  • $\begingroup$ @mark no it is not. For example when $g(1)=2$, $g(2)=1/2$ and $g(n)=1$ for all other $n$ you get the permutation $1\mapsto2$, $2\mapsto1$ and $n\mapsto n$ for all other $n$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 13, 2018 at 21:21
  • $\begingroup$ Yes, you are right.... $\endgroup$
    – mark
    Commented Feb 13, 2018 at 21:38
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If you are interested in the theorem itself (not in the Forsyth's proof of it), a very clear proof is given in many places, for example in I. N. Akhiezer, Elements of the theory of elliptic functions, AMS 1990. I mean the Weierstrass theorem about meromorphic functions satisfying an algebraic addition theorem.

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  • $\begingroup$ Weierstrass' proof, as presented by Phragmen and (later) Osgood, is very beautiful...but Forsyth's attempt is interesting, although apparently wrong...I thought it was worth resurrecting because of its novelty...but novelty, when wrong, is not enough.... $\endgroup$
    – mark
    Commented Feb 14, 2018 at 0:52
  • $\begingroup$ Weierstrass' proof does not apply to the function $e^(e^u)$ which is meromorphic and periodic but Forsyth's "proof" does...too bad it is invalid... $\endgroup$
    – mark
    Commented Feb 14, 2018 at 0:54
  • $\begingroup$ My arXiv paper arXiv:1212.6471v3 contains a detailed presentation of the history and proofs of Weierstrass' theorem and analytic functions of one and several variables which admit an algebraic addition theorem...however I will have to update the section on Forsyth's attempt. $\endgroup$
    – mark
    Commented Feb 14, 2018 at 4:16
  • $\begingroup$ @mark: First theorem in your paper is incorrect: not all periodic meromorphic functions satisfy algebraic addition theorems. An example is $\exp(e^z)$. $\endgroup$
    – Alexandre Eremenko
    Commented Feb 15, 2018 at 20:15

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