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If you consider $f=\frac{P}{Q}$ the quotient of two polynomial function (i.e. $P,Q\in \mathbb{C} [z]$) then $\frac{f'}{1+|f|^2}$ decrease like $\frac{1}{z}$. My question is, is the converse true? is an meromorphic function(define on the whole plane) which satisfies

$$\frac{f'}{1+|f|^2}=O\left(\frac{1}{z}\right)$$

as $z$ goes to infinity, is the quotient of two polynomial function?

Of course considering this quotient come from the metric of the sphere, and my question could be is any parametrization of the sphere with such a decreasing is of finite type?

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  • $\begingroup$ I don’t understand the question. If $f=P$ is a degree $d$ polynomial, then $f'/(1+|f|^2)$ is of order $z^{-(d+1)}$ as $z\to\infty$, so it decreases much faster than $1/z$. $\endgroup$
    – Emil Jeřábek
    Commented Oct 19, 2011 at 10:18
  • $\begingroup$ Yes, in fact i mean at least as $\frac{1}{z}$. $\endgroup$
    – Paul
    Commented Oct 19, 2011 at 10:50
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    $\begingroup$ I believe the question (including your question below) is completely answered in the thread of the identical question you asked on stackexchange. $\endgroup$
    – Lasse Rempe
    Commented Nov 1, 2011 at 1:02

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The answer seems to be no.The quantity $$\rho(f(z)):= \frac{|f'(z)|}{1+|f(z)|^2}$$ is called the spherical derivative of $f$. Since you're interested in the behaviour of $z\rho(f(z))$ near $\infty$, then you should really take a look at Lehto and Virtanen's article :

MR0087747 (19,404a) Lehto, Olli; Virtanen, K. I. On the behaviour of meromorphic functions in the neighbourhood of an isolated singularity. Ann. Acad. Sci. Fenn. Ser. A. I. 1957 (1957), no. 240, 9 pp. (Reviewer: A. J. Lohwater), 30.0X

From the abstract :

It is proved first that if $f(z)$ is single-valued and meromorphic in a neighborhood of the isolated essential singularity $z=∞$, then there exists a universal constant $k>0$ such that $$\limsup_{z→∞}|z|\rho(f(z))≥k$$ for all such $f(z)$, while, for arbitrary $\epsilon>0$, there exist functions for which

$$\limsup_{z \rightarrow \infty} |z|\rho(f(z)) < k+\epsilon.$$

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  • $\begingroup$ I will look to the paper, it looks to give my answer. Thank you $\endgroup$
    – Paul
    Commented Oct 19, 2011 at 16:32
  • $\begingroup$ In fact, it produces a counterexample to my question. But the construction is local around $\infty$. I don't understand if it is possible to get such a function, i.e a Julia exceptional function, without any other essential singularity. That is to say a meromorphic function on $\mathbb{C}$ with only one essential singularity at infinity satisfying $$ \frac{f'}{1+\vert f\vert^2}=O\left( \frac{1}{\vert z \vert}\right).$$ $\endgroup$
    – Paul
    Commented Oct 31, 2011 at 8:30
  • $\begingroup$ Or here: mathoverflow.net/questions/422531/… $\endgroup$
    – Alexandre Eremenko
    Commented May 14, 2022 at 21:12
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Some one give an almost complete answer on mathstackechange,

https://math.stackexchange.com/questions/73651/holomorphic-function-with-special-decreasing-property

the discussion should be continued there. thanks all.

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