holomorphic function with special decreasing property - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:52:21Z http://mathoverflow.net/feeds/question/78560 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78560/holomorphic-function-with-special-decreasing-property holomorphic function with special decreasing property Paul 2011-10-19T10:10:21Z 2011-11-01T09:37:02Z <p>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</p> <p>$$\frac{f'}{1+|f|^2}=O\left(\frac{1}{z}\right)$$</p> <p>as $z$ goes to infinity, is the quotient of two polynomial function?</p> <p>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?</p> http://mathoverflow.net/questions/78560/holomorphic-function-with-special-decreasing-property/78589#78589 Answer by Malik Younsi for holomorphic function with special decreasing property Malik Younsi 2011-10-19T15:56:14Z 2011-10-19T15:56:14Z <p>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 :</p> <p>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</p> <p>From the abstract :</p> <p>It is proved first that if $f(z)$ is single-valued and meromorphic in a neighborhood of the isolated essential singularity $z=ā$, then there exists a universal constant $k>0$ such that $$\limsup_{zāā}|z|\rho(f(z))ā„k$$ for all such $f(z)$, while, for arbitrary $\epsilon>0$, there exist functions for which </p> <p>$$\limsup_{z \rightarrow \infty} |z|\rho(f(z)) &lt; k+\epsilon.$$</p> http://mathoverflow.net/questions/78560/holomorphic-function-with-special-decreasing-property/79599#79599 Answer by Paul for holomorphic function with special decreasing property Paul 2011-10-31T08:30:04Z 2011-10-31T08:30:04Z <p>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 <strong>a Julia exceptional function</strong>, without any other essential singularity. That is to say a meromorphic function on $\mathbb{C}$ with only one essential singularity at infinity satisfying $$\frac{f'}{1+\vert f\vert^2}=O\left( \frac{1}{\vert z \vert}\right).$$</p> http://mathoverflow.net/questions/78560/holomorphic-function-with-special-decreasing-property/79692#79692 Answer by Paul for holomorphic function with special decreasing property Paul 2011-11-01T09:37:02Z 2011-11-01T09:37:02Z <p>Some one give an almost complete answer on mathstackechange, </p> <p><a href="http://math.stackexchange.com/questions/73651/holomorphic-function-with-special-decreasing-property" rel="nofollow">http://math.stackexchange.com/questions/73651/holomorphic-function-with-special-decreasing-property</a></p> <p>the discussion should be continued there. thanks all.</p>