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If you consider $f=\frac{P}{Q}$ the quotient of two polynomial function then $\frac{f'}{1+\vert f\vert^2}$ decrease like $\frac{1}{z}$. My question is, is the converse true? is an meromorphic function(define on the whole plane) which satisfies
$$\frac{f'}{1+\vert f\vert^2}=O\left(\frac{1}{z}\right)$$ as $z$ goes to infinity, is the quotient of two polynomial function?

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Welcome to math.stackexchange! My first instinct would be no - it certainly doesn't "feel" like it should be true. Being a "polynomial" is a different kind of property than having a certain type of growth condition - You can make all kinds of weird functions which grow similarly to a given polynomial just be changing it in some little way, such as adding $ + \sin x $ to the end. So perhaps we should look for a counterexample. But before we proceed any further, you should specify: Holomorphic where ? And the asymptotics are as z goes where? – Ragib Zaman Oct 18 '11 at 15:03
@Ragib, being polynomial is not quite as special a condition in the complex case as it is for reals. In particular, if a function is meromorphic on all of $\mathbb C$ and at infinity, then it has to be a quotient of polynomials; and having a pole (or a removable singularity) at infinity certainly feels like a kind of growth condition to me. – Henning Makholm Oct 18 '11 at 16:02
If you cross-post a question to MathOverflow, it would be good if you edited your question to include a link to there as well, and vice versa. I notice that someone gave a reference to the Lehto-Virtanen paper there the day before I posted my answer below. I do enjoy thinking about questions, but it is still somewhat annoying to spend time on one and then realize that the OP is no longer interested because they have already received a reference elsewhere ... – Lasse Rempe-Gillen Oct 26 '11 at 13:34
sorry for this, but i start to ask on mathstackexchange. Since there is no answer, i ask to mathoverflow...i will close the mathoverflow post. – Paul Nov 1 '11 at 9:33
Hi Paul, that's quite a common and sensible thing to do, it would just be good to edit your answer to include a link to the mathoverflow question, and vice versa. That way people can just check on the other thread before posting an answer that might be duplicated elsewhere. – Lasse Rempe-Gillen Nov 1 '11 at 16:42

EDIT. I have edited point 4) below to provide more details, and corrected the claim in Lehto's theorem.

This answer is positive if $f$ is entire, but negative for general meromorphic functions.

To simplify notation, let $f^{\#}(z)$ denote the spherical derivative, i.e. $f^{\#}(z) = |f'(z)|/(1+|f(z)|^2)$.

A little bit of further research on the web unsurprisingly showed that the result is not new. Indeed, this problem was introduced by Lehto and Virtanen. They prove most of what is needed for your question, but I don't have access to the paper. However, Lehto proved a stronger result in his paper "The spherical derivative of meromorphic functions in the neighbourhood of an isolated singularity", Comment. Math. Helv. 33, Number 1:

Theorem 1. If $f$ is meromorphic in a punctured neighborhood of infinity, and suppose that $f$ has an essential singularity at infinity. Then $$\limsup |z| f^{\#}(z) \geq \frac{1}{2}.$$ This inequality is best possible: there are meromorphic functions on the punctured plane for which equality holds (and these can be described explicitly as certain infinite products).

On the other hand, if $f$ is analytic (i.e. has no poles), then $$\limsup |z| f^{\#}(z) = \infty.$$

It may still be useful to give my account of (some weaker, but sufficient) versions of these results, particularly since the Lehto-Virtanen paper is not available online, and might or might not be in your local university library. Also, my account explicitly shows that you can get a function that is meromorphic in the whole plane which is a counterexample to your question, while, as you note, Lehto's example is not meromorphic at zero.

1) To prove the final statement in the theorem, we use the theory of normal families. Consider the functions $$f_{\lambda}(z) := f(\lambda z); \quad |\lambda|\geq 1.$$ If $\limsup |z| f^{\#}(z) < \infty$, then it follows that the spherical derivative of $f_{\lambda}$ remains locally bounded near every point $z$ with $|z|>0$ as $\lambda\to\infty$. Hence these maps form a normal family by Marty's theorem. Suppose that $f$ is analytic, and let $z_n\to\infty$ be a convergent sequence; we may assume without loss of generality that the limit $z_0$ is finite (as otherwise $f$ has a pole at infinity and there is nothing to prove). By passing to a subsequence, we can assume that the maps $f_{z_n}$ converge locally uniformly to a limit function $g$, which is defined on the punctured complex plane. Since $f_{z_n}(1)=f(z_n)\to z_0$, we have $f(1)=z_0$; in particular, $g$ is not constantly equal to infinity, and hence $g$ is analytic. It follows that the maximum modulus of $f$ on the circle $|z|=|z_n|$ remains bounded as $n\to\infty$. By the maximum modulus principle, $f$ is bounded near infinity, and hence $f$ has a removable singularity at infinity. This completes the proof.

2) A similar argument should allow us to prove that $$\limsup |z| f^{\#}(z) >0$$ when $f$ is meromorphic and the singularity is not removable. (In this case, all limit functions have to be constant, which should lead to a contradiction. Certainly one can get a contradiction using the essential singularity version of the Ahlfors five islands theorem, but that's quite a sledgehammer. Lehto's proof, mentioned below, is much easier.) Of course once this is proved, we also see that there is a universal constant $c>0$ such that $$\limsup |z| f^{\#}(z) >c$$ (again using normal families).

3) Finally, to construct a function $f$ such that $$\limsup |z| f^{\#}(z) <\infty,$$ just take a meromorphic function $f:\mathbb{C}\setminus\{0\}\to\overline{\mathbb{C}}$ such that $f(\lambda z)=f(z)$ for some complex $\lambda$, $|\lambda|>1$, and all $z$. In other words, $f$ is of the form $$f(e^z) = g(z),$$ where $g$ is an elliptic function with periods $2\pi i$ and $\log(\lambda)$, for example $g$ could be a Weierstrass $\wp$-function.

4) The above example is not meromorphic on the entire complex plane, but one could turn it into a function that is by using quasiconformal mappings.

To do so, construct the function $f$ using a Weierstrass function on a rectangular lattice (so $\lambda\in (1,\infty)$). Let $C$ be a circle centered at the origin and not passing through critical points, and let $\gamma$ be the image of $C$ under $f$. The properties of the $\wp$-function imply that $f$ is conformal and one-to-one in a neighborhood of $C$, so $\gamma$ is an analytic Jordan curve. Let $U$ be the complementary domain of $\gamma$ on the Riemann sphere for which $\gamma$ is the boundary of $D$ described in positive orientation. (Choosing $C$ appropriately, we could make sure that $U$ is the bounded complementary component of $\gamma$, but this doesn't matter for our purposes.)

Let $\phi:U\to \mathbb{D}$ be a conformal isomorphism, where $\mathbb{D}$ is the unit disk. Such an isomorphism exists by the Riemann mapping theorem, and extends analytically to a neighborhood of $U$ because $\gamma$ is an analytic curve (Schwarz reflection). Let $r$ be the radius of the circle $C$; then the map $$ \psi:S^1\to S^1; \quad z\mapsto \phi(f(rz))$$ is an orientation-preserving analytic circle diffeomorphism, and as such extends to a quasiconformal homeomorphism $\psi:\mathbb{D}\to\mathbb{D}$. (You can just define such an extension by hand, using $\psi$ as a function on the argument and fixing the modulus, or use something like the Douady-Earle extension.)

Now define $$\tilde{h}:\mathbb{C}\to \overline{\mathbb{C}}; \quad z\mapsto \begin{cases} f(z) & |z|\geq r \\ \phi^{-1}\left(\psi\left(\frac{z}{r}\right)\right) & |z|<r.\end{cases}$$ This is a quasiregular map. Using the measurable Riemann mapping theorem, we can write $\tilde{h}= h\circ\theta$, where $\theta$ is meromorphic and $\phi$ is a quasiconformal homeomorphism that is conformal near infinity. It follows that $h$ has the desired property.

This leaves open the question of whether Lehto's sharp examples can be realized (either exactly or asymptotically) by a function meromorphic in the complex plane. I suspect that this can be done by a similar method as above, but I have not thought about this.

5) Lehto's proof of the first statement in Theorem 1 above is actually very simple and elegant (though somewhat miraculous): Consider the function $$F(z) := f(z)\cdot \overline{f}(\overline{z}e^{2\pi i \theta}),$$ where $\theta$ is chosen so that $F$ also has an essential singularity. (There is at most one $\theta$ where this is not the case.) Then there exist points tending to infinity for which $F(z)$ is arbitrarily close to $1$, which means that the points $w_1 := f(z)$ and $w_2 := f(\overline{z} e^{2\pi i \theta})$ have the property that $w_1\cdot\overline{w_2}$ is close to $1$. This means that the two points are essentially on opposite points of the Riemann sphere. So if we map a circle $C$ centered at zero and passing through $z$ under $f$, the image must have length at least $\pi$ (up to a small error, tending to zero). Of course $C$ itself has length $2\pi |z|$, so there must be a point where the spherical derivative is at least $|z|/2$ (again up to an error tending to zero).

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Thank you for this detail answer, but i am not conveince by your extension 4). For me the Letho product are defined only on the disc (per example) and the extension by symmetry create an other essential singularity. And your argument with the quasi conformal maps looks strange when applied to exp(1/z), it seems that you can extend exp(1/z) in the interior of the disc without any essential singularity? – Paul Nov 1 '11 at 9:27
You are right that Lehto's functions are analytic only for $z\neq 0$; I missed that. However, I think the quasiconformal argument I give is ok. I am not sure what you mean about $\exp(1/z)$. Near infinity, this function just looks like $z\mapsto 1+1/z$; of course you can then use a similar glueing, and you will just get a M\"obius transformation. I can add some more details if you would like, but it will require some familiarity with QC techniques. – Lasse Rempe-Gillen Nov 1 '11 at 11:51
Yes, it will be great if I could understand this argument which allow to extend a meromorphic function defined in a neighbourhood of infinity to a meromorphic function on $\mathbb{C}$ without changing its property at infinity. have you any references? – Paul Nov 1 '11 at 12:03
I just edited my answer and hope that provides enough detail. – Lasse Rempe-Gillen Nov 1 '11 at 12:39
Very nice and detailed answer! I'm the one who posted the link to the Lehto-Virtanen paper on MathOverflow, but I couldn't find a version of it. Consequently, your detailed explanation of Lehto's results is very interesting. – Kalim Nov 1 '11 at 12:58

These are some quick thoughts, but only give a partial answer. Apologies if I am missing something!

If I understand correctly, you are assuming that f is a meromorphic function in the complex plane.

(Or perhaps you do mean entire holomorphic functions, as you state, in which case you should ask whether $f$ is a polynomial.)

Note that the quantity you are writing down is the spherical derivative, i.e. the derivative measured with respect to the Euclidean metric in the domain and the spherical metric in the range.

Let us assume for a moment that $f$ satisfies the stronger property that $$ \frac{|f'(z)|}{1+|f(z)|^2} = o\left(\frac{1}{z}\right).$$

Consider the image of any large circle around the origin under $f$. Then its image under $f$ has spherical length tending to zero. You should be able to show that this means the function extends continuously to infinity, and hence is rational.

With your stronger condition, the image of the curve described stays bounded, but no longer has to tend to zero. This seems like an odd property for a transcendental meromorphic function, but I do not see a simple argument that the function would have to be rational in this case (or a simple counterexample). A counterexample would have to have very small growth in the sense of Nevanlinna theory, so perhaps Nevanlinna theory could help you to investigate this further. I expect that experts in the field would know the answer, so perhaps there is one here to comment?

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Yes, i agree with your feeling with your condition. But my hypothesis is essential for my question. Thank you, for your answer – Paul Oct 18 '11 at 16:36
Could you say something more about how the question arises, and in particular whether you are interested in the entire or the meromorphic case? For meromorphic functions, I believe there are counterexamples. I will post details later. – Lasse Rempe-Gillen Oct 19 '11 at 11:59
You look at $u=\omega \circ f$ as a parametrization of the sphere where $\omega$ is the inverse of the stereographic projection. Then, i want to prove that if $\nabla u$ decreases at leats like $1/z$ then $f$ is a quotient of two polynomial functions. – Paul Oct 19 '11 at 12:03
to mathstribble, have you any modern and comprehensive reference about quasiconformalmap in which i can find the theorem you use like the existence of quasimorphism. thanks in advance – Paul Nov 6 '11 at 22:07
The classical references on quasiconformal maps are the books by Ahlfors (recently reprinted by the AMS, with additional commentary and appendices by experts in the field) and by Lehto and Virtanen. These do not count as modern, but they are good references nonetheless. – Lasse Rempe-Gillen Nov 7 '11 at 2:51

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