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The spherical derivative of a meromorphic function is defined as $$f^\#(z):=\frac{|2f'(z)|}{1+|f(z)|^2}.$$ The motivation is that given a piecewise smooth curve $\gamma$ in the complex plane, the length of $f\circ \gamma$ in the Riemann sphere is given by $\int_{\gamma} f^\#(z)d|z|$. When $f=az$, where $a$ is a nonzero complex number, then $f^\#(z)$ is radially symmetric and is strictly decreasing along the radial direction. Also, it is well known that $f_1^\#=f_2^\#$ is equivalent to the existence of complex numbers $p, q$ with $|p|^2+|q|^2=1$ such that $f_1=\frac{pf_2-\bar{q}}{qf_2+\bar{p}}$.


I did know from literature (see for example Lehto's 1961 paper) that the behavior of $f^\#$ near an essential singularity of $f$ can reflect certain value distributional properties of $f$ near the singularity, but I'm trying to understand how the property of $f^\#$ can determine certain rigidity property of $f$.

For example, here is a standard question: if generally $f^\#(z)$ is strictly decreasing along radial directions, is it ture that $f$ must be a Mobius transform? Are there any other possibilities?

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  • $\begingroup$ Why is this a "standard question"? Does this condition arise naturally anywhere? $\endgroup$ Jun 13, 2022 at 12:22

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The other possibility is that $f(z)=e^z$. Explicit computation gives $$\frac{d}{dr}f^\#(re^{i\phi})=2e^{r\cos\phi}\cos\phi(1-e^{2r\cos\phi}),$$ which is $\leq 0$ everywhere.

On the other hand, the exponential is the only exception among entire functions.

Indeed, if $f^\#$ is decreasing on each radius, then $f^\#$ has no zeros, this implies that $f'$ has no zeros. On the other hand, $f^\#(z)\leq f^\#(0)$ must be bounded, and a theorem of Clunie and Hayman implies that $f$ has at most exponential type (that is at most order 1, normal type). Then $f'$ has at most exponential type, and since it has no zeros, it must be $e^{az+b}$ by the Hadamard factorization theorem.

Refs. The original paper of Clunie and Hayman is

MR0192055 Clunie, J.; Hayman, W. K. The spherical derivative of integral and meromorphic functions. Comment. Math. Helv. 40 (1966), 117–148.

Their proof was much simplified (and generalized) in

MR2869124 Barrett, Matthew; Eremenko, Alexandre Generalization of a theorem of Clunie and Hayman. Proc. Amer. Math. Soc. 140 (2012), no. 4, 1397–1402.

Edit. One can describe all meromorphic functions with your property: there is probably nothing except $L(e^{az})$, where $L$ is linear-fractional transformation.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions).

Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f''+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.

See also A. Eremenko, A Toda lattice in dimension 2 and Nevanlinna theory, J. Math. Phys., Anal. Geom., 31 (2007) 39-46 for a generalization of this theorem of Nevanlinna.

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  • $\begingroup$ Professor Eremenko, Thank you very much for your contribution to several questions of mine. I'm a beginner in complex function theory. If I'm understanding correctly, you essentially proved that if $f$ is an entire function, $f'$ has no zeros and $f^\#$ is bounded ,then $f=ae^{bz}+c$, right? This is a quite strong result, but I do have a further questions: how to extend this type of results to meromorphic functions? (Clunie and Hayman's paper treats entire functions, but meromorphic function can have a sequence of poles converging to $\infty$. How to overcome this difficulty?) $\endgroup$
    – student
    Jun 13, 2022 at 15:19
  • $\begingroup$ The motivation of my post is from my way of thinking. I'm a pde guy, and often times I know a solution A exists and satisfies some global or local properties. Then I often would like to ask if a solution B satisfies some partial properties, is it true that it must be A? This motivates my question. I'd like to ask this types of questions to see if I really understand certain topics. $\endgroup$
    – student
    Jun 13, 2022 at 15:21
  • $\begingroup$ Professor Eremenko: I actually graduated form Purdue University, but by that time I was not into complex analysis. It was a great pity that I did not have much interaction with you during those years at West Lafayette. Only recently when I became a teacher, I fell in love with complex function theory. I'm deeply grateful to all your contributions to mathoverflow and having helped so many people including me. $\endgroup$
    – student
    Jun 13, 2022 at 15:35
  • $\begingroup$ @student: I will be glad to help further, for example by e-mail. $\endgroup$ Jun 13, 2022 at 19:58
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    $\begingroup$ @student: as a PDE guy you may appreciate the fact that the general solution of the PDE $\Delta u+e^{2u}=0$ in the plane is $u(z)=\log f^\#$, where $f$ is a meromorphic function, for which $f(z)\neq 0$ the very useful fact rarely used by PDE people:-) $\endgroup$ Jun 13, 2022 at 20:01

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