Bounded spherical derivative implies finite order

2011-10-27T13:48:56Z

Hi,

Let $f$ be an entire function. The spherical derivative $\rho(f)$ is defined by $$\rho(f)(z):= \frac{|f'(z)|}{1+|f(z)|^2}.$$

A result from Clunie and Hayman states that if $\rho(f)$ is bounded, then $f$ is of exponential type. The proof uses the machinery of Nevanlinna's theory of value distribution.

My question is the following :

Is there an elementary proof that if $\rho(f)$ is bounded, then $f$ is of finite order?

(Note that this is a weaker result, since I'm only asking for finite order here). Finite order means that there exists constants $K$ and $\alpha$ such that $$|f(z)| \leq Ke^{|z|^\alpha}$$ for all $z$.

Motivation : I'm interested in this because it would lead to a quick proof of Picard's little theorem. Indeed, if there exists a non-constant entire function which omits $0$ and $1$, then it is possible to obtain (using normal families techniques) a non-constant entire function $f$ which omits $0$ and $1$ and that has bounded spherical derivative. Write $f = e^g$ for some entire function $g$. Since $f$ is of finite order, $g$ is a polynomial. But $f$ does not take the value $1$, so $g$ must be constant, a contradiction.

Any reference is welcome. Thank you, Malik.

EDIT I asked the question on math.stackexchange.

Answer by Alexandre Eremenko for Bounded spherical derivative implies finite order

2012-08-03T19:00:16Z

Yes, of course. A bound on spherical derivative immediately gives T(r)=O(r^2) where T is the Nevanlinna characteristic. And that finite order of T implies finite order of f is proved in the beginning pages of any book on Nevanlinna theory.

BTW. Your idea on a simple proof of Picard theorem is not new. A proof based on this idea was published by Zalcman many years ago.

Bounded spherical derivative implies finite order - MathOverflow

Bounded spherical derivative implies finite order

Answer by Alexandre Eremenko for Bounded spherical derivative implies finite order