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](https://math.stackexchange.com/questions/77886/bounded-spherical-derivative-implies-finite-order/78182#78182)  on math.stackexchange.



Bounded spherical derivative implies finite order
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.


