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.


1 Answer 1


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.

  • $\begingroup$ Dear Prof. Eremenko, first of all, welcome on mathoverflow and thank you for answering my question. On the other hand, it was already answered on math.stackexchange on November 2, 2011 (see the link given in the question). The answer there is exactly what you suggest. My apologies though, I should have edited the question here to add a link to the answer on math.stackexchange. Also, you are absolutely right about the idea being not new : I'm aware of Zalcman's paper now (I read it a few weeks after asking the question, some time in november if I remember correctly) $\endgroup$ Aug 3, 2012 at 20:56

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