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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 : 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, Malik

NOTE: This is a duplicate of a question on MathOverflow. I'm posting it here too because I did not get any answer or comment.

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    I added a link to this page in the question on MathOverflow, as you asked. I already knew the nice proof in Bergweiler's article on Bloch's principle, but thank you for the reference.2011-11-03

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I will sketch a proof. This is obtained by isolating the relevant parts of Nevanlinna theory, and I am following Nevanlinna's treatment of the Ahlfors-Shimizu characteristic (Chapter VI, Section 3 in his book).

Let us assume without loss of generality that $f(0)=0$. Let $f^{\#}(z)$ denote the spherical derivative, as above. Set $A(r) := \int_{|z| so $A(r)$ is the spherical area of the image of the disk of radius $r$. If $f^{\#}(z)$ is bounded, then $A(r) = O(r^2).$

An application of Green's theorem gives $4 A(r) = r\cdot \frac{d}{dr} \int_{0}^{2\pi} \log(1+|f(re^{i\theta})|^2)d\theta.$ (I will leave it to you to check the details.)

We consider the expression $m(r) := \frac{1}{2\pi} \int_{0}^{2\pi}\log(1+f(re^{i\theta})|^2)d\theta.$ By the above, we have $\frac{d}{dr} m(r) = O(r), $ and hence $m(r) = O(r^2)$ as $r\to\infty$.

Since $\log(1+|f|^2)>2\log|f|$, we thus have $\int_0^{2\pi} \log|f(re^{i\theta})|d\theta \leq \operatorname{const}\cdot r^2$ for all sufficiently large $r$. By the Poisson integral formula for the harmonic function $\log|f|$, this implies that, for $|z|=r/2$, $\log|f(z)| \leq \operatorname{const}\cdot r^2=\operatorname{const}\cdot |z|^2.$ So $f$ has finite order, as claimed.

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    I do work broadly in the field of complex analysis (though I'm far from being an expert on Nevanlinna theory).2011-11-03