# Bounded spherical derivative implies finite order

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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There are elementary proofs of Picard's theorem along these lines. I will check my references to see whether the clunie-hayman thm gets used there. I'll also note though thatgetting finite order from bounded spherical derivative does not require that much Nevanlinna theory. You just need to know how the Ahlfors-Shimizu characteristic is related to the order. – Lasse Rempe-Gillen Nov 2 '11 at 10:22
PS could you add a link to the MO question? – Lasse Rempe-Gillen Nov 2 '11 at 10:23
Apologies for the preceding comment - the link did not show up on my phone for some reason. However, could you also add a link back here to the MO question? – Lasse Rempe-Gillen Nov 2 '11 at 10:53
For an elementary proof of Picard's theorem using normal family arguments, see Bergweiler's article on "Bloch's principle". The proof given there is due to Ros. – Lasse Rempe-Gillen Nov 2 '11 at 13:20
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. – Kalim Nov 3 '11 at 13:15

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 $$A(r) := \int_{|z|<r} (f^{\#}(z))^2dxdy,$$ 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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Nice proof! I already had a proof along the same lines, but yours is simpler. Indeed, not much of Nevanlinna's theory is needed. However, your proof shows that $f$ is of order at most $2$. I was wondering if there was a simpler proof that gives finite order, but perhaps without an explicit order in the end (since in the application to Picard's theorem, the order is not important). I will accept your answer though! – Kalim Nov 3 '11 at 13:23
By the way, thank you for answering my question. Are you a complex analyst? – Kalim Nov 3 '11 at 13:24
I do work broadly in the field of complex analysis (though I'm far from being an expert on Nevanlinna theory). – Lasse Rempe-Gillen Nov 3 '11 at 16:51