# Behavior of $|f'(z)|/(1+|f(z)|^2)$ as $|z| \rightarrow \infty$?

Let $$f(z)$$ be an entire holomorphic function in $$\mathbb{C}$$, and consider the real-valued function $$g_f(z)=\frac{|f'(z)|}{1+|f(z)|^2}.$$ If $$f(z)$$ is a polynomial, then it is easy to prove that $$\lim_{|z|\rightarrow \infty}g_f(z)=0$$.

When $$f$$ is transcendental, say $$f(z)=e^z$$, then $$g_f(z)=\frac{e^x}{1+e^{2x}}$$, which goes to zero when $$Re(z)$$ is going $$\infty$$, but remains a constant when $$z$$ is moving on any vertical line. So far I have not found any example such that the limit goes to zero when $$f$$ is transcendental.
My question is, can we rigorously prove that there is no transcendental entire function $$f$$ such that $$\lim_{|z|\rightarrow \infty}g_f(z)=0\; ?$$

• May 14, 2022 at 17:24
• @MartinR The linked post is great, but does it answer the question? May 14, 2022 at 20:31
• @MartinR: But what is not completely clear (to me) from the way this is quoted is if these functions are entire or just meromorphic on a neighborhood of $\infty$. May 14, 2022 at 20:47
• The results of Lehto and Virtantan say that $\limsup |z||f'|/(1+|f|^2)\geq 1/2$ for all meromorphic functions, and $=\infty$ for entire functions. The first result is best possible, while the second is not. See my ans. May 14, 2022 at 21:01
• @MartinR, the example constructed by Lehto and Virtanen is of the form $F(log z)$, where $F$ is a doubly-period function, with one period equal to $2\pi i$. Hence such function cannot be entire since doubly-period function cannot be entire unless a constant. May 15, 2022 at 8:19

This is not true. The optimal estimate from below for transcendental entire functions is $$\limsup_{z\to\infty}\frac{|z||f'(z)|}{\log|z|(1+|f(z)|^2)}=\infty,$$ and this is best possible,

J. Clunie and W. Hayman, The spherical derivative of integral and meromorphic functons, Comment Math. Helv., 40 (1966) 117-148.

More precisely, for every function $$\phi(r)\to+\infty$$, they constructed an example of transcendental entire function for which $$\frac{|z||f'(z)|}{\phi(|z|)\log|z|(1+|f(z)|^2)}$$ is bounded. The function in this example is of the form $$f(z)=\prod_{n=1}^\infty\left(1-\frac{z}{2^{k_n}}\right)^{k_n},$$ where $$k_n$$ is an increasing sequence of integers, which is choosen, depending on $$\phi$$.

• Thank you very much! May 15, 2022 at 7:37
• I'm wondering is it possible to have a very concrete example? May 15, 2022 at 7:39
• Take $k_n=n$ in the example that I wrote, to make it "concrete". May 15, 2022 at 13:55

A concrete example is given by $$f(z)=\cos{\sqrt{z}}$$. Then $$g_f(z) = \frac{|\sin\sqrt{z}|}{2|\sqrt{z}|(1+|\cos^2{\sqrt{z}}|)} .$$ Obviously, this is small for large $$|z|$$ if $$\sin{\sqrt{z}}$$ is not large, and if $$\sin\sqrt{z}$$ is large, then $$|\cos\sqrt{z}|$$ is of the same order of magnitude (since $$\sin^2 w+\cos^2 w=1$$), and again $$g_f$$ is small.

• $\sqrt{z}$ is discontinuous accross branch cut, and thus not entire. May 15, 2022 at 8:46
• @student: $\cos$ is an even function, and therefore $\cos\sqrt z$ is an entire function. If you prefer then you can define it via the power series $\sum_{n=0}^\infty (-1)^n z^n/(2n)!$ May 15, 2022 at 12:12
• @MartinR I see. Thank you very much! May 15, 2022 at 15:33
• Very nice! Did you just randomly guess this example, or was there some conceptual insight that led you to it? May 17, 2022 at 4:53
• @DanRomik: This may sound strange, but I don't remember very clearly how I got it (so certainly no conceptual insight). I tried for a while to prove the (false) statement, based on the fact that $f(z+w)$, $w\in\mathbb C$ is a normal family. May 17, 2022 at 17:15