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 $$ \lim_{|z|\rightarrow \infty}g_f(z)=0\; ? $$

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    $\begingroup$ See mathoverflow.net/a/78589/116247. $\endgroup$
    – Martin R
    May 14, 2022 at 17:24
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    $\begingroup$ @MartinR The linked post is great, but does it answer the question? $\endgroup$
    – GH from MO
    May 14, 2022 at 20:31
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    $\begingroup$ @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 $\infty$. $\endgroup$ May 14, 2022 at 20:47
  • $\begingroup$ 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. $\endgroup$ May 14, 2022 at 21:01
  • $\begingroup$ @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. $\endgroup$
    – student
    May 15, 2022 at 8:19

2 Answers 2


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 $\phi$.

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

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    $\begingroup$ $\sqrt{z}$ is discontinuous accross branch cut, and thus not entire. $\endgroup$
    – student
    May 15, 2022 at 8:46
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    $\begingroup$ @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 $\sum_{n=0}^\infty (-1)^n z^n/(2n)!$ $\endgroup$
    – Martin R
    May 15, 2022 at 12:12
  • $\begingroup$ @MartinR I see. Thank you very much! $\endgroup$
    – student
    May 15, 2022 at 15:33
  • $\begingroup$ Very nice! Did you just randomly guess this example, or was there some conceptual insight that led you to it? $\endgroup$
    – Dan Romik
    May 17, 2022 at 4:53
  • $\begingroup$ @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. $\endgroup$ May 17, 2022 at 17:15

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