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Alexandre Eremenko
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The other possibility is that $f(z)=e^z$. Explicit computation gives $$\frac{d}{dr}f^\#(re^{i\phi})=2e^{r\cos\phi}\cos\phi(1-e^{2r\cos\phi}),$$ which is $\leq 0$ everywhere.

On the other hand, the exponential is the only exception among entire functions.

Indeed, if $f^\#$ is decreasing on each radius, then $f^\#$ has no zeros, this implies that $f'$ has no zeros. On the other hand, $f^\#(z)\leq f^\#(0)$ must be bounded, and a theorem of Clunie and Hayman implies that $f$ has at most exponential type (that is at most order 1, normal type). Then $f'$ has at most exponential type, and since it has no zeros, it must be $e^{az+b}$ by the Hadamard factorization theorem.

Refs. The original paper of Clunie and Hayman is

MR0192055 Clunie, J.; Hayman, W. K. The spherical derivative of integral and meromorphic functions. Comment. Math. Helv. 40 (1966), 117–148.

Their proof was much simplified (and generalized) in

MR2869124 Barrett, Matthew; Eremenko, Alexandre Generalization of a theorem of Clunie and Hayman. Proc. Amer. Math. Soc. 140 (2012), no. 4, 1397–1402.

Edit. One can describe all meromorphic functions with your property: there is probably nothing except $L(e^{az})$, where $L$ is linear-fractional transformation. I can add this if desired.

Alexandre Eremenko
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