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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.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions).

Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f''+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.

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.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions).

Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f''+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.

See also A. Eremenko, A Toda lattice in dimension 2 and Nevanlinna theory, J. Math. Phys., Anal. Geom., 31 (2007) 39-46 for a generalization of this theorem of Nevanlinna.

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.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions).

Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f'+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.

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.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions).

Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f''+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.

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.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $\deg (P+2)/2$. $\deg P=0$ corresponds to the exponential, so it remains to consider orders $3/2$ and $2$. Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. This formula implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses a deeper theorem of Nevanlinna).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f'+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.

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.

Sketch of the proof for meromorphic functions. As before, we have that $f$ has no critical points (that is $f'(z)\neq 0$ and all poles are simple), but now $f$ is of order at most 2, normal type. (By definition of the order of a meromorphic function in terms of Nevanlinna characteristic). Now we use the theorem of R. Nevanlinna which describes all meromorphic functions of finite order without critical points. All such functions are ratios $f=w_1/w_2$ of two linearly independent solutions of the differential equation $$w''=Pw,$$ where $P$ is a polynomial. The order of the function is $(\deg P+2)/2$. The case $\deg P=0$ corresponds to an exponential, so it remains to consider orders $3/2$ (Airy functions) and $2$ (Weber functions). 

Now there is an asymptotic theory of these differential equations which gives a very precise asymptotic formula for solutions as $z\to\infty$. (In physics, this formula is called the WKB approximation). It implies that for orders $>1$, the spherical derivative of $f$ is unbounded. (This proof is independent of the previous proof for entire functions but uses deeper tools, like Nevanlinna theory).

Refs.

R. Nevanlinna, Uber Riemannsche Flachen mit endlich vielen Windungspunkten, Acta Math. 58 (1932) 295–373.

Expositions in English:

A. Eremenko, Entire and meromorphic solutions of ordinary differential equations, Chapter 6 in the book: Complex Analysis I, Encyclopaedia of Mathematical Sciences, vol. 85; Springer, NY, 1997, 141-153.

G. Gundersen, J. Heittokangas and A. Zemirni, Asymptotic integration theory for $f'+P(z)f=0$, Expositiones math., 40, 1 (2022) 94-126.