Meromorphic Functions

Abstract

According to Hayman (Ann Acad Sci Fenn Ser A I Math 7:65–71, (1982), [510]), Hilbert once told Nevanlinna: “You have made a hole in the wall of Mathematics. Other mathematicians will fill it.” Hayman continues: “If the hole means that many new problems were opened up, then this is indeed the case, and I am certain that Nevanlinna theory will continue to solve problems as it has done in the last 50 years.”

1.1 Preface by A. Eremenko

According to Hayman [510], Hilbert once told Nevanlinna: “You have made a hole in the wall of Mathematics. Other mathematicians will fill it.” Hayman continues: “If the hole means that many new problems were opened up, then this is indeed the case, and I am certain that Nevanlinna theory will continue to solve problems as it has done in the last 50 years.”

This chapter is dedicated to the problems on meromorphic functions stated by various authors during the period 1967–1989 and collected by Hayman and his collaborators.

In this book a “meromorphic function” means a function meromorphic in the complex plane. Most problems are about transcendental meromorphic functions (having an essential singularity at infinity).

The theory of meromorphic functions was mostly created by Nevanlinna in the 1920s, and he wrote two influential books on it [756, 757].

These books, especially the second one, contained many unsolved problems, and the present collection mentions only a few of them. This chapter reflects very well the development of Nevanlinna theory in the the second half of the 20th century.

Here I will try to give a very brief overview of the most important problems and their solutions. Of course, this selection reflects my own taste.

We use the definitions and notation introduced in the beginning of the chapter, and add to this $n_1(r, f)$, the counting function of critical points of a meromorphic function f, including multiplicities, and the averaged counting function $N_1(r, f)$, see Update 1.33. The Second Fundamental Theorem of Nevanlinna says that

$$\begin{aligned} \sum _{j=1}^qm(r,a_j, f)+N_1(r, f)\le 2T(r,f)+S(r, f), \end{aligned}$$

(1.1)

where $S(r,f)=o(T(r, f))$ outside an exceptional set of r of finite length. For functions of finite order, there is no exceptional set. (The fact that the exceptional set is in general really necessary was established by Hayman, see Update 1.22.)

This implies the defect relation:

$$\sum _a\delta (a, f)\le 2.$$

Problem 1.1 asks whether anything else can be said about deficiencies in general, and the answer is “no”: for every at most countable set of points $a_j$, and positive numbers $\delta _j$ whose sum is at most 2, there exists a meromorphic function f with $\delta (a_j, f)=\delta _j$ and $\delta (a, f)=0$ when $a\not \in \{ a_j\}$ (see Update 1.1).

The situation is quite different for the class of meromorphic functions of finite order. Problems 1.3, 1.6, 1.14, 1.29, 1.33 address the question: what are the restrictions on deficiencies for functions in this class. An almost complete answer is now known. Assume that f is of finite order. First of all, for functions of finite order,

$$\sum _a\delta (a, f)^{1/3}<\infty .$$

Second, if f is of finite order and

$$\sum _a\delta (a, f)=2,$$

then the number of deficient values is finite (see Updates 1.3, 1.33). Finally, if $\delta (b, f)=1$ for some b, then

$$\sum _a\delta (a, f)^{1/3-\epsilon }<\infty ,$$

where $\epsilon =2^{-264}$. Of course, this value is not the best possible.

On the other hand, for every sequence of points $a_j$ and numbers $\delta _j\in (0,1)$ whose sum is strictly less than 2 and

$$\sum \delta _j^{1/3}<\infty $$

there exists a meromorphic function of finite order with deficient values $a_j$ and deficiencies $\delta _j$.

If we further restrict the class of functions by prescribing the order $\rho$, the results are much less complete, see Problems 1.5, 1.7–1.8, 1.13, 1.15, 1.17, 1.26. Most of these problems are still unsolved, except the Edrei spread conjecture 1.15 and Paley conjecture 1.17. The method invented by Albert Baernstein for solving the spread conjecture found a lot of applications outside the theory of meromorphic functions.

Another very popular line of research stems from the paper of Hayman [510]: Picard-type theorems for derivatives and differential polynomials. Problems 1.18–1.20, 1.42 belong to this line. A remarkable application of these theorems is their use in the solution of an old question of Wiman about real zeros of derivatives of entire functions (Problem 2.64).

Let me also mention two recent major developments in Nevanlinna theory which are not reflected in this book. At the beginning of the theory, Nevanlinna had already asked whether the Second Fundamental Theorem can be generalised, if one uses “small functions” $a_j(z)$ instead of constants, where “small” means $T(r,a_j)=o(T(r, f))$. The weak form, without $N_1$, was proved by Charles Osgood [782] with some additional assumptions, and then by Osgood [783] and Steinmetz [921] in full generality (Steinmetz’s proof is much simpler than Osgood’s). The simple example $f(z)=\mathrm {e}^z+z$ shows that one cannot include the $N_1$ term as in (1.1). However, one can write a weaker form of (1.1) as

$$\begin{aligned} \sum _{j=1}^q\overline{N}(r, a_j)\ge (q-2)T(r,f)+S(r, f), \end{aligned}$$

(1.2)

and in this form the Second Fundamental Theorem generalises to small functions, as was proved by Yamanoi [1001].

The next major development was also made by Yamanoi [1002]: he proved Gol’dberg’s conjecture:

$$\overline{N}(r,f)\le N(r, 1/f'')+S(r, f),$$

which implies the conjecture of Mues:

$$\sum _{a\ne \infty }\delta (a, f')\le 1.$$

These two achievements, both made in the 21st century, show that the theory is very much alive.

1.2 Progress on Previous Problems

Notation We use the usual notations of Nevanlinna theory, see for example Nevanlinna [756, 757] and Hayman [493].

If f(z) is meromorphic in $|z|<R$, and $0<r<R$, we write

$$\begin{aligned} n(r,a)=n(r,a, f) \end{aligned}$$

for the number of roots of the equation $f(z)=a$ in $|z|\le r$, when multiple roots are counted according to multiplicity, and $\overline{n}(r, a)$ when multiple roots are counted only once. We also define

$$\begin{aligned} N(r, a)=\int ^r_0\frac{[ n(t, a)-n(0,a)] \,\mathrm {d}t}{t}+n(0,a)\log r, \end{aligned}$$

$$\begin{aligned} \overline{N}(r,a)=\int ^r_a\frac{[\overline{n}(t, a)-\overline{n}(0,a)]\,\mathrm {d}t}{t}+\overline{n}(0,a)\log r, \end{aligned}$$

$$\begin{aligned} m(r,f)=m(r, \infty , f)=\frac{1}{2\pi }\int ^{2\pi }_0\log ^+|f(r\mathrm {e}^{i\theta })|\,\mathrm {d}\theta , \end{aligned}$$

where $\log ^+x=\max \big \{\log x, 0\big \}$,

$$\begin{aligned} m(r,a, f)=m\left( r,\infty ,\frac{1}{f-a}\right) , a\ne \infty , \end{aligned}$$

and

$$\begin{aligned} T(r,f)=m(r,\infty ,f)+N(r,\infty , f). \end{aligned}$$

Then for every finite a, we have by the First Fundamental Theorem (see Hayman [493, p. 5]),

$$\begin{aligned} T(r,f)=m(r,a,f)+N(r,a, f)+O(1), \text { as }r\rightarrow R. \end{aligned}$$

(1.3)

We further define the deficiency,

$$\begin{aligned} \delta (a,f)=\liminf _{r\rightarrow R}\frac{m(r,a,f)}{T(r, f)}=1-\limsup _{r\rightarrow R}\frac{N(r,a,f)}{T(r, f)}, \end{aligned}$$

the Valiron deficiency,

$$\begin{aligned} \varDelta (a,f)=\limsup _{r\rightarrow R}\frac{m(r,a,f)}{T(r, f)}, \end{aligned}$$

and further,

$$\begin{aligned} \varTheta (a, f)=1-\limsup _{r\rightarrow R}\frac{\overline{N}(r,a,f)}{T(r, f)}. \end{aligned}$$

We then have the “defect relation” (see Hayman [493, p. 43]),

$$\begin{aligned} \sum _a \delta (a,f)\le \sum _a \varTheta (a, f)\le 2, \end{aligned}$$

(1.4)

provided that either $R=\infty$ and f(z) is not constant, or $R<+\infty$ and

$$\begin{aligned} \limsup _{r\rightarrow R}\frac{T(r, f)}{\log \big (1/(R-r)\big )}=+\infty . \end{aligned}$$

If $R=+\infty$ we also define the lower order $\lambda$ and order $\rho$,

$$\begin{aligned} \lambda =\liminf _{r\rightarrow R}\frac{\log T(r,f)}{\log r},\rho =\limsup _{r\rightarrow R}\frac{\log T(r, f)}{\log r}. \end{aligned}$$

If $\delta (a, f)>0$ the value a is called deficient. It follows from (1.4) that there are at most countably many deficient values if the conditions for (1.4) are satisfied.

Problem 1.1

Is (1.4) all that is true in general? In other words, can we construct a meromorphic function f(z) such that f(z) has an arbitrary sequence $a_n$ of deficient values and no others, and further that $\delta (a_n, f)=\delta _n$, where $\delta _n$ is an arbitrary sequence subject to $\sum \delta _n\le 2$? (If f(z) is an entire function $\delta (f,\infty )=1$, so that $\sum _{a\ne \infty }\delta (a, f)\le 1$. For a solution of the problem in this case, see Hayman [493, p. 80].)

Update 1.1

This problem has been completely settled by Drasin [255]. He constructs a meromorphic function f(z) with arbitrary deficiencies and branching indices on a preassigned sequence $a_n$ of complex numbers with f(z) growing arbitrarily slowly, subject to having infinite order.

Problem 1.2

How big can the set of Valiron deficiencies be for functions in the plane? It is known that

$$\begin{aligned} N(r,a)=T(r,f)+O\big (T(r, f)^{\frac{1}{2}+\varepsilon }\big ) \end{aligned}$$

(1.5)

as $r\rightarrow \infty$, for all a outside a set of capacity zero (see Nevanlinna [757, pp. 260–264]).

In the case $R<+\infty$ this is more or less best possible, but in the plane we only know from an example of Valiron [963] that the corresponding set of a can be non-countably infinite. It is also not known whether (1.5) can be sharpened.

Update 1.2

Hyllengren [564] has shown that all values of a set E can have Valiron deficiency greater than a positive constant for a function of finite order in the plane if and only if there exists a sequence of complex numbers $a_n$ and a positive k such that each point of E lies in infinitely many of the discs $\{z: |z-a_n|<\mathrm {e}^{-\mathrm {e}^{kn}}\}$.

Hayman [501] proved that all values of any $F_\sigma$ set of capacity zero can be Valiron deficiencies for an entire function of infinite order, and a little more.

Problem 1.3

If f(z) is meromorphic of finite order $\rho$ and $\sum \delta (a, f)=2$, it is conjectured that $\rho =n/2$, where n is an integer and $n\ge 2$, and all the deficiencies are rational. F. Nevanlinna [754] has proved this result on the condition that f(z) has no multiple values, so that $n(r,a)=\overline{n}(r, a)$ for every a (see also R. Nevanlinna [755]).

Update 1.3

Weitsman [982] proved that the number of deficiencies is at most twice the order in this case. The conjecture was completely proved by Drasin [256]. Eremenko [314] gave a simpler proof of a stronger result, see Problem 1.33.

Problem 1.4

Let f(z) be an entire function of finite order $\rho$, and let $n_1(r, a)$ denote the number of simple zeros of the equation $f(z)=a$. If

$$\begin{aligned} n_1(r, a)=O(r^c), n_1(r, b)=O(r^c), \text { as } r\rightarrow \infty , \end{aligned}$$

where $a\ne b, c<\rho$, is it true that $\rho$ is an integral multiple of $\frac{1}{2}$? More strongly, is this result true if $\varTheta (a)=\frac{1}{2}=\varTheta (b)$? (For a somewhat weaker result in this direction, see Gol’dberg and Tairova [416].)

Update 1.4

The answer is ‘no’, even in a very weak sense. Gol’dberg [405] has constructed an example of an entire function for which

$$\begin{aligned} n_1(r, a)=O((\log r)^{2+\varepsilon }), n_1(r, b)=O((\log r)^{2+\varepsilon }) \text { as }r\rightarrow \infty , \end{aligned}$$

but the order is not a multiple of $\frac{1}{2}$.

Problem 1.5

Under what conditions can $\sum \delta (a, f)$ be nearly 2 for an entire function of finite order $\rho$? Pfluger [799] proved that if $\sum \delta (a, f)=2$, then (see Hayman [493, p. 115]) $\rho$ is a positive integer q, the lower order $\lambda$ is such that $\lambda =\rho$ and all the deficiencies are integral multiples of 1/q. If further,

$$\begin{aligned} \sum \delta (a, f)>2-\varepsilon (\lambda ), \end{aligned}$$

where $\varepsilon (\lambda )$ is a positive quantity depending on $\lambda$, then Edrei and Fuchs [287, 288] proved that these results remain true ‘nearly’, in the sense that there exist ‘large’ deficiencies which are nearly positive integral multiplicities of 1/q, and whose sum of deficiencies is ‘nearly’ 2. Can there also be a finite or infinite number of small deficiencies in this case?

Update 1.5

No progress on this problem has been reported to us. Hayman suspects that the answer is ‘no’.

Problem 1.6

Arakelyan [41] has proved that, given $\rho >\frac{1}{2}$ and a countable set E, there exists an entire function f(z) of order $\rho$ for which all the points of E are deficient. Can E be the precise set of deficiencies of f in the sense that f has no other deficient values? It is also conjectured that if the $a_n$ are deficient values for an entire function of finite order, then

$$\begin{aligned} \sum \big (\log [1/\delta (a_n, f)]\big )^{-1}<+\infty . \end{aligned}$$

            (N.U. Arakelyan)

Update 1.6

Eremenko [305] has proved the first conjecture. He also proved [313] that the second conjecture is false: given $\rho >1/2$ and a sequence of complex numbers $(a_k)$, there is an entire function f of order $\rho$ with the property $\delta (a_k, f)>c^k, k=1,2,\ldots ,$ for some $c\in (0,1)$. On the other hand, Lewis and Wu [669] proved $\sum \delta (a_k, f)^\alpha <\infty$ for entire functions of finite order with an absolute constant $\alpha <1/3 - 2^{-264}$. The exact rate of decrease of deficiencies of an entire function of finite order remains unknown.

Problem 1.7

If f(z) is an entire function of finite order $\rho$ which is not an integer, it is known that (see Pfluger [799] and Hayman [493, p. 104])

$$\begin{aligned} \sum \delta (a, f)\le 2-K(\rho ), \end{aligned}$$

where $K(\rho )$ is a positive quantity depending on $\rho$. What is the best possible value for $K(\rho )$? Edrei and Fuchs [288] conjectured (see also Hayman [493, p. 104]) that if q is the integral part of $\rho$, and if $q\ge 1$, then

$$\begin{aligned} K(\rho )=\frac{|\sin (\pi \rho )|}{q+|\sin (\pi \rho )|}, q\le \rho <q+\frac{1}{2}, \end{aligned}$$

$$\begin{aligned} K(\rho )=\frac{|\sin (\pi \rho )|}{q+1}, q+\frac{1}{2}\le \rho <q+1. \end{aligned}$$

This result would be sharp.

If $\rho \le \frac{1}{2}$, there are no deficient values, so that $K(\rho )=1$. If $\frac{1}{2}<\rho <1$, Pfluger [799] proved that $K(\rho )=\sin (\pi \rho )$. See also Hayman [493, p. 104].

Update 1.7

For Problems 1.7 and 1.8 a better lower bound was found by Miles and Shea [728], who also obtained the exact lower bound for any order $\rho$ of

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{N(r, 0)+N(r,\infty )}{m_2(r)}, \end{aligned}$$

where

$$\begin{aligned} m_2(r)=\Big (\frac{1}{2\pi }\int ^{2\pi }_0(\log |f(r\mathrm {e}^{i\theta })|)^2\,\mathrm {d}\theta \Big )^{\frac{1}{2}}. \end{aligned}$$

Hellerstein and Williamson [534] have solved the problems completely for entire functions with zeros on a ray.

Problem 1.8

Following the notation in Problem 1.7, if f(z) is meromorphic in the plane of order $\rho$, it is conjectured by Pfluger [799] that, for $a\ne b$,

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{N(r,a)+N(r,b)}{T(r, f)}\ge K(\rho ). \end{aligned}$$

This is known to be true for $0<\rho \le 1$. If equality holds in the above inequality, it is conjectured that f(z) has regular growth, that is, $\rho =\lambda$.

Update 1.8

See Update 1.7.

Problem 1.9

If f(z) is an entire function of finite order $\rho$ which has a finite deficient value, find the best possible lower bound for the lower order $\lambda$ of f(z). (Edrei and Fuchs [288] showed that $\lambda$ is positive.)

Gol’dberg [400] showed that for every $\rho >1$, $\lambda \ge 1$ is possible.

Update 1.9

This has been settled by Gol’dberg [400].

Problem 1.10

If f(z) is a meromorphic function of finite order with more than two deficient values, is it true that if $\sigma >1$, then

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{T(\sigma r)}{T(r)}<+\infty . \end{aligned}$$

Update 1.10

No progress on this problem has been reported to us. (The update in [504] has been withdrawn.)

Problem 1.11

If f(z) is a meromorphic function of finite order with at least one finite deficient value, does the conclusion of Problem 1.10 hold?

Update 1.11

Drasin writes that Kotman [629] has shown that the answer to this question is ‘no’.

Problem 1.12

Edrei, Fuchs and Hellerstein [292] ask if f(z) is an entire function of infinite order with real zeros, is $\delta (0,f)>0$? More generally, is $\delta (0,f)=1$?

Update 1.12

This has been disproved by Miles [725], who showed that $\delta (0,f)=0$ is possible. However, Miles also showed that

$$\begin{aligned} \frac{N(r, 0)}{T(r, f)}\rightarrow 0 \end{aligned}$$

as $r\rightarrow \infty$ outside a fairly small set in this case.

Problem 1.13

If f(z) is an entire function of finite order $\rho$ and lower order $\lambda$ with real zeros, find the best possible bound $B=B(\rho ,\lambda )$ such that $\delta (0,f)\ge B$. From Edrei, Fuchs and Hellerstein [292] it is known that $B>0$ if $2<\rho <\infty$, and it is conjectured that $B\rightarrow 1$ as $\rho \rightarrow +\infty$.

Update 1.13

An affirmative answer with the exact value of B was given by Hellerstein and Shea [533].

Problem 1.14

If f(z) is a meromorphic function of finite order, then it is known (see Hayman [493, pp. 90, 98]) that $\sum \delta (a, f)^\alpha$ converges if $\alpha >\frac{1}{3}$, but may diverge if $\alpha <\frac{1}{3}$. What happens when $\alpha =\frac{1}{3}$?

Update 1.14

This has been completely settled by Weitsman [983], who proved that $\sum \big (\delta (a, f)\big )^{1/3}$ does indeed converge for any meromorphic function of finite order.

Problem 1.15

(Edrei’s spread conjecture) If f(z) is meromorphic in the plane and of lower order $\lambda$, and if $\delta =\delta (a, f)>0$, is it true that, for a sequence $r=r_\nu \rightarrow \infty$, f(z) is close to a on a part of the circle $|z|=r_\nu$ having angular measure at least

$$\begin{aligned} \frac{4}{\lambda }\sin ^{-1}\sqrt{\left( \frac{\delta }{2}\right) }+o(1)? \end{aligned}$$

(For a result in this direction, see Edrei [285].)

Update 1.15

This result was proved by Baernstein [54] by means of the function $T^*(r,\theta )$, where

$$\begin{aligned} T^*(r,\theta )=\sup _E\frac{1}{2\pi }\int _E\log |f(r\mathrm {e}^{i\varphi })|\,\mathrm {d}\varphi +N(r, f), \end{aligned}$$

and E runs over all sets of measure exactly $2\theta$. See also Baernstein [53, 61].

Problem 1.16

For any function f(z) in the plane, let $n(r)=\sup _a n(r, a)$ be the maximum number of roots of the equation $f(z)=a$ in $|z|<r$, and

$$\begin{aligned} A(r) = \frac{1}{\pi }\int \int _{|z|<r}\frac{|f'(z)|^2}{(1+|f(z)|^2)^2}\,\mathrm {d}x \,\mathrm {d}y = \frac{1}{\pi }\int \int _{|a|<\infty }\frac{n(r, a)\,|\mathrm {d}a|^2}{(1+|a|^2)^2}. \end{aligned}$$

Then $\pi A(r)$ is the area, with due count of multiplicity, of the image on the Riemann sphere of the disc $|z|<r$ under f, and A(r) is the average value of n(r, a) as a moves over the Riemann sphere. It is known (see Hayman [493, p. 14]) that

$$\begin{aligned} 1\le \liminf _{r\rightarrow \infty }\frac{n(r)}{A(r)}\le \mathrm {e}. \end{aligned}$$

Can $\mathrm {e}$ be replaced by any smaller quantity, and in particular, by 1?

Update 1.16

Toppila [951] has shown that $\mathrm {e}$ cannot be replaced by 1. He has constructed an example of a meromorphic function for which

$$\begin{aligned} \liminf _{r\rightarrow \infty }\frac{n(r)}{A(r)}\ge \frac{80}{79} \end{aligned}$$

for every sufficiently large r. The question remains open for entire functions. Among other examples, Toppila shows that for an entire function the following can occur:

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{n(r, 0)}{A(7r/6)}\ge \frac{9}{5} \end{aligned}$$

and

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{n(r)}{A(Kr)}=\infty \end{aligned}$$

for every K, $K\ge 1$.

Miles [727] gave a positive answer, by showing that for every meromorphic function

$$\begin{aligned} \liminf _{r\rightarrow \infty }\frac{\max _a n(r, a)}{A(r)}\le \mathrm {e}-10^{-28}. \end{aligned}$$

Problem 1.17

(Paley’s conjecture) For any entire function f(z) of finite order $\rho$ in the plane, we have

$$\begin{aligned} 1\le \liminf _{r\rightarrow \infty }\frac{\log M(r,f)}{T(r, f)}\le C(\rho ), \end{aligned}$$

where $C(\rho )$ depends on $\rho$ only. This follows very simply from Hayman [493, Theorem 1.6, p. 18]. Wahlund [975] has shown that the best possible value of $C(\rho )$ is $\pi \rho /\sin (\pi \rho )$ for $0<\rho <\frac{1}{2}$, and it is conjectured that $C(\rho )=\pi \rho$ is the corresponding result for $\rho >\frac{1}{2}$.

Update 1.17

This inequality has been proved by Govorov [438] for entire functions, and by Petrenko [797] for meromorphic functions.

Problem 1.18

Suppose that f(z) is meromorphic in the plane, and that f(z) and $f^{(l)}(z)$ have no zeros, for some $l\ge 2$. Prove that $f(z)=\mathrm {e}^{az+b}$ or $(Az+B)^{-n}$.

The result is known if f(z) has only a finite number of poles (see Clunie [209] and Hayman [493, p. 67]) or if f(z) has finite order and $f\ne 0, f'\ne 0, f''\ne 0$, and

$$\begin{aligned} \liminf _{r\rightarrow \infty }\frac{\log n(r, f)}{\log r}<+\infty , \end{aligned}$$

(see Hayman [490]), or if none of the derivatives of f(z) have any zeros and f(z) has unrestricted growth (see Polya [804], Hayman [493, p. 63]).

Update 1.18

The conjecture was proved by Mues [745] if f has finite order and $ff''\ne 0$ (instead of $ff'f''\ne 0$). For $l>2$, the conjecture was proved by Frank [352]. Since then Frank, Polloczek and Hennekemper [354] have obtained various extensions. Thus, if f and $f^{(l)}$ have only a finite number of zeros, and $l>2$, then

$$\begin{aligned} f(z)=\frac{p_1}{p_2}\mathrm {e}^{p_3}, \end{aligned}$$

where $p_1, p_2, p_3$ are polynomials. However, the paper [352] contains a gap in the proof of the case $l=2$.

The last case which remained unsolved, $l=2$, was settled by Langley [649] who proved that the only meromorphic functions f, for which $ff''$ is zero-free, are $f(z)=\mathrm {e}^{az+b}$ and $f(z)=(az+b)^{-n}$.

Problem 1.19

Suppose that f(z) is meromorphic in the plane and $f'(z)f(z)^n\ne 1$, where $n\ge 1$. Prove that f(z) is constant. Hayman [490] has shown this to be true for $n\ge 3$.

Update 1.19

The case $n=2$ has been settled by Mues [746]. The last case which remained unsolved, $n=1$, was settled by Bergweiler and Eremenko [110]: for every non-constant meromorphic function f, the equation $f'(z)f(z)=c$ has solutions for every c, $c\ne 0,\infty$. This was first proved by Bergweiler and Eremenko for functions of finite order; then, Bergweiler and Eremenko [110], Chen and Fang [204], and Zalcman [1016] independently noticed that a general method of Pang [789] permits an extension to arbitrary meromorphic functions. The proof actually applies whenever $n\ge 1$.

There were many extensions and generalisations of this result of Bergweiler and Eremenko. The strongest result so far is due to Jianming Chang [201]: Let f be a transcendental meromorphic function whose derivative is bounded on the set of zeros of f. Then the equation $f(z) = c$ has infinitely many solutions for every $c\in \mathbb {C}\setminus \{0\}$.

Problem 1.20

If f(z) is non-constant and meromorphic in the plane, and $n=3$ or $n=4$, prove that $\phi (z)=f'(z)-f(z)^n$ assumes all finite complex values. This is known to be true if f(z) is an entire function, or if $n\ge 5$ in the case where f(z) is meromorphic; see Hayman [490].

In connection to this, it would be most interesting to have general conditions under which a polynomial in f(z) and its derivatives can fail to take some complex value. When f(z) is a meromorphic rather than an entire function, rather little is known, see however Clunie [209, 210] and Hayman [493, Ch. 3].

Update 1.20

This question is closely related to Problem 1.19. Mues [746] proved that $\phi (z)$ may omit a finite non-zero value when $n=3$ or 4. He also showed that $\phi$ must have infinitely many zeros for $n=4$. The remaining case of zeros for $n=3$ was settled by Bergweiler and Eremenko [110].

Problem 1.21

If f(z) is non-constant in the plane, it is known (see Hayman [493, pp. 55–56]) that

$$ \alpha _f=\limsup _{r\rightarrow \infty }\frac{T(r,f)}{T(r, f')}\ge {\left\{ \begin{array}{ll} \frac{1}{2} &{} \text {if } f(z) \text { is meromorphic},\\ 1 &{} \text {if } f(z) \text { is an entire function}.\\ \end{array}\right. } $$

These inequalities are sharp. It is not known whether

$$\begin{aligned} \beta _f=\liminf _{r\rightarrow \infty }\frac{T(r,f)}{T(r, f')} \end{aligned}$$

can be greater than one, or even infinite. It is known that $\beta _f$ is finite if f(z) has finite order. Examples show that $\alpha _f$ may be infinite for entire functions of any order $\rho$, that is $0\le \rho \le \infty$, and that given any positive constants K, $\rho$ there exists an entire function of order at most $\rho$ such that

$$\begin{aligned} \frac{T(r,f)}{T(r, f')}>K \end{aligned}$$

on a set of r having positive lower logarithmic density. For this and related results, see Hayman [495].

Update 1.21

Let f be meromorphic in the plane. The relation between $T(r, f')$ and T(r, f) constitutes an old problem of Nevanlinna theory. It is classical that

$$\begin{aligned} m(r,f')\le m(r,f)+m\Big (r,\frac{f'}{f}\Big )\le m(r,f)+O(\log T(r, f)) \end{aligned}$$

outside an exceptional set. In particular, if f is entire so that $m(r,f)=T(r, f)$, we deduce that $T(r, f')<(1+o(1))T(r, f)$ outside an exceptional set.

The question of a corresponding result in the opposite direction had been open until fairly recently. Hayman [494] has shown that there exist entire functions of finite order $\rho$ for which $T(r,f)>KT(r, f')$ on a set having positive lower logarithmic density, for every positive $\rho$ and $K>1$. Toppila [952] has given a simple example for which

$$\begin{aligned} \frac{T(r,f)}{T(r, f')}\ge 1+\frac{7}{10^7} \end{aligned}$$

for all sufficiently large r. In this example, he takes for $f'$ the square of the sine product, having permuted the zeros in successive annuli to the positive or negative axis. The result is that $f'$ is sometimes large on each half-axis, and so f, the integral of $f'$, is always large on and near the real axis. Further, Langley [650] constructed an entire function of arbitrary order under $\rho >\frac{1}{2}$ for which $\beta _f>1$, where

$$\begin{aligned} \beta _f=\limsup _{r\rightarrow \infty }\frac{T(r,f)}{T(r, f')}. \end{aligned}$$

On the other hand, Hayman and Miles [518] have proved that $\beta _f\le 3e$ if f is meromorphic, and $\beta _f\le 2e$ if f is entire. Density estimates are also given to show that the previous examples are fairly sharp.

Problem 1.22

The defect relation (1.4) is a consequence of the inequality which is called the ‘Second Fundamental Theorem’ (see Hayman [493, formula (2.9), p. 43]),

$$\begin{aligned} \sum ^k_{\nu =1}\overline{N}(r, a_\nu , f)\ge \big (q-2+o(1)\big )T(r, f), \end{aligned}$$

(1.6)

which holds for any distinct numbers $a_\nu$ and $q\ge 3$, as $r\rightarrow \infty$ outside a set E of finite measure, if f(z) is meromorphic in the plane. The exceptional set E is known to be unnecessary if f(z) has finite order. Does (1.6) also hold as $r\rightarrow \infty$ without restriction if f(z) has infinite order?

Update 1.22

A negative answer to this question is provided by Hayman’s examples [501] discussed in connection with Problem 1.2. These show that the Second Fundamental Theorem fails to hold on a certain sequence $r=r_\nu$.

Problem 1.23

Under what circumstances does $f(z_0+z)$ have the same deficiencies as f(z)? It was shown by Dugué [265] that this need not be the case for meromorphic functions, and by Hayman [485] that it is not necessarily true for entire functions of infinite order. The case of functions of finite order remains open. Valiron [965] notes that a sufficient condition is

$$\begin{aligned} \frac{T(r+1,f)}{T(r, f)}\rightarrow 1, \text { as }r\rightarrow \infty , \end{aligned}$$

and this is the case in particular if $\rho -\lambda <1$. Since for entire functions of lower order $\lambda$, $\lambda \le \frac{1}{2}$ there are no deficiencies anyway, it follows that the result is true at any rate for entire functions of order $\rho <\frac{3}{2}$ and, since $\lambda \ge 0$ always, for meromorphic functions of order less than one.

Update 1.23

Gol’dberg and Ostrovskii [415] give examples of meromorphic functions of finite order for which the deficiency is not invariant under change of origin. See also Gol’dberg and Ostrovskii [415]; and Wittich [994, 996] for details.

Miles [726] provided a counterexample of an entire f of large finite order. Gol’dberg, Eremenko and Sodin [411] have constructed such f with preassigned order $\rho$, such that $5<\rho <\infty$.

Problem 1.24

If f is meromorphic in the plane, can n(r, a) be compared in general with its average value

$$\begin{aligned} A(r)=\frac{1}{\pi }\int \int _{|z|<r}\frac{|f'(z)|^2}{(1+|f(z)|^2)^2}\,\mathrm {d}x\, \mathrm {d}y \end{aligned}$$

in the same sort of way that N(r, a) can be compared with T(r)? In particular, is it true that

$$\begin{aligned} n(r, a)\sim A(r) \end{aligned}$$

as $r\rightarrow \infty$, outside an exceptional set of r, independent of a, and possibly an exceptional set of a? (Compare Problem 1.16.)

            (P. Erdős)

Update 1.24

Miles [723] has shown that

$$\begin{aligned} \lim _{r\rightarrow \infty ,\, r\notin E}\frac{n(r, a)}{A(r)}=1, \end{aligned}$$

for all a not in A, a set of inner capacity zero, and all r not in E, a set of finite logarithmic measure.

Problem 1.25

In the opposite direction to Problem 1.24, does there exist a meromorphic function such that for every pair of distinct values a, b, we have

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{n(r,a)}{n(r,b)}=\infty \text { and }\liminf _{r\rightarrow \infty }\frac{n(r,a)}{n(r, b)}=0? \end{aligned}$$

Note, of course, that either of the above limits for all distinct a, b implies the other.

(Compare the result (1.3) quoted in Problem 1.2, which shows that this certainly cannot occur for the N-function.) The above question can also be asked for entire functions.

            (P. Erdős)

Update 1.25

Both Gol’dberg [406] and Toppila [951] have produced examples of entire functions for which

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{n(r,a)}{n(r, b)}=\infty , \end{aligned}$$

for every finite unequal pair (a, b). A corresponding example for meromorphic functions has also been given by Toppila [951].

Problem 1.26

The analogue of Problem 1.7 may be asked for meromorphic functions. The proposers conjecture that in this case

$$\begin{aligned} \sum \delta (a, f)\le \max \{\varLambda _1(\rho ),\varLambda _2(\rho )\}, \end{aligned}$$

where for $\rho \ge 1$, $q=[2\rho ]$ we have

$$\begin{aligned} \varLambda _1(\rho )&=2-\frac{2\sin \big (\frac{1}{2}\pi (2\rho -q)\big )}{q+2\sin \big (\frac{1}{2}\pi (2\rho -q)\big )},\\ \varLambda _2(\rho )&=2-\frac{2\cos \big (\frac{1}{2}\pi (2\rho -q)\big )}{q+1}. \end{aligned}$$

Drasin and Weitsman [259] shows that this result would be sharp. The correct bound is known for $0\le \rho \le 1$.

            (D. Drasin and A. Weitsman)

Update 1.26

No progress on this problem has been reported to us.

Problem 1.27

Let E be the set for which $m(r, a)\rightarrow \infty$ as $r\rightarrow \infty$. How large can E be if:

• (a) f is entire and of order $\frac{1}{2}$ mean type,

• (b) f is meromorphic of order $\rho$, where $0\le \rho \le \frac{1}{2}$.

The proposers settled this problem in all other cases (see Updates 2.1(a) and 2.1(b) for more details).

            (D. Drasin and A. Weitsman)

Update 1.27(b)

Damodaran [232] proved the existence of meromorphic functions of growth $T(r, f)=O(\rho (r)(\log r)^3)$, where $\rho (r)\rightarrow \infty$ arbitrarily slowly, such that $m(r, a)\rightarrow \infty$ for all a in an arbitrarily prescribed set of capacity zero. Lewis [666] and Eremenko [303] independently improved this to $(\log r)^2$ in place of $(\log r)^3$. This is best possible, following from an old result of Tumura [958].

Problem 1.28

Are there upper bounds of any kind on the set of asymptotic values of a meromorphic function of finite order?

            (D. Drasin and A. Weitsman)

Update 1.28

A negative answer to this question has been given by Eremenko [304], who has constructed meromorphic functions of positive and of zero order, having every value in the closed plane as an asymptotic value. This has been improved by Canton, Drasin and Granados [182], who proved that for every $\phi (r)\rightarrow +\infty$ and every analytic (Suslin) set A, there exists a meromorphic function f with the property $T(r, f)=O(\phi (r)\log ^2r)$ and whose set of asymptotic values coincides with A.

Problem 1.29

Under what circumstances does there exist a meromorphic function f(z) of finite order $\rho$ with preassigned deficiencies $\delta _n=\delta (a_n, f)$ at a preassigned sequence of complex numbers? Weitsman has solved this problem (see Update 1.14) by showing that it is necessary that

$$\begin{aligned} \sum \delta _n^{\frac{1}{3}}<\infty , \end{aligned}$$

(1.7)

but the bound of the sum of the series depends on the largest term $\delta _1$. On the other hand, Hayman [493, p. 98] showed that the condition

$$\begin{aligned} \sum \delta _n^{\frac{1}{3}}<A, \end{aligned}$$

(1.8)

with $A=9^{-\frac{1}{3}}$, is sufficient to yield a meromorphic function of order 1 mean type, such that $\delta (a_n, f)\ge \delta _n$. It may be that (1.8) is a sufficient condition with a constant $A=A_1(\rho )$, and, with a larger constant $A=A_2(\rho )$ is also a necessary condition. The problem may be a little easier if $\rho$ is allowed to be arbitrary but finite.

Update 1.29

Eremenko [311] has constructed an example of a function of finite order, having preassigned deficiencies $\delta _n=\delta (a_n, f)$, subject to $0<\delta _n<1$, $\sum \delta _n<2$ and $\sum \delta _n^{\frac{1}{3}}<\infty$, and no other conditions. In view of the results reported in Updates 1.3 and 1.33, this result is a complete solution of the Inverse Problem in the class of functions of (unspecified) finite order.

Problem 1.30

Can one establish an upper bound on the number of finite asymptotic values of a meromorphic function f(z) in $\mathbb {C}$, taking into account both the order of f, and the angular measure of its tracts?

            (W. Al-Katifi)

Update 1.30

No progress on this problem has been reported to us.

Problem 1.31

Let the function f be meromorphic in the plane, and not rational, and satisfy the condition

$$\begin{aligned} \frac{T(r, f)}{(\log r)^3}\rightarrow \infty ,\text { as }r\rightarrow \infty , \end{aligned}$$

(1.9)

where T(r, f) is the Nevanlinna characteristic. A theorem of Yang Lo [1005] states that then there exists a direction $\theta _0\in [0,2\pi )$ such that for every positive $\varepsilon$, either f attains every finite value infinitely often in $D_\varepsilon =\{z:|\arg z - \theta _0|<\varepsilon \}$, or else $f^{(k)}$ attains every value, except possibly zero, infinitely often in $D_\varepsilon$ for all positive integers k. Can the condition (1.9) be dropped completely? Or, possibly, can it be replaced by the ‘more usual’ condition

$$\begin{aligned} \frac{T(r, f)}{(\log r)^2}\rightarrow \infty ,\text { as }r\rightarrow \infty \,? \end{aligned}$$

One cannot expect any more from Yang Lo’s method of finding $\theta _0$ through the use of ‘filling discs’. Rossi [849] has shown that (1.9) cannot be improved if $\theta _0$ is sought in this way.

            (D. Drasin; communicated by J. Rossi)

Update 1.31

Rossi writes that there is an incorrect paper of Zhu [1021] where he purports to use filling discs to solve this problem. However, Fenton and Rossi [340] remark that Zhu’s approach is wrong, and point to the example in Rossi [849]. Some work on this problem has been produced by Sauer [878].

Problem 1.32

Let f be meromorphic in $\mathbb {C}$, and let $f^{-1}$ denote any element of the inverse function that is analytic in a neighbourhood of a point w. A well-known theorem of Gross [446] states that $f^{-1}$ may be continued analytically along almost all rays beginning at w. Is it possible to refine the exceptional set in this theorem?

            (A. Eremenko)

Update 1.32

No progress on this problem has been reported to us.

Problem 1.33

Let f be a meromorphic function of finite order $\rho$. Does the condition

$$\begin{aligned} N(r, 1/f')+2N(r,f)-N(r,f')=o(T(r, f)),\text { as }r\rightarrow \infty , \end{aligned}$$

imply that $2\rho$ is an integer?

            (A. Eremenko)

Update 1.33

This is a slightly more precise conjecture than Problem 2.25. Both problems are solved completely by the following theorem of Eremenko [314]: suppose that f is a meromorphic function of finite lower order $\lambda$, and that

$$\begin{aligned} N_1(r,f):=N(r, 1/f')+2N(r,f)-N(r,f')=o(T(r, f)). \end{aligned}$$

Then

• (a) $2\lambda$ is an integer greater than or equal to 2.

• (b) $T(r, f)=r^\lambda l(r)$, where l is a slowly varying function in the sense of Karamata.

• (c) $\sum _a\delta (a, f)=2$, all deficient values are asymptotic, and all deficiencies are multiples of $1/\lambda$.

Problem 1.34

Let $n_1(r,a, f)$ denote the number of simple zeros of $f(z)-a$ in $\{|z|\le r\}$. Selberg [887] has shown that if:

• (a) f is a meromorphic function of finite order $\rho$, and

• (b) $n_1(r,a, f)=O(1)$, as $r\rightarrow \infty$ for four distinct values of a, then

$\rho$ is an integral multiple of $\frac{1}{2}$ or $\frac{1}{3}$.

Does this conclusion remain true if (b) is replaced by:

• (c) $n_1(r,a,f)=o(T(r, f))$, as $r\rightarrow \infty$, for four distinct values of a?

Gol’dberg [404] has constructed an entire function of arbitrary prescribed order which satisfies the condition (c) for two distinct values of a.

            (A. Eremenko)

Update 1.34

The answer is ‘no’. Künzi [646] has shown that $\rho$ can be arbitrary, subject to $1<\rho <\infty$, and Gol’dberg [404] has a counterexample for arbitrary positive $\rho$.

Problem 1.35

Determine the upper and lower estimates for the growth of entire and meromorphic solutions of algebraic ordinary differential equations (AODE). (This is a classical problem.)

For AODEs of first order, it is known that the meromorphic solutions f must have finite order (see Gol’dberg [399]) and that $(\log r)^2=O(T(r, f))$; see Eremenko [307, 309]. (The latter two references contain a general account of first-order AODEs, including modern proofs of some classical results.) The order of entire solutions of first-order AODEs is an integral multiple of $\frac{1}{2}$; see Malmquist [704]. For AODEs of second order, it is known that the order of entire solutions is positive; see Zimogljad [1023]. There is no upper estimate valid for all entire or meromorphic solutions of AODEs of order greater than one, but there is an old conjecture that $\log |f(z)|\le \exp _n(|z|)$ as $z\rightarrow \infty$ for entire solutions f of an AODE of order n.

            (A. Eremenko)

Update 1.35

Steinmetz [919, 920] proved that every meromorphic solution of a homogeneous algebraic differential equation of second order has the form $f=(g_1/g_2)\exp (g_3)$, where the $g_i$ are entire functions of finite order. Thus $T(r, f)=O(\exp (r^k))$ for some positive k. By Wiman–Valiron theory (see, for example [966]), it is known that ‘most’ algebraic differential equations do not have entire solutions of infinite order. A precise statement of this sort is contained in Hayman [512].

Problem 1.36

Let F be a polynomial in two variables, and let y be a meromorphic solution of the algebraic ordinary differential equation $F(y^{(n)}, y)=0$. Is it true that y must be an elliptic function, or a rational function of exponentials, or a rational function? This is known in the following cases:

• (a) $n=1$: a classical result, probably due to Abel;

• (b) $n=2$: an old result of Picard [801], and independently, Bank and Kaufman [72];

• (c) n is odd and y has at least one pole, Eremenko [308]; and

• (d) the genus of the curve $F(x_1,x_2)=0$ is at least equal to one, Eremenko [308].

            (A. Eremenko)

Update 1.36

This has been solved by Eremenko, Liao and Tuen-Wai Ng [325], who prove (c) for all n, and give an example of an entire solution which is neither rational, nor a rational function of exponentials.

Problem 1.37

Find criteria for and/or give explicit methods for the construction of meromorphic functions f in $\mathbb {C}$ with the following properties:

• (a) all poles of f are of odd multiplicity;

• (b) all zeros of f are of even multiplicity.

(Here ‘explicit methods’ means that all computations must be practicable.) The background of this problem lies in the question of meromorphic solutions of the differential equation $y''+A(z)y=0$ in the whole plane.

            (J. Winkler)

Update 1.37

We mention a result of Schmieder [884] which may be relevant: on every open Riemann surface, there exists an analytic function with prescribed divisors of zeros and critical points, subject to certain trivial restrictions.

Problem 1.38

• (a) Let f be non-constant and meromorphic in the open unit disc $\mathbb {D}$, with $\alpha <+\infty$, where

  $$\begin{aligned} \alpha = \limsup _{r\rightarrow 1}\frac{T(r, f)}{-\log (1-r)}, \end{aligned}$$

  (1.10)

  and set

  $$\begin{aligned} \varPsi =(f)^{m_0}(f')^{m_1}\ldots (f^{(k)})^{m_k}. \end{aligned}$$

  It is known that $\varPsi$ assumes all finite values, except possibly zero, infinitely often, provided that $m_0\ge 3$ and $\alpha >2/(m_0-2)$, (or $m_0\ge 2$ and $\alpha >2/(m_0-1)$, if f is analytic). For which smaller values of $\alpha$ does the same conclusion hold?

• (b) Let f be non-constant and meromorphic in $\mathbb {D}$, with $\alpha <+\infty$ in (1.10); assume also that f has only finitely many zeros and poles in $\mathbb {D}$. Let l be a positive integer, and write $\varPsi =\sum ^l_{\nu =0}a_\nu f^{(\nu )}$, where the $a_\nu$ are functions in $\mathbb {D}$ for which $T(r,a_\nu )=o(T(r, f))$ as $r\rightarrow 1$ (for each $\nu$). It is known that if $\varPsi$ is non-constant, then $\varPsi$ assumes every finite value, except possibly zero, infinitely often, provided that $\alpha >\frac{1}{2}l(l+1)+1$. For which smaller values of $\alpha$ does the same conclusion hold?

            (L.R. Sons)

Update 1.38(b)

Gunsul [454] provides a condition that establishes the same conclusion for smaller values of $\alpha$.

Problem 1.39

Let f be a function meromorphic in $\mathbb {D}$ for which $\alpha <+\infty$ in (1.10).

• (a) Shea and Sons [893, Theorem 5] have shown that if $f(z)\ne 0$, $\infty$ and $f'(z)\ne 1$ in $\mathbb {D}$, then $\alpha \le 2$. Is 2 best possible?

• (b) Shea and Sons [893] have shown that, if $f(z)\ne 0$ and $f'(z)\ne 1$ in $\mathbb {D}$, then $\alpha \le 7$. What is the best possible $\alpha$ in this case?

            (L.R. Sons)

Update 1.39

No progress on this problem has been reported to us.

Problem 1.40

Let f be a function meromorphic in $\mathbb {D}$ of finite order $\rho$. Shea and Sons [893] have shown that

$$\begin{aligned} \sum _{a\ne \infty }\delta (a, f)\le \delta (0,f')\big (1+k(f)\big )+\frac{2}{\lambda }(\rho +1), \end{aligned}$$

where

$$\begin{aligned} k(f)=\limsup _{r\rightarrow 1}\frac{\overline{N}(r,\infty ,f)}{T(r, f)+1}\quad \text { and }\quad \lambda (f)=\liminf _{r\rightarrow \infty }\frac{T(r, f)}{\log (1/(1-r))}. \end{aligned}$$

Can the factor 2 be eliminated? (If so, the result is then best possible.)

            (L.R. Sons)

Update 1.40

No progress on this problem has been reported to us.

Problem 1.41

Let f be a function meromorphic in $\mathbb {D}$ for which $\alpha =+\infty$ in (1.10). Then it is known that

$$\begin{aligned} \sum _{a\in \mathbb {C}\cup \{\infty \}}\delta (a, f)\le 2. \end{aligned}$$

Are there functions which have an ‘arbitrary’ assignment of deficiencies at an arbitrary sequence of complex numbers, subject only to these conditions?

For analytic functions, Girynk [394] has a result; whereas for arbitrary meromorphic functions, there is a result of Krutin [641].

            (L.R. Sons)

Update 1.41

No progress on this problem has been reported to us.

Problem 1.42

Let f be meromorphic in $\mathbb {C}$, and suppose that the function

$$\begin{aligned} F(z)=f^{(k)}(z)+\sum ^{k-2}_{j=0}a_j(z)f^{(j)}(z) \end{aligned}$$

is non-constant, where $k\ge 3$ and the coefficients $a_j$ are polynomials. Characterise those functions f for which f and F have no zeros.

The case where f is entire has been settled by Frank and Hellerstein [353]. If all the $a_j$ are constant, then the problem has also been solved by Steinmetz [922] using results from Frank and Hellerstein [353]. It seems possible that if the $a_j$ are not all constants, then the only solutions with infinitely many poles are of the form

$$\begin{aligned} f=(H')^{-\frac{1}{2}(k-1)}H^{-l}, \end{aligned}$$

(1.11)

where l is a positive integer, and $H''/H'$ is a polynomial.

            (G. Frank and J.K. Langley)

Update 1.42

This was solved by Brüggemann [169], who proved the following: let a linear differential operator

$$\begin{aligned} L(f)=f^{(k)}+\sum ^{k-2}_{j=0}a_j f^j \end{aligned}$$

with polynomial coefficients $a_j$ be given, with at least one non-constant $a_j$. Then the only meromorphic functions f with infinitely many poles, satisfying $fL(f)\ne 0$, are of the form (1.11). An extension to rational coefficients has been given by Langley [653].

Problem 1.43

Let f be a meromorphic function of lower order $\lambda$. Let

$$\begin{aligned} m_0(r, f)=\inf \{|f(z)|:|z|=r\} \end{aligned}$$

and

$$\begin{aligned} M(r, f)=\sup \{|f(z)|:|z|=r\} \end{aligned}$$

and suppose that

$$\begin{aligned} \log r=o(\log M(r, f)),\text { as }r\rightarrow \infty . \end{aligned}$$

Gol’dberg and Ostrovskii [415] proved that if $0<\lambda <\frac{1}{2}$, then

$$\begin{aligned} \limsup _{r\rightarrow \infty }\frac{\log m_0(r,f)}{\log M(r,f)}+\pi \lambda \sin (\pi \lambda )\limsup _{r\rightarrow \infty }\frac{N(r,f)}{\log M(r, f)}\ge \cos (\pi \lambda ). \end{aligned}$$

Does this inequality remain valid for $\frac{1}{2}\le \lambda <1$? See also Gol’dberg and Ostrovskii [415].

            (A.A. Gol’dberg and I.V. Ostrovskii)

Update 1.43

No progress on this problem has been reported to us.

1.3 New Problems

Problem 1.44

Consider the differential equation $\displaystyle { \dot{z} = \frac{\mathrm {d}z}{\mathrm {d}t} = f(z) }$, where f is a meromorphic function in the plane: see [455, 752] for fundamental results on such flows. King and Needham [752] showed that if f has a pole at infinity of order at least 2 then there exists at least one trajectory z(t) which tends to infinity in finite time, that is, $\lim _{t \rightarrow \beta -} z(t) = \infty$, where $\beta \in \mathbb {R}$ and $z'(t) = f(z(t))$. It is then natural to ask what happens when f is transcendental.

It turns out [652] that if f is a transcendental entire function then there are infinitely many trajectories tending to infinity in finite time, whereas a transcendental meromorphic function need not have any at all. On the other hand, such trajectories do exist [652] if the inverse of a meromorphic f has a logarithmic singularity over infinity: this means that there exist $M > 0$ and a component $U_M$ of the set $\{ z \in \mathbb {C} : M < |f(z)| \le \infty \}$ such that $v = \log f$ maps $U_M$ univalently onto the half-plane $\mathrm{Re } \, v > \log M$.

The problem is then to resolve what happens under the weaker hypothesis that the inverse of f has a direct singularity over infinity, which means that there exists a component $U_M$ of $\{ z \in \mathbb {C} : M < |f(z)| \le \infty \}$ on which f has no poles. In this case the Wiman–Valiron theory [505], which played a key role in the proof for entire functions in [652], is available in the version from [116], but the difficulty seems to be to show that the trajectories arising from the method of [652] do not all tend to poles of f outside $U_M$.

            (J.K. Langley)

Problem 1.45

Following Hinchliffe [548], call a plane domain D a Rubel domain if every unbounded analytic function f on D has a sequence $(z_n)$ in D such that $\lim _{n \rightarrow \infty } f^{(k)}(z_n) = \infty$ for all $k \ge 0$. The name arises because Rubel [594, 856] asked whether the unit disc has this property, which was subsequently proved to be the case by Gordon [435].

Evidently a Rubel domain must be bounded (otherwise take $f(z) = z$), but it is easy to construct simply connected bounded domains on which 1/z is bounded but $\log z$ is not [856]. Hinchliffe [548] extended Gordon’s method to show that a bounded quasidisc (that is, the image of the unit disc under a quasiconformal mapping which fixes infinity) is a Rubel domain, but nothing more appears to be known on this problem.

Determine necessary and/or sufficient conditions for a bounded plane domain to be a Rubel domain.

            (J.K. Langley)

Problem 1.46

A transcendental entire function is called pseudoprime if it cannot be written as a composition $f \circ g$ of transcendental entire functions f and g. Suppose that the Maclaurin series of a transcendental entire function F has large gaps: for example, Hadamard gaps

$$ F(z) = \sum _{n=0}^\infty a_n z^{\lambda _n} , \quad \lambda _{n+1} > q \lambda _n , $$

for some $q > 1$. Must F be pseudoprime?

            (J.K. Langley)

Problem 1.47

Let $k \ge 3$ and let $A_0, \ldots , A_{k-2}$ be entire functions. Let $f_1, \ldots , f_k$ be linearly independent solutions of

$$ w^{(k)} + \sum _{j=0}^{k-2} A_j(z) w^{(j)} = 0 $$

and let E be the product $E = f_1 \ldots f_k$. Must the order of growth of E be at least that of each $A_j$? Note that if $k=2$ then this follows immediately from the Bank–Laine formula [73].

            (J.K. Langley)