On a question of Beardon

Abstract

In this paper, we estimate the order of growth of the solutions of the equation $f(kz)=kf(z)f^{\prime }(z)$ and investigate the periodicity of the solutions in the case $k=3$, which give an answer to the question proposed by Beardon (Comput. Methods Funct. Theory 12(1):273-278, 2012).

MSC:30D35, 30D45.

1 Introduction and main results

As entire functions z, sinz, sinhz are solutions of the equation $f(2z)=2f(z)f^{\prime }(z)$, Beardon [1] studied entire solutions of the generalized functional equation

$$f(kz)=kf(z)f^{\prime }(z),f(0)=0,$$

(1.1)

where k is a non-zero complex number. Obviously, two formal power series f and g are linearly conjugate if there is a non-zero c such that $g(x)=cf(x/c)$, and if f satisfies (1.1).

Firstly, we define some notations as in the paper [1]. The formal series and ℐ are defined by $\mathcal{O}:=0+0z+0z^2+\cdots$ , $\mathcal{I}:=0+1z+0z^2+0z^3+\cdots$ . We also introduce sets ${\mathcal{K}}_p=\{z:z^p=p+2\}$ ($p=1,2,\dots$) and $\mathcal{K}={\mathcal{K}}_1\cup {\mathcal{K}}_2\cup \cdots$ . Thus, we have ${\mathcal{K}}_1=3$ and ${\mathcal{K}}_2=\{-2,2\}$. Obviously, ${\mathcal{K}}_p$ contains exactly p points which are equally spaced around the circle $|z|=R_p$, where $R_p={(p+2)}^{1/p}>1$ and $R_p\in {\mathcal{K}}_p$. Also, since $x^{-1}log(x+2)$ is decreasing when $x>1$, we see that $R_1=3>R_2=2>\cdots >1$, and $R_p\to 1$ as $p\to \mathrm{\infty }$. In particular, the sets ${\mathcal{K}}_p$ are mutually disjoint, and the derived set of is the unit circle $\{z:|z|=1\}$. Using the above notations, Beardon obtained the following two main results for the entire solutions of equation (1.1).

Theorem A Any transcendental solution f of (1.1) is of the form

$$f(z)=z+z(b{z}^p+\cdots ),$$

where p is a positive integer, $b\ne 0$ and $k\in {\mathcal{K}}_p$. In particular, if $k\notin \mathcal{K}$ then the only formal solutions of (1.1) are and ℐ.

Theorem B For each positive integer p, there is a unique real entire function

$$F_p(z)=z(1+z^p+b_2{z}^{2p}+b_3{z}^{3p}+\cdots )$$

which is a solution of (1.1) for each k in ${\mathcal{K}}_p$. Further, if $k\in {\mathcal{K}}_p$ then the only transcendental solutions of (1.1) are the linear conjugates of $F_p$.

Based on the above two known results, we use the value distribution theory in q-difference (see, e.g., [2–6]), which is analogue of the classical Nevanlinna theory of meromorphic functions (see, e.g., [7–9]), to study the properties of solutions of (1.1). We get the upper bound of the order of solutions (see [10]).

Theorem 1 Suppose that f is a transcendental solution of (1.1) for $k\in \mathcal{K}$, then the order $\lambda (f)\le \frac{log2}{log|k|}$.

In particular, when $k=3$, the order of solutions of $f(3z)=3f(z)f^{\prime }(z)$ is not more than $log2/log3$. In Section 3 of the paper [1], Beardon also studied the periodicity of the solutions of equation (1.1). Although the solutions of (1.1) are periodic when $k=\pm 2$ (that is, $p=2$), he proved that the periodicity fails when $p\ge 3$, see [[1], Theorem 3.1]. But the case $p=1$ (that is, $k=3$) remains open. Here we shall prove that the periodicity also fails for the remaining case.

Theorem 2 The solution f of equation (1.1) is not periodic when $k=3$.

From Theorem 1, we know that the order of the transcendental solution f is not more than 1 when $k=2$. This coincides with the fact that the transcendental solutions are sinz and sinhz, the order of which are 1. Naturally we ask: Is the order of transcendental solutions of equation (1.1) exactly $log2/log|k|$? That means we have to estimate the lower bound of the order of solutions. Unfortunately, we do not get the expected lower bound since we meet difficulties when using $T(r,f^{\prime })$ to bound $T(r,f)$, because for any given positive constant K, there exists an entire function f with order λ for which

$$\frac{T(r,f)}{T(r,f^{\prime })}>K$$

on a set E of positive lower logarithmic density; see Hayman [[11], p.98]. So the above question is open.

2 Some lemmas

In this paper we use the standard notations in the Nevanlinna theory (see, e.g., [7–9]). So, in the following we give some well-known results, which are needed for our proof, of the classical Nevanlinna theory without presenting proofs. Let $f(z)$ be a meromorphic function, and let $m(r,f)$, $N(r,f)$, $T(r,f)$ denote the proximity function, the counting function and characteristic function of $f(z)$, respectively, here $r=|z|$. $T(r,f)=m(r,f)+N(r,f)$ and for the entire function $N(r,f)=0$. Moreover, the order of growth of a meromorphic function $f(z)$ is defined by

$$\lambda (f):=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logT(r,f)}{logr}.$$

We denote by E a set of finite linear measure in $R^{+}$, not necessarily the same at each occurrence. For any non-constant meromorphic function $f(z)$, we denote by $S(r,f)$ any quantity satisfying $S(r,f)=o(T(r,f))$ ($r\to \mathrm{\infty }$, $r\notin E$). For two meromorphic functions $f(z)$ and $g(z)$, we have $m(r,fg)\le m(r,f)+m(r,g)$ and $T(r,fg)\le T(r,f)+T(r,g)$. In addition, the identity $m(r,\frac{f^{\prime }}{f})=S(r,f)$ is also a very important result in the Nevanlinna theory.

The first lemma on the relationship between $T(r,f(qz))$ and $T(|q|r,f(z))$ is due to Bergweiler et al. [[12], p.2].

Lemma 2.1 One case, see that

$$T(r,f(qz))=T(|q|r,f)+O(1)$$

(2.1)

holds for any meromorphic function f and any constant q.

Lemma 2.2 [13]

Let $\mathrm{\Phi }:(1,\mathrm{\infty })\to (0,\mathrm{\infty })$ be a monotone increasing function, and let f be a nonconstant meromorphic function. If for some real constant $\alpha \in (0,1)$, there exist real constants $K_1>0$ and $K_2\ge 1$ such that

$$T(r,f)\le K_1\mathrm{\Phi }(\alpha r)+K_{2}T(\alpha r,f)+S(\alpha r,f),$$

then the order of growth of f satisfies

$$\lambda (f)\le \frac{log{K}_2}{-log\alpha }+\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{log\mathrm{\Phi }(r)}{logr}.$$

Lemma 2.3 [[9], Lemma 5.1]

Suppose that a nonconstant meromorphic function f is periodic, that is, $f(z+\eta )=f(z)$ for nonzero complex number η. Then the order $\lambda (f)\ge 1$.

3 Proof of theorems

Proof of Theorem 1 By the definition of , we know that $|k|>1$. Thus, by Lemma 2.1 we have

$$T(r,f(kz))=T(|k|r,f(z))+O(1),$$

(3.1)

and by (1.1), we can get

$$T(r,f(kz))=T(r,kf(z)f^{\prime }(z))\le T(r,f(z))+T(r,f^{\prime }(z))+O(1).$$

(3.2)

Combining the two inequalities above and simplifying $T(r,f(z))$ by $T(r,f)$, we have

$$T(|k|r,f)\le T(r,f)+T(r,f^{\prime })+O(1).$$

(3.3)

By Theorem B, we know that the solution f is entire. Since for the entire function f its derivative is also entire, we have

$$\begin{array}{rcl}T(r,f^{\prime })& =& m(r,f^{\prime })=m(r,f\frac{f^{\prime }}{f})\le m(r,f)+m(r,\frac{f^{\prime }}{f})\\ =& m(r,f)+S(r,f)=T(r,f)+S(r,f).\end{array}$$

(3.4)

By (3.3) and (3.4), we have

$$T(|k|r,f)\le 2T(r,f)+S(r,f).$$

Set $\alpha =1/|k|$, thus, we get

$$T(r,f)\le 2T(\alpha r,f)+S(\alpha r,f).$$

Applying Lemma 2.2 yields

$$\lambda (f)\le \frac{log2}{log|k|}.$$

 □

Proof of Theorem 2 Theorem 2 follows from Theorem 1 and Lemma 2.3. □