Meromorphic solutions of q-difference equations

Ilham Eli and Niro Yanagihara

Abstract

Consider a linear q-difference equation: a(z)f(z) = b(z)f(qz)+c(z), 0 < |q| < 1, a(z), b(z), c(z) ∈ ℂ[z]. We suppose it admits a transcendental meromorphic solution f(z). Here we will determine the counting function N(r, f), and investigate deficiency property of f(z). Finally we give some examples.

Keywords and phrases. linear q-difference equation, transcendental meromorphic solution, Nevanlinna and Valiron deficiency. Mathematics Subject Classification. 39A13, 30D35

1 Introduction

Here we will consider a q-difference equation

(1.1)  a(z)f(z) = b(z)f(qz) + c(z), 0 < |q| < 1,
a(z) = ∑_{k=0}^{A} a_{k}z^{k}, b(z) = ∑_{k=0}^{B} b_{k}z^{k}, c(z) = ∑_{k=0}^{C} c_{k}z^{k}, (a(z), b(z), c(z)) = 1,

in which we assume that a(0) = a_{0} ≠ 0, b(z) ≢ 0, and A + B ≥ 1.

We suppose that (1.1) possesses a transcendental meromorphic solution f(z). If f(z) has a pole of order p at z = 0, then putting g(z) = z^{m}f(z), we get

a(z)g(z) = q^{−p}b(z)g(qz) + z^{m}c(z),

hence we suppose hereafter that f(z) is holomorphic at z = 0.

If f(z) has a zero of order m at z = 0, then obviously c(z) has also zero of order ≥ m at z = 0. Hence we can suppose that f(0) ≠ 0.

In the below, we denote by m(r, f), N(r, f), and T (r, f) the proximity function, counting function, and the Nevanlinna characteristic function, respectively, of f(z) [4], [5].

It has been proved [2] that

(1.2) m(r, f) = \frac{1}{2π} ∫_{0}^{2π} log^{+} |f(re^{iθ})|dθ = \frac{max(B − A, 0)}{−2 log |q|}(log r)^{2}(1 + o(1)).

Our first result in this note is the following:

Theorem 1.1 (1) Suppose f(z) has infinitely many poles. Then we have

(1.3)  N(r, f) = \frac{A^{∗}}{−2 log |q|}(log r)^{2}(1 + o(1)),

where 0 < A^{∗} ≤ A.

(2) When B ≤ A, there is no transcendental meromorphic solution with only finitely many poles.

Then we get

(1.4)  T (r, f) = \frac{max(B − A, 0) + A^{∗}}{−2 log |q|} (log r)^{2}(1 + o(1)).

To prove Theorem 1.1, we need

Lemma 1.2 (i) Suppose f(z) has poles. Let z_{0} ≠ 0 be a pole with the smallest modulus, which is of order p. Then a(z_{0}) = 0 with order p_{0} ≥ p.

(ii) Let zeros of a(z) be 𝔷_{1}, …, 𝔷_{A}, counted with multiplicity. Put 𝔄 = {q^{−k}𝔷_{j}; 1 ≤ j ≤ A, k ∈ ℕ, k ≥ 1}. If 𝔓 is the set of poles of f(z), then 𝔓 ⊂ 𝔄.

(iii) Suppose the solution f(z) of (1.1) has infinitely many poles of the form 𝔓(𝔷) = {q^{−m_{j}}𝔷}, m_{0} = 0 < m_{1} < m_{2} < · · · . with multiplicities μ_{j}, respectively. Then 𝔷 is a zero of a(z) with order μ′_{0} ≥ μ_{0}. Further we can suppose that

(1.5) if z^{∗} ∈ 𝔓(𝔷) and a(z^{∗}) = 0, then b(q^{−k}z^{∗}) ≠ 0 for any k ∈ ℕ, k ≥ 1.

By (1.4) we see that f(z) may have only one deficiency (in the sense of Nevanlinna), which is necessarily deficient in the sense of Valiron. Further, the deficient value is asymptotic value for f(z) [7] p.30. In this respect, we have the following theorem:

Theorem 1.3 (a) If A < B in (1.1), then ∞ is the only deficiency. (b) If A > B and c(z) = d(a(z) − b(z)) for some d ∈ ℂ, then d is the only deficiency.

(c) If either A = B, or A > B without such d, then f(z) has no deficiency.

The results in this note would be naturally generalized to higher order qdifference equations. Our results would be a first step for further studies.

2 Proof of Lemma 1.2

(i): Let a(z) have a zero at z = z_{0} with order p_{0} ≥ 0. Suppose p_{0} < p. Then by (1.1) we see that qz_{0} would be also a pole for f(z), a contradiction.

(ii): We can suppose a_{0} = a(0) = 1. Then we can put

a(z) = ∏_{j=1}^{A} (1 − \frac{z}{𝔷_{j}} ) , h(z) = ∏_{k=0}^{∞} a(q^{k}z).

Putting g(z) = h(z)f(z), we get g(z) = b(z)g(qz) + c(z) ∏_{k=1}^{∞} a(q^{k}z), which shows that g(z) can not have any pole. Hence we see that 𝔓 ⊂ 𝔄.

(iii): The first half is proved as in (i).

Suppose b(q^{−k_{j}}𝔷) = 0, 0 < k_{1} < ⋯ < k_{p}, with order ν_{j}, respectively. Put

f_{1}(z) = ∏_{k=0}^{k_{1}−1} (1 − \frac{q^{k}z}{𝔷} )^{μ_{0}} f(z)

and let, according to ν_{1} ≥ μ_{0} or ν_{1} < μ_{0},

a_{1}(z) = a(z) (1 − \frac{z}{𝔷} )^{−μ_{0}} , b_{1}(z) = b(z) (1 − \frac{q^{k_{1}}z}{𝔷} )^{−μ_{0}} ,

c_{1}(z) = c(z) ∏_{k=1}^{k_{1}−1} (1 − \frac{q^{k}z}{𝔷} )^{μ_{0}} or

a_{1}(z) = a(z) (1 − \frac{z}{𝔷} )^{−μ_{0}} (1 − \frac{q^{k_{1}}z}{𝔷} )^{μ_{0}−ν_{1}} , b_{1}(z) = b(z) (1 − \frac{q^{k_{1}}z}{𝔷} )^{−ν_{1}} ,

c_{1}(z) = c(z) ∏_{k=1}^{k_{1}−1} (1 − \frac{q^{k}z}{𝔷} )^{μ_{0}} (1 − \frac{q^{k_{1}}z}{𝔷} )^{μ_{0}−ν_{1}} ,

then deg[a_{1}(z)] = A − μ_{0} (or A − ν_{1}), deg[b_{1}(z)] = B − μ_{0} (or B − ν_{1}), and

a_{1}(z)f_{1}(z) = b_{1}(z)f_{1}(qz) + c_{1}(z).

There are poles q^{−m_{j}}𝔷 of f_{1}(z), in which we can suppose that q^{−m_{1}}𝔷 is the one with the smallest m. Further, it can be supposed to be of order μ_{1}. (Note that it may be m_{1} ≤ k_{1} − 1.) If m_{1} > k_{p} − 1, we are done. Hence we can suppose m_{1} ≤ k_{2} − 1. Then put

f_{2}(z) = ∏_{k=0}^{k_{2}−m_{1}−1} (1 − \frac{q^{m_{1}+k}z}{𝔷} )^{μ_{1}} f_{1}(z)

and let, according to ν_{2} ≥ μ_{1} or ν_{2} < μ_{1},

a_{2}(z) = a_{1}(z) (1 − \frac{q^{m_{1}}z}{𝔷} )^{−μ_{1}} , b_{2}(z) = b_{1}(z) (1 − \frac{q^{k_{2}}z}{𝔷} )^{−μ_{1}} ,

c_{2}(z) = c(z) ∏_{k=1}^{k_{2}−m_{1}−1} (1 − \frac{q^{m_{1}+k}z}{𝔷} )^{μ_{1}} or

a_{2}(z) = a_{1}(z) (1 − \frac{q^{m_{1}}z}{𝔷} )^{−μ_{1}} (1 − \frac{q^{k_{2}}z}{𝔷} )^{μ_{1}−ν_{2}} , b_{2}(z) = b_{1}(z) (1 − \frac{q^{k_{2}}z}{𝔷} )^{−ν_{2}} ,

c_{2}(z) = c_{1}(z) ∏_{k=1}^{k_{2}−m_{1}−1} (1 − \frac{q^{m_{1}+k}z}{𝔷} )^{μ_{1}} (1 − \frac{q^{k_{2}}z}{𝔷} )^{μ_{1}−ν_{2}} ,

then we get a_{2}(z)f_{2}(z) = b_{2}(z)f_{2}(qz) + c_{2}(z), and deg[b_{2}(z)] − deg[a_{2}(z)] = deg[b(z)] − deg[a(z)]. Repeating this procedure a finite number of times, we get the result.

3 Proof of Theorem 1.1

(1) By Lemma 1.2 (iii), we assume that (1.5) holds. Suppose f(z) has poles {z_{n}}, 0 < |z_{1}| ≤ |z_{2}| ≤ ⋯, counted with multiplicity. By lemma 1.2 (i), a(z_{1}) = 0, hence b(q^{−k}z_{1}) ≠ 0, k ≥ 1. Then q^{−1}z_{1} is a pole for b(z)f(qz), hence also a pole for f(z), as seen from (1.1). Similarly, q^{−k}z_{1}, k ∈ ℕ, are poles for f(z). Suppose 𝔷_{ℓ} = q^{−k_{ℓ}}z_{1} is a zero of a(z). Then q^{−k}𝔷_{ℓ} = q^{−k_{ℓ}−k}z_{1}, k ∈ ℕ, are also poles for f(z), together with q^{−k}z_{1}.

Let z_{m} be the smallest pole which is not of this form. Then a(z_{m}) = 0, since otherwise qz_{m} is also a pole, as seen from (1.1), a contradiction. Then q^{−k}z_{m} are poles for f(z). Repeating this procedure, we see that there are zeros 𝔷_{1}, ⋯, 𝔷_{A^{∗}} of a(z), for which q^{−k}𝔷_{j}, 1 ≤ j ≤ A^{∗}, k ∈ ℕ, are poles.

Thus we obtain that n(r, f) = \frac{A^{∗}}{− log |q|} log r(1 + o(1)), and hence

N(r, f) = \frac{A^{∗}}{−2 log |q|}(log r)^{2}(1 + o(1)).

(2) Suppose f(z) is a meromorphic solution of (1.1) with only finitely many poles. By the procedure shown in Lemma 1.2 (iii), we transform (1.1) into an equation (2.1), which satisfies (1.5). f(z) is also transformed to f_{1}(z), which has no poles, since otherwise f(z) would have infinitely many poles. Hence f_{1}(z) must be entire. Since we suppose B ≤ A, we have deg[b_{1}(z)] ≤ deg[a_{1}(z)]. Then, by [2], (2.1) can not have transcendental entire solution. Therefore f(z) can not be transcendental meromorphic.

4 Proof of Theorem 1.3

When A < B : By (1.5), it is obvious that ∞ is the only one Valiron deficiency. When A ≥ B : Since m(r, f) = o((log r)^{2}), we see that ∞ is not deficient. We can assume that (1.5) holds. Using notations in the last section, let

a^{∗}(z) = (1 − \frac{z}{𝔷_{1}} ) ⋯ (1 − \frac{z}{𝔷_{A^{∗}}}), ã(z) = \frac{a(z)}{a^{∗}(z)}, h(z) = ∏_{k=0}^{∞} a^{∗}(q^{−k}z).

Take d ∈ ℂ and let f_{d}(z) = f(z) − d = g_{d}(z)/h(z), g_{d}(z) = g(z) − h(z)d. then

a(z)g_{d}(z) = b(z)a^{∗}(z)g_{d}(qz) + c_{d}(z)h(z), c_{d}(z) = c(z) − d(a(z) − b(z)).

Suppose c_{d}(z) ≢ 0. We have, writing A_{1} = max(A, B +A^{∗}) and C_{d} = deg[c_{d}(z)],

M(r, h) ≤ Kr^{A_{1}−C_{d}}M(r, g_{d}), K = const.,
m(r, \frac{1}{f_{d}}) = \frac{1}{2π} ∫_{0}^{2π} log^{+} |\frac{h(re^{iθ})}{g(re^{iθ})}| dθ ≤ log^{+} \frac{M(r, h)}{L(r, g_{d})},

where L(r, g) = inf_{|z|=r} |g(z)|. For E ⊂ [0, ∞), denote by \overline{lgd}(E) the upper logarithmic density of E. By [1] p.477 Theorem 6, for any δ > 0, there are a set E and a const. K_{1} > 0 such that \overline{lgd}(E) < δ and for r ∉ E we have

M(r, g_{d}) ≤ K_{1}L(r, g_{d}), hence m(r, \frac{1}{f_{d}}) ≤ log^{+} \frac{M(r, h)}{M(r, g_{d})} = O(log r).

We can take r′_{n}, r′′_{n} ∉ E such that 0 < log(r′′_{n}/r′_{n}) < δ log r′′_{n} → ∞. Then (1 − δ) log r′′_{n} < log r′_{n} and

N(r′_{n}, 1/f_{d}) = A^{∗}(log r′_{n})^{2}(1 + o(1)), N(r′′_{n}, 1/f_{d}) = A^{∗}(log r′′_{n})^{2}(1 + o(1))

by (1.5). Hence for r′_{n} < r < r′′_{n}, we have

(1 − δ)^{2} \frac{N(r′_{n}, 1/f_{d})}{(log r′_{n})^{2}} ≤ \frac{N(r′_{n}, 1/f_{d})}{(log r′_{n})^{2}} \frac{(log r′_{n})^{2}}{(log r)^{2}} ≤ \frac{N(r, 1/f_{d})}{(log r)^{2}}

≤ \frac{N(r′′_{n}, 1/f_{d})}{(log r′′_{n})^{2}} \frac{(log r′′_{n})^{2}}{(log r)^{2}} ≤ \frac{N(r′′_{n}, 1/f_{d})}{(log r′′_{n})^{2}} \frac{1}{(1 − δ)^{2}}

Since δ > 0 is arbitrary, we see that d is not deficient.

Suppose c_{d}(z) ≡ 0. We get ã(z)g_{d}(z) = b(z)g_{d}(qz), ã(z) = a(z)/a^{∗}(z). Write B^{∗} = B − deg[ã(z)]. Obviously

N(r, 1/g_{d}) = \frac{B^{∗}}{−2 log |q|}(log r)^{2}(1 + o(1)).

When A = B, then B^{∗} = A^{∗} and d is not deficient. When A > B, then B^{∗} < A^{∗}, and d is deficient.

5 Some examples

Ex. 1. A zero of a(z) in (1.1) does not necessarily give a pole of f(z). In fact, solution f(z) of (1 − qz)f(z) = (1 − z)(1 + z)f(qz) is entire:

f(z) = (1 − z) ∏_{n=0}^{∞}(1 + q^{n}z).

Ex. 2. The equation (1 − z)f(z) = (1 − q^{m}z)(1 + z)f(qz) gives solution with only finitely many poles:

f(z) = ∏_{n=0}^{m−1}(1 − q^{n}z)^{−1} ∏_{n=0}^{∞}(1 + q^{n}z).

Now consider meromorphic solution f_{Q}(z) = ∏_{n=0}^{∞}{(1 − q^{n}z)/(1 + q^{n}z)} of the equation

(5.1)  (1 + z)f_{Q}(z) = (1 − z)f_{Q}(qz), 0 < q < 1.

Ex. 3. Suppose 0 < q < 1 in (5.1). Although the solution f_{Q}(z) has no deficiency by Theorem 1.3, it has no Julia direction. In fact, it is not difficult to see that |f_{Q}(z)| ≤ K in {| arg[z]| < 7π/8} with some K > 0, and |f_{Q}(z)| > 1 in {| arg[z] − π| < π/4}. f_{Q}(z) is one of “Juliasche Ausnahmefunktionen” [6].

We will show further, with respect to a result of Valiron [7] p.30, that f_{Q}(z) has no asymptotic value. In fact, suppose f_{Q}(z) tends to a on an asymptotic path C. Write simply f(z) for f_{Q}(z) and let f(z) = f_{1}(z)/f_{2}(z), with

f_{1}(z) = ∏_{n=0}^{∞}(1 − q^{n}z) and f_{2}(z) = ∏_{n=0}^{∞}(1 + q^{n}z).

Then M(r, f_{1}) = M(r, f_{2}). For 0 < δ < 1, there is a set E ⊂ [1, ∞) with \overline{lgd}(E) < δ such that L(r, f_{j})/M(r, f_{j}) ≥ K(δ) for r ∉ E [1] p.477 Theorem 6, where L(r, f_{j}) denotes the minimum modulus. Since for |z| = r ∉ E,

K(δ) = \frac{K(δ)M(r, f_{1})}{M(r, f_{2})} ≤ \frac{L(r, f_{1})}{M(r, f_{2})} ≤ |f(z)| ≤ \frac{M(r, f_{1})}{L(r, f_{2})} ≤ \frac{1}{K(δ)},

we have K(δ) ≤ |a| ≤ 1/K(δ).

Suppose the path C intersects with the imaginary axis infinitely many times. Then |a| = 1 since |f(iy)| = 1. Let C^{∗} be such that x + iy ∈ C^{∗} if and only if −x + iy ∈ C. Since f(−x + iy) = 1/\overline{f(x + iy)}, f(z) tends to 1/\overline{a} on C^{∗}. As a = 1/\overline{a}, f(z) tends to a in a region surrounded by C and C^{∗}, hence also on the imaginary axis, a contradiction.

Suppose C lies in the right half plane. Since |f(z)| < 1 there, by the Theorem of Lindelöf [3] p.19 Theorem 2.3 f(z) tends to a in any angular domain. Hence f(z) → a on the real axis, which is a conradiction since f(q^{−n}) = 0. Supposition that C lies in the left half plane leads also to a contradiction. □ Ex. 4. Suppose in (5.1) that q satisfies

q = |q|e^{2πλi}, 0 < |q| < 1, λ ∈ (0, 1) \ ℚ.

We will show, in contrast to the case λ ∈ ℚ as shown Ex.3, that meromorphic solution f_{Q}(z) of (5.1) takes any value a ∈ ℂ∪{∞} infinitely often in any sector

Ω_{αβ} = {z ; α < arg[z] < β}.

We write simply f(z) for f_{Q}(z). Let d_{n} = min{the decimal part of jλ ; 1 ≤ j ≤ n} and

n_{0} = min{n ; d_{n} < (β − α)/8}.

For |z| > |q|^{−2}, write z = q^{−k(z)}ζ(z), k(z) ∈ 2ℕ, |q| < |ζ(z)| ≤ |q|^{−1}. Then

(5.2)  f(z) = ∏_{n=0}^{∞} \frac{1 − q^{n−k(z)}ζ(z)}{1 + q^{n−k(z)}ζ(z)} = ∏_{n=0}^{∞} \frac{1 − q^{n}ζ(z)}{1 + q^{n}ζ(z)} · ∏_{n=1}^{k(z)} \frac{1 − q^{n}/ζ(z)}{1 + q^{n}/ζ(z)}.

Write

ϕ(ζ) = f(ζ)f(\frac{q}{ζ}) = f(ζ) ∏_{n=1}^{∞} \frac{1 − q^{n}/ζ}{1 + q^{n}/ζ} .

Our assertion is obvious for a = 0, ∞. Since a ≠ 0, ∞ is not deficient, there are {z_{m}} such that f(z_{m}) = a. If ζ(z_{m}) accumlate to ζ^{∗}, |q| ≤ |ζ^{∗}| ≤ |q|^{−1}, then ζ(z_{m}) ↛ −1, −q, −1/q, and ϕ(ζ^{∗}) = a. Such ζ^{∗} are finite in number, and write them as ζ_{j}^{∗}, 1 ≤ j ≤ J. Let γ_{j} be a circle with center ζ_{j}^{∗} and radius ρ_{j}, where ρ_{j} < |ζ_{j}^{∗}| sin \frac{β−α}{4}, such that |ϕ(ζ) − a| > 0 for ζ ∈ γ_{j}. Then |f(z) − a| ≥ η for some η > 0 if ζ(z) ∈ γ_{j}. We can suppose γ_{j} are outside γ_{j′}, j′ ≠ j.

By (5.2), |f(z)| ≤ K_{j} for some K_{j} > 0 if ζ(z) ∈ γ_{j}.

Put α′ = α + (β − α)/4, β′ = β − (β − α)/4. For z_{m} there is n_{1}, 0 ≤ n_{1} ≤ n_{0}^{2} such that q^{−n_{1}}z_{m} ∈ Ω_{α′β′}. Then

Δ(z_{m}) = {z ; |z − q^{−n_{1}}z_{m}| ≤ |q|^{−(n_{1}+k(z_{m}))}ρ_{j}} ⊂ Ω_{αβ}.

We have, noting that we can take n_{1} to be an even number,

f(q^{−n_{1}}z_{m}) = ∏_{k=1}^{n_{1}} \frac{1 − q^{−k}z_{m}}{1 + q^{−k}z_{m}} f(z_{m}) = ∏_{k=1}^{n_{1}} \frac{q^{−k}z_{m} − 1}{q^{−k}z_{m} + 1}f(z_{m})

Write Γ_{m} = ∂Δ(z_{m}). Take |z_{m}| so large that

|∏_{k=1}^{n_{1}} \frac{q^{−k}z − 1}{q^{−k}z + 1} − 1| K_{j} < \frac{η}{2}

for z ∈ Γ_{m}. Then by the Theorem of Rouché, f(z) takes a in Ω_{αβ}, which proves our assertion. □ Ex. 5. Consider the equation

(1 + z)g(z) = q(1 − z)g(qz), 0 < q < 1.

A solution is g(z) = z^{−1} ∏_{n=0}^{∞} \frac{1 − q^{n}z}{1 + q^{n}z} , which admits no deficiency by Theorem

1.3, while we will show, in contrast to Ex. 3, that g(z) has an asymptotic value 0 in {z ; −π+δ < arg[z] < π−δ} for any δ > 0, since for z = x+ikx (k = ± tan δ),

|zg(z)|^{2} = ∏_{n=0}^{∞} \frac{(1 − q^{n}x)^{2} + (kq^{n}x)^{2}}{(1 + q^{n}x)^{2} + (kq^{n}x)^{2}} = ∏_{n=0}^{∞} (1 + \frac{−4q^{n}x}{(1 + q^{n}x)^{2} + (kq^{n}x)^{2}} )

is bounded for x < 0, supposed k ≠ 0. In fact, write the product on the right hand side as Π_{1} · Π_{2}, ·Π_{3}, with Π_{j} = ∏_{E_{j}}, j = 1, 2, 3, where E_{j} = E_{j}(x) are

E_{1} = \{n ; q^{n}x ≤ −\frac{1}{q}\}, E_{2} = \{n ; −\frac{1}{q} < q^{n}x < −q\}, E_{3} = \{n ; q^{n}x ≥ −q\}.

(E_{2} may be empty.) Let n_{j} = n_{j}(x) be such that

n_{1} = max_{E_{1}} n, n_{2} ∈ E_{2}, n_{3} = min_{E_{3}} n.

For Π_{1} = ∏_{n=0}^{n_{1}}, writing n = n_{1} − ℓ, 0 ≤ ℓ ≤ n_{1}, we have

\frac{−4q^{n}x}{(1 + q^{n}x)^{2} + (kq^{n}x)^{2}} ≤ \frac{−4q^{n}x}{k^{2}(q^{n}x)^{2}} = \frac{4}{k^{2}} \frac{1}{−q^{n}x} = \frac{4}{k^{2}} \frac{q^{ℓ}}{q^{n_{1}}x} ≤ \frac{4}{k^{2}} q^{ℓ+1},

Hence

Π_{1} ≤ ∏_{ℓ=0}^{n_{1}} (1 + \frac{4}{k^{2}} q^{ℓ+1}) ≤ ∏_{ℓ=0}^{∞} (1 + \frac{4}{k^{2}} q^{ℓ+1}) = K_{1}.

For Π_{2}, we have E_{2} ⊂ {n_{2} − 1, n_{2}, n_{2} + 1}, since q^{n_{2}+2}x > −q, q^{n_{2}−2}x < −1/q. Then

\frac{−4q^{n}x}{(1 + q^{n}x)^{2} + (kq^{n}x)^{2}} ≤ \frac{4}{k^{2}} \frac{1}{−q^{n}x}
≤ \frac{4}{k^{2}} (n = n_{2} − 1); ≤ \frac{4}{k^{2}} \frac{1}{q} (n = n_{2}); ≤ \frac{4}{k^{2}} \frac{1}{q^{2}} (n = n_{2} + 1).

Hence

Π_{2} ≤ ∏_{ℓ=0}^{2} (1 + \frac{4}{k^{2}} \frac{1}{q^{ℓ}} ) = K_{2}.

For Π_{3} = ∏_{n=n_{3}}^{∞}, writing n = n_{3} + ℓ, 0 ≤ ℓ, we have, since 1 + q^{n}x ≥ 1 − q,

\frac{−4q^{n}x}{(1 + q^{n}x)^{2} + (kq^{n}x)^{2}} ≤ \frac{−4q^{n_{3}+ℓ}x}{(1 − q)^{2}} ≤ \frac{4q^{ℓ+1}}{(1 − q)^{2}} .

Hence

Π_{3} ≤ ∏_{ℓ=0}^{∞} (1 + \frac{4q^{ℓ+1}}{(1 − q)^{2}} ) = K_{3}.

Therefore, by the theorem of Gross and Iversen [3] p.101 Theorem 5.8, g(z) takes any value a, 0 < |a| ≤ ∞, infinitely often in sector {| arg[z] − π| < δ}, containing the direction {arg[z] = π}, for any δ > 0.

References

[1] P.D. Barry: The minimum modulus of small integral and subharmonic functions. Proc. London Math. Soc. (3) 12 (1962), 445–495.

[2] W. Bergweiler, K. Ishizaki, and N. Yanagihara: Growth of meromorphic solutions of some functional equations I. Aeq. Math. 63 (2002), 140–151.

[3] E.F. Collingwood and A.J. Lohwater: The Theory of Cluster Sets. Cambrigde Univ. Press 1966.

[4] W.K. Hayman: Meromorphic Functions. Oxford Univ. Press 1964.

[5] R. Nevanlinna: Analytic Functions. Springer-Verlag 1970.

[6] A. Ostrowski: Über Folgen analytischer Funktionen und einige Verschärfungen des Picardschen Satzes. Math. Zeitschr. 24 (1925), 215–258.

[7] G. Valiron: Sur les valeurs déficientes des fonctions algebroïdes. J. d’Analyse Math. 1 (1951), 28–42.

Ilham Eli Department of Mathematics, Kashgar Teacher’s College, Xinjiang, CHINA.

Present address of Professor Ilham Eli: Department of Mathematics, Faculty of Science, Chiba University, 1-33 Yayoicho, Inage-ku, Chiba City, Chiba-ken, 263-8522 JAPAN.

Niro Yanagihara Department of Mathematics, Faculty of Science, Chiba University, 1-33 Yayoicho, Inage-ku, Chiba City, Chiba-ken, 263-8522 JAPAN.