Entire solutions of q-difference equations

Ilham Eli and Niro Yanagihara

Abstract Let f(z) be a transcendental entire solution of a q-difference equation f(z) = a(z)f(qz)+b(z), with a(z), b(z) ∈ ℂ[z], deg[a(z)] ≥ 1, 0 < |q| < 1. Write q = |q|e^{2πiλ}. We will show here that, if λ ∉ ℚ, then, for any direction arg[z] = ω_{0}, f(z) takes any finite value infinitely often in any small sector {z ; | arg[z] − ω_{0}| < α} for any α. Further we will show that, if λ ∈ ℚ, then for any κ ∈ ℕ, κ ≥ 1, there is an equation for which solution f(z) has exactly κ direction(s) with this property.

Keywords and phrases. linear q-difference equation, transcendental entire solution, Julia direction. AMS Subject Classification. 39A13, 30D20

1 Introduction

Let f(z) be a transcendental entire solution, supposing that there exists any, of the equation

f(z) = a(z)f(qz) + b(z), 0 < |q| < 1, (1.1)

with a(z) = ∑_{k=0}^{A} a_{k}z^{k} (A ≥ 1) and b(z) = ∑_{k=0}^{B} b_{k}z^{k}. Wittich [7] p.72 proved that f(z) satisfies

(1.2) log M(r, f) = \frac{A}{−2 log |q|}(log r)^{2}(1 + o(1)), M(r, f) = max_{|z|=r} |f(z)|,

and does not satisfy any algebraic differential equation P (z, f, f′, ⋯, f^{(n)}) = 0. The relation (1.2) has been generalized to solutions of higher order linear qdifference equations [2].

It is known that any finite value is not deficient in the sense of Valiron [6] p.30 (5). Write

q = |q|e^{2πiλ}, 0 < |q| < 1, λ ∈ [0, 1). (1.3)

Let ω_{0} ∈ [0, 2π), α ∈ (0, π), and

d_{ω_{0}} = {z ; arg[z] = ω_{0}}, Ω(ω_{0}, α) = {z ; | arg[z] − ω_{0}| < α}. (1.4)

We will say that d_{ω_{0}} is a ∗-Julia direction if f(z) takes any finite value infinitely often in Ω(ω_{0}, α) for any α > 0.

Note that, in the classical definition of Julia direction [4], there may be at most one exceptional value, while there is no such value in our sense.

We will show in this note the following theorems.

Theorem 1.1 Suppose λ ∉ ℚ in (1.3). Let f(z) be a transcendental entire solution of (1.1). Then any direction is ∗-Julia for f(z).

Theorem 1.2 Suppose λ ∈ ℚ in (1.3). For any κ ∈ ℕ, κ ≥ 1, there is an equation (1.1) such that its trascendental entire solution f(z) has exactly κ directions which are ∗-Julia.

For the proof, we need the following Lemma:

Lemma 1.3 Let f(z) be transcendental entire. Suppose f(z) satisfies

log M(r, f) = O((log r)^{2}). (1.5)

Then, for any direction d_{ω_{0}}, there holds either

(i) d_{ω_{0}} is ∗-Julia for f(z), or

(ii) there is α > 0 such that f(z) → ∞ uniformly as z → ∞ in Ω(ω_{0}, α).

Hence, Julia direction is ∗-Julia for f(z) satisfying (1.5).

2 Proof of Lemma 1.3

Suppose d_{ω_{0}} is not ∗-Julia for f(z). There is α > 0 and a ∈ ℂ such that f(z) takes a only finitely many times in the sector Ω(ω_{0}, α). We can suppose a = 0. Then f(z) ≠ 0 for z ∈ Ω(ω_{0}, α) ∩ {|z| ≥ R_{0}} for an R_{0}.

Let n(r) = n(r, 1/f) be the number of zeros, with multiplicity, of f(z) in |z| ≤ r and

N(r) = ∫_{0}^{r} \frac{n(t)}{t}dt, Q(r) = r ∫_{r}^{∞} \frac{n(t)}{t^{2}}dt.

Let M(r) = M(r, f). For the Nevanlinna characteristic T (r, f) [5] p.16, we have

N(r) ≤ T (r, f) + O(1) ≤ 3M(2r) + O(1) = O((log r)^{2}), and
n(r) log r = n(r) ∫_{r}^{r^{2}} t^{−1}dt ≤ ∫_{r}^{r^{2}} t^{−1}n(t)dt ≤ N(r^{2}) = O((log r)^{2}),

hence we have n(r) = O(log r), and

Q(r) = O(r ∫_{r}^{∞} t^{−2} log tdt) = O(r^{1/2} log r ∫_{r}^{∞} t^{−3/2}dt) = O(log r).

Let σ > 1 be given. By the Boutroux–Cartan lemma we get [3] p.50,

(2.1) log |f(z)| ≥ n(σr)(\frac{1}{σ} + log \frac{H(r)}{σr}) + log M(σr) − (1 + \frac{1}{σ − 1})Q(σr)

outside a collection of discs which are at most n(σr) in number, contain zeros of f(z) in |z| ≤ σr, and the sum of whose radii is 2eH(σr), where H(σr) = o(σr) = o(r).

Hence (2.1) holds for z ∈ Ω(ω_{0}, α/2)∩{|z| > r/2} = Ω_{r}, since dis(Ω_{r}, Ω^{C}) ≥ K_{1}r for some K_{1} > 0, where Ω^{C} denotes the complement of Ω(ω_{0}, α).

Since n(r) ≤ Q(r), we get [3] p.50, and (2.1)

log |f(z)| ≥ log M(r) − Δ(r)Q(σr) (2.2)

holds for z ∈ Ω_{r}, where Δ(r) → ∞ arbitrarily slowly, which proves our assertion.

3 Proof of Theorem 1.1

Put s = q^{−1} and write the equation (1.1) in the form

f(sz) = a_{0}(z)f(z) + b_{0}(z), a_{0}(z) = a(sz), b_{0}(z) = b(sz). (3.1)

Then f(s^{2}z) = a_{0}(sz)f(sz)+b_{0}(sz) = a_{1}(z)f(z)+b_{1}(z) with a_{1}(z) = a_{0}(sz)a_{0}(z), b_{1}(z) = a_{0}(sz)b_{0}(z) + b_{0}(sz). In general

\begin{aligned}f(s^{1+m}z) & = a_{m−1}(sz)f(sz) + b_{m−1}(sz) = a_{m}(z)f(z) + b_{m}(z), \\ a_{m}(z) & = a_{m−1}(sz)a_{0}(z) = ∏_{k=0}^{m} a_{0}(s^{k}z) = ∏_{k=1}^{m+1} a(s^{k}z), \\ b_{m}(z) & = a_{m−1}(sz)b_{0}(z) + b_{m−1}(sz). \end{aligned} (3.2)

Hence deg[a_{m}(z)] = (m + 1)A, deg[b_{m}(z)] = mA + B (A = deg[a(z)], B = deg[b(z)]). Put p = B − A and

\begin{aligned}f(s^{1+m}z) & = a_{m}(z)𝔄_{m}(z, f), 𝔄_{m}(z, f) = f(z) + 𝔟_{m}(z), \\ 𝔟_{m}(z) & = b_{m}(z)/a_{m}(z) = Bz^{p}(1 + O(1/z)). \end{aligned} (3.3)

Since a(z) ≠ 0 for |z| > R_{0} for some R_{0}, we have a_{m}(z) ≠ 0 for |z| > R_{0}.

By Lemma 1.3, there is a ∗-Julia direction d_{ω^{∗}} for f(z). That is, for any ν ∈ ℕ, f(z) takes any finite value infinitly often in the sector Ω_{ν} = Ω(ω^{∗}, 2π/2^{ν}).

Let Ω(ω_{0}, α) be given. Take ν_{0} such that 2π/2^{ν_{0}} < α/4. We denote zeros of f(z) in Ω_{ν_{0}} as {z_{n}}. Suppose m be fixed, and let {z_{n}^{(m)}} be zeros of 𝔄_{m}(z, f). We will show later that, in obvious convention, we have

z_{n}^{(m)} = z_{n}(1 + o(1)), as z_{n} → ∞. (3.4)

Now the irrationality of λ in (1.3) plays its role.

Let m be fixed such that |(1 + m)λ + ω^{∗} − ω_{0}| < α/8 (mod 2π).

In (3.4), let z_{n} be so large that | arg[z_{n}^{(m)}] − arg[z_{n}]| < \frac{1}{8}α. Making use of (3.3), we get f(s^{1+m}z_{n}^{(m)}) = 0 for s^{1+m}z_{n}^{(m)} in Ω_{r} = Ω(ω_{0}, α/2) ∩ {|z| > r}. Hence (2.2) does not hold, and d_{ω_{0}} is ∗-Julia by Lemma 1.3.

Now we will show (3.4) and finish the proof. Hereafter, m is not fixed.

We can suppose f(0) ≠ 0. For simplicity, we suppose f(0) = 1. Then

f(z) = ∏_{n=0}^{∞}(1 − \frac{z}{z_{n}} ),  r = |z|, r_{n} = |z_{n}|,

\frac{|f(z)|}{M(r, f)} ≥ ∏_{n=0}^{n_{1}} \frac{|1 − r/r_{n}|}{1 + r/r_{n}} ∏_{n=n_{1}+1}^{n_{3}−1} \frac{|1 − z/z_{n}|}{1 + r/r_{n}} ∏_{n=n_{3}}^{∞} \frac{1 − r/r_{n}}{1 + r/r_{n}} = Π_{1} · Π_{2} · Π_{3},

where n_{1}, n_{3} are to be determined later.

For E ⊂ [1, ∞), let \overline{lgd}(E) be the upper logarithmic density of E. By [1] p.477 Theorem 6, for any δ < 1, there is a set E with \overline{lgd}(E) < δ such that

|f(z)| ≥ K_{δ}M(r, f) for |z| = r ∉ E, with a constant K_{δ}.

Take r′, r′′ ∉ E, 0 < log \frac{r′′}{r′} ≤ δ log r′′, with r′ > R_{0} and 2|B|(r′)^{p+1} < M(r′, f). Then f(z) and 𝔄_{m}(z, f) have the same number of zeros in {r′ < |z| < r′′}, by the Rouché theorem. Denote them z_{n} and z_{n}^{(m)}, 1 ≤ n ≤ μ, respectively. By (3.3) f(s^{1+m}z_{n}^{(m)}) = 0, that is, z_{n}^{[m]} = s^{1+m}z_{n}^{(m)}, 1 ≤ n ≤ μ, are zeros of f(z).

Thus, zeros z^{[m]} of f(z) are located as |s|^{1+m}r′ < |z^{[m]}| < |s|^{1+m}r′′. Since (1 − x)/(1 + x) is decreasing, we get, with obvious convention for ν_{3},

Π_{3} = ∏_{n=n_{3}}^{∞} \frac{1 − r/r_{n}}{1 + r/r_{n}} ≥ \{ ∏_{m=ν_{3}}^{∞} \frac{1 − r/(|s|^{m}r′)}{1 + r/(|s|^{m}r′)}\}^{μ} ≥ exp [− ∑_{m=ν_{3}}^{∞} \frac{4μr}{|s|^{m}} · \frac{1}{r′} ]

= exp [−\frac{4μr}{|s|^{ν_{3}}} · \frac{|s|}{|s| − 1} · \frac{1}{r′} ] ≥ exp [−\frac{2μ|s| log 2}{|s| − 1}] = K_{3},

supposed that 2r/(|s|^{ν_{3}}r′) ≤ log 2, because of 1 − x ≥ e^{−2x} and 1 + x < e^{2x} for 0 ≤ x ≤ (log 2)/2.

Similarly, with a convention for ν_{1},

Π_{1} = ∏_{n=0}^{n_{1}} \frac{r/r_{n} − 1}{r/r_{n} + 1} = ∏_{n=0}^{n_{1}} \frac{1 − r_{n}/r}{1 + r_{n}/r} ≥ \{ ∏_{m=0}^{ν_{1}} \frac{1 − (|s|^{m}r′′)/r}{1 + (|s|^{m}r′′)/r} \}^{μ}

≥ exp [−4μr′′ ∑_{m=0}^{ν_{1}} \frac{|s|^{m}}{r} ] ≥ exp [−\frac{4μr′′|s|^{ν_{1}}|s|}{r(|s| − 1)}] ≥ exp [−\frac{2μ|s| log 2}{|s| − 1}] = K_{1},

supposed that 2|s|^{ν_{1}}r′′/r ≤ log 2. Hence

ν_{3} − ν_{1} = ν^{∗} + 2, ν^{∗} = [\frac{log\{(4/(log 2)^{2})(r′′/r′))\}}{log |s|}] − 1,
(3.5)  n_{3} − n_{1} − 2 = n_{0} = μν^{∗},

where [x] denotes the greatest integer not exceeding x. Note that n_{0} does not depend on r, while n_{1}, n_{3} do. Thus we obtain

\frac{|f(z)|}{M(r, f)} ≥ K_{1}K_{3} \frac{|P (z)|}{∏_{n=n_{1}+1}^{n_{3}−1}(r + r_{n})}, P (z) = ∏_{n=n_{1}+1}^{n_{3}−1}(z − z_{n}),

where deg[P (z)] = n_{3} − n_{1} − 2 = n_{0}. By the Boutroux–Cartan Lemma [3] p.46, |P (z)| ≥ (H/e)^{n_{0}} outside a collection of discs the sum of whose radii is H. When H = ηr (η > 0), we have in the outside of exceptional discs,

|f(z)| ≥ K_{1}K_{3}(c^{∗}η)^{n_{0}}M(r, f), c^{∗} = r′ log 2/{e(2r′′ + r′ log 2)}, (3.6)

since r_{n_{3}−1} < |s|^{ν_{3}−1}r′′ = (r′′/r′)|s|^{ν_{3}−1}r′ < (r′′/r′)\frac{2}{log 2}r.

Let C_{n} = {|z−z_{n}| = |z_{n}| tan(2π/2^{ν_{0}})}, C_{n}^{∗} = {|z−z_{n}| = |z_{n}| tan(2π/2^{ν_{0}+1})}. Since the sum of radii of discs to be excluded is at most H = ηr, taking η sufficiently small, there is a closed curve L_{n} between C_{n} and C_{n}^{∗} such that L_{n} encloses a zero z_{n} of f(z) and is in the outside of exceptional discs. By the theorem of Rouché, we see that there is a zero z_{n}^{(m)} of 𝔄_{m}(z, f) in the inside enclosed by L_{n}, which proves (3.4).

4 Proof of Theorem 1.2

Suppose λ = τ/σ, τ, σ ∈ ℕ. As in (3.2), we obtain

f(z) = a_{2σ}(z)f(q^{2σ}z) + b_{2σ}(z).

Hence we can suppose 0 < q < 1 in (1.1).

For κ ≥ 1, consider the equation

f(z) = (1 − z^{κ})f(qz), 0 < q < 1. (4.1)

Then f(z) = ∏_{n=0}^{∞}(1 − q^{κn}z^{κ}). We have

f(z) = h(z^{κ}), h(z) = (1 − z)h(q^{κ}z) = ∏_{n=0}^{∞}(1 − q_{0}^{n}z), (4.2)

where we write q^{κ} = q_{0}.

For δ ∈ (0, π/2), write k_{0} = ± tan δ ≠ 0. When z = x + ik_{0}x, we will show that |h(x + ik_{0}x)| → ∞ as x → +∞.

Let t_{0} = min{t ; q_{0}^{t} < 1/3}. Put n_{1} = n_{1}(x) = max{n ; q_{0}^{n}x ≥ q_{0}^{−t_{0}}}, n_{3} = min{n ; q_{0}^{n}x ≤ q_{0}}. Then

|h(x + ik_{0}x)|^{2} = ∏_{n=0}^{n_{1}} · ∏_{n=n_{1}+1}^{n_{3}−1} · ∏_{n=n_{3}}^{∞}{(1 − q_{0}^{n}x)^{2} + k_{0}^{2}q_{0}^{2n}x^{2}} = Π_{1} · Π_{2} · Π_{3}.

For 0 ≤ n ≤ n_{1} we have (1 − q_{0}^{n}x)^{2} + k_{0}^{2}q_{0}^{2n}x^{2} ≥ (q_{0}^{n}x − 1)^{2} > 2^{2}, hence

Π_{1} > 2^{2n_{1}}.

For n_{1} + 1 ≤ n ≤ n_{3} − 1, we have (1 − q_{0}^{n}x)^{2} + k_{0}^{2}q_{0}^{2n}x^{2} ≥ k_{0}^{2}q_{0}^{2}, hence

Π_{2} ≥ (k_{0}^{2}q_{0}^{2})^{t_{0}+2}.

For n ≥ n_{3}, we have (1 − q_{0}^{n}x)^{2} + k_{0}^{2}q_{0}^{2n}x^{2} ≥ (1 − q_{0}^{n−n_{3}})^{2}, hence

Π_{3} ≥ ∏_{n=0}^{∞}(1 − q_{0}^{n})^{2}.

Since n_{1} → ∞ as x → ∞, we have that |h(x + ik_{0}x)| → ∞. We note h(z) ≠ 0 for z ∈ Ω(π, π − δ). Hence h(z) → ∞ in Ω(π, π − δ) uniformly.

Thus h(z) has the positive real axis as the only ∗-Julia direction. Therefore f(z) admits ∗-Julia ones which are exactly κ in number.

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] R.P. Boas, Jr.: Entire Functions. Academic Press 1954.

[4] G. Julia: Leçons sur les fonctions uniformes à point singulier essentiel isolé. Collection Borel, Paris 1924.

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

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

[7] H. Wittich: Neuere Untersuchungen über eindeutige analytische Funktionen. Ergebnisse der Math. Grenzgeb. Heft 8. Spriger–Verlag 1955.

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.