Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 19, 1994, 67–81

FURTHER RESULTS ON BOREL REMOVABLE SETS OF ENTIRE FUNCTIONS

Shengjian Wu

Peking University, Department of Mathematics Beijing 100871, People’s Republic of China

Abstract. In this paper, the following result is proved. Let (R_{m}) be a sequence of real numbers such that lim_{n→∞} R_{m+1}/R_{m} = +∞ and let (φ_{m}) be a sequence of real numbers such that 0 ≤ φ_{m} ≤ 2π. Suppose that η(0 < η < π) and S (> 1) are two constants. If E = ∪_{m=1}^{∞}D_{m}, where D_{m} = {z = re^{iθ}, R_{m} ≤ r ≤ SR_{m}} \ {z = re^{iθ}, φ_{m} − η < θ < φ_{m} + η} (m = 1, 2, …), then Borel’s theorem holds in C \ E for every entire function f(z) of positive order.

1. Introduction

Suppose that f(z) is meromorphic in the plane and that the order λ (0 < λ ≤ +∞) is defined by

λ = lim sup_{r→∞} \frac{log T (r.f)}{log r}.

Then Borel’s theorem asserts that

lim sup_{r→∞} \frac{log n(r, a)}{log r} = λ,

except for at most two values a for which the upper limit can be smaller, where we use standard notation from [3], as we shall do throughout.

Following the idea of Picard sets [5], L. Yang introduced the notation of Borel removable set [6]: a point set E ⊂ C is called a Borel removable set for a family ℱ of meromorphic functions, if for any meromorphic function f(z) ∈ ℱ with nonzero order λ_{f}, Borel’s theorem always holds in C \ E. That is, we have

lim sup_{r→∞} \frac{log n\{(|z| ≤ r) \ E, f = a\}}{log r} = λ_{f}

for all a ∈ \overline{C}, except for at most two values, where n\{D, f = a\} denotes the roots of f(z) − a = 0 in D including multiplicities.

1991 Mathematics Subject Classification: Primary 30D30.

Project supported by the Science Fund of the Chinese Academy of Sciences.

It is obvious that if E is a Borel removable set for ℱ, then E is also a Picard set for ℱ.

Since O. Lehto introduced the concept of Picard set in 1958, the Picard sets have been extensively studied for entire and meromorphic functions. But there is very little work done on Borel removable sets. In general, the Picard sets of entire functions can only contain a sequence of discs with extremely small radii. Since we need not consider the entire functions of order zero in the study of Borel removable sets, we would expect to obtain some results which have some important difference from the known Picard sets. Let ℱ be the family of entire functions of positive and finite order. The following result is proved in [8].

Theorem A. Let (a_{n}) be a sequence of complex numbers which satisfy the condition that |a_{n+1}| > |a_{n}|^{1+σ} for a positive constant σ and let (ε_{n}) be a sequence of positive numbers such that lim_{n→∞} ε_{n} = 0. Then the set

E = ⋃_{n=1}^{∞}{z; |z − a_{n}| < ε_{n}|a_{n}|}

is a Borel removable set for ℱ.

It is easily seen that in Theorem A if the sequence of (ε_{n}) is suitably chosen, then we may have lim_{n→∞} ε_{n}|a_{n}| = +∞.

In this paper, we shall prove that the Borel removable sets for all entire functions whose orders are finite or infinite can contain a sequence of large sectorial domains. Using simple examples, we shall show that these results are best possible in some sense. In the sequal for real numbers φ; η > 0; ϱ > 0 and S > 1, we shall define the angular domain | arg z − φ| < η by G(φ, η), and the sectorial domain {z = re^{iθ}, ϱ ≤ r ≤ Sϱ} \ G(φ, η) by D(ϱ, S, φ, η). We shall also use ℱ and ℱ_{∞} to stand for the family of all the entire functions of finite and nonzero order, and infinite order, respectively.

2. Statement and discussion of results

For ℱ, we have the following

Theorem 1. Let (R_{m}) be a sequence of real numbers such that

(2.1)  lim_{m→∞} \frac{R_{m+1}}{R_{m}} = ∞

and let (φ_{m}) be a sequence of real numbers. Suppose that η (0 < η < π) and S (> 1) are two constants. Then

E = ⋃_{m=1}^{∞} D_{m}

is a Borel removable set for ℱ, where D_{m} = D (R_{m}, S, φ_{m}, η) (m = 1, 2, …).

Remark 1. Theorem 1 is sharp in the following sense. If we relax the gap condition (2.1) by R_{m+1} ≥ KR_{m} for some K > 1, we may let S = K. In this case, if we let φ_{m} = 0, then the set E in Theorem 1 may be the angular domain G(π, π − η). Since for λ (0 < λ ≤ \frac{1}{2} ), the function

f(z) = ∏_{n=1}^{∞}(1 + \frac{z}{n^{1/λ}} )

is of order λ and f(z) → ∞ uniformly for | arg z| < π − δ for any positive δ, G(π, π − η) is not a Borel removable set for ℱ.

For ℱ_{∞}, the condition (2.1) can be weaker. Actually we have the following

Theorem 2. Let (R_{m}) be a sequence of real numbers such that

(2.2)  R_{m+1} > σR_{m},

where σ > 1 is a constant, and let (φ_{m}) be a sequence of real numbers such that 0 ≤ φ_{m} < 2π. Suppose that η (0 < η < π) and S (1 < S < σ) are two constants. Then

E_{∞} = ⋃_{m=1}^{∞} D′_{m}

is a Borel removable set for ℱ_{∞}, where D′_{m} = D(R_{m}, S, φ_{m}, η) (m = 1, 2, …).

Remark 2. Since there exists an entire function f(z) of infinite order bounded outside a half strip |y| < \frac{1}{2}π, x > 0, [3, p. 81] Theorem 2 is also sharp in the sense of Remark 1.

Whether E_{∞} in Theorem 2 is or is not a Borel removable set for ℱ, we are unable to solve this problem. But, combining Theorem 1 and 2, we have

Theorem 3. Let E be defined as in Theorem 1, then E is a Borel removable set of entire functions, i.e., E is a Borel removable set for ℱ ∪ ℱ_{∞}.

3. Lemmas required for Theorem 1

Lemma 1 [4, p. 129]. Suppose that f(z) is meromorphic in |z| < 1 and the equations f(z) = 0, ∞, 1 have there at most a finite number of roots a_{l} where l = 1 to L, b_{n} where n = 1 to N and c_{j} where j = 1 to J, respectively. We write

f_{0}(z) = f(z) ∏_{n=1}^{N}(\frac{z − b_{n}}{1 − \overline{b}_{n}z} ) / ∏_{l=1}^{L}(\frac{z − a_{l}}{1 − \overline{a}_{l}z} ).

Then if z_{1} = re^{iθ}, where 0 < r < 1, 0 ≤ θ ≤ 2π, and z_{0} = 0,

(3.1)  log^{+} |f_{0}(z_{1})| ≤ \frac{A_{0}}{1 − r} (log^{+} |f_{0}(z_{0})| + L + N + J + 1),

where A_{0} is a positive absolute constant.

70 Shengjian Wu

Lemma 2. Suppose further that f(z) is regular in |z| < 1 and that

(3.2)  log |f(0)| > 48A_{0}(L + J + 2) > 1,

where A_{0} , L and J are as in Lemma 1. Then

(3.3)  log |f(z)| > \frac{1}{8A_{0}} log |f(0)|

for all |z| < \frac{1}{2} , outside a set of circles (γ) the sum of whose radii is at most \frac{1}{64} .

Proof. By applying Lemma 1 with

f(\frac{z_{0} + z}{1 + \overline{z}_{0}z} ), f_{0}(\frac{z_{0} + z}{1 + \overline{z}_{0}z} ), \frac{z_{1} − z_{0}}{1 − \overline{z}_{0}z_{1}}

instead of

f(z), f_{0}(z), z_{1},

we see that the conclusion (3.1) holds for an arbitrary pair of points z_{0} , z_{1} in |z| < 1 with r = |(z_{1} − z_{0})/(1 − \overline{z}_{0}z_{1})|.

Without loss of generality, we suppose A_{0} ≥ 1. Let z_{1} = 0 and z_{0} = re^{iθ} in (3.1), we have

(3.4)  log^{+} |f_{0}(re^{iθ})| ≥ \frac{1 − r}{A_{0}} log^{+} |f_{0}(0)| − L − J − 1.

Since |f_{0}(0)| = |f(0)|/ ∏_{l=1}^{L} |a_{l}|, we have

(3.5)  log^{+} |f_{0}(0)| ≥ log^{+} |f(0)|.

Taking r = \frac{1}{2} and substituting (3.5) in (3.4), we deduce that

(3.6)  log^{+} |f_{0}(\frac{1}{2}re^{iθ})| ≥ \frac{1}{2A_{0}} log |f(0)| − L − J − 1 ≥ \frac{1}{4A_{0}} log |f(0)|.
Since there is no zero of f_{0}(z) for |z| ≤ \frac{1}{2} , we deduce from (3.6) that
|f_{0}(z)| ≥ |f(0)|^{1/4A_{0}}

for all |z| ≤ \frac{1}{2} . Thus

log |f(z)| = log |f_{0}(z)| + ∑_{l=1}^{L} log|\frac{z − a_{l}}{1 − \overline{a}_{l}z} |

≥ \frac{1}{4A_{0}} log |f(0)| − ∑_{l=1}^{L} log |1 − \overline{a}_{l}z| + ∑_{l=1}^{L} log |z − a_{l}|

≥ \frac{1}{4A_{0}} log |f(0)| − L log 2 + ∑_{l=1}^{L} log |z − a_{l}|

≥ \frac{1}{4A_{0}} log |f(0)| − L(log 128e)

≥ \frac{1}{4A_{0}} log |f(0)| − 6L ≥ \frac{1}{8A_{0}} log |f(0)|,

for |z| < \frac{1}{2} , outside a set of circles, the sum of whose radii is at most \frac{1}{64} , by Cartan’s lemma [1, p. 46].

Lemma 3. [7, p. 207] Let f(z) be a meromorphic function of order λ (0 < λ < +∞) in the plane and let (r < |z| < R) be an annular domain. Then there exists a point z_{j} ∈ (r < |z| < R) such that in (|z − z_{j}| < (100/q)|z_{j}|) f(z) takes every complex value at least

n_{1} = \frac{1}{900q^{2}} \frac{T (R, f)}{(log(R/r))^{2}}

times, except for at most some values which can be enclosed in two spherical circles with radii e^{−n_{1}} .

The above conclusion holds if the following conditions are satisfied. i. n_{1} and q are sufficiently large; ii. T (R, f) > max{c(f), 12T (r, f), (12T (kr, f)/ log k) log(R/r)}, where c(f) is

a positive constant depending only on f(z) and k is a constant satisfying 1 < k < R/r.

4. Proof of Theorem 1

The proof is indirect. We assume that E is not a Borel removable set for ℱ. Therefore there must exist an entire function of order λ (0 < λ < +∞) and two distinct finite complex values a_{1} and a_{2} such that

(4.1)  lim sup_{r→∞} \frac{log n\{(|z| < r) \ E, f = a_{i}\}}{log r} < τ < λ  (i = 1, 2).

Without loss of generality, we suppose a_{1} = 0 and a_{2} = 1.

According to a result of Valiron [9, p. 64], there exists a proximate order λ(r) of T (r, f) having the following properties:

(a) λ(r) is defined for r ≥ r_{0} > 0, continuous and nonnegative, and differentiable in adjacent intervals;

(b : 4.2)  lim_{r→∞} λ(r) = λ;

(c : 4.3)  lim_{r→∞}(rλ′(r) log r) = 0;

(d : 4.4)  lim sup_{r→∞} \frac{T (r.f)}{U(r)} = 1

where U(r) = r^{λ(r)} .

Now we take a sequence (r_{m}) of positive numbers such that

(4.5)  lim_{m→∞} \frac{T (r_{m}, f)}{U(r_{m})} = 1.

We next choose a fixed number k (0 < k < 1) such that

(4.6)  24k^{λ} < 1  and  \frac{24 log(1/k)}{log 2}(2k)^{λ} < 1

and distinguish two cases.

Case 1. There exists a subsequence (r_{m_{j}}) of (r_{m}) such that

(4.7)  (\frac{1}{2}kr_{m_{j}} ≤ |z| ≤ (1 + k)r_{m_{j}}) ∩ E = ϕ.

From (4.1) and (4.7), we have

(4.8) \begin{matrix}n\{ & (\frac{1}{2}kr_{m_{j}} ≤ |z| ≤ (1 + k)r_{m_{j}}), f = X\} \\ ≤ n\{(|z| ≤ (1 + k)r_{m_{j}}) \ E; f = X\} ≤ (1 + k)^{τ} r_{m_{j}}^{τ}  (X = 0, 1). \end{matrix}

From (4.2), it is easy to see that

(4.9)  lim_{r→∞} \frac{U(kr)}{U(r)} = k^{λ}.

Thus we deduce from (4.2), (4.4), (4.5), (4.6) and (4.8) that

12T (kr_{m_{j}}, f) < 16U(kr_{m_{j}}) < 20k^{λ}U(r_{m_{j}}) < 24k^{λ}T (r_{m_{j}}, f) < T (r_{m_{j}}, f)

and

\frac{12T (2kr_{m_{j}}, f)}{log 2} log \frac{r_{m_{j}}}{kr_{m_{j}}} ≤ \frac{16U(2kr_{m_{j}})}{log 2} log \frac{1}{k}

≤ \frac{24 log \frac{1}{k}}{log 2} (2k)^{λ}T (r_{m_{j}}f) < T (r_{m_{j}}, f).

We apply Lemma 3 to f(z) and (kr_{m_{j}} < |z| < r_{m_{j}}) in which we set q = 1/ log r_{m_{j}}. If j is sufficiently large, we conclude that there exists z_{j} ∈ (kr_{m_{j}} < |z| < r_{m_{j}}) such that in (|z − z_{j}| < (100/ log r_{m_{j}})|z_{j}|)f(z) takes every complex value at least

(4.10)  n_{j} = \frac{T (r_{m_{j}}, f)}{900(log r_{m_{j}})^{2}(log(1/k))^{2}} > \frac{U(r_{m_{j}}, f)}{1800(log r_{m_{j}})^{2}(log(1/k))^{2}}

times, except for at most some values which can be enclosed in two spherical circles with radii e^{−n_{j}}. Since

(|z − z_{j}| < \frac{100}{log r_{m_{j}}} |z_{j}|) ⊂ (\frac{1}{2}kr_{m_{j}} ≤ |z| ≤ (1 + k)r_{m_{j}})

and 0, 1 and ∞ cannot be all enclosed in the two spherical circles, there must exist a value, say 0, such that it is not enclosed in the two circles. For this value, we deduce from (4.2) and (4.10) that

(4.11)  \begin{matrix}n\{ & (\frac{1}{2}kr_{m_{j}} ≤ |z| ≤ (1 + k)r_{m_{j}}, f = 0\} \\ ≥ n\{(|z − z_{j}| < \frac{100}{log r_{m_{j}}} |z_{j}|, f = 0\} ≥ r_{m_{j}}^{λ−ε_{j}}, \end{matrix}

where ε_{j} → 0(j → ∞).

This contradicts (4.8).

Case 2. For (r_{m}), we have

(4.12)  (\frac{1}{2}kr_{m} ≤ |z| ≤ (1 + k)r_{m}) ∩ E ≠ ϕ.

In this case, we may suppose that

(4.13)  (\frac{1}{2}kr_{m} ≤ |z| ≤ (1 + k)r_{m}) ∩ D_{n_{m}} ≠ ϕ.

(4.13) implies that

(4.14)  D_{n_{m}} ⊂\{\frac{k}{2S} r_{m} ≤ |z| ≤ (1 + k)Sr_{m}\}.

We next prove that there exists a positive constant A not depending on m such that

(4.15)  T (\frac{k}{4S} r_{m}, f) ≥ AT (r_{m}, f)

for all sufficiently large m.

In order to prove (4.15), a disc train that we shall construct below is needed. Suppose that z_{0} = 4Sr_{m}e^{iθm} (in which we omit the subscript of z_{0,m} for the simplicity of notation) is such a point that

(4.16)  |f(z_{0})| = M(4Sr_{m}, f),

where M(r, f) = max_{0≤θ<2π} |f(re^{iθ})|. Then we get the first disc Γ_{0} : |z − z_{0}| < 5αr_{m}, where α < kη/625S is a positive constant. We next rotate Γ_{0} around the origin with the argument increasing by one α each time from z_{0} . With at most N′_{1} = 2[4Sπ/α] times of rotations we can obtain the discs Γ_{0}, Γ_{1}, …, Γ_{N_{1}} (N_{1} ≤ N′_{1} ) such that Γ_{N_{1}} is completely contained in G(φ_{m}, \frac{1}{2}η). Suppose that the center of Γ_{N_{1}} is 4Sr_{m}e^{iθ_{N_{1}}} . We continue to move Γ_{N_{1}} along the segment L = {re^{iφ_{N_{1}}} ; ((k/8S) − 2α)r_{m} ≤ r ≤ 4Sr_{m}} in succession with its distance from the origin decreasing αr_{m} each time and obtain the discs Γ_{N_{1}+1}, Γ_{N_{1}+2}, … Γ_{N_{1}+N_{2}} such that Γ_{N_{1}+N_{2}} is contained in |z| < ((k/8S) − α)r_{m}. It is not hard to see that N_{2} ≤ [4S/α] + 1. We set T_{m} = (⋃_{ν=1}^{N_{1}} Γ_{v}) ∪ {⋃_{μ=1}^{N_{2}} Γ_{N_{1}+μ}} which is the desired disc train. From the construction of T_{m}, we see that T_{m} ∩ E = ϕ. Thus we have

(4.17)  \begin{aligned}n_{m} & = n(T_{m}, f = 0\} + n(T_{m}, f = 1) \\ & ≤ n\{(|z| ≤ 5Sr_{m}) \ E, f = 0\} + n\{(|z| ≤ 5Sr_{m}) \ E, f = 1\} \\ & ≤ r_{m}^{τ},  τ < λ. \end{aligned}

Since f(z) is entire, we have

(4.18)  log |f(z_{0})| = M(4Sr_{m}, f) ≥ T (4Sr_{m}, f) > T (r_{m}, f).

We apply Lemma 2 to g(t) = f(z_{0} + 4αr_{m}t). Suppose that a_{l}^{ν} where l = 1 to L_{ν} and b_{j}^{ν} where j = 1 to J_{ν} are the roots of the equations f(z) = 0, 1 in Γ_{ν} (ν = 0, 1, 2, … N_{1}+N_{2} ), respectively. Then we have for every ν(0 ≤ ν ≤ N_{1}+N_{2})

(4.19)  L_{ν} + J_{ν} ≤ n_{m}.

On the other hand, we have for all large m,

(4.20)  log |g(0)| > T (r_{m}, f) > \frac{1}{2}U(r_{m}, f) = \frac{1}{2}r_{m}^{λ(r_{m})}.

From (4.17), (4.19) and (4.20) and by using Lemma 2, we conclude that

(4.21)  log |g(t)| > \frac{1}{8A_{0}} log |g(0)| = \frac{1}{8A_{0}} log |f(z_{0})| > \frac{1}{8A_{0}} T (r_{m}, f)

for all |t| ≤ \frac{1}{2} , outside a set of circles (γ′)_{0} the sum of whose radii is at most 1/64. (4.21) implies that

log |f(z)| > \frac{1}{8A_{0}} T (r_{m}, f)

for all |z − z_{0}| < 2αr_{m}, outside a set of circles (γ)_{0} the sum of whose radii is at most αr_{m}/16.

We next consider Γ_{1} . It is not hard to see that there exists a point z_{1} ∈ \frac{1}{5}Γ_{1} , in which for a disc Γ : |z − a| < t, δΓ refers to |z − a| < δt, such that

log |f(z_{1})| > \frac{1}{8A_{0}} T (r_{m}, f).

Since Γ′_{1} : |z − z_{1}| < \frac{9}{2}αr_{m} is contained in Γ_{1} and \frac{1}{2}Γ′_{1} ⊃ \frac{2}{5}Γ_{1} , by using Lemma 1 to f(z_{1} + 4αr_{m}t), we deduce that

log |f(z)| > \frac{1}{(8A_{0})^{2}} T (r_{m}, f)

for all z ∈ \frac{2}{5}Γ_{1} , outside a set of circles (γ)_{1} the sum of whose radii is at most αr_{m}/16.

Repeating the above arguments for Γ_{2}, Γ_{3}, …, Γ_{N_{1}+N_{2}} , we finally obtain that

(4.22)  log |f(z)| > \frac{1}{(8A_{0})^{N_{1}+N_{2}+1}} T (r_{m}, f)

for all z ∈ \frac{2}{5}Γ_{N_{1}+N_{2}} , outside a set of circles (γ)_{N_{1}+N_{2}} the sum of whose radii is at most αr_{m}/16.

It follows from (4.22) that there exists a point z with |z| < kr_{m}/8S such that

log |f(z)| > \frac{1}{(8A_{0})^{N_{1}+N_{2}+1}} T (r_{m}, f).

Since f(z) is entire, we have

T (\frac{k}{4S} r_{m}, f) > \frac{1}{3} log M(\frac{k}{8S} r_{m}, f) > \frac{1}{3(8A_{0})^{N_{0}+N_{1}+1}} T (r_{m}, f).

Setting A = 1/3(8A_{0})^{N_{1}+N_{2}+1} , we obtain (4.15).

Since

(2SR_{n_{m}−1} ≤ |z| ≤ \frac{k}{3S} r_{m}) ∩ E = ϕ

for all large m, we have

(4.23) \begin{aligned}n\{(2SR_{n_{m}−1} ≤ |z| ≤ \frac{k}{3S} r_{m}), f = X\} & ≤ n\{(|z| ≤ \frac{k}{3S} r_{m}) \ E, f = X\} \\ & < r_{m}^{τ}  (X = 0, 1). \end{aligned}

On the other hand, for ε > 0, if r is sufficiently large and R > r, we deduce from (4.3) that

|λ(R) − λ(r)| = |∫_{r}^{R} λ′(t) dt| = |∫_{r}^{R} \frac{λ′(t)t log t}{t log t} dt| ≤ \frac{ε}{log r} ∫_{r}^{R} \frac{dt}{t} = \frac{ε}{log r} log \frac{R}{r}.

Thus

\frac{U(R)}{U(r)} = (\frac{R}{r})^{λ(R)}e^{(λ(R)−λ(r)) log r} ≥ (\frac{R}{r})^{λ(R)}e^{−ε log(R/r)} ≥ (\frac{R}{r})^{λ/2}.

Therefore

(4.24)  \frac{T (6SR_{n_{m}−1}, f)}{(6SR_{n_{m}−1})^{λ/2}} ≤ \frac{U(6SR_{n_{m}−1})}{(6SR_{n_{m}−1})^{λ/2}} ≤ \frac{U(r_{m}, f)}{r_{m}^{λ/2}} ≤ \frac{2T (r_{m}, f)}{r_{m}^{λ/2}}.

It follows from (4.14) and (4.24) that

(4.25)  \begin{aligned}T (6SR_{n_{m}−1}, f) & ≤ 2(\frac{6SR_{n_{m}−1}}{r_{m}} )^{λ/2}T (r_{m}, f) \\ & ≤ 2(\frac{24S^{2}}{k} \frac{R_{n_{m}−1}}{R_{n_{m}}} )^{λ/2}T (r_{m}, f) \\ & ≤ A′(\frac{R_{n_{m}−1}}{R_{n_{m}}} )^{λ/2}T (\frac{k}{4S} r_{m}, f) \end{aligned}

where

A′ = \frac{2}{A}(\frac{24S^{2}}{k} )^{λ/2}.

We conclude from (2.1) and (4.25) that

12T (3SR_{n_{m}−1}, f) < T (\frac{k}{4S} r_{m}, f)

and

\frac{12T (6SR_{n_{m}−1}, f)}{log 2} log \frac{kr_{m}/4S}{6SR_{n_{m}−1}} ≤ \frac{12T (6SR_{n_{m}−1}, f)}{log 2}(log \frac{kR_{n_{m}}/16S^{2}}{6SR_{n_{m}−1}} )

≤ \frac{12A′}{log 2} (log \frac{kR_{n_{m}}/6S^{2}}{6SR_{n_{m}−1}} )(\frac{R_{n_{m}−1}}{R_{n_{m}}} )^{λ/2}T (\frac{k}{4S} r_{m}, f)

< T (\frac{k}{4S} r_{m}, f)

for all large m.

From Lemma 3, if m is sufficiently large, there exists a disc (|z − z_{m}| < 100|z_{m}|/ log r_{m}) ⊂ (2SR_{n_{m}−1} < |z| < kr_{m}/3S) such that f(z) takes every complex value in (|z − z_{m}| < 100|z_{m}|/ log r_{m}) at least

n_{m} > \frac{AT (r_{m}, f)}{900(log r_{m})^{4}} > \frac{AU(r_{m})}{1800(log r_{m})^{4}}

times, except for at most some values which can be enclosed in two spherical circles with radii e^{−n_{m}}. Thus we can similarily obtain a contradiction as we did in the proof of Case 1. The proof of the theorem is completed.

5. A lemma required for Theorem 2

Lemma 4 [7. p. 214]. Let f(z) be a meromorphic function of infinite order and let (r < |z| < R) be an annular domain. Then there exists a point z_{j} ∈ (r < |z| < R) such that in (|z − z_{j}| < (100/q)|z_{j}|) f(z) takes every complex value at least

n_{1} = \frac{1}{1800q^{2}} \frac{T (R − (2/T (R − 1, f)), f)}{log(R/r)}

times, except for at most some values which can be enclosed in two spherical circles with radii e^{−n_{1}} .

The above conclusion holds if the following conditions are satisfied. i. n_{1} and q are sufficiently large; ii. T (R−(2/T (R−1, f)), f) > max{c(f), 24T (r.f), 24(T (kr, f)/log k) log R/r},

where c(f) is a positive constant depending only on f(z) and k is a constant satisfying 1 < k < R/r.

6. Proof of Theorem 2

As in the proof of Theorem 1, the proof is indirect. We assume that E_{∞} is not a Borel removable set for ℱ_{∞}. Thus there exists an entire function f(z) with order λ = ∞ such that

(5.1)  lim sup_{r→∞} \frac{log n\{(|z| < r) \ E_{∞}, f = X\}}{log r} < B < +∞  (X = 0, 1).

From a result of Chuang [2. p. 178], there exists a proximate order λ(r) of T (r.f) which has the following properties.

(a) λ(r) is defined for r ≥ r_{0} > 0, and continuous and nondecreasing;

(b : 5.2)  lim_{r→∞} λ(r) = +∞;

(c : 5.3)  lim sup_{r→∞} \frac{T (r, f)}{U(r)} = 1,

where U(r) = r^{λ(r)} .

We choose a sequence (r_{m}) such that r_{m} < r_{m+1}/σ^{2} and

(5.4)  lim_{m→∞} \frac{T (r_{m}, f)}{U(r_{m})} = 1

and distinguish the following two cases.

Case 1. There exists a subsequence (r_{m_{j}}) of (r_{m}) and a fixed positive constant σ_{1} > 1 such that

(5.5)  (\frac{1}{σ_{1}} r_{m_{j}} ≤ |z| ≤ σ_{1}r_{m_{j}}) ∩ E_{∞} = ϕ.

It follows from (5.1) that

(5.6) \begin{aligned}n\{(r_{m_{j}}/σ_{1} ≤ |z| ≤ σ_{1}r_{m_{j}}), f = X\} & ≤ n\{(|z| ≤ σ_{1}r_{m_{j}}) \ E_{∞}, f = X\} \\ & < r_{m_{j}}^{2B}  (X = 0, 1). \end{aligned}

From the properties (a) and (b) of λ(r), if r is sufficiently large then we have

(5.7)  \frac{U(R)}{U(r)} = \frac{R^{λ(R)}}{r^{λ(r)}} ≥ (\frac{R}{r})^{λ(r)}

for all R > r.

Setting σ_{2} = σ_{1}^{1/4} , we deduce that

\begin{aligned}24T (\frac{1}{σ_{2}^{2}} r_{m_{j}}, f) & ≤ 25U(\frac{1}{σ_{2}^{2}} r_{m_{j}}) ≤ 25(\frac{1}{σ_{2}^{2}} )^{λ(r_{m_{j}})}U(r_{m_{j}}) \\ & ≤ 26(\frac{1}{σ_{2}^{2}} )^{λ(r_{m_{j}})}T (r_{m_{j}}, f) < T (r_{m_{j}}, f) \end{aligned}

and

\frac{24T (r_{m_{j}}/σ_{2}, f)}{log σ_{2}} log σ_{2}^{2} ≤ 49U(\frac{1}{σ_{2}} r_{m_{j}}) ≤ 49(\frac{1}{σ_{2}} )^{λ(r_{m_{j}})}U(r_{m_{j}}) < T (r_{m_{j}}, f).

By using Lemma 4, if j is sufficiently large, there exists a disc (|z − z_{j}| < (100/log r_{m_{j}})|z_{j}|) contained in (r_{m_{j}}/σ_{2}^{3} ≤ |z| < σ_{2}r_{m_{j}}) such that f(z) takes every complex value there at least

n_{j} = \frac{1}{1800(log r_{m_{j}})^{2}} \frac{T (r_{m_{j}}, f)}{log σ_{1}} > r_{m_{j}}^{λ(r_{m_{j}})−1}

times, except for at most some complex values which can be enclosed in two spherical circles with radii e^{−n_{j}}. Therefore we can obtain a contradiction as we did in the proof of Theorem 1.

Case 2. For any fixed σ_{1} > 1,

(\frac{1}{σ_{1}} r_{m} ≤ |z| ≤ σ_{1}r_{m}) ∩ E_{∞} ≠ ϕ

for all sufficiently large m.

In this case, we choose a fixed constant σ_{3} such that 1 < σ_{3}^{4} < min{S, σ/S}. In the following, we shall constuct a single connected domain T_{m} which is formed by two annular domains connected by a sectorial domain. The large annular domain G_{m} can be obtained in the following way. If (r_{m} ≤ |z| ≤ σ_{3}r_{m})∩E_{∞} = ϕ, we let G_{m} = (r_{m} ≤ |z| ≤ σ_{3}r_{m}). Otherwise we may suppose (r_{m} ≤ |z| ≤ σ_{3}r_{m}) ∩ D_{n_{m}} ≠ ϕ, then we let G_{m} = (σ_{3}SR_{n_{m}} ≤ |z| ≤ σ_{3}^{2}SR_{n_{m}}). If we write G_{m} = (r′_{m} ≤ |z| ≤ σ_{3}r′_{m}), then we have r′_{m} > r_{m} and G_{m} ∩ E_{∞} = ϕ.

The small annular domain and T_{m} can be obtained in the following way. If G_{m} = (r_{m} ≤ |z| ≤ σ_{2}r_{m}), then (r_{m}/σ_{3} ≤ |z| ≤ r_{m}) ∩ D_{n_{m}} ≠ ϕ for some

D_{n_{m}}, otherwise we have the Case 1. We let G′_{m} = (R_{n_{m}}/σ_{3}^{2} ≤ |z| ≤ R_{n_{m}})/σ_{3} and set T_{m} = G_{m} ∪ G′_{m} ∪ {G(φ_{n_{m}}, η/2) ∩ (R_{n_{m}}/σ_{3} ≤ |z| ≤ r_{m})}.

If G_{m} = (σ_{3}SR_{n_{m}} ≤ |z| ≤ σ_{3}^{2}SR_{n_{m}}) and (r_{m}/σ_{3} ≤ |z| ≤ r_{m}) ∩ E = ϕ, then we let G′_{m} = (r_{m}/σ_{3} ≤ |z| ≤ r_{m}) and T_{m} = G_{m} ∪ G′_{m} ∪ \{G(φ_{n_{m}}, \frac{η}{2} ) ∩ (r_{m} ≤ |z| ≤ σ_{3}SR_{n_{m}}). Finally we may have (r_{m}/σ_{3} ≤ |z| ≤ r_{m}) ∩ D_{n_{m}} ≠ ϕ, then we let G′_{m} = (R_{n_{m}}/σ_{3}^{2} ≤ |z| ≤ R_{n_{m}}/σ_{3}) and T_{m} = G_{m}∪G′_{m}∪{G(φ_{n_{m}}, η/2)∩(R_{n_{m}}/σ_{3} ≤ |z| ≤ σ_{3}SR_{n_{m}})}.

In all cases, if we write G′_{m} = (r′′_{m} ≤ |z| ≤ σ_{3}r′′_{m}), then we have σ_{3}r′′_{m} ≤ r_{m} and G′_{m} ∩ E_{∞} = ϕ. Thus we have T_{m} ∩ E_{∞} = ϕ.

Write σ_{4} = (σ_{3})^{1/4} . By using Lemma 2 for a disc train which can be similarly obtained as we did in the proof of Theorem 1 Case 2. We can show that there exists a positive constant A not depending on m such that

T (σ_{4}^{3}r′′_{m}, f) > AT (r_{m}, f).

Therefore by comparing T (σ_{4}r′′_{m}, f), T (σ_{4}^{2}r′′_{m}, f) and T (σ_{4}^{3}r′′_{m}, f), we conclude from Lemma 4 that there is a disc |z − z_{m}| < 100|z_{m}|/ log r_{m} contained in G′_{m} such that f(z) takes every value there at least

n_{m} > r_{m}^{λ(r_{m})−1}

times, except for some values which can be enclosed in two spherical circles with radii e^{−n_{m}}. Thus we can obtain a contradiction as we did before. The proof of Theorem 2 is completed.

Referee’s comments. The following examples show that Theorem 1 is essentially sharp.

Let

r_{n} = 16^{n}

and

f(z) = ∏_{n=1}^{∞}(1 + \frac{z}{r_{n}} )^{2^{n}}.

Let 2r_{n} ≤ |z| ≤ 8r_{n} = r_{n+1}/2. We have

log |f(z)| ≥ 2^{n−1} log 16 + 2^{n−2} log 16^{2} + 2^{n−3} log 16^{3} + 2^{n+1} log(1 − \frac{1}{2})

+ 2^{n+2} log(1 − \frac{1}{16}) + 2^{n+3} log(1 − \frac{1}{16^{2}} ) + ⋯

≥ 2^{n}(2 + 2 + 3/2) log 2 − 2^{n+1} log 2 − 2^{n+2} \frac{16}{15}(\frac{1}{16} + \frac{2}{16^{2}} + \frac{2^{2}}{16^{3}} + ⋯)

≥ 2^{n}(2 + 3/2) log 2 − 2^{n} \frac{4 · 16}{15 · 14} > 2^{n}.

Clearly

min{|f(z)| : |z| = r, | arg z| ≤ π} → ∞

as r → ∞.

If we choose

D′_{m} = D(\frac{1}{2}16^{m}, 4, 0, \frac{1}{2}π)

we deduce that Theorem 2 is not valid for entire functions of order \frac{1}{4} . Let n be a positive integer and

g(z) = f(z^{n}).

Choosing

D′_{m} = D((\frac{16^{m}}{2} )^{1/n}, 4^{1/n}, 0, \frac{π}{2n}),

we see that Theorem 2 does not hold for functions of order n/4.

Perhaps the following example is interesting, too.

Let r_{n} → ∞ and r_{n+1}/r_{n} → ∞ as n → ∞ and let 0 < ε_{n} < 1 for any n. We set

f(z) = ∑_{n=1}^{∞}(\frac{z}{r_{n}} )^{t_{n}}

where t_{n} is a positive integer such that

t_{n} log \frac{r_{n} + ε_{n}}{r_{n}} > 4t_{n−1} log r_{n}

Further results on Borel removable sets of entire functions 81

for all large n. If r_{n} + ε_{n} < |z| < r_{n+1} , then

|(z/r_{n})^{t_{n}} − f(z)| < \frac{1}{4}|z/r_{n}|^{t_{n}}

and we deduce that f(z) → ∞ as z → ∞ through these ring domains.

This implies that in Theorem 2, the sets D′_{m} cannot be replaced with the ring domains

r_{m} < |z| < r_{m} + ε_{m}.

Therefore, Theorem 2 is essentialy sharp, too.

Acknowledgement. The author wishes to thank the referee for his valuable comments and examples.

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Received 14 December 1992