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Nan Wu 1 and Zuxing Xuan 2
Academic Editor:Jaeyoung Chung
1, Department of Mathematics, School of Science, China University of Mining and Technology, Beijing 100083, China
2, Beijing Key Laboratory of Information Service Engineering, Department of General Education, Beijing Union University, No. 97 Bei Si Huan Dong Road, Chaoyang District, Beijing 100101, China
Received 4 January 2016; Accepted 10 April 2016
This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction and Results
For a meromorphic function f , disks of the form [figure omitted; refer to PDF] where z n [arrow right] ∞ and [straight epsilon] n [arrow right] 0 , are called filling disks of f if f takes every extended complex value with at most two exceptions infinitely often in any infinite subcollection of them, which are also called the filling disks of Julia type. We can prove that if a meromorphic function f satisfies [figure omitted; refer to PDF] then f must possess a sequence of filling disks of Julia type, and these filling disks can determine a Julia direction. A direction θ ∈ [ 0,2 π ] is said to be a Julia direction for a meromorphic function f , if, given [straight epsilon] > 0 , f takes all complex values infinitely often in the region D = ( z : ( a r g z - θ ) < [straight epsilon] ) except possibly two exceptions.
For a meromorphic function f of order 0 < ρ < ∞ , disks of the form (1), where z n [arrow right] ∞ and [straight epsilon] n [arrow right] 0 , are called filling disks of Borel type of f if f takes every extended complex value at least | z n | ρ - δ n times, except some complex values which can be contained in a disk whose spherical radius is e - ( z n ) ρ - δ n , where δ n [arrow right] ∞ . A meromorphic function with positive and finite order must possess a sequence of Borel type filling disks, which determine a ρ -order Borel direction, and a ρ -order Borel direction also determines a sequence of Borel type filling disks. A direction θ ∈ [ 0,2 π ] is said to be a Borel direction for a meromorphic function f of order ρ , if, given [straight epsilon] > 0 , [figure omitted; refer to PDF] for all complex values a , at most with two possible exceptions. Here and throughout the paper, n ( r , Z [straight epsilon] ( θ ) , f = a ) denotes the numbers of the roots of f = a in the region Z [straight epsilon] ( θ ) = { z : ( a r g z - θ ) < [straight epsilon] } .
For the case of Hayman direction, we pose a question whether there exist filling disks of Hayman type. In this paper, we mainly obtain the following two theorems.
Theorem 1.
Let f ( z ) be a meromorphic function in the plane satisfying [figure omitted; refer to PDF] Then, there exists a sequence of filling disks with the form (1), where z n [arrow right] ∞ and [straight epsilon] n [arrow right] 0 ; in each disk, f takes all complex values at least k n times or else there exists an integer number k such that f ( k ) takes all complex values, except possibly zero at least k n times, where limsup n [arrow right] ∞ [...] ( k n / log [...] ( z n ) ) = ∞ . Moreover, these filling disks can determine a Hayman direction a r g z = θ , such that for arbitrary small [straight epsilon] > 0 , positive integer k , and complex numbers a and b ≠ 0 we have [figure omitted; refer to PDF]
Theorem 2.
Let f ( z ) be a meromorphic function in the plane of order 0 < ρ < ∞ . Then, there exists a sequence of filling disks with the form (1), where z n [arrow right] ∞ and [straight epsilon] n [arrow right] 0 ; in each disk, f takes all complex values at least ( z n ) ρ - δ n times or else there exists an integer number k such that f ( k ) takes all complex values, except possibly zero at least ( z n ) ρ - δ n times, where δ n [arrow right] 0 . Moreover, these filling disks can determine a Hayman direction a r g z = θ , such that, for arbitrary small [straight epsilon] > 0 , positive integer k , and complex numbers a and b ≠ 0 , we have [figure omitted; refer to PDF]
Remark . The filling disks in Theorem 1 are called the filling disks of Hayman-Julia type. Moreover, if we add the growth condition 0 < ρ < ∞ to f , we can obtain the filling disks of Hayman-Borel type in Theorem 2. Hayman-Borel type filling disks are more precise than Hayman-Julia type filling disks.
2. Proof of Theorem 1
First of all, let us recall the definition of Ahlfors-Shimizu characteristic in an angle (see [1]). Let f ( z ) be a meromorphic function defined in an angle Ω = { z : α <= a r g z <= β } . Set Ω ( r ) = Ω ∩ { z : 1 < ( z ) < r } . Define [figure omitted; refer to PDF] To prove our theorems, we need the following lemma which was first established by Chen and Guo [2] and essentially comes from the Hayman inequality and the estimation of primary values appeared in the inequality and was used to confirm the existence of Hayman T directions of meromorphic functions by Zheng and the first author [3].
Lemma 3.
Let f be meromorphic in ( z ) < R and let [figure omitted; refer to PDF] for two complex numbers a and b with b ≠ 0 . Then, we have [figure omitted; refer to PDF] where C k is a positive constant depending only on k .
Lemma 3 can be obtained by using Lemma 8 of Chen and Guo [2] to the function ( b R k ) - 1 ( f ( z ) - a ) in the unit disk ( z : ( z ) < 1 ) and noticing the following: [figure omitted; refer to PDF] where d ω ( z ) is the area element on the Riemann sphere and ( x , z ) is the chordal distance between x and z .
The following lemma comes from [4], which is very useful for our study.
Lemma 4.
Let f be meromorphic in the plane, and let η , R , and M be arbitrary positive numbers with η < 1 and R > 2 . If f satisfies [figure omitted; refer to PDF] then there exists a positive number r > R such that [figure omitted; refer to PDF] where Γ ( ( 1 - η ) r , ( 1 + η ) r ) = { z : ( 1 - η ) r < ( z ) < ( 1 + η ) r } .
In view of Lemmas 3 and 4, we can prove Theorem 1.
Proof of Theorem 1.
For each n ∈ N , let M n > 0 , 0 < η n < 1 , and R n > 2 . Then, there exists r n > R n , such that [figure omitted; refer to PDF] Divide Γ ( ( 1 - η n ) r n , ( 1 + η n ) r n ) into K n domains D n j ( j = 1,2 , ... , K n ) , where [figure omitted; refer to PDF] Then, there exists at least one j 0 = j 0 ( n ) ( 1 <= j 0 <= K n ) , such that [figure omitted; refer to PDF] where z n = r n exp [...] ( i ( ( 2 j 0 + 1 ) / K n ) π ) . In view of [figure omitted; refer to PDF] for large enough n , we have [figure omitted; refer to PDF] Here we choose the proper K n , making M n / K n [arrow right] ∞ . The proof is complete.
Here we point out that, under the current technology, we cannot replace the growth condition (4) with (2). Rossi [4] also obtained that if the growth condition (4) was replaced by (2), then the inequality in Lemma 4 should be replaced by S ( Γ ( ( 1 - η ) r , ( 1 + η ) r ) , f ) > M . Unfortunately, the inequality of Lemma 3 has the term log [...] R ; owing to this term, we cannot replace the growth condition, because we notice that if the growth condition is (2) then the term log [...] R should be bounded. If we choose K n = r n , then the term log [...] R can be bounded but we cannot assure that M n / K n [arrow right] ∞ as n [arrow right] ∞ .
3. Proof of Theorem 2
The following lemma can be proved by the same method of Lemma 4, and we prove it for the completeness.
Lemma 5.
Let f be meromorphic in the plane with order 0 < ρ < ∞ , and let η , R , and M be arbitrary positive numbers with η < 1 and R > 2 . Then, for any [straight epsilon] > 0 , there exists a positive number r > R such that [figure omitted; refer to PDF]
Proof.
Suppose that the lemma is not true. Then, for every t > R , [figure omitted; refer to PDF] Choose r larger than 2 R and set β = ( 1 - η ) / ( 1 + η ) . Then, there exists α > 0 such that [figure omitted; refer to PDF] Let r 0 = r and r n = β r n - 1 . Combining (19) with (20), we have [figure omitted; refer to PDF] It follows from (20) that [figure omitted; refer to PDF] Substituting (22) into (21) leads to the order of f being ρ - [straight epsilon] by the fact that R and η are fixed.
Proof of Theorem 2.
For each n ∈ N , let M n > 0 , 0 < η n < 1 , [straight epsilon] n > 0 , and R n > 2 . Then, there exists r n > R n , such that [figure omitted; refer to PDF] Divide Γ ( ( 1 - η n ) r n , ( 1 + η n ) r n ) into K n domains D n j ( j = 1,2 , ... , K n ) , where [figure omitted; refer to PDF] Then, there exists at least one j 0 = j 0 ( n ) ( 1 <= j 0 <= K n ) , such that [figure omitted; refer to PDF] where z n = r n exp [...] ( i ( ( 2 j 0 + 1 ) / K n ) π ) . In view of [figure omitted; refer to PDF] when n is large enough, we have [figure omitted; refer to PDF] Choosing K n = M n , we can obtain the result. Hence, we complete the proof of Theorem 2. It is not difficult to see that these filling disks can determine a Borel direction of order ρ .
Zhang and Yang [5] also obtained a sequence of filling disks, whose result is as follows: let f be a meromorphic function of positive and finite order ρ . Then, there exists a sequence [figure omitted; refer to PDF] such that at least one of the following holds:
(1) f ( z ) in Γ j takes all complex values at least ( z j ) ρ - [straight epsilon] j times, with some exceptional values which can be contained in a spherical disk with center ∞ and radius e - ( z j ) ρ - [straight epsilon] j .
(2) For any fixed positive integer k , f ( k ) ( z ) in Γ j takes all complex values at least ( z j ) ρ - [straight epsilon] j times, with some exceptional values which can be contained in two spherical disks with center ∞ and 0 and radius e - ( z j ) ρ - [straight epsilon] j , where lim j [arrow right] ∞ [...] [straight epsilon] j = 0 .
We can see that our result is different from theirs.
4. Filling Disks on an Angular Region
In Yang's book [6], he said that a Borel direction of 0 < ρ < ∞ can determine a sequence of Borel type filling disks. In this paper, we can see that a ρ order Borel direction can determine not only a sequence of Borel type filling disks but also a sequence of Hayman type filling disks. Actually, if a meromorphic function f on any angular region Ω ( α , β ) satisfies some growth conditions, it possesses a sequence of filling disks. Zhang [7] obtained the following lemma, which established the growth condition on any angular domain Ω ( α , β ) containing the Borel direction a r g z = θ .
Lemma 6.
Let f be a meromorphic function in the plane of order 0 < ρ < ∞ . Then, a half line L : a r g z = θ is a ρ -order Borel direction if and only if it satisfies [figure omitted; refer to PDF]
In view of Lemma 6, we have the following result.
Theorem 7.
Let f ( z ) be a meromorphic function in the plane and let a r g z = θ be a Borel direction of order ρ . Then, there exists a sequence of disks of the form [figure omitted; refer to PDF] where z n = r n exp [...] [ i ( θ - δ n + ( ( 2 j 0 + 1 ) / K n ) δ n ) ] [arrow right] ∞ , K n [arrow right] ∞ . In C n , f takes all but possibly two extended complex values with exponent ( M n / K n ) ( z n ) ρ - [straight epsilon] n and f takes a and f ( k ) takes b with exponent ( M n / K n ) ( z n ) ρ - [straight epsilon] n , where M n [arrow right] ∞ and δ n [arrow right] 0 .
Proof.
For each n ∈ N , let M n > 0 , 0 < η n < 1 , [straight epsilon] n > 0 , and R n > 2 . It follows from (29) and Lemma 5 that there exists r n > R n , such that [figure omitted; refer to PDF] Divide Γ ( ( 1 - η n ) r n , ( 1 + η n ) r n , Z δ n ( θ ) ) into K n domains D n j ( j = 0,2 , ... , K n - 1 ) , where [figure omitted; refer to PDF] Then, there exists at least one j 0 = j 0 ( n ) ( 1 <= j 0 <= K n ) , such that [figure omitted; refer to PDF] where z n = r n exp [...] [ i ( θ - δ n + ( ( 2 j 0 + 1 ) / K n ) δ n ) ] . In view of [figure omitted; refer to PDF] when n is large enough, we have [figure omitted; refer to PDF] Choosing K n = M n , we can obtain the result.
At last, we pose a question:
: Does a Hayman direction of order 0 < ρ < ∞ have a sequence of filling disks of Hayman type ?
Acknowledgments
Nan Wu is supported in part by the grants of NSF of China (nos. 11231009, 11326086, and 11371363). Zuxing Xuan is supported in part by NNSFC (no. 91420202), Beijing Natural Science Foundation (no. 1132013), and the Project of Construction of Innovative Teams and Teacher Career Development for Universities and Colleges Under Beijing Municipality (CIT and TCD201504041, IDHT20140508).
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[3] J.-H. Zheng, N. Wu, "Hayman T directions of meromorphic functions," Taiwanese Journal of Mathematics , vol. 14, no. 6, pp. 2219-2228, 2010.
[4] J. Rossi, "A sharp result concerning cercles de remplissage," Annales Academiae Scientiarum Fennicae, Series A: I. Mathematica , vol. 20, no. 1, pp. 179-185, 1995.
[5] Q. D. Zhang, L. Yang, "New singular directions of meromorphic functions," Science in China, Series A , vol. 4, no. 26, pp. 352-366, 1984.
[6] L. Yang Value Distribution and New Research , Springer, Berlin, Germany, 1993.
[7] X. L. Zhang, "A fundamental inequality for meromorphic functions in an angular domain and its application," Acta Mathematica Sinica, New Series , vol. 10, no. 3, pp. 308-314, 1994.
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Abstract
[ProQuest: [...] denotes non US-ASCII text; see PDF]
We obtain the existence of the filling disks with respect to Hayman directions. We prove that, under the condition [subscript] limsup r [arrow right] ∞ [/subscript] [...] ( T ( r , f ) / [superscript] ( log [...] r ) 3 [/superscript] ) = ∞ , there exists a sequence of filling disks of Hayman type, and these filling disks can determine a Hayman direction. Every meromorphic function of positive and finite order ρ has a sequence of filling disks of Hayman type, which can also determine a Hayman direction of order ρ .
You have requested "on-the-fly" machine translation of selected content from our databases. This functionality is provided solely for your convenience and is in no way intended to replace human translation. Show full disclaimer
Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer