Research ArticleThe Singular Points of Analytic Functions with Finite 119883-OrderDefined by Laplace-Stieltjes Transformations

Hong-Yan Xu1 and Zu-Xing Xuan2

1Department of Informatics and Engineering Jingdezhen Ceramic Institute Jingdezhen Jiangxi 333403 China2Beijing Key Laboratory of Information Service Engineering Department of General Education Beijing Union UniversityNo 97 Bei Si Huan Dong Road Chaoyang District Beijing 100101 China

Correspondence should be addressed to Zu-Xing Xuan xuanzuxingssbuaaeducn

Received 19 April 2014 Accepted 18 July 2014

Academic Editor Seppo Hassi

Copyright copy 2015 H-Y Xu and Z-X Xuan This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We study the singular points of analytic functions defined by Laplace-Stieltjes transformations which converge on the right halfplane by introducing the concept of 119883-order functions We also confirm the existence of the finite 119883-order Borel points of suchfunctions and obtained the extension of the finite 119883-order Borel point of two analytic functions defined by two Laplace-Stieltjestransformations convergent on the right half plane The main results of this paper are improvement of some theorems given byShang and Gao

1 Introduction

For Laplace-Stieltjes transforms

119865 (119904) = int

+infin

0

119890minus119904119909

119889120572 (119909) 119904 = 120590 + 119894119905 (1)

where 120572(119909) is a bounded variation on any interval [0 119884] (0 lt119884 lt +infin) and120590 and 119905 are real variablesWe choose a sequence120582119899infin

119899=1

0 = 1205821lt 1205822lt 1205823lt sdot sdot sdot lt 120582

119899uarr +infin (2)

which satisfies the following conditions

lim sup119899rarr+infin

(120582119899+1minus 120582119899) lt +infin lim sup

119899rarr+infin

119899

120582119899

= 119863 lt infin (3)

lim sup119899rarr+infin

log119860lowast119899

120582119899

= 0 (4)

where

119860lowast

119899= sup120582119899lt119909le120582

119899+1minusinfinlt119905lt+infin

100381610038161003816100381610038161003816100381610038161003816

int

119909

120582119899

119890minus119894119905119910

119889120572 (119910)

100381610038161003816100381610038161003816100381610038161003816

(5)

Remark 1 Dirichlet series was regarded as a special exampleof Laplace-Stieltjes transformations a number of articles havefocused on the growth and the value distribution of analyticfunctions defined by Dirichlet series see [1ndash3] for somerecent results

In 1963 Yu [4] proved the Valiron-Knopp-Bohr formulaof the associated abscissas of bounded convergence absoluteconvergence and uniform convergence of Laplace-Stieltjes

Theorem A Suppose that Laplace-Stieltjes transformations(1) satisfy the first formula of (3) and lim sup

119899rarr+infin(log 119899

120582119899) lt +infin then

lim sup119899rarr+infin

log119860lowast119899

120582119899

le 120590119865

119906le lim sup119899rarr+infin

log119860lowast119899

120582119899

+ lim sup119899rarr+infin

log 119899120582119899

(6)

where 120590119865119906is called the abscissa of uniform convergence of 119865(119904)

Hindawi Publishing CorporationJournal of Function SpacesVolume 2015 Article ID 865069 9 pageshttpdxdoiorg1011552015865069

2 Journal of Function Spaces

By (3) (4) and Theorem A one can get that 120590119865119906= 0 that

is 119865(119904) is analytic in the right half plane Set

120583 (120590 119865) = max119899isin119873

119860lowast

119899119890minus120582119899120590

(120590 gt 0)

119872 (120590 119865) = supminusinfinlt119905lt+infin

|119865 (120590 + 119894119905)|

119872119906(120590 119865) = sup

0lt119909lt+infinminusinfinlt119905lt+infin

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(120590 gt 0)

(7)

Remark 2 From (4) for any 120590 gt 0 we have

lim sup119899rarr+infin

log119860lowast119899minus 120582119899120590

120582119899

= minus120590 lt 0

or lim sup119899rarr+infin

log119860lowast119899119890minus120582119899120590

= minusinfin

(8)

This shows that 120583(120590 119865) exists

In the past few decades many people studied someproblems of analytic functions defined by Laplace-Stieltjestransformations and obtained a number of interesting andimportant results (including [5ndash11]) In those papers thereare about two methods to control the growth of the maxi-mum modulus119872

119906(120590 119865) or the maximum term 120583(120590 119865) one

method is to replace the denominator in the definition ofgrowth order by using the technique of type function 119880(119909)(see [4 12ndash14]) and the other method is to take multiplelogarithm to119872

119906(120590 119865) or 120583(120590 119865) in the definition of growth

order (see [15 16]) For the second method as the logarithmfunction is a special function a question rises naturallywhether we can find a relatively general function to replacethe logarithm function to control the maximum growth rate

In this paper we investigate the above question and givea positive answer to this question Moreover we confirm theexistence of singular points for these functions by applyingthemain results of our paper To do this we introduce a com-pletely new technique based on the concept of119883(119909) which isdifferent with the type function 119880(119909) and more general thanlogarithm function and obtain the main theorems as follows

Theorem 3 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order has finite 119883-order and the sequence (2) satisfies(3) and (4) then one has

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

lArrrArr lim sup120590rarr0

+

119883(log+120583 (120590 119865))log (1120590)

= 120588lowast

(9)

Theorem 4 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) then onehas

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

lArrrArr lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899log+119860lowast

119899)= 120588lowast

(10)

where 0 lt 120588lowast lt infin

Remark 5 The definitions of 119883-order and the function inTheorems 3 and 4 119883(119909) will be introduced in Section 2

From Theorem 4 we further investigate the value distri-bution of analytic functions with finite 119883-order representedby Laplace-Stieltjes transformations convergent on the righthalf plane and obtain the following theorems

Theorem 6 Suppose that the sequence (2) satisfies (3) and (4)and Laplace-Stieltjes transformation 119865(119904) is of infinite orderLet 120572(119909) = 120572

1(119909) + 119894120572

2(119909) where 120572

1(119909) is a creasing function

and for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899 (log+119860lowast

119899))= 120588lowast

(0 lt 120588lowast

lt infin)

(11)

and then 119904 = 0 is the119883-point of 119865(119904)with119883-order 984858 ge 120588lowast thatis for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (12)

holds for any 119886 isin C except for one exception where 119899(120590 0120578 119865 = 119886) is the counting function of distinct zero of the function119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Theorem 7 Suppose that the sequence (2) satisfies (3) and (4)and Laplace-Stieltjes transformation 119865(119904) is of infinite orderLet 120572(119909) = int119909

0119903(119910)1198901198941199050119910119889119910 where 119903(119910) is continuous function

on 119910 isin [0 +infin) 119903(119910) ge 0 1199050is a positive real number and

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899log+119860lowast

119899)= 120588lowast

(0 lt 120588lowast

lt infin) (13)

then 119904 = 1198941199050is the119883-point of 119865(119904) with119883-order 984858 ge 120588lowast that is

for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (14)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is the counting function of distinct zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904) minus 119905

0| lt 120578

Journal of Function Spaces 3

The structure of this paper is as follows In Section 2we introduce the concept of 119883-order and give the proofs ofTheorems 3 and 4 Section 3 is devoted to provingTheorems 6and 7

Remark 8 FromTheorems 3ndash7 we assume that the 119883-order120588lowast of 119865(119904) is finite that is 0 lt 120588 lt infin For 120588lowast = +infin we havestudied the value distribution of analytic functions defined byLaplace-Stieltjes transformationswhich converge on the righthalf plane and obtained some results (see [17])

2 Proofs of Theorems 3 and 4

We first introduce the concept of 119883-order of such functionsas follows

Definition 9 (see [18]) If the Laplace-Stieltjes transform 119865(119904)satisfies 120590119865

119906= 0 (the sequence (2) satisfies (3) and (4)) and

lim sup120590rarr0

+

log+log+119872119906(120590 119865)

log (1120590)= infin (15)

then 119865(119904) is called a Laplace-Stieltjes transform of infiniteorder

To control the growth of themolecule119872119906(120590 119865) or 120583(120590 119865)

in the definition of order manymathematicians proposed thetype functions119880(119909) to enlarge the growth of the denominatorlog(1120590) or minus120590 (see [4 6 11 13 14]) In this paper wewill investigate the growth of Laplace-Stieltjes transform ofinfinite order by using a class of functions to reduce thegrowth of 119872

119906(120590 119865) or 120583(120590 119865) which is different with the

previous formThus we should give the definition of the newfunction as follows

Let F be the class of all functions 119883 which satisfies thefollowing conditions

(i) 119883(119909) is defined on [119886 +infin) 119886 gt 0 is positive strictlyincreasing and differential and tends to +infin as 119909 rarr+infin

(ii) 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin

Definition 10 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order satisfies

lim sup120590rarr0

+

119883(log119872119906(120590 119865))

log (1120590)= 120588lowast

(16)

where119883(119909) isin F then 120588lowast is called the119883-order of the Laplace-Stieltjes transform 119865(119904)

Remark 11 In particular if we take 119883(119909) = log119901119909 119901 ge 2 119901 isin

119873+ where log

1119909 = log119909 and log

119901119909 = log(log

119901minus1119909) 119883-order

is 119901-order of Laplace-Stieltjes transformations with infiniteorder

Remark 12 In addition 119883-order is more precise than 119901-order In fact for 119901(ge2) being a positive integer we can find

out function 119883(119909) isin F and a positive real function 119872(119909)satisfying

lim sup119909rarrinfin

119883 (119872(119909))

log119909= 119860 (0 lt 119860 lt infin)

lim sup119909rarrinfin

log119901119872(119909)

log119909= infin lim sup

119909rarrinfin

log119901+1119872(119909)

log119909= 0

(17)

For example letting 119872(119909) = exp119901(119905 log119909)1119889 119883(119909) =

(log119901119909)119889 where 119905 is a finite positive real constant and 0 lt 119889 lt

1

The following lemma is very important to study thegrowth of analytic functions represented by Laplace-Stieltjestransforms convergent on the right half plane which showthe relation between119872

119906(120590 119865) and 120583(120590 119865) of such functions

Lemma 13 (see [7 11]) If the abscissa 120590119865119906= 0 of uniform con-

vergence of Laplace-Stieltjes transformation and the sequence(2) satisfy (3) then for any given 120576 isin (0 1) and for 120590(gt0)sufficiently reaching 0 one has

1

3120583 (120590 119865) le 119872

119906(120590 119865) le 119870 (120576) 120583 ((1 minus 120576) 120590 119865)

1

120590 (18)

where119870(120576) is a constant depending on 120576 (3) and

log+119909 = log119909 119909 ge 1

0 119909 lt 1(19)

21 The Proof of Theorem 3 Firstly for any constant 1198701 we

will prove that

lim sup120590rarr0

+

119883(1198701log120583 (120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log 120583 (120590 119865))log (1120590)

(20)

Without loss of generality we suppose that 1198701gt 1 From

the Laplace-Stieltjes transformation 119865(119904) with infinite orderand Lemma 13 we have lim

120590rarr0+120583(120590 119865) = infin Then from

the Cauchy mean value theorem there exists 120578 (log120583(120590 119865) lt120578 lt 119870

1log120583(120590 119865)) satisfying

119883(1198701log 120583 (120590 119865)) minus 119883 (log 120583 (120590 119865))

log (1198701log 120583 (120590 119865)) minus loglog120583 (120590 119865)

=1198831015840(120578)

(log 120585)1015840= 1205781198831015840

(120578)

(21)

that is

119883(1198701log 120583 (120590 119865)) = 119883 (log120583 (120590 119865)) + log119870

11205781198831015840

(120578)

(22)

From 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin and the above equality wecan easily get (20)

Thus from (20) and Lemma 13 we can prove the conclu-sion of Theorem 3 easily

4 Journal of Function Spaces

22 The Proof of Theorem 4 (lArr) Suppose that

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899 log119860lowast

119899)= 120588lowast

(23)

We consider two steps in this case as follows

Step 1 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) ge 120588lowast From (23) and 0 lt 120588lowast lt infin there exists apositive sequence 119899

119896 for any 120576 (0 lt 120576 lt 120588lowast)

log119860lowast119899119896

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) 119896 = 1 2 3

(24)

From (18) and (23) we have

log119872119906(120590 119865) ge log119860lowast

119899119896

minus 120582119899119896

120590 + 119874 (1)

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

)

minus 120590120582119899119896

+ 119874 (1)

(25)

Setting 120590 = 120590119896and taking 120590

119896= (12) exp(minus1(120588lowast minus

120576))119883(120582119899119896

) (119896 = 1 2 3 ) then from (24) and (25) wehave

log119872119906(120590119896 119865) gt

1

2120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) (1 + 119900 (1))

(26)

that is

(1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865) gt 120582

119899119896

(27)

From 120590119896= (12) exp(minus1(120588lowast minus 120576))119883(120582

119899119896

) we have

119883(120582119899119896

) = (120588lowast

minus 120576) log 12120590119896

(28)

Thus from (27) (28) and 119883(119909) being strictly increasingfunction we have

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865)) gt (120588

lowast

minus 120576) log 12120590119896

(29)

Since 120590 gt 0 from the Cauchy mean value theorem thereexists 120577(log119872

119906(120590119896 119865) lt 120577 lt (1 + 119900(1))(1120590

119896) log119872

119906(120590119896 119865))

satisfying

119883((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus 119883 (log119872

119906(120590119896 119865))

log ((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus loglog119872

119906(120590119896 119865)

=1198831015840

(120577)

(log 120577)1015840= 1205771198831015840

(120577)

(30)

that is

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865))

= 119883 (log119872119906(120590119896 119865))

+ log [(1 + 119900 (1)) 1120590119896

] 1205771198831015840

(120577)

(31)

Since 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin from (23) and (31) we canget

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)ge 120588lowast

(32)

Step 2 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) le 120588lowastFrom 0 lt 120588lowast lt infin and (23) there exists a positive integer

1198990 for any 120576(gt 0) we have

log119860lowast119899lt 120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) 119899 gt 119899

0 (33)

From (3) for any 1205761015840(gt 0) and the above 120576 there exists apositive integer119873

1(gt 1198990) satisfying

1198731le (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590) lt 119873

1+ 1

as 120590 997888rarr 0+(34)

where119882(119909) and119883(119909) are two reciprocally inverse functionsSet 120582119899= 119882((120588

lowast+ 120576) log(2120590)) and then

infin

sum

119899=1198731+1

exp(120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590)

lt exp(minus120582119899120590

2)

lt

infin

sum

119899=1198731+1

exp( minus119899120590

2 (119863 + 1205761015840))

=

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))

(35)

From the above inequality and (34) we have

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))

=minus120590 (119873

1+ 1)

2 (119863 + 1205761015840)+ log 2

120590+ 119874 (1)

lt minus120590

2119882((120588

lowast

+ 120576) log 2120590) + log 2

120590+ 119874 (1) 120590 997888rarr 0

+

(36)

Journal of Function Spaces 5

Next we will prove that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(37)

By using the same argument as in (31) we can get that

119883((2

120590)

2

log 2120590) = 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

as 120590 997888rarr 0+(38)

Since 119865(119904) is infinite order and finite 119883-order from thedefinition of119883(119909) we can get the following equality easily

lim sup119909rarrinfin

119883 (119909)

119909= 0 (39)

From (38) (39) and 0 lt 120588lowast lt infin we have

(120588lowast

+ 120576) log 2120590gt 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

= 119883((2

120590)

2

log 2120590)

(40)

Thus from (40) we can deduce

lim sup120590rarr0

+

(minus1205902)119882 ((120588lowast+ 120576) log (2120590))

log (2120590)

= lim sup120590rarr0

+

minus2

120590997888rarr minusinfin

(41)

From (36) we can get that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(42)

That is

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))997888rarr 0 120590 997888rarr 0

+

(43)

Hence we haveinfin

sum

119899=1198731+1

exp(120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590) 997888rarr 0

120590 997888rarr 0+

(44)

Since 119883 isin F and 1199091205731015840(119909) rarr 0 as 119909 rarr infin then thereexists a positive integer119873

2(1198732gt 1198990) satisfying

1

2lt 1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899) lt 2 119899 gt 119873

2 (45)

Take

120582119899= 119882(minus (120588

lowast

+ 120576) log(120590(1 minus 1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

))

119899 gt 1198732

(46)

From (45) and119882(119909)being a strictly increasing function wecan get that119873

2lt 1198731 Then from (33) (45) (47) and 120590 gt 0

we have

max1198732le119899le119873

1

119860lowast

119899exp (minus120590120582

119899)

le exp120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120590120582

119899

le exp120582119899[120590(1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

minus 120590]

le expminus120582119899120590[1 minus (1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

]

le exp 120590120582119899 le exp 120590119882((120588lowast + 120576) log 2

120590)

(47)

For any 120590 gt 0 and any 119905 isin 119877 we have

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

le

119899minus1

sum

119896=1

119860lowast

119896(119890minus120582119896+1120590

+10038161003816100381610038161003816119890minus120582119896+1120590

minus 119890minus12058211989612059010038161003816100381610038161003816)

+ 119860lowast

119899(119890minus120590119909

+10038161003816100381610038161003816119890minus120590119909

minus 119890minus12058211989912059010038161003816100381610038161003816)

le

119899

sum

119896=1

119860lowast

119896119890minus120582119896120590

(48)

Thus from (33) (34) (45)ndash(47) and the above inequality wehave

119872119906(120590 119865) le

infin

sum

119899=1

119860lowast

119899119890minus120582119899120590

le (

1198732

sum

119899=1

+

1198731

sum

119899=1198732+1

+

infin

sum

119899=1198731+1

)119860lowast

119899exp (minus120590120582

119899)

le 119867 (1198732) + (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590)

times exp 120590119882((120588lowast + 120576) log 2120590) + 119900 (1)

120590 997888rarr 0+

(49)

where 119867(1198732) is the sum of finite items of the series

suminfin

119899=1119860lowast

119899119890minus120582119899120590 Hence from the above inequality we have

log119872119906(120590 119865) le (120590119882((120588

lowast

+ 120576) log 2120590)) (1 + 119900 (1))

120590 997888rarr 0+

(50)

6 Journal of Function Spaces

that is

119874 (1)1

120590log119872

119906(120590 119865) le 119882((120588

lowast

+ 120576) log 2120590) 120590 997888rarr 0

+

(51)

Then by using the same argument as in (31) we can deduce

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)le 120588lowast

(52)

From Steps 1 and 2 the sufficiency of the theorem is complet-ed

By using the same argument as in the above proof of thetheorem we can prove the necessity of the theorem easily

Thus we complete the proof of Theorem 4

3 Proofs of Theorems 6 and 7

Similar to the definition of 119883-order of Laplace-Stieltjestransformations in the right half plane we will give thedefinition of 119883-order of Laplace-Stieltjes transformations inthe level half-strip as follows

Definition 14 Let 119865(119904) be the analytic function with infiniteorder represented by Laplace-Stieltjes transformations con-vergent on the right half plane set 119878(119905

0 119897) = 120590 + 119894119905 120590 gt

0 |119905 minus 1199050| le 119897 where 119905

0is a real number and 119897 is a positive

number Let119883 isin F and

120591119883

119878= lim sup120590rarr0

+

119883(log+119872119878(120590 119865))

log (1120590) (53)

where119872119878(120590 119865) = sup

|119905minus1199050|le119897|119865(120590 + 119894119905)| then 120591119883

119878is called the

119883-order of 119865(119878) in the level half-strip 119878(1199050 119897)

To prove Theorems 6 and 7 we need some lemmas asfollows

Lemma 15 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) and120572(119909) = 120572

1(119909)+ 119894120572

2(119909) where 120572

1(119909) is a creasing function and

for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

(54)

then for any 120576 gt 0 one has

120588lowast

= lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120591119878120576

(55)

Proof We will prove this lemma by using the similar argu-ment as in [11] From the assumptions of Lemma 15 for any0 lt 119909 le infin

119872119878120576

(120590 119865) ge

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus120590119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

1198891205721(119910)

ge1

119870 + 1int

infin

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1int

119909

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(56)

Thus we have

119872(120590 119865) ge 119872119878120576

(120590 119865) ge1

119870 + 1119872119906(120590 119865) ge

1

119870 + 1119872 (120590 119865)

(57)

Since119865(119904) is an analytic function with infinite order fromthe above inequality and the definition of119883-order we can getthe conclusion of Lemma 15 easily

Lemma 16 (see [11 Lemma 24]) Let

119911 =1 minus sinh 1199041 + sinh 119904

119904 isin 119861 = 119904 R (119904) gt 0 |I (119904)| lt120587

2

(58)

Then one has the following

(i) this mapping maps the horizontal half-strip 119861 to theunit disc 119911 |119911| lt 1 and its inverse mapping is

119904 = Ψ (119911) = sinhminus1 1 minus 1199111 + 119911

(59)

(ii) min0le120579le2120587

R[Ψ(119903119890119894120579)] ge Ψ(119903) (0 lt 119903 lt 1)

(iii) max0le120579le1205874

R[Ψ(119903119890119894120579)] le Ψ(1199032) (0 lt 119903 lt 1)(iv) Ψ(119911 |119911| lt 119903) sube 119904 R(119904) gt Ψ(119903) |I(119904)| lt 1205872(0 lt 119903 lt 1)

Definition 17 (see [19 20]) Let 119891 be a meromorphic functionin D and lim

119903rarr1minus119879(119903 119891) = infin Then

119863(119891) = lim sup119903rarr1minus

119879 (119903 119891)

minus log (1 minus 119903)(60)

is called the (upper) index of inadmissibility of 119891 If 119863(119891) ltinfin 119891 is called nonadmissible if 119863(119891) = infin 119891 is calledadmissible

Lemma 18 (see [20 Theorem 2]) Let 119891 be a meromorphicfunction in D and lim

119903rarr1minus119879(119903 119891) = infin let 119902 be a positive

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Function Spaces

By (3) (4) and Theorem A one can get that 120590119865119906= 0 that

is 119865(119904) is analytic in the right half plane Set

120583 (120590 119865) = max119899isin119873

119860lowast

119899119890minus120582119899120590

(120590 gt 0)

119872 (120590 119865) = supminusinfinlt119905lt+infin

|119865 (120590 + 119894119905)|

119872119906(120590 119865) = sup

0lt119909lt+infinminusinfinlt119905lt+infin

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(120590 gt 0)

(7)

Remark 2 From (4) for any 120590 gt 0 we have

lim sup119899rarr+infin

log119860lowast119899minus 120582119899120590

120582119899

= minus120590 lt 0

or lim sup119899rarr+infin

log119860lowast119899119890minus120582119899120590

= minusinfin

(8)

This shows that 120583(120590 119865) exists

In the past few decades many people studied someproblems of analytic functions defined by Laplace-Stieltjestransformations and obtained a number of interesting andimportant results (including [5ndash11]) In those papers thereare about two methods to control the growth of the maxi-mum modulus119872

119906(120590 119865) or the maximum term 120583(120590 119865) one

method is to replace the denominator in the definition ofgrowth order by using the technique of type function 119880(119909)(see [4 12ndash14]) and the other method is to take multiplelogarithm to119872

119906(120590 119865) or 120583(120590 119865) in the definition of growth

order (see [15 16]) For the second method as the logarithmfunction is a special function a question rises naturallywhether we can find a relatively general function to replacethe logarithm function to control the maximum growth rate

In this paper we investigate the above question and givea positive answer to this question Moreover we confirm theexistence of singular points for these functions by applyingthemain results of our paper To do this we introduce a com-pletely new technique based on the concept of119883(119909) which isdifferent with the type function 119880(119909) and more general thanlogarithm function and obtain the main theorems as follows

Theorem 3 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order has finite 119883-order and the sequence (2) satisfies(3) and (4) then one has

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

lArrrArr lim sup120590rarr0

+

119883(log+120583 (120590 119865))log (1120590)

= 120588lowast

(9)

Theorem 4 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) then onehas

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

lArrrArr lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899log+119860lowast

119899)= 120588lowast

(10)

where 0 lt 120588lowast lt infin

Remark 5 The definitions of 119883-order and the function inTheorems 3 and 4 119883(119909) will be introduced in Section 2

From Theorem 4 we further investigate the value distri-bution of analytic functions with finite 119883-order representedby Laplace-Stieltjes transformations convergent on the righthalf plane and obtain the following theorems

Theorem 6 Suppose that the sequence (2) satisfies (3) and (4)and Laplace-Stieltjes transformation 119865(119904) is of infinite orderLet 120572(119909) = 120572

1(119909) + 119894120572

2(119909) where 120572

1(119909) is a creasing function

and for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899 (log+119860lowast

119899))= 120588lowast

(0 lt 120588lowast

lt infin)

(11)

and then 119904 = 0 is the119883-point of 119865(119904)with119883-order 984858 ge 120588lowast thatis for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (12)

holds for any 119886 isin C except for one exception where 119899(120590 0120578 119865 = 119886) is the counting function of distinct zero of the function119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Theorem 7 Suppose that the sequence (2) satisfies (3) and (4)and Laplace-Stieltjes transformation 119865(119904) is of infinite orderLet 120572(119909) = int119909

0119903(119910)1198901198941199050119910119889119910 where 119903(119910) is continuous function

on 119910 isin [0 +infin) 119903(119910) ge 0 1199050is a positive real number and

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899log+119860lowast

119899)= 120588lowast

(0 lt 120588lowast

lt infin) (13)

then 119904 = 1198941199050is the119883-point of 119865(119904) with119883-order 984858 ge 120588lowast that is

for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (14)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is the counting function of distinct zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904) minus 119905

0| lt 120578

Journal of Function Spaces 3

The structure of this paper is as follows In Section 2we introduce the concept of 119883-order and give the proofs ofTheorems 3 and 4 Section 3 is devoted to provingTheorems 6and 7

Remark 8 FromTheorems 3ndash7 we assume that the 119883-order120588lowast of 119865(119904) is finite that is 0 lt 120588 lt infin For 120588lowast = +infin we havestudied the value distribution of analytic functions defined byLaplace-Stieltjes transformationswhich converge on the righthalf plane and obtained some results (see [17])

2 Proofs of Theorems 3 and 4

We first introduce the concept of 119883-order of such functionsas follows

Definition 9 (see [18]) If the Laplace-Stieltjes transform 119865(119904)satisfies 120590119865

119906= 0 (the sequence (2) satisfies (3) and (4)) and

lim sup120590rarr0

+

log+log+119872119906(120590 119865)

log (1120590)= infin (15)

then 119865(119904) is called a Laplace-Stieltjes transform of infiniteorder

To control the growth of themolecule119872119906(120590 119865) or 120583(120590 119865)

in the definition of order manymathematicians proposed thetype functions119880(119909) to enlarge the growth of the denominatorlog(1120590) or minus120590 (see [4 6 11 13 14]) In this paper wewill investigate the growth of Laplace-Stieltjes transform ofinfinite order by using a class of functions to reduce thegrowth of 119872

119906(120590 119865) or 120583(120590 119865) which is different with the

previous formThus we should give the definition of the newfunction as follows

Let F be the class of all functions 119883 which satisfies thefollowing conditions

(i) 119883(119909) is defined on [119886 +infin) 119886 gt 0 is positive strictlyincreasing and differential and tends to +infin as 119909 rarr+infin

(ii) 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin

Definition 10 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order satisfies

lim sup120590rarr0

+

119883(log119872119906(120590 119865))

log (1120590)= 120588lowast

(16)

where119883(119909) isin F then 120588lowast is called the119883-order of the Laplace-Stieltjes transform 119865(119904)

Remark 11 In particular if we take 119883(119909) = log119901119909 119901 ge 2 119901 isin

119873+ where log

1119909 = log119909 and log

119901119909 = log(log

119901minus1119909) 119883-order

is 119901-order of Laplace-Stieltjes transformations with infiniteorder

Remark 12 In addition 119883-order is more precise than 119901-order In fact for 119901(ge2) being a positive integer we can find

out function 119883(119909) isin F and a positive real function 119872(119909)satisfying

lim sup119909rarrinfin

119883 (119872(119909))

log119909= 119860 (0 lt 119860 lt infin)

lim sup119909rarrinfin

log119901119872(119909)

log119909= infin lim sup

119909rarrinfin

log119901+1119872(119909)

log119909= 0

(17)

For example letting 119872(119909) = exp119901(119905 log119909)1119889 119883(119909) =

(log119901119909)119889 where 119905 is a finite positive real constant and 0 lt 119889 lt

1

The following lemma is very important to study thegrowth of analytic functions represented by Laplace-Stieltjestransforms convergent on the right half plane which showthe relation between119872

119906(120590 119865) and 120583(120590 119865) of such functions

Lemma 13 (see [7 11]) If the abscissa 120590119865119906= 0 of uniform con-

vergence of Laplace-Stieltjes transformation and the sequence(2) satisfy (3) then for any given 120576 isin (0 1) and for 120590(gt0)sufficiently reaching 0 one has

1

3120583 (120590 119865) le 119872

119906(120590 119865) le 119870 (120576) 120583 ((1 minus 120576) 120590 119865)

1

120590 (18)

where119870(120576) is a constant depending on 120576 (3) and

log+119909 = log119909 119909 ge 1

0 119909 lt 1(19)

21 The Proof of Theorem 3 Firstly for any constant 1198701 we

will prove that

lim sup120590rarr0

+

119883(1198701log120583 (120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log 120583 (120590 119865))log (1120590)

(20)

Without loss of generality we suppose that 1198701gt 1 From

the Laplace-Stieltjes transformation 119865(119904) with infinite orderand Lemma 13 we have lim

120590rarr0+120583(120590 119865) = infin Then from

the Cauchy mean value theorem there exists 120578 (log120583(120590 119865) lt120578 lt 119870

1log120583(120590 119865)) satisfying

119883(1198701log 120583 (120590 119865)) minus 119883 (log 120583 (120590 119865))

log (1198701log 120583 (120590 119865)) minus loglog120583 (120590 119865)

=1198831015840(120578)

(log 120585)1015840= 1205781198831015840

(120578)

(21)

that is

119883(1198701log 120583 (120590 119865)) = 119883 (log120583 (120590 119865)) + log119870

11205781198831015840

(120578)

(22)

From 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin and the above equality wecan easily get (20)

Thus from (20) and Lemma 13 we can prove the conclu-sion of Theorem 3 easily

4 Journal of Function Spaces

22 The Proof of Theorem 4 (lArr) Suppose that

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899 log119860lowast

119899)= 120588lowast

(23)

We consider two steps in this case as follows

Step 1 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) ge 120588lowast From (23) and 0 lt 120588lowast lt infin there exists apositive sequence 119899

119896 for any 120576 (0 lt 120576 lt 120588lowast)

log119860lowast119899119896

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) 119896 = 1 2 3

(24)

From (18) and (23) we have

log119872119906(120590 119865) ge log119860lowast

119899119896

minus 120582119899119896

120590 + 119874 (1)

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

)

minus 120590120582119899119896

+ 119874 (1)

(25)

Setting 120590 = 120590119896and taking 120590

119896= (12) exp(minus1(120588lowast minus

120576))119883(120582119899119896

) (119896 = 1 2 3 ) then from (24) and (25) wehave

log119872119906(120590119896 119865) gt

1

2120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) (1 + 119900 (1))

(26)

that is

(1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865) gt 120582

119899119896

(27)

From 120590119896= (12) exp(minus1(120588lowast minus 120576))119883(120582

119899119896

) we have

119883(120582119899119896

) = (120588lowast

minus 120576) log 12120590119896

(28)

Thus from (27) (28) and 119883(119909) being strictly increasingfunction we have

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865)) gt (120588

lowast

minus 120576) log 12120590119896

(29)

Since 120590 gt 0 from the Cauchy mean value theorem thereexists 120577(log119872

119906(120590119896 119865) lt 120577 lt (1 + 119900(1))(1120590

119896) log119872

119906(120590119896 119865))

satisfying

119883((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus 119883 (log119872

119906(120590119896 119865))

log ((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus loglog119872

119906(120590119896 119865)

=1198831015840

(120577)

(log 120577)1015840= 1205771198831015840

(120577)

(30)

that is

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865))

= 119883 (log119872119906(120590119896 119865))

+ log [(1 + 119900 (1)) 1120590119896

] 1205771198831015840

(120577)

(31)

Since 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin from (23) and (31) we canget

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)ge 120588lowast

(32)

Step 2 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) le 120588lowastFrom 0 lt 120588lowast lt infin and (23) there exists a positive integer

1198990 for any 120576(gt 0) we have

log119860lowast119899lt 120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) 119899 gt 119899

0 (33)

From (3) for any 1205761015840(gt 0) and the above 120576 there exists apositive integer119873

1(gt 1198990) satisfying

1198731le (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590) lt 119873

1+ 1

as 120590 997888rarr 0+(34)

where119882(119909) and119883(119909) are two reciprocally inverse functionsSet 120582119899= 119882((120588

lowast+ 120576) log(2120590)) and then

infin

sum

119899=1198731+1

exp(120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590)

lt exp(minus120582119899120590

2)

lt

infin

sum

119899=1198731+1

exp( minus119899120590

2 (119863 + 1205761015840))

=

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))

(35)

From the above inequality and (34) we have

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))

=minus120590 (119873

1+ 1)

2 (119863 + 1205761015840)+ log 2

120590+ 119874 (1)

lt minus120590

2119882((120588

lowast

+ 120576) log 2120590) + log 2

120590+ 119874 (1) 120590 997888rarr 0

+

(36)

Journal of Function Spaces 5

Next we will prove that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(37)

By using the same argument as in (31) we can get that

119883((2

120590)

2

log 2120590) = 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

as 120590 997888rarr 0+(38)

Since 119865(119904) is infinite order and finite 119883-order from thedefinition of119883(119909) we can get the following equality easily

lim sup119909rarrinfin

119883 (119909)

119909= 0 (39)

From (38) (39) and 0 lt 120588lowast lt infin we have

(120588lowast

+ 120576) log 2120590gt 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

= 119883((2

120590)

2

log 2120590)

(40)

Thus from (40) we can deduce

lim sup120590rarr0

+

(minus1205902)119882 ((120588lowast+ 120576) log (2120590))

log (2120590)

= lim sup120590rarr0

+

minus2

120590997888rarr minusinfin

(41)

From (36) we can get that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(42)

That is

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))997888rarr 0 120590 997888rarr 0

+

(43)

Hence we haveinfin

sum

119899=1198731+1

exp(120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590) 997888rarr 0

120590 997888rarr 0+

(44)

Since 119883 isin F and 1199091205731015840(119909) rarr 0 as 119909 rarr infin then thereexists a positive integer119873

2(1198732gt 1198990) satisfying

1

2lt 1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899) lt 2 119899 gt 119873

2 (45)

Take

120582119899= 119882(minus (120588

lowast

+ 120576) log(120590(1 minus 1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

))

119899 gt 1198732

(46)

From (45) and119882(119909)being a strictly increasing function wecan get that119873

2lt 1198731 Then from (33) (45) (47) and 120590 gt 0

we have

max1198732le119899le119873

1

119860lowast

119899exp (minus120590120582

119899)

le exp120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120590120582

119899

le exp120582119899[120590(1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

minus 120590]

le expminus120582119899120590[1 minus (1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

]

le exp 120590120582119899 le exp 120590119882((120588lowast + 120576) log 2

120590)

(47)

For any 120590 gt 0 and any 119905 isin 119877 we have

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

le

119899minus1

sum

119896=1

119860lowast

119896(119890minus120582119896+1120590

+10038161003816100381610038161003816119890minus120582119896+1120590

minus 119890minus12058211989612059010038161003816100381610038161003816)

+ 119860lowast

119899(119890minus120590119909

+10038161003816100381610038161003816119890minus120590119909

minus 119890minus12058211989912059010038161003816100381610038161003816)

le

119899

sum

119896=1

119860lowast

119896119890minus120582119896120590

(48)

Thus from (33) (34) (45)ndash(47) and the above inequality wehave

119872119906(120590 119865) le

infin

sum

119899=1

119860lowast

119899119890minus120582119899120590

le (

1198732

sum

119899=1

+

1198731

sum

119899=1198732+1

+

infin

sum

119899=1198731+1

)119860lowast

119899exp (minus120590120582

119899)

le 119867 (1198732) + (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590)

times exp 120590119882((120588lowast + 120576) log 2120590) + 119900 (1)

120590 997888rarr 0+

(49)

where 119867(1198732) is the sum of finite items of the series

suminfin

119899=1119860lowast

119899119890minus120582119899120590 Hence from the above inequality we have

log119872119906(120590 119865) le (120590119882((120588

lowast

+ 120576) log 2120590)) (1 + 119900 (1))

120590 997888rarr 0+

(50)

6 Journal of Function Spaces

that is

119874 (1)1

120590log119872

119906(120590 119865) le 119882((120588

lowast

+ 120576) log 2120590) 120590 997888rarr 0

+

(51)

Then by using the same argument as in (31) we can deduce

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)le 120588lowast

(52)

From Steps 1 and 2 the sufficiency of the theorem is complet-ed

By using the same argument as in the above proof of thetheorem we can prove the necessity of the theorem easily

Thus we complete the proof of Theorem 4

3 Proofs of Theorems 6 and 7

Similar to the definition of 119883-order of Laplace-Stieltjestransformations in the right half plane we will give thedefinition of 119883-order of Laplace-Stieltjes transformations inthe level half-strip as follows

Definition 14 Let 119865(119904) be the analytic function with infiniteorder represented by Laplace-Stieltjes transformations con-vergent on the right half plane set 119878(119905

0 119897) = 120590 + 119894119905 120590 gt

0 |119905 minus 1199050| le 119897 where 119905

0is a real number and 119897 is a positive

number Let119883 isin F and

120591119883

119878= lim sup120590rarr0

+

119883(log+119872119878(120590 119865))

log (1120590) (53)

where119872119878(120590 119865) = sup

|119905minus1199050|le119897|119865(120590 + 119894119905)| then 120591119883

119878is called the

119883-order of 119865(119878) in the level half-strip 119878(1199050 119897)

To prove Theorems 6 and 7 we need some lemmas asfollows

Lemma 15 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) and120572(119909) = 120572

1(119909)+ 119894120572

2(119909) where 120572

1(119909) is a creasing function and

for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

(54)

then for any 120576 gt 0 one has

120588lowast

= lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120591119878120576

(55)

Proof We will prove this lemma by using the similar argu-ment as in [11] From the assumptions of Lemma 15 for any0 lt 119909 le infin

119872119878120576

(120590 119865) ge

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus120590119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

1198891205721(119910)

ge1

119870 + 1int

infin

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1int

119909

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(56)

Thus we have

119872(120590 119865) ge 119872119878120576

(120590 119865) ge1

119870 + 1119872119906(120590 119865) ge

1

119870 + 1119872 (120590 119865)

(57)

Since119865(119904) is an analytic function with infinite order fromthe above inequality and the definition of119883-order we can getthe conclusion of Lemma 15 easily

Lemma 16 (see [11 Lemma 24]) Let

119911 =1 minus sinh 1199041 + sinh 119904

119904 isin 119861 = 119904 R (119904) gt 0 |I (119904)| lt120587

2

(58)

Then one has the following

(i) this mapping maps the horizontal half-strip 119861 to theunit disc 119911 |119911| lt 1 and its inverse mapping is

119904 = Ψ (119911) = sinhminus1 1 minus 1199111 + 119911

(59)

(ii) min0le120579le2120587

R[Ψ(119903119890119894120579)] ge Ψ(119903) (0 lt 119903 lt 1)

(iii) max0le120579le1205874

R[Ψ(119903119890119894120579)] le Ψ(1199032) (0 lt 119903 lt 1)(iv) Ψ(119911 |119911| lt 119903) sube 119904 R(119904) gt Ψ(119903) |I(119904)| lt 1205872(0 lt 119903 lt 1)

Definition 17 (see [19 20]) Let 119891 be a meromorphic functionin D and lim

119903rarr1minus119879(119903 119891) = infin Then

119863(119891) = lim sup119903rarr1minus

119879 (119903 119891)

minus log (1 minus 119903)(60)

is called the (upper) index of inadmissibility of 119891 If 119863(119891) ltinfin 119891 is called nonadmissible if 119863(119891) = infin 119891 is calledadmissible

Lemma 18 (see [20 Theorem 2]) Let 119891 be a meromorphicfunction in D and lim

119903rarr1minus119879(119903 119891) = infin let 119902 be a positive

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 3

The structure of this paper is as follows In Section 2we introduce the concept of 119883-order and give the proofs ofTheorems 3 and 4 Section 3 is devoted to provingTheorems 6and 7

Remark 8 FromTheorems 3ndash7 we assume that the 119883-order120588lowast of 119865(119904) is finite that is 0 lt 120588 lt infin For 120588lowast = +infin we havestudied the value distribution of analytic functions defined byLaplace-Stieltjes transformationswhich converge on the righthalf plane and obtained some results (see [17])

2 Proofs of Theorems 3 and 4

We first introduce the concept of 119883-order of such functionsas follows

Definition 9 (see [18]) If the Laplace-Stieltjes transform 119865(119904)satisfies 120590119865

119906= 0 (the sequence (2) satisfies (3) and (4)) and

lim sup120590rarr0

+

log+log+119872119906(120590 119865)

log (1120590)= infin (15)

then 119865(119904) is called a Laplace-Stieltjes transform of infiniteorder

To control the growth of themolecule119872119906(120590 119865) or 120583(120590 119865)

in the definition of order manymathematicians proposed thetype functions119880(119909) to enlarge the growth of the denominatorlog(1120590) or minus120590 (see [4 6 11 13 14]) In this paper wewill investigate the growth of Laplace-Stieltjes transform ofinfinite order by using a class of functions to reduce thegrowth of 119872

119906(120590 119865) or 120583(120590 119865) which is different with the

previous formThus we should give the definition of the newfunction as follows

Let F be the class of all functions 119883 which satisfies thefollowing conditions

(i) 119883(119909) is defined on [119886 +infin) 119886 gt 0 is positive strictlyincreasing and differential and tends to +infin as 119909 rarr+infin

(ii) 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin

Definition 10 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order satisfies

lim sup120590rarr0

+

119883(log119872119906(120590 119865))

log (1120590)= 120588lowast

(16)

where119883(119909) isin F then 120588lowast is called the119883-order of the Laplace-Stieltjes transform 119865(119904)

Remark 11 In particular if we take 119883(119909) = log119901119909 119901 ge 2 119901 isin

119873+ where log

1119909 = log119909 and log

119901119909 = log(log

119901minus1119909) 119883-order

is 119901-order of Laplace-Stieltjes transformations with infiniteorder

Remark 12 In addition 119883-order is more precise than 119901-order In fact for 119901(ge2) being a positive integer we can find

out function 119883(119909) isin F and a positive real function 119872(119909)satisfying

lim sup119909rarrinfin

119883 (119872(119909))

log119909= 119860 (0 lt 119860 lt infin)

lim sup119909rarrinfin

log119901119872(119909)

log119909= infin lim sup

119909rarrinfin

log119901+1119872(119909)

log119909= 0

(17)

For example letting 119872(119909) = exp119901(119905 log119909)1119889 119883(119909) =

(log119901119909)119889 where 119905 is a finite positive real constant and 0 lt 119889 lt

1

The following lemma is very important to study thegrowth of analytic functions represented by Laplace-Stieltjestransforms convergent on the right half plane which showthe relation between119872

119906(120590 119865) and 120583(120590 119865) of such functions

Lemma 13 (see [7 11]) If the abscissa 120590119865119906= 0 of uniform con-

vergence of Laplace-Stieltjes transformation and the sequence(2) satisfy (3) then for any given 120576 isin (0 1) and for 120590(gt0)sufficiently reaching 0 one has

1

3120583 (120590 119865) le 119872

119906(120590 119865) le 119870 (120576) 120583 ((1 minus 120576) 120590 119865)

1

120590 (18)

where119870(120576) is a constant depending on 120576 (3) and

log+119909 = log119909 119909 ge 1

0 119909 lt 1(19)

21 The Proof of Theorem 3 Firstly for any constant 1198701 we

will prove that

lim sup120590rarr0

+

119883(1198701log120583 (120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log 120583 (120590 119865))log (1120590)

(20)

Without loss of generality we suppose that 1198701gt 1 From

the Laplace-Stieltjes transformation 119865(119904) with infinite orderand Lemma 13 we have lim

120590rarr0+120583(120590 119865) = infin Then from

the Cauchy mean value theorem there exists 120578 (log120583(120590 119865) lt120578 lt 119870

1log120583(120590 119865)) satisfying

119883(1198701log 120583 (120590 119865)) minus 119883 (log 120583 (120590 119865))

log (1198701log 120583 (120590 119865)) minus loglog120583 (120590 119865)

=1198831015840(120578)

(log 120585)1015840= 1205781198831015840

(120578)

(21)

that is

119883(1198701log 120583 (120590 119865)) = 119883 (log120583 (120590 119865)) + log119870

11205781198831015840

(120578)

(22)

From 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin and the above equality wecan easily get (20)

Thus from (20) and Lemma 13 we can prove the conclu-sion of Theorem 3 easily

4 Journal of Function Spaces

22 The Proof of Theorem 4 (lArr) Suppose that

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899 log119860lowast

119899)= 120588lowast

(23)

We consider two steps in this case as follows

Step 1 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) ge 120588lowast From (23) and 0 lt 120588lowast lt infin there exists apositive sequence 119899

119896 for any 120576 (0 lt 120576 lt 120588lowast)

log119860lowast119899119896

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) 119896 = 1 2 3

(24)

From (18) and (23) we have

log119872119906(120590 119865) ge log119860lowast

119899119896

minus 120582119899119896

120590 + 119874 (1)

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

)

minus 120590120582119899119896

+ 119874 (1)

(25)

Setting 120590 = 120590119896and taking 120590

119896= (12) exp(minus1(120588lowast minus

120576))119883(120582119899119896

) (119896 = 1 2 3 ) then from (24) and (25) wehave

log119872119906(120590119896 119865) gt

1

2120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) (1 + 119900 (1))

(26)

that is

(1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865) gt 120582

119899119896

(27)

From 120590119896= (12) exp(minus1(120588lowast minus 120576))119883(120582

119899119896

) we have

119883(120582119899119896

) = (120588lowast

minus 120576) log 12120590119896

(28)

Thus from (27) (28) and 119883(119909) being strictly increasingfunction we have

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865)) gt (120588

lowast

minus 120576) log 12120590119896

(29)

Since 120590 gt 0 from the Cauchy mean value theorem thereexists 120577(log119872

119906(120590119896 119865) lt 120577 lt (1 + 119900(1))(1120590

119896) log119872

119906(120590119896 119865))

satisfying

119883((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus 119883 (log119872

119906(120590119896 119865))

log ((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus loglog119872

119906(120590119896 119865)

=1198831015840

(120577)

(log 120577)1015840= 1205771198831015840

(120577)

(30)

that is

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865))

= 119883 (log119872119906(120590119896 119865))

+ log [(1 + 119900 (1)) 1120590119896

] 1205771198831015840

(120577)

(31)

Since 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin from (23) and (31) we canget

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)ge 120588lowast

(32)

Step 2 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) le 120588lowastFrom 0 lt 120588lowast lt infin and (23) there exists a positive integer

1198990 for any 120576(gt 0) we have

log119860lowast119899lt 120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) 119899 gt 119899

0 (33)

From (3) for any 1205761015840(gt 0) and the above 120576 there exists apositive integer119873

1(gt 1198990) satisfying

1198731le (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590) lt 119873

1+ 1

as 120590 997888rarr 0+(34)

where119882(119909) and119883(119909) are two reciprocally inverse functionsSet 120582119899= 119882((120588

lowast+ 120576) log(2120590)) and then

infin

sum

119899=1198731+1

exp(120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590)

lt exp(minus120582119899120590

2)

lt

infin

sum

119899=1198731+1

exp( minus119899120590

2 (119863 + 1205761015840))

=

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))

(35)

From the above inequality and (34) we have

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))

=minus120590 (119873

1+ 1)

2 (119863 + 1205761015840)+ log 2

120590+ 119874 (1)

lt minus120590

2119882((120588

lowast

+ 120576) log 2120590) + log 2

120590+ 119874 (1) 120590 997888rarr 0

+

(36)

Journal of Function Spaces 5

Next we will prove that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(37)

By using the same argument as in (31) we can get that

119883((2

120590)

2

log 2120590) = 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

as 120590 997888rarr 0+(38)

Since 119865(119904) is infinite order and finite 119883-order from thedefinition of119883(119909) we can get the following equality easily

lim sup119909rarrinfin

119883 (119909)

119909= 0 (39)

From (38) (39) and 0 lt 120588lowast lt infin we have

(120588lowast

+ 120576) log 2120590gt 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

= 119883((2

120590)

2

log 2120590)

(40)

Thus from (40) we can deduce

lim sup120590rarr0

+

(minus1205902)119882 ((120588lowast+ 120576) log (2120590))

log (2120590)

= lim sup120590rarr0

+

minus2

120590997888rarr minusinfin

(41)

From (36) we can get that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(42)

That is

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))997888rarr 0 120590 997888rarr 0

+

(43)

Hence we haveinfin

sum

119899=1198731+1

exp(120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590) 997888rarr 0

120590 997888rarr 0+

(44)

Since 119883 isin F and 1199091205731015840(119909) rarr 0 as 119909 rarr infin then thereexists a positive integer119873

2(1198732gt 1198990) satisfying

1

2lt 1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899) lt 2 119899 gt 119873

2 (45)

Take

120582119899= 119882(minus (120588

lowast

+ 120576) log(120590(1 minus 1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

))

119899 gt 1198732

(46)

From (45) and119882(119909)being a strictly increasing function wecan get that119873

2lt 1198731 Then from (33) (45) (47) and 120590 gt 0

we have

max1198732le119899le119873

1

119860lowast

119899exp (minus120590120582

119899)

le exp120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120590120582

119899

le exp120582119899[120590(1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

minus 120590]

le expminus120582119899120590[1 minus (1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

]

le exp 120590120582119899 le exp 120590119882((120588lowast + 120576) log 2

120590)

(47)

For any 120590 gt 0 and any 119905 isin 119877 we have

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

le

119899minus1

sum

119896=1

119860lowast

119896(119890minus120582119896+1120590

+10038161003816100381610038161003816119890minus120582119896+1120590

minus 119890minus12058211989612059010038161003816100381610038161003816)

+ 119860lowast

119899(119890minus120590119909

+10038161003816100381610038161003816119890minus120590119909

minus 119890minus12058211989912059010038161003816100381610038161003816)

le

119899

sum

119896=1

119860lowast

119896119890minus120582119896120590

(48)

Thus from (33) (34) (45)ndash(47) and the above inequality wehave

119872119906(120590 119865) le

infin

sum

119899=1

119860lowast

119899119890minus120582119899120590

le (

1198732

sum

119899=1

+

1198731

sum

119899=1198732+1

+

infin

sum

119899=1198731+1

)119860lowast

119899exp (minus120590120582

119899)

le 119867 (1198732) + (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590)

times exp 120590119882((120588lowast + 120576) log 2120590) + 119900 (1)

120590 997888rarr 0+

(49)

where 119867(1198732) is the sum of finite items of the series

suminfin

119899=1119860lowast

119899119890minus120582119899120590 Hence from the above inequality we have

log119872119906(120590 119865) le (120590119882((120588

lowast

+ 120576) log 2120590)) (1 + 119900 (1))

120590 997888rarr 0+

(50)

6 Journal of Function Spaces

that is

119874 (1)1

120590log119872

119906(120590 119865) le 119882((120588

lowast

+ 120576) log 2120590) 120590 997888rarr 0

+

(51)

Then by using the same argument as in (31) we can deduce

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)le 120588lowast

(52)

From Steps 1 and 2 the sufficiency of the theorem is complet-ed

By using the same argument as in the above proof of thetheorem we can prove the necessity of the theorem easily

Thus we complete the proof of Theorem 4

3 Proofs of Theorems 6 and 7

Similar to the definition of 119883-order of Laplace-Stieltjestransformations in the right half plane we will give thedefinition of 119883-order of Laplace-Stieltjes transformations inthe level half-strip as follows

Definition 14 Let 119865(119904) be the analytic function with infiniteorder represented by Laplace-Stieltjes transformations con-vergent on the right half plane set 119878(119905

0 119897) = 120590 + 119894119905 120590 gt

0 |119905 minus 1199050| le 119897 where 119905

0is a real number and 119897 is a positive

number Let119883 isin F and

120591119883

119878= lim sup120590rarr0

+

119883(log+119872119878(120590 119865))

log (1120590) (53)

where119872119878(120590 119865) = sup

|119905minus1199050|le119897|119865(120590 + 119894119905)| then 120591119883

119878is called the

119883-order of 119865(119878) in the level half-strip 119878(1199050 119897)

To prove Theorems 6 and 7 we need some lemmas asfollows

Lemma 15 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) and120572(119909) = 120572

1(119909)+ 119894120572

2(119909) where 120572

1(119909) is a creasing function and

for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

(54)

then for any 120576 gt 0 one has

120588lowast

= lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120591119878120576

(55)

Proof We will prove this lemma by using the similar argu-ment as in [11] From the assumptions of Lemma 15 for any0 lt 119909 le infin

119872119878120576

(120590 119865) ge

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus120590119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

1198891205721(119910)

ge1

119870 + 1int

infin

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1int

119909

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(56)

Thus we have

119872(120590 119865) ge 119872119878120576

(120590 119865) ge1

119870 + 1119872119906(120590 119865) ge

1

119870 + 1119872 (120590 119865)

(57)

Since119865(119904) is an analytic function with infinite order fromthe above inequality and the definition of119883-order we can getthe conclusion of Lemma 15 easily

Lemma 16 (see [11 Lemma 24]) Let

119911 =1 minus sinh 1199041 + sinh 119904

119904 isin 119861 = 119904 R (119904) gt 0 |I (119904)| lt120587

2

(58)

Then one has the following

(i) this mapping maps the horizontal half-strip 119861 to theunit disc 119911 |119911| lt 1 and its inverse mapping is

119904 = Ψ (119911) = sinhminus1 1 minus 1199111 + 119911

(59)

(ii) min0le120579le2120587

R[Ψ(119903119890119894120579)] ge Ψ(119903) (0 lt 119903 lt 1)

(iii) max0le120579le1205874

R[Ψ(119903119890119894120579)] le Ψ(1199032) (0 lt 119903 lt 1)(iv) Ψ(119911 |119911| lt 119903) sube 119904 R(119904) gt Ψ(119903) |I(119904)| lt 1205872(0 lt 119903 lt 1)

Definition 17 (see [19 20]) Let 119891 be a meromorphic functionin D and lim

119903rarr1minus119879(119903 119891) = infin Then

119863(119891) = lim sup119903rarr1minus

119879 (119903 119891)

minus log (1 minus 119903)(60)

is called the (upper) index of inadmissibility of 119891 If 119863(119891) ltinfin 119891 is called nonadmissible if 119863(119891) = infin 119891 is calledadmissible

Lemma 18 (see [20 Theorem 2]) Let 119891 be a meromorphicfunction in D and lim

119903rarr1minus119879(119903 119891) = infin let 119902 be a positive

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Function Spaces

22 The Proof of Theorem 4 (lArr) Suppose that

lim sup119899rarrinfin

119883(120582119899)

log+ (120582119899 log119860lowast

119899)= 120588lowast

(23)

We consider two steps in this case as follows

Step 1 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) ge 120588lowast From (23) and 0 lt 120588lowast lt infin there exists apositive sequence 119899

119896 for any 120576 (0 lt 120576 lt 120588lowast)

log119860lowast119899119896

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) 119896 = 1 2 3

(24)

From (18) and (23) we have

log119872119906(120590 119865) ge log119860lowast

119899119896

minus 120582119899119896

120590 + 119874 (1)

gt 120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

)

minus 120590120582119899119896

+ 119874 (1)

(25)

Setting 120590 = 120590119896and taking 120590

119896= (12) exp(minus1(120588lowast minus

120576))119883(120582119899119896

) (119896 = 1 2 3 ) then from (24) and (25) wehave

log119872119906(120590119896 119865) gt

1

2120582119899119896

expminus 1

120588lowast minus 120576119883 (120582119899119896

) (1 + 119900 (1))

(26)

that is

(1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865) gt 120582

119899119896

(27)

From 120590119896= (12) exp(minus1(120588lowast minus 120576))119883(120582

119899119896

) we have

119883(120582119899119896

) = (120588lowast

minus 120576) log 12120590119896

(28)

Thus from (27) (28) and 119883(119909) being strictly increasingfunction we have

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865)) gt (120588

lowast

minus 120576) log 12120590119896

(29)

Since 120590 gt 0 from the Cauchy mean value theorem thereexists 120577(log119872

119906(120590119896 119865) lt 120577 lt (1 + 119900(1))(1120590

119896) log119872

119906(120590119896 119865))

satisfying

119883((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus 119883 (log119872

119906(120590119896 119865))

log ((1 + 119900 (1)) (1120590119896) log119872

119906(120590119896 119865)) minus loglog119872

119906(120590119896 119865)

=1198831015840

(120577)

(log 120577)1015840= 1205771198831015840

(120577)

(30)

that is

119883((1 + 119900 (1))1

120590119896

log119872119906(120590119896 119865))

= 119883 (log119872119906(120590119896 119865))

+ log [(1 + 119900 (1)) 1120590119896

] 1205771198831015840

(120577)

(31)

Since 1199091198831015840(119909) = 119900(1) as 119909 rarr +infin from (23) and (31) we canget

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)ge 120588lowast

(32)

Step 2 We will prove that lim sup120590rarr0

+(119883(log+119872119906(120590 119865))

log(1120590)) le 120588lowastFrom 0 lt 120588lowast lt infin and (23) there exists a positive integer

1198990 for any 120576(gt 0) we have

log119860lowast119899lt 120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) 119899 gt 119899

0 (33)

From (3) for any 1205761015840(gt 0) and the above 120576 there exists apositive integer119873

1(gt 1198990) satisfying

1198731le (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590) lt 119873

1+ 1

as 120590 997888rarr 0+(34)

where119882(119909) and119883(119909) are two reciprocally inverse functionsSet 120582119899= 119882((120588

lowast+ 120576) log(2120590)) and then

infin

sum

119899=1198731+1

exp(120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590)

lt exp(minus120582119899120590

2)

lt

infin

sum

119899=1198731+1

exp( minus119899120590

2 (119863 + 1205761015840))

=

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))

(35)

From the above inequality and (34) we have

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))

=minus120590 (119873

1+ 1)

2 (119863 + 1205761015840)+ log 2

120590+ 119874 (1)

lt minus120590

2119882((120588

lowast

+ 120576) log 2120590) + log 2

120590+ 119874 (1) 120590 997888rarr 0

+

(36)

Journal of Function Spaces 5

Next we will prove that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(37)

By using the same argument as in (31) we can get that

119883((2

120590)

2

log 2120590) = 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

as 120590 997888rarr 0+(38)

Since 119865(119904) is infinite order and finite 119883-order from thedefinition of119883(119909) we can get the following equality easily

lim sup119909rarrinfin

119883 (119909)

119909= 0 (39)

From (38) (39) and 0 lt 120588lowast lt infin we have

(120588lowast

+ 120576) log 2120590gt 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

= 119883((2

120590)

2

log 2120590)

(40)

Thus from (40) we can deduce

lim sup120590rarr0

+

(minus1205902)119882 ((120588lowast+ 120576) log (2120590))

log (2120590)

= lim sup120590rarr0

+

minus2

120590997888rarr minusinfin

(41)

From (36) we can get that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(42)

That is

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))997888rarr 0 120590 997888rarr 0

+

(43)

Hence we haveinfin

sum

119899=1198731+1

exp(120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590) 997888rarr 0

120590 997888rarr 0+

(44)

Since 119883 isin F and 1199091205731015840(119909) rarr 0 as 119909 rarr infin then thereexists a positive integer119873

2(1198732gt 1198990) satisfying

1

2lt 1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899) lt 2 119899 gt 119873

2 (45)

Take

120582119899= 119882(minus (120588

lowast

+ 120576) log(120590(1 minus 1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

))

119899 gt 1198732

(46)

From (45) and119882(119909)being a strictly increasing function wecan get that119873

2lt 1198731 Then from (33) (45) (47) and 120590 gt 0

we have

max1198732le119899le119873

1

119860lowast

119899exp (minus120590120582

119899)

le exp120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120590120582

119899

le exp120582119899[120590(1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

minus 120590]

le expminus120582119899120590[1 minus (1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

]

le exp 120590120582119899 le exp 120590119882((120588lowast + 120576) log 2

120590)

(47)

For any 120590 gt 0 and any 119905 isin 119877 we have

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

le

119899minus1

sum

119896=1

119860lowast

119896(119890minus120582119896+1120590

+10038161003816100381610038161003816119890minus120582119896+1120590

minus 119890minus12058211989612059010038161003816100381610038161003816)

+ 119860lowast

119899(119890minus120590119909

+10038161003816100381610038161003816119890minus120590119909

minus 119890minus12058211989912059010038161003816100381610038161003816)

le

119899

sum

119896=1

119860lowast

119896119890minus120582119896120590

(48)

Thus from (33) (34) (45)ndash(47) and the above inequality wehave

119872119906(120590 119865) le

infin

sum

119899=1

119860lowast

119899119890minus120582119899120590

le (

1198732

sum

119899=1

+

1198731

sum

119899=1198732+1

+

infin

sum

119899=1198731+1

)119860lowast

119899exp (minus120590120582

119899)

le 119867 (1198732) + (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590)

times exp 120590119882((120588lowast + 120576) log 2120590) + 119900 (1)

120590 997888rarr 0+

(49)

where 119867(1198732) is the sum of finite items of the series

suminfin

119899=1119860lowast

119899119890minus120582119899120590 Hence from the above inequality we have

log119872119906(120590 119865) le (120590119882((120588

lowast

+ 120576) log 2120590)) (1 + 119900 (1))

120590 997888rarr 0+

(50)

6 Journal of Function Spaces

that is

119874 (1)1

120590log119872

119906(120590 119865) le 119882((120588

lowast

+ 120576) log 2120590) 120590 997888rarr 0

+

(51)

Then by using the same argument as in (31) we can deduce

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)le 120588lowast

(52)

From Steps 1 and 2 the sufficiency of the theorem is complet-ed

By using the same argument as in the above proof of thetheorem we can prove the necessity of the theorem easily

Thus we complete the proof of Theorem 4

3 Proofs of Theorems 6 and 7

Similar to the definition of 119883-order of Laplace-Stieltjestransformations in the right half plane we will give thedefinition of 119883-order of Laplace-Stieltjes transformations inthe level half-strip as follows

Definition 14 Let 119865(119904) be the analytic function with infiniteorder represented by Laplace-Stieltjes transformations con-vergent on the right half plane set 119878(119905

0 119897) = 120590 + 119894119905 120590 gt

0 |119905 minus 1199050| le 119897 where 119905

0is a real number and 119897 is a positive

number Let119883 isin F and

120591119883

119878= lim sup120590rarr0

+

119883(log+119872119878(120590 119865))

log (1120590) (53)

where119872119878(120590 119865) = sup

|119905minus1199050|le119897|119865(120590 + 119894119905)| then 120591119883

119878is called the

119883-order of 119865(119878) in the level half-strip 119878(1199050 119897)

To prove Theorems 6 and 7 we need some lemmas asfollows

Lemma 15 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) and120572(119909) = 120572

1(119909)+ 119894120572

2(119909) where 120572

1(119909) is a creasing function and

for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

(54)

then for any 120576 gt 0 one has

120588lowast

= lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120591119878120576

(55)

Proof We will prove this lemma by using the similar argu-ment as in [11] From the assumptions of Lemma 15 for any0 lt 119909 le infin

119872119878120576

(120590 119865) ge

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus120590119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

1198891205721(119910)

ge1

119870 + 1int

infin

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1int

119909

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(56)

Thus we have

119872(120590 119865) ge 119872119878120576

(120590 119865) ge1

119870 + 1119872119906(120590 119865) ge

1

119870 + 1119872 (120590 119865)

(57)

Since119865(119904) is an analytic function with infinite order fromthe above inequality and the definition of119883-order we can getthe conclusion of Lemma 15 easily

Lemma 16 (see [11 Lemma 24]) Let

119911 =1 minus sinh 1199041 + sinh 119904

119904 isin 119861 = 119904 R (119904) gt 0 |I (119904)| lt120587

2

(58)

Then one has the following

(i) this mapping maps the horizontal half-strip 119861 to theunit disc 119911 |119911| lt 1 and its inverse mapping is

119904 = Ψ (119911) = sinhminus1 1 minus 1199111 + 119911

(59)

(ii) min0le120579le2120587

R[Ψ(119903119890119894120579)] ge Ψ(119903) (0 lt 119903 lt 1)

(iii) max0le120579le1205874

R[Ψ(119903119890119894120579)] le Ψ(1199032) (0 lt 119903 lt 1)(iv) Ψ(119911 |119911| lt 119903) sube 119904 R(119904) gt Ψ(119903) |I(119904)| lt 1205872(0 lt 119903 lt 1)

Definition 17 (see [19 20]) Let 119891 be a meromorphic functionin D and lim

119903rarr1minus119879(119903 119891) = infin Then

119863(119891) = lim sup119903rarr1minus

119879 (119903 119891)

minus log (1 minus 119903)(60)

is called the (upper) index of inadmissibility of 119891 If 119863(119891) ltinfin 119891 is called nonadmissible if 119863(119891) = infin 119891 is calledadmissible

Lemma 18 (see [20 Theorem 2]) Let 119891 be a meromorphicfunction in D and lim

119903rarr1minus119879(119903 119891) = infin let 119902 be a positive

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 5

Next we will prove that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(37)

By using the same argument as in (31) we can get that

119883((2

120590)

2

log 2120590) = 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

as 120590 997888rarr 0+(38)

Since 119865(119904) is infinite order and finite 119883-order from thedefinition of119883(119909) we can get the following equality easily

lim sup119909rarrinfin

119883 (119909)

119909= 0 (39)

From (38) (39) and 0 lt 120588lowast lt infin we have

(120588lowast

+ 120576) log 2120590gt 119883(log 2

120590) + 119900 (1) log( 2

120590)

2

= 119883((2

120590)

2

log 2120590)

(40)

Thus from (40) we can deduce

lim sup120590rarr0

+

(minus1205902)119882 ((120588lowast+ 120576) log (2120590))

log (2120590)

= lim sup120590rarr0

+

minus2

120590997888rarr minusinfin

(41)

From (36) we can get that

logexp ((minus120590 (119873

1+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus1205902 (119863 + 1205761015840))997888rarr minusinfin

120590 997888rarr 0+

(42)

That is

exp ((minus120590 (1198731+ 1)) (2 (119863 + 120576

1015840)))

1 minus exp (minus120590 (2 (119863 + 1205761015840)))997888rarr 0 120590 997888rarr 0

+

(43)

Hence we haveinfin

sum

119899=1198731+1

exp(120582119899exp(minus 1

120588lowast + 120576119883 (120582119899)) minus 120582

119899120590) 997888rarr 0

120590 997888rarr 0+

(44)

Since 119883 isin F and 1199091205731015840(119909) rarr 0 as 119909 rarr infin then thereexists a positive integer119873

2(1198732gt 1198990) satisfying

1

2lt 1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899) lt 2 119899 gt 119873

2 (45)

Take

120582119899= 119882(minus (120588

lowast

+ 120576) log(120590(1 minus 1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

))

119899 gt 1198732

(46)

From (45) and119882(119909)being a strictly increasing function wecan get that119873

2lt 1198731 Then from (33) (45) (47) and 120590 gt 0

we have

max1198732le119899le119873

1

119860lowast

119899exp (minus120590120582

119899)

le exp120582119899exp( minus1

120588lowast + 120576119883 (120582119899)) minus 120590120582

119899

le exp120582119899[120590(1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

minus 120590]

le expminus120582119899120590[1 minus (1 minus

1

120588lowast + 1205761205821198991198831015840

(120582119899))

minus1

]

le exp 120590120582119899 le exp 120590119882((120588lowast + 120576) log 2

120590)

(47)

For any 120590 gt 0 and any 119905 isin 119877 we have

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

le

119899minus1

sum

119896=1

119860lowast

119896(119890minus120582119896+1120590

+10038161003816100381610038161003816119890minus120582119896+1120590

minus 119890minus12058211989612059010038161003816100381610038161003816)

+ 119860lowast

119899(119890minus120590119909

+10038161003816100381610038161003816119890minus120590119909

minus 119890minus12058211989912059010038161003816100381610038161003816)

le

119899

sum

119896=1

119860lowast

119896119890minus120582119896120590

(48)

Thus from (33) (34) (45)ndash(47) and the above inequality wehave

119872119906(120590 119865) le

infin

sum

119899=1

119860lowast

119899119890minus120582119899120590

le (

1198732

sum

119899=1

+

1198731

sum

119899=1198732+1

+

infin

sum

119899=1198731+1

)119860lowast

119899exp (minus120590120582

119899)

le 119867 (1198732) + (119863 + 120576

1015840

)119882((120588lowast

+ 120576) log 2120590)

times exp 120590119882((120588lowast + 120576) log 2120590) + 119900 (1)

120590 997888rarr 0+

(49)

where 119867(1198732) is the sum of finite items of the series

suminfin

119899=1119860lowast

119899119890minus120582119899120590 Hence from the above inequality we have

log119872119906(120590 119865) le (120590119882((120588

lowast

+ 120576) log 2120590)) (1 + 119900 (1))

120590 997888rarr 0+

(50)

6 Journal of Function Spaces

that is

119874 (1)1

120590log119872

119906(120590 119865) le 119882((120588

lowast

+ 120576) log 2120590) 120590 997888rarr 0

+

(51)

Then by using the same argument as in (31) we can deduce

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)le 120588lowast

(52)

From Steps 1 and 2 the sufficiency of the theorem is complet-ed

By using the same argument as in the above proof of thetheorem we can prove the necessity of the theorem easily

Thus we complete the proof of Theorem 4

3 Proofs of Theorems 6 and 7

Similar to the definition of 119883-order of Laplace-Stieltjestransformations in the right half plane we will give thedefinition of 119883-order of Laplace-Stieltjes transformations inthe level half-strip as follows

Definition 14 Let 119865(119904) be the analytic function with infiniteorder represented by Laplace-Stieltjes transformations con-vergent on the right half plane set 119878(119905

0 119897) = 120590 + 119894119905 120590 gt

0 |119905 minus 1199050| le 119897 where 119905

0is a real number and 119897 is a positive

number Let119883 isin F and

120591119883

119878= lim sup120590rarr0

+

119883(log+119872119878(120590 119865))

log (1120590) (53)

where119872119878(120590 119865) = sup

|119905minus1199050|le119897|119865(120590 + 119894119905)| then 120591119883

119878is called the

119883-order of 119865(119878) in the level half-strip 119878(1199050 119897)

To prove Theorems 6 and 7 we need some lemmas asfollows

Lemma 15 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) and120572(119909) = 120572

1(119909)+ 119894120572

2(119909) where 120572

1(119909) is a creasing function and

for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

(54)

then for any 120576 gt 0 one has

120588lowast

= lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120591119878120576

(55)

Proof We will prove this lemma by using the similar argu-ment as in [11] From the assumptions of Lemma 15 for any0 lt 119909 le infin

119872119878120576

(120590 119865) ge

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus120590119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

1198891205721(119910)

ge1

119870 + 1int

infin

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1int

119909

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(56)

Thus we have

119872(120590 119865) ge 119872119878120576

(120590 119865) ge1

119870 + 1119872119906(120590 119865) ge

1

119870 + 1119872 (120590 119865)

(57)

Since119865(119904) is an analytic function with infinite order fromthe above inequality and the definition of119883-order we can getthe conclusion of Lemma 15 easily

Lemma 16 (see [11 Lemma 24]) Let

119911 =1 minus sinh 1199041 + sinh 119904

119904 isin 119861 = 119904 R (119904) gt 0 |I (119904)| lt120587

2

(58)

Then one has the following

(i) this mapping maps the horizontal half-strip 119861 to theunit disc 119911 |119911| lt 1 and its inverse mapping is

119904 = Ψ (119911) = sinhminus1 1 minus 1199111 + 119911

(59)

(ii) min0le120579le2120587

R[Ψ(119903119890119894120579)] ge Ψ(119903) (0 lt 119903 lt 1)

(iii) max0le120579le1205874

R[Ψ(119903119890119894120579)] le Ψ(1199032) (0 lt 119903 lt 1)(iv) Ψ(119911 |119911| lt 119903) sube 119904 R(119904) gt Ψ(119903) |I(119904)| lt 1205872(0 lt 119903 lt 1)

Definition 17 (see [19 20]) Let 119891 be a meromorphic functionin D and lim

119903rarr1minus119879(119903 119891) = infin Then

119863(119891) = lim sup119903rarr1minus

119879 (119903 119891)

minus log (1 minus 119903)(60)

is called the (upper) index of inadmissibility of 119891 If 119863(119891) ltinfin 119891 is called nonadmissible if 119863(119891) = infin 119891 is calledadmissible

Lemma 18 (see [20 Theorem 2]) Let 119891 be a meromorphicfunction in D and lim

119903rarr1minus119879(119903 119891) = infin let 119902 be a positive

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Function Spaces

that is

119874 (1)1

120590log119872

119906(120590 119865) le 119882((120588

lowast

+ 120576) log 2120590) 120590 997888rarr 0

+

(51)

Then by using the same argument as in (31) we can deduce

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)le 120588lowast

(52)

From Steps 1 and 2 the sufficiency of the theorem is complet-ed

By using the same argument as in the above proof of thetheorem we can prove the necessity of the theorem easily

Thus we complete the proof of Theorem 4

3 Proofs of Theorems 6 and 7

Similar to the definition of 119883-order of Laplace-Stieltjestransformations in the right half plane we will give thedefinition of 119883-order of Laplace-Stieltjes transformations inthe level half-strip as follows

Definition 14 Let 119865(119904) be the analytic function with infiniteorder represented by Laplace-Stieltjes transformations con-vergent on the right half plane set 119878(119905

0 119897) = 120590 + 119894119905 120590 gt

0 |119905 minus 1199050| le 119897 where 119905

0is a real number and 119897 is a positive

number Let119883 isin F and

120591119883

119878= lim sup120590rarr0

+

119883(log+119872119878(120590 119865))

log (1120590) (53)

where119872119878(120590 119865) = sup

|119905minus1199050|le119897|119865(120590 + 119894119905)| then 120591119883

119878is called the

119883-order of 119865(119878) in the level half-strip 119878(1199050 119897)

To prove Theorems 6 and 7 we need some lemmas asfollows

Lemma 15 If the Laplace-Stieltjes transformation 119865(119904) ofinfinite order and the sequence (2) satisfy (3) and (4) and120572(119909) = 120572

1(119909)+ 119894120572

2(119909) where 120572

1(119909) is a creasing function and

for any positive number 119870 gt 0 and |120575| 1205722(119909) satisfies

10038161003816100381610038161205722 (119909 + 120575) minus 1205722 (119909)1003816100381610038161003816 le 119870

10038161003816100381610038161205721 (119909 + 120575) minus 1205721 (119909)1003816100381610038161003816

0 le 119909 119909 + 120575 lt +infin

(54)

then for any 120576 gt 0 one has

120588lowast

= lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120591119878120576

(55)

Proof We will prove this lemma by using the similar argu-ment as in [11] From the assumptions of Lemma 15 for any0 lt 119909 le infin

119872119878120576

(120590 119865) ge

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus120590119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

1198891205721(119910)

ge1

119870 + 1int

infin

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1int

119909

0

119890minus120590119910 10038161003816100381610038161198891205721 (119910)

1003816100381610038161003816

ge1

119870 + 1

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus(120590+119894119905)119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(56)

Thus we have

119872(120590 119865) ge 119872119878120576

(120590 119865) ge1

119870 + 1119872119906(120590 119865) ge

1

119870 + 1119872 (120590 119865)

(57)

Since119865(119904) is an analytic function with infinite order fromthe above inequality and the definition of119883-order we can getthe conclusion of Lemma 15 easily

Lemma 16 (see [11 Lemma 24]) Let

119911 =1 minus sinh 1199041 + sinh 119904

119904 isin 119861 = 119904 R (119904) gt 0 |I (119904)| lt120587

2

(58)

Then one has the following

(i) this mapping maps the horizontal half-strip 119861 to theunit disc 119911 |119911| lt 1 and its inverse mapping is

119904 = Ψ (119911) = sinhminus1 1 minus 1199111 + 119911

(59)

(ii) min0le120579le2120587

R[Ψ(119903119890119894120579)] ge Ψ(119903) (0 lt 119903 lt 1)

(iii) max0le120579le1205874

R[Ψ(119903119890119894120579)] le Ψ(1199032) (0 lt 119903 lt 1)(iv) Ψ(119911 |119911| lt 119903) sube 119904 R(119904) gt Ψ(119903) |I(119904)| lt 1205872(0 lt 119903 lt 1)

Definition 17 (see [19 20]) Let 119891 be a meromorphic functionin D and lim

119903rarr1minus119879(119903 119891) = infin Then

119863(119891) = lim sup119903rarr1minus

119879 (119903 119891)

minus log (1 minus 119903)(60)

is called the (upper) index of inadmissibility of 119891 If 119863(119891) ltinfin 119891 is called nonadmissible if 119863(119891) = infin 119891 is calledadmissible

Lemma 18 (see [20 Theorem 2]) Let 119891 be a meromorphicfunction in D and lim

119903rarr1minus119879(119903 119891) = infin let 119902 be a positive

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 7

integer and let 1198861 1198862 119886

119902be pairwise distinct complex

numbers Then for 119903 rarr 1minus 119903 notin 119864

(119902 minus 2) 119879 (119903 119891) le

119902

sum

119895=1

119873(1199031

119891 minus 119886119895

) + log 1

1 minus 119903+ 119878 (119903 119891)

(61)

Remark 19 Wecan see that 119878(119903 119891) = 119900(log(1(1minus119903)))holds inLemma 18 without a possible exception set when 0 lt 119863(119891) ltinfin

Remark 20 The term log(1(1minus119903)) in Lemma 18 can enter theremainder 119878(119903 119891) = 119900(119879(119903 119891)) when the function 119891 satisfies119863(119891) = infin

FromLemmas 33 and 34 in [21] we can get the followingresult for nonadmissible functions in the unit disc which isused in this paper

Lemma 21 (see [19 page 282 (18)]) Let ℎ be analytic in thedisc |119911| = 119903 lt 1 then

119879 (119903 ℎ) le log119872(119903 ℎ) le 1 + 1199031 minus 119903

119879 (119903 ℎ) (62)

where119872(119903 ℎ) is the maximum modulus of ℎ in the disc |119911| =119903 lt 1

31 The Proof of Theorem 6 From the sequence (2) satisfying(3) and (4) Laplace-Stieltjes transformation 119865(119904) of infiniteorder and lim sup

119899rarrinfin(119883(120582119899)log+(120582

119899(log+119860lowast

119899))) = 120588

lowast

(0 lt 120588lowastlt infin) fromTheorem 4 we have

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= 120588lowast

(63)

and from Lemma 15 and (63) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)= 120588lowast

(64)

Thus we can get

lim sup120590rarr0

+

119883(log119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(65)

where 119878120576= 119904 R(119904) gt 0 |I(119904)| le 120576 and 119872(120590 119865 119878

120576) =

sup|119865(119904)| R(119904) ge 120590 119904 isin 119878120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911)) where Ψ(119911) is stated as inLemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205761

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(66)

where 0 lt 1205761lt 120576 Therefore from (66) and Lemma 16 we

have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(67)

From Lemma 21 by using the same argument as in (31) wecan get

119883(1 + 119903

1 minus 119903119879 (119903 119892)) = 119883 (119879 (119903 119892)) + 119900 (1) log 1 + 119903

1 minus 119903 (68)

From (67) (68) and Lemma 21 we can get that 119892 is anadmissible function in |119911| lt 1 and

lim sup119903rarr1minus

119883(119879 (119903 119892))

log (1 (1 minus 119903))= 120588lowast

(69)

Then fromLemma 18 we can get that there atmost exists oneexcept value 119886 satisfying

lim sup119903rarr1minus

119883(119899 (119903 119892 = 119886))

log (1 (1 minus 119903))= 984858 ge 120588

lowast

(70)

that is for any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 0 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (71)

holds for any 119886 isin C except for one exception where119899(120590 0 120578 119865 = 119886) is the counting function of zeros of thefunction 119865(119904) minus 119886 in the strip 119904 R(119904) gt 120590 |I(119904)| lt 120578

Thus we complete the proof of Theorem 6

32 The Proof of Theorem 7 Since 119903(119910) is a continuousfunction on 119910 isin [0 +infin) then we can get that 120572(119909) =int119909

0119903(119910)1198901198941199050119910119889119910 is a function of bounded variation on 119909 isin

[0 119883] (0 lt 119883 lt infin) Set

119878120576= 119904 R (119904) gt 0

1003816100381610038161003816I (119904) minus 11990501003816100381610038161003816 le 120576

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)| (72)

From the assumptions of Theorem 7 for any real number119909 (0 lt 119909 le infin) we have

119872119878120576

(120590 119865) = sup|119905minus1199050|le120576

|119865 (120590 + 119894119905)|

= sup|119905minus1199050|le120576

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

ge int

infin

0

119890minus120590119910

119903 (119910) 119889119910 ge int

119909

0

119890minus120590119910

119903 (119910) 119889119910

ge

10038161003816100381610038161003816100381610038161003816

int

119909

0

119890minus119904119910

119903 (119910) 1198901198941199050119910

119889119910

10038161003816100381610038161003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

int

infin

0

119890minus119904119910

119889120572 (119910)

10038161003816100381610038161003816100381610038161003816

(73)

that is

119872(120590 119865) le 119872119906(120590 119865) le 119872

119878120576

(120590 119865) le 119872 (120590 119865) (74)

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Function Spaces

From the assumption of Theorem 7 by (74) and Theorem 4we can get

lim sup120590rarr0

+

119883(log+119872119906(120590 119865))

log (1120590)= lim sup120590rarr0

+

119883(log+119872119878120576

(120590 119865))

log (1120590)

= lim sup120590rarr0

+

119883(log+119872(120590 119865))log (1120590)

= 120588lowast

(75)

From (75) for any 120576 gt 0 we have

lim sup120590rarr0

+

119883(log+119872(120590 119865 119878120576))

log (1120590)= 120588lowast

(76)

where119872(120590 119865 119878120576) = sup|119865(119904)| R(119904) ge 120590 119904 isin 119878

120576

Set 119892(119911) = 119865((2120576120587)Ψ(119911) + 1198941199050) whereΨ(119911) is stated as in

Lemma 16 then fromLemma 16 we have that119892(119911) is analyticin the unit disc |119911| lt 1 and satisfies

119872(2120576

120587Ψ (1199032

) 119865 1198781205762

) le 119872(119903 119892) le 119872(2120576

120587Ψ (119903) 119865 119878

120576)

(77)

where 0 lt 1205762lt 120576 Therefore from the above inequality and

Lemma 16 we have

lim sup119903rarr1minus

log+log+119872(119903 119892)log (1 (1 minus 119903))

= infin

lim sup119903rarr1minus

119883(log+119872(119903 119892))log (1 (1 minus 119903))

= 120588lowast

(78)

Then similar to the proof of Theorem 6 we can prove thatfor any 120578 gt 0 the inequality

lim sup120590rarr0

+

119883(119899 (120590 1198941199050 120578 119865 = 119886))

log (1120590)= 984858 ge 120588

lowast (79)

holds for any 119886 isin C except for one exception where 119899(120590 1198941199050

120578 119865 = 119886) is stated as in Theorem 7Thus we complete the proof of Theorem 7

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was supported by the National Natural ScienceFoundation of China (11301233 and 61202313) the NaturalScience Foundation of Jiangxi Province in China (20132BAB-211001) and the Foundation of Education Department ofJiangxi (GJJ14644) of China The second author is supportedin part by Beijing Natural Science Foundation (no 1132013)and The Project of Construction of Innovative Teams andTeacher Career Development for Universities and CollegesUnder Beijing Municipality (CIT and TCD20130513)

References

[1] M S Liu ldquoThe regular growth of Dirichlet series of finite orderin the half planerdquo Journal of Systems Science and MathematicalSciences vol 22 no 2 pp 229ndash238 2002

[2] D C Sun and J R Yu ldquoOn the distribution of values of randomDirichlet series IIrdquo Chinese Annals of Mathematics B vol 11 no1 pp 33ndash44 1990

[3] J R YuDirichlet Series and the RandomDirichlet Series SciencePress Beijing China 1997

[4] J R Yu ldquoBorelrsquos line of entire functions represented by Laplace-Stieltjes transformationrdquo Acta Mathematica Sinica vol 13 pp471ndash484 1963 (Chinese)

[5] C J K Batty ldquoTauberian theorems for the Laplace-Stieltjestransformrdquo Transactions of the American Mathematical Societyvol 322 no 2 pp 783ndash804 1990

[6] Y Y Kong andD C Sun ldquoOn the growth of zero order Laplace-Stieltjes transform convergent in the right half-planerdquo ActaMathematica Scientia B vol 28 no 2 pp 431ndash440 2008

[7] Y Y Kong and D C Sun ldquoThe analytic function in the righthalp plane defined by Laplace-Stieltjes transformsrdquo Journal ofMathematical Research amp Exposition vol 28 no 2 pp 353ndash3582008

[8] Y Y Kong and Y Hong On the Growth of Laplace-StieltjesTransforms and the Singular Direction of Complex AnalysisJinan University Press Guangzhou China 2010

[9] Y Y Kong and Y Yang ldquoOn the growth properties of theLaplace-Stieltjes transformrdquo Complex Variables and EllipticEquations vol 59 no 4 pp 553ndash563 2014

[10] A Mishkelyavichyus ldquoA Tauberian theorem for the Laplace-Stieltjes integral and the Dirichlet seriesrdquo Litovskij Matematich-eskij Sbornik vol 29 no 4 pp 745ndash753 1989 (Russian)

[11] L N Shang and Z S Gao ldquoValue distribution of analyticfunctions defined by Laplace-Stieltjes transformsrdquo Acta Math-ematica Sinica Chinese Series vol 51 no 5 pp 993ndash1000 2008

[12] Y Y Kong ldquoLaplace-Stieltjes transforms of infinite order in theright half-planerdquoActaMathematica Sinica vol 55 no 1 pp 141ndash148 2012

[13] Y Y Kong and D C Sun ldquoOn type-function and the growth ofLaplace-Stieltjes transformations convergent in the right half-planerdquoAdvances inMathematics vol 37 no 2 pp 197ndash205 2008(Chinese)

[14] L N Shang and Z S Gao ldquoThe growth of entire functions ofinfinite order represented by Laplace-Stieltjes transformationrdquoActa Mathematica Scientia A vol 27 no 6 pp 1035ndash1043 2007(Chinese)

[15] PWang ldquoThe119875(119877) type of the Laplace-Stieltjes transform in theright half-planerdquo Journal of Central China Normal Universityvol 21 no 1 pp 17ndash25 1987 (Chinese)

[16] P Wang ldquoThe 119875(119877) order of analytic functions defined by theLaplace-Stieltjes transformrdquo Journal of Mathematics vol 8 no3 pp 287ndash296 1988 (Chinese)

[17] H Y Xu and Z X Xuan ldquoThe growth and value distribution ofLaplace-Stieltjes transformations with infinite order in the righthalf-planerdquo Journal of Inequalities and Applications vol 2013article 273 15 pages 2013

[18] K Knopp ldquoUber die Konvergenzabszisse des Laplace-Inte-gralsrdquoMathematische Zeitschrift vol 54 pp 291ndash296 1951

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 9

[19] T Cao and H Yi ldquoThe growth of solutions of linear differentialequations with coefficients of iterated order in the unit discrdquoJournal of Mathematical Analysis and Applications vol 319 no1 pp 278ndash294 2006

[20] F Titzhoff ldquoSlowly growing functions sharing valuesrdquo Fizikosir Matematikos Fakulteto Mokslinio Seminaro Darbai vol 8 pp143ndash164 2005

[21] H X Yi and C C Yang Uniqueness Theory of MeromorphicFunctions Kluwer Academic Publishers Beijing China 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of