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Research ArticlePascu-Type Harmonic Functions with Positive CoefficientsInvolving Salagean Operator
K Vijaya G Murugusundaramoorthy and M Kasthuri
School of Advanced Sciences VIT University Vellore 632014 India
Correspondence should be addressed to K Vijaya kvijayavitacin
Received 25 November 2013 Accepted 21 February 2014 Published 6 April 2014
Academic Editor Remi Leandre
Copyright copy 2014 K Vijaya et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
Making use of a Salagean operator we introduce a new class of complex valued harmonic functionswhich are orientation preservingand univalent in the open unit disc Among the results presented in this paper including the coeffcient bounds distortion inequalityand covering property extreme points certain inclusion results convolution properties and partial sums for this generalized classof functions are discussed
1 Introduction and Preliminaries
A continuous function 119891 = 119906 + 119894V is a complex-valuedharmonic function in a complex domain G if both 119906 and Vare real and harmonic inG In any simply connected domainD sub G we can write 119891 = ℎ + 119892 where ℎ and 119892 are analyticin D We call ℎ the analytic part and 119892 the coanalytic partof 119891 A necessary and sufficient condition for 119891 to be locallyunivalent and orientation preserving in D is that |ℎ1015840(119911)| gt|1198921015840(119911)| inD (see [1])
Denote byH the family of functions
119891 = ℎ + 119892 (1)
which are harmonic univalent and orientation preserving inthe open unit discU = 119911 |119911| lt 1 so that119891 is normalized by119891(0) = 119891
119911(0)minus1 = 0Thus for119891 = ℎ+119892 isinH the functions ℎ
and 119892 are analytic inU and can be expressed in the followingforms
ℎ (119911) = 119911 +infin
sum119899=2
119886119899119911119899 119892 (119911) =
infin
sum119899=1
119887119899119911119899 (0 le 119887
1lt 1)
(2)
and 119891(119911) is then given by
119891 (119911) = 119911 +infin
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899 (0 le
100381610038161003816100381611988711003816100381610038161003816 lt 1) (3)
We note that the familyH of orientation preserving normal-ized harmonic univalent functions reduces to thewell-knownclass S of normalized univalent functions if the coanalyticpart of 119891 is identically zero that is 119892 equiv 0
For functions 119891 isinH Jahangiri et al [2] defined Salageanoperator on harmonic functions given by
119863ℓ119891 (119911) = 119863ℓℎ (119911) + (minus1)
ℓ119863ℓ119892 (119911) (4)
where
119863ℓℎ (119911) = 119911 +infin
sum119899=2
119899ℓ119886119899119911119899 119863ℓ119892 (119911) =
infin
sum119899=1
119899ℓ119887119899119911119899
(5)
In 1975 Silverman [3] introduced a new class T ofanalytic functions of the form 119891(119911) = 119911 minus sum
infin
119899=2|119886119899|119911119899
and opened up a new direction of studies in the theory ofunivalent functions as well as in harmonic functions withnegative coefficients [4] Uralegaddi et al [5] introducedanalogous subclasses of star-like convex functions withpositive coefficients and opened up a new and interestingdirection of research In fact they considered the functionswhere the coefficients are positive rather than negative realnumbers Motivated by the initial work of Uralegaddi et al[5] many researchers (see [6ndash9]) introduced and studiedvarious new subclasses of analytic functions with positivecoefficients but analogues results on harmonic univalent
Hindawi Publishing CorporationInternational Journal of AnalysisVolume 2014 Article ID 793709 10 pageshttpdxdoiorg1011552014793709
2 International Journal of Analysis
functions have not been explored in the literature Veryrecently Dixit and Porwal [10] attempted to fill this gap byintroducing a new subclass of harmonic univalent functionswith positive coefficients
Denote by VH the subfamily of H consisting of har-monic functions 119891 = ℎ + 119892 of the form
119891 (119911) = 119911 +infin
sum119899=2
119886119899119911119899 minusinfin
sum119899=1
119887119899119911119899
(119886119899ge 0 119887
119899ge 0 0 le
100381610038161003816100381611988711003816100381610038161003816 lt 1)
(6)
Motivated by the earlier works of [11ndash14] on the subjectof harmonic functions in this paper an attempt has beenmade to study the class of functions 119891 isinVH associated withSalagean operator on harmonic functions Further we obtaina sufficient coefficient condition for functions119891 isinH given by(3) and also show that this coefficient condition is necessaryfor functions 119891 isin VH the class of harmonic functions withpositive coefficients Distortion results and extreme pointsinclusion relations and convolution properties and results onpartial sums are discussed extensively
For 0 le 120582 le 1 1 lt 120574 le 43 we let PℓH(120582 120574) be anew subclass ofH consisting of all functions of the form (3)satisfying the condition
R((1 minus 120582)119863ℓ+1119891 (119911) + 120582119863ℓ+2119891 (119911)
(1 minus 120582)119863ℓ119891 (119911) + 120582119863ℓ+1119891 (119911)) lt 120574 (7)
where 119863ℓ119891(119911) is given by (4) (see [2]) Also let VℓH(120582 120574) =PℓH(120582 120574)⋂VH
2 Coefficient Bounds
In our first theorem we obtain a sufficient coefficient condi-tion for harmonic functions inPℓH(120582 120574)
Theorem 1 Let 119891 = ℎ + 119892 be given by (3) If
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(8)
where 1198861= 1 and 1 lt 120574 le 43 then 119891 isin PℓH(120582 120574)
Proof We let (8) hold for the coefficients of 119891 = ℎ + 119892 Itsuffices to show that
100381610038161003816100381610038161003816100381610038161003816
119860 (119911) 119861 (119911) minus 1
119860 (119911) 119861 (119911) minus (2120574 minus 1)
100381610038161003816100381610038161003816100381610038161003816lt 1 119911 isin U (9)
where
119860 (119911) = (1 minus 120582)119863ℓ+1119891 (119911) + 120582119863
ℓ+2119891 (119911)
= 119911 +infin
sum119899=2
119899ℓ+1 (1 minus 120582 + 119899120582) 119886119899119911119899
minus (minus1)ℓ
infin
sum119899=1
119899ℓ+1 (1 minus 120582 minus 119899120582) 119887119899119911119899
(10)
119861 (119911) = (1 minus 120582)119863ℓ119891 (119911) + 120582119863
ℓ+1119891 (119911)
= 119911 +infin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) 119886119899119911119899
+ (minus1)ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) 119887119899119911119899
(11)
Substituting for 119860(119911) and 119861(119911) in (9) we get100381610038161003816100381610038161003816100381610038161003816
119860 (119911) 119861 (119911) minus 1
119860 (119911) 119861 (119911) minus (2120574 minus 1)
100381610038161003816100381610038161003816100381610038161003816
le (infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 1)10038161003816100381610038161198861198991003816100381610038161003816 |119911|119899minus1 + (minus1)
ℓ
timesinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1)10038161003816100381610038161198871198991003816100381610038161003816 |119911|119899minus1)
times (2 (120574 minus 1) minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| [119899 minus (2120574 minus 1)]
times10038161003816100381610038161198861198991003816100381610038161003816 |119911|119899minus1 minus (minus1)
ℓ
timesinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| [119899 + (2120574 minus 1)]10038161003816100381610038161198871198991003816100381610038161003816 |119911|119899minus1)
minus1
le (infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 1)10038161003816100381610038161198861198991003816100381610038161003816
+ (minus1)ℓ
infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1)10038161003816100381610038161198871198991003816100381610038161003816)
times (2 (120574 minus 1) minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| [119899 minus (2120574 minus 1)]10038161003816100381610038161198861198991003816100381610038161003816
minus (minus1)ℓ
infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| [119899 + (2120574 minus 1)]10038161003816100381610038161198871198991003816100381610038161003816)
minus1
(12)
The above expression is bounded above by 1 if
2infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 1)10038161003816100381610038161198861198991003816100381610038161003816
+ 2infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1)10038161003816100381610038161198871198991003816100381610038161003816 le 2 (120574 minus 1)
(13)
International Journal of Analysis 3
which is equivalent to
infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(14)
But (8) is true by hypothesis Hence |(119860(119911)119861(119911) minus 1)(119860(119911)119861(119911)minus (2120574minus1))| lt 1 119911 isin U and the theorem is provedfor 119860(119911) and 119861(119911) is given by (10) and (11) respectively
Theorem 2 For 1198861= 1 and 1 lt 120574 le 43 119891 = ℎ + 119892 isin
VℓH (120582 120574) if and only if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(15)
Proof SinceVℓH(120582 120574) sub PℓH(120582 120574) we only need to prove theldquoonly if rdquo part of the theorem To this end for functions 119891 ofthe form (6) we notice that the condition
R((1 minus 120582)119863ℓ+1119891 (119911) + 120582119863ℓ+2119891 (119911)
(1 minus 120582)119863ℓ119891 (119911) + 120582119863ℓ+1119891 (119911)) lt 120574 (16)
Equivalently
R(((1 minus 120574) 119911 minusinfin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) (119899 minus 120574) 119886119899119911119899
minus (minus1)2ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) (119899 + 120574) 119887119899119911119899)
times (119911 minusinfin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) 119886119899119911119899
+ (minus1)2ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) 119887119899119911119899)
minus1
)
ge 0
(17)
The above required condition must hold for all values of 119911 inU Upon choosing the values of 119911 on the positive real axiswhere 0 le 119911 = 119903 lt 1 we must have
(1 minus 120574 minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574) 119886119899119903119899minus1
minusinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574) 119887119899119903119899minus1)
times (1 minusinfin
sum119899=2
119899ℓ |(1 minus 120582 + 119899120582)| 119886119899119903119899minus1
+infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| 119887119899119903119899minus1)
minus1
ge 0
(18)
If condition (15) does not hold then the numerator in (18)is negative for 119903 sufficiently close to 1 Hence there exists1199110= 1199030in (0 1) for which the quotient of (18) is negative
This contradicts the required condition for 119891 isin VℓH(120582 120574)This completes the proof of the theorem
3 Distortion Bounds and Extreme Points
By routine procedure (see [10ndash13]) we can easily prove thefollowing results hence we state the following theoremswithout proof for functions inVℓH(120582 120574)
Theorem 3 (distortion bounds) Let 119891 isin VℓH(120582 120574) Then for|119911| = 119903 lt 1 we have
(1 minus100381610038161003816100381611988711003816100381610038161003816) 119903 minus
1
2ℓ (1 + 120582)(120574 minus 1
2 minus 120574minus1 + 120574
2 minus 120574
100381610038161003816100381611988711003816100381610038161003816) 1199032 le
1003816100381610038161003816119891 (119911)1003816100381610038161003816
le (1 +100381610038161003816100381611988711003816100381610038161003816) 119903 +
1
2ℓ (1 + 120582)(120574 minus 1
2 minus 120574minus1 + 120574
(2 minus 120574)
100381610038161003816100381611988711003816100381610038161003816) 1199032
(19)
Corollary 4 (covering result) If 119891(119911) isinVℓH(120582 120574) then
119908 |119908| lt2ℓ+1 (1 + 120574) + 1 minus [2ℓ (1 + 120574) + 1] 120574
2ℓ (2 minus 120574) (1 + 120582)
minus2ℓ+1 (1 + 120574) minus 1 minus [2ℓ (1 + 120574) + 1] 120574
2ℓ (2 minus 120574) (1 + 120582) 1198871
sub 119891 (119880)
(20)
Next we state the extreme points of closed convex hulls ofVℓH(120582 120574) denoted by clcoVℓH(120582 120574)
Theorem 5 A function 119891(119911) isinVℓH(120582 120574) if and only if 119891(119911) =suminfin
119899=1(119883119899ℎ119899(119911) + 119884
119899119892119899(119911)) where ℎ
1(119911) = 119911 ℎ
119899(119911) = 119911 + ((120574 minus
1)(2ℓ|1 minus 120582 + 119899120582|(119899 minus 120574)))119911119899 (119899 ge 2) and 119892119899(119911) = 119911 + ((120574 minus
1)(2ℓ|1 minus 120582 minus 119899120582|(119899 + 120574)))119911119899 (119899 ge 1) also suminfin119899=1(119883119899+ 119884119899) =
1 119883119899ge 0 and 119884
119899ge 0 In particular the extreme points of
VℓH(120582 120574) are ℎ119899 and 119892119899
4 International Journal of Analysis
Theorem 6 The family VℓH(120582 120574) is closed under convexcombinations
4 Inclusion Results
Now we will examine the closure properties of the classVℓH(120582 120574) under the generalized Bernardi-Libera-Livingstonintegral operator L
119888(119891) which is defined by L
119888(119891) = ((119888 +
1)119911119888) int119911
0119905119888minus1119891(119905)119889119905 119888 gt minus1
Theorem 7 Let 119891(119911) isin VℓH(120582 120574) Then L119888(119891(119911)) isin
VℓH(120582 120574)
Lemma 8 (see [15]) Let 119891 = ℎ + 119892 be given by (3) If
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120572)
1 minus 120572
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120572)
1 minus 120572
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(21)
where 1198861= 1 and 0 le 120572 lt 1 then 119891 isinMlowastH(120582 120572)
Theorem 9 Let 119891 = ℎ + 119892 isin VℓH(120582 120572) be given by (6) Then119891 isinVℓH(120582 (4 minus 3120574)(3 minus 2120574))
Proof Since 119891 isinVℓH(120582 120574) then byTheorem 1 we must have
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(22)
To show that 119891 isin VℓH(120582 (4 minus 3120574)(3 minus 2120574)) by virtue ofLemma 8 we have to show that
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(23)
where 0 le (4 minus 3120574)(3 minus 2120574) lt 1 For this it is sufficient toprove that
|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
ge|1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 2 3 )
|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
ge|1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 1 2 3 )
(24)
or equivalently (120574 minus 1)|1 minus 120582 + 119899120582|(119899 minus 120574) ge 0 (119899 = 2 3 )and (120574 minus 1)|1 minus 120582 minus 119899120582|(119899 + 120574) ge 0 (119899 = 1 2 3 ) which istrue and the theorem is proved
Corollary 10 VℓH(120582 120574) subVℓH(120582 43) subVℓH
5 Convolution Properties
For functions 119891 isinH given by (3) and 119865 isinH given by
119865 (119911) = 119867 (119911) + 119866 (119911) = 119911 +infin
sum119899=2
119860119899119911119899 +infin
sum119899=1
119861119899119911119899 (25)
we recall the Hadamard product (or convolution) of 119891 and 119865by
(119891 lowast 119865) (119911) = 119911 +infin
sum119899=2
119886119899119860119899119911119899 +infin
sum119899=1
119887119899119861119899119911119899 (119911 isin U)
(26)
Let 119865119895(119911) isinVℓH(120582 120574) (119895 = 1 2 3 119901) be given by
119865119895(119911) = 119911 +
infin
sum119899=2
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899 + (minus1)
ℓ
infin
sum119899=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899 (27)
then the convolution is defined by
(1198651lowast sdot sdot sdot lowast 119865
119901) (119911) = 119911 minus
infin
sum119899=2
119901
prod119895=1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899
+ (minus1)ℓ
infin
sum119899=1
119901
prod119895=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899
(28)
Theorem 11 Let 119865119895(119911) isin VℓH(120582 120574119895) (119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(29)
Proof We use the principle of mathematical induction in ourproof Let 119865
1isin VℓH(120582 1205741) and 1198652 isin VℓH(120582 1205742) By using
Theorem 2 we haveinfin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(30)
International Journal of Analysis 5
then
[
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
timesinfin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
+ [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
timesinfin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
le 1
(31)
Thus by applying Cauchy-Schwarz inequality we have
infin
sum119899=2
(radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816)
2
le [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
infin
sum119899=1
(radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816)
2
le [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
(32)
Then we get
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
+infin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
le 1
(33)
Therefore if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120572)
120572 minus 1
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
leinfin
sum119899=2
radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120572)
120572 minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
leinfin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
(34)
that is if
radic1003816100381610038161003816119886119899111988611989921003816100381610038161003816 le
120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic1003816100381610038161003816119887119899111988711989921003816100381610038161003816 le
120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(35)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) By (30) we have
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 le 1
infin
sum119899=1
radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(36)
Hence we get
radic10038161003816100381610038161003816119886119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 + 119899120582| (119899 minus 120574119895)
radic10038161003816100381610038161003816119887119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 minus 119899120582| (119899 + 120574119895)
(37)
Consequently if
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 + 119899120582|2 (119899 minus 1205741) (119899 minus 120574
2)
le120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 minus 119899120582|2 (119899 + 1205741) (119899 + 120574
2)
le120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(38)
6 International Journal of Analysis
That is if
119899 minus 120572
120572 minus 1le|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
119899 + 120572
120572 minus 1le|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(39)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) Then we see that
120572 ge 1
+(119899 minus 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 + 119899120582| (119899 + 1205741) (119899 + 120574
2) + (1205741minus 1) (120574
2minus 1)
= 120601 (119899)
120572 ge 1
+(119899 + 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1205741) (119899 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
= 120595 (119899)
(40)
Since 120601(119899) for 119899 ge 2 and 120595(119899) for 119899 ge 1 are increasing
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(41)
120572 ge 1 +2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(42)
and also(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
le2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(43)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) where
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ (1 + 120582) (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(44)
Next we suppose that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901)(119911) isin VℓH(120582 120573)
where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(45)
We can show that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901+1)(119911) isin VℓH(120582 120575)
where
120575 ge 1 +(120573 minus 1) (120574
119901+1minus 1)
2ℓ (1 + 120582) (2 + 120573) (2 + 120574119901+1) + (120573 minus 1) (120574
119901+1minus 1)
(46)
Since
(120573 minus 1) (120572119901+1minus 1)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(120573 minus 2) (120572119901+1minus 2)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(47)
we have
120575 = 1 +prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901+1
119895=1(2 + 120574
119895) minus prod
119901+1
119895=1(120574119895minus 1)
(48)
Corollary 12 Let 119865119895(119911) isin VℓH(120582 120574)(119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +(120574 minus 1)
119901
2ℓ (1 + 120582) (2 + 120574)119901
minus (120574 minus 1)119901 (49)
6 Partial Sums Results
In 1985 Silvia [16] studied the partial sums of convexfunctions of order 120572 (0 le 120572 lt 1) Later on Silverman [17]and several researchers studied and generalized the resultson partial sums for various classes of analytic functionsonly but analogues results on harmonic functions have notbeen explored in the literature Very recently Porwal [18]and Porwal and Dixit [19] filled this gap by investigatinginteresting results on the partial sums of star-like harmonicunivalent functions Now in this section we discussed thepartial sums results for the class of harmonic functions withpositive coefficients based on Salagean operator of order120574 (1 lt 120574 le 43) on lines similar to Porwal [18]
LetPℓH(119860119899120575 119861119899120575) denote the subclass ofH consistingof functions 119891 = ℎ + 119892 of the form (3) which satisfy theinequality
infin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=1
119861119899
120575
10038161003816100381610038161198871198991003816100381610038161003816 le 1 (50)
where
119860119899
120575=119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
119861119899
120575=119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
(51)
and 120575 = 120574 minus 1 unless otherwise stated
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 International Journal of Analysis
functions have not been explored in the literature Veryrecently Dixit and Porwal [10] attempted to fill this gap byintroducing a new subclass of harmonic univalent functionswith positive coefficients
Denote by VH the subfamily of H consisting of har-monic functions 119891 = ℎ + 119892 of the form
119891 (119911) = 119911 +infin
sum119899=2
119886119899119911119899 minusinfin
sum119899=1
119887119899119911119899
(119886119899ge 0 119887
119899ge 0 0 le
100381610038161003816100381611988711003816100381610038161003816 lt 1)
(6)
Motivated by the earlier works of [11ndash14] on the subjectof harmonic functions in this paper an attempt has beenmade to study the class of functions 119891 isinVH associated withSalagean operator on harmonic functions Further we obtaina sufficient coefficient condition for functions119891 isinH given by(3) and also show that this coefficient condition is necessaryfor functions 119891 isin VH the class of harmonic functions withpositive coefficients Distortion results and extreme pointsinclusion relations and convolution properties and results onpartial sums are discussed extensively
For 0 le 120582 le 1 1 lt 120574 le 43 we let PℓH(120582 120574) be anew subclass ofH consisting of all functions of the form (3)satisfying the condition
R((1 minus 120582)119863ℓ+1119891 (119911) + 120582119863ℓ+2119891 (119911)
(1 minus 120582)119863ℓ119891 (119911) + 120582119863ℓ+1119891 (119911)) lt 120574 (7)
where 119863ℓ119891(119911) is given by (4) (see [2]) Also let VℓH(120582 120574) =PℓH(120582 120574)⋂VH
2 Coefficient Bounds
In our first theorem we obtain a sufficient coefficient condi-tion for harmonic functions inPℓH(120582 120574)
Theorem 1 Let 119891 = ℎ + 119892 be given by (3) If
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(8)
where 1198861= 1 and 1 lt 120574 le 43 then 119891 isin PℓH(120582 120574)
Proof We let (8) hold for the coefficients of 119891 = ℎ + 119892 Itsuffices to show that
100381610038161003816100381610038161003816100381610038161003816
119860 (119911) 119861 (119911) minus 1
119860 (119911) 119861 (119911) minus (2120574 minus 1)
100381610038161003816100381610038161003816100381610038161003816lt 1 119911 isin U (9)
where
119860 (119911) = (1 minus 120582)119863ℓ+1119891 (119911) + 120582119863
ℓ+2119891 (119911)
= 119911 +infin
sum119899=2
119899ℓ+1 (1 minus 120582 + 119899120582) 119886119899119911119899
minus (minus1)ℓ
infin
sum119899=1
119899ℓ+1 (1 minus 120582 minus 119899120582) 119887119899119911119899
(10)
119861 (119911) = (1 minus 120582)119863ℓ119891 (119911) + 120582119863
ℓ+1119891 (119911)
= 119911 +infin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) 119886119899119911119899
+ (minus1)ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) 119887119899119911119899
(11)
Substituting for 119860(119911) and 119861(119911) in (9) we get100381610038161003816100381610038161003816100381610038161003816
119860 (119911) 119861 (119911) minus 1
119860 (119911) 119861 (119911) minus (2120574 minus 1)
100381610038161003816100381610038161003816100381610038161003816
le (infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 1)10038161003816100381610038161198861198991003816100381610038161003816 |119911|119899minus1 + (minus1)
ℓ
timesinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1)10038161003816100381610038161198871198991003816100381610038161003816 |119911|119899minus1)
times (2 (120574 minus 1) minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| [119899 minus (2120574 minus 1)]
times10038161003816100381610038161198861198991003816100381610038161003816 |119911|119899minus1 minus (minus1)
ℓ
timesinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| [119899 + (2120574 minus 1)]10038161003816100381610038161198871198991003816100381610038161003816 |119911|119899minus1)
minus1
le (infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 1)10038161003816100381610038161198861198991003816100381610038161003816
+ (minus1)ℓ
infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1)10038161003816100381610038161198871198991003816100381610038161003816)
times (2 (120574 minus 1) minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| [119899 minus (2120574 minus 1)]10038161003816100381610038161198861198991003816100381610038161003816
minus (minus1)ℓ
infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| [119899 + (2120574 minus 1)]10038161003816100381610038161198871198991003816100381610038161003816)
minus1
(12)
The above expression is bounded above by 1 if
2infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 1)10038161003816100381610038161198861198991003816100381610038161003816
+ 2infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1)10038161003816100381610038161198871198991003816100381610038161003816 le 2 (120574 minus 1)
(13)
International Journal of Analysis 3
which is equivalent to
infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(14)
But (8) is true by hypothesis Hence |(119860(119911)119861(119911) minus 1)(119860(119911)119861(119911)minus (2120574minus1))| lt 1 119911 isin U and the theorem is provedfor 119860(119911) and 119861(119911) is given by (10) and (11) respectively
Theorem 2 For 1198861= 1 and 1 lt 120574 le 43 119891 = ℎ + 119892 isin
VℓH (120582 120574) if and only if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(15)
Proof SinceVℓH(120582 120574) sub PℓH(120582 120574) we only need to prove theldquoonly if rdquo part of the theorem To this end for functions 119891 ofthe form (6) we notice that the condition
R((1 minus 120582)119863ℓ+1119891 (119911) + 120582119863ℓ+2119891 (119911)
(1 minus 120582)119863ℓ119891 (119911) + 120582119863ℓ+1119891 (119911)) lt 120574 (16)
Equivalently
R(((1 minus 120574) 119911 minusinfin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) (119899 minus 120574) 119886119899119911119899
minus (minus1)2ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) (119899 + 120574) 119887119899119911119899)
times (119911 minusinfin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) 119886119899119911119899
+ (minus1)2ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) 119887119899119911119899)
minus1
)
ge 0
(17)
The above required condition must hold for all values of 119911 inU Upon choosing the values of 119911 on the positive real axiswhere 0 le 119911 = 119903 lt 1 we must have
(1 minus 120574 minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574) 119886119899119903119899minus1
minusinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574) 119887119899119903119899minus1)
times (1 minusinfin
sum119899=2
119899ℓ |(1 minus 120582 + 119899120582)| 119886119899119903119899minus1
+infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| 119887119899119903119899minus1)
minus1
ge 0
(18)
If condition (15) does not hold then the numerator in (18)is negative for 119903 sufficiently close to 1 Hence there exists1199110= 1199030in (0 1) for which the quotient of (18) is negative
This contradicts the required condition for 119891 isin VℓH(120582 120574)This completes the proof of the theorem
3 Distortion Bounds and Extreme Points
By routine procedure (see [10ndash13]) we can easily prove thefollowing results hence we state the following theoremswithout proof for functions inVℓH(120582 120574)
Theorem 3 (distortion bounds) Let 119891 isin VℓH(120582 120574) Then for|119911| = 119903 lt 1 we have
(1 minus100381610038161003816100381611988711003816100381610038161003816) 119903 minus
1
2ℓ (1 + 120582)(120574 minus 1
2 minus 120574minus1 + 120574
2 minus 120574
100381610038161003816100381611988711003816100381610038161003816) 1199032 le
1003816100381610038161003816119891 (119911)1003816100381610038161003816
le (1 +100381610038161003816100381611988711003816100381610038161003816) 119903 +
1
2ℓ (1 + 120582)(120574 minus 1
2 minus 120574minus1 + 120574
(2 minus 120574)
100381610038161003816100381611988711003816100381610038161003816) 1199032
(19)
Corollary 4 (covering result) If 119891(119911) isinVℓH(120582 120574) then
119908 |119908| lt2ℓ+1 (1 + 120574) + 1 minus [2ℓ (1 + 120574) + 1] 120574
2ℓ (2 minus 120574) (1 + 120582)
minus2ℓ+1 (1 + 120574) minus 1 minus [2ℓ (1 + 120574) + 1] 120574
2ℓ (2 minus 120574) (1 + 120582) 1198871
sub 119891 (119880)
(20)
Next we state the extreme points of closed convex hulls ofVℓH(120582 120574) denoted by clcoVℓH(120582 120574)
Theorem 5 A function 119891(119911) isinVℓH(120582 120574) if and only if 119891(119911) =suminfin
119899=1(119883119899ℎ119899(119911) + 119884
119899119892119899(119911)) where ℎ
1(119911) = 119911 ℎ
119899(119911) = 119911 + ((120574 minus
1)(2ℓ|1 minus 120582 + 119899120582|(119899 minus 120574)))119911119899 (119899 ge 2) and 119892119899(119911) = 119911 + ((120574 minus
1)(2ℓ|1 minus 120582 minus 119899120582|(119899 + 120574)))119911119899 (119899 ge 1) also suminfin119899=1(119883119899+ 119884119899) =
1 119883119899ge 0 and 119884
119899ge 0 In particular the extreme points of
VℓH(120582 120574) are ℎ119899 and 119892119899
4 International Journal of Analysis
Theorem 6 The family VℓH(120582 120574) is closed under convexcombinations
4 Inclusion Results
Now we will examine the closure properties of the classVℓH(120582 120574) under the generalized Bernardi-Libera-Livingstonintegral operator L
119888(119891) which is defined by L
119888(119891) = ((119888 +
1)119911119888) int119911
0119905119888minus1119891(119905)119889119905 119888 gt minus1
Theorem 7 Let 119891(119911) isin VℓH(120582 120574) Then L119888(119891(119911)) isin
VℓH(120582 120574)
Lemma 8 (see [15]) Let 119891 = ℎ + 119892 be given by (3) If
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120572)
1 minus 120572
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120572)
1 minus 120572
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(21)
where 1198861= 1 and 0 le 120572 lt 1 then 119891 isinMlowastH(120582 120572)
Theorem 9 Let 119891 = ℎ + 119892 isin VℓH(120582 120572) be given by (6) Then119891 isinVℓH(120582 (4 minus 3120574)(3 minus 2120574))
Proof Since 119891 isinVℓH(120582 120574) then byTheorem 1 we must have
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(22)
To show that 119891 isin VℓH(120582 (4 minus 3120574)(3 minus 2120574)) by virtue ofLemma 8 we have to show that
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(23)
where 0 le (4 minus 3120574)(3 minus 2120574) lt 1 For this it is sufficient toprove that
|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
ge|1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 2 3 )
|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
ge|1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 1 2 3 )
(24)
or equivalently (120574 minus 1)|1 minus 120582 + 119899120582|(119899 minus 120574) ge 0 (119899 = 2 3 )and (120574 minus 1)|1 minus 120582 minus 119899120582|(119899 + 120574) ge 0 (119899 = 1 2 3 ) which istrue and the theorem is proved
Corollary 10 VℓH(120582 120574) subVℓH(120582 43) subVℓH
5 Convolution Properties
For functions 119891 isinH given by (3) and 119865 isinH given by
119865 (119911) = 119867 (119911) + 119866 (119911) = 119911 +infin
sum119899=2
119860119899119911119899 +infin
sum119899=1
119861119899119911119899 (25)
we recall the Hadamard product (or convolution) of 119891 and 119865by
(119891 lowast 119865) (119911) = 119911 +infin
sum119899=2
119886119899119860119899119911119899 +infin
sum119899=1
119887119899119861119899119911119899 (119911 isin U)
(26)
Let 119865119895(119911) isinVℓH(120582 120574) (119895 = 1 2 3 119901) be given by
119865119895(119911) = 119911 +
infin
sum119899=2
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899 + (minus1)
ℓ
infin
sum119899=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899 (27)
then the convolution is defined by
(1198651lowast sdot sdot sdot lowast 119865
119901) (119911) = 119911 minus
infin
sum119899=2
119901
prod119895=1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899
+ (minus1)ℓ
infin
sum119899=1
119901
prod119895=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899
(28)
Theorem 11 Let 119865119895(119911) isin VℓH(120582 120574119895) (119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(29)
Proof We use the principle of mathematical induction in ourproof Let 119865
1isin VℓH(120582 1205741) and 1198652 isin VℓH(120582 1205742) By using
Theorem 2 we haveinfin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(30)
International Journal of Analysis 5
then
[
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
timesinfin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
+ [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
timesinfin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
le 1
(31)
Thus by applying Cauchy-Schwarz inequality we have
infin
sum119899=2
(radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816)
2
le [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
infin
sum119899=1
(radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816)
2
le [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
(32)
Then we get
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
+infin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
le 1
(33)
Therefore if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120572)
120572 minus 1
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
leinfin
sum119899=2
radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120572)
120572 minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
leinfin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
(34)
that is if
radic1003816100381610038161003816119886119899111988611989921003816100381610038161003816 le
120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic1003816100381610038161003816119887119899111988711989921003816100381610038161003816 le
120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(35)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) By (30) we have
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 le 1
infin
sum119899=1
radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(36)
Hence we get
radic10038161003816100381610038161003816119886119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 + 119899120582| (119899 minus 120574119895)
radic10038161003816100381610038161003816119887119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 minus 119899120582| (119899 + 120574119895)
(37)
Consequently if
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 + 119899120582|2 (119899 minus 1205741) (119899 minus 120574
2)
le120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 minus 119899120582|2 (119899 + 1205741) (119899 + 120574
2)
le120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(38)
6 International Journal of Analysis
That is if
119899 minus 120572
120572 minus 1le|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
119899 + 120572
120572 minus 1le|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(39)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) Then we see that
120572 ge 1
+(119899 minus 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 + 119899120582| (119899 + 1205741) (119899 + 120574
2) + (1205741minus 1) (120574
2minus 1)
= 120601 (119899)
120572 ge 1
+(119899 + 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1205741) (119899 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
= 120595 (119899)
(40)
Since 120601(119899) for 119899 ge 2 and 120595(119899) for 119899 ge 1 are increasing
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(41)
120572 ge 1 +2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(42)
and also(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
le2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(43)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) where
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ (1 + 120582) (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(44)
Next we suppose that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901)(119911) isin VℓH(120582 120573)
where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(45)
We can show that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901+1)(119911) isin VℓH(120582 120575)
where
120575 ge 1 +(120573 minus 1) (120574
119901+1minus 1)
2ℓ (1 + 120582) (2 + 120573) (2 + 120574119901+1) + (120573 minus 1) (120574
119901+1minus 1)
(46)
Since
(120573 minus 1) (120572119901+1minus 1)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(120573 minus 2) (120572119901+1minus 2)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(47)
we have
120575 = 1 +prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901+1
119895=1(2 + 120574
119895) minus prod
119901+1
119895=1(120574119895minus 1)
(48)
Corollary 12 Let 119865119895(119911) isin VℓH(120582 120574)(119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +(120574 minus 1)
119901
2ℓ (1 + 120582) (2 + 120574)119901
minus (120574 minus 1)119901 (49)
6 Partial Sums Results
In 1985 Silvia [16] studied the partial sums of convexfunctions of order 120572 (0 le 120572 lt 1) Later on Silverman [17]and several researchers studied and generalized the resultson partial sums for various classes of analytic functionsonly but analogues results on harmonic functions have notbeen explored in the literature Very recently Porwal [18]and Porwal and Dixit [19] filled this gap by investigatinginteresting results on the partial sums of star-like harmonicunivalent functions Now in this section we discussed thepartial sums results for the class of harmonic functions withpositive coefficients based on Salagean operator of order120574 (1 lt 120574 le 43) on lines similar to Porwal [18]
LetPℓH(119860119899120575 119861119899120575) denote the subclass ofH consistingof functions 119891 = ℎ + 119892 of the form (3) which satisfy theinequality
infin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=1
119861119899
120575
10038161003816100381610038161198871198991003816100381610038161003816 le 1 (50)
where
119860119899
120575=119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
119861119899
120575=119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
(51)
and 120575 = 120574 minus 1 unless otherwise stated
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Stochastic AnalysisInternational Journal of
International Journal of Analysis 3
which is equivalent to
infin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(14)
But (8) is true by hypothesis Hence |(119860(119911)119861(119911) minus 1)(119860(119911)119861(119911)minus (2120574minus1))| lt 1 119911 isin U and the theorem is provedfor 119860(119911) and 119861(119911) is given by (10) and (11) respectively
Theorem 2 For 1198861= 1 and 1 lt 120574 le 43 119891 = ℎ + 119892 isin
VℓH (120582 120574) if and only if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(15)
Proof SinceVℓH(120582 120574) sub PℓH(120582 120574) we only need to prove theldquoonly if rdquo part of the theorem To this end for functions 119891 ofthe form (6) we notice that the condition
R((1 minus 120582)119863ℓ+1119891 (119911) + 120582119863ℓ+2119891 (119911)
(1 minus 120582)119863ℓ119891 (119911) + 120582119863ℓ+1119891 (119911)) lt 120574 (16)
Equivalently
R(((1 minus 120574) 119911 minusinfin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) (119899 minus 120574) 119886119899119911119899
minus (minus1)2ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) (119899 + 120574) 119887119899119911119899)
times (119911 minusinfin
sum119899=2
119899ℓ (1 minus 120582 + 119899120582) 119886119899119911119899
+ (minus1)2ℓ
infin
sum119899=1
119899ℓ (1 minus 120582 minus 119899120582) 119887119899119911119899)
minus1
)
ge 0
(17)
The above required condition must hold for all values of 119911 inU Upon choosing the values of 119911 on the positive real axiswhere 0 le 119911 = 119903 lt 1 we must have
(1 minus 120574 minusinfin
sum119899=2
119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574) 119886119899119903119899minus1
minusinfin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574) 119887119899119903119899minus1)
times (1 minusinfin
sum119899=2
119899ℓ |(1 minus 120582 + 119899120582)| 119886119899119903119899minus1
+infin
sum119899=1
119899ℓ |1 minus 120582 minus 119899120582| 119887119899119903119899minus1)
minus1
ge 0
(18)
If condition (15) does not hold then the numerator in (18)is negative for 119903 sufficiently close to 1 Hence there exists1199110= 1199030in (0 1) for which the quotient of (18) is negative
This contradicts the required condition for 119891 isin VℓH(120582 120574)This completes the proof of the theorem
3 Distortion Bounds and Extreme Points
By routine procedure (see [10ndash13]) we can easily prove thefollowing results hence we state the following theoremswithout proof for functions inVℓH(120582 120574)
Theorem 3 (distortion bounds) Let 119891 isin VℓH(120582 120574) Then for|119911| = 119903 lt 1 we have
(1 minus100381610038161003816100381611988711003816100381610038161003816) 119903 minus
1
2ℓ (1 + 120582)(120574 minus 1
2 minus 120574minus1 + 120574
2 minus 120574
100381610038161003816100381611988711003816100381610038161003816) 1199032 le
1003816100381610038161003816119891 (119911)1003816100381610038161003816
le (1 +100381610038161003816100381611988711003816100381610038161003816) 119903 +
1
2ℓ (1 + 120582)(120574 minus 1
2 minus 120574minus1 + 120574
(2 minus 120574)
100381610038161003816100381611988711003816100381610038161003816) 1199032
(19)
Corollary 4 (covering result) If 119891(119911) isinVℓH(120582 120574) then
119908 |119908| lt2ℓ+1 (1 + 120574) + 1 minus [2ℓ (1 + 120574) + 1] 120574
2ℓ (2 minus 120574) (1 + 120582)
minus2ℓ+1 (1 + 120574) minus 1 minus [2ℓ (1 + 120574) + 1] 120574
2ℓ (2 minus 120574) (1 + 120582) 1198871
sub 119891 (119880)
(20)
Next we state the extreme points of closed convex hulls ofVℓH(120582 120574) denoted by clcoVℓH(120582 120574)
Theorem 5 A function 119891(119911) isinVℓH(120582 120574) if and only if 119891(119911) =suminfin
119899=1(119883119899ℎ119899(119911) + 119884
119899119892119899(119911)) where ℎ
1(119911) = 119911 ℎ
119899(119911) = 119911 + ((120574 minus
1)(2ℓ|1 minus 120582 + 119899120582|(119899 minus 120574)))119911119899 (119899 ge 2) and 119892119899(119911) = 119911 + ((120574 minus
1)(2ℓ|1 minus 120582 minus 119899120582|(119899 + 120574)))119911119899 (119899 ge 1) also suminfin119899=1(119883119899+ 119884119899) =
1 119883119899ge 0 and 119884
119899ge 0 In particular the extreme points of
VℓH(120582 120574) are ℎ119899 and 119892119899
4 International Journal of Analysis
Theorem 6 The family VℓH(120582 120574) is closed under convexcombinations
4 Inclusion Results
Now we will examine the closure properties of the classVℓH(120582 120574) under the generalized Bernardi-Libera-Livingstonintegral operator L
119888(119891) which is defined by L
119888(119891) = ((119888 +
1)119911119888) int119911
0119905119888minus1119891(119905)119889119905 119888 gt minus1
Theorem 7 Let 119891(119911) isin VℓH(120582 120574) Then L119888(119891(119911)) isin
VℓH(120582 120574)
Lemma 8 (see [15]) Let 119891 = ℎ + 119892 be given by (3) If
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120572)
1 minus 120572
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120572)
1 minus 120572
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(21)
where 1198861= 1 and 0 le 120572 lt 1 then 119891 isinMlowastH(120582 120572)
Theorem 9 Let 119891 = ℎ + 119892 isin VℓH(120582 120572) be given by (6) Then119891 isinVℓH(120582 (4 minus 3120574)(3 minus 2120574))
Proof Since 119891 isinVℓH(120582 120574) then byTheorem 1 we must have
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(22)
To show that 119891 isin VℓH(120582 (4 minus 3120574)(3 minus 2120574)) by virtue ofLemma 8 we have to show that
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(23)
where 0 le (4 minus 3120574)(3 minus 2120574) lt 1 For this it is sufficient toprove that
|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
ge|1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 2 3 )
|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
ge|1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 1 2 3 )
(24)
or equivalently (120574 minus 1)|1 minus 120582 + 119899120582|(119899 minus 120574) ge 0 (119899 = 2 3 )and (120574 minus 1)|1 minus 120582 minus 119899120582|(119899 + 120574) ge 0 (119899 = 1 2 3 ) which istrue and the theorem is proved
Corollary 10 VℓH(120582 120574) subVℓH(120582 43) subVℓH
5 Convolution Properties
For functions 119891 isinH given by (3) and 119865 isinH given by
119865 (119911) = 119867 (119911) + 119866 (119911) = 119911 +infin
sum119899=2
119860119899119911119899 +infin
sum119899=1
119861119899119911119899 (25)
we recall the Hadamard product (or convolution) of 119891 and 119865by
(119891 lowast 119865) (119911) = 119911 +infin
sum119899=2
119886119899119860119899119911119899 +infin
sum119899=1
119887119899119861119899119911119899 (119911 isin U)
(26)
Let 119865119895(119911) isinVℓH(120582 120574) (119895 = 1 2 3 119901) be given by
119865119895(119911) = 119911 +
infin
sum119899=2
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899 + (minus1)
ℓ
infin
sum119899=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899 (27)
then the convolution is defined by
(1198651lowast sdot sdot sdot lowast 119865
119901) (119911) = 119911 minus
infin
sum119899=2
119901
prod119895=1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899
+ (minus1)ℓ
infin
sum119899=1
119901
prod119895=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899
(28)
Theorem 11 Let 119865119895(119911) isin VℓH(120582 120574119895) (119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(29)
Proof We use the principle of mathematical induction in ourproof Let 119865
1isin VℓH(120582 1205741) and 1198652 isin VℓH(120582 1205742) By using
Theorem 2 we haveinfin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(30)
International Journal of Analysis 5
then
[
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
timesinfin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
+ [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
timesinfin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
le 1
(31)
Thus by applying Cauchy-Schwarz inequality we have
infin
sum119899=2
(radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816)
2
le [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
infin
sum119899=1
(radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816)
2
le [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
(32)
Then we get
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
+infin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
le 1
(33)
Therefore if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120572)
120572 minus 1
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
leinfin
sum119899=2
radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120572)
120572 minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
leinfin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
(34)
that is if
radic1003816100381610038161003816119886119899111988611989921003816100381610038161003816 le
120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic1003816100381610038161003816119887119899111988711989921003816100381610038161003816 le
120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(35)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) By (30) we have
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 le 1
infin
sum119899=1
radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(36)
Hence we get
radic10038161003816100381610038161003816119886119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 + 119899120582| (119899 minus 120574119895)
radic10038161003816100381610038161003816119887119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 minus 119899120582| (119899 + 120574119895)
(37)
Consequently if
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 + 119899120582|2 (119899 minus 1205741) (119899 minus 120574
2)
le120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 minus 119899120582|2 (119899 + 1205741) (119899 + 120574
2)
le120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(38)
6 International Journal of Analysis
That is if
119899 minus 120572
120572 minus 1le|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
119899 + 120572
120572 minus 1le|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(39)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) Then we see that
120572 ge 1
+(119899 minus 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 + 119899120582| (119899 + 1205741) (119899 + 120574
2) + (1205741minus 1) (120574
2minus 1)
= 120601 (119899)
120572 ge 1
+(119899 + 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1205741) (119899 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
= 120595 (119899)
(40)
Since 120601(119899) for 119899 ge 2 and 120595(119899) for 119899 ge 1 are increasing
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(41)
120572 ge 1 +2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(42)
and also(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
le2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(43)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) where
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ (1 + 120582) (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(44)
Next we suppose that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901)(119911) isin VℓH(120582 120573)
where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(45)
We can show that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901+1)(119911) isin VℓH(120582 120575)
where
120575 ge 1 +(120573 minus 1) (120574
119901+1minus 1)
2ℓ (1 + 120582) (2 + 120573) (2 + 120574119901+1) + (120573 minus 1) (120574
119901+1minus 1)
(46)
Since
(120573 minus 1) (120572119901+1minus 1)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(120573 minus 2) (120572119901+1minus 2)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(47)
we have
120575 = 1 +prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901+1
119895=1(2 + 120574
119895) minus prod
119901+1
119895=1(120574119895minus 1)
(48)
Corollary 12 Let 119865119895(119911) isin VℓH(120582 120574)(119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +(120574 minus 1)
119901
2ℓ (1 + 120582) (2 + 120574)119901
minus (120574 minus 1)119901 (49)
6 Partial Sums Results
In 1985 Silvia [16] studied the partial sums of convexfunctions of order 120572 (0 le 120572 lt 1) Later on Silverman [17]and several researchers studied and generalized the resultson partial sums for various classes of analytic functionsonly but analogues results on harmonic functions have notbeen explored in the literature Very recently Porwal [18]and Porwal and Dixit [19] filled this gap by investigatinginteresting results on the partial sums of star-like harmonicunivalent functions Now in this section we discussed thepartial sums results for the class of harmonic functions withpositive coefficients based on Salagean operator of order120574 (1 lt 120574 le 43) on lines similar to Porwal [18]
LetPℓH(119860119899120575 119861119899120575) denote the subclass ofH consistingof functions 119891 = ℎ + 119892 of the form (3) which satisfy theinequality
infin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=1
119861119899
120575
10038161003816100381610038161198871198991003816100381610038161003816 le 1 (50)
where
119860119899
120575=119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
119861119899
120575=119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
(51)
and 120575 = 120574 minus 1 unless otherwise stated
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 International Journal of Analysis
Theorem 6 The family VℓH(120582 120574) is closed under convexcombinations
4 Inclusion Results
Now we will examine the closure properties of the classVℓH(120582 120574) under the generalized Bernardi-Libera-Livingstonintegral operator L
119888(119891) which is defined by L
119888(119891) = ((119888 +
1)119911119888) int119911
0119905119888minus1119891(119905)119889119905 119888 gt minus1
Theorem 7 Let 119891(119911) isin VℓH(120582 120574) Then L119888(119891(119911)) isin
VℓH(120582 120574)
Lemma 8 (see [15]) Let 119891 = ℎ + 119892 be given by (3) If
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120572)
1 minus 120572
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120572)
1 minus 120572
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(21)
where 1198861= 1 and 0 le 120572 lt 1 then 119891 isinMlowastH(120582 120572)
Theorem 9 Let 119891 = ℎ + 119892 isin VℓH(120582 120572) be given by (6) Then119891 isinVℓH(120582 (4 minus 3120574)(3 minus 2120574))
Proof Since 119891 isinVℓH(120582 120574) then byTheorem 1 we must have
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(22)
To show that 119891 isin VℓH(120582 (4 minus 3120574)(3 minus 2120574)) by virtue ofLemma 8 we have to show that
infin
sum119899=2
2ℓ |1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198861198991003816100381610038161003816
+infin
sum119899=1
2ℓ |1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
10038161003816100381610038161198871198991003816100381610038161003816 le 1
(23)
where 0 le (4 minus 3120574)(3 minus 2120574) lt 1 For this it is sufficient toprove that
|1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
ge|1 minus 120582 + 119899120582| [119899 minus ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 2 3 )
|1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
ge|1 minus 120582 minus 119899120582| [119899 + ((4 minus 3120574) (3 minus 2120574))]
1 minus ((4 minus 3120574) (3 minus 2120574))
(119899 = 1 2 3 )
(24)
or equivalently (120574 minus 1)|1 minus 120582 + 119899120582|(119899 minus 120574) ge 0 (119899 = 2 3 )and (120574 minus 1)|1 minus 120582 minus 119899120582|(119899 + 120574) ge 0 (119899 = 1 2 3 ) which istrue and the theorem is proved
Corollary 10 VℓH(120582 120574) subVℓH(120582 43) subVℓH
5 Convolution Properties
For functions 119891 isinH given by (3) and 119865 isinH given by
119865 (119911) = 119867 (119911) + 119866 (119911) = 119911 +infin
sum119899=2
119860119899119911119899 +infin
sum119899=1
119861119899119911119899 (25)
we recall the Hadamard product (or convolution) of 119891 and 119865by
(119891 lowast 119865) (119911) = 119911 +infin
sum119899=2
119886119899119860119899119911119899 +infin
sum119899=1
119887119899119861119899119911119899 (119911 isin U)
(26)
Let 119865119895(119911) isinVℓH(120582 120574) (119895 = 1 2 3 119901) be given by
119865119895(119911) = 119911 +
infin
sum119899=2
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899 + (minus1)
ℓ
infin
sum119899=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899 (27)
then the convolution is defined by
(1198651lowast sdot sdot sdot lowast 119865
119901) (119911) = 119911 minus
infin
sum119899=2
119901
prod119895=1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 119911119899
+ (minus1)ℓ
infin
sum119899=1
119901
prod119895=1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 119911119899
(28)
Theorem 11 Let 119865119895(119911) isin VℓH(120582 120574119895) (119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(29)
Proof We use the principle of mathematical induction in ourproof Let 119865
1isin VℓH(120582 1205741) and 1198652 isin VℓH(120582 1205742) By using
Theorem 2 we haveinfin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816
+infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(30)
International Journal of Analysis 5
then
[
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
timesinfin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
+ [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
timesinfin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
le 1
(31)
Thus by applying Cauchy-Schwarz inequality we have
infin
sum119899=2
(radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816)
2
le [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
infin
sum119899=1
(radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816)
2
le [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
(32)
Then we get
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
+infin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
le 1
(33)
Therefore if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120572)
120572 minus 1
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
leinfin
sum119899=2
radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120572)
120572 minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
leinfin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
(34)
that is if
radic1003816100381610038161003816119886119899111988611989921003816100381610038161003816 le
120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic1003816100381610038161003816119887119899111988711989921003816100381610038161003816 le
120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(35)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) By (30) we have
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 le 1
infin
sum119899=1
radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(36)
Hence we get
radic10038161003816100381610038161003816119886119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 + 119899120582| (119899 minus 120574119895)
radic10038161003816100381610038161003816119887119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 minus 119899120582| (119899 + 120574119895)
(37)
Consequently if
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 + 119899120582|2 (119899 minus 1205741) (119899 minus 120574
2)
le120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 minus 119899120582|2 (119899 + 1205741) (119899 + 120574
2)
le120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(38)
6 International Journal of Analysis
That is if
119899 minus 120572
120572 minus 1le|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
119899 + 120572
120572 minus 1le|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(39)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) Then we see that
120572 ge 1
+(119899 minus 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 + 119899120582| (119899 + 1205741) (119899 + 120574
2) + (1205741minus 1) (120574
2minus 1)
= 120601 (119899)
120572 ge 1
+(119899 + 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1205741) (119899 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
= 120595 (119899)
(40)
Since 120601(119899) for 119899 ge 2 and 120595(119899) for 119899 ge 1 are increasing
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(41)
120572 ge 1 +2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(42)
and also(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
le2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(43)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) where
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ (1 + 120582) (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(44)
Next we suppose that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901)(119911) isin VℓH(120582 120573)
where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(45)
We can show that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901+1)(119911) isin VℓH(120582 120575)
where
120575 ge 1 +(120573 minus 1) (120574
119901+1minus 1)
2ℓ (1 + 120582) (2 + 120573) (2 + 120574119901+1) + (120573 minus 1) (120574
119901+1minus 1)
(46)
Since
(120573 minus 1) (120572119901+1minus 1)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(120573 minus 2) (120572119901+1minus 2)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(47)
we have
120575 = 1 +prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901+1
119895=1(2 + 120574
119895) minus prod
119901+1
119895=1(120574119895minus 1)
(48)
Corollary 12 Let 119865119895(119911) isin VℓH(120582 120574)(119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +(120574 minus 1)
119901
2ℓ (1 + 120582) (2 + 120574)119901
minus (120574 minus 1)119901 (49)
6 Partial Sums Results
In 1985 Silvia [16] studied the partial sums of convexfunctions of order 120572 (0 le 120572 lt 1) Later on Silverman [17]and several researchers studied and generalized the resultson partial sums for various classes of analytic functionsonly but analogues results on harmonic functions have notbeen explored in the literature Very recently Porwal [18]and Porwal and Dixit [19] filled this gap by investigatinginteresting results on the partial sums of star-like harmonicunivalent functions Now in this section we discussed thepartial sums results for the class of harmonic functions withpositive coefficients based on Salagean operator of order120574 (1 lt 120574 le 43) on lines similar to Porwal [18]
LetPℓH(119860119899120575 119861119899120575) denote the subclass ofH consistingof functions 119891 = ℎ + 119892 of the form (3) which satisfy theinequality
infin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=1
119861119899
120575
10038161003816100381610038161198871198991003816100381610038161003816 le 1 (50)
where
119860119899
120575=119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
119861119899
120575=119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
(51)
and 120575 = 120574 minus 1 unless otherwise stated
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 5
then
[
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
timesinfin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
+ [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
timesinfin
sum119899=1
(radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
le 1
(31)
Thus by applying Cauchy-Schwarz inequality we have
infin
sum119899=2
(radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816)
2
le [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1)
1205741minus 1
100381610038161003816100381611988611989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=2
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
2)
1205742minus 1
100381610038161003816100381611988611989921003816100381610038161003816)
2
]
]
12
infin
sum119899=1
(radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816)
2
le [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
1)
1205741minus 1
100381610038161003816100381611988711989911003816100381610038161003816)
2
]
]
12
times [
[
infin
sum119899=1
(radic119899ℓ|1 minus 120582 + 119899120582| (119899 + 120574
2)
1205742minus 1
100381610038161003816100381611988711989921003816100381610038161003816)
2
]
]
12
(32)
Then we get
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
+infin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1)
1205741minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
le 1
(33)
Therefore if
infin
sum119899=2
119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120572)
120572 minus 1
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
leinfin
sum119899=2
radic1198992ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119886119899111988611989921003816100381610038161003816
infin
sum119899=1
119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120572)
120572 minus 1
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
leinfin
sum119899=1
radic1198992ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
1003816100381610038161003816119887119899111988711989921003816100381610038161003816
(34)
that is if
radic1003816100381610038161003816119886119899111988611989921003816100381610038161003816 le
120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic1003816100381610038161003816119887119899111988711989921003816100381610038161003816 le
120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(35)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) By (30) we have
infin
sum119899=2
radic119899ℓ|1 minus 120582 + 119899120582| (119899 minus 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988611989911989510038161003816100381610038161003816 le 1
infin
sum119899=1
radic119899ℓ|1 minus 120582 minus 119899120582| (119899 + 120574
119895)
120574119895minus 1
1003816100381610038161003816100381611988711989911989510038161003816100381610038161003816 le 1
(36)
Hence we get
radic10038161003816100381610038161003816119886119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 + 119899120582| (119899 minus 120574119895)
radic10038161003816100381610038161003816119887119899119895
10038161003816100381610038161003816 le120574119895minus 1
|1 minus 120582 minus 119899120582| (119899 + 120574119895)
(37)
Consequently if
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 + 119899120582|2 (119899 minus 1205741) (119899 minus 120574
2)
le120572 minus 1
119899 minus 120572radic|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
radic(1205741minus 1) (120574
2minus 1)
1198992ℓ|1 minus 120582 minus 119899120582|2 (119899 + 1205741) (119899 + 120574
2)
le120572 minus 1
119899 + 120572radic|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(38)
6 International Journal of Analysis
That is if
119899 minus 120572
120572 minus 1le|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
119899 + 120572
120572 minus 1le|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(39)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) Then we see that
120572 ge 1
+(119899 minus 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 + 119899120582| (119899 + 1205741) (119899 + 120574
2) + (1205741minus 1) (120574
2minus 1)
= 120601 (119899)
120572 ge 1
+(119899 + 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1205741) (119899 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
= 120595 (119899)
(40)
Since 120601(119899) for 119899 ge 2 and 120595(119899) for 119899 ge 1 are increasing
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(41)
120572 ge 1 +2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(42)
and also(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
le2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(43)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) where
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ (1 + 120582) (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(44)
Next we suppose that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901)(119911) isin VℓH(120582 120573)
where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(45)
We can show that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901+1)(119911) isin VℓH(120582 120575)
where
120575 ge 1 +(120573 minus 1) (120574
119901+1minus 1)
2ℓ (1 + 120582) (2 + 120573) (2 + 120574119901+1) + (120573 minus 1) (120574
119901+1minus 1)
(46)
Since
(120573 minus 1) (120572119901+1minus 1)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(120573 minus 2) (120572119901+1minus 2)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(47)
we have
120575 = 1 +prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901+1
119895=1(2 + 120574
119895) minus prod
119901+1
119895=1(120574119895minus 1)
(48)
Corollary 12 Let 119865119895(119911) isin VℓH(120582 120574)(119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +(120574 minus 1)
119901
2ℓ (1 + 120582) (2 + 120574)119901
minus (120574 minus 1)119901 (49)
6 Partial Sums Results
In 1985 Silvia [16] studied the partial sums of convexfunctions of order 120572 (0 le 120572 lt 1) Later on Silverman [17]and several researchers studied and generalized the resultson partial sums for various classes of analytic functionsonly but analogues results on harmonic functions have notbeen explored in the literature Very recently Porwal [18]and Porwal and Dixit [19] filled this gap by investigatinginteresting results on the partial sums of star-like harmonicunivalent functions Now in this section we discussed thepartial sums results for the class of harmonic functions withpositive coefficients based on Salagean operator of order120574 (1 lt 120574 le 43) on lines similar to Porwal [18]
LetPℓH(119860119899120575 119861119899120575) denote the subclass ofH consistingof functions 119891 = ℎ + 119892 of the form (3) which satisfy theinequality
infin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=1
119861119899
120575
10038161003816100381610038161198871198991003816100381610038161003816 le 1 (50)
where
119860119899
120575=119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
119861119899
120575=119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
(51)
and 120575 = 120574 minus 1 unless otherwise stated
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
Submit your manuscripts athttpwwwhindawicom
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Stochastic AnalysisInternational Journal of
6 International Journal of Analysis
That is if
119899 minus 120572
120572 minus 1le|1 minus 120582 + 119899120582| (119899 minus 120574
1) (119899 minus 120574
2)
(1205741minus 1) (120574
2minus 1)
119899 + 120572
120572 minus 1le|1 minus 120582 minus 119899120582| (119899 + 120574
1) (119899 + 120574
2)
(1205741minus 1) (120574
2minus 1)
(39)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) Then we see that
120572 ge 1
+(119899 minus 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 + 119899120582| (119899 + 1205741) (119899 + 120574
2) + (1205741minus 1) (120574
2minus 1)
= 120601 (119899)
120572 ge 1
+(119899 + 1) (120574
1minus 1) (120574
2minus 1)
119899ℓ |1 minus 120582 minus 119899120582| (119899 + 1205741) (119899 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
= 120595 (119899)
(40)
Since 120601(119899) for 119899 ge 2 and 120595(119899) for 119899 ge 1 are increasing
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(41)
120572 ge 1 +2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(42)
and also(1205741minus 1) (120574
2minus 1)
2ℓ |1 + 120582| (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
le2 (1205741minus 1) (120574
2minus 1)
|1 minus 2120582| (1 + 1205741) (1 + 120574
2) minus (1205741minus 1) (120574
2minus 1)
(43)
then (1198651lowast 1198652)(119911) isinVℓH(120582 120572) where
120572 ge 1 +(1205741minus 1) (120574
2minus 1)
2ℓ (1 + 120582) (2 + 1205741) (2 + 120574
2) + (1205741minus 1) (120574
2minus 1)
(44)
Next we suppose that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901)(119911) isin VℓH(120582 120573)
where
120573 = 1 +prod119901
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(45)
We can show that (1198651lowast 1198652lowast sdot sdot sdot lowast 119865
119901+1)(119911) isin VℓH(120582 120575)
where
120575 ge 1 +(120573 minus 1) (120574
119901+1minus 1)
2ℓ (1 + 120582) (2 + 120573) (2 + 120574119901+1) + (120573 minus 1) (120574
119901+1minus 1)
(46)
Since
(120573 minus 1) (120572119901+1minus 1)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(120573 minus 2) (120572119901+1minus 2)
=prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901
119895=1(2 + 120574
119895) minus prod
119901
119895=1(120574119895minus 1)
(47)
we have
120575 = 1 +prod119901+1
119895=1(120574119895minus 1)
2ℓ (1 + 120582)prod119901+1
119895=1(2 + 120574
119895) minus prod
119901+1
119895=1(120574119895minus 1)
(48)
Corollary 12 Let 119865119895(119911) isin VℓH(120582 120574)(119895 = 1 2 3 119901) then
(1198651lowast sdot sdot sdot lowast 119865
119901)(119911) isinVℓH(120582 120573) where
120573 = 1 +(120574 minus 1)
119901
2ℓ (1 + 120582) (2 + 120574)119901
minus (120574 minus 1)119901 (49)
6 Partial Sums Results
In 1985 Silvia [16] studied the partial sums of convexfunctions of order 120572 (0 le 120572 lt 1) Later on Silverman [17]and several researchers studied and generalized the resultson partial sums for various classes of analytic functionsonly but analogues results on harmonic functions have notbeen explored in the literature Very recently Porwal [18]and Porwal and Dixit [19] filled this gap by investigatinginteresting results on the partial sums of star-like harmonicunivalent functions Now in this section we discussed thepartial sums results for the class of harmonic functions withpositive coefficients based on Salagean operator of order120574 (1 lt 120574 le 43) on lines similar to Porwal [18]
LetPℓH(119860119899120575 119861119899120575) denote the subclass ofH consistingof functions 119891 = ℎ + 119892 of the form (3) which satisfy theinequality
infin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=1
119861119899
120575
10038161003816100381610038161198871198991003816100381610038161003816 le 1 (50)
where
119860119899
120575=119899ℓ |1 minus 120582 + 119899120582| (119899 minus 120574)
120574 minus 1
119861119899
120575=119899ℓ |1 minus 120582 minus 119899120582| (119899 + 120574)
120574 minus 1
(51)
and 120575 = 120574 minus 1 unless otherwise stated
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
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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
International Journal of Analysis 7
Now we discuss the ratio of a function of the form (6)with 119887
1= 0 being
119891119898(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +infin
sum119899=1
119887119899119911119899
119891119896(119911) = 119911 +
infin
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
119891119898119896(119911) = 119911 +
119898
sum119899=2
119886119899119911119899 +
119896
sum119899=2
119887119899119911119899
(52)
We first obtain the sharp bounds forR119891(119911)119891119898(119911)
Theorem 13 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R119891 (119911)
119891119898(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (53)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(54)
The result (53) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (55)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
120575[119891 (119903119890119894120579)
119891119898(119903119890119894120579)
minus119860119898+1
minus 120575
119860119898+1
]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(56)
It suffices to show that |119908(119911)| le 1 Now from (56) we canwrite
119908 (119911) = (119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (2 + 2(infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
+119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
minus1
(57)
Hence we obtain
|119908 (119911)|
le(119860119898+1120575) (sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
2 minus 2 [sum119898
119899=2
10038161003816100381610038161198861198991003816100381610038161003816 + suminfin
119899=2
10038161003816100381610038161198871198991003816100381610038161003816] minus (119860119898+1120575)sum
infin
119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
(58)
Now |119908(119911)| le 1 ifinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (59)
From condition (50) it is sufficient to show thatinfin
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816
leinfin
sum119899=2
119860119899
120575
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
119861119899
120575
10038161003816100381610038161198861198991003816100381610038161003816
(60)
which is equivalently to119898
sum119899=2
(119860119899minus 120575
120575)10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
(119861119899minus 120575
120575)10038161003816100381610038161198871198991003816100381610038161003816
+infin
sum119899=119898+1
(119860119899minus 119860119899+1
120575)10038161003816100381610038161198861198991003816100381610038161003816 ge 0
(61)
To see that the function given by (55) gives the sharp resultwe observe that for 119911 = 119903119890119894120587119899
119891 (119911)
119891119898(119911)
= 1 +120575
119860119898+1
119911119898 997888rarr 1 minus120575
119860119898+1
=119860119898+1
minus 120575
119860119898+1
when 119903 997888rarr 1minus
(62)
We next determine bounds forR119891119898(119911)119891(119911)
Theorem 14 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (63)
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Mathematical PhysicsAdvances in
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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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 International Journal of Analysis
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(64)
The result (63) is sharp with the function given by
119891 (119911) = 119911 +120575
119860119898+1
119911119898+1 (65)
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119860119898+1
+ 120575
120575[119891119898(119903119890119894120579)
119891 (119903119890119894120579)minus
119860119898+1
119860119898+1
+ 120575]
= (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
120575(infin
sum119899=119898+1
119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(66)
Hence we obtain
|119908 (119911)| le (119860119898+1
+ 120575
120575(infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816))
times (2 minus 2 [119898
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816]
minus ((119860119898+1
minus 120575) 120575)infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816)
minus1
le 1
(67)
The last inequality is equivalent to
ℓ
sum119899=2
10038161003816100381610038161198861198991003816100381610038161003816 +infin
sum119899=2
10038161003816100381610038161198871198991003816100381610038161003816 +119860119898+1
120575
infin
sum119899=119898+1
10038161003816100381610038161198861198991003816100381610038161003816 le 1 (68)
Making use of (50) and the condition (64) we obtain (61)Finally equality holds in (63) for the extremal function 119891(119911)given by (65)
We next turns to ratios for R1198911015840(119911)1198911015840119898(119911) and
R1198911015840119898(119911)1198911015840(119911)
Theorem 15 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (69)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge 120575 119894119891 119899 = 2 3
(70)
The result (69) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=
119860119898+1
(119898 + 1) 120575[1198911015840 (119911)
1198911015840119898(119911)
minus119860119898+1
minus (119898 + 1) 120575
119860119898+1
]
= (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
+infin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579
minus119860119898+1
(119898 + 1) 120575(infin
sum119899=119898+1
119899119886119899119903119899minus1119890119894(119899minus1)120579))
times (1 +119898
sum119899=2
119899119886119899119903119899minus1119890119894(119899minus1)120579
minusinfin
sum119899=2
119899119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(71)
The result (69) follows by using the techniques as used inTheorem 13
Proceeding exactly as in the proof of Theorem 14 we canprove the following theorem
Theorem 16 If 119891 of the form (6) with 1198871= 0 satisfies the
condition (50) then
R1198911015840119898(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (72)
The result is sharp with the function given by 119891(119911) = 119911 +
(120575119860119898+1)119911119898+1
We next determine bounds for R119891(119911)119891119896(119911) and
R119891119896(119911)119891(119911)
Theorem 17 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (73)
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
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
International Journal of Analysis 9
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(74)
The result (73) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem 18 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (75)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge 120575 119894119891 119899 = 2 3
(76)
The result (75) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Proof Define the function 119908(119911) by
1 + 119908 (119911)
1 minus 119908 (119911)=119861119896+1+ 120575
120575[119891119896(119903119890119894120579)
119891 (119903119890119894120579)minus
119861119896+1
119861119896+1+ 120575]
= (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579 +
119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579
minus119861119896+1
119861119896+1+ 120575
119896+1
sum119899=2
119887119899119903119899minus1119890minus119894(119899minus1)120579)
times (1 +infin
sum119899=2
119886119899119903119899minus1119890119894(119899minus1)120579
+119896
sum119899=2
119887119899119903119899minus1119890minus119894(119899+1)120579)
minus1
(77)
We omit the details of proof because it runs parallel to thatfromTheorem 14
Theorem 19 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119860119898+1
minus 120575
119860119898+1
(119911 isin U) (78)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898119898 + 1
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(79)
The result (78) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem20 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891 (119911)
119891119898119896(119911) ge
119861119896+1minus 120575
119861119896+1
(119911 isin U) (80)
where
119861119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
119860119899ge
120575 119894119891 119899 = 2 3 119896
119861119896+1 119894119891 119899 = 119896 + 1 119896 + 2
(81)
Theorem21 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119860119898+1
119860119898+1
+ 120575 (119911 isin U) (82)
Theorem22 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R119891119898119896(119911)
119891 (119911) ge
119861119896+1
119861119896+1+ 120575 (119911 isin U) (83)
The result (83) is sharp with the function given by 119891(119911) = 119911 +(120575119861119896+1)119911119896+1
Theorem23 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840 (119911)
1198911015840119898(119911) ge
119860119898+1
minus (119898 + 1) 120575
119860119898+1
(119911 isin U) (84)
where
119860119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
119861119899ge
120575 119894119891 119899 = 2 3 119898
119860119898+1 119894119891 119899 = 119898 + 1119898 + 2
(85)
The result (84) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Theorem24 If119891 of the form (6)with 1198871= 0 satisfies condition
(50) then
R1198911015840119898119896(119911)
1198911015840 (119911) ge
119860119898+1
119860119898+1
+ (119898 + 1) 120575 (119911 isin U) (86)
The result (86) is sharp with the function given by 119891(119911) = 119911 +(120575119860119898+1)119911119898+1
Concluding Remarks By choosing 120582 = 0 (or 120582 = 1) andℓ = 0 (or ℓ = 1) the various results presented in this paperwould provide interesting extensions and generalizations ofthe subclasses of harmonic star-like functions with positivecoefficients of order 120574 (1 lt 120574 le 43) based on Salageanoperator and similarly for convex functions The detailsinvolved in the derivations of such specializations of theresults presented in this paper are fairly straight-forward andhence omitted
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
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
10 International Journal of Analysis
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
We record our sincere thanks to the referees for their valuablesuggestions
References
[1] J Clunie and T Sheil-Small ldquoHarmonic univalent functionsrdquoAnnales Academiae Scientiarum Fennicae A vol 9 pp 3ndash251984
[2] J M Jahangiri G Murugusundaramoorthy and K VijayaldquoSalagean-type harmonic univalent functionsrdquo Southwest Jour-nal of Pure and Applied Mathematics no 2 pp 77ndash82 2002
[3] H Silverman ldquoUnivalent functions with negative coefficientsrdquoProceedings of the American Mathematical Society vol 51 pp109ndash116 1975
[4] H Silverman ldquoHarmonic univalent functions with negativecoefficientsrdquo Journal of Mathematical Analysis and Applicationsvol 220 no 1 pp 283ndash289 1998
[5] B A Uralegaddi M D Ganigi and S M Sarangi ldquoUnivalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 25 no 3 pp 225ndash230 1994
[6] K K Dixit and V Chandra ldquoOn subclass of univalent functionswith positive coefficientsrdquoThe Aligarh Bulletin of Mathematicsvol 27 no 2 pp 87ndash93 2008
[7] K K Dixit and A L Pathak ldquoA new class of analytic functionswith positive coefficientsrdquo Indian Journal of Pure and AppliedMathematics vol 34 no 2 pp 209ndash218 2003
[8] S Porwal and K K Dixit ldquoAn application of certain convolu-tion operator involving hypergeometric functionsrdquo Journal ofRajasthan Academy of Physical Sciences vol 9 no 2 pp 173ndash186 2010
[9] S Porwal K K Dixit V Kumar and P Dixit ldquoOn a subclass ofanalytic functions defined by convolutionrdquo General Mathemat-ics vol 19 no 3 pp 57ndash65 2011
[10] K K Dixit and S Porwal ldquoA subclass of harmonic univalentfunctions with positive coefficientsrdquo Tamkang Journal of Math-ematics vol 41 no 3 pp 261ndash269 2010
[11] J M Jahangiri ldquoHarmonic functions starlike in the unit diskrdquoJournal of Mathematical Analysis and Applications vol 235 no2 pp 470ndash477 1999
[12] G Murugusundaramoorthy and K Vijaya ldquoA subclass ofharmonic functions associated with wright hypergeometricfunctionsrdquoAdvanced Studies inContemporaryMathematics vol18 no 1 pp 87ndash95 2009
[13] G Murugusundaramoorthy K Vijaya and R K Raina ldquoAsubclass of harmonic functions with varying arguments definedby Dziok-Srivastava operatorrdquo Archivum Mathematicum vol45 no 1 pp 37ndash46 2009
[14] S Porwal and K K Dixit ldquoNew subclasses of harmonic starlikeand convex functionsrdquo Kyungpook Mathematical Journal vol53 no 3 pp 467ndash478 2013
[15] K Vijaya Studies on certain subclasses of Harmonic functions[PhD thesis] VIT University Vellore India 2007
[16] E M Silvia ldquoOn partial sums of convex functions of order 120572rdquoHouston Journal ofMathematics vol 11 no 3 pp 397ndash404 1985
[17] H Silverman ldquoPartial sums of starlike and convex functionsrdquoJournal of Mathematical Analysis and Applications vol 209 no1 pp 221ndash227 1997
[18] S Porwal ldquoPartial sums of certain harmonic univalent func-tionsrdquo Lobachevskii Journal of Mathematics vol 32 no 4 pp366ndash375 2011
[19] S Porwal and K K Dixit ldquoPartial sums of starlike harmonicunivalent functionsrdquo Kyungpook Mathematical Journal vol 50no 3 pp 433ndash445 2010
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