# New Class of Close-to-Convex Harmonic Functions Defined by a Fourth-Order Differential Inequality

Authors: Mohammad Faisal Khan; Khaled Matarneh; Shahid Khan; Saqib Hussain; Maslina Darus
Journal: Journal of Mathematics
 (2022)

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## Abstract

In the recent past, various new subclasses of normalized harmonic functions have been defined in open unit diskU which satisfy second-order and third-order differential inequalities. Here, in this study, we define a new class of normalized harmonic functions in open unit disk U which is satisfying a fourth-order differential inequality. We investigate some useful results such as close-to-convexity, coefficient bounds, growth estimates, sufficient coefficient condition, and convolution for the functions belonging to this new class of harmonic functions. In addition, under convex combination and convolution of its members, we prove that this new class is closed, and we also give some lemmas to prove our main results.


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## Body

## 1. Introduction and Definitions

Letℋ represent the class of all harmonic functions f=s+v¯ in open unit disk U=z∈ℂ:z<1, and all these harmonic functions are normalized by(1)f0=fz0−1=0,and f can be expressed as f=s+v¯, where s is the analytic part and v is the coanalytic part of f in U, and also, these functions have a series of the form(2)sz=z+∑j=2∞ajzj,vz=∑j=2∞bjzj.Harmonic functionsf is locally univalent and sense-preserving in U, if it satisfies a necessary sufficient condition s′z>v′z [1, 2]. If coanalytic part of f is zero, then class ℋ of complex valued harmonic functions reduces to class A of normalized analytic functions.LetS denote the family of analytic univalent and normalized functions in U and also S⊂Sℋ which are defined as(3)S=f∈Sℋ:v=0inU.Also, let classℋ0 define as(4)ℋ0=f∈ℋ:fz¯0=0,Sℋ0=f∈Sℋ:fz¯0=0,where class Sℋ represent the class of functions f which are harmonic, univalent, and sense-preserving in open unit disk U.We can see that classSℋ0 is compact and normal, but class Sℋ is only normal. Let Sℋ0,∗,Kℋ0, and Cℋ0 are the subclasses of Sℋ0 which map open unit disk U onto starlike, convex, and close-to-convex domains for harmonic functions, respectively. We can observe that(5)S∗⊂Sℋ0,∗,K⊂Kℋ0,C⊂Cℋ0.These subclassesS∗, K, and C map open unit disk U onto their respective domains.For [2], Ponnusamy et al. defined the class of harmonic functions f∈ℋ0 which satisfy the condition(6)Refzz>fz¯z,forz∈U.In this class, they studied about close-to-convexity of harmonic functions. After that, Li and Ponnusamy [3, 4] discussed univalency and convexity of the abovementioned class. A class Wℋ0 of harmonic functions f=s+v¯∈ℋ0 defined by Nagpal and Ravichandran in [5] and the functions in this class satisfy the condition(7)Res′z+zs″z>v′z+zs″z,forz∈U.Note that(8)Wℋ0⊂Sℋ0,∗,and members of Wℋ0 are fully starlike in U.Recently, Ghosh and Vasudevarao [6] for α≥0 defined a new class Wℋ0α for harmonic functions f=s+v¯∈ℋ0 satisfying the condition(9)Res′z+αzs″z>v′z+αzs″z,forz∈U.Rajbala and Prajapat [7] for δ≥0, 0≤λ<1, defined a new class Wℋ0δ,λ of harmonic functions which satisfy the following inequality:(10)Res′z+δzs″z−λ>s′z+δzv″z.For this class, authors used Gaussian hypergeometric function and created harmonic polynomials for the classWℋ0δ,λ.Very recently, for the functionsf=s+v¯∈ℋ0, Yaşar and Yalçın [8] introduced the class ℛℋ0λ,δ which satisfy the following inequality:(11)Res′z+λzs″z+δz2s″′z>v′z+λzv″z+δz2s″′z,for λ≥δ≥0. For further study about harmonic functions, refer [2, 5, 9–11].By taking the inspiration from the abovementioned work, we define new class of harmonic functions inU which satisfy the fourth-order differential inequality.Definition 1.
Forλ≥δ≥γ≥0, let ℛℋ0λ,δ,γ denote the class of functions f=s+v¯∈ℋ0 and satisfy the condition(12)Res′z+λzs″z+δz2s″′z+γz3s″″z>v′z+λzv″z+δz2v″′z+γz3v″″z.Definition 2.
Forλ≥δ≥γ≥0, let ℛλ,δ,γ denote a class of functions f∈A if it satisfies the inequality(13)Ref′z+λzf″z+δz2f″′z+γz3f″″z>0.
Note that(14)ℛλ,δ,γ⊂ℛℋ0λ,δ,γ.
Special cases are(1)
ℛℋ0λ,δ,γ=ℛℋ0λ,δ, defined by Yaşar and Yalçın [8].(2)
ℛℋ01,0,0=Wℋ0, discussed by Nagpal and Ravichandran [5].(3)
ℛℋ0α,0,0=Wℋ0α, defined by Ghosh and Vasudevarao [6].(4)
ℛℋ0δ,1−2λ,0=Wℋ0δ,λ, defined by Rajbala and Prajapat [7].
In this section, we prove that all the members of the class are close-to-convex. We will derive coefficient bounds, growth estimates, and sufficient coefficient condition for the classℛℋ0λ,δ,γ. Furthermore, we investigate example of harmonic polynomial belonging to ℛℋ0λ,δ,γ.


## 2. Main Results

Lemma 1 (see [1]).
Lets and v are analytic functions in open unit disk U along with s′0<v′0 and ℒμ=s+μv is close-to-convex for each μ∈ℂμ=1. Then,(15)f=s+v¯is close to convex inU.Theorem 1.
Letf=s+v¯∈ℛℋ0λ,δ,γ if and only if ℒμ=s+μv∈ℛλ,δ,γ for each μ∈ℂμ=1.Proof.
Supposef=s+v¯∈ℛℋ0λ,δ,γ. For each μ=1,(16)Reℒμ′z+λzℒμ″z+δz2ℒμ″′z+γz3ℒμ″″z=Res′z+λzs″z+δz2s″′z+γz3s″″z+μv′z+λzv″z+δz2v″′z+γz3v″″z>Res′z+λzs″z+δz2s″′z+γz3s″″z−v′z+λzv″z+δz2s″′z+γz3v″″z>0,z∈U.
Thus,ℒμ∈ℛλ,δ,γ for each μμ=1. Conversely, let ℒμ=s+μv∈ℛℋλ,δ,γ, then(17)Res′z+λzs″z+δz2s″′z+γz3s″″z>Re−μv′z+λzv″z+δz2v″′z+γz3v″″z,z∈U. By choosing μμ=1, we get(18)Res′z+λzs″z+δz2s″′z+γz3s″″z>v′z+λzv″z+δz2v″′z+γz3v″″z,z∈U,
Hence,f∈ℛℋ0λ,δ,γ.Lemma 2 (see [12]).
Letz0≠0, z0∈U, with r0=z0. Let φz defined by(19)φz=Cjzj+Cj+1zj+1+⋯,be analytic in U, such that(20)φz0=maxz≤z0φz.Then, forn∈ℝ, n≥j≥1, such that(21)z0φ′z0φz0=n,Rez0″z0φ′z0≥n−1,Re1+3z0φ″z0φ′z0+z02φ″′z0φ′z0≥n2,or(22)Rez02φ″′z0φ′z0≥n2−3n+2. Then,(23)Re1+7z0φ″z0φ′z0+6z02φ″′z0φ′z0+z03φ″″z0φ′z0≥n3.Lemma 3.
Ifℒ∈ℛλ,δ,γ with λ≥δ≥γ≥0, then(24)Reℒ′z>0,and hence,(25)ℒzis close−to−convex inU.Proof.
Letℒ∈ℛλ,δ,γ, and we have(26)ℒ′z+λzℒ″z+δz2ℒ″′z+γz3ℒ″″z=ϑz.
Then,(27)Reϑz>0,forz∈U.
Letφ be an analytic function in U with(28)φ0=0,ℒ′z=1+φz1−φz,φz≠1.
We have to show thatφz<1, ∀z∈U. Then,(29)ϑz=ℒ′z+λzℒ″z+δz2ℒ″′z+γz3ℒ″″z,ϑz=1+φz1−φz+2λzφ′z1−φz2+2δzφ′z1−φz2zφ″zφ′z+4δzφ′z21−φz3+2γzφ′z1−φz2z2φ′″zφ′z+12γzφ′z1−φz3zφ″zφ′z+12γzφ′z31−φz4.
Sinceφ is analytic in U,(30)φ0=0,if∃z0∈U,such that(31)maxz≤z0φz=φz0=1.
Then, by using Lemma2, we may write(32)φz0=eiθ,z0φ′z0=nφz0=neiθ,n≥1,0<θ<2π,Rez0φ″z0φ′z0≥n−1,Rez02φ″z0φ′z0≥n2−3n+2.
For allz0∈U, we get(33)Reϑz0=Re1+φz1−φz+2λzφ′z1−φz2+2δzφ′z1−φz2zφ″zφ′z+4δzφ′z21−φz3+2γzφ′z1−φz2z2φ″′zφ′z+12γzφ′z1−φz3zφ″zφ′z+12γzφ′z31−φz4=Re1+eiθ1−eiθ+2λneiθ1−eiθ2+2δneiθ1−eiθ2z0φ″z0φ′z0+4δneiθ21−eiθ3+2γneiθ1−eiθ2z02φ″′z0φ′z0+12γneiθ1−eiθ3z0φ″z0φ′z0+12γneiθ31−eiθ4,=−λn1−cosθ−δn1−cosθRez0φ″z0φ′z0+δn21−cosθ−γn1−cosθRez0φ″′z0φ′z0−3γn1−cosθRez0φ″z0φ′z0−3γn321−cosθ≤−λn1−cosθ+δn1−cosθn−1+δn21−cosθ−γn1−cosθn2−3n+2−3γn1−cosθn−1−3γn321−cosθ,Reϑz0≤−λn1−cosθ+δn1−cosθ−γn5n2−121−cosθ≤δ−λn1−cosθ−γn5n2−121−cosθ≤0,which opposes the hypothesis. Hence, there is no z0∈U, such that(34)φz0=1.
Hence,φz<1 for all z∈U. So, we get(35)Reℒ′z>0.Theorem 2.
A functionf=s+v¯∈ℛℋ0λ,δ,γ is close-to-convex in U.Proof.
According to Lemma3, we derive that ℒμ=s+μv∈ℛλ,δ,γ are close-to-convex in U, ∀μμ=1. Therefore, in the light of Theorem 1 and Lemma 1, we get ℒμ∈ℛℋ0λ,δ,γ are close-to-convex in U.Theorem 3.
Letf=s+v¯∈ℛℋ0λ,δ,γ. Then,(36)bj≤1j1+j−1λ+j−2δ+γj−3,forj≥2.The equality is satisfied for(37)fz=z+1j1+j−1λ+j−2δ+γj−3z¯j.Proof.
Letf=s+v¯∈ℛℋ0λ,δ,γ. Applying the series of vz, we get(38)rj−1j+jλj−1+δjj−1j−2+jγj−1j−2j−3bj≤12π∫02πv′reiθ+λreiθv″reiθ+δr2e2iθv″′reiθ+γr3e3iθv″″reiθdθ<12π∫02πRes′reiθ+λreiθs″reiθ+δr2e2iθs″′reiθ+γr3e3iθs″″reiθdθ=12π∫02πRe1+∑j=2∞j1+j−1λ+j−2δ+γj−3×ajrj−1eij−1θdθ=1.
Takingr⟶1−, we get required result.Theorem 4.
Let forj≥2, f=s+v¯∈ℛℋ0λ,δ,γ. Then,(39)1aj+bj≤2j1+j−1λ+j−2δ+γj−3,2aj−bj≤2j1+j−1λ+j−2δ+γj−3,3aj≤2j1+j−1λ+j−2δ+γj−3.The equality holds in each case for the function(40)fz=z+∑j=2∞2j1+j−1λ+j−2δ+γj−3zj.Proof.
Let(41)f=s+v¯∈ℛℋ0λ,δ,γ.
Then, from Theorem1, we have(42)ℒμ=s+μv∈ℛλ,δ,γ,for each μμ=1. Thus, for each μ=1, we have(43)ReΦs,v>0,for z∈U, where(44)Φs,v=s+μv′+λzs+μv″+δz2s+μv″′+γz3s+μv″″.
Then, there exists an analytic functionp of the form(45)pz=1+∑j=1∞pjzj,with(46)Repz>0,such that(47)Φ1s,v+μΦ2s,v=pz,where(48)Φ1s,v=s′z+λzs″z+δz2s″′z+γz3s″″z,Φ2s,v=v′z+λzv″z+δz2v″′z+γz3v″″z.
Comparing coefficients on both sides of (47), we yield(49)j1+j−1λ+j−2δ+γj−3aj+μbj=pj−1,j≥2.
Sincepj≤2 for j≥1 and μμ=1 is arbitrary, first part of Theorem 4 is complete. Similarly, we can prove (2) and (3).
Now, we investigate sufficient condition forf∈ℛℋ0λ,δ,γ.Theorem 5.
Letf=s+v¯∈ℋ0 with(50)∑j=2∞j1+j−1λ+j−2δ+γj−3aj+bj≤1.
Then,f∈ℛℋ0λ,δ,γ.Proof.
Letf=s+v¯∈ℋ0. Then by using (50), we have(51)Res′z+λzs″z+δz2s″′z+γz3s″″z=Re1+∑j=2∞j1+j−1λ+j−2δ+γj−3ajzj−1>1−∑j=2∞j1+j−1λ+j−2δ+γj−3aj≥∑j=2∞j1+j−1λ+j−2δ+γj−3bj>∑j=2∞j1+j−1λ+j−2δ+γj−3bjzj−1=v′z+λzv″z+δz2v″′z+γz3v″″z.
Hence,f∈ℛℋ0λ,δ,γ.Corollary 1.
Letf=s+v¯∈ℋ0. If(52)∑j=2∞j23−j+λj−1aj+bj≤2,then f∈ℛℋ0λ,λ−1/2,0 with λ≥1.Example 1.
By takingλ=0.5 and δ=0.05, γ=0 in the light of Theorem 5, then harmonic polynomials(53)f1z=z+0.15z¯3,f1z=z−0.079z3+0.079z¯3,belong to ℛℋ00.5,0.05,0.Example 2.
By takingλ=3 and δ=1, γ=0 in the light of Theorem 5, then harmonic polynomials(54)f3z=z−116z¯2+154z¯3,f1z=z−0.079z3+0.079z¯3,belong to ℛℋ03,1,0.Remark 1.
The results which we obtained above lead to the results of the classesℛℋ00,0,0 and ℛℋ0λ,0,0 which are defined and studied in [2–4, 6], respectively.Remark 2.
The classℛλ,λ−1/2,0 with λ≥1 is a special case of the class defined in [13] and also our results lead the results of [13].
Now, we investigate convex combinations and convolutions for the classℛℋ0λ,δ,γ.Theorem 6.
The classℛℋ0λ,δ,γ of harmonic functions is closed under convex combinations.Proof.
Let, fork=1,2,…,j and fk=sk+vk¯∈ℛℋ0λ,δ,γ,(55)∑k=1jtk=1,0≤tk≤1,and convex combination of functions fkk=1,2,…,j can be defined as(56)fz=∑k=1jtkfkz=sz+vz¯,where(57)sz=∑k=1jtkskz,vz=∑k=1jtkvkz,and s and v are the analytic in U with(58)s0=v0=s′0−1=v′0=0.
Now,(59)Res′z+λzs″z+δz2s″′z+γz3s″″z=Re∑i=1jtksk′z+λzsk″z+δz2sk″′z+γz3sk″″z>∑k=1jtkvk′z+λzvk″z+δz2vk″′z+γz3vk″″z≥v′z+λzv″z+δz2v″′z+γz3v″″z,showing that f∈ℛℋ0λ,δ,γ.
A sequenceBjj=0∞ of nonnegative real numbers is said to be a convex null sequence, if Bj⟶0 as j⟶∞ and(60)B0−B1≥B1−B2≥B2−B3≥⋯≥Bj−1−Bj⋯≥0.
We need Lemma4 and Lemma [14] to prove results for convolution.Lemma 4 (see [15]).
IfBjj=0∞ is a convex null sequence, then(61)Bz=B02+∑j=1∞Bjzj,is analytic and(62)ReBz>0,inU.Lemma 5.
Letℒ∈ℛλ,δ,γ. Then, Reℒz/z>1/2.Proof.
Supposeℒ∈ℛλ,δ,γ be given by ℒz=z+∑j=2∞Ajzj. Then,(63)Re1+∑j=2∞j1+j−1λ+j−2δ+γj−3Ajzj−1>0,which is equivalent to(64)Repz>12,inU,where (65)pz=1+12∑j=2∞j1+j−1λ+j−2δ+γj−3Ajzj−1.
Now, forj≥2, we consider a sequence Bjj=0∞ defined by(66)B0=1,Bj−1=2j1+j−1λ+j−2δ+γj−3.
SinceBjj=0∞ is a convex null sequence and using Lemma 4, the function(67)Bz=1+∑j=2∞2j1+j−1λ+j−2δ+γj−3zj−1,is analytic and ReBz>1/2 in U. Writing(68)ℒzz=pz∗1+∑j=2∞2j1+j−1λ+j−2δ+γj−3zj−1,and using Lemma [14], we get Reℒz/z>1/2.Lemma 6.
Letℒi∈ℛλ,δ,γ, for k=1,2. Then, ℒ1∗ℒ2∈ℛλ,δ,γ.Proof.
Supposeℒ1z=z+∑j=2∞Ajzj and ℒ2z=z+∑j=2∞Bjzj. Then, the convolution of ℒ1z and ℒ2z is defined by(69)ℒz=ℒ1∗ℒ2z=z+∑j=2∞AjBjzj.
Since(70)ℒ′z=ℒ1′z∗ℒ2zz,zℒ″z=zℒ1″z∗ℒ2zz,z2ℒ″′z=z2ℒ1″′z∗ℒ2zz,z3ℒ″″z=z3ℒ1″″z∗ℒ2zz.
Then, we have(71)ℒ′z+λzℒ1″z+δz2ℒ″′z+γz3ℒ″″z=ℒ1′z+λzℒ1″z+δz2ℒ1″′z+γz3ℒ1″″z∗ℒ2zz. Since ℒ1∈ℛλ,δ,γ,(72)Reℒ1′z+λzℒ1″z+δz2ℒ1″′z+γz3ℒ1″″z>0,and using Lemma 5, Reℒz/z>1/2 in U.
Now applying Lemma [14] on (71) yields(73)Reℒ′z+λzℒ″z+δz2ℒ″′z+γz3ℒ″″z>0,in U. Thus, ℒ=ℒ1∗ℒ2∈ℛλ,δ,γ.
Now, by using Lemma6, let us prove that the class ℛℋ0λ,δ,γ is closed under convolutions of its members.Theorem 7.
Letfk∈ℛℋ0λ,δ,γ, for k=1,2. Then,(74)f1∗f2∈ℛℋ0λ,δ,γ.Proof.
Let(75)fk=sk+vk¯∈ℛℋ0λ,δ,γ,k=1,2,
Then, the convolution off1z and f2z is defined by(76)f1∗f2=s1∗s2+v1∗v2¯,in order to show that(77)f1∗f2∈ℛℋ0λ,δ,γ.
We have to prove that(78)ℒμ=s1∗s2+μv1∗v2∈ℛℋλ,δ,γ,for each μμ=1. By Lemma 6, the class ℛℋλ,δ,γ is closed under convolutions for each μμ=1, sk+μvk∈ℛℋλ,δ,γ for k=1,2. Then, both ℒ1 and ℒ1 given by(79)ℒ1=s1−v1∗s2−μv2,ℒ2=s1+v1∗s2+μv2,belong to ℛℋλ,δ,γ. Since ℛℋλ,δ,γ is closed under convex combinations, then(80)ℒ=12ℒ1+ℒ2=s1∗s2+μv1∗v2,belongs to ℛℋλ,δ,γ. Hence, ℛℋ0λ,δ,γ is closed under convolution.
Goodloe [16] defined the Hadamard product of a harmonic function as follows:(81)f∗˜φ=s∗φ+v∗φ¯,where f=s+v¯∈ℋ and φ∈A. By considering this Hadamard product of a harmonic function, we investigate following result.Theorem 8.
Letf∈ℛℋ0λ,δ,γ and φ∈A be such that Reφz/z>1/2, for z∈U. Then, f∗˜φ∈ℛℋ0λ,δ,γ.Proof.
Suppose(82)f=s+v¯∈ℛℋ0λ,δ,γ.
Then,(83)ℒμ=s+μv∈ℛℋλ,δ,γ,for each μμ=1. Using Theorem 1 and in order to show that f∗˜φ∈ℛℋ0λ,δ,γ, we need to show that(84)G=s∗φ+μv∗φ∈ℛℋλ,δ,γ,for each μμ=1. Write G=ℒμ∗φ, and(85)G′z+λzG″z+δz2G″′z+γz3G″″z=ℒμ′z+λzℒμ″z+δz2ℒμ″′z+γz3ℒμ″″z∗φzz. Since Reφz/z>1/2 and(86)Reℒμ′z+λzℒμ″+δz2ℒμ″′z+γz3ℒμ″″z>0,in U. Lemma [14] proves that G∈ℛℋ0λ,δ,γ.Corollary 2.
Letf∈ℛℋ0λ,δ,γ and φ∈K. Then, f∗˜φ∈ℛℋ0λ,δ,γ.Proof.
Supposeφ∈K. Then, Reφz/z>1/2 for z∈U. Theorem 8 concludes that f∗˜φ∈ℛℋ0λ,δ,γ.


## 3. Conclusion

Various new subclasses of normalized harmonic functions have been defined in open unit diskU, satisfying second-order and third-order differential inequalities. In this study, we defined a new class of normalized harmonic functions in open unit disk U, satisfying a fourth-order differential inequality. We gave some useful results such that close-to-convexity, coefficient bounds, growth estimates, sufficient coefficient condition, and convolution for the functions belong to this new class of harmonic functions. Further using the concepts of fourth-order differential inequality, all these problems can be studied for classes of meromorphic harmonic functions, Bazilevic harmonic functions, and for p valent harmonic functions as well.



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