# On the Inequality Theorem for a Wider Class of Analytic Functions with Hadamard Product

Authors: Ma’moun I. Y. Alharayzeh; Maslina Darus
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

In this paper, we discuss a well known class studied by Ramesha in 1995 and later by Janteng in 2006, and we then extend the class to a wider class of functionsf denoted by nαβ,γ which are normalized and univalent, in the unit disk D=z∈ℂ:z<1 satisfying the condition Reαz2f′′z/gz+zf′z/gz>β,0≤α<1,0≤β<1, where g is analytic function in D, such that gz≠0, with a new condition that is introduced. The main purpose of this paper is to give an estimate for the same a3−μa22 when f belongs to the class nαβ,γ.


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## Body

## 1. Introduction and Definition

LetS denote the class of normalized analytic univalent functions f of the form(1)fz=z+∑n=2∞anzn,anis complex number,where z∈D=z:z<1. Also, let gz=z+∑n=2∞bnzn, ϕz=z+∑n=2∞knzn, and   ψz=z+∑n=2∞dnzn be analytic functions in D where bn,kn,dn>0 and  kn>dn. We define the Hadamard product as follows:(2)gz∗ϕz=z+∑n=2∞bnknzn,gz∗ψz=z+∑n=2∞bndnzn.Earlier in 1933, Fekete and Szego [1] states that for f∈S and given by (1),(3)a3−μa22≤1+2exp−2μ1−μ,for 0≤μ≤1 and the inequality is sharp.The Fekete–Szego problems for the subclass ofS consisting of the families, convex functions C, starlike functions S∗, and close-to-convex functions CC have been completely solved in the literature. Among others are Keogh and Merkes [2], Koepf [3], Darus and Thomas [4, 5], Frasin and Darus [6], Ebadian et al. [7], and Mohammed et al. [8]. In particular, for f∈CC and be given by (1), Keogh and Merkes [2] showed that(4)a3−μa22≤3−4μ,ifμ≤13,13+49μ,if13≤μ≤231,if23≤μ≤1,4μ−3,ifμ≥1,and for each μ, there is a function in CC for which equality holds.Moreover, an estimate is given for the same functional for the new classnαβ,γ defined as follows.Definition 1.
For0≤α<1, 0≤β<1, and 0≤γ<1, let the function f be given by (1). Then, the function f∈nαβ,γ if and only if there exists g analytic function, gz≠0, such that for z∈D,(5)Reαz2f″zgz+zf′zgz>β,(6)Regz∗ϕzgz∗ψz>γ,gz∗ψz≠0,forz∈D.
This class is extended from Ramesha et al. [10] and Janteng [11], for suitable choices of ϕ, ψ, we easily obtain the various subclasses of S.
For example, ifϕz=z/1−z2 and ψz=z/1−z satisfy (6), we have the class of starlike functions g of order γ0≤γ<1 denoted by S∗γ. Also if ϕz=z+z2/1−z3 and ψz=z/1−z2 satisfy (6), we get the class of convex functions g of order γ0≤γ<1 denoted by Cγ.
To establish our main theorems in this paper, we first state some preliminary lemmas, required in proving of our theorem.


## 2. Preliminary Results

Lemma 1 (see [9]).
Leth be analytic in D with Rehz>0 and be given by(7)hz=1+c1z+c2z2+…,forz∈D,then,cn≤2wheren≥1,c2−c122≤2−c122.Lemma 2 (see [2]).
Letg∈S∗, the starlike function with gz=z+b2z2+b3z3…. Then, for μ real,(8)b3−μb22≤max1,3−4μ.
The first result for the classnαβ,γ is as follows.Lemma 3.
Let the functionf given by (1) belong to the class nαβ,γ. Then,(9)α+1a2≤k2−d21−β+1−γk2−d2,32α+1a3≤41−γ2d2k3−d3k2−d2+41−γ1−βk2−d2+21−γk3−d3+21−β,where 0≤α<1.Proof.
From (6), we get(10)gz∗ϕz=pz1−γ+γgz∗ψz,for z∈D, with Repz>0 given by pz=1+p1z+p2z2+p3z3+…, where p1,p2,…∈ℂ.
From (10), we have(11)z+b2k2z2+b3k3z3+…=1+p11−γz+p21−γz2+p31−γz3+…z+b2d2z2+b3d3z3+….
Now, equating coefficients, we get(12)b2k2−d2=p11−γ,(13)b3k3−d3=b2d2p11−γ+p21−γ.
It also follows from (5) that(14)αz2f″z+zf′z=gzhz1−β+β,where Rehz>0, and writing hz=1+c1z+c2z2…, where c1,c2,…∈ℂ, and equating coefficients give(15)2α+1a2=c11−β+b2,(16)32α+1a3=1−βc2+b2c1+b3.
The result now follows after applying the classical inequalities:p1≤2,p2≤2, c1≤2, c2≤2, and the inequalities b2≤21−γ/k2−d2 and b3≤4d21−γ2/k2−d2k3−d3+21−γ/k3−d3 which follow from (12) and (13).
Now we will display the main result for the classnαβ,γ.


## 3. Main Result

Theorem 1.
Let the functionf be given by (1) and belong to the class nαβ,γ. Then, for 0≤α<1,(17)32α+1a3−μa22≤41−γ2d2k3−d3k2−d2+21−γk3−d3+21−β+41−γ1−βk2−d2−2μ1−γ+1−βk2−d22μ1k2−d22ifμ≤μ•;41−γ2d2k3−d3k2−d2−41−γ2k2−d22+21−γk3−d3+21−β+2μ11−γ2μk2−d22ifμ•≤μ≤μ1;41−γ2d2k3−d3k2−d2−21−γ2k2−d22+21−γk3−d3+21−βifμ1≤μ≤μ2;−41−γ2d2k3−d3k2−d2−21−γk3−d3−21−β−41−γ1−βk2−d2+2μ1−γ+1−βk2−d22μ1k2−d22ifμ2≤μ;where(18)μ•=μ11−γ1−γ+1−βk2−d2,μ1=2α+1232α+1,μ2=μ•k2−d221−γ+1−βk2−d241−γd2k3−d3k2−d2+2k3−d3+21−β1−γ+21−βk2−d2−1−γk2−d22.The inequalities are sharp for all cases. However, the proofs for the caseμ=μ2 are still unsolved.Proof.
Write(19)32α+1a3−μa22=32α+1a3−32α+1μa22.
From (15) and (16), we have 32α+1a3−μa22=c21−β+b2c11−β+b3−32α+1μc121−β2+b22+2c1b21−β/4α+12. Then, we get(20)32α+1a3−μa22=b3−32α+1μb224α+12+c21−β+c121−β22α+12−32α+1μ4α+12−12+b2c11−β1−32α+1μ2α+12.
From (20), we have(21)32α+1a3−μa22≤b3−32α+1μb224α+12+c21−β−12c121−β2+14α+122α+12−32α+1μc121−β2+b2c11−β2α+122α+12−32α+1μ.
Now, consider the first case for all(22)μ≤2α+1232α+1,that is, having b3−32α+1μb22/4α+12>0 and 2α+12−32α+1μ>0.
Inequalities(23)b2≤21−γk2−d2,b3≤41−γ2d2k3−d3k2−d2+21−γk3−d3,and(24)c21−β−c121−β22≤21−β−c121−β22,follow immediately from Lemma 1. Now we get(25)32α+1a3−μa22≤41−γ2d2k3−d3k2−d2+21−γk3−d3+21−β−32α+11−γ2μα+12k2−d22−32α+11−β2μc124α+12+1−γ1−β2α+12−32α+1μc1α+12k2−d2=ϕx,say with,x=c1.
Now, after doing some operations, the functionϕ achieves its maximum value at(26)x°=22α+12−3μ2α+11−γ3μ2α+1k2−d21−β.
Next, we find(27)ϕx°=41−γ2d2k3−d3k2−d2−41−γ2k2−d22+21−γk3−d3+21−β+41−γ2α+1232α+1k2−d22μ.
Finally, we get our results:(28)32α+1a3−μa22≤41−γ2d2k3−d3k2−d2−41−γ2k2−d22+21−γk3−d3+21−β+41−γ2α+1232α+1k2−d22μ.
Now since we knowx°≤2, we get the interval(29)2α+121−γ32α+11−γ+1−βk2−d2≤μ.
Hence, result (28) concludes for the case(30)2α+121−γ32α+11−γ+1−βk2−d2≤μ≤2α+1232α+1.
We shall be certain that the result of this case is sharp. Letc1=21−γ2α+12−32α+1μ/31−β2α+1μk2−d2, c2=2, p1=2, p2=2, b2=21−γ/k2−d2 and b3=41−γ2d2/k3−d3k2−d2+21−γ/k3−d3, in (20).
Secondly, we consider the caseμ≤μ•=2α+121−γ/32α+11−γ+1−βk2−d2.
Write(31)a3−μa22=a3−μ•a22+μ•−μa22.
Now we say that with the aid of the inequalitya2≤1−γ+1−βk2−d2/k2−d2α+1, we get the following:(32)32α+1a3−μa22≤32α+1a3−μ•a22+32α+1μ•−μa22,=41−γ2d2k3−d3k2−d2+21−γk3−d3+41−γ1−βk2−d2,+21−β−32α+1μ1−γ+1−βk2−d22α+12k2−d22.
Thus, the proof for this case is complete.
Here, to find32α+1a3−μ•a22 we use the previous result of (28), for μ=μ•=2α+121−γ/32α+11−γ+1−βk2−d2.
After that, the result is sharp for this case. Upon choosingc1=c2=p1=p2=2,b2=21−γ/k2−d2, and b3=41−γ2d2/k3−d3k2−d2+21−γ/k3−d3 and substituting in (20), we get the result sharp.
Next we consider the case2α+12/32α+1≤μ≤μ2.
The results for this case are correct and sharp, but we have problems to solve for the valuea3−μ2a22. As for μ1=2α+12/32α+1, it is very easy and can be directly obtained from the previous result (28).
Without any doubt, we can assure that(33)32α+1a3−μ2a22≤41−γ2d2k3−d3k2−d2−21−γ2k2−d22+21−γk3−d3+21−β,which is left as conjecture for the readers to solve. Without loss of generality, we would suggest that in the case μ1≤μ≤μ2,(34)32α+1a3−μa22≤41−γ2d2k3−d3k2−d2−21−γ2k2−d22+21−γk3−d3+21−β.
This is sharp on choosingc1=p1=0,p2=2, c2=41−γ2d2/k3−d3k2−d21−β−21−γ2/k2−d221−β+2 , b2=0, and b3=21−γ/k3−d3 in (20). Thus, we get the desired equality.
Finally, for the caseμ2≤μ, we achieve the following easily.
Write(35)a3−μa22=a3−μ2a22+μ2−μa22,and by using the inequality a2≤1−γ+1−βk2−d2/k2−d2α+1, we obtain the following:(36)32α+1a3−μa22≤32α+1a3−μ2a22+32α+1μ2−μa22=−41−γ2d2k3−d3k2−d2−21−γk3−d3−21−β−41−γ1−βk2−d2+32α+1μ1−γ+1−βk2−d22α+12k2−d22,and hence the proof is complete for this case.
Here(37)32α+1a3−μ2a22≤41−γ2d2k3−d3k2−d2−21−γ2k2−d22+21−γk3−d3+21−β.
We used result (34) for(38)μ=μ2.
At last, it remains to show that the result is sharp, upon choosingc1=p1=2i, c2=p2=−2, b2=2i1−γ/k2−d2, and b3=−41−γ2d2/k3−d3k2−d2−21−γ/k3−d3 and substituting in (20).


## 4. Conclusion

This article aims at finding a new class of analytic univalent functions on the open unit disc defined by Hadamard product. Further to study their inequality theorem, one of the main requirements needed to satisfy certain classes. This approach, for example, can provide several many fascinating features.



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