# Fekete-Szegö Coefficient Functional for Certain Subclasses of Close-to-Star Functions

Authors: B. S. Mehrok; Gagandeep Singh
Journal: Geometry
 (2013)

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## Abstract

We introduce some subclasses of close-to-star functions defined by subordination and obtain sharp upper bounds of the functional|a3-μa22|, μ real, for an analytic function f(z)=z+∑n=2∞‍anzn, |z|<1, belonging to these sub-classes.


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## Body

## 1. Introduction

LetΩ be the class of bounded functions of the following form:
(1)w(z)=∑n=1∞dnzn,
which are analytic in the unit disc E={z:|z|<1} and satisfy the conditions w(0)=0 and |w(z)|<1.It is known (see [1]) that
(2)|d1|≤1,|d2|≤1-|d1|2.LetA denote the class of functions of the following form:
(3)f(z)=z+∑n=2∞anzn
which are analytic in E. Let S be the class of functions of the form (3) which are analytic univalent in E.We will concentrate on the coefficient problem for the subclasses ofA. In 1916, Bieberbach [2] proved that |a2|≤2 for f(z)∈S as an elementary corollary area theorem. He conjectured that, for each function f(z)∈S, |an|≤n; the equality holds for the Koebe function k(z)=z/(1-z)2, which maps the unit disc E onto the entire complex plane minus the slit along the negative real axis from -1/4 to -∞. De Branges [3] solved the Bieberbach conjecture in 1985. The contribution of Löwner [4] in proving that |a3|≤3 for the class S was huge.With the known estimates|a2|≤2 and |a3|≤3, it was natural to seek some relation between a3 and a22 for the class S. This thought prompted Fekete and Szegö [5] and they used Löwner's method to prove the following well-known result for the class S: if f(z)∈S, then
(4)|a3-μa22|≤{3-4μifμ≤0,1+2exp(-2μ1-μ)if0≤μ≤1,4μ-3ifμ≥1.The inequality (4) plays a very important role in determining estimates of higher coefficients for some subclasses of S (see Chichra [6], Babalola [7]).Next, we define some subclasses ofA and obtain the analogy of (4).We denote byS* the class of univalent star-like functions g(z)=z+∑n=2∞‍bnzn∈A which satisfy the following condition:
(5)ℜ(zg′(z)g(z))>0,z∈E.We denote byK the class of convex univalent functions h(z)=z+∑n=2∞‍cnzn∈A which satisfy the condition
(6)ℜ((zh′(z))′h′(z))>0,z∈E.Letf and g be two analytic functions in E. Then, f is said to be subordinate to g (symbolically f≺g) if there exists a bounded function w(z)∈Ω, such that f(z)=g(w(z)).A functionf(z)∈A is said to be close to star if there exists a function g(z)∈S* such that
(7)ℜ(f(z)g(z))>0,z∈E.The class of close to star functions was introduced by Reade [8] and is denoted by C*. A close to star function need not be univalent. The immediate shoot of C* is the following subclass:
(8)C1*={f(z)∈A:ℜ(f(z)h(z))>0,h(z)∈K,z∈E}.In this paper, we consider the following subclasses of close-to-star functions:(9)C*(A,B)={f(z)∈A:f(z)g(z)≺1+Az1+Bz,g(z)∈S*,-1≤B<A≤1,z∈EA:f(z)g(z)≺1+Az1+Bz}.The classC*(A,B) was introduced and studied by Mehrok et al. [9]. In particular, C*(1,-1)≡C*,
(10)C1*(A,B)={f(z)∈A:f(z)h(z)≺1+Az1+Bz,h(z)∈K,-1≤B<A≤1,z∈Ef(z)h(z)≺1+Az1+Bz},
is the class introduced and studied by Mehrok et al. [10]. Obviously, C1*(1,-1)≡C1*.Hummel [11] proved the conjecture of V. Singh that |c3-c22|≤1/3 for the class K. Keogh and Merkes [12] obtained the estimates (4) for the classes S* and K. We obtain sharp upper bounds of the functional |a3-μa22| for functions belonging to the classes C*(A,B) and C1*(A,B).


## 2. Preliminary Lemmas

Lemma 1.
Letg(z)∈S*. Then,
(11)|b3-μb22|≤{3-4μifμ≤12,1if12≤μ≤1,4μ-3ifμ≥1.This lemma is due to Keogh and Merkes [12].Lemma 2.
Leth(z)∈K. Then,
(12)|c3-μc22|≤{1-μifμ≤23,13if23≤μ≤43,μ-1ifμ≥43.This result was proved by Keogh and Merkes [12].


## 3. Main Results

Theorem 3.
Letf(z)∈C*(A,B). Then,
(13)|a3-μa22|≤{[3+(A-B)(2-B)]-[2+(A-B)]2μifμ≤-B2+(A-B)(B<0,A+B<0)orμ≤-B(A-B)(B>0),[3+(A-B)-4μ]+(A-B)(1-2μ)2(1+B)+μ(A-B)if-B2+(A-B)≤μ≤12or-B2+(A-B)≤μ≤-B(A-B)(B<0),1+(A-B)+(A-B)(2μ-1)2(1-B)-μ(A-B)if12≤μ≤(2-B)2+(A-B),(A-B)(4+A-B)μ-[(A-B)(2-B)-1]if2-B2+(A-B)≤μ≤1-BA-B,[2+(A-B)]2μ-[3+(A-B)(2-B)]ifμ≥1-BA-B.These results are sharp.Proof.
By definition ofC*(A,B),
(14)f(z)=g(z)1+Aw(z)1+Bw(z),g∈S*,w(z)∈Ω.
Identifying terms in the above expansion,(15)a2=b2+(A-B)d1,a3=b3+(A-B)b2d1+(A-B)(d2-Bd12).
From (15) and by using (2), it is easily established that
(16)|a3-μa22|≤(A-B)+|b3-μb22|+(A-B)|1-2μ|xy+(A-B)(|B+μ(A-B)|-1)x2,
where x=|d1|≤1 and y=|b2|≤2.
Two cases arise: (a)μ<-B/(A-B) and (b) μ≥-B/(A-B).
Consider the caseμ<-B/(A-B).
Case I. Suppose that μ≤1/2 so that A+B≤0.
By Lemma1, (16) can be written as
(17)|a3-μa22|≤[3+(A-B)]-4μ+2(A-B)×(1-2μ)x-(A-B)[(1+B)+μ(A-B)]x2=H1(x),say,H1′(x)=2(A-B)(1-2μ)-2(A-B)×[(1+B)+μ(A-B)]x,H1′′(x)=-2(A-B)[(1+B)+μ(A-B)],H1′(x)=0 implies that x=(1-2μ)/((1+B)+μ(A-B))=x1.
Subcase I(i). For μ<0, H1′(x)>0 provided μ≤-(1+B)/(A-B)<-B/(A-B).
WhereH1(x) is an increasing function in [0,1] and maxH1(x)=H1(1).
Subcase I(ii). Suppose μ≥0 and -(1+B)/(A-B)≤μ≤-B/(2+(A-B)).
Ifμ<-B/(2+(A-B)),thenH1′(x)>0 and maxH1(x)=H1(1).
Combining the above two subcases, we obtain the first result of (13). For x1>1,H1′′(x)<0 which holds for μ<-(1+B)/(A-B).
For-(1+B)/(A-B)≤μ≤-B/(2+(A-B))(B<0) or -B/(2+(A-B))≤μ≤1/2 (for all A and B), maxH1(x)=H1(x1) from which the second result of (13) follows.
We now consider the caseμ≥-B/(A-B).
Case II. Suppose that 1/2≤μ≤1. Then by Lemma 1, (16) takes the following form:
(18)|a3-μa22|≤1+(A-B)+2(A-B)(2μ-1)x-(A-B)[(1-B)-μ(A-B)]x2=H2(x),
Subcase II(i). For 1/2≤μ≤(2-B)/(2+(A-B)),
(19)H2′(x)=2(A-B)(2μ-1)-2(A-B)[(1-B)-μ(A-B)]x,H2′′(x)=-2(A-B)[(1-B)-μ(A-B)],
where H2′(x)=0 implies that x=(2μ-1)/((1-B)-μ(A-B))=x2.
Forμ≤(2-B)/(2+(A-B)), we have x2<1 and H2′(x)>0. So maxH2(x)=H2(x2), which gives the third result of (13).
Subcase II(ii). For (2-B)/(2+(A-B))≤μ≤(1-B)/(A-B). For x2≥1,(2-B)/(2+(A-B))≤μ≤(1-B)/(A-B),H2′(x)>0, and maxH2(x)=H2(1), which leads us to the fourth result of (13).
Case III. For μ≥(1-B)/(A-B)≥1. By Lemma 1, (16) can be put in the following form:
(20)|a3-μa22|≤(A-B)+(4μ-3)+2(A-B)(2μ-1)x-(A-B)[(1-B)-μ(A-B)]x2=H3(x),say,H3′(x)=2(A-B)(2μ-1)-2(A-B)×[(1-B)-μ(A-B)]x,H3′′(x)=-2(A-B)[(1-B)-μ(A-B)],
where H3′(x)=0 shows that x=(2μ-1)/((1-B)-μ(A-B))=x3.
Ifx3>1,  then (2-B)/(2+(A-B))<μ, which is true.
H
3
′
(
x
)
>
0 for μ≥(1-B)/(A-B)≥1. So maxH3(x)=H3(1), which takes us straight to the fifth result of (13).
The equality holds in the first and fifth results of (13) for the function f1(z) defined by
(21)f1(z)=z(1-z)2(1+Az1+Bz).
The equality holds in the second result of (13) for the function f2(z) defined by
(22)f2(z)=z(1-z)2(1+(A+x1)z+Az21+(B+x1)z+Bz2),
where x1=(1-2μ)/((1+B)+μ(A-B)).
The equality holds in the third result of (13) for the function f3(z) defined by
(23)f3(z)=z(1-z)2(1+(A+x2)z+Az21+(B+x2)z+Bz2),
where x2=(2μ-1)/((1-B)-μ(A-B)).
The equality holds in the fourth result of (13) for the function f4(z) defined by
(24)f4(z)=(∫0z‍[1+(M2-2LM)z]M2/(M2-2L)dz),
where L=3[(A-B)(2-B)-1] and M=2[(A-B)(4+A-B)]1/2 for |(M2-2L)/M|≤1.
Proof of Theorem3 is completed.ForA=1 and B=-1, Theorem 3 gives the following result.Corollary 4.
Iff(z)∈C*, then,
(25)|a3-μa22|≤{9-16μifμ≤14,1+1μif14≤μ≤12,3+(2μ-1)21-μif12≤μ≤34,12μ-5if34≤μ≤1,16μ-9ifμ≥1.Theorem 5.
Letf(z)∈C1*(A,B). Then,(26)|a3-μa22|≤{[1+(A-B)(1-B)]-[1+A-B]2μifμ≤-(1+2B)2(1+A-B)(A+B>0),[1+(A-B)-μ]+(A-B)(1-2μ)24[(1+B)+μ(A-B)]if-(1+2B)2(1+A-B)≤μ≤-BA-Bor-(1+2B)2(1+A-B)≤μ≤12,[1+(A-B)-μ]+(A-B)(2μ-1)24[(1-B)-μ(A-B)]if-BA-B≤μ≤23,13[1+3(A-B)]+(A-B)(2μ-1)24[(1-B)-μ(A-B)]if23≤μ≤3-2B2(1+A-B),13-(A-B)(1-B)+(A-B)(2+A-B)μif3-2B2(1+A-B)≤μ≤43,[1+A-B]2μ-[1+(A-B)(1-B)]ifμ≥43.The results are sharp.Proof.
Proceeding as in Theorem3, we have
(27)|a3-μa22|≤(A-B)+|c3-μc22|+(A-B)|1-2μ|xy+(A-B)(|B+μ(A-B)|-1)x2,
where x=|d1|≤1 and y=|c2|≤1.
Two cases arise: (a)μ<-B/(A-B),  (b) μ≥-B/(A-B).
Consider the caseμ<-B/(A-B).
By Lemma2, (27) can be written as
(28)|a3-μa22|≤[1+(A-B)]-μ+(A-B)|1-2μ|x-(A-B)[(1+B)+μ(A-B)]x2.
Subcase I(i). For μ<0,
(29)|a3-μa22|≤[1+(A-B)]-μ+(A-B)(1-2μ)x-(A-B)×[(1+B)+μ(A-B)]x2=H1(x),H1′(x)=(A-B)(1-2μ)-2(A-B)[(1+B)+μ(A-B)]x,H1′′(x)=-2(A-B)[(1+B)+μ(A-B)].
Forμ<0, H1′(x)>0 which shows that H1(x) is an increasing function in [0,1] and maxH1(x)=H1(1).
Subcase I(ii). Suppose μ≥0.
H
1
′
(
x
)
=
0 implies that x=(1-2μ)/(2[(1+B)+μ(A-B)])=x1.
Forx1≥1,-(1+B)/(A-B)≤μ≤-(1+2B)/(2(1+(A-B))). So H1′(x)>0 and maxH1(x)=H1(1).
Combining the above two subcases, we obtain the first result of (26).
Forx1>1,H1′′(x)<0, which holds for -(1+2B)/(2(1+(A-B)))≤μ≤-B/(A-B).
SomaxH1(x)=H1(x1), which leads us to the second result of (26).
We now consider the caseμ≥-B/(A-B).
Subcase I(iii). For 1/2≤-B/(A-B)≤μ≤(2/3)(B<0,2A+B>0), (27) takes the following form:
(30)|a3-μa22|≤1+(A-B)-μ+(A-B)(2μ-1)x-(A-B)×[(1-B)-μ(A-B)]x2=H2(x),H2′(x)=(A-B)(2μ-1)-2(A-B)[(1-B)-μ(A-B)]x,H2′′(x)=-2(A-B)[(1-B)-μ(A-B)],
where H2′(x)=0 implies that x=(2μ-1)/(2[(1-B)-μ(A-B)])=x2.
It is easy to verify thatH2′(x)<0. So maxH2(x)=H2(x2), which gives the third result of (26).
Case II. For 2/3≤μ≤4/3.
Subcase II(i). For 2/3≤μ≤(3-2B)/(2(1+A-B)), (27) reduces to
(31)|a3-μa22|≤13+(A-B)+(A-B)(2μ-1)x-(A-B)[(1-B)-μ(A-B)]x2=H3(x),H3′(x)=(A-B)(2μ-1)-2(A-B)[(1-B)-μ(A-B)]x,H3′′(x)=-2(A-B)[(1-B)-μ(A-B)],
where H3′(x)=0 implies that x=(2μ-1)/(2[(1-B)-μ(A-B)])=x3.
Forx3<1, (2/3)≤μ≤(3-2B)/(2(1+A-B)), H3′′(x)<0, and maxH3(x)=H3(x3), which leads us to the fourth result of (26).
Subcase II(ii). For (3-2B)/(2(1+A-B))≤μ≤1, (27) reduces to the following form:
(32)|a3-μa22|≤13+(A-B)+(A-B)(2μ-1)x-(A-B)[(1-B)-μ(A-B)]x2=H4(x),H4′(x)=(A-B)(2μ-1)-2(A-B)[(1-B)-μ(A-B)]x;H4′(x)=0 implies that x=(2μ-1)/(2[(1-B)-μ(A-B)]). maxH4(x)=H4(1).
Combining the above two subcases,maxH4(x)=H4(1), which take us straight to the fifth result of (26).
Case III. For μ≥4/3. By Lemma 1, (27) can be put in the form
(33)|a3-μa22|≤(A-B)-1+μ+(A-B)(2μ-1)x-(A-B)[(1-B)-μ(A-B)]x2=H5(x),(34)H5′(x)=(A-B)(2μ-1)-2(A-B)[(1-B)-μ(A-B)]x,
where H5′(x)=0 shows that x=(2μ-1)/(2[(1-B)-μ(A-B)])=x5.
x
5
>
1 and maxH5(x)=H5(1), which yield the sixth result of (26).
The equality holds in the first and sixth results of (26) for the function f1(z) defined by
(35)f1(z)=z1-z(1+Az1+Bz).
The equality holds in the second result of (26) for the function f2(z) defined by
(36)f2(z)=z1-z(1+(A+x1)z+Az21+(B+x1)z+Bz2),
where x1=(1-2μ)/(2[(1+B)+μ(A-B)]).
The equality holds in the third and fourth results of (26) for the function f3(z) defined by
(37)f3(z)=z1-z(1+(A+x2)z+Az21+(B+x2)z+Bz2),
where x2=(2μ-1)/(2[(1-B)-μ(A-B)]).
The equality holds in the fifth result of (26) for the function f4(z) defined by
(38)f4(z)=(∫0z‍[1+(M2-2LM)z]M2/(M2-2L)dz),
where L=1-3(A-B)(1-B) and M=2[(A-B)(2+A-B)]1/2 for |(M2-2L)/M|≤1.
Proof of Theorem5 is completed.ForA=1,B=-1, Theorem 5 gives the following result.Corollary 6.
Iff(z)∈C1*, then
(39)|a3-μa22|≤{5-9μifμ≤16,2+14μif16≤μ≤12,3-μ+(2μ-1)24(1-μ)if12≤μ≤23,73+(2μ-1)24(1-μ)if23≤μ≤56,8μ-113if56≤μ≤43,9μ-5ifμ≥43.The results of Corollaries4 and 6 were proved by the authors of [13].



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