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Kyungpook Mathematical Journal 2020; 60(3): 477-484

Published online September 30, 2020

Copyright © Kyungpook Mathematical Journal.

On the Fekete-SzegÖ Problem for Starlike Functions of Complex Order

Hanan Darwish and Abdel-Moniem Lashin, Bashar Al Saeedi*

Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516 Egypt
e-mail : darwish333@yahoo.com and aylashin@mans.edu.eg
Department of Mathematics, Faculty of Science & Engineering, Manchester Metropolitan University, Manchester, M15GD, UK
e-mail : basharfalh@yahoo.com

Received: January 28, 2016; Revised: May 3, 2020; Accepted: May 4, 2020

For a non-zero complex number b and for m and n in let Ψn,m(b) denote the class of normalized univalent functions f satisfying the condition [1+1b(Dn+mf(z)Dnf(z)-1)]>0 in the unit disk U, where Dnf(z) denotes the Salagean operator of f. Sharp bounds for the Fekete-Szegö functional a3-μa22 are obtained.

Keywords: coefficient estimates, Salagean operator, Fekete Szego problem, starlike functions of complex order

Fekete and Szegö proved the remarkable result that the estimate

|a3-λa22|1+2exp(-2λ1-λ)

holds with 0 ≤ λ ≤ 1 for any normalized univalent function

f(z)=z+a2z2+a3z3+

in the open unit disk U. This inequality is sharp for all λ (see [7]). The coefficient functional

φ(f(z))=a3-λa22=16(f(0)-3λ2[f(0)]2)

in the unit disk represents various geometric quantities. For example, when λ = 1, φ(f(z))=a3-λa22 becomes Sf (0)/6 where Sf denotes the Schwarzian derivative (f/f′)′ − (f/f′)2/2. Note that, if we consider the nth root transform [f(zn)]1/n = z + cn+1zn+1 + c2n+1z2n+1 + … of f with the power series (1.1), then cn+1 = a2/n and c2n+1=a3/n+(n-1)a22/2n2, so that

a3-λa22=n(c2n+1-μcn+12),

where μ = λn + (n − 1)/2. Moreover, φ (f(z)) behaves well with respect to the rotation, namely φ ((ef(ez)) = e2φ(f), for θR.

This is quite natural to discuss the behavior of φ (f(z)) for subclasses of normalized univalent functions in the unit disk. This is called the Fekete-Szegö problem. Many authors have considered this problem for typical classes of univalent functions (see, for instance [1, 2]).

We denote by A the set of all functions of the from (1.1) that are normalized analytic and univalent in the unit disk U. Also, for 0 ≤ α < 1, let S*(α) and C(α) denote the classes of starlike and convex univalent functions of order α, respectively, i.e., let

S*(α)={f(z)A:(zf(z)f(z))>α,zU}

and

C(α)={f(z)A:(1+zf(z)f(z))>α,zU}.

The notions of α-starlikeness and α-convexity were generalized onto a complex order by Nasr and Aouf [4], Wiatrowski [8], and Nasr and Aouf [3].

Observe that S* (0) = S* and C(0) = C represent standard starlike and convex univalent functions, respectively.

Let f(z)=z+k=2akzk and g(z)=z+k=2bkzk be analytic functions in U. The Hadamard product (convolution) of f and g, denoted by f * g is defined by

(f*g)(z)=z+k=2akbkzk,zU.

For a function f(z) in A, we define

D0f(z)=f(z),D1f(z)=Df(z)=zf(z)

and

Dnf(z)=D(Dn-1f(z)),nN={0,1,2,}.

With the above operator Dn, we say that a function f(z) belonging to A is in the class A(n, m, α) if and only if

{Dn+mf(z)Dnf(z)}>α(n,mN0=N{0})

for some α (0 ≤ α < 1), and for all zU.

We note that A(0, 1, α) = S*(α) is the class of starlike functions of order α, A(1, 1, α) = C(α) is the class of convex functions of order α, and that A(1, 0, α) = An(α) is the class of functions defined by Salagean [6].

Definition 1.1

Let b be a nonzero complex number, and let f be an univalent function of the form (1.1), such that Dn+mf(z) ≠ 0 for zU\ {0}. We say that f belongs to Ψn,m(b) if

{1+1b(Dn+mf(z)Dnf(z)-1)}>0,zU.

By giving specific values to n,m and b, we obtain the following important subclasses studied by various researchers in earlier works, for instance, Ψ1,0(b) = S*(1 − b) (Nasr and Aouf [4]), Ψ1,1(b) = C(1 − b) (Wiatrowski [8], Nasr and Aouf [3]).

We denote by ℘ class of the analytic functions in U with p(0) = 1 and ℜ(p(z)) > 0. We shall require the following:

Lemma 2.1.([5], p.166)

Let p ∈ ℘ with p(z) = 1+c1z + c2z2 + …, then

cn2,         forn1.

If |c1| = 2 then p(z) ≡ p1(z) = (1+γ1z)/(1−γ1z) with γ1 = c1/2. Conversely, if p(z) ≡ p1(z) for some 1| = 1, then c1 = 2γ1 and |c1| = 2. Furthermore we have

|c2=c122|2-c12.

If |c1| < 2 and |c2-c122|=2-c12, then p(z) ≡ p2(z), where

p2(z)=1+γ2z+γ11+γ¯1+γ2z1-γ2z+γ11+γ¯1+γ2z,

and γ1 = c1/2, γ2=2c2-c124-c12. Conversely if p(z) ≡ p2(z) for some 1| < 1 and 2| = 1, then γ1 = c1/2, γ2=2c2-c124-c12 and |c2-c122|=2-c12.

Theorem 2.2

Let n,m ≥ 0 and let b be non-zero complex number. If f of the form (1.1) is in Ψn,m(b), then

a22b2n(2m-1),a32b3n(3m-1)max {1,(2m-1)+2b(2m-1)},

and

|a3-22n(2m-1)3n(3m-1)a22|2b3n(3m-1).

Equality in (2.1) holds if Dn+mf(z)/Dnf(z) = 1+b [p1(z) − 1] , and in (2.2) if Dn+mf(z)/Dnf(z) = 1+b [p2(z) − 1] , where p1, p2are given in Lemma 2.1.

Proof

Denote F(z) = Dnf(z) = z + A2z2 + A3z3 + …,DmF(z) = Dn+mf(z) = z + 2mA2z2 + 3mA3z3 + …, where A2 = 2na2 and A3 = 3na3, …., then

2mA2=2n+ma2,         3mA3=3n+ma3.

By the definition of the class Ψn,m(b) there exists p ∈ ℘ such that DmF(z)F(z)=1-b+bp(z), so that

z+2mA2z2+3mA3z3+z+A2z2+A3z3+=1-b+b(1+c1z+c2z2+),

which implies the equality

z+2mA2z2+3mA3z3+=z+(A2+bc1)z2+(A3+bc1A2+bc2)z3+

Equating the coefficients of both sides we have

A2=b2m-1c1,         A3=b3m-1[c2+bc122m-1],

so that, on account of Dn+mf(z)

a2=bc12n(2m-1),         a3=b3n(3m-1)[c2+b(2m-1)c12].

Taking into account (2.5) and Lemma 2.1, we obtain

a2=|b2n(2m-1)c1|2b2n(2m-1),

and

a3=|b3n(3m-1)[c2-c122+(2m-1)+2b2(2m-1)c12]|b3n(3m-1)[2-c122+(2m-1)+2b2(2m-1)c12]=b3n(3m-1)[2+(2m-1)+2b-(2m-1)2(2m-1)c12]2b3n(3m-1)max{1,1+(2m-1)+2b-(2m-1)(2m-1)}.

Thus

a32b3n(3m-1)max {1,(2m-1)+2b(2m-1)}.

Moreover

|a3-22n(2m-1)3n(3m-1)a22|=|b3n(3m-1)[c2+b(2m-1)c12]-b2c12[2n(2m-1)]222n(2m-1)3n(3m-1)|=|bc23n(3m-1)|2b3n(3m-1),

as asserted.

Remark 2.3

Putting b = 1−α in the above theorem we get the following corollary.

Corollary 2.4

Let n,m ≥ 0 . If f of the form (1.1) is in A(n, m, α), then

a22(1-α)2n(2m-1),a32(1-α)3n(3m-1)max {1,(2m+1-2α)(2m-1)},

and

|a3-22n(2m-1)3n(3m-1)a22|2(1-α)3n(3m-1).

Remark 2.5

In the above Theorem 2.2 and Corollary 2.4 a special case of Fekete- Szegö problem e.g. for real μ = 22n (2m − 1) /3n(3m − 1) occurred very naturally and simple estimate was obtained.

Now, we consider functional a3-μa22 for complex μ.

Theorem 2.6

Let b be a non-zero complex number and let f ∈ Ψn,m(b). Then for μ ∈ ℂ

a3-μa222b3n(3m-1)max {1,|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|}.

For each μ there is a function in Ψn,m(b) such that equality holds.

Proof

Applying (2.5) we have

a3-μa22=b3n(3m-1)[c2+b(2m-1)c12]-μb2c12[2n(2m-1)]2=b3n(3m-1)[c2+b(2m-1)c12-μb3n(3m-1)[2n(2m-1)]2c12]=b3n(3m-1)[c2-c122+c122(1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2)].

Then, with the aid of Lemma 2.1, we obtain

a3-μa22b3n(3m-1)[2-c122+c122|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|]=b3n(3m-1)[2+c122(|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|-1)]2b3n(3m-1)max{1,|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|}.

An examination of the proof shows that equality is attained for the first case, when c1 = 0, c2 = 2, then the functions in Ψn,m(b) is given by

Dn+mf(z)Dnf(z)=1+(2b-1)z1-z,

and, for the second case, when c1 = c2 = 2, so that

Dn+mf(z)Dnf(z)=1+(2b-1)z21-z,

respectively.

Remark 2.7

Putting b = 1 − α in the above theorem we get the following corollary.

Corollary 2.8

Let fA(n, m, α) . Then for μ ∈ ℂ

a3-μa222(1-α)3n(3m-1)max {1,|1+2(1-α)(2m-1)-2μ(1-α)3n(3m-1)[2n(2m-1)]2|}.

For each μ there is a function in A(n, m, α) such that equality holds.

We next consider the case, when μ and b are real. Then we have:

Theorem 2.9

Let b > 0 and let f ∈ Ψn,m(b) . Then for μR we have

a3-μa22{2b3n(3m-1)[1+2b(2m-1)(1-μ3n(3m-1)22n(2m-1))]ifμ22n(2m-1)3n(3m-1),b3n(3m-1)if22n(2m-1)3n(3m-1)μ22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)],2b3n(3m-1)[2μb3n(3m-1)[2n(2m-1)]2-1-2b(2m-1)]ifμ22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)].

For each there is a function in Ψn,m(b) such that equality holds.

Proof

First, let μ22n(2m-1)3n(3m-1) In this case, (2.5) and Lemma 2.1 give

a3-μa22b3n(3m-1)[2-c122+c122(1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2)]2b3n(3m-1)[1+2b(2m-1)(1-μ3n(3m-1)22n(2m-1))].

Let, now 22n(2m-1)3n(3m-1)μ22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)]. Then, using the above calculations, we obtain

a3-μa22b3n(3m-1).

Finally, if μ22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)], then

a3-μa22b3n(3m-1)[2-c122+c122(2μb3n(3m-1)[2n(2m-1)]2-1-2b(2m-1))]=b3n(3m-1)[2+c122(2μb3n(3m-1)[2n(2m-1)]2-2-2b(2m-1))]2b3n(3m-1)[2μb3n(3m-1)[2n(2m-1)]2-1-2b(2m-1)].

Equality is attained for the second case on choosing c1 = 0, c2 = 2 in (2.7) and in (2.8) by choosing c1 = 2, c2 = 2 and c1 = 2i, c2 = −2 for the first and third case, respectively. Thus the proof is complete.

Remark 2.10

Put b = 1 − α in the above theorem we get the following corollary.

Corollary 2.11

Let fA(n, m, α) . Then for μR we have

a3-μa22{2(1-α)3n(3m-1)[1+2(1-α)(2m-1)(1-μ3n(3m-1)22n(2m-1))]ifμ22n(2m-1)3n(3m-1),(1-α)3n(3m-1)if22n(2m-1)3n(3m-1)μ22n(2m-1)[2m+1-2α]2(1-α)[3n(3m-1)],2(1-α)3n(3m-1)[2μ(1-α)3n(3m-1)[2n(2m-1)]2-1-2(1-α)(2m-1)]ifμ22n(2m-1)[2m+1-2α]2(1-α)[3n(3m-1)].

For each there is a function in A(n, m, α) such that equality holds.

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