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Bounds for the second Hankel determinant of certain univalent functions

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Abstract

The estimates for the second Hankel determinant a 2 a 4 a 3 2 of the analytic function f(z)=z+ a 2 z 2 + a 3 z 3 + , for which either z f (z)/f(z) or 1+z f (z)/ f (z) is subordinate to a certain analytic function, are investigated. The estimates for the Hankel determinant for two other classes are also obtained. In particular, the estimates for the Hankel determinant of strongly starlike, parabolic starlike and lemniscate starlike functions are obtained.

MSC:30C45, 30C80.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let A denote the class of all analytic functions

f(z)=z+ a 2 z 2 + a 3 z 3 +
(1)

defined on the open unit disk D:={zC:|z|<1}. The Hankel determinants H q (n) (n=1,2, , q=1,2,) of the function f are defined by

H q (n):=[ a n a n + 1 a n + q 1 a n + 1 a n + 2 a n + q a n + q 1 a n + q a n + 2 q 2 ]( a 1 =1).

Hankel determinants are useful, for example, in showing that a function of bounded characteristic in D, i.e., a function which is a ratio of two bounded analytic functions with its Laurent series around the origin having integral coefficients, is rational [1]. For the use of Hankel determinants in the study of meromorphic functions, see [2], and various properties of these determinants can be found in [[3], Chapter 4]. In 1966, Pommerenke [4] investigated the Hankel determinant of areally mean p-valent functions, univalent functions as well as of starlike functions. In [5], he proved that the Hankel determinants of univalent functions satisfy

| H q ( n ) | <K n ( 1 2 + β ) q + 3 2 (n=1,2,,q=2,3,),

where β>1/4000 and K depends only on q. Later, Hayman [6] proved that | H 2 (n)|<A n 1 / 2 (n=1,2, ; A an absolute constant) for areally mean univalent functions. In [79], the estimates for the Hankel determinant of areally mean p-valent functions were investigated. ElHosh obtained bounds for Hankel determinants of univalent functions with a positive Hayman index α [10] and of k-fold symmetric and close-to-convex functions [11]. For bounds on the Hankel determinants of close-to-convex functions, see [1214]. Noor studied the Hankel determinant of Bazilevic functions in [15] and of functions with bounded boundary rotation in [1619]. In the recent years, several authors have investigated bounds for the Hankel determinant of functions belonging to various subclasses of univalent and multivalent functions [2027]. The Hankel determinant H 2 (1)= a 3 a 2 2 is the well-known Fekete-Szegö functional. For results related to this functional, see [28, 29]. The second Hankel determinant H 2 (2) is given by H 2 (2)= a 2 a 4 a 3 2 .

An analytic function f is subordinate to an analytic function g, written f(z)g(z), if there is an analytic function w:DD with w(0)=0 satisfying f(z)=g(w(z)). Ma and Minda [30] unified various subclasses of starlike ( S ) and convex functions (C) by requiring that either of the quantity z f (z)/f(z) or 1+z f (z)/ f (z) is subordinate to a function φ with a positive real part in the unit disk D, φ(0)=1, φ (0)>0, φ maps D onto a region starlike with respect to 1 and symmetric with respect to the real axis. He obtained distortion, growth and covering estimates as well as bounds for the initial coefficients of the unified classes.

The bounds for the second Hankel determinant H 2 (2)= a 2 a 4 a 3 2 are obtained for functions belonging to these subclasses of Ma-Minda starlike and convex functions in Section 2. In Section 3, the problem is investigated for two other related classes defined by subordination. In proving our results, we do not assume the univalence or starlikeness of φ as they were required only in obtaining the distortion, growth estimates and the convolution theorems. The classes introduced by subordination naturally include several well-known classes of univalent functions and the results for some of these special classes are indicated as corollaries.

Let P be the class of functions with positive real part consisting of all analytic functions p:DC satisfying p(0)=1 and Rep(z)>0. We need the following results about the functions belonging to the class P.

Lemma 1 [31]

If the function pP is given by the series

p(z)=1+ c 1 z+ c 2 z 2 + c 3 z 3 +,
(2)

then the following sharp estimate holds:

| c n |2(n=1,2,).
(3)

Lemma 2 [32]

If the function pP is given by the series (2), then

2 c 2 = c 1 2 +x ( 4 c 1 2 ) ,
(4)
4 c 3 = c 1 3 +2 ( 4 c 1 2 ) c 1 x c 1 ( 4 c 1 2 ) x 2 +2 ( 4 c 1 2 ) ( 1 | x | 2 ) z,
(5)

for some x, z with |x|1 and |z|1.

2 Second Hankel determinant of Ma-Minda starlike/convex functions

Subclasses of starlike functions are characterized by the quantity z f (z)/f(z) lying in some domain in the right half-plane. For example, f is strongly starlike of order β if z f (z)/f(z) lies in a sector |argw|<βπ/2, while it is starlike of order α if z f (z)/f(z) lies in the half-plane Rew>α. The various subclasses of starlike functions were unified by subordination in [30]. The following definition of the class of Ma-Minda starlike functions is the same as the one in [30] except for the omission of starlikeness assumption of φ.

Definition 1 Let φ:DC be analytic, and let the Maclaurin series of φ be given by

φ(z)=1+ B 1 z+ B 2 z 2 + B 3 z 3 +( B 1 , B 2 R, B 1 >0).
(6)

The class S (φ) of Ma-Minda starlike functions with respect to φ consists of functions fA satisfying the subordination

z f ( z ) f ( z ) φ(z).

For the function φ given by φ α (z):=(1+(12α)z)/(1z) , 0<α1, the class S (α):= S ( φ α ) is the well-known class of starlike functions of order α. Let

φ PAR (z):=1+ 2 π 2 ( log 1 + z 1 z ) 2 .

Then the class

S P := S ( φ PAR )= { f A : Re ( z f ( z ) f ( z ) ) > | z f ( z ) f ( z ) 1 | }

is the parabolic starlike functions introduced by Rønning [33]. For a survey of parabolic starlike functions and the related class of uniformly convex functions, see [34]. For 0<β1, the class

S β := S ( ( 1 + z 1 z ) β ) = { f A : | arg ( z f ( z ) f ( z ) ) | < β π 2 }

is the familiar class of strongly starlike functions of order β. The class

S L := S ( 1 + z )= { f A : | ( z f ( z ) f ( z ) ) 2 1 | < 1 }

is the class of lemniscate starlike functions studied in [35].

Theorem 1 Let the function f S (φ) be given by (1).

  1. 1.

    If B 1 , B 2 and B 3 satisfy the conditions

    | B 2 | B 1 , | 4 B 1 B 3 B 1 4 3 B 2 2 | 3 B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | B 1 2 4 .
  1. 2.

    If B 1 , B 2 and B 3 satisfy the conditions

    | B 2 | B 1 , | 4 B 1 B 3 B 1 4 3 B 2 2 | B 1 | B 2 |2 B 1 2 0,

or the conditions

| B 2 | B 1 , | 4 B 1 B 3 B 1 4 3 B 2 2 | 3 B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | 1 12 | 4 B 1 B 3 B 1 4 3 B 2 2 | .
  1. 3.

    If B 1 , B 2 and B 3 satisfy the conditions

    | B 2 |> B 1 , | 4 B 1 B 3 B 1 4 3 B 2 2 | B 1 | B 2 |2 B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | B 1 2 12 ( 3 | 4 B 1 B 3 B 1 4 3 B 2 2 | 4 B 1 | B 2 | + 4 B 1 2 B 2 2 | 4 B 1 B 3 B 1 4 3 B 2 2 | 2 B 1 | B 2 | B 1 2 ) .

Proof Since f S (φ), there exists an analytic function w with w(0)=0 and |w(z)|<1 in D such that

z f ( z ) f ( z ) =φ ( w ( z ) ) .
(7)

Define the functions p 1 by

p 1 (z):= 1 + w ( z ) 1 w ( z ) =1+ c 1 z+ c 2 z 2 +,

or, equivalently,

w(z)= p 1 ( z ) 1 p 1 ( z ) + 1 = 1 2 ( c 1 z + ( c 2 c 1 2 2 ) z 2 + ) .
(8)

Then p 1 is analytic in D with p 1 (0)=1 and has a positive real part in D. By using (8) together with (6), it is evident that

φ ( p 1 ( z ) 1 p 1 ( z ) + 1 ) =1+ 1 2 B 1 c 1 z+ ( 1 2 B 1 ( c 2 c 1 2 2 ) + 1 4 B 2 c 1 2 ) z 2 +.
(9)

Since

z f ( z ) f ( z ) =1+ a 2 z+ ( a 2 2 + 2 a 3 ) z 2 + ( 3 a 4 3 a 2 a 3 + a 2 3 ) z 3 +,
(10)

it follows by (7), (9) and (10) that

a 2 = B 1 c 1 2 , a 3 = 1 8 [ ( B 1 2 B 1 + B 2 ) c 1 2 + 2 B 1 c 2 ] , a 4 = 1 48 [ ( 4 B 2 + 2 B 1 + B 1 3 3 B 1 2 + 3 B 1 B 2 + 2 B 3 ) c 1 3 a 4 = + 2 ( 3 B 1 2 4 B 1 + 4 B 2 ) c 1 c 2 + 8 B 1 c 3 ] .

Therefore

a 2 a 4 a 3 2 = B 1 96 [ c 1 4 ( B 1 3 2 + B 1 2 B 2 + 2 B 3 3 B 2 2 2 B 1 ) + 2 c 2 c 1 2 ( B 2 B 1 ) + 8 B 1 c 1 c 3 6 B 1 c 2 2 ] .

Let

d 1 = 8 B 1 , d 2 = 2 ( B 2 B 1 ) , d 3 = 6 B 1 , d 4 = B 1 3 2 + B 1 2 B 2 + 2 B 3 3 B 2 2 2 B 1 , T = B 1 96 .
(11)

Then

| a 2 a 4 a 3 2 | =T | d 1 c 1 c 3 + d 2 c 1 2 c 2 + d 3 c 2 2 + d 4 c 1 4 | .
(12)

Since the function p( e i θ z) (θR) is in the class P for any pP, there is no loss of generality in assuming c 1 >0. Write c 1 =c, c[0,2]. Substituting the values of c 2 and c 3 respectively from (4) and (5) in (12), we obtain

| a 2 a 4 a 3 2 | = T 4 | c 4 ( d 1 + 2 d 2 + d 3 + 4 d 4 ) + 2 x c 2 ( 4 c 2 ) ( d 1 + d 2 + d 3 ) + ( 4 c 2 ) x 2 ( d 1 c 2 + d 3 ( 4 c 2 ) ) + 2 d 1 c ( 4 c 2 ) ( 1 | x | 2 ) z | .

Replacing |x| by μ and substituting the values of d 1 , d 2 , d 3 and d 4 from (11) yield

| a 2 a 4 a 3 2 | T 4 [ c 4 | 2 B 1 3 + 8 B 3 6 B 2 2 B 1 | + 4 | B 2 | μ c 2 ( 4 c 2 ) + μ 2 ( 4 c 2 ) ( 2 B 1 c 2 + 24 B 1 ) + 16 B 1 c ( 4 c 2 ) ( 1 μ 2 ) ] = T [ c 4 4 | 2 B 1 3 + 8 B 3 6 B 2 2 B 1 | + 4 B 1 c ( 4 c 2 ) + | B 2 | ( 4 c 2 ) μ c 2 + B 1 2 μ 2 ( 4 c 2 ) ( c 6 ) ( c 2 ) ] F ( c , μ ) .
(13)

Note that for (c,μ)[0,2]×[0,1], differentiating F(c,μ) in (13) partially with respect to μ yields

F μ =T [ | B 2 | ( 4 c 2 ) + B 1 μ ( 4 c 2 ) ( c 2 ) ( c 6 ) ] .
(14)

Then, for 0<μ<1 and for any fixed c with 0<c<2, it is clear from (14) that F μ >0, that is, F(c,μ) is an increasing function of μ. Hence, for fixed c[0,2], the maximum of F(c,μ) occurs at μ=1, and

maxF(c,μ)=F(c,1)G(c).

Also note that

G(c)= B 1 96 [ c 4 4 ( | 2 B 1 3 + 8 B 3 6 B 2 2 B 1 | 4 | B 2 | 2 B 1 ) + 4 c 2 ( | B 2 | B 1 ) + 24 B 1 ] .

Let

P = 1 4 ( | 2 B 1 3 + 8 B 3 6 B 2 2 B 1 | 4 | B 2 | 2 B 1 ) , Q = 4 ( | B 2 | B 1 ) , R = 24 B 1 .
(15)

Since

max 0 t 4 ( P t 2 + Q t + R ) ={ R , Q 0 , P Q 4 ; 16 P + 4 Q + R , Q 0 , P Q 8  or  Q 0 , P Q 4 ; 4 P R Q 2 4 P , Q > 0 , P Q 8 ,
(16)

we have

| a 2 a 4 a 3 2 | B 1 96 { R , Q 0 , P Q 4 ; 16 P + 4 Q + R , Q 0 , P Q 8  or  Q 0 , P Q 4 ; 4 P R Q 2 4 P , Q > 0 , P Q 8 ,

where P, Q, R are given by (15). □

Remark 1 When B 1 = B 2 = B 3 =2, Theorem 1 reduces to [[24], Theorem 3.1].

Corollary 1

  1. 1.

    If f S (α), then | a 2 a 4 a 3 2 | ( 1 α ) 2 .

  2. 2.

    If f S L , then | a 2 a 4 a 3 2 |1/16=0.0625.

  3. 3.

    If f S P , then | a 2 a 4 a 3 2 |16/ π 4 0.164255.

  4. 4.

    If f S β , then | a 2 a 4 a 3 2 | β 2 .

Definition 2 Let φ:DC be analytic, and let φ(z) be given as in (6). The class C(φ) of Ma-Minda convex functions with respect to φ consists of functions f satisfying the subordination

1+ z f ( z ) f ( z ) φ(z).

Theorem 2 Let the function fC(φ) be given by (1).

  1. 1.

    If B 1 , B 2 and B 3 satisfy the conditions

    B 1 2 +4| B 2 |2 B 1 0, | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | 4 B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | B 1 2 36 .
  1. 2.

    If B 1 , B 2 and B 3 satisfy the conditions

    B 1 2 +4| B 2 |2 B 1 0,2 | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | B 1 3 4 B 1 | B 2 |6 B 1 2 0,

or the conditions

B 1 2 +4| B 2 |2 B 1 0, | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | 4 B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | 1 144 | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | .
  1. 3.

    If B 1 , B 2 and B 3 satisfy the conditions

    B 1 2 +4| B 2 |2 B 1 >0,2 | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | B 1 3 4 B 1 | B 2 |6 B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | B 1 2 576 ( 16 | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | 12 B 1 3 48 B 1 | B 2 | 36 B 1 2 B 1 4 8 B 1 2 | B 2 | 16 B 2 2 | 6 B 1 B 3 + B 1 2 B 2 B 1 4 4 B 2 2 | B 1 3 4 B 1 | B 2 | 2 B 1 2 ) .

Proof Since fC(φ), there exists an analytic function w with w(0)=0 and |w(z)|<1 in D such that

1+ z f ( z ) f ( z ) =φ ( w ( z ) ) .
(17)

Since

1+ z f ( z ) f ( z ) =1+2 a 2 z+ ( 4 a 2 2 + 6 a 3 ) z 2 + ( 8 a 2 3 18 a 2 a 3 + 12 a 4 ) z 3 +,
(18)

equations (9), (17) and (18) yield

a 2 = B 1 c 1 4 , a 3 = 1 24 [ ( B 1 2 B 1 + B 2 ) c 1 2 + 2 B 1 c 2 ] , a 4 = 1 192 [ ( 4 B 2 + 2 B 1 + B 1 3 3 B 1 2 + 3 B 1 B 2 + 2 B 3 ) c 1 3 a 4 = + 2 ( 3 B 1 2 4 B 1 + 4 B 2 ) c 1 c 2 + 8 B 1 c 3 ] .

Therefore

a 2 a 4 a 3 2 = B 1 768 [ c 1 4 ( 4 3 B 2 + 2 3 B 1 1 3 B 1 3 1 3 B 1 2 + 1 3 B 1 B 2 + 2 B 3 4 3 B 2 2 B 1 ) + 2 3 c 2 c 1 2 ( B 1 2 4 B 1 + 4 B 2 ) + 8 B 1 c 1 c 3 16 3 B 1 c 2 2 ] .

By writing

d 1 = 8 B 1 , d 2 = 2 3 ( B 1 2 4 B 1 + 4 B 2 ) , d 3 = 16 3 B 1 , d 4 = 4 3 B 2 + 2 3 B 1 1 3 B 1 3 1 3 B 1 2 + 1 3 B 1 B 2 + 2 B 3 4 3 B 2 2 B 1 , T = B 1 768 ,
(19)

we have

| a 2 a 4 a 3 2 | =T | d 1 c 1 c 3 + d 2 c 1 2 c 2 + d 3 c 2 2 + d 4 c 1 4 | .
(20)

Similar as in Theorems 1, it follows from (4) and (5) that

| a 2 a 4 a 3 2 | = T 4 | c 4 ( d 1 + 2 d 2 + d 3 + 4 d 4 ) + 2 x c 2 ( 4 c 2 ) ( d 1 + d 2 + d 3 ) + ( 4 c 2 ) x 2 ( d 1 c 2 + d 3 ( 4 c 2 ) ) + 2 d 1 c ( 4 c 2 ) ( 1 | x | 2 ) z | .

Replacing |x| by μ and then substituting the values of d 1 , d 2 , d 3 and d 4 from (19) yield

| a 2 a 4 a 3 2 | T 4 [ c 4 | 4 3 B 1 3 + 4 3 B 1 B 2 + 8 B 3 16 3 B 2 2 B 1 | + 2 μ c 2 ( 4 c 2 ) ( 2 3 B 1 2 + 8 3 | B 2 | ) + μ 2 ( 4 c 2 ) ( 8 3 B 1 c 2 + 64 3 B 1 ) + 16 B 1 c ( 4 c 2 ) ( 1 μ 2 ) ] = T [ c 4 3 | B 1 3 + B 1 B 2 + 6 B 3 4 B 2 2 B 1 | + 4 B 1 c ( 4 c 2 ) + 1 3 μ c 2 ( 4 c 2 ) ( B 1 2 + 4 | B 2 | ) + 2 B 1 3 μ 2 ( 4 c 2 ) ( c 4 ) ( c 2 ) ] F ( c , μ ) .
(21)

Again, differentiating F(c,μ) in (21) partially with respect to μ yields

F μ =T [ c 2 3 ( 4 c 2 ) ( B 1 2 + 4 | B 2 | ) + 4 B 1 3 μ ( 4 c 2 ) ( c 4 ) ( c 2 ) ] .
(22)

It is clear from (22) that F μ >0. Thus F(c,μ) is an increasing function of μ for 0<μ<1 and for any fixed c with 0<c<2. So, the maximum of F(c,μ) occurs at μ=1 and

maxF(c,μ)=F(c,1)G(c).

Note that

G ( c ) = T [ c 4 3 ( | B 1 3 + B 1 B 2 + 6 B 3 4 B 2 2 B 1 | B 1 2 4 | B 2 | 2 B 1 ) + 4 3 c 2 ( B 1 2 + 4 | B 2 | 2 B 1 ) + 64 3 B 1 ] .

Let

P = 1 3 ( | B 1 3 + B 1 B 2 + 6 B 3 4 B 2 2 B 1 | B 1 2 4 | B 2 | 2 B 1 ) , Q = 4 3 ( B 1 2 + 4 | B 2 | 2 B 1 ) , R = 64 3 B 1 .
(23)

By using (16), we have

| a 2 a 4 a 3 2 | B 1 768 { R , Q 0 , P Q 4 ; 16 P + 4 Q + R , Q 0 , P Q 8  or  Q 0 , P Q 4 ; 4 P R Q 2 4 P , Q > 0 , P Q 8 ,

where P, Q, R are given in (23). □

Remark 2 For the choice of φ(z)=(1+z)/(1z), Theorem 2 reduces to [[24], Theorem 3.2].

3 Further results on the second Hankel determinant

Definition 3 Let φ:DC be analytic, and let φ(z) be as given in (6). Let 0γ1 and τC{0}. A function fA is in the class R γ τ (φ) if it satisfies the following subordination:

1+ 1 τ ( f ( z ) + γ z f ( z ) 1 ) φ(z).

Theorem 3 Let 0γ1, τC{0}, and let the function f as in (1) be in the class R γ τ (φ). Also, let

p= 8 9 ( 1 + γ ) ( 1 + 3 γ ) ( 1 + 2 γ ) 2 .
  1. 1.

    If B 1 , B 2 and B 3 satisfy the conditions

    2| B 2 |(1p)+ B 1 (12p)0, | B 1 B 3 p B 2 2 | p B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | | τ | 2 B 1 2 9 ( 1 + 2 γ ) 2 .
  1. 2.

    If B 1 , B 2 and B 3 satisfy the conditions

    2| B 2 |(1p)+ B 1 (12p)0,2 | B 1 B 3 p B 2 2 | 2(1p) B 1 | B 2 | B 1 0,

or the conditions

2| B 2 |(1p)+ B 1 (12p)0, | B 1 B 3 p B 2 2 | B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | | τ | 2 8 ( 1 + γ ) ( 1 + 3 γ ) | B 3 B 1 p B 2 2 | .
  1. 3.

    If B 1 , B 2 and B 3 satisfy the conditions

    2| B 2 |(1p)+ B 1 (12p)>0,2 | B 1 B 3 p B 2 2 | 2(1p) B 1 | B 2 | B 1 2 0,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | | τ | 2 B 1 2 32 ( 1 + γ ) ( 1 + 3 γ ) × ( 4 p | B 3 B 1 p B 2 2 | 4 ( 1 p ) B 1 [ | B 2 | ( 3 2 p ) + B 1 ] 4 B 2 2 ( 1 p ) 2 B 1 2 ( 1 2 p ) 2 | B 3 B 1 p B 2 2 | ( 1 p ) B 1 ( 2 | B 2 | + B 1 ) ) .

Proof For f R γ τ (φ), there exists an analytic function w with w(0)=0 and |w(z)|<1 in D such that

1+ 1 τ ( f ( z ) + γ z f ( z ) 1 ) =φ ( w ( z ) ) .
(24)

Since f has the Maclaurin series given by (1), a computation shows that

1 + 1 τ ( f ( z ) + γ z f ( z ) 1 ) = 1 + 2 a 2 ( 1 + γ ) τ z + 3 a 3 ( 1 + 2 γ ) τ z 2 + 4 a 4 ( 1 + 3 γ ) τ z 3 + .
(25)

It follows from (24), (9) and (25) that

a 2 = τ B 1 c 1 4 ( 1 + γ ) , a 3 = τ B 1 12 ( 1 + 2 γ ) [ 2 c 2 + c 1 2 ( B 2 B 1 1 ) ] , a 4 = τ 32 ( 1 + 3 γ ) [ B 1 ( 4 c 3 4 c 1 c 2 + c 1 3 ) + 2 B 2 c 1 ( 2 c 2 c 1 2 ) + B 3 c 1 3 ] .

Therefore

a 2 a 4 a 3 2 = τ 2 B 1 c 1 128 ( 1 + γ ) ( 1 + 3 γ ) [ B 1 ( 4 c 3 4 c 1 c 2 + c 1 3 ) + 2 B 2 c 1 ( 2 c 2 c 1 2 ) + B 3 c 1 3 ] τ 2 B 1 2 144 ( 1 + 2 γ ) 2 [ 4 c 2 2 + c 1 4 ( B 2 B 1 1 ) 2 + 4 c 2 c 1 2 ( B 2 B 1 1 ) ] = τ 2 B 1 2 128 ( 1 + γ ) ( 1 + 3 γ ) { [ ( 4 c 1 c 3 4 c 1 2 c 2 + c 1 4 ) + 2 B 2 c 1 2 B 1 ( 2 c 2 c 1 2 ) + B 3 B 1 c 1 4 ] 8 9 ( 1 + γ ) ( 1 + 3 γ ) ( 1 + 2 γ ) 2 [ 4 c 2 2 + c 1 4 ( B 2 B 1 1 ) 2 + 4 c 2 c 1 2 ( B 2 B 1 1 ) ] } ,

which yields

| a 2 a 4 a 3 2 | = T | 4 c 1 c 3 + c 1 4 [ 1 2 B 2 B 1 p ( B 2 B 1 1 ) 2 + B 3 B 1 ] 4 p c 2 2 4 c 1 2 c 2 [ 1 B 2 B 1 + p ( B 2 B 1 1 ) ] | ,
(26)

where

T= | τ | 2 B 1 2 128 ( 1 + γ ) ( 1 + 3 γ ) andp= 8 9 ( 1 + γ ) ( 1 + 3 γ ) ( 1 + 2 γ ) 2 .

It can be easily verified that p[ 64 81 , 8 9 ] for 0γ1.

Let

d 1 = 4 , d 2 = 4 [ 1 B 2 B 1 + p ( B 2 B 1 1 ) ] , d 3 = 4 p , d 4 = 1 2 B 2 B 1 p ( B 2 B 1 1 ) 2 + B 3 B 1 .
(27)

Then (26) becomes

| a 2 a 4 a 3 2 | =T | d 1 c 1 c 3 + d 2 c 1 2 c 2 + d 3 c 2 2 + d 4 c 1 4 | .
(28)

It follows that

| a 2 a 4 a 3 2 | = T 4 | c 4 ( d 1 + 2 d 2 + d 3 + 4 d 4 ) + 2 x c 2 ( 4 c 2 ) ( d 1 + d 2 + d 3 ) + ( 4 c 2 ) x 2 ( d 1 c 2 + d 3 ( 4 c 2 ) ) + 2 d 1 c ( 4 c 2 ) ( 1 | x | 2 ) z | .

Application of the triangle inequality, replacement of |x| by μ and substituting the values of d 1 , d 2 , d 3 and d 4 from (27) yield

| a 2 a 4 a 3 2 | T 4 [ 4 c 4 | B 3 B 1 p B 2 2 B 1 2 | + 8 | B 2 B 1 | μ c 2 ( 4 c 2 ) ( 1 p ) + ( 4 c 2 ) μ 2 ( 4 c 2 + 4 p ( 4 c 2 ) ) + 8 c ( 4 c 2 ) ( 1 μ 2 ) ] = T [ c 4 | B 3 B 1 p B 2 2 B 1 2 | + 2 c ( 4 c 2 ) + 2 μ | B 2 B 1 | c 2 ( 4 c 2 ) ( 1 p ) + μ 2 ( 4 c 2 ) ( 1 p ) ( c α ) ( c β ) ] F ( c , μ ) ,
(29)

where α=2, β=2p/(1p)>2.

Similarly as in the previous proofs, it can be shown that F(c,μ) is an increasing function of μ for 0<μ<1. So, for fixed c[0,2], let

maxF(c,μ)=F(c,1)G(c),

which is

G ( c ) = T { c 4 [ | B 3 B 1 p B 2 2 B 1 2 | ( 1 p ) ( 2 | B 2 B 1 | + 1 ) ] + 4 c 2 [ 2 | B 2 B 1 | ( 1 p ) + 1 2 p ] + 16 p } .

Let

P = | B 3 B 1 p B 2 2 B 1 2 | ( 1 p ) ( 2 | B 2 B 1 | + 1 ) , Q = 4 [ 2 | B 2 B 1 | ( 1 p ) + 1 2 p ] , R = 16 p .
(30)

Using (16), we have

| a 2 a 4 a 3 2 | T{ R , Q 0 , P Q 4 ; 16 P + 4 Q + R , Q 0 , P Q 8  or  Q 0 , P Q 4 ; 4 P R Q 2 4 P , Q > 0 , P Q 8 ,

where P, Q, R are given in (30). □

Remark 3 For the choice φ(z):=(1+Az)/(1+Bz) with 1B<A1, Theorem 3 reduces to [[36], Theorem 2.1].

Definition 4 Let φ:DC be analytic, and let φ(z) be as given in (6). For a fixed real number α, the function fA is in the class G α (φ) if it satisfies the following subordination:

(1α) f (z)+α ( 1 + z f ( z ) f ( z ) ) φ(z).

Al-Amiri and Reade [37] introduced the class G α := G α ((1+z)/(1z)) and they showed that G α S for α<0. Univalence of the functions in the class G α was also investigated in [38, 39]. Singh et al. also obtained the bound for the second Hankel determinant of functions in G α . The following theorem provides a bound for the second Hankel determinant of the functions in the class G α (φ).

Theorem 4 Let the function f given by (1) be in the class G α (φ), 0α1. Also, let

p= 8 9 ( 1 + 2 α ) ( 1 + α ) .
  1. 1.

    If B 1 , B 2 and B 3 satisfy the conditions

    B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) 0 , | B 1 4 α ( 2 α 1 p α ) + α B 1 2 B 2 ( 3 2 p ) + ( α + 1 ) B 1 B 3 p B 2 2 | p B 1 2 0 ,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | B 1 2 9 ( 1 + α ) 2 .
  1. 2.

    If B 1 , B 2 and B 3 satisfy the conditions

    B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) 0 , 2 | B 1 4 α ( 2 α 1 p α ) + α B 1 2 B 2 ( 3 2 p ) + ( α + 1 ) B 1 B 3 p B 2 2 | B 1 3 α ( 3 2 p ) 2 ( 1 + α p ) B 1 | B 2 | ( α + 1 ) B 1 2 0 ,

or

B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) 0 , | B 1 4 α ( 2 α 1 p α ) + α B 1 2 B 2 ( 3 2 p ) + ( α + 1 ) B 1 B 3 p B 2 2 | p B 1 2 0 ,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | | B 1 4 α ( 2 α 1 p α ) + α B 1 2 B 2 ( 3 2 p ) + ( α + 1 ) B 1 B 3 p B 2 2 | 8 ( 1 + α ) ( 1 + 2 α ) .
  1. 3.

    If B 1 , B 2 and B 3 satisfy the conditions

    B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) > 0 , 2 | B 1 4 α ( 2 α 1 p α ) + α B 1 2 B 2 ( 3 2 p ) + ( α + 1 ) B 1 B 3 p B 2 2 | B 1 3 α ( 3 2 p ) 2 ( 1 + α p ) B 1 | B 2 | ( α + 1 ) B 1 2 0 ,

then the second Hankel determinant satisfies

| a 2 a 4 a 3 2 | B 1 2 32 ( 1 + α ) ( 1 + 2 α ) × [ 4 p [ B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) ] 2 | B 1 4 α ( 2 α 1 p α ) + α B 1 2 B 2 ( 3 2 p ) + ( α + 1 ) B 1 B 3 p B 2 2 | B 1 3 α ( 3 2 p ) ( 1 + α p ) B 1 ( 2 | B 2 | + B 1 ) ] .

Proof For f G α (φ), a calculation shows that

| a 2 a 4 a 3 2 | = T | 4 ( 1 + α ) B 1 c 1 c 3 + c 1 4 [ 3 α B 1 2 + α ( 2 α 1 ) B 1 3 + B 1 ( 1 + α ) + 3 α B 1 B 2 + ( 1 + α ) ( B 3 2 B 2 ) p ( α B 1 2 B 1 + B 2 ) 2 B 1 ] 4 p B 1 c 2 2 + 2 c 1 2 c 2 [ 2 ( 1 + α ) B 1 + 3 α B 1 2 + 2 ( 1 + α ) B 2 2 p ( α B 1 2 B 1 + B 2 ) ] | ,
(31)

where

T= B 1 128 ( 1 + α ) ( 1 + 2 α ) andp= 8 9 ( 1 + 2 α ) ( 1 + α ) .

It can be easily verified that for 0α1, p[ 8 9 , 4 3 ]. Let

d 1 = 4 ( 1 + α ) B 1 , d 2 = 2 [ 2 ( 1 + α ) B 1 + 3 α B 1 2 + 2 ( 1 + α ) B 2 2 p ( α B 1 2 B 1 + B 2 ) ] , d 3 = 4 p B 1 , d 4 = 3 α B 1 2 + α ( 2 α 1 ) B 1 3 + B 1 ( 1 + α ) + 3 α B 1 B 2 d 4 = + ( 1 + α ) ( B 3 2 B 2 ) p ( α B 1 2 B 1 + B 2 ) 2 B 1 .
(32)

Then

| a 2 a 4 a 3 2 | =T | d 1 c 1 c 3 + d 2 c 1 2 c 2 + d 3 c 2 2 + d 4 c 1 4 | .
(33)

Similarly as in earlier theorems, it follows that

| a 2 a 4 a 3 2 | = T 4 | c 4 ( d 1 + 2 d 2 + d 3 + 4 d 4 ) + 2 x c 2 ( 4 c 2 ) ( d 1 + d 2 + d 3 ) + ( 4 c 2 ) x 2 ( d 1 c 2 + d 3 ( 4 c 2 ) ) + 2 d 1 c ( 4 c 2 ) ( 1 | x | 2 ) z | T [ c 4 | B 1 3 α ( 2 α 1 p α ) + α B 1 B 2 ( 3 2 p ) + ( α + 1 ) B 3 p B 2 2 B 1 | + μ c 2 ( 4 c 2 ) [ B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) ] + 2 c ( 4 c 2 ) B 1 ( 1 + α ) + μ 2 ( 4 c 2 ) B 1 ( 1 + α p ) ( c 2 ) ( c 2 p 1 + α p ) ] F ( c , μ ) ,
(34)

and for fixed c[0,2], maxF(c,μ)=F(c,1)G(c) with

G ( c ) = T [ c 4 [ | B 1 3 α ( 2 α 1 p α ) + α B 1 B 2 ( 3 2 p ) + ( α + 1 ) B 3 p B 2 2 B 1 | B 1 2 α ( 3 2 p ) ( 1 + α p ) ( 2 | B 2 | + B 1 ) ] + 4 c 2 [ B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) ] + 16 p B 1 ] .

Let

P = | B 1 3 α ( 2 α 1 p α ) + α B 1 B 2 ( 3 2 p ) + ( α + 1 ) B 3 p B 2 2 B 1 | P = B 1 2 α ( 3 2 p ) ( 1 + α p ) ( 2 | B 2 | + B 1 ) , Q = 4 [ B 1 2 α ( 3 2 p ) + 2 | B 2 | ( 1 + α p ) + B 1 ( 1 + α 2 p ) ] , R = 16 p B 1 .
(35)

By using (16), we have

| a 2 a 4 a 3 2 | T{ R , Q 0 , P Q 4 ; 16 P + 4 Q + R , Q 0 , P Q 8  or  Q 0 , P Q 4 ; 4 P R Q 2 4 P , Q > 0 , P Q 8 ,

where P, Q, R are given in (35). □

Remark 4 For α=1, Theorem 4 reduces to Theorem 2. For 0α<1, let φ(z):=(1+(12α)z)/(1z). For this function φ, B 1 = B 2 = B 3 =2(1α). In this case, Theorem 4 reduces to [[40], Theorem 3.1].

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Acknowledgements

The work presented here was supported in part by research grants from Universiti Sains Malaysia (FRGS grants) and University of Delhi as well as MyBrain MyPhD programme of the Ministry of Higher Education, Malaysia.

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Correspondence to See Keong Lee.

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Lee, S.K., Ravichandran, V. & Supramaniam, S. Bounds for the second Hankel determinant of certain univalent functions. J Inequal Appl 2013, 281 (2013) doi:10.1186/1029-242X-2013-281

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Keywords

  • Analytic Function
  • Convex Function
  • Unit Disk
  • Meromorphic Function
  • Univalent Function