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Hankel inequalities for a subclass of bi-univalent functions based on Salagean type q-difference operator

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Lately, in [2] the Sălăgean type q-differential operator has been introduced as given by

( ) = ( ) ( ) = ( )

( ) = ( ( ))

( ) = + ∑ [ ] ( ∈ ℕ , ∈ Δ). (3)

For → 1 , we get

( ) = +  ( ∈ ℕ , ∈ Δ)

the familiar Sălăgean derivative [3].

Noonan and Thomas [4] introduced the Hankel determinant of function by

( ) =

⋮ ⋮ ⋮ ⋮

( ≥ 1).

In particular,

(1) = = − = −

and

(2) = = − .

Then, Fekete and Szegö [5] obtained estimates of | (1)| = | − | for is real. That is, if ∈ , then

| − | ≤

4 − 3 ≥ 1

1 + 2 ( ) 0 ≤ ≤ 1

3 − 4 ≤ 0

.

Furthermore, Keogh and Merkes [6] derived sharp estimates for | (1)| when is starlike, convex and close-to-convex in Δ.

Next, according to the Koebe One Quarter Theorem [7], every univalent function has an inverse satisfying ( ( )) = , ( ∈△) and

( ( )) = (| | < ( ), ( ) ≥ ). A function ∈ is said to be bi-

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univalent in △ if both and are univalent in △. Let Σ indicate the class of bi-univalent functions defined in the unit disk △. Since ∈ Σ has the Taylor representation given by Eq. (1), computation shows that = has the following representation:

( ) = ( ) = − + (2 − ) + ⋯. (4)

Several researchers have introduced new subclasses of bi-univalent functions and derived non-sharp the initial coefficients (see [8-18]).

Now, by using the Sălăgean type q-differential operator for functions of the form Eq. (4), we define:

( ) = − [2] + (2 − )[3] + ⋯ (5)

and introduce a new subclass of Σ to acquire the estimates of the initial Taylor- Maclaurin coefficients. Then, by using the values of and , we derive the Fekete-Szego and Hankel inequalities.

2 Bi-Univalent Function Class ( , )

In this section, we will give the following new subclass involving the Sălăgean type q-difference operator and also its related classes.

Definition 2.1. A function ∈ Σ given by Eq. (1) is said to be in the class ℱΣ ( , ) (0 ≤ < 1,0 ≤ ≤ 1, , ∈ ∆)

if the following conditions hold:

ℜ (1 − ) ( )+ ( ) >

and

ℜ (1 − ) ( )+ ( ) > .

Example 2.2. A function ∈ Σ, members of which are given by Eq. (1) and 1. for = 0, let ℱΣ (0, ) =: ℛΣ ( ) denote the subclass of Σ and the

following conditions hold

( ) > and ℜ ( ) >

2. for = 1, let ℱΣ (1, ) =: ℋΣ ( ) denote the subclass of Σ and satisfy the following conditions

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ℜ ( ) > and ℜ ( ) > .

3 Hankel Inequalities for ( , )

In this section, we will determine the functional − for the functions

∈ ℱΣ ( , ) due to Altınkaya and Yalçın [19]. Now, we recall the following lemmas:

Lemma 3.1. (See [4]) Let be the well-known class of Carathéodory functions, that is ( ) ∈ with the power series expansion

( ) = 1 + ∑ ( ∈ Δ) (6)

and ℜ( ( )) > 0. Then

| | ≤ 2 ( = 1,2,3, . . . ) and is sharp for each . Indeed,

( ) = = 1 + ∑ 2 (∀ ≥ 1).

Lemma 3.2. (See [20]) If ∈ , then

2 = + (4 − ), (7)

4 = + 2(4 − ) − (4 − ) + 2(4 − )(1 − | | ) for some complex numbers , with | | ≤ 1 and | | ≤ 1.

Lemma 3.3. (See [5]) The power series for converges in Δ to a function in if and only if the Toeplitz determinants

=

2 ⋯

2 ⋯

⋮ ⋮ ⋮ ⋮ ⋮

⋯ 2

   (   = 1,2,3, … )

and = are all nonnegative. They are exactly positive except for ( ) = ( ),   > 0,     real

and ≠ ( ≠ ). In this case > 0 ( < − 1) and = 0 ( ≥ ).

Next, we designate the second Hankel coefficient estimates for ∈ ℱΣ ( , ).

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Theorem 3.4. Let ∈ ℱΣ ( , ). Then

| − |

(2), ( , , , ) ≥ 0, ( , , , ) ≥ 0

max 4(1 − )

(1 + 2 ) [3] , (2) , ( , , , ) > 0, ( , , , ) < 0 4(1 − )

(1 + 2 ) [3] , ( , , , ) ≤ 0, ( , , , ) ≤ 0 max { ( ), (2)} , ( , , , ) < 0, ( , , , ) > 0 ,

where

(2) = 16(1 − )

(1 + ) [2] + 4(1 − )

(1 + )(1 + 3 )[2] [4] ,

= − ( , , , )

( , , , ) = 4(1 − )

(1 + 2 ) [3] − ( , , , ) 4 ( , , , ),

( , , , ) = (1 − )

(1 + ) [2] − (1 − )

4(1 + ) (1 + 2 )[2] [3]

− (1 − )

2(1 + )(1 + 3 )[2] [4] + (1 − ) 4(1 + 2 ) [3]

( , , , ) = (1 − )

(1 + ) (1 + 2 )[2] [3] + 3(1 − ) (1 + )(1 + 3 )[2] [4]

− 2(1 − ) (1 + 2 ) [3] .

Proof. Suppose that ∈ ℱΣ ( , ). There are two functions , ∈ satisfying the conditions of Lemma 3.1 such that

(1 − ) ( )+ ( ) = + (1 − ) ( ), (8)

(1 − ) ( )+ ( ) = + (1 − ) ( ), (9)

where

( ) = 1 + + + + ⋯,

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( ) = 1 + + + + ⋯.

Now, by comparing the corresponding coefficients in Eq. (8) and Eq. (9), we get

(1 + )[2] = (1 − ) , (10)

(1 + 2 )[3] = (1 − ) , (11)

(1 + 3 )[4] = (1 − ) (12)

and

−(1 + )[2] = (1 − ) , (13)

(1 + 2 )[3] (2 − ) = (1 − ) , (14)

−(1 + 3 )[4] (5 − 5 + ) = (1 − ) . (15)

From Eq. (10) and Eq. (13), we get

=( )[ ] = −

( )[ ] , (16)

which implies

= − .

Now from Eq. (11) and Eq. (14), we obtain

= ( )

( ) [ ] + ( )

( )[ ] ( − ).

On the other hand, subtracting Eq. (15) from Eq. (12) and using Eq. (16), we get

= ( )(( ))[ ] [ ] ( − ) + ( ( )[ ]) ( − ).

Thus, we establish that

| − | =

( )

( ) [ ]

+ ( ) (( ))[ ] [ ] ( − )

+ ( )

( )( )[ ] [ ] ( − ) − ( )

( )[ ] ( − ) . (17)

Now, by Lemma 3.2, we get

2 = + (4 − ) and 2 = + (4 − ), (18) and hence, by Eq. (18), we have

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− = ( − ). (19) Further, we get

4 = + 2(4 − ) − (4 − ) + 2(4 − )(1 − | | ) , 4 = + 2(4 − ) − (4 − ) + 2(4 − )(1 − | | ) and thus, we acquire

− = + ( + ) − ( + ) (20)

+4 −

2 [(1 − | | ) − (1 − | | ) ].

Using Eq. (19) – Eq. (20) in Eq. (17), we get

− =

−(1 − )

(1 + ) [2] + (1 − )

4(1 + )(1 + 3 )[2] [4]

+ ( )

( ) ( )[ ] [ ]

( )( − )

+ ( )

( )( )[ ] [ ] ( + )

( )

( )( )[ ] [ ] ( + )

= + ( )

( )( )[ ] [ ] [(1 − | | ) − (1 − | | ) ]

( ( ) [ ]) ( ) ( − ) .

Since ∈ , we find that | | ≤ 2. Thus, letting | | = ∈ [0,2] and applying triangle inequality on Eq. (21), we get

− ≤ ( )

( ) [ ] + ( )

( )( )[ ] [ ]

+ ( )

( )( )[ ] [ ] (4 − )

+ ( ) (( ))[ ] [ ] + ( )(( ))[ ] [ ] ( )(| | + | |)

+ ( )

( )( )[ ] [ ]

( )( )(| | + | | )

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( ( ) [ ]) ( ) (| | + | |) . (21) For = | | ≤ 1 and = | | ≤ 1, we get

| − | ≤ + ( + ) + ( + ) (22)

( + ) = Ψ( , ), where

= ( ) = (1 − )

(1 + ) [2] + (1 − )

4(1 + )(1 + 3 )[2] [4]

+ (1 − )

2(1 + )(1 + 3 )[2] [4] (4 − ) ≥ 0,

= ( ) = (1 − )

4(1 + ) (1 + 2 )[2] [3]

+ (1 − )

2(1 + )(1 + 3 )[2] [4]

(4 − ) 2 ≥ 0,

= ( ) = (1 − )

2(1 + )(1 + 3 )[2] [4]

( − 2)(4 − )

4 ≤ 0,

= ( ) = (1 − ) 4(1 + 2 ) [3]

(4 − ) 4 ≥ 0.

Next, we will find the maximum of (Ψ( , )) in Γ = {( , ): 0 ≤ ≤ 1,0 ≤

≤ 1}. Since the coefficients of Ψ( , ) have dependent variable , we should maximize Ψ( , ) for the cases = 0, = 2 and ∈ (0,2).

1. Let = 0. Thus, from (22), we may write Ψ( , ) =( ( ) [ ]) ( + ) .

2. We can find that the maximum of Ψ( , ) occurs at = = 1 and we find max { Ψ( , ): 0 ≤ ≤ 1,0 ≤ ≤ 1} =( ( ) [ ]) .

3. Let = 2. Thus, Ψ( , ) is a constant function Ψ( , ) = ( )

( ) [ ] + ( )

( )( )[ ] [ ] .

4. Let ∈ (0,2). If we change + = and . = , then

Ψ( , ) = ( ) + ( ) + [ ( ) + ( )] − 2 ( )

= ( , ),  0 ≤ ≤ 2,0 ≤ ≤ 1.

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Presently, we try to get maximum of ( , ) in

= {( , ): 0 ≤ ≤ 2,0 ≤ ≤ 1}.

From the definition of ( , ), we get

( , ) = ( ) + 2[ ( ) + ( )] = 0, ( , ) = −2 ( ) = 0.

We deduce that the function doesn’t have any critical point in . Thus, Ψ( , ) doesn’t have any critical point in square Γ and so the function doesn’t get maximum value in Γ.

Next, we inspect the maximum of Ψ( , ) on the boundary of Γ.

Firstly, let = 0,0 ≤ ≤ 1 (or let = 0,0 ≤ ≤ 1). Then, we may write Ψ(0, ) = ( ) + ( ) + [ ( ) + ( )] = ( ).

Thus,

( ) = ( ) + 2[ ( ) + ( )] .

Case (i): If ( ) + ( ) ≥ 0, then ( ) > 0. The function is increasing and so the maximum occurs at = 1.

Case (ii): Let ( ) + ( ) < 0. Since ( ) + 2[ ( ) + ( )] > 0, ( ) + 2[ ( ) + ( )] ≥ ( ) + 2[ ( ) + ( )] holds for all

∈ [0,1]. So, ( ) > 0. Hence, ( ) is an increasing function. Thus, the maximum occurs at = 1,

max { Ψ(0, ): 0 ≤ ≤ 1} = ( ) + ( ) + ( ) + ( ).

Secondly, let = 1,0 ≤ ≤ 1 (similarly, = 1,0 ≤ ≤ 1). Then Ψ(1, ) = ( ) + ( ) + ( ) + ( )

+[ ( ) + 2 ( )] + [ ( ) + ( )]

= ( ).

It can be stated that ( ) is an increasing function like case (i). In that way, max { Ψ(1, ): 0 ≤ ≤ 1} = ( ) + 2[ ( ) + ( )] + 4 ( ).

Also, for every ∈ (0,2), we can easily see that

( ) + 2[ ( ) + ( )] + 4 ( ) > ( ) + ( ) + ( ) + ( ).

Therefore, we find that

max { Ψ( , ): 0 ≤ ≤ 1,0 ≤ ≤ 1} = ( ) + 2[ ( ) + ( )] + 4 ( ).

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Since (1) ≤ (1) for ∈ [0,2], max Ψ ( , ) = Ψ(1,1) on the boundary of Γ. So, the maximum of Ψ occurs at = 1 and = 1 in the Γ.

Let us define ℋ: (0,2) → ℝ as

ℋ( ) = max Ψ ( , ) = Ψ(1,1)

= 2[ ( ) + ( )] + ( ) + 4 ( ). (23)

Therefore, from Eq. (23), we obtain ℋ( ) = 4(1 − )

(1 + 2 ) [3] + ( , , , ) + 2 ( , , , ) , where

( , , , ) = (1 − )

(1 + ) [2] − (1 − )

4(1 + ) (1 + 2 )[2] [3]

− (1 − )

2(1 + )(1 + 3 )[2] [4] + (1 − ) 4(1 + 2 ) [3]

( , , , ) = (1 − )

(1 + ) (1 + 2 )[2] [3] + 3(1 − ) (1 + )(1 + 3 )[2] [4]

− 2(1 − ) (1 + 2 ) [3] .

Now, we try to get the maximum value of ℋ( ) in (0,2). After some basic calculations, we have

ℋ ( ) = 4 ( , , , ) + 2 ( , , , ) .

Next, we examine the different cases of ( , , , ) and ( , , , ) as follows:

Case 1: Let ( , , , ) ≥ 0 and ( , , , ) ≥ 0, then ℋ ( ) ≥ 0. Hence, the maximum point has to be on the boundary of ∈ [0,2], that is = 2.

Thus,

max { Ψ( , ): 0 ≤ ≤ 1,0 ≤ ≤ 1}

= ℋ(2)

=

( () [ ])

+

( )(( )[ ] [ ]) (24)

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Case 2: If ( , , , ) > 0 and ( , , , ) < 0, = ( , , , )

( , , , ) is a critical point of ℋ( ). Since ℋ ( ) < 0, the maximum value of function ℋ( ) occurs at = and

ℋ( ) = 4(1 − )

(1 + 2 ) [3] + ( , , , ) + 2 ( , , , )

= 4(1 − )

(1 + 2 ) [3] −3 ( , , , ) 4 ( , , , ). In this case, ℋ( ) <( ( ) [ ]) . Therefore,

max { Ψ( , ): 0 ≤ ≤ 1,0 ≤ ≤ 1}

= max ( ( ) [ ]) ,( () [ ]) +( )(( )[ ] [ ]) . (25) Case 3: If ( , , , ) ≤ 0 and ( , , , ) ≤ 0, ℋ( ) is decreasing in (0,2).

Therefore,

max { Ψ( , ): 0 ≤ ≤ 1,0 ≤ ≤ 1} =( ( ) [ ]) (26) Case 4: If ( , , , ) < 0 and ( , , , ) > 0, is a critical point of ℋ( ).

Since ℋ ( ) < 0, the maximum value of ℋ( ) occurs at = and

( )

( ) [ ] < ℋ( ).

Therefore,

max { Ψ( , ): 0 ≤ ≤ 1,0 ≤ ≤ 1}

= max ℋ( ), ( )

( ) [ ] + ( )

( )( )[ ] [ ] (27)

Thus, from Eqs. (24-26) and Eq. (27), the proof is completed.

Remark 3.5. For = 0 (and = 1) in Theorem 3.4, we can confirm the Hankel inequalities for the function classes ℛΣ ( ), ℋΣ ( ), respectively.

Acknowledgement

The authors are grateful to the referees of this article for their valuable comments and advice.

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References

[1] Jackson, F.H., On Q-Functions and a Certain Difference Operator, Transactions of the Royal Society of Edinburgh, 46, pp. 253-281,1908.

[2] Govindaraj, M. & Sivasubramanian, S., On a Class of Analytic Functions Related to Conic Domains Involving Q-Calculus, Analysis Math., 43(3), pp. 475-487, 2017.

[3] Sălăgean, G.S., Subclasses of Univalent Functions, Complex Analysis – Fifth Romanian Finish Seminar, Bucharest, pp. 362-372, 1983.

[4] Noonan, J.W. & Thomas, D.K., On the Second Hankel Determinant of A Really Mean P-Valent Functions, Trans. Amer. Math. Soc., 223(2), pp.

337-346, 1976.

[5] Fekete, M. & Szegö, G., Eine Bemerkung über Ungerade Schlichte Funktionen, J. Lond. Math. Soc., 2, pp. 85-89, 1933. (Text in Germany) [6] Keogh, F.R. & Merkes, E.P., A Coefficient Inequality for Certain Classes

of Analytic Functions, Proc. Amer. Math. Soc., 20, pp. 8-12, 1969.

[7] Duren, P.L., Univalent Functions, in: Grundlehren der Mathematischen Wissenchaften, 259, Springer, New York, 1983.

[8] Brannan, D.A., Clunie, J. & Kirwan, W.E., Coefficient Estimates for A Class of Starlike Functions, Canad. J. Math., 22, pp. 476-485, 1970.

[9] Brannan, D.A. & Taha, T.S., On Some Classes of Bi-Univalent Functions, Studia Univ. BabeŞ-Bolyai Math., 31(2), pp. 70-77, 1986.

[10] Frasin, B.A. & Aouf, M.K., New Subclasses of Bi-Univalent Functions, Appl. Math. Lett., 24, pp. 1569-1573, 2011.

[11] Hayami, T. & Owa, S., Coefficient Bounds for Bi-Univalent Functions, Pan Amer. Math. J., 22(4), pp. 15-26, 2012.

[12] Lewin, M., On a Coefficient Problem for Bi-Univalent Functions, Proc.

Amer. Math. Soc., 18, pp. 63-68, 1967.

[13] Li, X-F. & Wang, A-P., Two New Subclasses of Bi-Univalent Functions, International Mathematical Forum, 7(30), pp. 1495-1504, 2012.

[14] Panigarhi, T. & Murugusundaramoorthy, G., Coefficient Bounds for Bi- Univalent Functions Analytic Functions Associated with Hohlov Operator, Proc. Jangjeon Math. Soc., 16(1), pp. 91-100, 2013.

[15] Srivastava, H.M., Mishra, A.K. & Gochhayat, P., Certain Subclasses of Analytic and Bi-Univalent Functions, Appl. Math. Lett., 23(10), pp.

1188-1192, 2010.

[16] Xu, Q-H., Srivastava, H.M., & Li, Z., A Certain Subclass of Analytic and Closed-To Convex Functions, Appl. Math. Lett., 24, pp. 396-401, 2011.

[17] Xu, Q-H., Gui, Y-C. & Srivastava, H.M., Coefficinet Estimates for A Certain Subclass of Analytic and Bi-Univalent Functions, Appl. Math.

Lett., 25, pp. 990-994, 2012.

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[18] Zaprawa, P. On the Fekete-Szegö Problem for Classes of Bi-Univalent Functions, Bull. Belg. Math. Soc. Simon Stevin, 21(1), pp. 169-178, 2014.

[19] Altınkaya, Ş. & Yalçın, S., Upper Bounds of Second Hankel Determinant for Bi-Bazilevic Functions, Mediterranean Journal of Mathematics (MJOM), 13(6), pp. 4081-4090, 2016.

[20] Libera, R.J. & Zlotkiewicz, E.J., Coefficient Bounds for the Inverse of A Function with Derivative in , Proc. Amer. Math. Soc., 87(2), pp. 251- 257, 1983.

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