Some Properties of a Subclass of Harmonic Univalent Functions Defined By Salagean Operator

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1 Int. J. Open Problems Complex Analysis, Vol. 8, No. 2, July 2016 ISSN ; Copyright c ICSRS Publication, Some Properties of a Subclass of Harmonic Univalent Functions Defined By Salagean Operator R. M. El-Ashwah and W. Y. Kota Department of Mathematics, Faculty of science, Damietta University New Damietta, Egypt r elashwah@yahoo.com Department of Mathematics, Faculty of science, Damietta University New Damietta, Egypt wafaa kota@yahoo.com Received 1 February 2016; Accepted 12 May 2016 Abstract In this paper, we introduced the classes M l H (m, n, φ, ψ; γ) and V l H (m, n, φ, ψ; γ), l = {0, 1} consisting of harmonic univalent functions f = h + g. We studied the coefficients estimate, distortion theorem, extreme points, convex combination and family of integral operators. Also, we established some results concerning the convolution. In proving our results certain conditions on the coefficients of φ and ψ are considered which lead various well-known results proved earlier. Keywords: Harmonic functions, univalent functions, Salagean operator, extreme points, convolution Mathematical Subject Classification: 30C45, 30C50. 1 Introduction A continuous complex-valued function f = u+iv defined in a simply connected domain D is said to be harmonic in D if both u and v are real harmonic in D, that is u, v satisfy, respectively, the Laplace equations. It follows that

2 22 R. M. El-Ashwah and W. Y. Kota every analytic function is a complex-valued harmonic function. In any simply connected domain D, we can write f = h + g, where h and g are analytic in D. We call h the analytic part and g the co-analytic part of f. A necessary and sufficient condition for f to be locally univalent and sense-preserving in D is that h (z) > g (z), z D (see[5]). Denote by S H the class of functions f = h + g that are harmonic univalent and sense-preserving in the unit disk U = {z : z < 1} for which f(0) = f z(0) 1 = 0. Then for f = h + g S H we may express the analytic functions h and g as h(z) = z + a k z k, g(z) = b k z k, b 1 < 1. (1) We, note that S H reduces to the class of normalized analytic functions if the co-analytic part of f is identically zero; that is g = 0, then f(z) = z + a k z k. (2) In 1984 Clunie and Sheil-Small [5] investigated the class S H as well as its geometric subclasses and some coefficient bounds for functions in S H were obtained. Also, various subclasses of S H were investigated by several authors (see [1], [2], [3], [7] and [13]). Definition 1.1 A function f(z) S H is said to be in the class of harmonic starlike functions of order α denote by SH (α) if it satisfies the following condition ( arg(f(re iθ )) ) > α θ which is equivalent to the condition { } zh (z) zg Re (z) α, (0 α < 1; z U). h (z) + g (z) Definition 1.2 A function f(z) S H is said to be in the class of harmonic convex functions of order α denote by K H (α) if it satisfies the following condition ( ( )) arg θ θ f(reiθ ) > α which is equivalent to the condition { } Re 1 + z2 h (z) + 2zg (z) + z 2 g (z) zh (z) zg (z) α, (0 α < 1; z U).

3 Some Properties of a Subclass of Harmonic Univalent 23 The classes S H (α) and K H(α) were introduced and studied by Jahangiri [9]. From the above conditions, we see that f(z) K H (α) zf (z) S H(α). A necessary condition for a function f(z) = h + g, where h(z) and g(z) are of the form (1) belong to the classes S H (α) and K H(α) (k α) a k + (k + α) b k 1 α f(z) SH(α), and k(k α) a k + k(k + α) b k 1 α f(z) K H (α). For 1 < γ 4/3 and z U, Porwal and Dixit [14] introduced and studied the classes M H (γ) of harmonic functions f = h + g of the form (1) satisfying the condition (argf(z)) = Re θ { } zh (z) zg (z) γ, h(z) + g(z) (z U), and L H (γ) the class of harmonic functions of the form (1) satisfying the condition { ( )} { } arg θ θ f(z) = Re 1 + z2 h (z) + 2zg (z) + z 2 g (z) γ, (z U). zh (z) zg (z) Further, let V H and U H be the subclasses of S H consisting of functions of the form f(z) = z + a k z k b k z k, (3) and f(z) = z + a k z k + b k z k, (4) respectively. Let V H (γ) M H (γ) V H, and U H (γ) L H (γ) U H. A necessary and sufficient condition for a function f(z) is of the form (3) belongs to the classes V H (γ) and (k γ) a k + (k + γ) b k, (5)

4 24 R. M. El-Ashwah and W. Y. Kota where 1 < γ 4 3 [14]. A necessary and sufficient condition for a function f(z) is of the form (4) belongs to the classes U H (γ) and k(k γ) a k + k(k + γ) b k, (6) where 1 < γ 4 3 [14]. We note that for g = 0 the classes M H(γ) M(γ), L H (γ) L(γ), V H (γ) V (γ) and U H (γ) U(γ) were studied in [18]. For f = h + g with h and g are of the form (1), the modified differential Salagean operator D n for n N 0 = N {0}, is given by where D n h(z) = z + D n f(z) = D n h(z) + ( 1) n D n g(z), (7) k n a k z k, D n g(z) = k n b k z k. Sharma et al. [15] define a generalized class SH l (m, n, φ, ψ; α) of functions f = h + g S H satisfying for l {0, 1}, the condition { } R D m h(z) φ(z) + ( 1) m+l D m g(z) ψ(z) D n h(z) + ( 1) n D n g(z) > α, where m, n N 0, m n, 0 α < 1, φ(z) = z + β kz k and ψ(z) = z + µ kz k are analytic in U with the conditions β k, µ k 1. Further denote by T SH l (m, n, φ, ψ; α), a subclass of Sl H (m, n, φ, ψ; α) consisting of functions f = h + ḡ S H such that h and g are of the form h(z) = z a k z k, g(z) = ( 1) m+l 1 b k z k, b 1 < 1. (8) It is interesting to note that by specializing the parameters m, n, l and the functions φ and ψ we obtain the following known subclasses of S H studied earlier. (i) SH 0 (m, n, z, z ; α) = S 1 z 1 z H(m, n; α) and TSH 0 (m, n, z studied by [19]. (ii) SH 0 (n + 1, n, z, z ; α) = S 1 z 1 z H(n; α) and TSH 0 (n + 1, n, z TS H (n; α) studied by [10]. ; α)= 1 z TS H (α) stud- (iii) SH 0 (1, 0, z ied by [9]. 1 z ; α) = S H (α) and TS0 H (1, 0, z 1 z ; α) = TS H(m, n; α) 1 z ; α) =

5 Some Properties of a Subclass of Harmonic Univalent 25 1 z ; α) = TK H(α) stud- (iv) SH 0 (2, 1, z ied by [9]. 1 z ; α) = K H(α) and TS 0 H (2, 1, z (v) SH 1 (0, 0, φ, ψ; α) = S H(φ, ψ; α) and TSH 0 (0, 0, φ, ψ; α) = TS1 H (0, 0, φ, ψ; α) = TS H (φ, ψ, α) studied by [8]. (vi) SH 0 (2, 1, z, z ; 0) = K 1 z 1 z H, TSH 0 (2, 1, z, z ; 0) = TK 1 z 1 z H, SH 0 (1, 0, z SH, and TS0 H (1, 0, z, z ; 0) = 1 z 1 z TS H studied by [16] and [17]. Using the operator D n, we define a generalized class MH l (m, n, φ, ψ; γ) of functions f = h + g S H satisfying for l {0, 1}, the condition { } R D m h(z) φ(z) + ( 1) m+l D m g(z) ψ(z) D n h(z) + ( 1) n D n g(z) γ, (9) where m, n N 0, m n, 1 < γ 4, and φ(z) = z + 3 β kz k and ψ(z) = z + µ kz k are analytic in U with the conditions β k, µ k 1. Further denote by VH l (m, n, φ, ψ; γ), a subclass of M H l (m, n, φ, ψ; γ) consisting of functions f = h + g S H such that h and g are of the form h(z) = z + a k z k, g(z) = ( 1) m+l b k z k, b 1 < 1. (10) It is interesting to note that by specializing the parameters m, n, l and the functions φ and ψ we obtain the following known subclasses of S H studied earlier. (i) VH 0(1, 0, z by [14]. (ii) VH 0(2, 1, z by [14]. 1 z ; γ) = V H(γ) and M 0 H (1, 0, z 1 z ; γ) = V H(γ) and M 0 H (2, 1, z 1 z ; γ) = M H(γ) studied 1 z ; γ) = M H(γ) studied In the present paper, we prove a number of coefficients estimate, distortion theorem, extreme points, a family of integral operators, convolution properties and convex combination for functions in MH l (m, n, φ, ψ; γ) and V H l (m, n, φ, ψ; γ) under certain conditions on the coefficients of φ and ψ. 1 z ; 0) = 2 Coefficients Estimate In this section, we studies a sufficient coefficient condition for functions in MH l (m, n, φ, ψ; γ) under certain conditions on the coefficients of φ and ψ.

6 26 R. M. El-Ashwah and W. Y. Kota Theorem 2.1 Let f(z) = h(z) + g(z), where h and g are given by (1) and satisfy the condition where (β k k m γk n ) a k + (µ k k m ( 1) m+l n γk n ) b k, (11) k β kk m γk n, k µ kk m ( 1) m+l n γk n for k 2. (12) l {0, 1}, m N, n N 0, m > n, β k, µ k 1, k 1, 1 γ < 4/3. Then f(z) is sense-preserving, harmonic univalent in U and f(z) MH l (m, n, φ, ψ; γ). Proof. To show that f is sense-preserving in U. h (z) = 1 + k a k z k 1 1 k a k r k 1 > 1 k a k β k k m γk n µ k k m ( 1) m+l n γk n > 1 a k > b k > k b k > k b k r k 1 g (z). To show that f is univalent in U, suppose z 1, z 2 U such that z 1 z 2, then we have f(z 1 ) f(z 2 ) h(z 1 ) h(z 2 ) = 1 g(z 1 ) g(z 2 ) h(z 1 ) h(z 2 ) = 1 b k(z2 k z1) k (z 2 z 1 ) + a k(z1 k z2) k > 1 k b k 1 k a k > 1 µ k k m ( 1) m+l n γk n 1 b γ 1 k β k k m γk n a γ 1 k > 0. Now, we show that f MH l (φ, ψ, m, n; γ). We only need to show that if (11) holds then the condition (9) is satisfied, then we want to proof that D m h(z) φ(z)+( 1) m+l D m g(z) ψ(z) 1 D n h(z)+( 1) n D n g(z) < 1, z U. (2) D n h(z)+( 1) n D n g(z) D m h(z) φ(z)+( 1) m+l D m g(z) ψ(z)

7 Some Properties of a Subclass of Harmonic Univalent 27 D m h(z) φ(z) + ( 1) m+l D m g(z) ψ(z) 1 D n h(z) + ( 1) n D n g(z) = (β kk m k n )a k z k + (µ kk m ( 1) m+l n k n )b k z k z + kn a k z k + ( 1) n, kn b k z k D m h(z) φ(z) + ( 1) m+l D m g(z) ψ(z) (2) D n h(z) + ( 1) n D n g(z) = z + β kk m a k z k + ( 1) m+l µ kk m b k z k z + kn a k z k + ( 1) n (2) kn b k z k = (2 2γ)z + (β kk m (2)k n )a k z k + (( 1)m+l µ k k m ( 1) n (2)k n )b k z k z + kn a k z k + ( 1) n, kn b k z k we have D m h(z) φ(z)+( 1) m+l D m g(z) ψ(z) 1 D n h(z)+( 1) n D n g(z) (2) D n h(z)+( 1) n D n g(z) [β kk m k n ] a k + ( 1) m+l µ k k m ( 1) n k n bk 2() [β kk m (2)k n ] a k ( 1)m+l µ k k m ( 1) n (2)k n b k. D m h(z) φ(z)+( 1) m+l D m g(z) ψ(z) The last expression is bounded above by 1, if [β k k m k n ] a k + 2() ( 1) m+l µ k k m ( 1) n k n b k [β k k m (2)k n ] a k which is equivalent to [β k k m γk n ] a k + ( 1) m+l µ k k m ( 1) n (2)k n b k, [µ k k m ( 1) m+l n γk n ] b k. (13) But (13) is true by hypothesis and then Theorem 2.1 is proved. Sharpness of the coefficient inequality (11) can be seen by the function f(z) = z + β k k m γk n x kz k + µ k k m ( 1) m+l n γk n y kz k, where l {0, 1}, m N, n N 0, m n, β k, µ k 1, k 1, 1 γ < 4/3 and x k + y k = 1. In the following theorem, it is shown that the condition (11) is also necessary for function f(z) given by (10) belong to the class VH l (m, n, φ, ψ; γ).

8 28 R. M. El-Ashwah and W. Y. Kota Theorem 2.2 let f(z) be given by (10), then f(z) VH l (m, n, φ, ψ; γ) if and only if (β k k m γk n ) a k + (µ k k m ( 1) m+l n γk n ) b k. (14) Proof. Since VH l (m, n, φ, ψ; γ) M H l (m, n, φ, ψ; γ), we only need to prove the only if part of the theorem. To prove the only if, let f(z) VH l (m, n, φ, ψ; γ) then we have { } R D m h(z) φ(z) + ( 1) m+l D m g(z) ψ(z) D n h(z) + ( 1) n D n g(z) γ. { (z + R km a k z k ) (z + β kz k ) + ( 1) 2m+2l ( km b k z k ) (z + } µ kz k ) z + kn a k z k + ( 1) n+m+l γ. kn b k z k This is equivalent to { ()z R [β kk m γk n ] a k z k ( 1) 2m+2l [µ kk m ( 1) m+l n γk n ] b k z k } z + kn a k z k + ( 1) n+m+l kn b k z k 0. The above condition must hold for all values of z U, so that on taking z = r < 1, the above inequality reduces to () [β kk m γk n ] a k r k 1 [µ kk m ( 1) m+l n γk n ] b k r k kn a k r k 1 + ( 1) n+m+l kn b k r k 1 0. (15) If the condition (11) does not hold then the numerator of (15) is negative for r and sufficintly close to 1. This contradicts the required condition for f(z) (m, n, φ, ψ; γ). This complete the proof of Theorem 2.2. V l H Remark 2.3 z (i) Putting φ = ψ = 1 z, l = 0, m = 1, n = 0 in Theorem 2.1, we obtain the result obtained in [14 Theorem 2.1.]; z (ii) Putting φ = ψ = 1 z, l = 0, m = 2, n = 1 in Theorem 2.1, we obtain the result obtained in [14 Theorem 2.4.]. 3 Distortion Theorem Now, we give the distortion theorem for functions in VH l (m, n, φ, ψ; γ), which yield a covering result for VH l (m, n, φ, ψ; γ).

9 Some Properties of a Subclass of Harmonic Univalent 29 Theorem 3.1 Let f = h+g with h and g are of the form (10) belongs to the class V l H (m, n, φ, ψ; γ) for functions φ and ψ with non-decreasing sequences {β k}, {µ k } satisfying β k, µ k β 2, k 2, and A k β k k m γk n, B k µ k k m ( 1) m+l n γk n for k 2, C = min{a 2, B 2 }. Then for z = r < 1, we have and f(z) (1 + b 1 )r + C f(z) (1 b 1 )r C [ 1 1 ] ( 1)m+l n γ b 1 r 2, (16) [ 1 1 ] ( 1)m+l n γ b 1 r 2. (17) The equalities in (16) and (17) are attained for the functions f(z) given by f(z) = (1 + b 1 )z + C [ 1 1 ] ( 1)m+l n γ b 1 z 2, and f(z) = (1 b 1 )z C [ 1 1 ] ( 1)m+l n γ b 1 z 2. where b 1 γ 1 1 ( 1) m+l n γ. Proof. Let f VH l (m, n, φ, ψ; γ) then on taking the absolute value of f, we get for z = r < 1 f(z) h(z) + g(z) r + a k r k + b k r k, b 1 < 1 = (1 + b 1 )r + ( a k + b k )r k (1 + b 1 )r + r 2 ( a k + b k ) (1 + b 1 )r + r2 () C (1 + b 1 )r + r2 () C () (1 + b 1 )r + C (1 + b 1 )r + [ C C () a k + C () b k β k k m γk n a k + µ kk m ( 1) m+l n γk n b k (1 µ 1 ( 1) m+l n γ 1 1 ( 1)m+l n γ b 1 b 1 )r 2 ] r 2, which proves the assertion (16) of Theorem 3.1. The proof of the assertion (17) is similar, thus, we omit it.

10 30 R. M. El-Ashwah and W. Y. Kota Corollary 3.2 Let the function f(z) given by (10) be in the class VH l (m, n, φ, ψ; γ), γ 1 where b 1 1 ( 1) m+l n γ and A k β k k m γk n, B k µ k k m ( 1) m+l n γk n for k 2, C = min{a 2, B 2 }. Then for z = r < 1, we have { ( w : w < 1 ) ( 1 ( 1) m+l n ) } γ + 1 b 1 f(u). C C 4 Extreme Points Theorem 4.1 Let f(z) be given by (10). Then f(z) clcovh l (m, n, φ, ψ; γ) if and only if f(z) = (x k h k (z) + y k g k (z)), (18) where h 1 (z) = z, h k (z) = z + β k k m γk n zk, (k 2), ( 1) m+l () g k (z) = z + µ k k m ( 1) m+l n γk n zk, (k 1), where x k 0, y k 0 and (x k + y k ) = 1. In particular, the extreme points of V l H (m, n, φ, ψ; γ) are {h k} and {g k }, respectively. Proof. Suppose that Then Since f(z) = z + f(z) = (x k h k (z) + y k g k (z)). β k k m γk n x kz k + ( 1) m+l ()y k z k µ k k m ( 1) m+l n γk n β k k m γk n β k k m γk n x µ k k m ( 1) m+l n γk n k + µ k k m ( 1) m+l n γk n y k = x k + y k = 1 x 1 1. Thus, f(z) clcovh l (m, n, φ, ψ; γ). Conversely, assume that f(z) clcovh l (m, n, φ, ψ; γ). Set x k = β kk m γk n a k, (0 x k 1; k 2),

11 Some Properties of a Subclass of Harmonic Univalent 31 and y k = µ kk m ( 1) m+l n γk n b k, (0 y k 1; k 1), where x 1 = 1 x k y k. Therefore, f(z) = z + = z + a k z k + ( 1) m+l b k z k β k k m γk n x kz k + = z + (h k (z) z)x k + (g k (z) z)y k ( 1) m+l ()y k z k µ k k m ( 1) m+l n γk n ( ) = z 1 x k y k + h k (z)x k + g k (z)y k = (h k (z)x k + g k (z)y k ). This completes the proof of Theorem A family of integral operators In this section, we examine a closure property of the class VH l (m, n, φ, ψ; γ) under the generalized Bernardi-Libera Livingston integral operator (see [4], [11] and [12]). Bernardi defined the integral operator J c by J c (f) = c + 1 z c z 0 t c 1 f(t)dt, (f V l H(m, n, φ, ψ; γ), c > 1). Now, we define the Bernardi integral operator J c (f) on the class S H of harmonic univalent functions of the form (1) as follows:[6] J c (f) = J c (h) + J c (g) c + 1 = z + c + n a nz n c c + n b nz n. n=2 Theorem 5.1 Let the function f(z) defined by (10) be in the class VH l (m, n, φ, ψ; γ) and c be a real number such that c > 1, then the function F (z) defined by F (z) = c + 1 z c z 0 t c 1 h(t)dt + ( 1) m+l c + 1 z c also belongs to the class VH l (m, n, φ, ψ; γ). n=1 z 0 t c g(t)dt, (c > 1), (19)

12 32 R. M. El-Ashwah and W. Y. Kota Proof. Let the function f(z) be defined by (10), then from the representation (19) of F (z), it follows that where Therefore, we have F (z) = z + ζ k = ζ k z k + ( 1) m+l η k z k, ( ) c + 1 a k and η k = c + k ( ) c + 1 b k. c + k β k k m γk n µ k k m ( 1) m+l n γk n ζ k + η k β k k m γk n ( ) c + 1 µ k k m ( 1) m+l n γk n = a k + c + k β k k m γk n µ k k m ( 1) m+l n γk n a k + b k 1, ( ) c + 1 b k c + k since f(z) VH l (m, n, φ, ψ; γ). Hence F (z) V H l (m, n, φ, ψ; γ). This completes the proof of Theorem convex combinations The convex combination properties of the class VH l (m, n, φ, ψ; γ) is given in the following theorem. Theorem 6.1 The class VH l (m, n, φ, ψ; γ) is closed under convex combination. Proof. For j = 1, 2,..., suppose that f j V l H (m, n, φ, ψ; γ) where f j(z) is given by f j (z) = z + Then, by Theorem 2.1, we have a j,k z k + ( 1) m+l b j,k z k. β k k m γk n µ k k m ( 1) m+l n γk n a j,k + b j,k 1. For j=1 t j = 1, 0 t j 1, the convex combination of f j (z) may be written as t j f j (z) = z + t j a j,k z k + ( 1) m+l j=1 j=1 j=1 t j b j,k z k.

13 Some Properties of a Subclass of Harmonic Univalent 33 Now β k k m γk n µ k k m ( 1) m+l n γk n t j a j,k + t j b j,k j=1 j=1 ( ) β k k m γk n µ k k m ( 1) m+l n γk n = t j a j,k + b j,k j=1 1, we have j=1 t jf j (z) VH l (m, n, φ, ψ; γ). Theorem 6.2 If f VH l (m, n, φ, ψ; γ) then, f is convex in the disc { z min k ()(1 b 1 ) k[ (1 ( 1) m+l n γ) b 1 ] } 1 k 1, k = 2, 3, 4,.... Proof. If f VH l (m, n, φ, ψ; γ) and let 0 < r < 1 be fixed. Then r 1 f(rz) VH l (m, n, φ, ψ; γ) and we have k 2 ( a k + b k )r k 1 = k( a k + b k )(kr k 1 ) ( βk k m γk n a k + µ kk m ( 1) m+l n γk n ) b k kr k 1 1 b 1 provided which is true if { r min k kr k 1 1 b ( 1)m+l n γ γ 1 b 1 (1 b 1 )() k[() (1 ( 1) m+l n γ) b 1 ] } 1 k 1 k = 2, 3, Convolution properties For harmonic functions f(z) = z + a k z k + ( 1) m+l b k z k, and F (z) = z + A k z k + ( 1) m+l B k z k,

14 34 R. M. El-Ashwah and W. Y. Kota we define the convolution of f and F as (f F )(z) = f(z) F (z) = z + a k A k + ( 1) m+l b k B k z k. In the following theorem we examine the convolution properties of the class VH l (m, n, φ, ψ; γ). Theorem 7.1 If f VH l (m, n, φ, ψ; γ) and F V H l (m, n, φ, ψ; γ) then f F VH l (m, n, φ, ψ; γ). Proof. let f(z) = z + F (z) = z + Then, by Theorem 2.1, we have a k z k + ( 1) m+l b k z k, A k z k + ( 1) m+l B k z k. β k k m γk n µ k k m ( 1) m+l n γk n a k + b k 1 β k k m γk n µ k k m ( 1) m+l n γk n A k + B k 1 so for f F, we may write β k k m γk n µ k k m ( 1) m+l n γk n a k A k + b k B k β k k m γk n µ k k m ( 1) m+l n γk n a k + b k 1. Thus f F VH l (m, n, φ, ψ; γ). Theorem 7.2 Let f j (z) = z + a k,j z k + ( 1) m+l b k,j z k be in the class VH l (m, n, φ, ψ; α j) for all (j = 1, 2,..., p), then (f 1 f 2... f p )(z) VH l (m, n, φ, ψ; δ) where (2 m β 2 2 n ) p j=1 δ = 1 + (α j 1) p j=1 (2m β 2 2 n α j ) + 2 n p j=1 (α (20) j 1). Proof. We use the principle of mathematical induction in our proof. Let f 1 (z) = z+ a k,1 z k +( 1) m+l b k,1 z k and f 2 (z) = z+ a k,2 z k +( 1) m+l b k,2 z k

15 Some Properties of a Subclass of Harmonic Univalent 35 be in the class VH l (m, n, φ, ψ; α 1) and VH l (m, n, φ, ψ; α 2), respectively. Then by Theorem 2.1, we have (β k k m α 1 k n ) a k,1 (µ k k m ( 1) m+l n α 1 k n ) b k,1 + 1 α 1 1 α 1 1 (β k k m α 2 k n ) a k,2 α (µ k k m ( 1) m+l n α 2 k n ) b k,2 α 2 1 We need to find the largest δ such that β k k m δk n µ k k m ( 1) m+l n δk n a k,1 a k,2 + b k,1 b k,2 1. (21) δ 1 δ 1 Then (β k k m α 1 k n ) a k,1 α (β k k m α 2 k n ) a k,2 α (µ k k m ( 1) m+l n α 1 k n ) b k,1 α thus, by applying Cauchy-Schwarz inequality, we have (β k k m α 1 k n )(β k k m α 2 k n ) a k,1 a k, (µ k k m ( 1) m+l n α 2 k n ) b k,2 α β k k m α 1 k n ) a k,1. (β k k m α 2 k n ) a k,2 α 1 1 α 2 1 (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) b k,1 b k,2 (µ k k m ( 1) m+l n α 1 k n ) b k,1 α /2 (22) (µ k k m ( 1) m+l n α 2 k n ) b k,2 α 2 1 Then, we get (β k k m α 1 k n )(β k k m α 2 k n ) a k,1 a k,2 (µ k k + m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) b k,1 b k,2 1, (23) (24) 1/2.

16 36 R. M. El-Ashwah and W. Y. Kota then, we get (β k k m δk n ) a k,1 a k,2 (β k k m α 1 k n )(β k k m α 2 k n ) a k,1 a k,2 δ 1 (µ k k m ( 1) m+l n δk n ) b k,1 b k,2 δ 1 (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) b k,1 b k,2, that is, if (β k k m δk n ) (β k k a k,1 a k,2 m α 1 k n )(β k k m α 2 k n ) a k,1 a k,2 δ 1 (µ k k m ( 1) m+l n δk n ) (µ k k b k,1 b k,2 m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) b k,1 b k,2, δ 1 (26) hence that, a k,1 a k,2 b k,1 b k,2 δ 1 (β k k m α 1 k n )(β k k m α 2 k n ) β k k m δk n δ 1 µ k k m ( 1) m+l n δk n (25) (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ). (α 1 1)(α 1 1) (27) We know that (β k k m α 1 k n )(β k k m α 2 k n ) a k,1 a k,2 1 (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) b k,1 b k,2 1, then a k,1 a k,2 b k,1 b k,2 (β k k m α 1 k n )(β k k m α 2 k n ), (α 1 1)(α 1 1) (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ). (28)

17 Some Properties of a Subclass of Harmonic Univalent 37 Consequently, from Eqs. (27) and (28) we obtain (β k k m α 1 k n )(β k k m α 2 k n ) (δ 1) (β k k m α 1 k n )(β k k m α 2 k n ) (β k k m δk n, ) (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) (δ 1) (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) (µ k k m ( 1) m+l n δk n. ) Then we see that δ (β kk m α 1 k n )(β k k m α 2 k n ) + β k k m (β k k m α 1 k n )(β k k m α 2 k n ) + k n δ (µ kk m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) + µ k k m (µ k k m ( 1) m+l n α 1 k n )(µ k k m ( 1) m+l n α 2 k n ) + k n, then (β k k m k n ) δ 1 + [β k k m α 1 k n ][β k k m α 2 k n ] + k n = ζ(k), (µ k k m k n ) δ 1 + [µ k k m ( 1) m+l n α 1 k n ][µ k k m ( 1) m+l n α 2 k n ] + ( 1) m+l n k n = η(k), since ζ(k) for k 2 and η(k) for k 1 are increasing, then (f 1 f 2 )(z) VH l (m, n, φ, ψ; δ) where (β 2 2 m 2 n ) δ = 1 + [β 2 2 m α 1 2 n ][β 2 2 m α 2 2 n ] + 2 n Next, we suppose that (f 1 f 2... f p )(z) VH l (m, n, φ, ψ; γ) then (β 2 2 m 2 n ) p j=1 γ = 1 + (α j 1) p j=1 [β 22 m α j 2 n ] + 2 n p j=1 (α j 1) we show that (f 1 f 2... f p+1 )(z) VH l (m, n, φ, ψ; δ) then since we have δ = 1 + (β 2 2 m 2 n )()(α p+1 1) [β 2 2 m γ2 n ][β 2 2 m α p+1 2 n ] + 2 n ()(α p+1 1) ()(α p+1 1) = (β 2 2 m γ2 n )(β m 2 α p+1 2 n ) = δ = 1 + (β 2 2 m 2 n ) p+1 j=1 (α j 1) p j=1 (β 22 m α j 2 n ) + 2 n p j=1 (α j 1). (β 2 2 m 2 n ) p+1 j=1 (β 22 m α j 2 n ) p j=1 (β 22 m α j 2 n ) + 2 n p j=1 (α j 1), (β 2 2 m 2 n ) p+1 j=1 (α j 1) p+1 j=1 (β 22 m α j 2 n ) + 2 n p+1 j=1 (α j 1). (29) (30)

18 38 R. M. El-Ashwah and W. Y. Kota Theorem 7.3 Let f j (z) = z + a k,j z k + ( 1) m+l b k,j z k be in the class VH l (m, n, φ, ψ; α) for all (j = 1, 2,..., p) then (f 1 f 2... f p )(z) VH l (m, n, φ, ψ; δ) where (β 2 2 m 2 n )(α 1) p δ = 1 + (β 2 2 m α2 n ) p + 2 n (α 1) p (31). 8 Inclusion Results To prove our next theorem, we need to the following lemma. Lemma 8.1 [15] Let f = h + g be given by (8), then f T SH l (m, n, φ, ψ; α) if and only if (β k k m αk n ) a k + (µ k k m ( 1) m+l n αk n ) b k 1 α, where l {0, 1}, m N, n N 0, m n, β k, µ k 1, k 1, 0 α < 1. Theorem 8.2 Let f VH l (m, n, φ, ψ; γ), then f T Sl H (m, n, φ, ψ; (4 3γ)/(3 2γ)). Proof. Since f VH l (m, n, φ, ψ; γ), then by Theorem 2.1, we have (β k k m γk n ) a k + (µ k k m ( 1) m+l n γk n ) b k. To show that f(z) T SH l (m, n, φ, ψ; (4 3γ)/(3 2γ)) by Lemma 8.1, we have to show that (β k k m 4 3γ 3 2γ kn ) a k + (µ k k m m+l n 4 3γ ( 1) 3 2γ kn ) b k 1 4 3γ 3 2γ, where 0 4 3γ 3 2γ < 1. For this, it is sufficient to prove that β k k m γk n 3 2γ kn β kk m 4 3γ 1 4 3γ 3 2γ, (k = 2, 3, 4,...), µ k k m ( 1) m+l n γk n µ kk m ( 1) m+l n 4 3γ 1 4 3γ 3 2γ 3 2γ kn, (k = 1, 2, 3,...), or equivalently 2()β k k m 4()k n 0, and 2()µ k k m 4( 1) m+l n (γ 1)k n 0, which is true and the theorem is proved.

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