CHAPTER 3 Application of Differential Subordination, Ruscheweyh Derivative and Linear Operator for Analytic and Meromorphic Functions

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1 CHAPTER 3 Application of Differential Subordination, Ruscheweyh Derivative and Linear Operator for Analytic and Meromorphic Functions Introduction This Chapter is mainly concerned with the study of application of differential subordination, linear operator and Ruscheweyh derivative to analytic and meromorphic functions. The study of differential subordination in the complex plane C is the generalization of a differential inequality on the real line. The concept of differential subordination plays a very important role in functions of real or complex variable. This concept is also used in finding the information about the range of original function. In the theory of complex-valued function, there are several differential applications in which a characterization of a function is determined from a differential condition. Miller and Mocanu [16] have contributed number of papers on differential subordination. The study of differential subordination stems out from text books by Duren [7], Goodman [9] and Pommerenke [22]. Firstly, we have studied the application of linear operator on meromorphically univalent functions involving differential subordination, by making use of the Briot-Bouquet differential subordination, we study several inclusion relationships and other interesting properties and characteristics of the 121

2 subclasses of meromorphically univalent functions in the punctured unit disk U = {z C :0< z < 1} = U \{0} which are defined here by a linear operator. We have also obtained here result about the convolution properties for functions belonging to the class Σ λ,k m (A, B). Such type of study was carried out by several authors e.g. Patel and Sahoo [20], Srivastava and Patel [27], Miller and Mocanu [15]. Secondly, we have studied the application of subordination and Ruscheweyh derivative for p-valent functions with negative coefficients having Taylor series of the form: f(z) =z p a k z k, (a k 0,p IN), which are analytic in the unit disk U and satisfying: z(d λ+p 1 f(z)) D λ+p 1 f(z) p +(pb +(A B)(p α))z, z U, 1+Bz where 1 B<A 1, 0 α<p,λ> p. We have investigated the properties of the class A p (A, B, α, λ) ofp-valent functions with negative coefficients defined by Ruscheweyh derivative and subordination. We obtain coefficient bounds, convolution properties for the class A p (A, B, α, λ) under the assumption 1 B<0. We also obtain the class preserving operator of the form: G c (z) = c + p z c z 0 s c 1 f(s)ds, c > p for the class A p (A, B, α, λ). Such type of study was carried out by various authors e.g. Aouf [4] has studied Ruscheweyh derivative and subordination for analytic univalent functions. Obradović and Owa [18] have also contributed in the same field stated above. Patel and Rout [19] have also tackled application of differential subordination. 122

3 Joshi and Srivastava [10] have studied p-valent meromorphic functions by making use of Ruscheweyh derivative and subordination. Motivated by these researcherswehaveobtainedniceresults. Thirdly, we have discussed the application of differential subordination for certain subclass of meromorphically functions with positive coefficients defined by linear operator having Taylor series of the form: f(z) =z p + a k z k (for any m p, p IN, a k 0), which are analytic and p-valent in the punctured unit disk U and satisfying the condition: zp+1 (L n f(z)) p 1+Az 1+Bz, (n IN 0). We obtain the coefficient bounds, δ-neighbourhood and integral representation.also we have obtained convex combination, arithmetic mean and convolution properties. 3.1 Application of Linear Operator on Meromorphically Univalent Functions Involving Differential Subordination For any integer m > 1, let Σ m denote the class of all meromorphic functions f(z) normalized by f(z) = 1 z + a n z n, (3.1) which are analytic and univalent in the punctured unit disk U. Let Σ s m and Σc m denote the subclasses of Σ m defined respectively as follows: n=m 123

4 A function f of the form (3.1) is in the class Σ s m if and only if { } Re zf (z) > 0 (z U = {z C : z < 1}). f(z) A function f of the form (3.1) is in the class Σ c m if and only if { ( )} Re 1+ zf (z) > 0, (z U). f (z) The class Σ s 0 and similar other classes have been extensively studied by Clunie [6], Pommerenke [21], Royster [23] and others. For functions f(z) Σ m given by (3.1), and g(z) Σ m defined by g(z) = 1 z + b n z n (m> 1). (3.2) n=m We define the Hadamard product (or convolution) of f(z) andg(z) by (f g)(z) = 1 z + a n b n z n =(g f)(z) (m> 1,z U). (3.3) n=m Following the recent work of Liu and Srivastava [13], for a function f(z) in the class Σ m given by (3.1), the linear operator denoted by L k is defined as follows: L 0 f(z) =f(z), L 1 f(z) = 1 z + (n +2)a n z n = (z2 f(z)), z n=m and (in general) L k f(z) =L(L k 1 f(z)) = 1 z + (n +2) k a n z n = (z2 L k 1 f(z)) n=m z (k IN). (3.4) It is easily verified from (3.4) that z(l k f(z)) = L k+1 f(z) 2L k f(z) (f Σ m ; k IN 0 = IN {0}). (3.5) 124

5 The case m = 0 of the linear operator L k was considered earlier by Uralegaddi and Somanatha [28]. Making use of the principle of differential subordination as well as the linear operator L k, we introduce a subclass of the function class Σ m as follows: For a function f Σ m,wesaythatf is in the class Σ λ,k m (A, B) if and only if satisfies z2 ((L k f(z)) + λz(l k f(z)) ) 1 2λ 1+Az 1+Bz (z U), (3.6) where 0 λ< 1 2, 1 B < A 1,k IN 0 = IN {0},m > 1. For convenience, we write Σ λ,k m (1 2α, 1) = Σλ,k m (α), where Σ λ,k m (α) denotes the class of functions in Σ m satisfying the following inequality: Re{ z 2 ((L k f(z)) + λz(l k f(z)) )} > (1 2λ)α (0 λ< 1 2, 0 α<1,k IN 0,z U). We write Σ 0,0 0 (α) = Σ(α) (0 α<1). It is known [17] functions in Σ(α) are meromorphically univalent in U.Ifλ = 0, then the class Σ λ,k m (α) reduces to the class Σ k m (α) contains the functions in Σ m satisfying Re{ z 2 (L k f(z)) } >α. If m = 0 and k = 0, then the class Σ λ,k m (A, B) reduces to the class Σ λ (A, B), studied by Patel and Sahoo [20]. Theorem :Let0<λ< 1 2 then z 2 (L k f(z)) Q(z) 1+Az 1+Bz and 1 B<A 1. If f Σλ,k m (A, B), 125 (z U), (3.7)

6 where the function Q(z) is given by Q(z) = { A B +(1 A B )(1 + Bz) 1 2F 1 ( 1, 1; 1 λ+mλ λ(1+m) ; Bz 1+Bz ), B 0 1+ (1 2λ)A λ(1+m)+1 z B =0, and Q(z) is the best dominant of (3.7) and 2 F 1 is the Gaussian hypergeometric function given by the Definition Furthermore, Σ λ,k m (A, B) Σλ,k m (σ(λ, A, B)), (3.8) where σ(λ, A, B) = { A The result is the best possible. ( ) +(1 A )(1 B B B) 1 2F 1 1, 1; 1 λ+mλ ; B, B 0 λ(1+m) B λ A, B =0,. λ(1+m)+1 Proof : Let the function φ(z) defined by φ(z) = z 2 (L k f(z)) (z U). (3.9) Then φ(z) is of the form (0.15) and is analytic in U. Differentiating (3.9), we get φ(z)+ λzφ (z) 1 2λ = ((L k f(z)) + λz(l k f(z)) ) z2 1+Az 1 2λ 1+Bz (z U). By using Lemma (0.2.2) for γ = 1 2λ, we deduce that λ z 2 (L k f(z)) Q(z) = 1 2λ 1 2λ z { λ(1+m)} z λ(1 + m) = ( ) 1+At dt 1+Bt t { λ(1+m)} 1 1 2λ 0 { ( A +(1 A)(1 + B B Bz) 1 2F 1 1, 1; 1 λ+mλ; Bz λ(1+m) 1+Bz 1+ (1 2λ)A λ(1+m)+1 ), (B 0) z, (B =0), 126

7 by change of variables followed by the use of the identities (0.18), (0.19) and (0.20) (with b =1andc = a + 1). This proves the assertion (3.7) of Theorem Next, to prove (3.8), it suffices to show that inf {Re(Q(z))} = Q( 1). (3.10) z <1 We have for z r<1, { } 1+Az Re 1+Bz 1 Ar 1 Br ( z r<1). Setting G(s, z) = 1+Asz 1+Bsz and dµ(s) = 1 2λ λ(1 + m) s 1 3λ λm λ(1+m) ds, 0 s 1, which is a positive measure on the closed interval [0, 1], we get Q(z) = 1 0 G(s, z)dµ(s), so that Re{Q(z)} Letting r ( ) 1 Asr dµ(s) =Q( r) 1 Bsr ( z r<1). in the above inequality, we obtain the assertion (3.8) of Theorem Finally, the result in (3.8) is best possible as the function Q(z) is the best dominant of (3.7). Corollary : For 0 λ 2 <λ 1 < 1, 1 B<A 1,m > 1 and 2 k IN 0,wehave Proof :Letf Σ λ 1,k m m (A, B) Σλ 2,k m (A, B). Σ λ 1,k (A, B). Then by Theorem 3.1.1, f Σ0,k(A, B). Since z2 {(L k f(z)) + λ 2 z(l k f(z)) } 1 2λ { 2 = 1 λ 2(1 2λ 1 ) λ 1 (1 2λ 2 ) { z2 {(L k f(z)) + λ 1 z(l k f(z)) } 1 2λ 1 } ( z 2 (L k f(z)) )+ λ 2(1 2λ 1 ) λ 1 (1 2λ 2 ) } Az 1+Bz, m

8 we see that f Σ λ 2,k m (A, B). Putting A =1 2α (0 α<1) and B = 1 in Theorem 3.1.1, we have Corollary :For0<λ< 1 2,m> 1 andk IN 0,wehave Σ λ,k m (α) Σλ,k m (σ(λ, α)), where { ( σ(λ, α) =α +(1 α) 2F 1 1, 1; 1 λ + mλ λ(1 + m) ; 1 ) } 1. 2 The result is best possible. The above corollaries are new results and not found in the literature. Putting A =1 2α (0 α<1),b = 1,m=0andk =0inTheorem 3.1.1, we have Corollary [20] :For0<λ< 1 2,wehave Σ λ (α) Σ λ (σ(λ, α)), where { σ(λ, α) =α +(1 α) 2F 1 (1, 1; 1 λ λ ; 1 } 2 ) 1. The result is the best possible. If m =0andk =0,weget Corollary [20] :For0 λ 2 <λ 1 < 1, 1 B<A 1, we have 2 Σ λ 1 (A, B) Σ λ 2 (A, B). Letting m =0andk = 0 in Theorem 3.1.1, we have Corollary [20] : Let 0 <λ< 1 2 and 1 B < A 1. If f Σ λ (A, B), then z 2 f (z) Q(z) 1+Az 1+Bz 128 (z U), ( )

9 where the function Q(z) given by Q(z) = { A +(1 A)(1 + B B Bz) 1 2F 1 (1, 1; 1 λ; Bz λ 1+Bz B 0 1+ (1 2λ)A 1+λ B =0, and Q(z) is the best dominant of ( ). Furthermore, Σ λ (A, B) Σ λ (σ(λ, A, B)), where σ(λ, A, B) = The result is best possible. Theorem :Letf Σ k m R = The result is best possible. { A +(1 A)(1 B B B) 1 2F 1 (1, 1; 1 λ; B ), B 0 λ B λ A B =0,. 1+λ (α). Then f Σλ,k m (α) in z <R,where ( λ2 (1 + m) 2 +(1 2λ) 2 λ(1 + m) 1 2λ Proof :Sincef Σ k m(α), we can write ) 1 1+m. z 2 (L k f(z)) = α +(1 α)u(z), (3.11) where u(z) is of the form (0.15), is analytic and has a positive real part in U. Differentiating (3.11) followed by a simple calculation, it follows that { } { Re z2 ((L k f(z)) + λz(l k f(z)) )+α = Re u(z)+ λ } (1 2λ)(1 α) 1 2λ zu (z). (3.12) Now, using the well-known estimate [14] zu (z) Re{u(z)} 2(1 + m)rm+1 1 r 2(m+1) ( z = r<1) (3.13) 129

10 in (3.13), deduce that { } { Re z2 ((L k f(z)) + λz(l k f(z)) )+α Re{u(z)} 1 (1 2λ)(1 α) 2λ(1 + m)r m+1 (1 2λ)(1 r 2(m+1) ) (3.14) It is easily seen that the right-hand side of (3.14) is positive, when z <R for R given as in Theorem Hence f Σ λ,k m (α) for z <R. To show that the bound R is best possible, we consider the function f Σ m defined by z 2 (L k f(z)) = α +(1 α) 1+zm+1 1 z m+1 (0 α<1, z U). Noting that }. z2 ((L k f(z)) + λz(l k f(z)) )+α (1 2λ)(1 α) = (1 2λ)(1 z2(m+1) )+2λ(m +1)z m+1 (1 2λ)(1 z) 2(m+1) =0 for z = R exp ( iπ 1+m), we complete the proof of Theorem Putting m =0andk = 0 in Theorem 3.1.2, we get Corollary [20] :Iff Σ(α), then f Σ λ (α) in z <R,where R = λ2 +(1 2λ) 2 λ. 1 2λ The result is best possible. Theorem : Let f Σ λ,k m (A, B), (0 λ< 1 2 ). Then the function F δ defined by satisfies F δ (f)(z) = δ z t δ f(t)dt (δ >0; z U ), (3.15) z δ+1 0 z2 ((L k F δ (f)(z)) + λz(l k F δ (f)(z)) ) 1 2λ 130 Θ(z) 1+Az 1+Bz (z U), (3.16)

11 where the function Θ(z) is given by Θ(z) = { A B +(1 A B )(1 + Bz) 1 2F 1 (1, 1; δ m+1 +1; Bz Bz+1 1+ Aδ δ+m+1 ) (B 0) z (B =0) and Θ(z) is the best dominant of (3.16). Furthermore, where σ(δ, A, B) = F δ Σ λ,k m (σ(δ, A, B)), (3.17) { A +(1 A)(1 B B B) 1 δ 2F 1 (1, 1; +1; B ) (B 0) m+1 B 1 1 Aδ (B =0). δ+m+1 The result is the best possible. Proof :Let φ(z) = z2 ((L k F δ (f)(z)) + λz(l k F δ (f)(z)) ) 1 2λ (z U). (3.18) We note that φ(z) is of the form (0.15) and is analytic in U with φ(0) = 1. Using the following operator identity: z((l k F δ (f)(z)) )=δl k f(z) (δ +1)L k F δ (f)(z) (3.19) in (3.18), and differentiating the resulting equation with respect to z, we find that φ(z)+ zφ (z) δ = z2 ((L k f(z)) + λz(l k f(z)) ) 1 2λ 1+Az 1+Bz, (z U), which with the aid of Lemma for γ = δ, yields φ(z) Θ(z) = δ m +1 z z δ m+1 0 ( ) 1+At m+1 1 dt. 1+Bt The assertions (3.16) and (3.17) can now be deduced on the same lines that t δ proved in Theorem This completes the proof of Theorem

12 Letting λ =0,A =1 2α (0 α<1) and B = 1 in Theorem 3.1.3, we get Corollary :Iff Σ k m (α), then the function F δ defined by (3.15) is in the class Σ k m(η(δ, α)), where { ) } δ η(δ, α) =α +(1 α) 2F 1 (1, 1; m +1 +1; The result is best possible. The above corollary is new result and not found in the literature. Letting λ =0,A =1 2α (0 α<1),b = 1,m =0andk =0in Theorem 3.1.3, we get Corollary [20] :Iff Σ(α), then the function F δ defined by (3.15) is in the class Σ(η(δ, α)), where { η(δ, α) =α +(1 α) The result is best possible. ( 2F 1 1, 1; δ +1; 1 ) } 1. 2 Theorem :LetF δ Σ λ,k m (α). Then the function f, given by (3.15) is in the class Σ λ,k m (α) for z < R, where δ2 +(m +1) R = 2 (m +1). δ The result is best possible. Proof :SinceF δ Σ λ,k m (α), we have z2 ((L k F δ (f)(z)) + λz(l k F δ (f)(z)) ) 1 2λ = α +(1 α)u(z), (3.20) where u(z) is analytic,u(0) = 0, and Re{u(z)} > 0 in U. Using (3.19) in (3.20) and differentiating the resulting equation, we obtain { } Re z2 ((L k f(z)) + λz(l k f(z)) )+α = Re (1 2λ)(1 α) 132 { u(z)+ zu (z) δ }. (3.21)

13 By using (3.13) in (3.21) and proceeding as in Theorem 3.1.2, we get the assertion of Theorem The bound R can be seen to be sharp by taking f Σ m defined by (3.15) and z2 ((L k F δ (f)(z)) + λz(l k F δ (f)(z)) ) 1 2λ = α +(1 α) 1+zm+1 1 z m+1 (z U). (3.22) Letting m =0andk = 0 in Theorem 3.1.4, we have Corollary [20] : If F δ (f)(z) Σ λ (α), then the function f, givenby (3.15) is in the class Σ λ (α) for z < R, where R = δ δ The result is best possible. Theorem :Letf Σ λ,k m (A, B) andletg(z) Σ m satisfy the following inequality: Re(zg(z)) > 1 2 (z U). Then the function (f g)(z) Σ λ,k m (A, B). Proof :Wecanwrite z2 ((L k (f g)(z)) + λz(l k (f g)(z)) ) 1 2λ = z2 ((L k f(z)) + λz(l k f(z)) ) zg(z) 1 2λ (3.23) since Re(zg(z)) > 1 2 in U, and {z2 ((L k f(z)) + λz(l k f(z)) )}/(1 2λ) is subordinate to the convex (univalent) function (1 + Az)/(1 + Bz), it follows from (3.23) and Lemma that z2 ((L k (f g)(z)) + λz(l k (f g)(z)) ) 1 2λ Az 1+Bz (z U).

14 This proves the theorem. 3.2 Application of Subordination and Ruscheweyh Derivative for p-valent Functions with Negative Coefficients Let T (p) denote the subclass of A(p) consisting of functions f of the form: f(z) =z p a k z k, (a k 0,p IN), (3.24) which are analytic and p-valent in the unit disk U. Definition :Letf(z) T (p) is said to be a member of the D(δ, β, λ, p) if f(z) satisfies the inequality { } z(d λ+p 1 f(z)) Re >δ z(d λ+p 1 f(z)) D λ+p 1 f(z) D λ+p 1 f(z) p + β, (3.25) where 0 β<p,δ 0,λ > p, p IN, z U and the operator D λ+p 1 denotes Ruscheweyh derivative operator [24], satisfying the condition D λ+p 1 f(z) = and if f(z) be defined by (3.24), then D λ+p 1 f(z) =z p z p f(z) (1 z) p+λ Γ(λ + k) Γ(λ + p)(k p)! a kz k. The family D(δ, β, λ, p) is of special interest to us for it contains many wellknown as well as classes of analytic functions. In particular, for δ =0,p=1 and 0 λ 1 it provides a transition from starlike functions to convex functions. More specifically, D(0, β,0, 1) is the family of functions starlike of order β and D(0,β,1, 1) is the family of functions convex of order β. For 134

15 D(δ, 0, 0, 1) we obtain the class of uniformly δ-starlike functions introduced by Kanas and Wiśniowska [11] which can begeneralized to D(δ, β, 0, 1), the class of uniformly δ-starlike functions of order β. Generally speaking, D(δ, β, λ, p), consists of functions G(z) =D λ+p 1 f(z) which are uniformly δ-starlike of order β in U. Definition :LetM(A, B, α, p) consist of all analytic functions h in U for which h(0) = p and h(z) p +[pb +(A B)(p α)]z, 1+Bz where 1 B<A 1, 0 <A 1, 0 α<p. Definition : For A, B fixed, 1 B<A 1and0 α<p,let A p (A, B, α, λ) denote the class of functions f D(δ, β, λ, p) of the form (3.24) for which z(dλ+p 1 f(z)) D λ+p 1 f(z) M(A, B, α, p) and z(d λ+p 1 f(z)) D λ+p 1 f(z) p +(pb +(A B)(p α))z, z U (3.26) 1+Bz where denotes subordination. From the Definition, it follows that f(z) A p (A, B, α, λ) ifandonly if there exists a function w(z) analyticinu and satisfying w(0) = 0 and w(z) < 1forz U, such that z(d λ+p 1 f(z)) D λ+p 1 f(z) = p +[pb +(A B)(p α)]w(z), z U. (3.27) 1+Bw(z) This condition (3.27) is equivalent to z(d λ+p 1 f(z)) p D λ+p 1 f(z) pb +(A B)(p α) B z(dλ+p 1 f(z)) < 1,z U. (3.28) D λ+p 1 f(z) 135

16 When λ =1 p, then the class A p (A, B, α, λ) reduces to the class A p (A, B, α, 1 p) studied in [3]. Lemma [1] :LetA(z) =p + k=p M(A, B, α, p). Then for k p, A k (A B)(p α). A k z k be analytic in U and belong to Lemma [1] :TheclassM(A, B, α, p) is a convex set. Theorem :Letf T(p). Then f is in D(δ, β, λ, p) if and only if Γ(λ + k) [k(1 + δ) (β + δp)] Γ(λ + p)(k p)! a k <p β. (3.29) Proof : Suppose that f D(δ, β, λ, p). Using the fact that for real γ, Re(w) >δ w p + β if and only if Re[w(1 + δe iγ ) pδe iγ ] >βand letting w = z(dλ+p 1 f(z)) D λ+p 1 f(z) in (3.25), we obtain { } z(d λ+p 1 f(z)) Re (1 + δe iγ ) pδe iγ >β, D λ+p 1 f(z) which is equivalent to (p β) Γ(λ+k) (k β) Γ(λ+p)(k p)! a kz k p δe iγ Γ(λ+k) (k p) Γ(λ+p)(k p)! a kz k p Re 1 > 0. Γ(λ+k) Γ(λ+p)(k p)! a kz k p The above inequality must hold for all z in U. Letting z 1 yields (p β) Γ(λ+k) (k β) a Γ(λ+p)(k p)! k δe iγ Γ(λ+k) (k p) Re 1 Γ(λ+k) Γ(λ+p)(k p)! a k and so by the mean value theorem, we have Γ(λ + k) Re (p β) (k β) Γ(λ + p)(k p)! a k δe iγ Γ(λ+p)(k p)! a k > 0 Γ(λ + k) (k p) Γ(λ + p)(k p)! a k > 0. Therefore, we obtain Γ(λ + k) [k(1 + δ) (β + δp)] Γ(λ + p)(k p)! a k <p β. 136

17 Conversely, let (3.29) hold. We will show that (3.25) is satisfied and so f D(δ, β, λ, p). Using the fact that Re(w) >δif and only if w (p + δ) < w +(p δ), it is sufficient to show that ( ) z(d λ+p 1 f(z)) p + δ z(d λ+p 1 f(z)) D λ+p 1 f(z) p D λ+p 1 f(z) + β ( ) < z(d λ+p 1 f(z)) + p δ z(d λ+p 1 f(z)) D λ+p 1 f(z) p D λ+p 1 f(z) β. For letting e iθ = Dλ+p 1 f(z) D λ+p 1 f(z) E = ( ) z(d λ+p 1 f(z)) + p δ z(d λ+p 1 f(z)) D λ+p 1 f(z) p D λ+p 1 f(z) β = 1 D λ+p 1 f(z) z(dλ+p 1 f(z)) +(p β)d λ+p 1 f(z) δe iθ z(d λ+p 1 f(z)) pd λ+p 1 f(z) > z p D λ+p 1 f(z) [ (2p β) ] Γ(λ + k) (k + p β + δ(k p)) Γ(λ + p)(k p)! a k and F = = ( ) z(d λ+p 1 f(z)) p + δ z(d λ+p 1 f(z)) D λ+p 1 f(z) p D λ+p 1 f(z) + β 1 D λ+p 1 f(z) z(dλ+p 1 f(z)) (p + β)d λ+p 1 f(z) δe iθ z(d λ+p 1 f(z)) pd λ+p 1 f(z) < z p D λ+p 1 f(z) [ β + ] Γ(λ + k) (k p β + δ(k p)) Γ(λ + p)(k p)! a k. It is easy to verify that E F >0 if (3.29) holds and so the proof is complete. Remark :Letf D(δ, β, λ, p). Then a k < (p β)γ(k + p)(k p)!, k = p +1,p+2,. [k(1 + δ) (β + δp)]γ(λ + k) 137

18 Remark : Theorem for the special cases D(0, β,0, 1) and D(0, β,1, 1) leads to results obtained by Silverman [26]. Remark :Since D(δ, β, λ 1,p) D(δ, β, λ 2,p). Γ(λ 2 +k) Γ(λ 2 +p)(k p)! < Γ(λ 1+k) Γ(λ 1 +p)(k p)! for λ 2 <λ 1,wenotethat Now, we state a necessary and sufficient coefficient inequality. Theorem :Letf(z) =z p a k z k,a k 0, be analytic in U. Then f A p (A, B, α, λ) if and only if Γ(λ + k) [(k p)(1 B)+(A B)(p α)] Γ(λ + p)(k p)! a k (A B)(p α). The result is sharp. Proof :Let z =1. Then z(d λ+p 1 f(z)) pd λ+p 1 f(z) (3.30) [pb +(A B)(p α)]d λ+p 1 f(z) Bz(D λ+p 1 f(z)) = Γ(λ + k) (k p) Γ(λ + p)(k p)! a kz k (A B)(p Γ(λ + k) α)zp [(A B)(p α) B(k p)] Γ(λ + p)(k p)! a kz k Γ(λ + k) [(k p)(1 B)+(A B)(p α)] Γ(λ + p)(k p)! a k (A B)(p α) 0 by (3.30). Hence by the principle of maximum modulus f(z) A p (A, B, α, λ). To prove the converse, let us assume that f(z) =z p A p (A, B, α, λ). Then z(d λ+p 1 f(z)) D λ+p 1 f(z) p pb +(A B)(p α) B z(dλ+p 1 f(z)) D λ+p 1 f(z) 138 a k z k,a k 0isin

19 = (A B)(p α)zp Γ(λ+k) (k p) a Γ(λ+p)(k p)! kz k Γ(λ+k) [(A B)(p α) (k p)b] a Γ(λ+p)(k p)! kz k < 1 for all z U. Using the fact that Re(z) z for all z, it follows that Γ(λ+k) (k p) Γ(λ+p)(k p)! a kz k p Re (A B)(p α) < 1,z U. Γ(λ+k) [(A B)(p α) (k p)b] Γ(λ+p)(k p)! a kz k p (3.31) Now choose the values of z on the real axis so that z(d λ+p 1 f(z)) /D λ+p 1 f(z) is real. Upon clearing the denominator in (3.31) and letting z 1 through positive values, we obtain Γ(λ + k) (k p) Γ(λ + p)(k p)! a k (A B)(p α) which implies that Γ(λ + k) [(A B)(p α) (k p)b] Γ(λ + p)(k p)! a k, Γ(λ + k) [(k p)(1 B)+(A B)(p α)] Γ(λ + p)(k p)! a k (A B)(p α). The function f(z) =z p (A B)(p α)γ(λ + p)(k p)! ((k p)(1 B)+(A B)(p α))γ(λ + k) zk, shows that the inequality (3.30) is sharp. Remark : Let f(z) =z p Then a k z k,a k 0, is in A p (A, B, α, λ). a k (A B)(p α)γ(λ + p)(k p)! [(k p)(1 B)+(A B)(p α)]γ(λ + k), k p +1,p IN 139

20 with equality for each k, for functions of the form f k (z) =z p (A B)(p α)γ(λ + p)(k p)! [(k p)(1 B)+(A B)(p α)]γ(λ + k) zk,k p +1,p IN (3.32) 1 B Theorem : D(δ, β, λ, p) =A p (A, B, α, λ) if 1+δ. (A B)(p α) p β Proof : By Definition we have A p (A, B, α, λ) D(δ, β, λ, p). Assume that f be an arbitrary element of D(δ, β, λ, p). Then we have [k(1 + δ) (β + δp)] p β Γ(λ+k) Γ(λ+p)(k p)! a k < 1 by Theorem for showing f A p (A, B, α, λ) it is sufficient to show that (3.30) hold true, but (3.30) holds true, when 1+ which is equivalent to (k p)(1 B) (A B)(p α) 1 B 1+δ (A B)(p α) p β k(1 + δ) (δ + pβ), p β and this completes the proof. Now, if we set δ = λ =0,p=1,thenf D(0,β,0, 1) is starlike of order β and if we put δ = λ 1=0,p=1,thenf D(0,β,1, 1) is convex of order β, therefore we have the following corollary. Corollary :Let 1 B (A B)(p α) 1, p =1 (3.33) p β (1) Let λ =0. Thenf is starlike of order β if and only if zf (z) f(z) 1+[B +(A B)(1 α)]z. 1+Bz (2) Let λ =1. Thenf is convex of order β if and only if 1+ zf (z) f (z) 1+[B +(A B)(1 α)]z. 1+Bz 140

21 Theorem :Let0 α 2 <α 1 <p.thena p (A, B, α 1,λ) A p (A, B, α 2,λ). Proof :Letf A p (A, B, α 1,λ). Then [(k p)(1 B)+(A B)(p α 1 )]Γ(λ + k) a k 1, (A B)(p α 1 )Γ(λ + p)(k p)! we must prove [(k p)(1 B)+(A B)(p α 2 )]Γ(λ + k) a k 1. (3.34) (A B)(p α 2 )Γ(λ + p)(k p)! But (3.34) holds true if (k p)(1 B)+(A B)(p α 2 ) (A B)(p α 2 ) (k p)(1 B)+(A B)(p α 1). (A B)(p α 1 ) Then by hypothesis the preceding inequality definitely holds true. Theorem :Letf A p (A, B, α, λ). Then the function (λ + p) Dλ+p f(z) λ is in M(A, B, α, p). D λ+p 1 f(z) Proof : By the easily verified identity z(d λ+p 1 f(z)) =(λ + p)d λ+p f(z) λd λ+p 1 f(z), we have z(d λ+p 1 f(z)) D λ+p 1 f(z) and since f A p (A, B, α, λ), then =(λ + p) Dλ+p f(z) D λ+p 1 f(z) λ (λ + p) Dλ+p f(z) λ M(A, B, α, p). D λ+p 1 f(z) Remark :Iff A p (A, B, α, λ), then by Lemma for every t (0 <t<1), we have (1 t) z(dλ+p 1 f(z)) D λ+p 1 f(z) [ ] + t (λ + p) Dλ+p f(z) D λ+p 1 f(z) λ M(A, B, α, p). 141

22 Theorem :Letf A p (A, B, α, λ). Then D λ+p 1 f(z)isina p (A, B, α, λ 1 ) if p <λ 1 0. Proof :Sincef A p (A, B, α, λ), then [(1 B)(k p)+(a B)(p α)]γ(λ + k) a k 1 (3.35) (A B)(p α)γ(λ + p)(k p)! also we have L(z) =D λ+p 1 f(z) =z p Γ(λ + k) Γ(λ + p)(k p)! a kz k. We must show that z(dλ 1 +p 1 L(z)) M(A, B, α, p), equivalently, we must D λ 1 +p 1 L(z) show that z(d λ1+p 1 L(z)) pd λ1+p 1 L(z) y(d λ 1+p 1 L(z)) Bz(D λ 1+p 1 L(z)) < 1, (3.36) where y = pb +(A B)(p α), therefore, we have z(d λ1+p 1 L(z)) pd λ1+p 1 L(z) y(d λ 1+p 1 L(z)) Bz(D λ 1+p 1 L(z)) (k p) Γ(λ 1+k)Γ(λ+k)a k z k 1 Γ(λ 1 +p)γ(λ+p)[(k p)!] 2 = (y pb) (y kb) Γ(λ 1+k)Γ(λ+k)a k z k 1 Γ(λ 1 +p)γ(λ+p)[(k p)!] 2 (A B)(p α) so with respect to (3.35) if (k p) Γ(λ 1 +k)γ(λ+k) Γ(λ 1 a +p)γ(λ+p)[(k p)!] 2 k (B(p k)+(a B)(p α)) Γ(λ 1 +k) Γ(λ 1 +p)(k p)!, Γ(λ 1 +k)γ(λ+k)a k Γ(λ 1 +p)γ(λ+p)[(k p)!] 2 p, then (3.36) holds true, but λ 1 + p Γ(λ 1+k) Γ(λ 1 <prequires that λ +p)(k p)! 1 0, therefore (3.36) holds true if p <λ 1 0 and the proof is complete. Theorem :Letf A p (A, B, α, λ). Then the function z(d λ+p 1 f(z)) is in A p (A, B, α, λ 1 ), when λ 1 inf k [ ] 1 Γ(λ + k) 2 Γ(λ + p)(k p)! p. 142

23 Proof :Wehave also we have we want to prove [ ] (1 B)(k p)+(a B)(p α) Γ(λ + k) (A B)(p α) Γ(λ + p)(k p)! a k 1 z(d λ+p 1 f(z)) = z p [ (1 B)(k p)+(a B)(p α) (A B)(p α) Therefore (3.37) holds true if Γ(λ + k)a k k Γ(λ + p)(k p)! zk, ] ka k Γ(λ 1 + k) Γ(λ 1 + p)(k p)! Γ(λ 1 + k) k Γ(λ 1 + p)(k p)! Γ(λ + k) Γ(λ + p)(k p)! Γ(λ+k) 1. (3.37) or equivalently λ 1 1 p, this completes the proof. 2 Γ(λ+p)(k p)! Theorem :Letf(z) isina p (A, B, α, λ). Then the integral operator is in A p (A, B, α, λ). Proof :Letf(z) =z p G c (z) = c + p z c G c (z) = c + p z c = c + p z c = z p z 0 s c 1 f(s)ds, c > p (3.38) a k z k and G c (z) be given by (3.38). Then z 0 z 0 s [s c 1 p ( s p+c 1 ( ) c + p a k z k. c + k a k s k ] ds a k s k+c 1 ) ds 143

24 Since f(z) A p (A, B, α, λ), we have Γ(λ + k) [(1 B)(k p)+(a B)(p α)] Γ(λ + p)(k p)! a k (A B)(p α). Thus ( ) Γ(λ + k) c + p [(1 B)(k p)+(a B)(p α)] a k Γ(λ + p)(k p)! c + k Γ(λ + k) [(1 B)(k p)+(a B)(p α)] Γ(λ + p)(k p)! a k (A B)(p α), which implies that G c (z) A p (A, B, α, λ). This proves the theorem. Theorem : Let c be a real number such that c> p. If G c (z) A p (A, B, α, λ), then the function f(z) defined by (3.38) is p-valently starlike of order δ(0 δ<p)in z <R,where R =inf k (3.32). { } 1 (p δ)(c + p)[(1 B)(k p)+(a B)(p α)]γ(λ + k) k p. (k δ)(c + k)(a B)(p α)γ(λ + p)(k p)! The bound for z is sharp for each k with the function being of the form Proof : Let G c (z) =z p Theorem 3.2.2, we have d k z k,d k 0, be in A p (A, B, α, λ). Then by Γ(λ + k) [(1 B)(k p)+(a B)(p α)]d k Γ(λ + p)(k p)! (A B)(p α). (3.39) Also, it follows from (3.38) that f(z) = (zc G c (z)) z 1 c p + c = z p ( ) c + k d k z k. c + p 144

25 In order to obtain the required result it is sufficient to show that zf (z) f(z) p p δ in z <R. (3.40) Now zf (z) f(z) p = (k p)(c+k) d (c+p) k z k p (k p)(c+k) d (c+p) k z k p 1 ( ) c+k d c+p k z k p 1 ( ). c+k d c+p k z k p The last expression will be bounded above by (p δ) if (k δ)(c + k) (p δ)(c + p) d k z k p 1. (3.41) In view of (3.39), it follows that (3.40) is true if (k δ)(c + k) (p δ)(c + p) z k p or if z [(1 B)(k p)+(a B)(p α)]γ(λ + k),k p +1,p IN (A B)(p α)γ(λ + p)(k p)! { } 1 (p δ)(c + p)[(1 B)(k p)+(a B)(p α)]γ(λ + k) k p (k δ)(c + k)(a B)(p α)γ(λ + p)(k p)! (3.42) for k p +1,p IN. Letting z = R in (3.41), the result follows. This completes the proof of theorem. For two power series f(z) =z p a k z k and g(z) =z p we define the convolution as the power series (f g)(z) =z p b k z k, a k b k z k. Here, we prove some convolution results for functions belonging to the class A p (A, B, α, λ). Theorem :Letf(z) =z p a k z k and g(z) =z p b k z k are in A p (A, B, α, λ) and(k p)(1 B) (A B)(p α) 0, (k p +1). Then F (z) =z p (a 2 k + b2 k )zk is also in A p (A, B, α, λ). 145

26 Proof :Sincef(z) A p (A, B, α, λ), we have so that Similarly Hence Γ(λ + k) [(k p)(1 B)+(A B)(p α)] Γ(λ + p)(k p)! a k (A B)(p α), [ ] 2 [(k p)(1 B)+(A B)(p α)] 2 a 2 Γ(λ + k) k [(A B)(p α)] 2. Γ(λ + p)(k p)! [ ] 2 [(k p)(1 B)+(A B)(p α)] 2 b 2 Γ(λ + k) k [(A B)(p α)] 2. Γ(λ + p)(k p)! 1 2 [ ((k p)(1 B)+(A B)(p α))γ(λ + k) (A B)(p α)γ(λ + p)(k p)! In view of Theorem 3.2.2, it is sufficient to show that [ ((k p)(1 B)+(A B)(p α))γ(λ + k) (A B)(p α)γ(λ + p)(k p)! Thus (3.43) will be satisfied if, for k p +1,p IN, ((k p)(1 B)+(A B)(p α))γ(λ + k) (A B)(p α)γ(λ + p)(k p)! [ ((k p)(1 B)+(A B)(p α))γ(λ + k) ] 2 (a 2 k + b 2 k) 1. ] (a 2 k + b2 k ) 1. (3.43) ] (A B)(p α)γ(λ + p)(k p)! or if (k p)(1 B) (A B)(p α) 0, (k p +1). (3.44) The left hand side of (3.44) is an increasing function of k, hence (3.44) is satisfied for all k p +1if(1 B) (A B)(p α) 0, which is certainly true by our assumption. Hence the result. 146

27 Theorem :Letf(z) =z p a k z k and g(z) =z p in A p (A, B, α, λ). Then (f g)(z) A p (A, B, α, λ), where { [ ] 2 (1 B)+(A B)(p α) Γ(λ + k) λ 1 inf p} k (A B)(p α) Γ(λ + p)(k p)! Proof : By the hypothesis, we can write [ ] ((k p)(1 B)+(A B)(p α))γ(λ + k) a k 1 (A B)(p α)γ(λ + p)(k p)! and [ ((k p)(1 B)+(A B)(p α))γ(λ + k) (A B)(p α)γ(λ + p)(k p)! ] b k 1 and by applying the Cauchy-Schwarz inequality, we have [ ] ((k p)(1 B)+(A B)(p α))γ(λ + k) ak b k (A B)(p α)γ(λ + p)(k p)! ( [ ] ) 1/2 ((k p)(1 B)+(A B)(p α))γ(λ + k) a k (A B)(p α)γ(λ + p)(k p)! ( [ ] ) 1/2 ((k p)(1 B)+(A B)(p α))γ(λ + k) b k. (A B)(p α)γ(λ + p)(k p)! However, we obtain b k z k be. [ ] ((k p)(1 B)+(A B)(p α))γ(λ + k) ak b k 1. (3.45) (A B)(p α)γ(λ + p)(k p)! Now we want to prove [ ] ((k p)(1 B)+(A B)(p α))γ(λ1 + k) a k b k 1. (3.46) (A B)(p α)γ(λ 1 + p)(k p)! Let (3.46) holds true. Then we have [ (k p)(1 B)+(A B)(p α) (A B)(p α) ] ak Γ(λ + k) b k ak b k Γ(λ + p)(k p)! 147 Γ(λ 1+k) Γ(λ 1+p)(k p)! Γ(λ+k) Γ(λ+p)(k p)! 1. (3.47)

28 Therefore (3.47) (consequently (3.46)) holds true if ak b k But from (3.45) we conclude that ak b k Γ(λ+k) Γ(λ+p)(k p)! Γ(λ 1 +k). (3.48) Γ(λ 1 +p)(k p)! (A B)(p α)γ(λ + p)(k p)! [(1 B)+(A B)(p α)]γ(λ + k). (3.49) In view of (3.49) the inequality (3.48) holds true if (A B)(p α)γ(λ + p)(k p)! [(1 B)+(A B)(p α)]γ(λ + k) Γ(λ+k) Γ(λ+p)(k p)! Γ(λ 1 +k) Γ(λ 1 +p)(k p)! or equivalently λ 1 + p [ [(1 B)+(A B)(p α)] (A B)(p α) Γ(λ+k) Γ(λ+p)(k p)! ] 2 and this inequality gives the required result. 3.3 Application of Differential Subordination for Certain Subclass of Meromorphically p-valent Functions with Positive Coefficients Defined by Linear Operator Let L(p, m) be a class of all meromorphic functions f(z) oftheform: f(z) =z p + a k z k, (for any m p, p IN,a k 0, k m), (3.50) which are analytic and p-valent in the punctured unit disk U. Following the recent work of Liu and Srivastava [13], we now define a linear operator L n as follows: 148

29 Definition :Letf(z) be a function in the class L(p, m) given by (3.50), we define a linear operator L n by L 0 f(z) =f(z),l 1 f(z) =z p + andingeneral (p + k +1)a k z k = (zp+1 f(z)) z p L n f(z) =L(L n 1 f(z)) = z p + (p+k +1) n a k z k = (zp+1 L n 1 f(z)) z p, (n IN). (3.51) It is easily verified from (3.51) that z(l n f(z)) = L n+1 f(z) (p +1)L n f(z), (f L(p, m), n IN 0 = IN {0}). Definition :LetA and B ( 1 B<A 1,B 0) be fixed parameters, we say that a function f(z) L(p, m) isintheclassl(p, m, n, A, B), if it satisfies the following subordination condition: zp+1 (L n f(z)) p 1+Az 1+Bz (n IN 0; z U). (3.52) By the definition of differential subordination, (3.52) is equivalent to the following condition: z p+1 (L n f(z)) + p Bz p+1 (L n f(z)) + pa < 1, (z U). We can write L(p, m, n, 1 2β, 1) = L(p, m, n, β), where L(p, m, n, β) denotes the class of functions in L(p, m) satisfying the p following: Re{ z p+1 (L n f(z)) } >β (0 β<p; z U). The following theorem gives a sufficient and necessary condition for a function f(z) tobeintheclassl(p, m, n, A, B). 149

30 Theorem : Let the function f(z) of the form (3.50), be in L(p, m). Then the function f(z) isintheclassl(p, m, n, A, B) if and only if k(1 B)(p + k +1) n a k < (A B)p, (3.53) where 1 B<A 1,B 0,p IN,n IN 0,m p. The result is sharp for the function f(z) givenby f(z) =z p + (A B)p k(1 B)(p + k +1) n zk, k m. Proof : Assume that the condition (3.53) is true. We must show that f L(p, m, n, A, B), or equivalently we prove that z p+1 (L n f(z)) + p Bz p+1 (L n f(z)) + Ap < 1, (3.54) we have z p+1 (L n f(z)) + p Bz p+1 (L n f(z)) + Ap = The last inequality is true by (3.53). = z p+1 ( pz (p+1) + k(p + k +1) n a k z k 1 )+p Bz p+1 ( pz (p+1) + k(p + k +1) n a k z k 1 )+Ap k(p + k +1) n a k z k+p (A B)p + B k(p + k +1) n a k z k+p k(p + k +1) n a k (A B)p + B < 1. k(k + p +1) n a k Conversely, suppose that f(z) L(p, m, n, A, B). We must show that the condition (3.53) holds true. We have z p+1 (L n f(z)) + p Bz p+1 (L n f(z)) + Ap < 1, 150

31 hence we have k(p + k +1) n a k z k+p (A B)p + B < 1. k(p + k +1) n a k z k+p Since Re(z) z, sowehave k(p + k +1) n a k z k+p Re (A B)p + B < 1. k(p + k +1) n a k z k+p We choose the values of z on the real axis and letting z 1.Thenwehave k(p + k +1) n a k (A B)p + B < 1, k(p + k +1) n a k then k(1 B)(p + k +1) n a k < (A B)p and the proof is complete. Remark :Letf(z) L(p, m, n, A, B). Then we have a k < (A B)p k(1 B)(p + k +1) n, k m. We also note that if 0 <n 1 n 2,thenL(p, m, n 2,A,B) L(p, m, n 1,A,B). Definition :Let 1 B<A 1,B 0,m p, n IN 0,p IN and δ 0, we define the δ - neighbourhood of a function f L(p, m) by { N δ (f) = g L(p, m) :g(z) =z p + b k z k, } k(1 B)(p + k +1) n and a k b k δ. (A B)p (3.55) Goodman [8], Ruscheweyh [25] and Altintas and Owa [2] have investigated neighbourhood for analytic univalent functions. We consider this concept for the class L(p, m, n, A, B). 151

32 Theorem : Let the function f(z) defined by (3.50) be in L(p, m, n, A, B). For every complex number µ with µ <δ,δ 0, let f(z)+µz p L(p, m, n, A, B). 1+µ Then N δ (f) L(p, m, n, A, B),δ 0. Proof : Since f L(p, m, n, A, B), f satisfies (3.53) we can write for γ C, γ =1,that [ ] z p+1 (L n f(z)) + p γ, (3.56) Bz p+1 (L n f(z)) + pa equivalently, we must have (f Q)(z) z p 0, z U, (3.57) where Q(z) =z p + e k z k, such that e k = γk(1 B)(p+k+1)n (A B)p e k k(1 B)(p+k+1)n and k m, p IN,n IN (A B)p 0. then Since f(z)+µz p Now assume 1 1+µ L(p, m, n, A, B), therefore by (3.57) 1+µ ( ) 1 f(z)+µz p Q(z) 0, z p 1+µ (f Q)(z) z p f Q z p + µ satisfying ( ) 1 (f Q)(z)+µz p 0. (3.58) z p 1+µ <δ. Then by (3.58), we have 1+µ µ 1+µ 1 1+µ that is a contradiction by µ <δ, therefore Let g(z) =z p + b k z k N δ (f). Then (f Q)(z) z p (f Q)(z) z p δ. > µ δ 1+µ 0, δ (g Q)(z) z p ((f g) Q)(z) z p (a k b k )e k z k [ ] k(1 B)(p + k +1) a k b k e k z k < z m n a k b k δ, (A B)p 152

33 therefore (g Q)(z) 0andwegetg(z) L(p, m, n, A, B), z p so N δ (f) L(p, m, n, A, B). Theorem :Letf(z) be defined by (3.50) and the partial sums S 1 (z) and S q (z) be defined by S 1 (z) =z p and Also suppose that S q (z) =z p + m+q 2 C k a k 1, where a k z k, q 2, m p, p IN. C k = k(1 B)(p + k +1)n. (A B)p Then (i) f L(p, m, n, A, B) { } (ii) Re f(z) S q(z) > 1 1 C m+q 1, (3.59) { } Re Sq(z) > C m+q 1 f(z) 1+C m+q 1, z U, q 2. (3.60) Proof : (i) since z p +µz p = z p L(p, m, n, A, B), µ < 1, then by Theorem 3.3.2, we have N 1 (z p ) L(p, m, n, A, B),p IN(N 1 (z p )denoting 1+µ the 1-neighbourhood). Now since C k a k 1, then f N 1 (z p )and f L(p, m, n, A, B). (ii) Since {C k } is an increasing sequence, we obtain Setting m+q 2 a k + C m+q 1 k=q+m 1 a k C k a k 1. (3.61) ( ( f(z) G 1 (z) =C m+q 1 S q (z) 1 1 C m+q 1 )) = C m+q 1 k=q+m 1 1+ m+q 2 a k z k+p a k z k+p

34 From (3.61)we get G 1 (z) 1 G 1 (z)+1 = 2 2 m+q m+q 2 C m+q 1 a k k=q+m 1 a k C m+q 1 C m+q 1 k=q+m 1 a k z k+p + C m+q 1 k=q+m 1 a k 1. a k z k+p k=q+m 1 a k z k+p { } This proves (3.59). Therefore, Re(G 1 (z)) > 0andweobtainRe f(z) S q(z) > 1 1 C m+q 1. Now, by the same way, we can prove the assertion (3.60), with setting ( Sq (z) G 2 (z) =(1+C m+q 1 ) f(z) This completes the proof. C m+q 1 1+C m+q 1 In the next theorem, we obtain the integral representation for L n f(z). Theorem :Letf L(p, m, n, A, B). Then z L n p(aψ(t) 1) f(z) = t p+1 (1 Bψ(t)) dt, where ψ(z) < 1,z U. 0 Proof :Letf(z) L(p, m, n, A, B). Letting zp+1 (L n f(z)) = y(z). Then we p have y(z) 1+Az or we can write y(z) 1 1+Bz By(z) A < 1, consequently we have y(z) 1 By(z) A = ψ(z), ψ(z) < 1, z U. ). So we can write Therefore we have z p+1 (L n f(z)) p (L n f(z)) = = 1 Aψ(z) 1 Bψ(z). p(aψ(z) 1) z p+1 (1 Bψ(z)), 154

35 hence L n f(z) = and this gives the required result. z 0 p(aψ(t) 1) t p+1 (1 Bψ(t)) dt, In this theorem, we find the convex combination for the class L(p, m, n, A, B). Theorem :Let f i (z) =z p + a k,i z k, (a k,i 0,i=1, 2,,l,k m, m p) be in L(p, m, n, A, B). Then F (z) = l c i f i (z) L(p, m, n, A, B), where l c i =1. i=1 Proof : By Theorem 3.3.1, we can write for every i {1, 2,,l} k(1 B)(p + k +1) n a k,i < 1, (A B)p therefore F (z) = But i=1 ) l l c i (z p + a k,i z k = z p + ( c i a k,i )z k. i=1 i=1 k(1 B)(p + k +1) n l ( c i a k,i ) (A B)p i=1 [ l ] k(1 B)(p + k +1) n = a k,i c i < 1, (A B)p i=1 then F (z) L(p, m, n, A, B), so the proof is complete. In the following theorem, we shall prove that the class L(p, m, n, A, B) is closed under arithmetic mean. Theorem :Letf 1 (z),f 2 (z),,f l (z) defined by f i (z) =z p + a k,i z k, (a k,i 0,i=1, 2,,l,k m, m p) (3.62) 155

36 be in the class L(p, m, n, A, B). Then the arithmetic mean of f i (z) (i =1, 2,,l) defined by h(z) = 1 l l f i (z) (3.63) i=1 is also in the class L(p, m, n, A, B). Proof : By (3.62), (3.63) we can write ( ) h(z) = 1 l z p + a k,i z k = z p + l i=1 ( 1 l ) l a k,i z k. Since f i (z) L(p, m, n, A, B) for every i =1, 2,,l,sobyusingTheorem i= , we prove that ( ) k(1 B)(p + k +1) n 1 l a k,i l i=1 ( = 1 l ) k(1 B)(p + k +1) n a k,i < 1 l l i=1 The proof is complete. l (A B)p. i=1 Theorem :Letf(z) andg(z) beintheclassl(p, m, n, A, B) such that f(z) =z p + a k z k, g(z) =z p + b k z k. (3.64) Then T (z) =z p + (a 2 k + b2 k )zk is in the class L(p, m, n, A 1,B 1 ) such that A 1 (1 B 1 )µ 2 + B 1,where µ = 2p(A B) m(m + p +1)n (1 B). Proof :Sincef,g L(p, m, n, A, B) so Theorem yields ([k(1 B)(p + k +1) n /(A B)p]a k ) 2 < 1 156

37 and ([k(1 B)(p + k +1) n /(A B)p]b k ) 2 < 1. We obtain from the last two inequalities 1 2 [k(1 B)(p + k +1)n /(A B)p] 2 (a 2 k + b 2 k) < 1. (3.65) But T (z) L(p, m, n, A 1,B 1 ) if and only if [k(1 B 1 )(p + k +1) n /(A 1 B 1 )p](a 2 k + b2 k ) < 1, (3.66) where 1 B 1 <A 1 1andB 1 0, however, (3.65) implies (3.66) if [k(1 B 1 )(p + k +1) n /(A 1 B 1 )p] 1 2 [k(1 B)(p + k +1)n /(A B)p] 2. Hence if 1 B 1 A 1 B 1 k(p + k +1)n α 2, 2p k m where α = 1 B A B. This is equivalent to A 1 B 1 1 B 1 2p k(p + k +1) n α 2. So we can write A 1 B 1 1 B 1 2p(A B) 2 m(m + p +1) n (1 B) 2 = µ2. (3.67) Hence T (z) L(p, m, n, A 1,B 1 )ifa 1 (1 B 1 )µ 2 + B 1. Theorem :Letf(z) andg(z) of the form (3.64) be in L(p, m, n, A, B). Then convolution (or Hadamard product) of two functions f and g is in the class, that is, (f g)(z) L(p, m, n, A 1,B 1 ), where A 1 (1 B 1 )v + B 1 and v = p(a B) 2 m(1 B) 2 (m + p +1) n. 157

38 Proof : Since f, g L(p, m, n, A, B). By using Cauchy-Schwarz inequality and Theorem 3.3.1, we obtain ( k(1 B)(p + k +1) n ) 1/2 k(1 B)(p + k +1) n ak b k a k (A B)p (A B)p ( ) 1/2 k(1 B)(p + k +1) n b k < 1. (3.68) (A B)p We must find the values of A 1,B 1 so that k(1 B 1 )(p + k +1) n a k b k < 1, (3.69) (A 1 B 1 )p therefore by (3.68), (3.69) holds true if by (3.68) we have a k b k < ak b k < (1 B)(A 1 B 1 ),k m, m p, (3.70) (1 B 1 )(A B) (A B)p k(1 B)(p+k+1) n, therefore (3.70) holds true if k(1 B 1 )(p + k +1) n (A 1 B 1 )p [ ] k(1 B)(p + k +1) n 2. (A B)p Hence if 1 B 1 k(1 B)2 (p + k +1) n, k m A 1 B 1 (A B) 2 p which is equivalent to so we can write A 1 B 1 1 B 1 A 1 B 1 1 B 1 HencewegetA 1 v(1 B 1 )+B 1. p(a B) 2 k(1 B) 2 (p + k +1) n, p(a B) 2 m(1 B) 2 (m + p +1) n = v. This completes the proof. 158

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