Di_erential subordinations and superordinations for a comprehensive class of analytic functions

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Differential subordinations and superordinations for

a comprehensive class of analytic functions

N. Magesh and G. Murugusundaramoorthy

(Received April 5, 2008; Revised June 12, 2008)

Abstract. _{In the present investigation, we obtain some subordination and} superordination results involving Hadamard product for certain normalized an-alytic functions in the open unit disk. Our results extend corresponding previ-ously known results.

AMS 2000 Mathematics Subject Classification. 30C45; 30C80.

Key words and phrases. Univalent functions, starlike functions, convex func-tions, differential subordination, differential superordination, Hadamard prod-uct (convolution), Dziok-Srivastava linear operator.

§1. Introduction

Let H be the class of analytic functions in U := {z : |z| < 1} and H(a, n) be the subclass of H consisting of functions of the form

f (z) = a + anzn+ an+1zn+1+ · · · . Let A be the subclass of H consisting of functions of the form

f (z) = z + ∞ X

n=2 anzn.

Let p, h ∈ H and let φ(r, s, t; z) : C3_{×U → C. If p and φ(p(z), zp}0

(z), z2_p00 (z); z) are univalent and if p satisfies the second order superordination

(1.1) h(z) ≺ φ(p(z), zp0

(z), z2p00 (z); z),

then p is a solution of the differential superordination (1.1). (If f is subordi-nate to F , then F is superordisubordi-nate to f .) An analytic function q is called a

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subordinant _{if q ≺ p for all p satisfying (1.1). A univalent subordinant e}q that satisfies q ≺ eq for all subordinants q of (1.1) is said to be the best subordinant. Recently Miller and Mocanu[12] obtained conditions on h, q and φ for which the following implication holds:

h(z) ≺ φ(p(z), zp0

(z), z2p00

(z); z) ⇒ q(z) ≺ p(z). For two functions f (z) = z +P∞

n=2anzn and g(z) = z +P ∞

n=2bnzn, the Hadamard product (or convolution) of f and g is defined by

(f ∗ g)(z) := z + ∞ X n=2

anbnzn=: (g ∗ f )(z).

For αj ∈ C (j = 1, 2, . . . , l) and βj ∈ C \ {0, −1, −2, . . .} (j = 1, 2, . . . m), the generalized hypergeometric functionlFm(α1, . . . , αl; β1, . . . , βm; z) is defined by the infinite series

lFm(α1, . . . , αl; β1, . . . , βm; z) := ∞ X n=0 (α1)n. . . (αl)n (β1)n. . . (βm)n zn n! (l ≤ m + 1; l, m ∈ N0 := {0, 1, 2, . . .}), where (a)n is the Pochhammer symbol defined by

(a)n:= Γ(a + n) Γ(a) = 1, (n = 0); a(a + 1)(a + 2) . . . (a + n − 1), (n ∈ N := {1, 2, 3 . . .}). Corresponding to the function

h(α1, . . . , αl; β1, . . . , βm; z) := z lFm(α1, . . . , αl; β1, . . . , βm; z), the Dziok-Srivastava operator [6] (see also [7, 20]) Hl

m(α1, . . . , αl; β1, . . . , βm) is defined by the Hadamard product

Hml (α1, . . . , αl; β1, . . . , βm)f (z) := h(α1, . . . , αl; β1, . . . , βm; z) ∗ f (z) = z + ∞ X n=2 (α1)n−1. . . (αl)n−1 (β1)n−1. . . (βm)n−1 anzn (n − 1)!. (1.2)

For brevity, we write

Hml [α1]f (z) := Hml (α1, . . . , αl; β1, . . . , βm)f (z). It is easy to verify from (1.2) that

(1.3) z(Hl [α ]f (z))0

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Special cases of the Dziok-Srivastava linear operator includes the Hohlov linear operator [8], the Carlson-Shaffer linear operator L(a, c) [5], the Ruscheweyh derivative operator Dn _{[18], the generalized Bernardi-Libera-Livingston linear} integral operator (cf. [2], [9], [10]) and the Srivastava-Owa fractional derivative operators (cf. [16], [17]).

Using the results of Miller and Mocanu[12], Bulboac˘a [4] considered certain classes of first order differential superordinations as well as superordination-preserving integral operators (see [3]). Recently many authors [1, 13, 14, 19] have used the results of Bulboac˘a [4] and shown some sufficient conditions applying first order differential subordinations and superordinations.

The main object of the present paper is to find sufficient condition for certain normalized analytic functions f (z) in U such that (f ∗ Ψ)(z) 6= 0 and f to satisfy

q1(z) ≺

(f ∗ Φ)(z)

(f ∗ Ψ)(z) ≺ q2(z),

where q1, q2 are given univalent functions in U and Φ(z) = z + ∞ P n=2 λnzn, Ψ(z) = z + ∞ P n=2

µnzn are analytic functions in U with λn ≥ 0, µn ≥ 0 and λn≥ µn. Further the results are extended to Dziok-Srivastava linear operator. Also we obtain number of known results as special cases.

§2. Subordination and Superordination Results For our present investigation, we shall need the following:

Definition 2.1. [12] Denote by Q, the set of all functions f that are analytic and injective on U − E(f ), where

E(f ) = {ζ ∈ ∂U : lim

z→ζf (z) = ∞} and are such that f0

(ζ) 6= 0 for ζ ∈ ∂U − E(f ).

Lemma 2.2. [11] Let q be univalent in the unit disk U and θ and φ be analytic in a domainD containing q(U ) with φ(w) 6= 0 when w ∈ q(U ). Set

ψ(z) := zq0

(z)φ(q(z)) and h(z) := θ(q(z)) + ψ(z). Suppose that

1. ψ(z) is starlike univalent in U and

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If p is analytic with p(0) = q(0), p(U ) ⊆ D and (2.1) θ(p(z)) + zp0 (z)φ(p(z)) ≺ θ(q(z)) + zq0 (z)φ(q(z)), then p(z) ≺ q(z) and q is the best dominant.

Lemma 2.3. [4] Let q be convex univalent in the unit disk U and ϑ and ϕ be analytic in a domain D containing q(U ). Suppose that

1. Re{ϑ0

(q(z))/ϕ(q(z))} > 0 for z ∈ U and 2. ψ(z) = zq0

(z)ϕ(q(z)) is starlike univalent in U . If p(z) ∈ H[q(0), 1] ∩ Q, with p(U ) ⊆ D, and ϑ(p(z)) + zp0

(z)ϕ(p(z)) is univa-lent in U and

(2.2) ϑ(q(z)) + zq0

(z)ϕ(q(z)) ≺ ϑ(p(z)) + zp0

(z)ϕ(p(z)), then q(z) ≺ p(z) and q is the best subordinant.

Using Lemma 2.2, we first prove the following theorem.

Theorem 2.4. Let Φ, Ψ ∈ A, γ 6= 0 and α, β be the complex numbers and q(z) be convex univalent in U with q(0) = 1. Further assume that

(2.3) Re βq(z) γ − zq0 (z) q(z) + 1 +zq 00 (z) q0_(z) > 0 (z ∈ U ). If f ∈ A satisfies Υ1(f, Φ, Ψ, α, β, γ) ≺ α + βq(z) + γ zq0 (z) q(z) , (2.4) where Υ1(f, Φ, Ψ, α, β, γ) := α + β (f ∗ Φ)(z) (f ∗ Ψ)(z) + γ z(f ∗ Φ)0 (z) (f ∗ Φ)(z) − z(f ∗ Ψ)0 (z) (f ∗ Ψ)(z) , (2.5) (f ∗ Φ)(z) 6= 0 and (f ∗ Ψ)(z) 6= 0, then (f ∗ Φ)(z) (f ∗ Ψ)(z) ≺ q(z)

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Proof. Define the function p(z) by

(2.6) p(z) := (f ∗ Φ)(z)

(f ∗ Ψ)(z) (z ∈ U ).

Then the function p(z) is analytic in U and p(0) = 1. Therefore, by making use of (2.6), we obtain α + β(f ∗ Φ)(z) (f ∗ Ψ)(z)+ γ z(f ∗ Φ)0 (z) (f ∗ Φ)(z) − z(f ∗ Ψ)0 (z) (f ∗ Ψ)(z) = α + βp(z) + γzp 0 (z) p(z) . By using (2.7) in (2.4), we have (2.7) α + βp(z) + γzp 0 (z) p(z) ≺ α + βq(z) + γ zq0 (z) q(z) . By setting θ(w) := α + βω and φ(ω) := γ ω,

it can be easily observed that θ(w) and φ(w) are analytic in C − {0} and that φ(w) 6= 0. Hence the result now follows by an application of Lemma 2.2.

Taking p(z) = Hml [α1+1](f ∗Φ)(z) Hl m[α1](f ∗Ψ)(z) and p(z) = Hl m[α1](f ∗Φ)(z) Hl m[α1+1](f ∗Ψ)(z) respectively we

obtain the following two theorems.

Theorem 2.5. Let Φ, Ψ ∈ A, γ 6= 0 and α, β be the complex numbers and q(z) be convex univalent in ∆ with q(0) = 1. Further assume that (2.3) holds true. If f ∈ A satisfies Υ2(f, Φ, Ψ, α, β, γ) ≺ α + βq(z) + γ zq0 (z) q(z) , (2.8) where Υ2(f, Φ, Ψ, α, β, γ) :=      α + βHml[α1+1](f ∗Φ)(z) Hl m[α1](f ∗Ψ)(z) +γh(α1+ 1)H l m[α1+2](f ∗Φ)(z) Hl m[α1](f ∗Φ)(z) − α1 Hl m[α1+1](f ∗Ψ)(z) Hl m[α1](f ∗Ψ)(z) − 1 i , (2.9) then Hl m[α1+ 1](f ∗ Φ)(z) Hl m[α1](f ∗ Ψ)(z) ≺ q(z)

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Theorem 2.6. Let Φ, Ψ ∈ A, γ 6= 0 and α, β be the complex numbers and q(z) be convex univalent in ∆ with q(0) = 1. Further assume that (2.3) holds true. If f ∈ A satisfies Υ3(f, Φ, Ψ, α, β, γ) ≺ α + βq(z) + γ zq0 (z) q(z) , (2.10) where Υ3(f, Φ, Ψ, α, β, γ) :=      α + β Hml [α1](f ∗Φ)(z) Hl m[α1+1](f ∗Ψ)(z) +γhα1H l m[α1+1](f ∗Φ)(z) Hl m[α1](f ∗Φ)(z) − (α1+ 1) Hl m[α1+2](f ∗Ψ)(z) Hl m[α1+1](f ∗Ψ)(z) + 1 i , (2.11) then Hl m[α1](f ∗ Φ)(z) Hl m[α1+ 1](f ∗ Ψ)(z) ≺ q(z)

and q is the best dominant.

When l = 2, m = 1, α1 = a, α2= 1 and β1 = c in Theorem 2.5 and Theorem 2.6, we state the following corollaries for Carlson-Shaffer linear operator L(a, c) [5].

Corollary 2.7. Let Φ, Ψ ∈ A, γ 6= 0 and α, β be the complex numbers and q(z) be convex univalent in ∆ with q(0) = 1. Further assume that (2.3) holds true. If f ∈ A satisfies Υ4(f, Φ, Ψ, α, β, γ) ≺ α + βq(z) + γ zq0 (z) q(z) , (2.12) where Υ4(f, Φ, Ψ, α, β, γ) :=      α + βL(a+1,c)(f ∗Φ)(z)_{L(a,c)(f ∗Ψ)(z)}

+γh(a + 1)L(a+2,c)(f ∗Φ)(z)_{L(a,c)(f ∗Φ)(z)} − aL(a+1,c)(f ∗Ψ)(z)_{L(a,c)(f ∗Ψ)(z)} − 1i, (2.13)

then

L(a + 1, c)(f ∗ Φ)(z)

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Corollary 2.8. Let Φ, Ψ ∈ A, γ 6= 0 and α, β be the complex numbers and q(z) be convex univalent in ∆ with q(0) = 1. Further assume that (2.3) holds true. If f ∈ A satisfies Υ5(f, Φ, Ψ, α, β, γ) ≺ α + βq(z) + γ zq0 (z) q(z) , (2.14) where Υ5(f, Φ, Ψ, α, β, γ) :=      α + β_{L(a+1,c)(f ∗Ψ)(z)}L(a,c)(f ∗Φ)(z)

+γhaL(a+1,c)(f ∗Φ)(z)_{L(a,c)(f ∗Φ)(z)} − (a + 1)L(a+2,c)(f ∗Ψ)(z)_{L(a+1,c)(f ∗Ψ)(z)} + 1i, (2.15)

then

L(a, c)(f ∗ Φ)(z)

L(a + 1, c)(f ∗ Ψ)(z) ≺ q(z) and q is the best dominant.

By fixing Φ(z) = z

(1−z)2 and Ψ(z) = _1−zz in Theorem 2.4, we obtain the

following corollary.

Corollary 2.9. Let γ 6= 0, α, β be the complex numbers and q be convex univalent in U with q(0) = 1 and (2.3) holds true. If f ∈ A satisfies

α + (β − γ)zf 0 (z) f (z) + γ 1 + zf 00 (z) f0_(z) ≺ α + βq(z) + γzq 0 (z) q(z) , then zf0 (z) f (z) ≺ q(z) and q is the best dominant.

Specializing the values of α = 1, β = 0, q(z) = _(1−z)1 2b (b ∈ C − {0}),

γ = 1_b, Φ(z) = _1−zz and Ψ(z) = z in Theorem 2.4, we have the following corollary as stated in [21].

Corollary 2.10. Let b be a non zero complex number. If f ∈ A and

1 +1 b zf0 (z) f (z) − 1 ≺ 1 + z 1 − z, then f (z) z ≺ 1 (1 − z)2b and _(1−z)1 _2b is the best dominant.

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Similarly for α = 1, β = 0, γ = 1_b, q(z) = _(1−z)1 2b (b ∈ C − {0}), Φ(z) =

z

(1−z)2 and Ψ(z) = z in Theorem 2.4, we have the following corollary as stated

in [21].

Corollary 2.11. Let b be a non zero complex number. If f ∈ A and 1 + 1 b zf00 (z) f0_(z) ≺ 1 + z 1 − z, then f0 (z) ≺ 1 (1 − z)2b and 1

(1−z)2b is the best dominant.

Remark 2.12. For the choices Φ(z) = _(1−z)z 2, Ψ(z) =

z

(1−z), α = 0, β > −1, γ = 1 and q(z) = _k+zk (k > 1) in Theorem 2.4, we get the result obtained by Obradovic et.al., [15].

By taking l = 2, m = 1, α1 = 1, α2 = 1 and β1 = 1 in Theorem 2.5 and Theorem 2.6, we state the following corollaries.

Corollary 2.13. Let Φ, Ψ ∈ A, γ 6= 0, α, β be the complex numbers and q be convex univalent inU with q(0) = 1 and (2.3) holds true. If f ∈ A satisfies

(α + γ) + βz(f ∗ Φ) 0 (z) (f ∗ Ψ)(z) + γ z(f ∗ Φ)00 (z) (f ∗ Φ)0_(z) − z(f ∗ Ψ)0 (z) (f ∗ Ψ)(z) ≺ α + βq(z) + γzq 0 (z) q(z) with (f ∗ Ψ)(z) 6= 0 and (f ∗ Φ)0 (z) 6= 0, then z(f ∗ Φ)0 (z) (f ∗ Ψ)(z) ≺ q(z) and q is the best dominant.

Corollary 2.14. Let Φ, Ψ ∈ A and γ 6= 0, α, β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.3) holds true. If f ∈ A satisfies

(α − γ) + β (f ∗ Φ)(z) z(f ∗ Ψ)0_(z) + γ z(f ∗ Φ)0 (z) (f ∗ Φ)(z) − z(f ∗ Ψ)00 (z) (f ∗ Ψ)0_(z) ≺ α + βq(z) + γzq 0 (z) q(z) with (f ∗ Φ)(z) 6= 0 and (f ∗ Ψ)0 (z) 6= 0, then (f ∗ Φ)(z) z(f ∗ Ψ)0_(z) ≺ q(z)

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By fixing Φ(z) = _1−zz and Ψ(z) = _1−zz in Theorem 2.5 and Theorem 2.6 we obtain the following corollaries.

α + βH l m[α1+ 1]f (z) Hl m[α1]f (z) + γ (α1+ 1) Hl m[α1+ 2]f (z) Hl m[α1]f (z) − α1 Hl m[α1+ 1]f (z) Hl m[α1]f (z) − 1 ≺ α + βq(z) + γzq 0 (z) q(z) , then Hl m[α1+ 1]f (z) Hl m[α1]f (z) ≺ q(z)

α + β H l m[α1]f (z) Hl m[α1+ 1]f (z) + γ α1 Hl m[α1+ 1]f (z) Hl m[α1]f (z) − (α1+ 1) Hl m[α1+ 2]f (z) Hl m[α1+ 1]f (z) + 1 ≺ α + βq(z) + γzq 0 (z) q(z) , then Hl m[α1]f (z) Hl m[α1+ 1]f (z) ≺ q(z)

By fixing Φ(z) = _1−zz and Ψ(z) = _1−zz in Corollary 2.13, Corollary 2.14 and also l = 2, m = 1, α1 = 1, α2 = 1 and β1= 1 in Corollary 2.15, Corollary 2.16 we obtain the following corollaries.

(α + γ) + βzf 0 (z) f (z) + γ zf00 (z) f0_(z) − zf0 (z) f (z) ≺ α + βq(z) + γzq 0 (z) q(z) ,

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then

zf0 (z)

f (z) ≺ q(z) and q is the best dominant.

α + β f (z) zf0 (z)− γ 1 +zf 00 (z) f0 (z) − zf0 (z) f (z) ≺ α + βq(z) + γzq 0 (z) q(z) , then f (z) zf0_(z) ≺ q(z) and q is the best dominant.

Theorem 2.19. Let Φ, Ψ ∈ A and γ 6= 0, α, β be the complex numbers. Let q be convex univalent in U with q(0) = 1. Assume that

(2.16) Re {γβq(z)} > 0.

Let f ∈ A, (f ∗Φ)(z)_{(f ∗Ψ)(z)} ∈ H[q(0), 1] ∩ Q. Let Υ1(f, Φ, Ψ, α, β, γ) be univalent in U and

(2.17) α + βq(z) + γzq 0

(z)

q(z) ≺ Υ1(f, Φ, Ψ, α, β, γ),

whereΥ1(f, Φ, Ψ, α, β, γ) is given by (2.5) with (f ∗Φ)(z) 6= 0 and (f ∗Ψ)(z) 6= 0, then

q(z) ≺ (f ∗ Φ)(z) (f ∗ Ψ)(z) and q is the best subordinant.

Proof. Define the function p(z) by

(2.18) p(z) := (f ∗ Φ)(z)

(f ∗ Ψ)(z). Simple computation from (2.18), we get,

Υ1(f, Φ, Ψ, α, β, γ) = α + βp(z) + γ zp0 (z) p(z) , then α + βq(z) + γzq 0 (z) ≺ α + βp(z) + γzp 0 (z) .

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By setting ϑ(ω) = α + βω and φ(ω) = γ_ω, it is easily observed that ϑ(ω) is analytic in C. Also, φ(ω) is analytic in C − {0} and that φ(ω) 6= 0.

Since q(z) is convex univalent function, it follows that

Re ϑ0 (q(z)) φ(q(z)) = < {γβq(z)} > 0, z ∈ U.

Now Theorem 2.19 follows by applying Lemma 2.3.

Theorem 2.20. Let Φ, Ψ ∈ A. Let γ 6= 0, α and β be the complex num-bers. Let q be convex univalent in U with q(0) = 1. Assume that (2.16) holds true. Let f ∈ A, Hml[α1+1](f ∗Φ)(z) Hl m[α1](f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q. Let Υ2(f, Φ, Ψ, α, β, γ) be univalent in U and (2.19) α + βq(z) + γzq 0 (z) q(z) ≺ Υ2(f, Φ, Ψ, α, β, γ), where Υ2(f, Φ, Ψ, α, β, γ) is given by (2.9), then

q(z) ≺ H l

m[α1+ 1](f ∗ Φ)(z) Hl

m[α1](f ∗ Ψ)(z) and q is the best subordinant.

Theorem 2.21. Let Φ, Ψ ∈ A. Let γ 6= 0, α and β be the complex num-bers. Let q be convex univalent in U with q(0) = 1. Assume that (2.16) holds true. Let f ∈ A, Hml [α1](f ∗Φ)(z) Hl m[α1+1](f ∗Ψ)(z) ∈ H[q(0), 1] ∩ Q. Let Υ3(f, Φ, Ψ, α, β, γ) be univalent in U and (2.20) α + βq(z) + γzq 0 (z) q(z) ≺ Υ2(f, Φ, Ψ, α, β, γ), where Υ3(f, Φ, Ψ, α, β, γ) is given by (2.11), then

q(z) ≺ H l

m[α1](f ∗ Φ)(z) Hl

m[α1+ 1](f ∗ Ψ)(z) and q is the best subordinant.

For the Choices of p(z) = Hml [α1+1](f ∗Φ)(z)

Hl m[α1](f ∗Ψ)(z) and p(z) = Hl m[α1](f ∗Φ)(z) Hl m[α1+1](f ∗Ψ)(z), the

proofs of Theorem 2.20 and Theorem 2.21 are lines similar to the proof of Theorem 2.19, so we omitted the proofs of Theorems 2.20 and 2.21.

When l = 2, m = 1, α1 = a, α2 = 1 and β1 = c in Theorem 2.20 and Theorem 2.21, we state the following corollary.

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Corollary 2.22. Let Φ, Ψ ∈ A. Let γ 6= 0, α and β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.16) holds true. If f ∈ A and

α + βq(z) + γzq 0

(z)

q(z) ≺ Υ4(f, Φ, Ψ, α, β, γ), where Υ4(f, Φ, Ψ, α, β, γ) is given by (2.13), then

q(z) ≺ L(a + 1, c)(f ∗ Φ)(z) L(a, c)(f ∗ Ψ)(z) and q is the best subordinant.

Corollary 2.23. Let Φ, Ψ ∈ A. Let γ 6= 0, α and β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.16) holds true. If f ∈ A and

α + βq(z) + γzq 0

(z)

q(z) ≺ Υ5(f, Φ, Ψ, α, β, γ), where Υ5(f, Φ, Ψ, α, β, γ) is given by (2.15), then

q(z) ≺ L(a, c)(f ∗ Φ)(z) L(a + 1, c)(f ∗ Ψ)(z) and q is the best subordinant.

When l = 2, m = 1, α1 = 1, α2 = 1 and β1 = 1 in Theorem 2.20 and Theorem 2.21, we derive the following corollaries.

Corollary 2.24. Let Φ, Ψ ∈ A. Let γ 6= 0, α and β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.16) holds true. If f ∈ A and α + βq(z) + γzq 0 (z) q(z) ≺ (α + γ) + β z(f ∗ Φ)0 (z) (f ∗ Ψ)(z) + γ z(f ∗ Φ)00 (z) (f ∗ Φ)0_(z) − z(f ∗ Ψ)0 (z) (f ∗ Ψ)(z) with (f ∗ Ψ)(z) 6= 0 and (f ∗ Φ)0 (z) 6= 0, then q(z) ≺ z(f ∗ Φ) 0 (z) (f ∗ Ψ)(z) and q is the best subordinant.

Corollary 2.25. Let Φ, Ψ ∈ A. Let γ 6= 0, α and β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.16) holds true. If f ∈ A and α + βq(z) + γzq 0 (z) ≺ (α − γ) + β (f ∗ Φ)(z) + γ z(f ∗ Φ)0 (z) −z(f ∗ Ψ) 00 (z)

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with (f ∗ Φ)(z) 6= 0 and (f ∗ Ψ)0

(z) 6= 0, then

q(z) ≺ (f ∗ Φ)(z) z(f ∗ Ψ)0_(z) and q is the best subordinant.

By Taking l = 2, m = 1, α1 = 1, α2 = 1 and β1 = 1 in Theorem 2.20 and Theorem 2.21 and by fixing Φ(z) = Ψ(z) = _1−zz in Corollary 2.24 and 2.25, we obtain the following corollaries.

Corollary 2.26. Let γ 6= 0, α and β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.16) holds true. If f ∈ A and

α + βq(z) + γzq 0 (z) q(z) ≺ (α + γ) + β zf0 (z) f (z) + γ zf00 (z) f0_(z) − zf0 (z) f (z) , then q(z) ≺ zf 0 (z) f (z) and q is the best subordinant.

Corollary 2.27. Let γ 6= 0, α and β be the complex numbers. Let q be convex univalent in U with q(0) = 1 and (2.16) holds true. If f ∈ A and

α + βq(z) + γzq 0 (z) q(z) ≺ α + β f (z) zf0_(z)− γ 1 + zf 00 (z) f0_(z) − zf0 (z) f (z) , then q(z) ≺ f (z) zf0 (z) and q is the best subordinant.

We Conclude this paper by stating the following sandwich results.

§3. Sandwich Results

Theorem 3.1. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose _{(f ∗Ψ)(z)}(f ∗Φ)(z) ∈ H[1, 1] ∩ Q and Υ1(f, Φ, Ψ, α, β, γ) is univalent in U. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ Υ1(f, Φ, Ψ, α, β, γ) ≺ α + βq2(z) + γ zq0 2(z) q2(z) ,

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whereΥ1(f, Φ, Ψ, α, β, γ) is given by (2.5) with (f ∗Φ)(z) 6= 0 and (f ∗Ψ)(z) 6= 0, then

q1(z) ≺

(f ∗ Φ)(z)

(f ∗ Ψ)(z) ≺ q2(z)

and q1, q2 are respectively the best subordinant and best dominant. By taking q1(z) = 1+A_1+B1₁zz (−1 ≤ B1 < A1 ≤ 1) and q2(z) =

1+A2z

1+B2z (−1 ≤

B2 < A2 ≤ 1) in Theorem 3.1 we obtain the following result. Corollary 3.2. Let Φ, Ψ ∈ A. If f ∈ A, (f ∗Φ)(z)_{(f ∗Ψ)(z)} ∈ H[1, 1] ∩ Q and Υ1(f, Φ, Ψ, α, β, γ) is univalent in U. Further α + β 1 + A1z 1 + B1z + γ(A1− B1)z (1 + A1z)(1 + B1z) ≺ Υ1(f, Φ, Ψ, α, β, γ) ≺ α + β 1 + A2z 1 + B2z + γ(A2− B2)z (1 + A2z)(1 + B2z) whereΥ1(f, Φ, Ψ, α, β, γ) is given by (2.5) with (f ∗Φ)(z) 6= 0 and (f ∗Ψ)(z) 6= 0, then 1 + A1z 1 + B1z ≺ (f ∗ Φ)(z) (f ∗ Ψ)(z) ≺ 1 + A2z 1 + B2z and 1+A1z 1+B1z, 1+A2z

1+B2z are respectively the best subordinant and best dominant.

Theorem 3.3. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose Hml[α1+1](f ∗Φ)(z) Hl m[α1](f ∗Ψ)(z) ∈ H[1, 1]∩Q and Υ2(f, Φ, Ψ, α, β, γ) is univalent inU. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ Υ2(f, Φ, Ψ, α, β, γ) ≺ α + βq2(z) + γ zq0 2(z) q2(z) ,

where Υ2(f, Φ, Ψ, α, β, γ) is given by (2.9), then

q1(z) ≺ Hl m[α1+ 1](f ∗ Φ)(z) Hl m[α1](f ∗ Ψ)(z) ≺ q2(z)

and q1, q2 are respectively the best subordinant and best dominant.

Theorem 3.4. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose Hml [α1](f ∗Φ)(z) Hl m[α1+1](f ∗Ψ)(z) ∈ H[1, 1]∩Q and Υ3(f, Φ, Ψ, α, β, γ) is univalent inU. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q (z) ≺ Υ3(f, Φ, Ψ, α, β, γ) ≺ α + βq2(z) + γ zq0 2(z) q (z) ,

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where Υ3(f, Φ, Ψ, α, β, γ) is given by (2.11), then q1(z) ≺ H_ml [α1](f ∗ Φ)(z) Hl m[α1+ 1](f ∗ Ψ)(z) ≺ q2(z)

By making use of Corollaries 2.7 and 2.22, we state the following corollary. Corollary 3.5. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose L(a+1,c)(f ∗Φ)(z)_{L(a,c)(f ∗Ψ)(z)} ∈ H[1, 1] ∩ Q and Υ4(f, Φ, Ψ, α, β, γ) is univalent inU. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ Υ4(f, Φ, Ψ, α, β, γ) ≺ α + βq2(z) + γ zq0 2(z) q2(z) ,

q1(z) ≺

L(a + 1, c)(f ∗ Φ)(z)

L(a, c)(f ∗ Ψ)(z) ≺ q2(z)

By making use of Corollaries 2.8 and 2.23, we state the following corollary. Corollary 3.6. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose _{L(a+1,c)(f ∗Ψ)(z)}L(a,c)(f ∗Φ)(z) ∈ H[1, 1] ∩ Q and Υ5(f, Φ, Ψ, α, β, γ) is univalent inU. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ Υ5(f, Φ, Ψ, α, β, γ) ≺ α + βq2(z) + γ zq0 2(z) q2(z) ,

q1(z) ≺

L(a, c)(f ∗ Φ)(z)

L(a + 1, c)(f ∗ Ψ)(z) ≺ q2(z)

By making use of Corollaries 2.13 and 2.24, we state the following corollary. Corollary 3.7. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies

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(2.16). Moreover suppose z(f ∗Φ)_{(f ∗Ψ)(z)}0(z) ∈ H[1, 1] ∩ Q and (α + γ) + βz(f ∗Φ)_{(f ∗Ψ)(z)}0(z) + γhz(f ∗Φ)_{(f ∗Φ)}000_(z)(z) − z(f ∗Ψ)0_(z) (f ∗Ψ)(z) i is univalent inU. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ (α + γ) + βz(f ∗ Φ) 0 (z) (f ∗ Ψ)(z) + γ z(f ∗ Φ)00 (z) (f ∗ Φ)0 (z) − z(f ∗ Ψ)0 (z) (f ∗ Ψ)(z) ≺ α + βq2(z) + γzq 0 2(z) q2(z) , with (f ∗ Ψ)(z) 6= 0 and (f ∗ Φ)0 (z) 6= 0, then q1(z) ≺ z(f ∗ Φ)0 (z) (f ∗ Ψ)(z) ≺ q2(z)

By making use of Corollaries 2.14 and 2.25, we state the following corollary. Corollary 3.8. Let q1 and q2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Let Φ, Ψ ∈ A. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose _{z(f ∗Ψ)}(f ∗Φ)(z)0_(z) ∈ H[1, 1] ∩ Q and (α − γ) + β

(f ∗Φ)(z) z(f ∗Ψ)0_(z) + γhz(f ∗Φ)_{(f ∗Φ)(z)}0(z)−z(f ∗Ψ)_{(f ∗Ψ)}000_(z)(z) i is univalent in U. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ (α − γ) + β (f ∗ Φ)(z) z(f ∗ Ψ)0_(z)+ γ z(f ∗ Φ)0 (z) (f ∗ Φ)(z) − z(f ∗ Ψ)00 (z) (f ∗ Ψ)0_(z) ≺ α + βq2(z) + γ zq0 2(z) q2(z) , with (f ∗ Φ)(z) 6= 0 and (f ∗ Ψ)0 (z) 6= 0, then q1(z) ≺ (f ∗ Φ)(z) z(f ∗ Ψ)0_(z) ≺ q2(z)

By making use of Corollaries 2.17 and 2.26, we state the following corollary. Corollary 3.9. Let q1 andq2 be convex univalent in U, γ 6= 0 and α, β be the complex numbers. Suppose q satisfies (2.3) and q satisfies (2.16). Moreover

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suppose zf_f(z)0(z) ∈ H[1, 1]∩Q and (α+γ)+βzf_f(z)0(z)+γhzf_f000_(z)(z) − zf0_(z) f(z) i is univalent in U. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ (α + γ) + βzf 0 (z) f (z) + γ zf00 (z) f0_(z) − zf0 (z) f (z) ≺ α + βq2(z) + γ zq0 2(z) q2(z) , then q1(z) ≺ zf0 (z) f (z) ≺ q2(z)

By making use of Corollaries 2.18 and 2.27, we state the following corollary. Corollary 3.10. Letq1andq2 be convex univalent inU, γ 6= 0 and α, β be the complex numbers. Suppose q2 satisfies (2.3) and q1 satisfies (2.16). Moreover suppose _zff(z)0_(z) ∈ H[1, 1]∩Q and α+β f(z) zf0_(z)−γ h 1 +zf_f000_(z)(z) − zf0_(z) f(z) i is univalent in U. If f ∈ A satisfies α + βq1(z) + γ zq0 1(z) q1(z) ≺ α + β f (z) zf0_(z) − γ 1 + zf 00 (z) f0_(z) − zf0 (z) f (z) ≺ α + βq2(z) + γ zq0 2(z) q2(z) , then q1(z) ≺ f (z) zf0_(z) ≺ q2(z)

Acknowledgement

The authors would like to thank the referee for his valuable comments and suggestions.

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N.Magesh

Department of Mathematics, Adhiyamaan College of Engineering (Autonomous), Hosur - 635 109, India.

E-mail: nmagi [email protected] G.Murugusundaramoorthy

School of Science and Humanities, VIT University, Vellore - 632 014, India.