SUBORDINATIONS AND SUPERORDINATIONS USING THE DZIOK-SRIVASTAVA

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SUBORDINATIONS AND SUPERORDINATIONS USING THE DZIOK-SRIVASTAVA

LINEAR OPERATOR

GEORGIA IRINA OROS

Using properties of the Dziok-Srivastava linear operator, we obtain differential subordinations and superordinations for functions from classA. A sandwich-type result is also given.

AMS 2000 Subject Classification: 30C45, 30A10, 30C80.

Key words: univalent function, starlike function, convex function, differential subordination, differential superordination, Dziok-Srivastava linear operator.

1. INTRODUCTION AND PRELIMINARIES

LetU denote the unit disc U ={z ∈C: |z|<1} of the complex plane and U ={z∈C: |z| ≤1}.

LetH(U) denote the space of holomorphic functions in U, let A_n={f ∈ H(U), f(z) =z+a_n+1zⁿ⁺¹+· · ·, z ∈U} with A1 =A, and let

H[a, n] ={f ∈ H(U), f(z) =a+anzⁿ+an+1zⁿ⁺¹+· · · , z ∈U}, S ={f ∈A; f is univalent in U}.

Let

K =

f ∈A, Rezf⁰⁰(z)

f⁰(z) + 1>0, z ∈U

denote the class of normalized convex function in U and S^∗=

f ∈A: Rezf⁰(z)

f(z) >0, z∈U

the class of starlike functions in U.

MATH. REPORTS11(61),2 (2009), 155–163

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Iff andgare analytic functions inU, then we say thatf is subordinate to g, and write f ≺g, if there is a functionw analytic in U, with w(0) = 0,

|w(z)|<1 for allz∈U, such that f(z) =g[w(z)], for z∈U.

Ifgis univalent, then f ≺g if and only iff(0) =g(0) andf(U)⊆g(U).

The method of differential subordinations (also known as the admissi- ble functions method) was introduced and developed by Mocanu and Miller [2, 3, 4].

Let Ω and ∆ be any sets inC. Letpbe an analytic function in the unit disk with p(0) =aand ψ(r, s, t;z) :C³×U →Ca function. This theory deals with generalizations of the implication

(i){ψ(p(z), zp⁰(z), z²p⁰⁰(z);z)|z∈U} ⊂Ω implies p(U)⊂∆.

Definition1 [4, p. 320]. Letψ:C³×U →Cand lethbe univalent inU. If pis analytic inU and satisfies the (second-order) differential subordination

(ii)ψ(p(z), zp⁰(z), z²p⁰⁰(z);z)≺h(z),z∈U,

then p is called a solution of the differential subordination. The univalent functionqis called a dominant of the solutions of the differential subordination, or, simply, a dominant, if p ≺ q for all p satisfying (ii). A dominant eq that satisfiesqe≺qfor all dominantsqof (ii) is said to be the best dominant of (ii).

(Note that the best dominant is unique up to a rotation of U.)

In [5] the authors introduced the dual problem of the differential subordination which they call differential superordination.

Definition 2 [5]. Let f, F ∈ H(U) and let F be univalent in U. The function F is said to be superordinate to f, or f is subordinate toF, and we write f ≺F, iff(0) =F(0) andf(U)⊂F(U).

Let Ω and ∆ be any sets in C. Let p be an analytic function in the unit disk and ϕ(r, s, t;z) : C³×U → C a function. This theory deals with generalizations of the implication:

(iii) Ω⊂ {ϕ(p(z), zp⁰(z), z²p⁰⁰(z);z)|z∈U} implies ∆⊂p(U).

Definition 3 [5]. Let ϕ : C³×U → C and let h be analytic in U. If p andϕ(p(z), zp⁰(z), z²p⁰⁰(z);z) are univalent inU and satisfy the (second-order) differential superordination

(iv)h(z)≺ϕ(p(z), zp⁰(z), z²p⁰⁰(z);z),

thenpis called a solution of the differential superordination. An analytic functionqis called a subordinant of the solutions of the differential superordination or, simply, a subordinant if q ≺ p for all p satisfying (iv). A univalent subordinant qethat satisfies q ≺ eq for all subordinants q of (iv) is said to be the best subordinant. (Note that the best subordinant is unique up to a rotation of U.)

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Definition 4 [4, Definition 2.2b, p. 21]. We denote by Qthe set of functions f that are analytic and injective onU \E(f), where

E(f) =

ζ ∈∂U; lim

z→ζf(z) =∞

and are such that f⁰(ζ) 6= 0 forζ ∈∂U \E(f). The subclass of Q for which f(0) =ais denoted byQ(a).

In order to prove new results we shall use the lemmas below.

Lemma A [4, Theorem 3.4h, p. 132]. Let q be univalent in U and let θ and φ be analytic in a domain D containing q(U), with φ(w) 6= 0, when w∈q(U). Set

Q(z) :=zq⁰(z)φ[q(z)], h(z) :=θ[q(z)] +Q(z) and suppose that either

(j)h is convex, or (jj)Q is starlike.

In addition, assume that (jjj) Re^zh_Q(z)⁰^(z) = Reh_θ0[q(z)]

φ[q(z)] +^zQ_Q(z)⁰^(z)i

>0.

If p is analytic in U, with p(0) =q(0), p(U)⊂D and

θ[p(z)] +zp⁰(z)φ[p(z)]≺θ[q(z)] +zq⁰(z)φ[q(z)] =h(z), then p≺q, andq is the best dominant.

Lemma B[6, Theorem 3]. Let q be univalent inU with q(0) =a, and θ and ϕ be analytic in a domain Dcontaining q(U). Define

Q(z) =zq⁰(z)ϕ[q(z)], h(z) =θ[q(z)] +Q(z).

Suppose that (i) Re h_θ0[q(z)]

ϕ[q(z)]

i

>0 and

(ii)Q is starlike univalent in U.

Ifp∈ H[a,1]∩Q withp(U)⊂D, and θ[p(z)] +zp⁰(z)ϕ[p(z)]is univalent in U, then

θ[q(z)] +zq⁰(z)ϕ[q(z)]≺θ[p(z)] +zp⁰(z)ϕ[p(z)]

implies q(z)≺p(z), andq(z) is the best subordinant.

In [1] there was defined the Dziok-Srivastava operator H_m^l (α₁, α₂, . . . , α_l;β₁, β₂, . . . , β_m)f(z) (1)

=z+

∞

X

n=2

(α1)n−1(α2)n−1. . .(αl)n−1

(β1)n−1(β2)n−1. . .(β_l)n−1

·an· zⁿ (n−1)!, where αi ∈C,i= 1,2, . . . , l,βj ∈C\ {0,−1,−2, . . .},j= 1,2, . . . , m.

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For simplicity we write

H_m^l [α₁]f(z) =H_m^l (α₁, α₂, . . . , α_l;β₁, β₂, . . . , β_m)f(z).

For this operator we have the property

(2) α1H_m^l [α1+ 1]f(z) =z{H_m^l [α1]f(z)}⁰+ (α1−1)H_m^l [α1]f(z).

2. MAIN RESULTS

Theorem 1. Let l, m ∈ N0 = {0,1,2, . . .}, l ≤ m+ 1, αi ∈ C, i = 1,2, . . . , l, βj ∈ C \ {0,−1,−2, . . .}, j = 1,2, . . . , m, and H_m^l [α1]f(z) the Dziok-Srivastava linear operator given by (1).

If f ∈ A, α₁ ≥ 0, and ^H^m^l ^[α_z¹^]f(z) 6= 0 in U and verifies the differential subordination

(3) z{H_m^l [α₁+ 1]f(z)}⁰

H_m^l [α1+ 1]f(z) + (1−α₁)H_m^l [α₁+ 1]f(z)

H_m^l [α1]f(z) +α₁−1≺h(z), where

h(z) = 1 +z

1−z + 2z

(1−z)[1 +γ+ (1−γ)z], 0≤γ <1, z ∈U, then

H_m^l [α1+ 1]f(z)

H_m^l [α₁]f(z) ≺ 1 +z

1−z, z∈U, and

g(z) = 1 +z 1−z is the best dominant.

Proof. Let

(4) H_m^l [α1+ 1]f(z)

H_m^l [α1]f(z) =p(z),

p(z) = 1 +A₁z+A₂z²+· · ·, p(0) = 1 andp∈ H[1,1].

Differentiating (4), we obtain

(5) zp⁰(z)

p(z) = z{H_m^l [α₁+ 1]f(z)}⁰

H_m^l [α₁+ 1]f(z) −z{H_m^l [α₁]f(z)}⁰ H_m^l [α₁]f(z) . Using (2), (5) becomes

(6) zp⁰(z)

p(z) = z{H_m^l [α₁+ 1]f(z)}⁰

H_m^l [α₁+ 1]f(z) −α₁H_m^l [α₁+ 1]f(z)

H_m^l [α₁]f(z) +α₁−1.

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Using (4) and (6), we get (7) p(z) +zp⁰(z)

p(z) = z{H_m^l [α1+ 1]f(z)}⁰

H_m^l [α₁+ 1]f(z) + (1−α1)H_m^l [α1+ 1]f(z)

H_m^l [α₁]f(z) +α1−1.

Using (7), the differential subordination (2) becomes (8) p(z) +zp⁰(z)

p(z) ≺h(z) = 1 +z

1−z + 2z

(1−z)[1 +γ+ (1−γ)z], z∈U.

To prove the theorem, we use Lemma A. For that, letq(z) = ^1+z_1−z,q(0) = 1, q(U) = {w ∈ C : Rew > 0}. Define the functions θ : D ⊃ q(U) → C, θ(w) =w,and

φ:D⊃q(U)→C, φ(w) = 1 w+γ, with φ(w)6= 0, 0≤γ <1. Then we have

θ[q(z)] =q(z) = 1 +z 1−z, φ[q(z)] = 1

q(z) +γ = 1 1 +z 1−z+γ

= 1−z

1 +γ+ (1−γ)z, Q(z) =zq⁰(z)φ[q(z)] = 2z

(1−z)² · 1−z

1 +γ+ (1−γ)z = 2z

(1−z)[1 +γ+ (1−γ)z], h(z) =θ[q(z)] +Q(z) = 1 +z

1−z+ 2z

(1−z)[1 +γ+ (1−γ)z]. We now calculate

RezQ⁰(z) Q(z) = Re

1 + z

1−z − (1−γ)z 1 +γ+ (1−γ)z

= 1

2+ (1−γ) 1 + cost+γ(1−cost)

[1 +γ+ (1−γ) cost]²+ (1−γ)²sin²t >0, Reθ⁰[q(z)]

φ(z) = 2Re1 +γ+ (1−γ)z

1−z = 2Re

−1 +γ+ 2 1−z

= 2γ >0, Rezh⁰(z)

Q(z) = Reθ⁰[q(z)]

φ(z) + RezQ⁰(z)

Q(z) >0, z∈U.

SinceQis starlike and Re^zh_Q(z)⁰^(z) >0,z∈U,γ ≥0, by Lemma A we have p(z)≺q(z), i.e.,

H_m^l [α₁+ 1]f(z)

H_m^l [α₁]f(z) ≺ 1 +z

1−z =q(z), z∈U, and q is the best dominant.

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Remark1. Using the properties of the Dziok-Srivastava linear operator, for m= 0, l= 1, α₁ = 1, we have H₀¹[1]f(z) =f(z), H₀¹[2]f(z) =zf⁰(z), and using this in Theorem 1 we obtain the result below.

Corollary 1. If f ∈A, α1≥0 and zf⁰⁰(z)

f⁰(z) + (1−α₁)zf⁰(z) f(z) +α₁

≺ 1 +z

1−z + 2z

(1−z)[1 +γ+ (1−γ)z] =h(z), z∈U, 0≤γ <1, then

zf⁰(z)

f(z) ≺ 1 +z

1−z =q(z), z∈U.

Remark 2. Since q(z) = ^1+z_1−z, z ∈ U, is univalent and q(U) = {w ∈ C, Rew >0}, from Theorem 1 we get

Corollary 2. If f ∈A, α₁≥0, and zf⁰⁰(z)

f⁰(z) + (1−α₁)zf⁰(z) f(z) +α₁

≺h(z) = 1 +z

1−z + 2z

[1 +γ+ (1−γ)z](1−z), z∈U, 0≤γ <1, then

Rezf⁰(z)

f(z) >0, z∈U, i.e. f ∈S^∗. Remark3. Ifα₁= 1 then from Theorem 1 we get Corollary 3. If f ∈A and

zf⁰⁰(z)

f⁰(z) + 1≺ 1 +z

1−z + 2z

(1−z)[1 +γ+ (1−γ)z] =h(z), z∈U, 0≤γ <1, then

Rezf⁰(z)

f(z) >0, z∈U, i.e. f ∈S^∗. Remark4. Ifα₁= 0 then from Theorem 1 we get Corollary 4. If f ∈A and

zf⁰⁰(z)

f⁰(z) +zf⁰(z)

f(z) ≺h(z)

= 1 +z

1−z + 2z

(1−z)[1 +γ+ (1−γ)z], z∈U, 0≤γ <1,

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then

Rezf⁰(z)

f(z) >0, z∈U, i.e., f ∈S^∗.

Theorem 2. Let l, m∈N0 ={0,1,2, . . .}, l≤m+ 1, αi ∈C, i= 1,2, . . . , l, βj ∈ C\ {0,−1,−2, . . .}, j = 1,2, . . . , m and H_m^l [α1]f(z) the Dziok- Srivastava linear operator given by (1).

If p∈ H[1,1]∩Q, p(U)⊂Dand z{H_m^l [α1+ 1]f(z)}⁰

H_m^l [α1+ 1]f(z) + (1−α1)H_m^l [α1+ 1]f(z)

H_m^l [α1]f(z) +α1−1 is univalent in U, then

(9) 1 +z+ z

1 +γ+z ≺ z{H_m^l [α₁+ 1]f(z)}⁰ H_m^l [α₁]f(z) +(1−α₁)H_m^l [α₁+ 1]f(z)

H_m^l [α₁]f(z) +α−1 implies

1 +z≺p(z) := H_m^l [α₁+ 1]f(z)

H_m^l [α1]f(z) , z∈U, 0≤γ <1, and q(z) = 1 +z is the best subordinant.

Proof. Using (4) and (7), subordination (9) becomes

(10) 1 +z+ z

1 +γ+z ≺p(z) + zp⁰(z)

p(z) , z∈U, 0≤γ <1.

In order to prove the theorem, we use Lemma B. For that, letq(z) = 1+z, q(0) = 1, q(U) ={w∈C: |w−1|<1},q(U)⊂D.

Define the functionsθ:D⊃q(U)→C,θ(w) =w, and ϕ:D⊃q(U)→C, ϕ(w) = 1

w+γ, with φ(w)6= 0, w ∈q(U).We have

θ[q(z)] =q(z) = 1 +z, z∈U, and

ϕ[q(z)] = 1

q(z) +γ = 1

1 +γ+z, z∈U, 0≤γ <1.

We deduce that

Q(z) =zq⁰(z)ϕ(q(z)) = z

1 +γ+z, z∈U, 0≤γ <1,

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and

h(z) =θ[q(z)] +Q(z) = 1 +z+ z

1 +γ+z, z∈U, 0≤γ <1.

We now calculate RezQ⁰(z)

Q(z) = Re

1− z

1 +γ+z

= Re 1 +γ 1 +γ+z

= Re (1 +γ) 1 +γ+ cost

(1 +γ)²+ 2(1 +γ) cost+ 1 >0, z∈U, 0≤γ <1, hence Qis starlike and univalent in U.

We evaluate Reθ⁰[q(z)]

ϕ[q(z)] = Re (1 +γ+z) = 1 +γ+ Rez >0, z∈U, 0≤γ <1.

Since the conditions in Lemma B are satisfied, by using it we obtain 1 +z≺p(z) = H_m^l [α₁+ 1]f(z)

H_m^l [α1]f(z) , and so q(z) = 1 +z is the best subordinant.

Remark5. Using Theorems 1 and 2 we can state the sandwich-type result below.

Theorem 3. Let l, m∈N0 ={0,1,2, . . .}, l≤m+ 1, α_i∈C, p_j ∈C\ {0,−1,−2, . . .}, j = 1,2, . . . , m and H_m^l [α1]f(z) the Dziok-Srivastava linear operator given by (1).

If f ∈A, α₁ ≥0, ^H^m^l ^[α_z¹^]f^(z) 6= 0 in U and 1 +z+ z

1 +γ+z ≺ z{H_m^l [α1+ 1]f(z)}⁰

H_m^l [α1]f(z) + (1−α1)H_m^l [α1+ 1]f(z)

H_m^l [α1]f(z) +α1−1

≺ 1 +z

1−z + 2z

(1−z)[1 +γ+ (1−γ)z], z∈U, 0≤γ <1, then

1 +z≺ H_m^l [α1+ 1]f(z)

H_m^l [α₁]f(z) ≺ 1 +z

1−z, z∈U.

REFERENCES

[1] J. Dziok and H.M. Srivastava,Classes of analytic functions associated with the genera- lized hypergeometric function. Appl. Math. Comput.103(1999), 1–3.

[2] S.S. Miller and P.T. Mocanu,Second order differential inequalities in the complex plane.

J. Math. Anal. Appl.65(1978), 298–305.

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[3] S.S. Miller and P.T. Mocanu,Differential subordinations and univalent functions. Michi- gan Math. J.28(1981), 157–171.

[4] S.S. Miller and P.T. Mocanu, Differential Subordinations. Theory and Applications.

Marcel Dekker, Inc., New York, 2000.

[5] S.S. Miller and P.T. Mocanu, Subordinants of differential superordinations. Complex Var. Theory Appl.48(2003), 815–826.

[6] S.S. Miller and P.T. Mocanu,Briot-Bouquet differential superordinations and sandwich theorems. J. Math. Anal. Appl.329(2007), 327–335.

Received 1 Octomber 2008 University of Oradea

Department of Mathematics Universit˘at¸ii No.1 410087 Oradea, Romania georgia oros [email protected]