Let ¥ be the class of all analytic univalent functions having power series extensions of the form

in the open unit disk U={z∈ℂ: |z|<1} when normalized with the condition f(0)=f′(z)−1=0.

The Hadamard product (or convolution) (f1∗f2)(z) of the functions

is given by

For two functions f1 and f2 which are analytic in U, we say that the function f1 is subordinate to f2, and denote this by

if there exists a Schwarz function ζ(z) analytic in U with ζ(0)=0 and |ζ(z)|<1(z∈U) such that f1(z)=f2(ζ(z)), for z∈U. See [3]. In particular, if f2(z) is univalent in U, we get the following equivalence

Now, for function f(z) of the form (1.1) we define the series expansion of the linear operator ℨτ,σx(u,v,y)f(z) of the function as

for x∈ℤ,σ>−1,τ>0,y>0 and Re(u)>Re(v)>−y.

The operator ℨτ,σx(u,v,y)f(z) was introduced in [9] generalizes some known operators as follows

(i) ℨ1,0x(u,u,y)=Dx(x∈ℕ0={0,1,...})(Salagean [10]),

(ii) ℨτ,0x(u,u,y)=Dτx(x∈ℕ0)(Al-Oboudi [1]),

(iii) ℨτ,σ0(u,v+t,1)=Gut(t>0,u>−1)(Gao et al. [4], Jung et al.[5]),

(iv) ℨτ,σx(u,0,1)=Jx(τ,u,σ)(x∈ℕ0)(Catas [2]),

(v) ℨ1,u−x(u,u,y)=Lu+1x(x∈ℕ0,u≥0)(Komatu [6]),

(vi) ℨτ,0x(u,v,1)=Dτx(u+1,v+1)(x∈ℕ0)(Selvaraj and Karthikeyan [11]).

From (1.5), it is easy to show that

Using the concept of subordination in (1.4) and the operator ℨτ,σx(u,v,y)f(z), we investigate the subclass of ¥ defined as follows.

Definition 1.1. A function f∈¥ is in the class Tℜτ,σx(u,v,y;γ;V,W) if it satisfies

for V,W∈ℝ with −1≤V<W≤1 and 0≤γ<1.

The following lemmas will be useful in deriving our results.

Lemma 1.2. ([8]) If −1≤V<W≤1, λ>0 and η∈ℂ is restricted by

then the differential equation

has a univalent solution given by

If ψ is regular in U and satisfies the differential subordination

then ψ(z)≺g(z)≺1+Vz1+Wz and g is the best dominant of the above subordination.

Lemma 1.3.([7]) Let g be univalent in U and Ψ, ϑ be analytic functions in a domain D containing g(D) with Ψ(ρ)≠0 for ρ∈g(D). Set T(z)=zg′(z)Ψ(g(z)) and g(z)=ϑ(g(z))+T(z). Suppose that

(i) T is starlike univalent in U,

(ii) Rezg′(z)T(z)>0, z∈U.

If h is analytic in U for h(0)=g(0), h(U)⊆D and

then

and g is the best dominant.

For x∈ℤ, σ>−1, τ>0, y>0, −1≤V<W≤1 and 0≤γ<1, the following theorems are obtained.

Theorem 2.1. Let f∈Tℜτ,σx(u+1,v,y;γ;V,W) such that ℨτ,σx(u,v,y)f(z)≠0, ∀z∈U∗=U\{0} and (1−γ)(V−1)≤(1−W)γ+uy, then

where

and

Further, g1(z) is the best dominant of (2.1).

Proof. Since f∈Tℜτ,σx(u+1,v,y;γ;V,W), if we consider the function

then ψ(z) is analytic in U and ψ(0)=1. Using identity (1.6),yields

then differentiating (2.5) with respect to z and multiplying by z, we have

Applying Lemma 1.2 for λ=1−γ and η=γ+uy, we get

where g1(z) defined in (2.2) and it is the best dominant. Hence, the proof is completed.

Theorem 2.2. If f∈Tℜτ,σx+1(u,v,y;γ;V,W) is such that ℨτ,σx(u,v,y)f(z)≠0, ∀z∈U∗=U\{0} and (1−γ)(V−1)≤(1−W)σ+1τ−1+γ, then

such that

and

Further, g2(z) is the best dominant of (2.6).

Proof. Suppose that f∈Tℜτ,σx+1(u,v,y;γ;V,W), define

Thus, the function ψ(z) is analytic in U and ψ(0)=1. From (1.7), the equation (2.9) becomes

then differentiating (2.10) with respect to z and multiplying by z, we obtain

By make use of Lemma 1.2 for λ=1−γ and η=σ+1τ−1+γ, we get

where g2(z) defined in (2.7) and it is the best dominant. Thus, the proof is completed.

Theorem 2.3. If f∈Tℜτ,σx(u,v,y;γ;V,W) such that ℨτ,σx(u,v+1,y)f(z)≠0, ∀z∈U∗=U\{0} and (1−γ)(V−1)≤(1−W)vy+γ, then

where

and

Further, g3(z) is the best dominant of (2.11).

Proof. Suppose that f∈Tℜτ,σx(u,v,y;γ;V,W) and consider that the function

Subsequently, ψ(z) is analytic in U together ψ(0)=1. Applying (1.8), yields

By differentiation (2.15) with respect to z and multiplying by z, we have

Thus, from Lemma 1.2 for λ=1−γ and η=vy+γ, we obtain

where g3(z) defined in (2.12) and it is the best dominant.

Theorem 2.4. Suppose that g(z) is a univalent function in U with g(0)=1 and

Let f∈¥ satisfy the condition

where θ, κ∈ℂ and θ+κ≠0. If

then

for δ∈ℂ\{0}, and g(z) is the best dominant of (2.17).

Proof. From Lemma 1.3, we can prove the results above for

Since T′(0)=g′(0)≠0, from (2.16) the function T is a starlike univalent in U and

Hence, both (i) and (ii) of Lemma 1.3 are satisfied. Consider the function h be defined by

subsequently, the function h is analytic in U, h(0)=g(0)=1, and

from (2.20),(2.17) becomes

equivalent to

Thus, applying Lemma 1.3 it follows that h(z)≺g(z) and that g is the best dominant of (2.17). Taking θ=0, κ=1 and g(z)=1+Vz1+Wz(−1≤V<W≤1), it is easy to show

From Theorem 2.4, we obtain the following result.

Corollary 2.5. Let f∈¥ satisfy the condition

If

then

with δ∈ℂ\{0} and 1+Vz1+Wz is the best dominant of (2.21).

We took V=1 and W=−1 in Corollary 2.5 as a special case and we got:

Corollary 2.6. Let f∈¥ satisfies the condition

If

then

For δ∈ℂ\{0} and 1+z1−z is the best dominant of (2.22).

Taking θ=1 and κ=0 and g(z)=1+Vz1+Wz (−1≤V<W≤1) in Theorem 2.4, it follows that.

Corollary 2.7. Suppose that f∈¥ satisfies the condition

If

then

For δ∈ℂ\{0} and 1+Vz1+Wz is the best dominant of (2.23).

We took V=1 and W=−1 in Corollary 2.7 as a special case and we got:

Corollary 2.8. Let f∈¥ satisfies the condition

If

then

For δ∈ℂ\{0} and 1+z1−z is the best dominant of (2.24).

In the present work, we were able to obtain the best results, or best dominants of the subordination. Our main results give an interesting process for the study of many analytic univalent classes earlier defined by several authors. These classes expand and generalize many of those defined by many specialists in this field. Furthermore, the general subordination theorems lead us to some special cases that were used to determine new results connected with the classes we investigated.