Paper10-Acta25-2011. 125KB Jun 04 2011 12:03:15 AM

Acta Universitatis Apulensis
ISSN: 1582-5329

No. 25/2011
pp. 133-144

FIRST ORDER LINEAR DIFFERENTIAL SUBORDINATIONS FOR
A GENERALIZED OPERATOR

Afaf Ali Abubaker and Maslina Darus
Abstract. In this work, we obtain some results for first order linear strong
differential subordination for new subclass of generalized operator.
2000 Mathematics Subject Classification: 30C45.
1. Introduction And Preliminaries
Let H = H(U ) denote the class of functions analytic in U . For n, a positive
integer and a ∈ C, and let
H[a, n] = {f ∈ H; f (z) = a + an z n + . . . , z ∈ U }.
Let A be the class of functions f of the form
f (z) = z + a2 z 2 + . . . , z ∈ U
which are analytic in the unit disk. Let S ∗ and C be the class of starlike and convex
functions respectively given by

S ∗ = {f ∈ A, ℜ

zf ′ (z)
> 0}
f (z)

and
C = {f ∈ A, ℜ{1 +

zf ′′ (z)
} > 0}.
f ′ (z)

For f ∈ A, Ruscheweyh [6] considered the following generalized integral operator
δ+γ
f (z) = Iδ,γ (F )(z) = [ γ
z

Zz

1

tγ−1 F δ (t) dt] δ , (δ > 0, ℜ γ > 0).

0

We may also see some other results for other types of integral operators studied
by different authors, namely Pescar and Breaz [9], El-Ashwah and Aouf[10], few to
mention.
133

A. A. Abubaker and M. Darus - First order linear differential subordinations...

Let
k
Dα,β,λ,µ
f (z)

=

∞
X

[β(n − 1)(λ − α) + 1]k G(µ, n) an z n

n=2

where z ∈ U for α ≥ 0, β ≥ 0, λ ≥ 0 and k ∈ N0 = N ∪ {0}, µ ≥ 0.
Remark 1.3.
1. When α = 0, β = 1 and λ = 1 ,µ = 0, we get S˜
al˜
agean differential operator
(see [8]).
2. When k = 0 gives Ruscheweyh (see [7]).
3. When α = 0, β = 1 and λ = 1 and µ = 0, we get Al-Oboudi differential
operator (see [3]).
4. When µ = 0, the operator reduces to Darus and Ibrahim (see [2]).
k
Definition 1.4.Let f ∈ A, then f ∈ Sα,β,λ,µ
if and only if for z ∈ U :

ℜ{

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

} > 0, .

In our present investigation of the first order linear strong differential subordination for new subclass of generalized operator, we need the following definitions and
lemmas:
¯ and let f analytic and univalent
Definition 1.5. [1]Let H(z, ξ) be analytic in U × U
in U . The function H(z, ξ) is strongly subordinate to f , written H(z, ξ) ≺≺ f (z) if
¯ , the function of z,H(z, ξ) is subordinate to f .
for each ξ ∈ U
Remark 1.6. Since f is analytic and univalent, Definition 1.4 is equivalent to

¯ ) ⊂ f (U ).
H(0, ξ) = f (0) and H(U × U
Definition 1.7. [4]We denote by Q the set of functions f that are analytic and
¯ \E(f ),where
injective in U
E(f ) = {ζ ∈ ∂U ; lim f (z) = ∞}
z→ζ

and are such that f ′ (ζ) 6= 0 for ζ ∈ ∂U \E(f ). The subclass of Q for which f (0) = a
is denoted by Q(a).
134

A. A. Abubaker and M. Darus - First order linear differential subordinations...

Definition 1.8. [5] Let Ω be a set in C, q ∈ Q and n be a positive integer. The
¯ that
class of admissible functions ψn [Ω, q] consists of those functions ψ : C 2 × U × U
satisfy the admissibility condition:

ψ(r, s; z, ξ) ∈

/Ω
¯ , ζ ∈ ∂U \E(f ) and m ≥ n.
whenever r = q(ζ), s = mζq ′ (ζ), z ∈ U, ξ ∈ U
Lemma 1.9. [4] Let q ∈ Q(a), with q(0) = a and let p(z) = a + an z n + · · · be
analytic in U , with p(z) 6= a and n ≥ 1. If p is not subordinate to q, then there exist
points z0 = r0 eiθ0 ∈ U and ζ0 ∈ ∂U \E(q), and an m ≥ n ≥ 1
1. p(z0 ) = q(ζ0 )
2. z0 p′ (z0 ) = mζ0 q(ζ0 ).
2.Main Results
By referring to Definition 1.5, we state the first order linear strong differential subordination for new subclass of generalized operator as follows:
Definition 2.1.A strong differential subordination for some subclasses of generalized
operator of the form

A(z, ξ)[

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ

f (z)

B(z, ξ)

where p(z) =

+

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ

f (z)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z) )2

¯
≺≺ h(z), z ∈ U, ξ ∈ U

]+

(1)

and ψ(r, s; z, ξ) = A(z, ξ)zp′ (z) + B(z, ξ)p(z) is analytic

¯ and h(z) is analytic in U is called first order linear strong
in U for all ξ ∈ U
differential subordination for new subclass of generalized operator.
Remark 2.2. If A(z, ξ) = 1, B(z, ξ) = 0 and h(z) is convex function then (1)
becomes
k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

+

k
z(Dα,β,λ,µ
f (z))′

135

k
Dα,β,λ,µ
f (z)

A. A. Abubaker and M. Darus - First order linear differential subordinations...

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ

f (z))2

≺ h(z), z ∈ U.

k
z(Dα,β,λ,µ
f (z))′

¯ → C, B : U × U
¯ → C with
∈ H[0, n], A : U × U
¯ and
ψ(r, s; z, ξ) analytic in U for all , ξ ∈ U

Theorem 2.3. Let

k
Dα,β,λ,µ
f (z)

ℜ[nA(z, ξ) + B(z, ξ)] ≥ 1, ℜA(z, ξ) ≥ 0.
If
A(z, ξ)[

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

] + B(z, ξ)

+

k
z(Dα,β,λ,µ
f (z))′

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

k
Dα,β,λ,µ
f (z)

¯ , M > 0 (2)
≺≺ (n2 + 1)M z , z ∈ U, ξ ∈ U

then
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

≺ M z.

¯→ C
Proof. Let ψ : C 2 × U × U
ψ(r, s; z, ξ) = A(z, ξ)[

−

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

] + B(z, ξ)

+

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

k
f (z))′
z(Dα,β,λ,µ
k
Dα,β,λ,µ
f (z)

,

and (2) becomes
ψ(r, s; z, ξ) ≺≺ (n2 + 1)M z

(3)

since h(z) = (n2 + 1)M z, it gives h(z) = U (0, (n2 + 1)M ). In this case (3) is
equivalent to

136

A. A. Abubaker and M. Darus - First order linear differential subordinations...

ψ(r, s; z, ξ) ∈ U (0, (n2 + 1)M ).

(4)

Suppose that
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

is not subordinated to q(z) = M z. Then by using Lemma 1.9, we have that there
exist z0 ∈ U and ζ0 ∈ ∂U such that
k
z0 (Dα,β,λ,µ
f (z0 ))′
k
Dα,β,λ,µ
f (z0 )

= q(z0 ) = M eiθ0 ,

where θ0 ∈ R when |ζ0 | = 1 and
k
z02 (Dα,β,λ,µ
f (z0 ))′′
k
Dα,β,λ,µ
f (z0 )

−

k
f (z0 ))′2
z02 (Dα,β,λ,µ
k
(Dα,β,λ,µ
f (z0 ))2

+

k
z0 (Dα,β,λ,µ
f (z0 ))′
k
Dα,β,λ,µ
f (z0 )

= m ζ0 h′ (ζ0 ) = K eiθ0 , K ≥ nM .

Hence we obtain






|ψ(r, s; z0 , ξ)| = A(z0 , ξ) K eiθ0 + B(z0 , ξ) M eiθ0 
=
≥

|A(z0 , ξ) K + B(z0 , ξ) M |

ℜ |A(z0 , ξ) K + B(z0 , ξ) M | ≥ K ℜA(z0 , ξ) + M ℜB(z0 , ξ)
≥ M [ n ℜA(z0 , ξ) + ℜB(z0 , ξ) ] ≥ M,

Since this result contradicts (4), we conclude that the assumption made concerning
the subordination relation between p and q is false, hence
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

≺ M z, z ∈ U.

137

A. A. Abubaker and M. Darus - First order linear differential subordinations...

When φ(z) = 1 in Theorem 2.3 we get:
Corollary 2.4. Let

A(z, ξ)[

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

+

¯ → C, B : U × U
¯ → C with
∈ H[0, n], A : U × U

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

] + B(z, ξ)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

¯ and
analytic function in U for all , ξ ∈ U
ℜ[nA(z, ξ) + B(z, ξ)] ≥ 1, ℜA(z, ξ) ≥ 0.
If
A(z, ξ)[

k
z 2 (Dα,β,λ,µ
f (z))′′

+B(z, ξ)

k
Dα,β,λ,µ
f (z)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

+

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

]

¯, M > 0
≺≺ (n2 + 1)M z , z ∈ U, ξ ∈ U

(5)

then
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

≺ M z.

For k = 0, we get the following corollary.
Corollary 2.5. Let
A(z, ξ)[

z f ′ (z)
f (z)

¯ → C, B : U × U
¯ → C with
∈ H[0, n] and A : U × U

z 2 f ′′ (z) z f ′ (z)
z 2 (f ′ (z))2
zf ′ (z)
]
+
B(z,
ξ)
+
−
,
f (z)
f (z)
(f (z))2
f (z)

¯ and
analytic function in U for all , ξ ∈ U
ℜ[nA(z, ξ) + B(z, ξ)] ≥ 1, ℜA(z, ξ) ≥ 0.
If

A(z, ξ)[

z 2 f ′′ (z) z f ′ (z)
z 2 (f ′ (z))2
+
−
]
f (z)
f (z)
(f (z))2
138

,

A. A. Abubaker and M. Darus - First order linear differential subordinations...

+B(z, ξ)

zf ′ (z)
¯ , M > 0,
≺≺ (n2 + 1) M z, z ∈ U, , ξ ∈ U
f (z)

(6)

then
z f ′ (z)
≺ M z.
f (z)
k
z(Dα,β,λ,µ
f (z))′

¯ → C, B : U × U
¯→ C
∈ H[1, n] and A : U × U
¯ and
with ψ(r, s; z, ξ) a function of z, analytic in for any ξ ∈ U
Theorem 2.6. Let

k
Dα,β,λ,µ
f (z)

ℜA(z, ξ) ≥ 0, ℑB(z, ξ) ≤ n ℜA(z, ξ).
If
ℜ{A(z, ξ)[

−

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

k
z(Dα,β,λ,µ
f (z))′

+

] + B(z, ξ)

k
Dα,β,λ,µ
f (z)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

}>0

(7)

then
ℜ

k
f (z))′
z(Dα,β,λ,µ
k
Dα,β,λ,µ
f (z)

> 0, z ∈ U.

¯ → C,
Proof. Let ψ : C 2 × U × U
ψ(r, s; z, ξ) = A(z, ξ)[

−

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

k
f (z))′2
z 2 (Dα,β,λ,µ
k
(Dα,β,λ,µ
f (z))2

] + B(z, ξ)

+

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

k
f (z))′
z(Dα,β,λ,µ
k
Dα,β,λ,µ
f (z)

,

In this case, (7) becomes
¯.
ℜψ(r, s; z, ξ) > 0, z ∈ U, ξ ∈ U
139

(8)

A. A. Abubaker and M. Darus - First order linear differential subordinations...

Since h(z) =

1+Az
1+Bz ,

−1 ≤ B < A ≤ 1, and
h(U ) = {ω ∈ C : ℜω(z) > 0}

from which we have that (8) becomes
ψ(r, s; z, ξ) ≺
Suppose ℜ{
h(z) =

k
z(Dα,β,λ,µ
f (z))′

1+Az
1+Bz ,

k
Dα,β,λ,µ
f (z)

1 + Az
, −1 ≤ B < A ≤ 1.
1 + Bz

} < 0, meaning p(z) =

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

is not subordinate to

−1 ≤ B < A ≤ 1. Using Lemma 1.5, we have that there exist z0 ∈ U

and ζ0 ∈ ∂U with |ζ0 | = 1 such that

k
z0 (Dα,β,λ,µ
f (z0 ))′
k
φ(z0 )Dα,β,λ,µ
f (z0 )

k
z02 (Dα,β,λ,µ
f (z0 ))′′
k
Dα,β,λ,µ
f (z0 )

−

+

k
z02 (Dα,β,λ,µ
f (z0 ))′2
k
(Dα,β,λ,µ
f (z0 ))2

= h(ζ0 ) = ρi, and

k
z0 (Dα,β,λ,µ
f (z0 ))′
k
Dα,β,λ,µ
f (z0 )

= m ζ0 h′ (ζ0 ) = σ,

where ρ, σ ∈ R and σ ≤ −n(1 + ρ2 ), n ≥ 1. Then we obtain:
ℜψ(r, s; z0 , ξ) = ℜψ(ρi, σ; z0 , ξ)
= ℜ[A(z0 , ξ) σ + B(z0 , ξ)ρi] = ℜ{A(z0 , ξ) σ + [B(z0 , ξ) + i B(z0 , ξ)]ρi}
= σ ℜA(z0 , ξ) − ρℑB(z0 , ξ) ≤ −n(1 + ρ2 )ℜA(z0 , ξ) − ρℑB(z0 , ξ)
n
n
≤ − ρ2 ℜA(z0 , ξ) − ρℑB(z0 , ξ) − ≤ 0.
2
2
Hence ℜψ(r, s; z0 , ξ) ≤ 0 which contradicts (2.8) and we conclude that
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

> 0, z ∈ U.

When k = 0, we get
140

A. A. Abubaker and M. Darus - First order linear differential subordinations...
′ (z)
¯ → C, B : U × U
¯ → C with
∈ H[1, n] and A : U × U
Corollary 2.7. Let zff(z)
¯ and
ψ(r, s; z, ξ) a function of z, analytic in for any ξ ∈ U

ℜA(z, ξ) ≥ 0, ℑB(z, ξ) ≤ n ℜA(z, ξ).
If
A(z, ξ)[

z 2 f ′′ (z) z f ′ (z)
z 2 (f ′ (z))2
zf ′ (z)
+
−
]
+
B(z,
ξ)
>0
f (z)
f (z)
(f (z))2
f (z)

then
zf ′ (z)
> 0, z ∈ U.
f (z)
k
z(Dα,β,λ,µ
f (z))′

¯ → C, B : U × U
¯ → C with
∈ H[0, n], A : U × U
¯ and
ψ(r, s; z, ξ) analytic in U for all ξ ∈ U

Theorem 2.8. Let

k
Dα,β,λ,µ
f (z)

ℜ A(z, ξ) ≥ 0, ℑB(z, ξ) ≤ nℜ A(z, ξ)[−nℜ A(z, ξ) + z].
If

A(z, ξ)[

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

] + B(z, ξ)

+

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

¯, M > 0
≺≺ M z, z ∈ U, , ξ ∈ U

then
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

≺

1 + Az
, −1 ≤ B < A ≤ 1, z ∈ U.
1 + Bz

¯ → C, and
Proof. Let ψ : C 2 × U × U
ψ(r, s; z, ξ) = A(z, ξ)[

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

141

(9)

+

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

A. A. Abubaker and M. Darus - First order linear differential subordinations...

−

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

] + B(z, ξ)

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

,

then (9) becomes
ψ(r, s; z, ξ) ≺≺ M z

(10)

since h(z) = M z, we have h(z) = U (0, M ). Thus
¯.
ψ(r, s; z, ξ) ⊂ U, z ∈ U, , ξ ∈ U
Suppose that

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

is not subordinated to q(z) =

(11)
1+Az
1+Bz ,

−1 ≤ B < A ≤ 1.

then ,by using Lemma 1.9, we have that there exist z0 ∈ U and ζ0 ∈ ∂U such that
k
z(Dα,β,λ,µ
f (z))′

= q(ζ0 ) = υi,

k
Dα,β,λ,µ
f (z)

k
z 2 (Dα,β,λ,µ
f (z))′′
k
Dα,β,λ,µ
f (z)

−

+

k
z 2 (Dα,β,λ,µ
f (z))′2
k
(Dα,β,λ,µ
f (z))2

where υ, ρ ∈ R and ρ ≤

n
2 (1

k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

] = mζ0 q ′ (ζ0 ) = ρ

+ υ 2 ), n ≥ 1. Then we obtain

ℜ ψ(r, s; z0 , ξ) = ℜ ψ(υi, ρ; z0 , ξ)
= ℜ [A(z0 , ξ) ρ + B(z0 , ξ) υi] = ρℜ A(z0 , ξ) − υ ℑB(z0 , ξ)

n
≤ − (1 + υ 2 )ℜ A(z0 , ξ) − υ ℑB(z0 , ξ)
2
n
n
≤ − ℜ A(z0 , ξ) − υ ℑB(z0 , ξ) − υℜ A(z0 , ξ) ≤ −1.
2
2
Hence, we have
ℜ ψ(r, s; z0 , ξ) ≤ −1,
142

A. A. Abubaker and M. Darus - First order linear differential subordinations...

which contradicts (11) and we conclude that
k
z(Dα,β,λ,µ
f (z))′
k
Dα,β,λ,µ
f (z)

≺

1 + Az
, −1 ≤ B < A ≤ 1, z ∈ U.
1 + Bz

When k = 0 , M = 1 and q(z) =
Corollary 2.9. Let
A(z, ξ) [

z f ′ (z)
f (z)

1+z
1−z

in Theorem 2.8 we get:

∈ H[1, n] and

z 2 (f ′ (z))2
zf ′ (z)
z 2 f ′′ (z) z f ′ (z)
+
−
]
+
B(z,
ξ)
f (z)
f (z)
(f (z))2
f (z)

¯ and
analytic function in U for all ξ ∈ U
ℜ A(z, ξ) ≥ 0, ℑB(z, ξ) ≤ nℜ A(z, ξ)[−nℜ A(z, ξ) + z].
If
A(z, ξ) [

z 2 (f ′ (z))2
zf ′ (z)
z 2 f ′′ (z) z f ′ (z)
¯
+
−
]
+
B(z,
ξ)
≺≺ z, z ∈ U, , ξ ∈ U
f (z)
f (z)
(f (z))2
f (z)

then
z f ′ (z)
1+z
≺
, z ∈ U.
f (z)
1−z
Acknowledgement: This work is partially supported by UKM-ST-06-FRGS01072009, MOHE Malaysia.

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A. A. Abubaker and M. Darus - First order linear differential subordinations...

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Afaf Abubaker
School of Mathematical Sciences
Faculty of Science and Technology
Universiti Kebangsaan Malaysia
Bangi 43600 Selangor D. Ehsan, Malaysia
email:m.afaf48@yahoo.com
Maslina Darus
School of Mathematical Sciences
Faculty of Science and Technology
Universiti Kebangsaan Malaysia
Bangi 43600 Selangor D. Ehsan, Malaysia
email:maslina@ukm.my

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