axioms Article
Subordination Properties for Multivalent Functions Associated with a Generalized Fractional Differintegral Operator Hanaa M. Zayed 1, * 1 2
*
ID
, Mohamed Kamal Aouf 2 and Adela O. Mostafa 2
Department of Mathematics, Faculty of Science, Menofia University, Shebin Elkom 32511, Egypt Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt;
[email protected] (M.K.A.);
[email protected] (A.O.M.) Correspondence:
[email protected]
Received: 19 January 2018; Accepted: 17 April 2018; Published: 24 April 2018
Abstract: Using of the principle of subordination, we investigate some subordination and convolution properties for classes of multivalent functions under certain assumptions on the parameters involved, which are defined by a generalized fractional differintegral operator under certain assumptions on the parameters involved. Keywords: differential subordination; p-valent functions; generalized fractional differintegral operator JEL Classification: 30C45; 30C50
1. Introduction and Definitions Denote by A( p) the class of analytic and p-valent functions of the form: f (z) = z p +
∞
∑ a p+n z p+n ( p ∈ N = {1, 2, ...}; z ∈ U = {z ∈ C : |z| < 1}).
(1)
n =1
For functions f , g analytic in U, f is subordinate to g, written f (z) ≺ g(z) if there exists a function w, analytic in U with w(0) = 0 and |w(z)| < 1, such that f (z) = g(w(z)), z ∈ U. If g is univalent in U, then (see [1,2]): f (z) ≺ g(z) ⇔ f (0) = g(0) and f (U) ⊂ g(U). If ϕ(z) is analytic in U and satisfies: H ( ϕ(z), zϕ0 (z)) ≺ h(z),
(2)
then ϕ is a solution of (2). The univalent function q is called dominant, if ϕ(z) ≺ q(z) for all ϕ. A dominant qe is called the best dominant, if qe(z) ≺ q(z) for all dominants q. Let 2 F1 (a, b; c; z) (c 6= 0, −1, −2, . . . ) be the well-known (Gaussian) hypergeometric function defined by: ∞ ( a)n (b)n n z , z ∈ U, 2 F1 ( a, b; c; z ) : = ∑ ( c ) n (1) n n =0 where:
(λ)n :=
Γ (λ + n) . Γ (λ)
We will recall some definitions that will be used in our paper.
Axioms 2018, 7, 27; doi:10.3390/axioms7020027
www.mdpi.com/journal/axioms
Axioms 2018, 7, 27
2 of 13
Definition 1. For f (z) ∈ A( p), the fractional integral and fractional derivative operators of order λ are defined by Owa [3] (see also [4]) as: Dz−λ f (z) := Dzλ f (z) :=
1 Γ(λ)
Z z
d 1 Γ(1 − λ) dz
0
f (ζ ) dζ ( z − ζ )1− λ
Z z 0
f (ζ ) dζ (z − ζ )λ
( λ > 0),
(0 ≤ λ < 1),
where f is an analytic function in a simply-connected region of the complex z-plane containing the origin, and the multiplicity of (z − ζ )λ−1 ((z − ζ )−λ ) is removed by requiring log(z − ζ ) to be real when z − ζ > 0. Definition 2. For f (z) ∈ A( p) and in terms of 2 F1 , the generalized fractional integral and generalized fractional derivative operators defined by Srivastava et al. [5] (see also [6]) as: Z z−λ−µ z ζ λ −1 := (z − ζ ) f (ζ ) 2 F1 µ + λ, −η; λ; 1 − dζ (λ > 0, µ, η ∈ R) , Γ(λ) 0 z ( ) Rz zλ−µ 0 (z−ζ )−λ f (ζ )2 F1 µ−λ,1−η;1−λ;1− ζz dζ d (0 ≤ λ < 1), dz Γ (1− λ ) λ,µ,η J0,z f (z) := n λ−n,µ,η d J f (z) (n ≤ λ < n + 1; n ∈ N), dzn 0,z
λ,µ,η I0,z f (z)
where f (z) is an analytic function in a simply-connected region of the complex z−plane containing the origin with the order f (z) = O(|z|ε ), z → 0 when ε > max{0, µ − η } − 1, and the multiplicity of (z − ζ )λ−1 ((z − ζ )−λ ) is removed by requiring log(z − ζ ) to be real when z − ζ > 0. We note that: λ,−λ,η
I0,z
f (z) = Dz−λ f (z) (λ > 0) and J0,z
λ,λ,η
f (z) = Dzλ f (z) (0 ≤ λ < 1),
where Dz−λ f (z) and Dzλ f (z) are the fractional integral and fractional derivative operators studied by Owa [3]. Goyal and Prajapat [7] (see also [8]) defined the operator:
λ,µ,η,p S0,z f (z)
=
Γ( p+1−µ)Γ( p+1−λ+η ) µ λ,µ,η z J0,z f (z) Γ ( p +1) Γ ( p +1− µ + η )
Γ( p+1−µ)Γ( p+1−λ+η ) µ −λ,µ,η z I0,z f (z) Γ ( p +1) Γ ( p +1− µ + η )
(0 ≤ λ < η + p + 1; z ∈ U), (−∞ < λ < 0; z ∈ U).
For f (z) ∈ A( p), we have: λ,µ,η,p
S0,z
f (z)
p
= z 3 F2 (1, 1 + p, 1 + p + η − µ; 1 + p − µ, 1 + p + η − λ; z) ∗ f (z) = zp +
(p
∈
∞
( p + 1) n ( p + 1 − µ + η ) n a p+n z p+n ( p + 1 − µ)n ( p + 1 − λ + η )n n =1
∑
N; µ, η ∈ R; µ < p + 1; −∞ < λ < η + p + 1),
where “∗” stands for convolution of two power series, and q Fs (q ≤ s + 1; q, s ∈ N0 = N ∪ {0}) is the well-known generalized hypergeometric function.
Axioms 2018, 7, 27
3 of 13
Let: G λp,η,µ (z)
(p and:
= zp + ∈
∞
( p + 1) n ( p + 1 − µ + η ) n z p+n ( p + 1 − µ ) ( p + 1 − λ + η ) n n n =1
∑
N; µ, η ∈ R; µ < p + 1; −∞ < λ < η + p + 1).
h i −1 G λp,η,µ (z) ∗ G λp,η,µ (z) =
zp (δ > − p; z ∈ U). (1 − z ) δ + p
λ,δ Tang et al. [9] (see also [10–15]) defined the operator H p,η,µ : A( p) → A( p), where:
λ,δ H p,η,µ f (z)
(p
= zp + ∈
∞
(δ + p)n ( p + 1 − µ)n ( p + 1 − λ + η )n a p+n z p+n (1) n ( p + 1) n ( p + 1 − µ + η ) n n =1
∑
N, δ > − p, µ, η ∈ R, µ < p + 1, −∞ < λ < η + p + 1).
It is easy to verify that:
and:
0 λ,δ λ,δ+1 λ,δ z H p,η,µ f (z) = (δ + p) H p,η,µ f (z) − δH p,η,µ f ( z ),
(3)
0 λ,δ λ+1,δ λ+1,δ f (z) − (η − λ) H p,η,µ f ( z ). z H p,η,µ f (z) = ( p + η − λ) H p,η,µ
(4)
λ,δ By using the operator H p,η,µ , we introduce the following class. λ,δ Definition 3. For A, B (−1 ≤ B < A ≤ 1), f ∈ A( p) is in the class T p,η,µ ( A, B) if λ,δ ( H p,η,µ f (z))0 1 + Az ≺ (z ∈ U; p ∈ N), p − 1 1 + Bz pz
which is equivalent to: λ,δ ( H p,η,µ f (z))0 − 1 pz p−1 < 1 (z ∈ U). λ,δ 0 ( H p,η,µ f ( z )) B − A p − 1 pz λ,δ λ,δ , − 1 = T p,η,µ (ξ ) (0 ≤ ξ < p), which satisfies For convenience, we write T p,η,µ 1 − 2ξ p the inequality: ( λ,δ ) ( H p,η,µ f (z))0 < > ξ (0 ≤ ξ < p ). z p −1 In this paper, we investigate some subordination and convolution properties for classes of multivalent functions, which are defined by a generalized fractional differintegral operator. The theory of subordination received great attention, particularly in many subclasses of univalent and multivalent functions (see, for example, [13,15–17]). 2. Preliminaries To prove our main results, we shall need the following lemmas. Lemma 1. [18]. Let h be an analytic and convex (univalent) function in U with h(0) = 1. Additionally, let φ given by: φ(z) = 1 + cn zn + cn+1 zn+1 + ... (5)
Axioms 2018, 7, 27
4 of 13
be analytic in U. If: φ(z) +
zφ0 (z) ≺ h(z) ( 0; σ 6= 0), σ
then: φ(z) ≺ ψ(z) =
σ −σ z n n
Z z 0
t n −1 h(t)dt ≺ h(z), σ
(6)
(7)
and ψ is the best dominant of (6). Denote by P(ς) the class of functions Φ given by: Φ(z) = 1 + c1 z + c1 z2 + ...,
(8)
which are analytic in U and satisfy the following inequality:
< {Φ(z)} > ς (0 ≤ ς < 1). Using the well-known growth theorem for the Carathéodory functions (cf., e.g., [19]), we may easily deduce the following result: Lemma 2. [19]. If Φ ∈ P(ς). Then
< {Φ(ς)} > 2ς − 1 +
2(1 − ς ) (0 ≤ ς < 1). 1 + |z|
Lemma 3. [20]. For 0 ≤ ς 1 , ς 2 < 1, P(ς 1 ) ∗ P(ς 2 ) ⊂ P(ς 3 ) (ς 3 = 1 − 2(1 − ς 1 )(1 − ς 2 )). The result is the best possible. Lemma 4. [21]. Let ϕ be such that ϕ(0) = 1 and ϕ(z) 6= 0 and A, B ∈ C, with A 6= B, | B| ≤ 1, ν ∈ C∗ . (i ) If ν( AB− B) − 1 ≤ 1 or ν( AB− B) + 1 ≤ 1, B 6= 0 and ϕ(z) satisfies: 1+ then:
zϕ0 (z) 1 + Az ≺ , νϕ(z) 1 + Bz
ϕ(z) ≺ (1 + Bz)ν(
A− B B
)
and this is the best dominant. (ii ) If B = 0 and |νA| < π and if ϕ satisfies: 1+
zϕ0 (z) ≺ 1 + Az, νϕ(z)
then: ϕ(z) ≺ eνAz , and this is the best dominant.
Axioms 2018, 7, 27
5 of 13
Lemma 5. [2]. Let Ω ⊂ C , b ∈ C, < (b) > 0 and ψ : C2 × U → C satisfy ψ (ix, y; z) ∈ / Ω for all x, y ≤ −
|b−ix |2 2 0 in U. Lemma 6. [22]. Let ψ (z) be analytic in U with ψ(0) = 1 and ψ(z) 6= 0 for all z. If there exist two points z1 , z2 ∈ U such that:
−
π π ρ = arg{ψ(z1 )} < arg{ψ(z)} < ρ2 = arg{ψ(z2 )}, 2 1 2
(9)
for some ρ1 and ρ2 (ρ1 , ρ2 > 0) and for all z (|z| < |z1 | = |z2 |) , then: z1 ψ 0 ( z1 ) = −i ψ ( z1 ) where:
ρ1 + ρ2 κ 2
and
z2 ψ 0 ( z2 ) =i ψ ( z2 )
ρ2 − ρ1 ρ2 + ρ1
1 − | a| and a = i tan κ≥ 1 + | a|
ρ1 + ρ2 κ , 2
(10)
.
(11)
λ,δ 3. Properties Involving H p,η,µ
Unless otherwise mentioned, we assume throughout this paper that p ∈ N, δ > − p, µ, η ∈ R, µ < p + 1, −∞ < λ < η + p + 1, −1 ≤ B < A ≤ 1, θ > 0, and the powers are considered principal ones. Theorem 1. Let f ∈ A( p) satisfy:
(1 − θ ) Then:
λ,δ H p,η,µ f (z) θ
Z1
λ,δ+1 H p,η,µ f (z)
0
≺
pz p−1
u
δ+ p θ −1
0
1 − Au 1 − Bu
1 + Az . 1 + Bz
(12)
τ1
du , τ ≥ 1.
(13)
The estimate in (13) is sharp. Proof. Let:
φ(z) =
λ,δ H p,η,µ f (z)
0
(z ∈ U).
pz p−1
(14)
Then, φ is analytic in U. After some computations, we get:
(1 − θ )
λ,δ H p,η,µ f (z)
0
pz p−1
+θ
λ,δ+1 H p,η,µ f (z)
0
= φ(z) +
pz p−1
θzφ0 (z) 1 + Az ≺ . δ+p 1 + Bz
Now, by using Lemma 1, we deduce that:
λ,δ H p,η,µ f (z)
pz p−1
0
≺
δ + p − δ+ p z θ θ
Zz 0
t
δ+ p θ −1
1 + At 1 + Bt
dt,
(15)
Axioms 2018, 7, 27
6 of 13
or, equivalently,
λ,δ H p,η,µ f (z)
0 δ+p = θ
pz p−1 and so:
λ,δ H p,η,µ f (z) θ
u
1 + Auw(z) 1 + Buw(z)
δ+ p θ −1
0
du,
1 − Au 1 − Bu
du .
1 1 < χ τ ≥ (< (χ)) τ (χ ∈ C, < {χ} ≥ 0, τ ≥ 1) .
(16)
(17)
The inequality (13) now follows from (16) and (17). To prove that the result is sharp, let:
λ,δ H p,η,µ f (z)
0
=
pz p−1
δ+p θ
Z1
u
δ+ p θ −1
0
1 + Auz 1 + Buz
du.
(18)
Now, for f (z) defined by (18), we have:
(1 − θ )
λ,δ H p,η,µ f (z)
0
pz p−1
+θ
λ,δ+1 H p,η,µ f (z)
0
=
pz p−1
1 + Az (z ∈ U) , 1 + Bz
Letting z → −1, we obtain:
λ,δ H p,η,µ f (z)
0 δ+p → θ
pz p−1
Z1
u
δ+ p θ −1
0
1 − Au 1 − Bu
du,
which ends our proof. Putting θ = 1 and using Lemma 1 for Equation (15) in Theorem 1, we obtain the following example. Example 1. Let the function f (z) ∈ A( p). Then, following containment property holds, λ,δ+1 λ,δ T p,µ,η ( A, B) ⊂ T p,µ,η ( A, B).
Using (4) instead of (3) in Theorem 1, one can prove the following theorem. Theorem 2. Let f ∈ A( p) satisfy
(1 − θ )
λ+1,δ H p,η,µ f (z)
pz p−1
0
+θ
λ,δ H p,η,µ f (z)
0
≺
pz p−1
1 + Az . 1 + Bz
Then:
λ+1,δ H p,η,µ f (z) θ
Z1 0
u
p+η −λ −1 θ
1 − Au 1 − Bu
The result is sharp. Putting θ = 1 in Theorem 2, we obtain the following example.
τ1 du , τ ≥ 1.
(19)
Axioms 2018, 7, 27
7 of 13
Example 2. Let the function f (z) ∈ A( p). Then, following inclusion property holds λ,δ λ+1,δ T p,µ,η ( A, B) ⊂ T p,µ,η ( A, B).
For a function f ∈ A( p), the generalized Bernardi–Libera–Livingston integeral operator Fp,γ is defined by (see [23]): Fp,γ f (z)
=
Zz
γ+p zp
tγ−1 f (t)dt
0 ∞
=
γ+p zp + ∑ z p+k γ + p+k k =1
!
∗ f (z) (γ > − p)
(20)
= z p 3 F2 (1, 1, γ + p; 1, γ + p + 1; z) ∗ f (z). Lemma 7. If f ∈ A ( p) , prove that: λ,δ λ,δ (i) H p,η,µ Fp,γ f = Fp,γ H p,η,µ f , (ii) λ,δ z H p,η,µ Fp,γ f (z)
0
λ,δ λ,δ = ( p + γ) H p,η,µ f (z) − γH p,η,µ Fp,γ f (z).
(21)
Proof. Since λ,δ H p,η,µ Fp,γ f
= [z p 3 F2 (δ + p, p + 1 − µ, p + 1 − λ + η; p + 1, p + 1 − µ + η; z)] ∗ Fp,γ f
= [z p 3 F2 (δ + p, p + 1 − µ, p + 1 − λ + η; p + 1, p + 1 − µ + η; z)] ∗ [z p 3 F2 (1, 1, γ + p; 1, γ + p + 1; z) ∗ f (z)] , and: λ,δ Fp,γ H p,η,µ f
λ,δ = z p 3 F2 (1, 1, γ + p; 1, γ + p + 1; z) ∗ H p,η,µ f
= z p 3 F2 (1, 1, γ + p; 1, γ + p + 1; z) ∗ [z p 3 F2 (δ + p, p + 1 − µ, p + 1 − λ + η; p + 1, p + 1 − µ + η; z) ∗ f (z)] . Now, the first part of this lemma follows. Furthermore, z Fp,γ f (z)
0
= ( p + γ) f (z) − γFp,γ f (z).
(22)
λ,δ If we replace f (z) by H p,η,µ f (z) and using the first part of this lemma, we get (21) . λ,δ Theorem 3. Suppose that p + γ > 0, f ∈ T p,η,µ ( A, B) and Fp,γ defined by (20). Then:
λ,δ H p,η,µ Fp,γ f (z) ( p + γ)
Z1
u p + γ −1
0
1 − Au 1 − Bu
τ1 du , τ ≥ 1.
(23)
The result is sharp. Proof. Let:
φ(z) =
λ,δ H p,η,µ Fp,γ f (z)
pz p−1
0
(z ∈ U).
(24)
Axioms 2018, 7, 27
8 of 13
Then, φ is analytic in U. After some calculations, we have: λ,δ ( H p,η,µ f (z))0 zφ0 (z) 1 + Az = φ(z) + ≺ . p − 1 p+γ 1 + Bz pz
Employing the same technique that was used in proving Theorem 1, the remaining part of the theorem can be proven. Theorem 4. Let −1 ≤ Bi < Ai ≤ 1 (i = 1, 2). If each of the functions f i ∈ A( p) satisfies: λ,δ λ,δ+1 H p,η,µ H p,η,µ f i (z) f i (z) 1 + Ai z (1 − θ ) +θ ≺ (i = 1, 2), zp zp 1 + Bi z
then:
(1 − θ )
λ,δ λ,δ+1 H p,η,µ H p,η,µ F (z) F (z) 1 + (1 − 2$)z + θ ≺ , p p z z 1−z
(25)
(26)
where: λ,δ F (z) = H p,η,µ ( f 1 ∗ f 2 )(z)
and: $ = 1−
(27)
1 δ+p 1 4( A1 − B1 )( A2 − B2 ) 1 − 2 F1 1, 1; + 1; . (1 − B1 )(1 − B2 ) 2 θ 2
(28)
The result is possible when B1 = B2 = −1. Proof. Suppose that f i ∈ A( p) (i = 1, 2) satisfy the condition (25). Setting: p i ( z ) = (1 − θ )
λ,δ λ,δ+1 H p,η,µ f i (z) H p,η,µ f i (z) + θ (i = 1, 2), p p z z
we have:
pi ( z ) ∈ P ( ς i )
(29)
1 − Ai , i = 1, 2 . ςi = 1 − Bi
Thus, by making use of the identity (3) in (29), we get: λ,δ H p,η,µ f i (z) =
δ + p p− δ+ p θ z θ
Z z 0
t
δ+ p θ −1
pi (t)dt (i = 1, 2),
(30)
which, in view of F given by (27) and (30), yields: λ,δ H p,η,µ F (z) =
δ + p p− δ+ p θ z θ
Z z 0
t
δ+ p θ −1
F (t)dt,
(31)
where: F ( z ) = (1 − θ )
Z λ,δ λ,δ+1 H p,η,µ F (z) H p,η,µ F (z) δ + p − δ + p z δ + p −1 θ t θ ( p1 ∗ p2 )(t)dt. + θ = z zp zp θ 0
(32)
Since pi (z) ∈ P(ς i ) (i = 1, 2), it follows from Lemma 3 that:
( p1 ∗ p2 )(z) ∈ P(ς 3 ) (ς 3 = 1 − 2(1 − ς 1 )(1 − ς 2 )).
(33)
Axioms 2018, 7, 27
9 of 13
Now, by using (33) in (32) and then appealing to Lemma 2, we have:
< { F (z)} = > > = =
δ + p 1 δ + p −1 u θ < {( p1 ∗ p2 )(uz)} du θ 0 Z 2(1 − ς 3 ) δ + p 1 δ + p −1 2ς 3 − 1 + u θ du θ 1 + u |z| 0 Z 2(1 − ς 3 ) δ + p 1 δ + p −1 u θ 2ς 3 − 1 + du θ 1+u 0 Z 1 δ+ p 4( A1 − B1 )( A2 − B2 ) δ+ p 1− 1− θ u θ −1 (1 + u)−1 du (1 − B1 )(1 − B2 ) 0 4( A1 − B1 )( A2 − B2 ) 1 1 δ+ p 1− 1 − 2 F1 1, 1; θ + 1; = $. (1 − B1 )(1 − B2 ) 2 2 Z
When B1 = B2 = −1, we consider the functions f i (z) ∈ A( p) (i = 1, 2), which satisfy (25), are defined by: δ+ p Z z δ+ p 1 + Ai t δ+ p p− − 1 λ,δ θ dt (i = 1, 2). H p,η,µ f i (z) = θ z t θ 1−t 0 Thus, it follows from (32) that: Z 1
δ+ p θ −1
(1 + A1 )(1 + A2 ) du (1 − uz) 0 z = 1 − (1 + A1 )(1 + A2 ) + (1 + A1 )(1 + A2 )(1 − z)−1 2 F1 1, 1; δ+θ p + 1; z−1 1 1 → 1 − (1 + A1 )(1 + A2 ) + (1 + A1 )(1 + A2 ) 2 F1 1, 1; δ+θ p + 1; as z → −1, 2 2 F (z) =
δ+ p θ
u
1 − (1 + A1 )(1 + A2 ) +
which evidently ends the proof. Theorem 5. Let υ ∈ C∗ , and let A, B ∈ C with A 6= B and | B| ≤ 1. Suppose that: υ(δ+ p)( A− B) A− B) − 1 ≤ 1 or υ(δ+ p)( + 1 ≤ 1 B B
if B 6= 0,
|υ(δ + p) A| ≤ π
if B = 0.
λ,δ If f ∈ A( p) with H p,η,µ f (z) 6= 0 for all z ∈ U∗ = U\{0}, then: λ,δ+1 H p,η,µ f (z) λ,δ H p,η,µ f (z)
≺
1 + Az , 1 + Bz
implies: λ,δ H p,η,µ f (z) p z
where:
( q(z) =
!υ
≺ q ( z ),
(1 + Bz)υ(δ+ p)( A− B)/B eυ(δ+ p) Az
if B 6= 0, if B = 0,
is the best dominant. Proof. Putting: ∆(z) =
λ,δ H p,η,µ f (z) p z
!υ
(z ∈ U).
(34)
Axioms 2018, 7, 27
10 of 13
Then, ∆ is analytic in U, ∆(0) = 1 and ∆(z) 6= 0 for all z ∈ U. Taking the logarithmic derivatives on both sides of (34) and using (3), we have: λ,δ+1 H p,η,µ f (z) 1 + Az z∆0 (z) ≺ 1+ = λ,δ . υ(δ + p)∆(z) 1 + Bz H p,η,µ f (z)
Now, the assertions of Theorem 5 follow by Lemma 4. Theorem 6. Let 0 ≤ α ≤ 1, ζ > 1. If f (z) ∈ A ( p) satisfies: 0
0 λ,δ+1 H f ( z ) p,η,µ < (1 − α ) 0 + α 0 < ζ, λ,δ+1 λ,δ H p,η,µ f (z) H p,η,µ f (z)
λ,δ+2 H p,η,µ f (z)
then:
