Hacettepe Journal of Mathematics and Statistics Volume 40 (5) (2011), 725 – 736
SUBORDINATION RESULTS OF MULTIVALENT FUNCTIONS DEFINED BY CONVOLUTION A. O. Mostafa∗†, Mohamed K. Aouf∗ and Teodor Bulboac˘a‡
Received 23 : 09 : 2010 : Accepted 23 : 03 : 2011
Abstract Using the method of differential subordination, we investigate some properties of certain classes of multivalent functions, which are defined by means of convolution. Keywords: Analytic functions, Multivalent functions, Convolution product, Differential subordination. 2000 AMS Classification: 30 C 45.
1. Introduction Let An (p) denote the class of functions of the form (1.1)
f (z) = z p +
∞ X
ak+p z k+p ,
p, n ∈ N = {1, 2, . . . } ,
k=n
which are analytic and p–valent in the unit disc U = {z ∈ C : |z| < 1}. If f and g are analytic functions in U, we say that f is subordinate to g, written f (z) ≺ g(z), if there exists a Schwarz function w, which (by definition) is analytic in U, with w(0) = 0, and |w(z)| < 1 for all z ∈ U, such that f (z) = g(w(z)), z ∈ U. Furthermore, if the function g is univalent in U, then we have the equivalence (cf., e.g., [18] and [19]) f (z) ≺ g(z) ⇔ f (0) = g(0) and f (U) ⊂ g(U). For functions f given by (1.1) and g ∈ An (p) given by g(z) = z p +
∞ X
bk+p z k+p ,
k=n ∗
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt. E-Mail: (A. O. Mostafa)
[email protected] (M. K. Aouf)
[email protected] † Corresponding Author. ‡ Faculty of Mathematics and Computer Science, Babe¸s-Bolyai University, 400084 ClujNapoca, Romania. E-mail:
[email protected]
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A. O. Mostafa, M. K. Aouf, T. Bulboac˘ a
the Hadamard product (convolution) of f and g is defined by (f ∗ g)(z) = z p +
∞ X
ak+p bk+p z k+p = (g ∗ f )(z).
k=n m For the functions f, g ∈ An (p) we define the linear operator Dλ,p : An (p) → An (p), where λ ≥ 0, p ∈ N, m ∈ N0 = N ∪ {0}, by 0 Dλ,p h(z) = h(z), 1 Dλ,p h)(z) = (1 − λ)h(z) +
and (1.2)
λz (h(z))′ , p
m 1 m−1 Dλ,p h(z) = Dλ,p Dλ,p h(z) m ∞ X p + λk p =z + ak+p bk+p z k+p , m ∈ N, p k=n
where h = f ∗ g. From (1.2) we may easily deduce that ′ λz m m+1 m Dλ,p (f ∗ g)(z) = Dλ,p (f ∗ g)(z) − (1 − λ)Dλ,p (f ∗ g)(z), (1.3) p
for λ ≥ 0 and m ∈ N0 . m For the special case p = 1, the operator Dλ,p (f ∗ g) was introduced and studied by m Aouf and Mostafa [4], while for different choices of the function g, the operator Dλ,p (f ∗g) reduces to several interesting operators as follows: z p+n (i) For bk+p = 1 for all k ≥ n or ge(z) = z p + , we have 1 − zm P p+λk m m Dλ,p (f ∗ e g )(z) ≡ Dλ,p f (z) = z p + ∞ ak+p z k+p , λ ≥ 0. k=n p
m Taking in this special case λ = 1, we have D1,p (f ∗ ge) ≡ Dpm f , where Dpm is the p–valent S˘ al˘ agean operator introduced and studied by Kamali and Orhan [14] (see also [3]); (ii) For m = 0 and s ∞ X p + l + λ(k − p) (1.4) g∗ (z) = z p + z k , (λ ≥ 0, p ∈ N, l, s ∈ N0 ) , p+l k=n+p
0 we see that Dλ,p (f ∗ g∗ ) = f ∗ g∗ = Ip (s, λ, l)f , where Ip (s, λ, l) is the generalized multiplier transformation which was introduced and studied by C˘ atas [7]. The operator Ip (s, λ, l) contains as special cases the multiplier transformation (see [8]), the generalized S˘ al˘ agean operator introduced and studied by Al-Oboudi [1], which in turn contains as a special case the S˘ al˘ agean operator (see [24]). For p = 1 and P Γ(k + 1)Γ(2 − β) g∗∗ (z) = z + ∞ (1 + λ(k − 1)) zk , k=2 Γ(k + 1 − β)
m where 0 ≤ β < 1, λ ≥ 0, we see that Dλ,1 (f ∗ g∗∗ ) ≡ Dλm,β f is the fractional differential multiplier operator defined and studied by Al-Oboudi and Al-Amoudi in [2]. (iii) For m = 0 and
(1.5)
g ∗ (z) = z p +
∞ X (α1 )k · . . . · (αl )k z k+p , (β1 )k · . . . · (βs )k (1)k k=n
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where αi ∈ C, i = 1, l, and βj ∈ C\{0, −1, −2, . . . }, j = 1, s, with l ≤ s+1, l, s ∈ 0 N0 , we see that Dλ,p (f ∗ g ∗ ) = f ∗ g ∗ ≡ Hpl,s (α1 , . . . αl ; β1 , . . . βs )f = Hpl,s [α1 ]f , p where Hl,s [α1 ] is the Dziok-Srivastava operator introduced and studied by Dziok and Srivatava [9] (see also [10] and [11]). The operator Hpl,s [α1 ] contains in turn many interesting operators, such as the Hohlov linear operator (see [13]), the Carlson-Shaffer linear operator (see [6] and [23]), the Ruscheweyh derivative operator (see [22]), the Bernardi-Libera-Livingston operator (see [5], [15] and [16]), and the Owa-Srivastava fractional derivative operator (see [20]). m Using the linear operator Dλ,p , we define a new subclass of the class An (p) as follows:
1.1. Definition. For fixed parameters A and B, with −1 ≤ B < A ≤ 1, for λ > 0, p ∈ N, m m ∈ N0 and g ∈ An (p), we say that a function f ∈ An (p) is in the class Tp,n (λ; A, B), if it satisfies the following subordination condition ′ m Dλ,p (f ∗ g)(z) 1 + Az ≺ . (1.6) pz p−1 1 + Bz A function f analytic in U is said to be convex of order η, η < 1, if f ′ (0) 6= 0 and zf ′′ (z) Re 1 + ′ > η, z ∈ U. f (z) If η = 0, then the function f is convex. It is easy to check that, if h(z) =
zh′′ (z) 1 + Az , then h′ (0) 6= 0 and Re 1 + ′ = 1 + Bz h (z)
1 − Bz > 0, z ∈ U, whenever |B| ≤ 1 and A 6= B, hence h is convex in the unit disc. 1 + Bz If B 6= −1, from the fact that h(z) = h(z), z ∈ U, we deduce that the image h(U) is symmetric with respect to the real axis, and that h maps the unit disc U onto the disc w − 1 − AB < A − B . If B = −1, the function h maps the unit disc U onto the half 1 − B2 1 − B2 1−A , hence we obtain: plane Re w > 2 Re
m 1.2. Remark. The function f ∈ A(p) is in the class Tp,n (λ; A, B) if and only if Dm (f ∗ g)(z) ′ 1 − AB A−B λ,p − , z ∈ U, for B 6= −1, < p−1 2 pz 1−B 1 − B2
and
Re
m Dλ,p (f ∗ g)(z) pz p−1
′
>
1−A , z ∈ U, for B = −1. 2
m Denoting by Tp,n (λ; γ) the class of functions f ∈ An (p) that satisfy the inequality
Re
m (Dλ,p (f ∗ g)(z))′ > γ, z ∈ U z p−1
(0 ≤ γ < p),
m m where g ∈ An (p), we have Tp,n (λ; γ) = Tp,n λ; 1 −
2γ , −1 p
.
In the present paper, we derive several inclusion relationships for the function class m Tp,n (λ; A, B).
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2. Preliminaries To prove our main results, we need the following lemmas. 2.1. Lemma. [12] Let h be a convex function in U with h(0) = 1. Suppose also that the function ϕ given by (2.1)
ϕ(z) = 1 + cp+n z n + cn+1 z n+1 + . . . ,
is analytic in U. Then ϕ(z) +
zϕ′ (z) ≺ h(z) γ
implies (2.2)
ϕ(z) ≺ ψ(z) =
γ − nγ z n
(Re γ ≥ 0, γ 6= 0) , Z
z
γ
t n −1 h(t) d t ≺ h(z),
0
and ψ is the best dominant of (2.2). 2.2. Lemma. [25] Let Φ be analytic in U, with 1 , z ∈ U. 2 Then, for any function F analytic in U, the set (Φ ∗ F )(U) is contained in the convex hull of F (U), i.e. (Φ ∗ F ) (U) ⊂ co F (U). Φ(0) = 1
and
Re Φ(z) >
For real or complex numbers a, b and c, the Gauss hypergeometric function is defined by 2 F1 (a, b, c; z)
=1+
(2.3) =
a·b z a(a + 1) · b(b + 1) z 2 + + ··· c 1! c(c + 1) 2!
∞ X (a)k (b)k z k , (c)k k!
a, b ∈ C, c ∈ C \ {0, −1, −2, . . . },
k=0
where (d)k = d(d + 1) . . . (d + k − 1) and (d)0 = 1. The series (2.3) converges absolutely for z ∈ U, hence it represents an analytic function in U (see [26, Chapter 14]). Each of the following identities are fairly well-known: 2.3. Lemma. [26, Chapter 14] For all real or complex numbers a, b and c, with c 6= 0, −1, −2, . . . , the following equalities hold: Z 1 Γ(b)Γ(c − b) tb−1 (1 − t)c−b−1 (1 − tz)−a d t = 2 F1 (a, b, c; z) Γ(c) (2.4) 0 where Re c > Re b > 0, z −a (2.5) , 2 F1 (a, b, c; z) = (1 − z) 2 F1 a, c − b, c; z−1 and (2.6)
2 F1 (a, b, c; z)
=2 F1 (b, a, c; z).
3. Main Results Unless otherwise mentioned, we assume throughout this paper that α > 0, −1 ≤ B < A ≤ 1, λ ≥ 0, p ∈ N, m ∈ N0 .
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3.1. Theorem. Let g ∈ An (p) be a given function, and suppose that the function f ∈ An (p) satisfies the subordination condition ′ ′ m+1 m (1 − α) Dλ,p h)(z) + α Dλ,p h(z) 1 + Az ≺ , pz p−1 1 + Bz where h = f ∗ g, and λ > 0. Then ′ m Dλ,p h(z) 1 + Az ≺ Q(z) ≺ , (3.1) pz p−1 1 + Bz where (3.2)
Q(z) =
(
A B
+ (1 −
1+
A )(1 B
p Az, αλn+p
+ Bz)−1 2 F1 1, 1,
p αλn
+ 1;
Bz 1+Bz
if B 6= 0,
,
if B = 0,
is the best dominant of (3.1). Furthermore, ′ m Dλ,p h(z) > η, z ∈ U (0 ≤ η < 1), (3.3) Re pz p−1 where (3.4)
η=
(
A B
+ (1 −
1−
A )(1 B
p A, αλn+p
− B)−1 2 F1 1, 1,
p αλn
+ 1;
B B−1
,
if B 6= 0, if B = 0.
The inequality (3.3) is the best possible. Proof. If we let (3.5)
ϕ(z) =
′ m Dλ,p h(z) , z ∈ U, pz p−1
then ϕ is of the form (2.1), and it is analytic in U. Applying the identity (1.3) in (3.5) and differentiating the resulting equation with respect to z, we get ′ ′ m+1 m (1 − α) Dλ,p h(z) + α Dλ,p h(z) αλzϕ′ (z) 1 + Az = ϕ(z) + ≺ . p−1 pz p 1 + Bz p , we deduce that Using Lemma 2.1 for γ = αλ ′ m Dλ,p h(z) p−1 pz ≺ Q(z) Z z p p − p 1 + At = z αλn t αλn −1 dt αλn 1 + Bt 0 =
(
A B
+ (1 −
1+
A )(1 B
p Az, αλn+p
+ Bz)−1 2 F1 1, 1,
p αλn
+ 1;
Bz 1+Bz
,
if B 6= 0, if B = 0,
where we have also made a change of variables followed by the use of the identities (2.4), (2.5), and (2.6). Next we will show that inf {Re Q(z) : |z| < 1} = Q(−1). Indeed, for |z| ≤ r < 1 we have Re
1 + Az 1 − Ar ≥ . 1 + Bz 1 − Br
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A. O. Mostafa, M. K. Aouf, T. Bulboac˘ a
Setting G(z, s) =
1 + Azs 1 + Bzs
and d ν(s) =
p p αλn −1 d s, 0 ≤ s ≤ 1, s αλn
which is a positive measure on the closed interval [0, 1], we have Q(z) = and thus Re Q(z) ≥
Z1
G(s, z) d ν(s),
0
Z
1 0
1 − Asr d ν(s) = Q(−r), |z| ≤ r < 1. 1 − Bsr
Letting r → 1− in the above inequality, we obtain the assertion (3.2). Finally, the estimate (3.3) is the best possible as the function Q is the best dominant of (3.1), which completes the proof of the theorem. Taking g(z) = z p +
z p+n in Theorem 3.1, we have the following result: 1−z
3.2. Corollary. If the function f ∈ An (p) satisfy the subordination condition ′ ′ m m+1 (1 − α) Dλ,p f (z) + α Dλ,p f (z) 1 + Az ≺ , pz p−1 1 + Bz with λ > 0, then ′ m Dλ,p f (z) 1 + Az ≺ Q(z) ≺ , pz p−1 1 + Bz where Q is given by (3.2), and it is the best dominant. Furthermore, ′ m Dλ,p f (z) > η, z ∈ U, (3.6) Re pz p−1 where η is given by (3.4), and the inequality (3.6) is the best possible.
∗
For m = 0 and g = g given by (1.5), using the identity ′ z Hpl,s [α1 ]f (z) = α1 Hpl,s [α1 + 1]f (z) + (p − α1 ) Hpl,s [α1 ]f (z),
Theorem 3.1 reduces to the next result:
3.3. Corollary. Let λ > 0, let αi ∈ C, i = 1, l, and βj ∈ C \ {0, −1, −2, . . . }, j = 1, s, with l ≤ s+1, l, s ∈ N0 , and suppose that the function f ∈ An (p) satisfy the subordination condition ′ ′ 1 1 1 − λαα Hpl,s [α1 ]f (z) + λαα Hpl,s [α1 + 1]f (z) p p 1 + Az ≺ . pz p−1 1 + Bz Then ′ Hpl,s [α1 ]f (z) 1 + Az ≺ Q(z) ≺ , pz p−1 1 + Bz where Q is given by (3.2), and it is the best dominant. Furthermore, ′ Hpl,s [α1 ]f (z) (3.7) Re > η, z ∈ U, pz p−1
Multivalent Functions defined by Convolution
where η is given by (3.4), and the inequality (3.7) is the best possible.
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Taking in Theorem 3.1 the parameter m = 0 and g = g∗ of the form (1.4), and using the identity (see [7]) λz (Ip (s, λ, l)f (z))′ = (p +l) Ip (s+1, λ, l)f (z)−[p(1−λ)+l] Ip (s, λ, l)f (z), λ ≥ 0, we deduce the following result: 3.4. Corollary. Let λ > 0, p ∈ N, and l, s ∈ N0 , and suppose that the function f ∈ An (p) satisfy the subordination condition i h ′ ′ 1 − α 1 + pl Ip (s, λ, l)f (z) + α 1 + pl Ip (s + 1, λ, l)f (z) 1 + Az ≺ . pz p−1 1 + Bz Then 1 + Az (Ip (s, λ, l)f (z))′ ≺ Q(z) ≺ , pz p−1 1 + Bz where Q is given by (3.2), and it is the best dominant. Furthermore, (3.8)
Re
(Ip (s, λ, l)f (z))′ > η, z ∈ U, pz p−1
where η is given by (3.4), and the inequality (3.8) is the best possible. 3.5. Theorem. Let g ∈ An (p) be a given function, and suppose that f ∈ (0 ≤ η < p). Then ′ ′ m+1 m (1 − α) Dλ,p h(z) + α Dλ,p h(z) Re > η, |z| < R, z p−1
m Tp,n (λ; η),
where h = f ∗ g, and
(3.9)
R=
"p
(αλn)2
+ p
p2
#1 − αλn n
.
The result is the best possible. m Proof. Since f ∈ Tp,n (λ; η), we write ′ m Dλ,p h(z) (3.10) = η + (p − η)u(z). z p−1
Then the function u is of the form (2.1), analytic in U, and has a positive real part in U. Substituting the relation (1.3) in (3.10), and differentiating the resulting equation with respect to z, we have ′ ′ m+1 m (1 − α) D h(z) + α D h(z) λ,p λ,p 1 αλ ′ (3.11) − η = u(z) + zu (z). p−η z p−1 p Applying the following well-known estimate [17] |zu′ (z)| 2nr n ≤ , |z| = r < 1, Re u(z) 1 − r 2n
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A. O. Mostafa, M. K. Aouf, T. Bulboac˘ a
in (3.11), we get
′ ′ m m+1 (1 − α) D h(z) + α D h(z) λ,p λ,p 1 − η Re p−η z p−1
2λαnr n ≥ Re u(z) 1 − , |z| = r < 1. p (1 − r 2n )
It is easy to see that the right-hand side of the inequality (??) is positive whenever r < R, where R is given by (3.9). In order to show that the bound R is the best possible, we consider the function f ∈ An (p) such that, for the given function g ∈ An (p) we have ′ m Dλ,p h(z) 1 + zn = η + (p − η) , z ∈ U (0 ≤ η < p). p−1 z 1 − zn Noting that ′ ′ m m+1 1 (1 − α) Dλ,p h(z) + α Dλ,p h(z) p(1 − z 2n ) − 2αλnz n − η = = 0, p−η z p−1 p (1 − z n )2 for z = R exp
iπ n
, the proof of the Theorem 3.5 is complete.
Putting α = 1 in Theorem 3.5, we obtain the following result: m 3.6. Corollary. Let g ∈ An (p) be a given function, and suppose that f ∈ Tp,n (λ; η), m+1 ∗ (0 ≤ η < p). Then f ∈ Tp,n (λ; η) for |z| < R , where
R∗ =
"p
(λn)2
+ p
p2
#1 − λn n
.
The result is the best possible.
Now we define the integral operator Fδ,p : An (p) → An (p) by Z δ + p z δ−1 Fδ,p (f )(z) = t f (t) d t, z ∈ U (δ > −p). zδ 0
m 3.7. Theorem. Let g ∈ An (p) be a given function, and suppose that f ∈ Tp,n (λ; A, B). Then ′ m Dλ,p Fδ,p h(z) 1 + Az (3.12) ≺ Θ(z) ≺ , pz p−1 1 + Bz where h = f ∗ g, and the function Θ is given by ( A A Bz + (1 − B )(1 + Bz)−1 2 F1 1, 1, δ+p + 1; 1+Bz , if B 6= 0, B n Θ(z) = δ+p if B = 0, 1 + p+n+δ Az,
and it is the best dominant of (3.12). Furthermore, ′ m Dλ,p Fδ,p h(z) Re > ϑ, z ∈ U (δ > −p), pz p−1 where ( A A + (1 − B )(1 − B)−1 2 F1 1, 1, δ+p + 1; B n ϑ= δ+p 1 − p+n+δ A,
B B−1
,
if B 6= 0, if B = 0.
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The result is the best possible. Proof. Letting (3.13)
′ m Dλ,p Fδ,p h(z) , z ∈ U, ϕ(z) = pz p−1
then ϕ is of the form (2.1), and analytic in U. Using the operator identity ′ m m m z Dλ,p Fδ,p h(z) = (p + δ)Dλ,p h(z) − δDλ,p Fδ,p h(z)
in (3.13), and differentiating the resulting equation with respect to z, we have ′ m Dλ,p h(z) zϕ′ (z) 1 + Az = ϕ(z) + ≺ . p−1 pz p+δ 1 + Bz Now the remaining part of Theorem 3.7 follows by employing the techniques that were used in the proof of Theorem 3.1. It is easy to see that, ′ Z z m ′ Dλ,p Fδ,p h(z) p+δ m = tδ Dλ,p h(t) d t, z ∈ U, p−1 p+δ pz pz 0
whenever f ∈ An (p) with δ > −p. In view of the above identity, Theorem 3.7 for the special case A = 1 − 2η (0 ≤ η < 1) and B = −1 yields the next result: 3.8. Corollary. Let g ∈ An (p) be a given function, and suppose that f ∈ An (p) satisfies the inequality ′ m Dλ,p h(z) Re > η, z ∈ U (0 ≤ η < 1), pz p−1 where h = f ∗ g, and λ > 0. Then Z z ′ p+δ δ m t D h(t) d t Re λ,p pz p+δ 0 p+δ 1 > η + (1 − η) 2 F1 1, 1, + 1; − 1 , z ∈ U (δ > −p). n 2 The result is the best possible.
3.9. Theorem. Let g ∈ An (p) be a given function, and suppose that the function H ∈ An (p) satisfies the inequality m Dλ,p H(z) > 0, z ∈ U. zp If the function f ∈ An (p) satisfies the inequality Dm h(z) λ,p − 1 < 1, z ∈ U, m Dp,λ H(z)
Re
where h = f ∗ g, then Re
m z Dλ,p h(z) m Dλ,p h(z)
where (3.14)
R0 =
p
′
> 0, |z| < R0 ,
9n2 + 4p(p + n) − 3n . 2(p + n)
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Proof. Letting (3.15)
ϕ(z) =
m Dλ,p h(z) − 1 = en z n + en+1 z n+1 + · · · , z ∈ U, m Dλ,p H(z)
then ϕ is analytic in U, with ϕ(0) = 0 and |ϕ(z)| < 1 for all z ∈ U. Defining the function ψ by ϕ(z) n , z ∈ U \ {0}, z ψ(z) = (n) ϕ (0) , z = 0, n! then ψ is analytic in U \ {0} and continuous in U, hence it is analytic in the whole unit disc U. If r ∈ (0, 1) is an arbitrary number, since |ϕ(z)| < 1 for all z ∈ U, we deduce ϕ(z) |ϕ(z)| 1 |ψ(z)| ≤ max n ≤ max < n , |z| ≤ r < 1. |z|=r |z|=r |z|n z r By letting r → 1− in the above inequality, we get |ψ(z)| < 1 for all z ∈ U, i.e. ϕ(z) = z n ψ(z), where the function ψ is analytic in U, and |ψ(z)| < 1, z ∈ U. Therefore, (3.15) leads to m m Dλ,p h(z) = Dλ,p H(z) [1 + z n ψ(z)] ,
and differentiating logarithmically with respect to z the above relation, we obtain ′ ′ m m z Dλ,p h(z) z Dλ,p H(z) z n [nψ(z) + zψ ′ (z)] = + . (3.16) m m Dλ,p h(z) Dλ,p H(z) 1 + z n ψ(z) m Dλ,p H(z) , we see that the function χ is of the form (2.1), analytic in U zp with Re χ(z) > 0 for all z ∈ U, and ′ m z Dλ,p H(z) zχ′ (z) = + p, m Dλ,p H(z) χ(z)
Letting χ(z) =
so we find from (3.16) that ′ ′ n m zχ (z) z [nψ(z) + zψ ′ (z)] z Dλ,p h(z) − , z ∈ U. (3.17) Re ≥ p − m χ(z) Dλ,p h(z) 1 + z n ψ(z)
Using the following known estimates [21] (see also [17]) ′ n−1 χ (z) nψ(z) + zψ ′ (z) n ≤ 2nr and χ(z) 1 + z n ψ(z) ≤ 1 − r n , |z| = r < 1, 1 − r 2n from (3.17) we deduce that ′ m z Dλ,p h(z) p − 3nr n − (p + n)r 2n Re ≥ , |z| = r < 1, m Dλ,p h(z) 1 − r 2n
and the right-hand side fraction is positive provided that r < R0 , where R0 is given by (3.14). 3.10. Theorem. Let g ∈ An (p) be a given function, and suppose that the function H ∈ An (p) satisfies the inequality H(z) 1 > , z ∈ U. zp 2 m If f ∈ Tp,n (λ; A, B), then (3.18)
Re
m f ∗ H ∈ Tp,n (λ; A, B).
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Proof. A simple calculation shows that ′ ′ m m Dλ,p (f ∗ g)(z) Dλ,p (f ∗ g ∗ H)(z) H(z) = ∗ p . pz p−1 pz p−1 z Using the assumption (3.18) and the fact that the function Lemma 2.2 follows m Dλ,p (f ∗ g ∗ H)(z) pz p−1
′
m that is f ∗ H ∈ Tp,n (λ; A, B).
≺
1 + Az is convex in U, from 1 + Bz
1 + Az , 1 + Bz
3.11. Remark. Specializing in the above results the parameters λ and m, and the function g, we obtain new results corresponding to the operators defined in the introduction.
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