Paper20-Acta26-2011. 120KB Jun 04 2011 12:03:16 AM
Acta Universitatis Apulensis
ISSN: 1582-5329
No. 26/2011
pp. 189-196
ON CERTAIN SUBCLASS OF P-VALENT FUNCTIONS
INVOLVING THE DZIOK-SRIVASTAVA OPERATOR
Ali Muhammad
Abstract. In this paper, we introduce a class Tk (λ, α1 , p, q, s, ρ). We investigate a number of inclusion relationships, radius problem and some other interesting
properties of p-valent functions which are defined here by means of a certain linear
integral operator Hp,q,s (α1 ).
2000 Mathematics Subject Classification: 30C45, 30C50.
Keywords: Analytic functions, Hadamard product (or convolution), Generalized hypergeometric functions, Functions with positive real part, Dziok-Srivastava
operator.
1. Introduction
Let A(p) denote the class of functions f (z) normalized by
p
f (z) = z +
∞
X
ap+k z p+k (p ∈ N = {1, 2, 3 · · · } ,
(1)
k=1
which are analytic and p-valent in the unit disk E = {|z| : z ∈ C, |z| < 1}.
For functions fj (z) ∈ A(p), given by we define the Hadamard product (or convolution) of f1 (z) and f2 (z) by
p
fj (z) = z +
∞
X
ap+k,j z p+k (j = {1, 2, 3 · · · } ,
(2)
k=1
we define the Hadamard product (or convolution) of f1 (z) and f2 (z) by
(f1 ∗ f2 )(z) = z p +
∞
X
ap+k,1 ap+k,2 z p+k = (f2 ∗ f1 )(z), (z ∈ E).
k=1
189
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A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
Let Pk (ρ) be the class of functions p(z) analytic in E satisfying the properties p(0) =
1 and
Z2π
Rep(z) − ρ
(4)
1 − ρ dθ ≤ kπ, ,
0
reiθ , k
where z =
≥ 2 and 0 ≤ ρ < 1. This class has been introduced in [10]. We note,
for ρ = 0, we obtain the class Pk defined and studied in [11], and for ρ = 0, k = 2,
we have the well-known class P of functions with positive real part. The case k = 2
gives the class P (ρ) of functions with positive real part greater than ρ. From (4) we
can easily deduce that p ∈ Pk (ρ) if and only if, there exists p1 , p2 ∈ P (ρ) such that
for z ∈ E,
k 1
k 1
p(z) =
+
−
p1 (z) −
p2 (z).
(5)
4 2
4 2
Making use of the Hadamard product (or convolution) given by (3), we now
define the Dziok-Srivastava operator,
Hp (α1 , · · · , αq ; β 1 , · · · β q ) : A(p) → A(p).
which was introduced and studied in a series of recent papers by Dziok and Srivastava
[1], [2], see also [5], [6]. Indeed, for complex parameters
α1 , · · · αq and β 1 , · · · , β s , (β j ∈
/ Z0− = {0, −1, −2, −3, · · · }; j = 1, · · · s),
the generalized hypergeometric function
q Fs (α1 , · · ·
, αq ; β 1 , · · · , β s ; z)
is given by
∞
X
(α1 )n · · · (αq )n n
z
q Fs (α1 , · · · , αq ; β 1 , · · · , β s ; z) =
(β 1 , · · · , β s )n n!
(6)
n=0
(q ≤ s+1; q, s ∈ N0 = N∪{0}; N = {1, 2, · · · }; z ∈ E, where (v)k is the Pochhammer
symbol (or the shifted factorial) defined in (terms of the Gamma function) by
Γ(v + k)
1
if k = 0, v ∈ C\{0}
(v)k =
=
v(v + 1), · · · (v + k − 1)
if k ∈ N, v ∈ C.
Γ(v)
Corresponding to a function
Fp (α1 , · · · αq ; β 1 , · · · , β s ; z)
defined by
Fp (α1 , · · · αq ; β 1 , · · · , β s ; z) = z pq Fs (α1 , · · · , αq ; β 1 , · · · , β s ; z).
190
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
Dziok and Srivastava [1] considered a linear operator defined by the following Hadamard
product (or convolution):
Hp (α1 , · · · αq ; β 1 , · · · , β s )f (z) = Fp (α1 , · · · αq ; β 1 , · · · , β s ; z) ∗ f (z).
(7)
For convenience, we write
Hp,q,s (α1 ) = Hp (α1 , · · · αq ; β 1 , · · · , β s ).
(8)
Thus after some calculations, we have
z(Hp,q,s (α1 )f (z))′ = α1 Hλ,q,s (α1 + 1)f (z) − (α1 − p)Hp,q,s (α1 )f (z).
(9)
Many interesting subclasses of analytic functions, associated with the Dziok-Srivastava
operator Hp,q,s (α1 ) and its many special cases, were investigated recently by Dziok
and Srivastava [1], [2], Gangadharan et. al [3], Liu and Srivastava [5], [6], see also
[5], [9], [13].
Definition 1.1. Let f ∈ A(p). Then f ∈ Tk (λ, α1 , p, q, s, ρ), if and only if
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)f (z)
(1 − λ)
∈ Pk (ρ), z ∈ E,
+λ
zp
zp
where λ > 0, k ≥ 2 and 0 ≤ ρ < p.
2. Preliminary Results
Lemma 2.1.[12] If p(z) is analytic in E with p(0) = 1, and if λ1 is a convex
number satisfying Re(λ1 ) ≥ 0, (λ1 6= 0), then
Re p(z) + λ1 zp′ (z) > β (0 ≤ β < 1)
implies
Rep(z) > β + (1 − β)(2γ − 1),
where γ is given by
γ = γ(Reλ1 =
Z1
−1
(1 + tReλ1 ) dt,
0
which is an increasing function of Re(λ1 ) and
the sense that the bound cannot be improved.
1
2
≤ γ < 1. The estimate is sharp in
Lemma 2.2.[14] If p(z) is analytic in E, p(0) = 1 and Rep(z) > 12 , z ∈ E, then
for any function F analytic in E, the function p ∗ F takes the value in the convex
hull of the image of E under F .
191
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
3. Main Results
Theorem 3.1. Let Reα1 > 0. Then Tk (λ, α1 , p, q, s, ρ) ⊂ Tk (0, α1 , p, q, s, ρ1 ),
where ρ1 is given by
ρ1 = ρ + (1 − ρ)(2γ − 1),
(10)
and
Z1
1+t
−1
Re αλ
1
dt.
0
Proof. Let f ∈ Tk (λ, α1 , p, q, s, ρ) and set
Hp,q,s (α1 )f (z)
k 1
k 1
= h(z) =
+
−
h1 (z) −
h2 (z)
zp
4 2
4 2
(11)
Then h(z) is analytic in E with h(0) = 1. By a simple computation, we have
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
λzh′ (z)
∈ Pk (ρ)
= h(z) +
+λ
(1 − λ)
zp
zp
α1
for z ∈ E.
o
n
λzh′ (z)
> ρ, i = 1, 2.
This implies that Re hi (z) + αi1
Using Lemma 2.1, we see that Rehi (z) > ρ1 , where ρ1 is given by (10). Consequently h ∈ Pk (ρ1 ), where ρ1 is given by (10) for z ∈ E and proof is complete.
Theorem 3.2. Let f ∈ Tk (0, α1 , p, q, s, ρ) for z ∈ E. Then f ∈ Tk (λ, α1 , p, q, s, ρ)
for |z| < R(α1 , λ), where
|α1 |
R(α1 , λ) =
λ+
q
2
(λ + |α1
.
(12)
|2 )
Proof. Set
Hp,q,s (α1 )f (z)
= (p − ρ)h(z) + ρ,
h ∈ Pk .
zp
Now proceeding as in Theorem 3.1, we have
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
λzh′ (z)
(1 − λ)
+λ
− ρ = (p − ρ) h(z) +
zp
zp
α1
k 1
λzh′1 (z)
λzh′2 (z)
k 1
− ( − ) h2 (z) +
(13)
= (p − ρ) ( + ) h1 (z) +
4 2
α1
4 2
α1
192
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
where we have used (5) and h1 , h2 ∈ P, z ∈ E. Using the following well-known
estimates, see [7]
|zh′i (z)| ≤
2r
Rehi (z), (|z| = r < 1), i = 1, 2,
1 − r2
we have
λzh′i (z)
λzh′i (z)
}
≥ Re hi (z) +
}
Re hi (z) +
|α1 |
|α1 |
2λr
≥ Rehi (z) 1 −
.
|α1 |(1 − r2 )
The right hand side of this inequality is positive if r < R(α1 , λ), where R(α1 , λ)
is given by (12). Consequently it follows from (13) that f ∈ Tk (λ, α, p, q, s, ρ) for
1+z
in (13),
|z| < R(α1 , λ). Sharpness of this result follows by taking hi (z) = 1−z
i = 1, 2.
Theorem 3.3. Tk (λ1 , α1 , p, q, s, ρ) ⊂ Tk (λ2 , α1 , p, q, s, ρ) for 0 ≤ λ2 < λ1 .
Proof. For λ2 = 0 the proof is immediate. Let λ2 > 0 and let f ∈ Tk (λ1 , α1 , p, q, s, ρ).
Then there exist two functions H1 , H2 ∈ Pk (ρ) such that, from Definition 1.1 and
Theorem 3.1,
(1 − λ1 )
and
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)
+ λ1
= H1 (z),
p
z
zp
Hp,q,s (α1 )
= H2 (z).
zp
Hence
(1 − λ2 )
Hp,q,s (α1 + 1)
Hp,q,s (α2 )f (z)
λ2
λ2
+ λ2
=
H1 (z) + (1 −
)H2 (z).
p
p
z
z
λ1
λ1
(14)
Since the class Pk (ρ) is a convex set, see [8], it follows that the right hand side of
(14) belongs to Pk (ρ) and this proves the result.
Theorem 3.4. Let f ∈ Tk (λ, α1 , p, q, s, ρ) and let φ ∈ C(ρ) is the class of
p-valent convex functions. Then φ ∗ f ∈ Tk (λ, α1 , p, q, s, ρ).
Proof. Let F = φ ∗ f . Then we have
Hp,q,s (α1 + 1)F (z)
Hp,q,s (α1 )F (z)
φ(z)
= p ∗ G(z),
+λ
(1 − λ)
p
p
z
z
z
193
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
where
G(z) =
(1 − λ)
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
+λ
zp
zp
∈ Pk (ρ).
Therefore,we have
k 1
k 1
φ(z)
φ(z)
φ(z)
∗G(z)
=
(
+
)
(p
−
ρ)
∗
g
(z)
+
ρ
−(
−
)
(p
−
ρ)
+
ρ
1
zp
4 2
zp
4 2
z p ∗ g2 (z)
where g1 , g2 ∈ P .
o
n
> 21 , z ∈ E, and so using Lemma 2.2, we conclude
Since φ ∈ C(p), Re φ(z)
p
z
that F = φ ∗ f ∈ Tk (λ, α1 , p, q, s, ρ).
Theorem 3.5. Let f (z) ∈ A(p) and define the one-parameter integral operator
Jc (c > −p) by
c+p
Jc f (z) =
zc
Zz
tc−1 f (t)dt (f ∈ A(p); c > p).
(15)
0
If
Hp,q,s (α1 )f (z)
Hp,q,s (α1 )Jc f (z)
+λ
(1 − λ)
p
z
zp
then
∈ Pk (ρ).
(16)
Hp,q,s (α1 )Jc f (z)
∈ Pk (ρ2 ),
zp
where ρ2 is given by
ρ2 = ρ + (1 − ρ)(2γ 1 − 1),
and
γ1 =
Z1
1+t
λ
Re( (c+p)
0
−1
(17)
dt.
Proof. First of all it follows from the Definition 3.6, that
Let
z(Hp,q,s (α1 )Jc f (z))′ = (c + p)Hλ,q,s (α1 )f (z) − cHp,q,s (α1 )Jc f (z).
(18)
Hp,q,s (α1 Jc f (z)
k 1
k 1
= h(z) = ( + )h1 (z) − ( − )h2 (z),
p
z
4 2
4 2
(19)
194
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
then the hypothesis (16) in conjection with (18) would yield
Hp,q,s (α1 )f (z)
Hp,q,s (α1 )Jc f (z)
λzh′ (z)
= h(z) +
+λ
∈ Pk (ρ) forz ∈ E.
(1 − λ)
zp
zp
c+p
Consequently
λzh′i (z)
hi (z) +
c+p
Using Lemma 2.1 with λ1 =
and the proof is complete.
∈ P (ρ), i = 1, 2, 0 ≤ ρ < p, andz ∈ E.
λ
(c+p) ,
we have Rehi (z) > ρ2 , where ρ2 is given by (17),
Acknowledgments. The author express deep gratitude to Vice Chancellor
Imtiaz Gillani and Dean Azamul Asar University of Engineering and Technology
Peshawar Pakistan for his support and providing excellent research facilities.
References
[1] J. Dziok and H. M. Srivastava, Classes of analytic functions associated with
the generalized hypergeometric function, Appl. Math. Comput, 103(1999), 1-13.
[2] J.Dziok and H. M. Srivastava, Certain subclasses of analytic functions associated with the generalized hypergeometric function, Integral Trans. Spec. Funct.
14(2003), 7-18.
[3] A. Gangadharan, T. N. Shanmugam and H. M. Srivastava, Generalized hypergeometric functions associated with k-uniformly convex functions, Comput. Math.
Appl., 44(2002), 1515-1526.
[4] Y. C. Kim and H. M. Srivastava, Fractional integral and other linear operators with the Gaussian hypergeometric function, Complex vaiables theory, Appl.,
34(1997), 293-312.
[5] J. L. Liu, Strongly starlike functions associated with the Dziok-Srivastava
operator, Tamakang J. Math., 35(2004), 37-42.
[6] J. L. Liu, and H. M. Srivastava, Classes of meromorphically multivalent functions associated with the generalized hypergeometric function, Math. Comput. Modelling 39(2004), 21-34.
[7] T. H. MacGregor, Radius of univalence of certain analytic functions, Proc.
Amer. Math. Soc., 14(1963), 514-520.
[8] K. I. Noor, On subclasses of close-to-convex functions of higher order, Internat. J. Math. and Math Sci., 15(1992), 279-290.
[9] S. Owa and H. M. Srivastava, Univalent and starlike generalized hypergeometric functions, Canad. J. Math., 39(1987), 1057-1077.
195
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
[10] K. Padmanabhan and R. Parvatham, Properties of a class of functions with
bounded boundary rotation, Ann. Polon. Math., 31(1975), 311-323.
[11] B. Pinchuk, Functions with bounded boundary rotation, Isr. J. Math., 10(1971),
7-16.
[12] S. Ponnusamy, Differential subordination and Bazilevic functions, Proc. Ind.
Acad. Sci., 105(1995), 169-186.
[13] H. M. Srivstava, S. Owa (Eds), Current topics in analytic function theory,
World Scientific Publishing Company, Singapore, New Jersey, London and Hong
Kong, 1992.
[14] R . Singh and S. Singh, Convolution properties of a class of starlike functions,
Proc. Amer. Math. Soc., 106(1989), 145-152.
Ali Muhammad,
Department of Basic Sciences
University of Engineering and Technology Peshawar Pakistan.
email: [email protected]
196
ISSN: 1582-5329
No. 26/2011
pp. 189-196
ON CERTAIN SUBCLASS OF P-VALENT FUNCTIONS
INVOLVING THE DZIOK-SRIVASTAVA OPERATOR
Ali Muhammad
Abstract. In this paper, we introduce a class Tk (λ, α1 , p, q, s, ρ). We investigate a number of inclusion relationships, radius problem and some other interesting
properties of p-valent functions which are defined here by means of a certain linear
integral operator Hp,q,s (α1 ).
2000 Mathematics Subject Classification: 30C45, 30C50.
Keywords: Analytic functions, Hadamard product (or convolution), Generalized hypergeometric functions, Functions with positive real part, Dziok-Srivastava
operator.
1. Introduction
Let A(p) denote the class of functions f (z) normalized by
p
f (z) = z +
∞
X
ap+k z p+k (p ∈ N = {1, 2, 3 · · · } ,
(1)
k=1
which are analytic and p-valent in the unit disk E = {|z| : z ∈ C, |z| < 1}.
For functions fj (z) ∈ A(p), given by we define the Hadamard product (or convolution) of f1 (z) and f2 (z) by
p
fj (z) = z +
∞
X
ap+k,j z p+k (j = {1, 2, 3 · · · } ,
(2)
k=1
we define the Hadamard product (or convolution) of f1 (z) and f2 (z) by
(f1 ∗ f2 )(z) = z p +
∞
X
ap+k,1 ap+k,2 z p+k = (f2 ∗ f1 )(z), (z ∈ E).
k=1
189
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A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
Let Pk (ρ) be the class of functions p(z) analytic in E satisfying the properties p(0) =
1 and
Z2π
Rep(z) − ρ
(4)
1 − ρ dθ ≤ kπ, ,
0
reiθ , k
where z =
≥ 2 and 0 ≤ ρ < 1. This class has been introduced in [10]. We note,
for ρ = 0, we obtain the class Pk defined and studied in [11], and for ρ = 0, k = 2,
we have the well-known class P of functions with positive real part. The case k = 2
gives the class P (ρ) of functions with positive real part greater than ρ. From (4) we
can easily deduce that p ∈ Pk (ρ) if and only if, there exists p1 , p2 ∈ P (ρ) such that
for z ∈ E,
k 1
k 1
p(z) =
+
−
p1 (z) −
p2 (z).
(5)
4 2
4 2
Making use of the Hadamard product (or convolution) given by (3), we now
define the Dziok-Srivastava operator,
Hp (α1 , · · · , αq ; β 1 , · · · β q ) : A(p) → A(p).
which was introduced and studied in a series of recent papers by Dziok and Srivastava
[1], [2], see also [5], [6]. Indeed, for complex parameters
α1 , · · · αq and β 1 , · · · , β s , (β j ∈
/ Z0− = {0, −1, −2, −3, · · · }; j = 1, · · · s),
the generalized hypergeometric function
q Fs (α1 , · · ·
, αq ; β 1 , · · · , β s ; z)
is given by
∞
X
(α1 )n · · · (αq )n n
z
q Fs (α1 , · · · , αq ; β 1 , · · · , β s ; z) =
(β 1 , · · · , β s )n n!
(6)
n=0
(q ≤ s+1; q, s ∈ N0 = N∪{0}; N = {1, 2, · · · }; z ∈ E, where (v)k is the Pochhammer
symbol (or the shifted factorial) defined in (terms of the Gamma function) by
Γ(v + k)
1
if k = 0, v ∈ C\{0}
(v)k =
=
v(v + 1), · · · (v + k − 1)
if k ∈ N, v ∈ C.
Γ(v)
Corresponding to a function
Fp (α1 , · · · αq ; β 1 , · · · , β s ; z)
defined by
Fp (α1 , · · · αq ; β 1 , · · · , β s ; z) = z pq Fs (α1 , · · · , αq ; β 1 , · · · , β s ; z).
190
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
Dziok and Srivastava [1] considered a linear operator defined by the following Hadamard
product (or convolution):
Hp (α1 , · · · αq ; β 1 , · · · , β s )f (z) = Fp (α1 , · · · αq ; β 1 , · · · , β s ; z) ∗ f (z).
(7)
For convenience, we write
Hp,q,s (α1 ) = Hp (α1 , · · · αq ; β 1 , · · · , β s ).
(8)
Thus after some calculations, we have
z(Hp,q,s (α1 )f (z))′ = α1 Hλ,q,s (α1 + 1)f (z) − (α1 − p)Hp,q,s (α1 )f (z).
(9)
Many interesting subclasses of analytic functions, associated with the Dziok-Srivastava
operator Hp,q,s (α1 ) and its many special cases, were investigated recently by Dziok
and Srivastava [1], [2], Gangadharan et. al [3], Liu and Srivastava [5], [6], see also
[5], [9], [13].
Definition 1.1. Let f ∈ A(p). Then f ∈ Tk (λ, α1 , p, q, s, ρ), if and only if
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)f (z)
(1 − λ)
∈ Pk (ρ), z ∈ E,
+λ
zp
zp
where λ > 0, k ≥ 2 and 0 ≤ ρ < p.
2. Preliminary Results
Lemma 2.1.[12] If p(z) is analytic in E with p(0) = 1, and if λ1 is a convex
number satisfying Re(λ1 ) ≥ 0, (λ1 6= 0), then
Re p(z) + λ1 zp′ (z) > β (0 ≤ β < 1)
implies
Rep(z) > β + (1 − β)(2γ − 1),
where γ is given by
γ = γ(Reλ1 =
Z1
−1
(1 + tReλ1 ) dt,
0
which is an increasing function of Re(λ1 ) and
the sense that the bound cannot be improved.
1
2
≤ γ < 1. The estimate is sharp in
Lemma 2.2.[14] If p(z) is analytic in E, p(0) = 1 and Rep(z) > 12 , z ∈ E, then
for any function F analytic in E, the function p ∗ F takes the value in the convex
hull of the image of E under F .
191
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
3. Main Results
Theorem 3.1. Let Reα1 > 0. Then Tk (λ, α1 , p, q, s, ρ) ⊂ Tk (0, α1 , p, q, s, ρ1 ),
where ρ1 is given by
ρ1 = ρ + (1 − ρ)(2γ − 1),
(10)
and
Z1
1+t
−1
Re αλ
1
dt.
0
Proof. Let f ∈ Tk (λ, α1 , p, q, s, ρ) and set
Hp,q,s (α1 )f (z)
k 1
k 1
= h(z) =
+
−
h1 (z) −
h2 (z)
zp
4 2
4 2
(11)
Then h(z) is analytic in E with h(0) = 1. By a simple computation, we have
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
λzh′ (z)
∈ Pk (ρ)
= h(z) +
+λ
(1 − λ)
zp
zp
α1
for z ∈ E.
o
n
λzh′ (z)
> ρ, i = 1, 2.
This implies that Re hi (z) + αi1
Using Lemma 2.1, we see that Rehi (z) > ρ1 , where ρ1 is given by (10). Consequently h ∈ Pk (ρ1 ), where ρ1 is given by (10) for z ∈ E and proof is complete.
Theorem 3.2. Let f ∈ Tk (0, α1 , p, q, s, ρ) for z ∈ E. Then f ∈ Tk (λ, α1 , p, q, s, ρ)
for |z| < R(α1 , λ), where
|α1 |
R(α1 , λ) =
λ+
q
2
(λ + |α1
.
(12)
|2 )
Proof. Set
Hp,q,s (α1 )f (z)
= (p − ρ)h(z) + ρ,
h ∈ Pk .
zp
Now proceeding as in Theorem 3.1, we have
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
λzh′ (z)
(1 − λ)
+λ
− ρ = (p − ρ) h(z) +
zp
zp
α1
k 1
λzh′1 (z)
λzh′2 (z)
k 1
− ( − ) h2 (z) +
(13)
= (p − ρ) ( + ) h1 (z) +
4 2
α1
4 2
α1
192
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
where we have used (5) and h1 , h2 ∈ P, z ∈ E. Using the following well-known
estimates, see [7]
|zh′i (z)| ≤
2r
Rehi (z), (|z| = r < 1), i = 1, 2,
1 − r2
we have
λzh′i (z)
λzh′i (z)
}
≥ Re hi (z) +
}
Re hi (z) +
|α1 |
|α1 |
2λr
≥ Rehi (z) 1 −
.
|α1 |(1 − r2 )
The right hand side of this inequality is positive if r < R(α1 , λ), where R(α1 , λ)
is given by (12). Consequently it follows from (13) that f ∈ Tk (λ, α, p, q, s, ρ) for
1+z
in (13),
|z| < R(α1 , λ). Sharpness of this result follows by taking hi (z) = 1−z
i = 1, 2.
Theorem 3.3. Tk (λ1 , α1 , p, q, s, ρ) ⊂ Tk (λ2 , α1 , p, q, s, ρ) for 0 ≤ λ2 < λ1 .
Proof. For λ2 = 0 the proof is immediate. Let λ2 > 0 and let f ∈ Tk (λ1 , α1 , p, q, s, ρ).
Then there exist two functions H1 , H2 ∈ Pk (ρ) such that, from Definition 1.1 and
Theorem 3.1,
(1 − λ1 )
and
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)
+ λ1
= H1 (z),
p
z
zp
Hp,q,s (α1 )
= H2 (z).
zp
Hence
(1 − λ2 )
Hp,q,s (α1 + 1)
Hp,q,s (α2 )f (z)
λ2
λ2
+ λ2
=
H1 (z) + (1 −
)H2 (z).
p
p
z
z
λ1
λ1
(14)
Since the class Pk (ρ) is a convex set, see [8], it follows that the right hand side of
(14) belongs to Pk (ρ) and this proves the result.
Theorem 3.4. Let f ∈ Tk (λ, α1 , p, q, s, ρ) and let φ ∈ C(ρ) is the class of
p-valent convex functions. Then φ ∗ f ∈ Tk (λ, α1 , p, q, s, ρ).
Proof. Let F = φ ∗ f . Then we have
Hp,q,s (α1 + 1)F (z)
Hp,q,s (α1 )F (z)
φ(z)
= p ∗ G(z),
+λ
(1 − λ)
p
p
z
z
z
193
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
where
G(z) =
(1 − λ)
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
+λ
zp
zp
∈ Pk (ρ).
Therefore,we have
k 1
k 1
φ(z)
φ(z)
φ(z)
∗G(z)
=
(
+
)
(p
−
ρ)
∗
g
(z)
+
ρ
−(
−
)
(p
−
ρ)
+
ρ
1
zp
4 2
zp
4 2
z p ∗ g2 (z)
where g1 , g2 ∈ P .
o
n
> 21 , z ∈ E, and so using Lemma 2.2, we conclude
Since φ ∈ C(p), Re φ(z)
p
z
that F = φ ∗ f ∈ Tk (λ, α1 , p, q, s, ρ).
Theorem 3.5. Let f (z) ∈ A(p) and define the one-parameter integral operator
Jc (c > −p) by
c+p
Jc f (z) =
zc
Zz
tc−1 f (t)dt (f ∈ A(p); c > p).
(15)
0
If
Hp,q,s (α1 )f (z)
Hp,q,s (α1 )Jc f (z)
+λ
(1 − λ)
p
z
zp
then
∈ Pk (ρ).
(16)
Hp,q,s (α1 )Jc f (z)
∈ Pk (ρ2 ),
zp
where ρ2 is given by
ρ2 = ρ + (1 − ρ)(2γ 1 − 1),
and
γ1 =
Z1
1+t
λ
Re( (c+p)
0
−1
(17)
dt.
Proof. First of all it follows from the Definition 3.6, that
Let
z(Hp,q,s (α1 )Jc f (z))′ = (c + p)Hλ,q,s (α1 )f (z) − cHp,q,s (α1 )Jc f (z).
(18)
Hp,q,s (α1 Jc f (z)
k 1
k 1
= h(z) = ( + )h1 (z) − ( − )h2 (z),
p
z
4 2
4 2
(19)
194
A. Muhammad - On certain subclass of p-valent functions involving the Dziok...
then the hypothesis (16) in conjection with (18) would yield
Hp,q,s (α1 )f (z)
Hp,q,s (α1 )Jc f (z)
λzh′ (z)
= h(z) +
+λ
∈ Pk (ρ) forz ∈ E.
(1 − λ)
zp
zp
c+p
Consequently
λzh′i (z)
hi (z) +
c+p
Using Lemma 2.1 with λ1 =
and the proof is complete.
∈ P (ρ), i = 1, 2, 0 ≤ ρ < p, andz ∈ E.
λ
(c+p) ,
we have Rehi (z) > ρ2 , where ρ2 is given by (17),
Acknowledgments. The author express deep gratitude to Vice Chancellor
Imtiaz Gillani and Dean Azamul Asar University of Engineering and Technology
Peshawar Pakistan for his support and providing excellent research facilities.
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Ali Muhammad,
Department of Basic Sciences
University of Engineering and Technology Peshawar Pakistan.
email: [email protected]
196