Volume 10 (2009), Issue 1, Article 13, 17 pp.
ON CERTAIN SUBCLASSES OF MEROMORPHICALLY MULTIVALENT
FUNCTIONS ASSOCIATED WITH THE GENERALIZED HYPERGEOMETRIC
FUNCTION
JAGANNATH PATEL AND ASHIS KU. PALIT
D EPARTMENT OF M ATHEMATICS
U TKAL U NIVERSITY, VANI V IHAR
B HUBANESWAR -751004, I NDIA
jpatelmath@yahoo.co.in
D EPARTMENT OF M ATHEMATICS
B HADRAK I NSTITUTE OF E NGINEERING AND T ECHNOLOGY
B HADRAK -756 113, I NDIA
ashis_biet@rediffmail.com
Received 12 September, 2008; accepted 20 February, 2009
Communicated by N.E. Cho
A BSTRACT. In the present paper, we investigate several inclusion relationships and other interesting properties of certain subclasses of meromorphically multivalent functions which are
defined here by means of a linear operator involving the generalized hypergeometric function.
Some interesting applications on Hadamard product concerning this and other classes of integral
operators are also considered.
Key words and phrases: Meromorphic function, p-valent, Subordination, Hypergeometric function, Hadamard product.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
For any integer m > 1 − p, let
(1.1)
f (z) = z
P
−p
p,m
+
be the class of functions of the form:
∞
X
ak z k
(p ∈ N = {1, 2, . . . }),
k=m
∗
which are analytic and p-valent
in the
C : 0 < |z|P< 1} =
P
P punctured unit disk U = {z ∈ P
U \ {0}. We also denote p,1−p = p . For 0 5 α < p, we denote by S (p; α),
K (p; α)
P
P
and C (p; α), the subclasses of p consisting of all meromorphic functions which are, respectively, p-valently starlike of order α, p-valently convex of order α and p-valently close-to-convex
of order α.
If f and g are analytic in U, we say that f is subordinate to g, written f ≺ g or (more
precisely) f (z) ≺ g(z) z ∈ U, if there exists a function ω, analytic in U with ω(0) = 0 and
250-08
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JAGANNATH PATEL AND A SHIS K U . PALIT
|ω(z)| < 1 such that f (z) = g(ω(z)), z ∈ U. In particular, if g is univalent in U, then we have
the following equivalence:
f (z) ≺ g(z) (z ∈ U) ⇐⇒ f (0) = g(0) and f (U) ⊂ g(U).
P
P
P
k
For a function f ∈ p,m , given by (1.1) and g ∈ p,m defined by g(z) = z −p + ∞
k=m bk z ,
we define the Hadamard product (or convolution) of f and g by
∞
X
−p
f (z) ∗ g(z) = (f ∗ g)(z) = z +
ak bk z k (p ∈ N).
k=m
For real or complex numbers
α1 , α2 , . . . , αq
β1 , β2 , . . . , βs βj ∈
/ Z−
0 = {0, −1, −2, . . . }; j = 1, 2, . . . , s ,
and
we consider the generalized hypergeometric function q Fs (see, for example, [17]) defined as
follows:
∞
X
(α1 )k · · · (αq )k z k
(1.2)
F
(α
,
.
.
.
,
α
;
β
,
.
.
.
,
β
;
z)
=
q s
1
q
1
s
(β1 )k · · · (βs )k k!
k=0
(q 5 s + 1; q, s ∈ N0 = N ∪ {0}; z ∈ U) ,
where (x)k denotes the Pochhammer symbol (or the shifted factorial) defined, in terms of the
Gamma function Γ, by
(
Γ(x + k)
x(x + 1)(x + 2) · · · (x + k − 1) (k ∈ N);
(x)k =
=
Γ(x)
1
(k = 0).
Corresponding to the function φp (α1 , . . . , αq ; β1 , . . . , βs ; z) given by
(1.3)
φp (α1 , . . . , αq ; β1 , . . . , βs ; z) = z −p q Fs (α1 , . . . , αq ; β1 , . . . , βs ; z),
we introduce a function φp,µ (α1 , . . . , αq ; β1 , . . . , βs ; z) defined by
φp (α1 , . . . , αq ; β1 , . . . , βs ; z) ∗ φp,µ (α1 , . . . , αq ; β1 , . . . , βs ; z)
1
= p
(µ > −p; z ∈ U∗ ).
µ+p
z (1 − z)
P
P
m,µ
We now define a linear operator Hp,q,s (α1 , . . . , αq ; β1 , . . . , βs ) : p,m −→ p,m by
(1.4)
(1.5)
m,µ
Hp,q,s
(α1 , . . . , αq ; β1 , . . . , βs )f (z) = φp,µ (α1 , . . . , αq ; β1 , . . . , βs ; z) ∗ f (z)
X
∗
αi , βj ∈ C \ Z−
;
i
=
1,
2
.
.
.
,
q;
j
=
1,
2,
.
.
.
,
s;
µ
>
−p;
f
∈
;
z
∈
U
.
0
p,m
For convenience, we write
m,µ
m,µ
(α1 , . . . , αq ; β1 , . . . , βs ) = Hp,q,s
(α1 ) and
Hp,q,s
1−p,µ
µ
Hp,q,s
(α1 ) = Hp,q,s
(α1 ) (µ > −p).
If f is given by (1.1), then from (1.5), we deduce that
∞
X
(µ + p)p+k (β1 )p+k · · · (βs )p+k
m,µ
−p
Hp,q,s (α1 )f (z) = z +
(1.6)
ak z k
(α
)
·
·
·
(α
)
1 p+k
q p+k
k=m
(µ > −p; z ∈ U∗ ).
and it is easily verified from (1.6) that
0
m,µ
m,µ+1
m,µ
(1.7)
z Hp,q,s
(α1 )f (z) = (µ + p) Hp,q,s
(α1 )f (z) − (µ + 2p) Hp,q,s
(α1 )f (z)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
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O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
3
and
(1.8)
0
m,µ
m,µ
m,µ
z Hp,q,s
(α1 + 1)f (z) = α1 Hp,q,s
(α1 )f (z) − (p + α1 )Hp,q,s
(α1 )f (z).
m,µ
We note that the linear operator Hp,q,s
(α1 ) is closely related to the Choi-Saigo-Srivastava
operator [5] for analytic functions and is essentially motivated by the operators defined and
0,µ
studied in [3]. The linear operator H1,q,s
(α1 ) was investigated recently by Cho and Kim [2],
1−p
whereas Hp,2,1 (c, 1; a; z) = Lp (a, c) (c ∈ R, a ∈
/ Z−
0 ) is the operator studied in [7]. In
particular, we have the following observations:
Z z
p
m,0
(i) Hp,s+1,s (p + 1, β1 , . . . , βs ; β1 , . . . , βs )f (z) = 2p
t2p−1 f (t) dt;
z
0
m,0
m,1
(ii) Hp,s+1,s
(p, β1 , ..., βs ; β1 , ..., βs )f (z) = Hp,s+1,s
(p + 1, β1 , ..., βs ; β1 , ..., βs )f (z) = f (z);
m,1
(iii) Hp,s+1,s
(p, β1 , . . . , βs ; β1 , . . . , βs )f (z) =
zf 0 (z) + 2pf (z)
;
p
m,2
(iv) Hp,s+1,s
(p + 1, β1 , . . . , βs ; β1 , . . . , βs )f (z) =
zf 0 (z) + (2p + 1)f (z)
;
p+1
1
= Dn+p−1 f (z)
− z)n+p
(n is an integer > −p), the operator studied in [6], and
Z z
δ
m,1−p
(vi) Hp,s+1,s (δ + 1, β2 , . . . , βs , 1; δ, β2 , . . . , βs )f (z) = δ+p
tδ+p−1 f (t) dt
z
0
(δ > 0; z ∈ U∗ ), the integral operator defined by (3.6).
1−p,n
(v) Hp,s+1,s
(β1 , β2 , . . . , βs , 1; β1 , . . . , βs )f (z) =
z p (1
Let Ω be the class of all functions φ which are analytic, univalent in U and for which φ(U) is
convex with φ(0) = 1 and < {φ(z)} > 0 in U.
m,µ
Next,
P by making use of the linear operator Hp,q,s (α1 ), we introduce the following subclasses
of p,m .
P
Definition 1.1. A function f ∈ p,m is said to be in the class MS µ,m
p,α1 (q, s; η; φ), if it satisfies
the following subordination condition:
(
)
0
m,µ
z Hp,q,s
(α1 )f (z)
1
(1.9)
−
+ η ≺ φ(z)
m,µ
p−η
Hp,q,s
(α1 )f (z)
(φ ∈ Ω, 0 5 η < p, µ > −p; z ∈ U).
In particular, for fixed parameters A and B (−1 5 B < A 5 1), we set
1 + Az
µ,m
MS p,α1 q, s; η;
= MS µ,m
p,α1 (q, s; η; A, B) .
1 + Bz
It is easy to see that
MS µ,0
1,α1 (q, s; η; φ) = MS µ+1,α1 (q, s; η; φ) and
MS µ,0
1,α1 (q, s; η; A, B) = MS µ+1,α1 (q, s; η; A, B)
are the function classes studied by Cho and Kim [2].
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
4
JAGANNATH PATEL AND A SHIS K U . PALIT
P
Definition 1.2. For fixed parameters A and B, a function f ∈ p,m is said to be in the class
MC µ,m
p,α1 (q, s; λ; A, B), if it satisfies the following subordination condition:
m,µ+1
m,µ
(α1 )f )0 (z) + λ(Hp,q,s
(α1 )f )0 (z)
z p+1 (1 − λ)(Hp,q,s
1 + Az
(1.10)
−
≺
p
1 + Bz
(−1 5 B < A 5 1, λ = 0, µ > −p; z ∈ U) .
2η
µ,m
To make the notation simple, we write MC p,α1 q, s; 0; 1 − p , −1 = MC µ,m
p,α1 (q, s; η), the
P
class of functions f ∈ p,m satisfying the condition:
n
0 o
m,µ
−< z p+1 Hp,q,s
(α1 )f (z) > η (0 5 η < p; z ∈ U).
Meromorphically multivalent functions have been extensively studied by (for example) Liu
and Srivastava [7], Cho et al. [4], Srivastava and Patel [18], Cho and Kim [2], Aouf [1], Srivastava et al. [19] and others.
The object of the present paper is to investigate several inclusion relationships and other interesting properties of certain subclasses of meromorphically multivalent functions which are
m,µ
defined here by means of the linear operator Hp,q,s
(α1 ) involving the generalized hypergeometric function. Some interesting applications of the Hadamard product concerning this and other
classes of integral operators are also considered. Relevant connections of the results presented
here with those obtained by earlier workers are also mentioned.
2. P RELIMINARIES
To prove our results, we need the following lemmas.
Lemma 2.1 ([8], see also [10]). Let the function h be analytic and convex(univalent) in U with
h(0) = 1. Suppose also that the function φ given by
(2.1)
φ(z) = 1 + cn z n + cn+1 z n+1 + · · ·
(n ∈ N)
is analytic in U. If
zφ0 (z)
≺ h(z) (<(κ) = 0, κ 6= 0; z ∈ U),
κ
then
Z
κ − κ z κ −1
φ(z) ≺ q(z) = z n
t n h(t) dt ≺ h(z) (z ∈ U)
n
0
and q is the best dominant.
φ(z) +
The following identities are well-known [21, Chapter 14].
Lemma 2.2. For real or complex numbers a, b, c (c ∈
/ Z−
0 ), we have
Z 1
Γ(b)Γ(c − b)
(2.2)
tb−1 (1 − t)c−b−1 (1 − tz)−a dt =
2 F1 (a, b; c; z) (<(c) > <(b) > 0)
Γ(c)
0
(2.3)
2 F1 (a, b; c; z) = 2 F1 (b, a; c; z)
z
−a
(2.4)
2 F1 (a, b; c; z) = (1 − z) 2 F1 a, c − b; c;
z−1
(2.5)
(b + 1)2 F1 (1, b; b + 1; z) = (b + 1) + bz 2 F1 (1, b + 1; b + 2; z)
and
(2.6)
2 F1
1
1, 1; 2;
2
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
= 2 ln 2.
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O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
5
We denote by P(γ), the class of functions ψ of the form
ψ(z) = 1 + c1 z + c2 z 2 + · · · ,
(2.7)
which are analytic in U and satisfy the inequality:
<{ψ(z)} > γ
(0 5 γ < 1; z ∈ U).
It is known [20] that if fj ∈ P(γj ) (0 5 γj < 1; j = 1, 2), then
(2.8)
(f1 ∗ f2 )(z) ∈ P(γ3 )
(γ3 = 1 − 2(1 − γ1 )(1 − γ2 )) .
The result is the best possible.
We now state
Lemma 2.3 ([12]). If the function ψ, given by (2.7) belongs to the class P(γ), then
<{ψ(z)} = 2γ − 1 +
2(1 − γ)
1 + |z|
(0 5 γ < 1; z ∈ U).
Lemma 2.4 ([8, 10]). Let the function Ψ : C2 ×U −→ C satisfy the condition < {Ψ(ix, y; z)} 5
ε for ε > 0, all real x and y 5 −n(1 + x2 )/2, where n ∈ N. If φ defined by (2.1) is analytic in
U and < {Ψ (φ(z), zφ0 (z); z)} > ε, then <{φ(z)} > 0 in U.
We now recall the following result due to Singh and Singh [16].
Lemma 2.5. Let the function Φ be analytic in U with Φ(0) = 1 and <{Φ(z)} > 1/2 in U.
Then for any function F , analytic in U, (Φ ∗ F )(U) is contained in the convex hull of F (U).
Lemma 2.6 ([13]). The function (1 − z)β = eβ log(1−z) , β 6= 0 is univalent in U, if β satisfies
either |β + 1| 5 1 or |β − 1| 5 1.
Lemma 2.7 ([9]). Let q be univalent in U, θ and Φ be analytic in a domain D containing q(U)
with Φ(w) 6= 0 when w ∈ q(U). Set Q(z) = zq 0 (z)φ(q(z)), h(z) = θ(q(z)) + Q(z) and
suppose that
(i) Q is starlike(univalent) in U with Q(0) = 0, Q0 (0) 6= 0 and
(ii) Q and h satisfy
0
zh(z)
Q (q(z)) zQ0 (z)
<
=<
+
> 0.
Q(z)
Φ(q(z))
Q(z)
If φ is analytic in U with φ(0) = q(0), φ(U) ⊂ D and
(2.9)
θ (φ(z)) + zφ0 (z)Φ (φ(z)) ≺ θ (q(z)) + zq 0 (z)Φ (q(z)) = h(z)
(z ∈ U),
then φ ≺ q and q is the best dominant of (2.9).
3. M AIN R ESULTS
Unless otherwise mentioned, we assume throughout the sequel that
α1 > 0, αi , βj ∈ R \ Z−
0 (i = 2, 3, . . . , q; j = 1, 2, . . . , s),
λ > 0, µ > −p
and
− 1 5 B < A 5 1.
Following the lines of proof of Cho and Kim [2] (see, also [4]), we can prove the following
theorem.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
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6
JAGANNATH PATEL AND A SHIS K U . PALIT
Theorem 3.1. Let φ ∈ Ω with
max < {φ(z)} < min {(µ + 2p − η)/(p − η), (α1 + p − η)/(p − η)}
(0 5 η < p).
z∈U
Then
µ,m
MS µ+1,m
(q, s; η; φ) ⊂ MS µ,m
p,α1
p,α1 (q, s; η; φ) ⊂ MS p,α1 +1 (q, s; η; φ) .
By carefully choosing the function φ in the above theorem, we obtain the following interesting consequences.
Example 3.1. The function
φ(z) =
1 + Az
1 + Bz
α
(0 < α 5 1; z ∈ U)
is analytic and convex univalent in U. Moreover,
α
α
1−A
1+A
< <{φ(z)} <
05
1−B
1+B
(0 < α 5 1, −1 < B < A 5 1; z ∈ U).
Thus, by Theorem 3.1 , we deduce that, if
α
1+A
µ + 2p − η α1 + p − η
< min
,
1+B
p−η
p−η
(0 < α 5 1, −1 < B < A 5 1),
then
µ,m
MS µ+1,m
(q, s; η; φ) ⊂ MS µ,m
p,α1
p,α1 (q, s; η; φ) ⊂ MS p,α1 +1 (q, s; η; φ) .
Example 3.2. The function
√ 2
1+ αz
2
√
φ(z) = 1 + 2 log
π
1− αz
(0 < α < 1; z ∈ U)
is in the class Ω (cf. [14]) and satisfies
√ 2
2
1+ α
√
<{φ(z)} < 1 + 2 log
π
1− α
(z ∈ U).
Thus, by using Theorem 3.1, we obtain that, if
√ 2
2
µ + 2p − η α1 + p − η
1+ α
√
1 + 2 log
< min
,
π
p−η
p−η
1− α
(0 < α < 1),
then
µ,m
MS µ+1,m
(q, s; η; φ) ⊂ MS µ,m
p,α1
p,α1 (q, s; η; φ) ⊂ MS p,α1 +1 (q, s; η; φ) .
Example 3.3. The function
∞ X
β+1
φ(z) = 1 +
αk z k (0 < α < 1, β = 0; z ∈ U)
β+k
k=1
belongs to the class Ω (cf. [15]) and satisfies
∞ X
β+1
<{φ(z)} < 1 +
αk
β
+
k
k=1
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
(0 < α < 1, β = 0).
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O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
Thus, by Theorem 3.1, if
∞ X
µ + 2p − η α1 + p − η
β+1
k
1+
α < min
,
β
+
k
p
−
η
p−η
k=1
7
(0 < α < 1, β = 0),
then
µ,m
MS µ+1,m
(q, s; η; φ) ⊂ MS µ,m
p,α1
p,α1 (q, s; η; φ) ⊂ MS p,α1 +1 (q, s; η; φ) .
Theorem 3.2. If f ∈ MC µ,m
p,α1 (q, s; λ; A, B), then
0
m,µ
z p+1 Hp,q,s
(α1 )f (z)
1 + Az
(3.1)
−
≺ ψ(z) ≺
(z ∈ U),
p
1 + Bz
where the function ψ given by

µ+p
A
A
Bz
B
+ 1− B
(1 + Bz)−1 2 F1 1, 1; λ(p+m)
+ 1; 1+Bz
(B 6= 0);
ψ(z) =

(µ+p)A
1 + µ+p+λ(p+m)
z
(B = 0)
is the best dominant of (3.1). Further,
f ∈ MC µ,m
p,α1 (q, s; pρ) ,
(3.2)
where
ρ=

µ+p
A
A
B
B
+ 1− B
(1 − B)−1 2 F1 1, 1; λ(p+m)
+ 1; B−1

(µ+p)A
1 − µ+p+λ(p+m)
The result is the best possible.
(B 6= 0);
(B = 0).
Proof. Setting
0
m,µ
z p+1 Hp,q,s
(α1 )f (z)
(3.3)
ϕ(z) = −
(z ∈ U),
p
we note that ϕ is of the form (2.1) and is analytic in U. Making use of the identity (1.7) in (3.3)
and differentiating the resulting equation, we get
n
0
0 o
p+1
m,µ
m,µ+1
z
(1
−
λ)
H
(α
)f
(z)
+
λ
H
(α
)f
(z)
0
1
1
p,q,s
p,q,s
zϕ (z)
(3.4) ϕ(z) +
=−
(µ + p)/λ
p
1 + Az
≺
(z ∈ U).
1 + Bz
Now, by applying Lemma 2.1 (with κ = (µ + p)/λ) in (3.4), we deduce that
0
m,µ
z p+1 Hp,q,s
(α1 )f (z)
−
p
µ+p
µ+p Z z
µ + p − λ(p+m)
1
+
Az
−1
≺ ψ(z) =
z
t λ(p+m)
dt
λ(p + m)
1 + Bz
0

A
A
µ+p
Bz

−1


(B 6= 0)
 B + 1 − B (1 + Bz) 2 F1 1, 1; λ(p + m) + 1; 1 + Bz
=

(µ + p)A


z
(B = 0)
1 +
µ + p + λ(p + m)
by a change of variables followed by the use of the identities (2.2), (2.3), (2.4) and (2.5), respectively. This proves the assertion (3.3).
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
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8
JAGANNATH PATEL AND A SHIS K U . PALIT
To prove (3.2), we follow the lines of proof of Theorem 1 in [18]. The result is the best
possible as ψ is the best dominant. This completes the proof of Theorem 3.2.
Setting A = 1 − (2η/p) , B = −1, µ = 0, m = 1 − p, α1 = λ = p and αi+1 = βi (i =
1, 2, . . . , s) in TheoremP
3.2 followed by the use of the identity (2.6), we get
Corollary 3.3. If f ∈ p satisfies
−< z p+1 ((p + 2)f 0 (z) + zf 00 (z)) > η (0 5 η < p; z ∈ U),
then
−<{z p+1 f 0 (z)} > η + 2(p − η)(ln 2 − 1) (z ∈ U).
The result is the best possible.
Putting A = 1 − (2η/p) , B = −1, µ = 0, m = 2 − p, α1 = λ = p and
1, 2, . . . , s) in TheoremP
3.2, we obtain the following result due to Pap [11].
Corollary 3.4. If f ∈ p,2−p satisfies
p(π − 2)
−< z p+1 ((p + 2)f 0 (z) + zf 00 (z)) > −
4−π
αi+1 = βi (i =
(z ∈ U),
then
−<{z p+1 f 0 (z)} > 0
(z ∈ U).
The result is the best possible.
The proof of the following result is much akin to that of Theorem 2 in [18] and we choose to
omit the details.
Theorem 3.5. If f ∈ MC µ,m
p,α1 (q, s; η) (0 5 η < p), then
h
n
0
0 oi
p+1
m,µ
m,µ+1
−< z
(1 − λ) Hp,q,s (α1 )f (z) + λ Hp,q,s (α1 )f (z) > η
(|z| < R(p, µ, λ, m)),
where
"p
R(p, µ, λ, m) =
(µ + p)2 + λ2 (p + m)2 − λ(p + m)
µ+p
1
# p+m
.
The result is the best possible.
m,µ
Upon replacing ϕ(z) by z p Hp,q,s
(α1 )f (z) in (3.3) and using the same techniques as in the
proof of Theorem 3.2, we get the following result.
P
Theorem 3.6. If f ∈ p,m satisfies
1 + Az
m,µ
m,µ+1
z p (1 − λ) Hp,q,s
(α1 )f (z) + λ Hp,q,s
(α1 )f (z) ≺
1 + Bz
(z ∈ U),
then
m,µ
z p Hp,q,s
(α1 )f (z) ≺ ψ(z) ≺
1 + Az
1 + Bz
(z ∈ U)
and
m,µ
< z p Hp,q,s
(α1 )f (z) > ρ
(z ∈ U),
where ψ and ρ are given as in Theorem 3.2. The result is the best possible.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
9
Letting
A=
2 F1
1
p
1, 1;
+ 1;
λ(p + m)
2
−1
2 − 2 F1 1, 1;
1
p
+ 1;
λ(p + m)
2
−1
,
B = −1, µ = 0, α1 = p and αi+1 = βi (i = 1, 2, . . . , s) in Theorem 3.6, we obtain
P
Corollary 3.7. If f ∈ p,m satisfies
p
1
3 − 2 2 F1 1, 1; λ(p+m) + 1; 2
λ
o (z ∈ U),
(3.5)
< (1 + λ)f (z) + z p+1 f 0 (z) > n
p
2 2 − F 1, 1; p + 1; 1
2
then
<{z p f (z)} >
1
2
1
λ(p+m)
2
(z ∈ U).
The result is the best possible.
P
For a function f ∈ p,m , we consider the integral operator Fδ,p defined by
Z z
δ
(3.6)
tδ+p−1 f (t) dt
Fδ,p (z) = Fδ,p (f )(z) = δ+p
z
0
!
∞
X
δ
−p
k
z ∗ f (z) (δ > 0; z ∈ U∗ ).
= z +
δ+p+k
k=m
P
It follows from (3.6) that Fδ,p (f ) ∈ p,m and
0
m,µ
m,µ
m,µ
(3.7)
z Hp,q,s
(α1 )Fδ,p (f ) (z) = δ Hp,q,s
(α1 )f (z) − (δ + p) Hp,q,s
(α1 )Fδ,p (f )(z).
Using (3.7) and the lines of proof of Theorem 1 [2], we obtain the following inclusion relation.
Theorem 3.8. Let φ ∈ Ω with maxz∈U < {φ(z)} < (δ + p − η)/(p − η) (0 5 η < p; δ > 0).
µ,m
If f ∈ MS µ,m
p,α1 (q, s; η; φ), then Fδ,p (f ) ∈ MS p,α1 (q, s; η; φ).
P
Theorem 3.9. If f ∈ p,m and the function Fδ,p (f ), defined by (3.6) satisfies
n
0
0 o
m,µ
m,µ
z p+1 (1 − λ) Hp,q,s
(α1 )Fδ,p (f ) (z) + λ Hp,q,s
(α1 )f (z)
1 + Az
−
≺
(z ∈ U),
p
1 + Bz
then
(
−<
)
0
m,µ
z p+1 Hp,q,s
(α1 )Fδ,p (f ) (z)
>%
p
(z ∈ U),
where
%=

A
δ
B
A
B
+ 1− B
(1 − B)−1 2 F1 1, 1; λ(p+m)
+ 1; B−1

1−
(B 6= 0)
δA
µ+p+λ(p+m)
(B = 0).
The result is the best possible.
Proof. If we let
(3.8)
0
m,µ
z p+1 Hp,q,s
(α1 )Fδ,p (f ) (z)
ϕ(z) = −
p
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
(z ∈ U),
http://jipam.vu.edu.au/
10
JAGANNATH PATEL AND A SHIS K U . PALIT
then ϕ is of the form (2.1) and is analytic in U. Using the identity (3.7) in (3.8) followed by
differentiation of the resulting equation, we get
zϕ0 (z)
1 + Az
ϕ(z) +
≺
(z ∈ U).
δ/λ
1 + Bz
The proof of the remaining part follows by employing the techniques that proved Theorem
3.2.
Upon setting A = 1 − (2η/p) , B = −1, λ = µ = 1, α1 = p + 1 and αi+1 = βi (i =
1, 2, . . . , s) in Theorem 3.9, we have
P
Corollary 3.10. If P
f ∈
C (p; η) (0 5 η < p), then the function Fδ,p (f ) defined by (3.6)
belongs to the class C (p; κ), where
δ
1
κ = η + (p − η) 2 F1 1, 1;
+ 1;
−1 .
p+m
2
The result is the best possible.
Remark 1. Under the hypothesis of Theorem 3.9 and using the fact that
Z z
0
0
δ
p+1
m,µ
m,µ
z
Hp,q,s (α1 )Fδ,p (f ) (z) = δ
tδ+p Hp,q,s
(α1 )f (t) dt (δ > 0; z ∈ U),
z 0
we obtain
Z z
0
δ
δ+p
m,µ
−<
t
Hp,q,s (α1 )f (t) dt > % (z ∈ U),
zδ 0
where % is given as in Theorem 3.9.
Following the same lines of proof as in Theorem 3.9, we obtain
P
Theorem 3.11. If f ∈ p,m and the function Fδ,p (f ) defined by (3.6) satisfies
1 + Az
m,µ
m,µ
(z ∈ U),
z p (1 − λ) Hp,q,s
(α1 )Fδ,p (f )(z) + λ Hp,q,s
(α1 )f (z) ≺
1 + Bz
then
m,µ
< z p Hp,q,s
(α1 )Fδ,p (f )(z) > % (z ∈ U),
where % is given as in Theorem 3.9. The result is the best possible.
In the special case when A = 1 − 2η, B = −1, λ = 1, µ = 1 − p, α1 = δ + 1, β1 =
δ, αi = βi (i = 2, 3, . . . ,P
s) and αs+1 = 1 in Theorem 3.11, we get
Corollary 3.12. If f ∈ p,m satisfies
<{z p f (z)} > η
(0 5 η < 1; z ∈ U),
then
<
δ
zδ
Z
z
δ+p−1
t
f (t) dt > η + (1 − η) 2 F1 1, 1;
0
δ
1
+ 1;
p+m
2
−1
(δ > 0; z ∈ U).
The result is the best possible.
P
Theorem 3.13. Let −1 5 Bj < Aj 5 1 (j = 1, 2). If fj ∈
p satisfies the following
subordination condition:
1 + Aj z
µ
µ+1
(3.9)
z p (1 − λ)Hp,q,s
(α1 )fj (z) + λHp,q,s
(α1 )fj (z) ≺
1 + Bj z
(j = 1, 2; z ∈ U),
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
11
then
(3.10)
µ
µ+1
< z p (1 − λ)Hp,q,s
(α1 )g(z) + λHp,q,s
(α1 )g(z) > τ
(z ∈ U),
where
µ
g(z) = Hp,q,s
(α1 )(f1 ∗ f2 )(z)
(3.11)
(z ∈ U∗ )
and
4(A1 − B1 )(A2 − B2 )
τ =1−
(1 − B1 )(1 − B2 )
1
µ+p
1
1 − 2 F1 1, 1;
+ 1;
.
2
λ
2
The result is the best possible when B1 = B2 = −1.
Proof. Setting
µ
µ+1
ϕj (z) = z p (1 − λ)Hp,q,s
(α1 )fj (z) + λHp,q,s
(α1 )fj (z)
(3.12)
(j = 1, 2; z ∈ U),
we note that ϕj is of the form (2.7) for each j = 1, 2 and using (3.9), we obtain
1 − Aj
; j = 1, 2
ϕj ∈ P(γj )
γj =
1 − Bj
so that by (2.8),
(3.13)
ϕ1 ∗ ϕ2 ∈ P(γ3 )
(γ3 = 1 − 2(1 − γ1 )(1 − γ2 )) .
Using the identity (1.7) in (3.12), we conclude that
µ
Hp,q,s
(α1 )fj (z)
Z
µ + p −p− µ+p z µ+p −1
λ
=
z
t λ
ϕj (t)dt
λ
0
(j = 1, 2; z ∈ U∗ )
which, in view of (3.11) yields
µ
Hp,q,s
(α1 )g(z)
µ + p −p− µ+p
λ
=
z
λ
Z
z
t
µ+p
−1
λ
ϕ0 (t)dt
(z ∈ U∗ ),
0
where, for convenience
(3.14)
µ
µ+1
ϕ0 (z) = z p (1 − λ)Hp,q,s
(α1 )g(z) + λHp,q,s
(α1 )g(z)
Z
µ + p − µ+p z µ+p −1
=
z λ
t λ
(ϕ1 ∗ ϕ2 )(t) dt (z ∈ U).
λ
0
Now, by using (3.13) in (3.14) and by appealing to Lemma 2.3 and Lemma 2.5, we get
Z
µ + p 1 µ+p −1
<{ϕ0 (z)} =
s λ
<(ϕ1 ∗ ϕ2 )(sz) ds
λ
0
Z
2(1 − γ3 )
µ + p 1 µ+p −1
λ
=
s
2γ3 − 1 +
ds
λ
1 + s|z|
0
Z
2(1 − γ3 )
µ + p 1 µ+p −1
>
s λ
2γ3 − 1 +
ds
1+s
λ
0
Z
4(A1 − B1 )(A2 − B2 )
µ + p 1 µ+p −1
−1
=1−
1−
s λ
(1 + s) ds
(1 − B1 )(1 − B2 )
λ
0
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
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12
JAGANNATH PATEL AND A SHIS K U . PALIT
4(A1 − B1 )(A2 − B2 )
=1−
(1 − B1 )(1 − B2 )
= τ (z ∈ U).
1
µ+p
1
1 − 2 F1 1, 1;
+ 1;
2
λ
2
This proves the assertion (3.10).
P
When B1 = B2 = −1, we consider the functions fj ∈ p defined by
Z
µ + p −p− µ+p z µ+p −1 1 + Aj t
µ
λ
Hp,q,s (α1 )fj (z) =
z
t λ
dt
λ
1−t
0
(j = 1, 2; z ∈ U∗ ).
Then it follows from (3.14) that and Lemma 2.2 that
Z
µ + p 1 µ+p −1
(1 + A1 )(1 + A2 )
ϕ0 (z) =
s λ
1 − (1 + A1 )(1 + A2 ) +
ds
λ
1 − sz
0
µ+p
z
−1
= 1 − (1 + A1 )(1 + A2 ) + (1 + A1 )(1 + A2 )(1 − z) 2 F1 1, 1;
+ 1;
λ
z−1
1
µ+p
1
+ 1;
as z → 1− ,
−→ 1 − (1 + A1 )(1 + A2 ) + (1 + A1 )(1 + A2 ) 2 F1 1, 1;
2
λ
2
which evidently completes the proof of Theorem 3.13.
By taking Aj = 1 − 2ηj , Bj = −1 (j = 1, 2), µ = 0, α1 = p and αi+1 = βi (i =
1, 2, . . . , s) in Theorem 3.13, we get the following result which refines the corresponding work
of Yang [22, Theorem 4].
P
Corollary 3.14. If each of the functions fj ∈ p satisfies
λ 0
p
< z (1 + λ) fj (z) + zfj (z)
> ηj
p
(0 5 ηj < 1, j = 1, 2; z ∈ U),
then
λ
p
0
< z (1 + λ) (f1 ∗ f2 )(z) + z(f1 ∗ f2 ) (z)
>σ
p
(z ∈ U),
where
1
p
1
σ = 1 − 4(1 − η1 )(1 − η2 ) 1 − 2 F1 1, 1; + 1;
.
2
λ
2
The result is the best possible.
For Aj = 1 − 2ηj , Bj = −1 (j = 1, 2), µ = 0, λ = 1, α1 = p + 1 and αi+1 = βi (i =
1, 2, . . . , s) in Theorem 3.13, we obtain
P
Corollary 3.15. If each of the functions fj ∈ p satisfies
< {z p fj (z)} > ηj
(0 5 ηj < 1, j = 1, 2; z ∈ U),
then
1
1
< {z (f1 ∗ f2 )(z)} > 1 − 4(1 − η1 )(1 − η2 ) 1 − 2 F1 1, 1; p + 1;
2
2
p
(z ∈ U).
The result is the best possible.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
13
Theorem 3.16. Let −1 5 Bj < Aj 5 1 (j = 1, 2). If each of the functions fj ∈
m,µ+1
z p Hp,q,s
(α1 )fj (z) ≺
(3.15)
1 + Aj z
1 + Bj z
P
p,m
satisfies
(j = 1, 2; z ∈ U),
m,µ
then the function h = Hp,q,s
(α1 )(f1 ∗ f2 ) satisfies
m,µ+1
Hp,q,s (α1 )h(z)
<
>0
m,µ
Hp,q,s
(α1 )h(z)
(z ∈ U),
provided
(3.16)
(A1 − B1 )(A2 − B2 )
<
(1 − B1 )(1 − B2 )
2µ + 3p + m
n
.
o2
µ+p 1
2 (p + m) 2 F1 1, 1; p+m ; 2 − 2 + 2(µ + p)
Proof. From (3.15), we have
z
p
m,µ+1
Hp,q,s
(α1 )fj (z)
∈ P(γj )
Thus, it follows from (2.8) that
(
< z
(3.17)
p
m,µ+1
Hp,q,s
(α1 )h(z)
1 − Aj
γj =
; j = 1, 2 .
1 − Bj
)
0
m,µ+1
z z p Hp,q,s
(α1 )h (z)
+
µ+p
m,µ+1
m,µ+1
= < z p Hp,q,s
(α1 )f1 (z) ∗ z p Hp,q,s
(α1 )f2 (z)
>1−
2(A1 − B1 )(A2 − B2 )
(1 − B1 )(1 − B2 )
(z ∈ U),
which in view of Lemma 2.1 for
(A1 − B1 )(A2 − B2 )
,
(1 − B1 )(1 − B2 )
B = −1, n = p + m and κ = µ + p
A = −1 + 4
yields
m,µ+1
(3.18) < z p Hp,q,s
(α1 )h(z)
(A1 − B1 )(A2 − B2 )
>1+
(1 − B1 )(1 − B2 )
2 F1
µ+p 1
1, 1;
;
p+m 2
− 2 (z ∈ U).
From (3.18), by using Theorem 3.6 for
(A1 − B1 )(A2 − B2 )
A = −1 − 4
(1 − B1 )(1 − B2 )
B = −1 and λ = 1,
2 F1
µ+p 1
1, 1;
;
p+m 2
−2 ,
we deduce that
(A1 − B1 )(A2 − B2 )
(3.19) < {z ϑ(z)} > 1 − 2
(1 − B1 )(1 − B2 )
p
2 F1
µ+p 1
1, 1;
;
p+m 2
2
−2
(z ∈ U),
m,µ
where ϑ(z) = z p Hp,q,s
(α1 )h(z). If we set
ϕ(z) =
m,µ+1
Hp,q,s
(α1 )h(z)
m,µ
Hp,q,s (α1 )h(z)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
(z ∈ U),
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14
JAGANNATH PATEL AND A SHIS K U . PALIT
then ϕ is of the form (2.1), analytic in U and a simple calculation gives
(3.20) z
p
m,µ+1
Hp,q,s
(α1 )h(z)
0
m,µ+1
z z p Hp,q,s
(α1 )h (z)
+
µ+p
zϕ0 (z)
2
= ϑ(z) (ϕ(z)) +
= Ψ (ϕ(z), zϕ0 (z); z) (z ∈ U),
µ+p
where Ψ(u, v; z) = ϑ(z) {u2 + (v/(µ + p))}. Thus, by applying (3.17) in (3.20), we get
< {Ψ (ϕ(z), zϕ0 (z); z)} > 1 − 2
(A1 − B1 )(A2 − B2 )
(1 − B1 )(1 − B2 )
(z ∈ U).
Now, for all real x, y 5 −(p + m)(1 + x2 )/2, we have
y
2
< {Ψ(ix, y; z)} =
− x <{ϑ(z)}
µ+p
p+m
2(µ + p) 2
2
5−
1+x +
x <{ϑ(z)}
2(µ + p)
p+m
(A1 − B1 )(A2 − B2 )
p+m
5−
<{ϑ(z)} 5 1 − 2
2(µ + p)
(1 − B1 )(1 − B2 )
(z ∈ U),
by (3.16) and (3.19). Thus, by Lemma 2.4, we get <{ϕ(z)} > 0 in U. This completes the proof
of Theorem 3.16.
Taking Aj = 1 − 2ηj , Bj = −1 (j = 1, 2), µ = 0, λ = 1, α1 = p + 1 and αi+1 =
βi (i = 1, 2, . . . , s) in Theorem 3.16, we have P
Corollary 3.17. If each of the functions fj ∈ p,m satisfies
<{z p fj (z)} > ηj
(0 5 ηj < 1, j = 1, 2; z ∈ U),
then
<
z 2p (f1 ∗ f2 )(z)
Rz
t2p−1 (f1 ∗ f2 )(t) dt
0
>0
(z ∈ U),
provided
3p + m
.
n
o2
p
1
2 (p + m) 2 F1 1, 1; p+m ; 2 − 2 + 2p
P
Theorem 3.18. If f ∈ MC µ,m
p,α1 (q, s; λ; A, B) and g ∈
p,m satisfies (3.5), then f ∗ g ∈
µ,m
MC p,α1 (q, s; λ; A, B).
(1 − η1 )(1 − η2 ) <
Proof. From Corollary 3.7, it follows that <{g(z)} > 1/2 in U. Since
m,µ
m,µ+1
z p+1 (1 − λ)(Hp,q,s
(α1 )(f ∗ g))0 (z) + λ(Hp,q,s
(α1 )(f ∗ g))0 (z)
−
p
p+1
m,µ
m,µ+1
z
(1 − λ)Hp,q,s (α1 )f )0 (z) + λ(Hp,q,s
(α1 )f )0 (z)
=
∗ g(z)
p
(z ∈ U)
and the function (1 + Az)/(1 + Bz) is convex(univalent) in U, the assertion of the theorem
follows from (1.10) and Lemma 2.5.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
15
Theorem 3.19. Let 0P6= β ∈ C and 0 < γ 5 p be such that either |1 + 2βγ| 5 1 or
|1 − 2βγ| 5 1. If f ∈ p satisfies
µ+1
Hp,q,s (α1 )f (z)
γ
(3.21)
<
<1+
(z ∈ U),
µ
Hp,q,s (α1 )f (z)
µ+p
then
β
µ
(α1 )f (z) ≺ q(z) = (1 − z)2βγ
z p Hp,q,s
(z ∈ U)
and q is the best dominant.
Proof. Letting
(3.22)
β
µ
ϕ(z) = z p Hp,q,s
(α1 )f (z)
(z ∈ U)
and choosing the principal branch in (3.22), we note that ϕ is analytic in U with ϕ(0) = 1.
Differentiating (3.22) logarithmically, we deduce that
(
)
0
µ
0
z
H
(α
)f
(z)
zϕ (z)
1
p,q,s
=β p+
(z ∈ U),
µ
ϕ(z)
Hp,q,s (α1 )f (z)
which in view of the identities (1.7) and (3.21) give
2γ
1− 1−
z
zϕ0 (z)
p
(3.23)
−p +
≺ −p
β ϕ(z)
1−z
(z ∈ U).
If we take q(z) = (1 − z)2βγ , θ(z) = −p, Φ(z) = 1/βz in Lemma 3.11, then by Lemma 2.6,
q is univalent in U. Further, it is easy to see that q, θ and Φ satisfy the hypothesis of Lemma 2.7.
Since
2γ z
Q(z) = zq 0 (z)Φ(q(z)) = −
1−z
is starlike (univalent) in U,
0 −p + (p − 2γ)z
zh (z)
1
h(z) =
and <
=<
> 0 (z ∈ U),
1−z
Q(z)
1−z
it is readily seen that the conditions (i) and (ii) of Lemma 2.7 are satisfied. Thus, the assertion
of the theorem follows from (3.23) and Lemma 2.7.
Putting µ = 0, γ = p(1 − η), β = −1/2γ, α1 = p and αi+1 = βi (i = 1, 2, . . . , s) in
Theorem 3.19, we deduce
Pthat
Corollary 3.20. If f ∈ p satisfies
0 zf (z)
−<
> p η (0 5 η < 1; z ∈ U),
f (z)
then
1
1
< {z p f (z)} 2p(1 − η) >
2
−
(z ∈ U).
The result is the best possible.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
16
JAGANNATH PATEL AND A SHIS K U . PALIT
R EFERENCES
[1] M.K. AOUF, Certain subclasses of meromorphically multivalent functions associated with generalized hypergeometric function, Comput. Math. Appl., 55 (2008), 494–509.
[2] N.E. CHO AND I.H. KIM, Inclusion properties of certain classes of meromorphic functions associated with the generalized hypergeometric function, Appl. Math. Comput., 187 (2007), 115–121
[3] N.E. CHO AND K.I. NOOR, Inclusion properties for certain classes of meromorphic functions
associated with Choi-Saigo-Srivastava operator, J. Math. Anal. Appl., 320 (2006), 779–786.
[4] N.E. CHO, O.S. KWON AND H.M. SRIVASTAVA, Inclusion relationships for certain subclasses of
meromorphic functions associated with a family of multiplier transformations, Integral Transforms
Spec. Funct., 16 (2005), 647–659.
[5] J.H. CHOI, M. SAIGO AND H.M. SRIVASTAVA, Some inclusion properties of a certain family of
integral operators, J. Math. Anal. Appl., 276 (2002), 432–445.
[6] M.R. GANIGI AND B.A. URALEGADDI, New criteria for meromorphic univalent functions, Bull.
Math. Soc. Sci. Math. Roumanie (N.S.) 33(81) (1989), 9–13.
[7] J.-L. LIU AND H.M. SRIVASTAVA, A linear operator and associated families of meromorphically
multivalent functions, J. Math. Anal. Appl., 259 (2000), 566–581.
[8] S.S. MILLER AND P.T. MOCANU, Differential subordinations and univalent functions, Michigan
Math. J., 28 (1981), 157–171.
[9] S.S. MILLER AND P.T. MOCANU, On some classes of first-order differential subordinations,
Michigan Math. J., 32 (1985), 185–195.
[10] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and Applications, Series
on Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker, New
York and Basel, 2000.
[11] M. PAP, On certain subclasses of meromorphic m-valent close-to-convex functions, Pure Math.
Appl., 9 (1998), 155–163.
[12] D.Ž. PASHKOULEVA, The starlikeness and spiral-convexity of certain subclasses of analytic
functions, in: H.M. Srivastava and S. Owa(Editors), Current Topics in Analytic Function Theory,
pp.266-273, World Scientific Publishing Company, Singapore, New Jersey, London and Hongkong,
1992.
[13] M.S. ROBERTSON, Certain classes of starlike functions, Michigan Math. J., 32 (1985), 134–140.
[14] F. RONNING, Uniformly convex functions and a corresponding class of starlike functions, Proc.
Amer. Math. Soc., 118 (1993), 189–196.
[15] S. RUSCHEWEYH, New criteria for univalent functions, Proc. Amer. Math. Soc., 49 (1975), 109–
115.
[16] R. SINGH AND S. SINGH, Convolution properties of a class of starlike functions, Proc. Amer.
Math. Soc., 106 (1989), 145–152.
[17] H.M. SRIVASTAVA AND P.W. KARLSSON, Multiple Gaussian Hypergeometric Series, Halsted
Press, Ellis Horwood Limited, Chichester, John Wiley and Sons, New York, Chichester, Brisbane,
Toronto, 1985.
[18] H.M. SRIVASTAVA AND J. PATEL, Applications of differential subordinations to certain subclasses of meromorphically multivalent functions, J. Ineq. Pure Appl. Math., 6(3) (2005), Art. 88.
[ONLINE http://jipam.vu.edu.au/article.php?sid=752].
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/
O N CERTAIN SUBCLASSES OF MEROMORPHICALLY.....
17
[19] H.M. SRIVASTAVA, D.-G. YANG AND N-ENG XU, Some subclasses of meromorphically multivalent functions associated with a linear operator, Appl. Math. Comput., 195 (2008), 11–23.
[20] J. STANKIEWICZ AND Z. STANKIEWICZ, Some applications of the Hadamard convolution in
the theory of functions, Ann. Univ. Mariae Curie-Skłodowska Sect.A, 40 (1986), 251–265.
[21] E.T. WHITTAKER AND G.N. WATSON, A Course on Modern Analysis: An Introduction to the
General Theory of Infinite Processes and Analytic Functions; With an Account of the Principal
Transcendental Functions, Fourth Ed.(Reprinted), Cambridge University Press, Cambridge, 1927.
[22] D.-G. YANG, Certain convolution operators for meromorphic functions, Southeast Asian Bull.
Math., 25 (2001), 175–186.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 13, 17 pp.
http://jipam.vu.edu.au/