Mathematical and Computer Modelling 49 (2009) 955–965
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Mathematical and Computer Modelling journal homepage: www.elsevier.com/locate/mcm
Some classes of analytic and multivalent functions involving a linear operator N-Eng Xu a,∗ , Ding-Gong Yang b a
Department of Mathematics, Changshu Institute of Technology, Changshu, Jiangsu 215500, China
b
Department of Mathematics, Suzhou University, Suzhou, Jiangsu 215006, China
article
a b s t r a c t
info
Article history: Received 8 January 2008 Accepted 25 April 2008
By using a linear operator, which is defined here by means of the Hadamard product (or convolution), the authors introduce some new classes of analytic and multivalent functions in the open unit disk and investigate their inclusion relationships and convolution properties. Integral transforms of functions in these classes are also discussed. © 2008 Elsevier Ltd. All rights reserved.
Keywords: Analytic function Multivalent function Convex univalent function Linear operator Hadamard product (or convolution) Subordination Integral operator
1. Introduction and preliminaries Let the functions f (z ) =
∞ X
an z p+n
and
g (z ) =
∞ X
n =0
bn z p+n
(p ∈ N = {1, 2, 3, . . .})
n=0
be analytic in the open unit disk U = {z : |z | < 1}. Then the Hadamard product (or convolution) (f ∗ g )(z ) of f (z ) and g (z ) is defined by
(f ∗ g )(z ) =
∞ X
an bn z p+n = (g ∗ f )(z ).
(1.1)
n=0
Let Ap denote the class of functions f (z ) normalized by f (z ) = z p +
∞ X
an z p+n
(p ∈ N ),
(1.2)
n=1
which are analytic in U. A function f (z ) ∈ Ap is said to be in the class Sp∗ (α) if it satisfies
Re
∗
zf 0 (z ) pf (z )
> α (z ∈ U )
Corresponding author. E-mail addresses:
[email protected],
[email protected] (N.-E. Xu).
0895-7177/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2008.04.005
(1.3)
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N.-E. Xu, D.-G. Yang / Mathematical and Computer Modelling 49 (2009) 955–965
for some α(α < 1). Note that, for 0 ≤ α < 1, Sp∗ (α) is the class of p-valently starlike functions of order α in U. Also we write A1 = A and S1∗ (α) = S ∗ (α). A function f (z ) ∈ A is said to be prestarlike of order α in U if z
(1 − z )2(1−α)
∗ f (z ) ∈ S ∗ (α) (α < 1).
(1.4)
We denote this class by R(α) (see [7]). Clearly, a function f (z ) ∈ A is in the class R(0) if and only if f (z ) is convex univalent in U and
1
R
2
= S∗
1
2
.
We now define the function ϕp (a, c ; z ) by
ϕp (a, c ; z ) = z p +
∞ X (a)n n=1
(c )n
z p+n
(z ∈ U ),
(1.5)
where c 6∈ {0, −1, −2, . . .}
and (b)n = b(b + 1) · · · (b + n − 1)
(n ∈ N ).
Corresponding to the function ϕp (a, c ; z ), Saitoh [8] introduced and studied a linear operator Lp (a, c ) on Ap by the following Hadamard product (or convolution): Lp (a, c )f (z ) = ϕp (a, c ; z ) ∗ f (z )
(f (z ) ∈ Ap ).
(1.6)
For p = 1, L1 (a, c ) on A was first defined by Carlson and Shaffer [1]. It is known [8] that z (Lp (a, c )f )0 (z ) = aLp (a + 1, c )f (z ) − (a − p)Lp (a, c )f (z )
(f (z ) ∈ Ap ).
(1.7)
Let P be the class of analytic functions h(z ) with h(0) = 1, which are convex and univalent in U and for which Re h(z ) > 0 (z ∈ U ).
(1.8)
Let f (z ) and g (z ) be analytic in U. The function f (z ) is subordinate to g (z ), written f (z ) ≺ g (z ), if g (z ) is univalent in U, f (0) = g (0) and f (U ) ⊂ g (U ). Throughout our present investigation, we assume that p, k ∈ N , c 6∈ {0, −1, −2, . . .},
εk = exp
2π i k
,
(1.9)
and fp,k (a, c ; z ) =
k −1 1 X −jp ε (Lp (a, c )f )(εkj z ) = z p + · · · k j=0 k
(f (z ) ∈ Ap ).
(1.10)
Obviously, for k = 1, we have fp,1 (a, c ; z ) = Lp (a, c )f (z ). In this paper we introduce and investigate the following subclasses of the class Ap : Tp,k (a, c ; h),
Kp,k (a, c ; h) and
Cp,k (α; a, c ; h)
(h(z ) ∈ P ).
Definition 1. A function f (z ) ∈ Ap is said to be in the class Tp,k (a, c ; h) if it satisfies z (Lp (a, c )f )0 (z )
≺ h(z ),
pfp,k (a, c ; z )
(1.11)
where h(z ) ∈ P
and fp,k (a, c ; z ) 6= 0 (0 < |z | < 1).
Note that
Tp,1
1, 1;
1 + (1 − 2α)z 1−z
= Sp∗ (α) (0 ≤ α < 1).
Furthermore, the class
T1,2
1, 1;
1+z 1−z
= Ss∗
has been studied by several authors (see Sakaguchi [9], Owa et al. [4], Yang and Liu [11], and others).
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Definition 2. A function f (z ) ∈ Ap is said to be in the class Kp,k (a, c ; h) if it satisfies z (Lp (a, c )f )0 (z ) pgp,k (a, c ; z )
≺ h( z )
(1.12)
for some g (z ) ∈ Tp,k (a, c ; h), where h(z ) ∈ P and gp,k (a, c ; z ) is defined as in (1.10). If we let k=a=c=1
and h(z ) =
1+z 1−z
,
then Kp,k (a, c ; h) reduces to the familiar class of p-valently close-to-convex functions in U. Definition 3. A function f (z ) ∈ Ap is said to be in the class Cp,k (δ; a, c ; h) if it satisfies
(1 − δ)
z (Lp (a, c )f )0 (z ) pgp,k (a, c ; z )
+δ
(z (Lp (a, c )f )0 (z ))0 (z ) ≺ h(z ) pgp0 ,k (a, c ; z )
(1.13)
for some δ(δ ≥ 0) and g (z ) ∈ Tp,k (a, c ; h), where h( z ) ∈ P
gp0 ,k (a, c ; z ) 6= 0 (0 < |z | < 1).
and
In order to derive our results, we need the following lemmas. Lemma 1. Let f (z ) ∈ Tp,k (a, c ; h). Then zfp0,k (a, c ; z ) pfp,k (a, c ; z )
≺ h(z ).
(1.14)
Proof. For f (z ) ∈ Ap , we find from (1.10) that k−1 1 X −mp ε (Lp (a, c )f )(εkm+j z ) k m=0 k
j
fp,k (a, c ; εk z ) =
k−1 εkjp X
=
k
εk−(m+j)p (Lp (a, c )f )(εkm+j z )
m=0
= εkjp fp,k (a, c ; z ) (j ∈ {0, 1, . . . , k − 1}) and fp0,k (a, c ; z ) =
k−1 1 X j(1−p) (Lp (a, c )f )0 (εkj z ). ε k j =0 k
Hence zfp0,k (a, c ; z ) pfp,k (a, c ; z )
=
=
j k−1 j(1−p) 1 X εk z (Lp (a, c )f )0 (εk z )
pfp,k (a, c ; z )
k j =0
j k−1 j 1 X εk z (Lp (a, c )f )0 (εk z )
k j =0
j
pfp,k (a, c ; εk z )
(z ∈ U ).
(1.15)
Since f (z ) ∈ Tp,k (a, c ; h), we have
εkj z (Lp (a, c )f )0 (εkj z ) j
pfp,k (a, c ; εk z )
≺ h(z )
(1.16)
for j ∈ {0, 1, . . . , k − 1}. In view of h(z ) being convex univalent in U, the subordination (1.14) follows immediately from (1.15) and (1.16). By Lemma 1 we see that, if f (z ) ∈ Tp,k (a, c ; h), then fp,k (a, c ; z ) ∈ Sp∗ (0).
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Lemma 2 (Ruscheweyh [7]). Let α < 1, f (z ) ∈ R(α) and g (z ) ∈ S ∗ (α). Then, for any analytic function F (z ) in U, f ∗ (gF ) f ∗g
(U ) ⊂ co(F (U )),
where co(F (U )) denotes the convex hull of F (U ). Lemma 3 (Miller and Mocanu [3]). Let h(z ) be analytic and convex univalent in U and let w(z ) be analytic in U with Re w(z ) ≥ 0(z ∈ U ). If q(z ) is analytic in U and q(0) = h(0), then the subordination q(z ) + w(z )zq0 (z ) ≺ h(z ) implies that q(z ) ≺ h(z ). Lemma 4 (Miller and Mocanu [2]). Let β (β 6= 0) and γ be complex numbers and let h(z ) be analytic and convex univalent in U with Re{β h(z ) + γ } > 0 (z ∈ U ). If q(z ) is analytic in U with q(0) = h(0), then the subordination q(z ) +
zq0 (z )
β q(z ) + γ
≺ h(z )
implies that q(z ) ≺ h(z ). 2. Inclusion relationships Theorem 1. Let h(z ) ∈ P and Re h(z ) > β
(z ∈ U ; 0 ≤ β < 1).
0 < a1 < a2
and
(2.1)
If a2 ≥ 2p(1 − β),
(2.2)
then Tp,k (a2 , c ; h) ⊂ Tp,k (a1 , c ; h). Proof. Let us define
ψ(z ) = z +
∞ X (a1 )n n=1
(a2 )n
z n+1
(z ∈ U ; 0 < a1 < a2 ).
Then
ϕp (a1 , a2 ; z ) z p−1
= ψ(z ) ∈ A,
(2.3)
where ϕp (a1 , a2 ; z ) is defined as in (1.5), and z
(1 − z )
a2
∗ ψ(z ) =
z
(1 − z )a1
.
(2.4)
By (2.4) we have z
(1 − z )
a2
a1 a2 ∗ ψ(z ) ∈ S ∗ 1 − ⊂ S∗ 1 − 2
2
for 0 < a1 < a2 , which implies that
a2 ψ(z ) ∈ R 1 − .
(2.5)
2
For f (z ) ∈ Ap it is easy to see that Lp (a1 , c )f (z ) = ϕp (a1 , a2 ; z ) ∗ Lp (a2 , c )f (z ),
(2.6)
z (Lp (a1 , c )f ) (z ) = ϕp (a1 , a2 ; z ) ∗ (z (Lp (a2 , c )f ) (z )) 0
0
(2.7)
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959
and k−1 1 X −jp ε ϕp (a1 , a2 ; z ) ∗ (Lp (a2 , c )f )(εkj z ) k j =0 k
fp,k (a1 , c ; z ) =
= ϕp (a1 , a2 ; z ) ∗ fp,k (a2 , c ; z ).
(2.8)
Making use of (2.3) and (2.6)–(2.8), we deduce that z (Lp (a1 , c )f )0 (z ) pfp,k (a1 , c ; z )
=
(z p−1 ψ(z )) ∗ (z (Lp (a2 , c )f )0 (z )) p(z p−1 ψ(z )) ∗ fp,k (a2 , c ; z )
=
ψ(z ) ∗ (w(z )F (z )) (f (z ) ∈ Ap ), ψ(z ) ∗ w(z )
(2.9)
where
w(z ) =
fp,k (a2 , c ; z ) z p−1
∈A
and F (z ) =
z (Lp (a2 , c )f )0 (z ) pfp,k (a2 , c ; z )
.
Let f (z ) ∈ Tp,k (a2 , c ; h). Then F (z ) ≺ h(z )
(2.10)
and it follows from Lemma 1 that z w 0 (z )
w(z )
=
zfp0,k (a2 , c ; z ) fp,k (a2 , c ; z )
+ 1 − p ≺ ph(z ) + 1 − p.
(2.11)
Noting that (2.1), (2.2) and (2.11), we obtain
Re
z w 0 (z )
w(z )
> pβ + 1 − p ≥ 1 −
a2 2
(z ∈ U ),
that is,
a2 . w(z ) ∈ S ∗ 1 −
(2.12)
2
Now, in view of (2.5), (2.9), (2.10) and (2.12), an application of Lemma 2 yields z (Lp (a1 , c )f )0 (z ) pfp,k (a1 , c ; z )
≺ h(z )
because h(z ) is convex univalent in U. Therefore f (z ) ∈ Tp,k (a1 , c ; h) and the proof of Theorem 1 is completed. Theorem 2. Let h(z ) ∈ P and Re h(z ) > β
(z ∈ U ; 0 ≤ β < 1).
0 < a1 < a2
and
If a2 ≥ 2p(1 − β),
then Kp,k (a2 , c ; h) ⊂ Kp,k (a1 , c ; h). Proof. Let f (z ) ∈ Kp,k (a2 , c ; h). Then there exists a function g (z ) ∈ Tp,k (a2 , c ; h) such that z (Lp (a2 , c )f )0 (z ) pgp,k (a2 , c ; z )
≺ h(z ) (h(z ) ∈ P ).
Furthermore, it follows from Lemma 1 that zgp0 ,k (a2 , c ; z ) pgp,k (a2 , c ; z )
≺ h(z ),
and Theorem 1 leads to g (z ) ∈ Tp,k (a1 , c ; h).
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By virtue of the proof of Theorem 1, we find that z (Lp (a1 , c )f )0 (z )
ϕp (a1 , a2 ; z ) ∗ (z (Lp (a2 , c )f )0 (z )) pϕp (a1 , a2 ; z ) ∗ gp,k (a2 , c ; z ) ψ(z ) ∗ (w(z )F (z )) , = ψ(z ) ∗ w(z )
=
pgp,k (a1 , c ; z )
where
ϕp (a1 , a2 ; z )
a2 , ∈R 1− 2 a2 gp,k (a2 , c ; z ) ∈ S∗ 1 − w(z ) = p−1 ψ(z ) =
z p−1
z
2
and F (z ) =
z (Lp (a2 , c )f )0 (z ) pgp,k (a2 , c ; z )
≺ h(z ).
Consequently, an application of Lemma 2 yields z (Lp (a1 , c )f )0 (z )
≺ h(z ),
pgp,k (a1 , c ; z )
which shows that f (z ) ∈ Kp,k (a1 , c ; h) and the proof of Theorem 2 is completed.
Taking a1 = a > 0 and a2 = a + 1 in Theorems 1 and 2, we have: Corollary 1. Let h(z ) ∈ P and a > 0 satisfy
Re h(z ) > max 0, 1 −
a+1
2p
(z ∈ U ).
Then Tp,k (a + 1, c ; h) ⊂ Tp,k (a, c ; h) and
Kp,k (a + 1, c ; h) ⊂ Kp,k (a, c ; h).
For appropriate choice of the function h(z ) involved in Theorems 1 and 2, we can obtain a number of useful consequences. Corollary 2. Let 0 < γ ≤ 1, −1 ≤ B < A ≤ 1 and h(z ) =
1 + Az
γ
1 + Bz
(z ∈ U ).
(2.13)
If 0 < a1 < a2
and
a2 ≥ 2p 1 −
1−A 1−B
γ
,
then Tp,k (a2 , c ; h) ⊂ Tp,k (a1 , c ; h) and
Kp,k (a2 , c ; h) ⊂ Kp,k (a1 , c ; h).
Proof. The analytic function h(z ) defined by (2.13) is convex univalent in U (see [10]), h(0) = 1, and h(U ) is symmetric with respect to the real axis. Thus h(z ) ∈ P and Re h(z ) > β = h(−1) =
1−A 1−B
γ
≥ 0 (z ∈ U ).
Hence we have the corollary by using Theorems 1 and 2.
Corollary 3. Let 0 < ρ ≤ 1, γ ≥ 0 and h(z ) = 1 +
∞ X γ +1 ρ n z n (z ∈ U ). γ + n n=1
(2.14)
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If 0 < a1 < a2
and
a2 ≥ 2p
∞ X
(−1)n+1
n=1
γ +1 ρn, γ +n
then Tp,k (a2 , c ; h) ⊂ Tp,k (a1 , c ; h) and
Kp,k (a2 , c ; h) ⊂ Kp,k (a1 , c ; h).
Proof. The function h(z ) defined by (2.14) belongs to the class P (cf. [6]) and satisfies h(z ) = h(z ). Thus Re h(z ) > β = h(−1) = 1 +
∞ X
(−1)n
n=1
γ +1 ρ n > 1 − ρ ≥ 0 (z ∈ U ). γ +n
Therefore the desired result follows from Theorems 1 and 2 at once.
Corollary 4. Let γ = arccos λ, 0 < λ < 1, −λ ≤ µ < 1, and
√ 2γ √ 2γ 1+ z π 1− z π h( z ) = 1 + + − 2 √ √ 1− z 1+ z 2 sin2 γ 1−µ
(z ∈ U ).
(2.15)
If
0 < a1 < a2
and
a2 ≥ 2p
1−µ
1+λ
,
then Tp,k (a2 , c ; h) ⊂ Tp,k (a1 , c ; h) and
Kp,k (a2 , c ; h) ⊂ Kp,k (a1 , c ; h).
Proof. The function h(z ) defined by (2.15) is in the class P (see [12]) and satisfies h(z ) = h(z ). Thus Re h(z ) > β = h(−1) = 1 +
1−µ sin2 γ
(cos γ − 1) =
λ+µ ≥ 0 (z ∈ U ). 1+λ
Therefore the corollary follows immediately from Theorems 1 and 2.
Corollary 5. Let 0 < γ ≤ 1 and h( z ) = 1 +
2
π2
log
√ 2 γz √ 1 − γz 1+
(z ∈ U ).
(2.16)
If 0 < a1 < a2
and
a2 ≥
16p
π
arctan
2
√ 2 γ ,
then Tp,k (a2 , c ; h) ⊂ Tp,k (a1 , c ; h) and
Kp,k (a2 , c ; h) ⊂ Kp,k (a1 , c ; h).
Proof. The function h(z ) defined by (2.16) is in the class P (cf. [5]) and satisfies h(z ) = h(z ). Thus Re h(z ) > β = h(−1) = 1 −
8
π
2
arctan
√ 2 1 γ ≥ (z ∈ U ; 0 < γ ≤ 1).
Hence, by Theorems 1 and 2, we have the desired result.
2
For a1 = a > 0 and a2 = a + 1, Corollary 5 leads to: Corollary 6. Let h(z ) be defined as in Corollary 5. If a > 0 and 0 < α ≤
π
2
tan √ 4 p
,
then Tp,k (a + 1, c ; h) ⊂ Tp,k (a, c ; h) and
Kp,k (a + 1, c ; h) ⊂ Kp,k (a, c ; h).
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Theorem 3. Let 0 ≤ δ1 < δ2 . Then Cp,k (δ2 ; a, c ; h) ⊂ Cp,k (δ1 ; a, c ; h). Proof. For f (z ) ∈ Cp,k (δ2 ; a, c ; h), there exists a function g (z ) ∈ Tp,k (a, c ; h) such that
(1 − δ2 )
z (Lp (a, c )f )0 (z )
+ δ2
pgp,k (a, c ; z )
(z (Lp (a, c )f )0 (z ))0 (z ) ≺ h(z ) (h(z ) ∈ P ). pgp0 ,k (a, c ; z )
(2.17)
Let us define the function q(z ) by q(z )gp,k (a, c ; z ) =
1 p
z (Lp (a, c )f )0 (z )
(z ∈ U ).
(2.18)
Then q(z ) is analytic in U with q(0) = 1. Differentiating both sides of (2.18) we get q(z ) +
gp,k (a, c ; z ) zgp0 ,k (a, c ; z )
zq0 (z ) =
(z (Lp (a, c )f )0 (z ))0 (z ) . pgp0 ,k (a, c ; z )
(2.19)
Making use of (2.17)–(2.19), we arrive at q(z ) + w(z )zq0 (z ) ≺ h(z ),
(2.20)
where
w(z ) = δ2
zgp0 ,k (a, c ; z )
−1
.
gp,k (a, c ; z )
In view of Lemma 1 and δ2 > 0, we see that w(z ) is analytic in U and Re w(z ) > 0(z ∈ U ). Hence, using (2.20), an application of Lemma 3 yields q(z ) ≺ h(z ).
(2.21)
Now, from (2.17), (2.18) and (2.21), we deduce that
(1 − δ1 )
z (Lp (a, c )f )0 (z ) pgp,k (a, c ; z )
+ δ1
(z (Lp (a, c )f )0 (z ))0 (z ) pgp0 ,k (a, c ; z )
(z (Lp (a, c )f )0 (z ))0 (z ) + δ2 (1 − δ2 ) pgp,k (a, c ; z ) pgp0 ,k (a, c ; z )
δ1 = δ2
z (Lp (a, c )f )0 (z )
≺ h(z )
!
δ1 + 1− q(z ) δ2 (2.22)
because h(z ) is convex univalent in U and 0 ≤ δ1 < δ2 . Thus f (z ) ∈ Cp,k (δ1 ; a, c ; h) and the proof of Theorem 3 is completed. 3. Convolution properties Theorem 4. Let h(z ) ∈ P and α < 1 satisfy
Re h(z ) > max 0, 1 −
1−α p
(z ∈ U ).
(3.1)
If f (z ) ∈ Tp,k (a, c ; h), g (z ) ∈ Ap
and
g (z ) z p−1
∈ R(α) (α < 1),
then
(f ∗ g )(z ) ∈ Tp,k (a, c ; h). Proof. Let f (z ) ∈ Tp,k (a, c ; h) and suppose that
w(z ) =
fp,k (a, c ; z ) z p−1
.
(3.2)
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963
Then z (Lp (a, c )f )0 (z )
F (z ) =
≺ h(z ),
pfp,k (a, c ; z )
(3.3)
w(z ) ∈ A and z w 0 (z )
≺ ph(z ) + 1 − p
w(z )
(3.4)
(see (2.11) used in the proof of Theorem 1). By (3.1) and (3.4), we easily have
w(z ) ∈ S ∗ (α).
(3.5)
Noting that j
Lp (a, c )(f ∗ g )(εk z ) z p−1
=
g (z ) z p−1
∗
(Lp (a, c )f )(εkj z )
(j ∈ {0, 1, . . . , k − 1})
z p−1
and z (Lp (a, c )(f ∗ g ))0 (z ) z p−1
=
g (z ) z p−1
∗
z (Lp (a, c )f )0 (z ) z p−1
,
we deduce that z (Lp (a, c )(f ∗ g ))0 (z )
Φ (z ) = p k
=
=
k−1
P
εk−jp (Lp (a, c )(f ∗ g ))(εkj z )
j =0
g (z ) z p−1
z (Lp (a,c )f )0 (z ) pz p−1
∗
g (z ) z p−1
∗
g (z ) z p−1 g (z ) z p−1
fp,k (a,c ;z ) z p−1
∗ (w(z )F (z )) ∗ w(z )
(z ∈ U )
(3.6)
for g (z ) ∈ Ap . Since h(z ) is convex univalent in U, it follows from (3.2), (3.3), (3.5) and (3.6) and Lemma 2 that Φ (z ) ≺ h(z ). Hence we conclude that the function (f ∗ g )(z ) belongs to Tp,k (a, c ; h). Theorem 5. Let h(z ) ∈ P and α < 1 satisfy
Re h(z ) > max 0, 1 −
1−α
p
(z ∈ U ).
If f (z ) ∈ Kp,k (a, c ; h), g (z ) ∈ Ap
and
g (z ) z p−1
∈ R(α) (α < 1),
then
(f ∗ g )(z ) ∈ Kp,k (a, c ; h). The proof of Theorem 5 is similar to that of Theorem 4 and so we omit it. For α = 0 and α = 12 , Theorems 4 and 5 reduce to the following. Corollary 7. Let h(z ) ∈ P and g (z ) ∈ Ap satisfy either of the following conditions:
(i)
g (z ) z p−1
is convex univalent in U and
Re h(z ) > 1 − or
1 p
(z ∈ U )
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(ii)
g (z ) z p−1
∈ S ∗ ( 12 ) and 1
Re h(z ) > 1 −
(z ∈ U ).
2p
If f (z ) ∈ Tp,k (a, c ; h)(Kp,k (a, c ; h)), then
(f ∗ g )(z ) ∈ Tp,k (a, c ; h) (Kp,k (a, c ; h)). 4. Integral operator Theorem 6. Let h(z ) ∈ P and Reλ
Re h(z ) > max 0, −
p
(z ∈ U ),
(4.1)
where λ is a complex number such that Re λ > −p. If f (z ) ∈ Tp,k (a, c ; h), then the function F (z ) =
λ+p
z
Z
zλ
t λ−1 f (t )dt
(4.2)
0
is also in the class Tp,k (a, c ; h), provided that Fp,k (a, c ; z ) 6= 0 (0 < |z | < 1) where Fp,k (a, c ; z ) is defined as in (1.10). Proof. Let f (z ) ∈ Tp,k (a, c ; h). Then from (4.2) and Re λ > −p we see that F (z ) ∈ Ap and
(λ + p)Lp (a, c )f (z ) = λLp (a, c )F (z ) + z (Lp (a, c )F )0 (z ).
(4.3)
Also, by (4.3), we have k−1 1 X −jp εk λ(Lp (a, c )F )(εkj z ) + εkj z (Lp (a, c )F )0 (εkj z ) k j =0
(λ + p)fp,k (a, c ; z ) =
= λFp,k (a, c ; z ) + zFp0 ,k (a, c ; z ).
(4.4)
If we let zFp0 ,k (a, c ; z )
w(z ) =
pFp,k (a, c ; z )
,
(4.5)
then w(z ) is analytic in U, with w(0) = 1, and it follows from (4.4) and (4.5) that pw(z ) + λ = (λ + p)
fp,k (a, c ; z ) Fp,k (a, c ; z )
.
(4.6)
Differentiating both sides of (4.6) logarithmically and using (4.5) and Lemma 1, we obtain z w 0 (z )
w(z ) +
pw(z ) + λ
=
zfp0,k (a, c ; z ) pfp,k (a, c ; z )
≺ h(z ).
(4.7)
Furthermore, (4.1) and (4.7) and Lemma 4 lead to
w(z ) ≺ h(z ).
(4.8)
Suppose now that q(z ) =
z (Lp (a, c )F )0 (z ) pFp,k (a, c ; z )
.
(4.9)
Then q(z ) is analytic in U, with q(0) = 1, and it follows from (4.3) and (4.9) that Fp,k (a, c ; z )q(z ) =
λ+p p
Lp (a, c )f (z ) −
λ p
Lp (a, c )F (z ).
(4.10)
Differentiating both sides of (4.10) and using (4.9), we arrive at zq0 (z ) +
zFp0 ,k (a, c ; z )
z (Lp (a, c )f )0 (z ) + λ q(z ) = (λ + p) . Fp,k (a, c ; z ) pFp,k (a, c ; z )
(4.11)
N.-E. Xu, D.-G. Yang / Mathematical and Computer Modelling 49 (2009) 955–965
965
Making use of (4.4), (4.5) and (4.11), we deduce that q(z ) +
zq0 (z ) pw(z ) + λ
= =
λ+p
z (Lp (a, c )f )0 (z )
p
λFp,k (a, c ; z ) + zFp0 ,k (a, c ; z )
z (Lp (a, c )f )0 (z ) pfp,k (a, c ; z )
≺ h(z )
(4.12)
because f (z ) ∈ Tp,k (a, c ; h). Also, by (4.1) and (4.8), we have Re {pw(z ) + λ} > 0 (z ∈ U ). Therefore, from (4.12), (4.13) and Lemma 3 we conclude that q(z ) ≺ h(z ), which shows that F (z ) ∈ Tp,k (a, c ; h).
(4.13)
By an argument similar to that used in the proof of Theorem 6, the following theorem can be proved. We omit the details involved. Theorem 7. Let h(z ) ∈ P and
Re h(z ) > max 0, −
Re λ
p
(z ∈ U ; Re λ > −p).
If f (z ) ∈ Kp,k (a, c ; h) with respect to g (z ) ∈ Tp,k (a, c ; h), then the function F (z ) =
λ+p zλ
z
Z
t λ−1 f (t )dt
0
belongs to the class Kp,k (a, c ; h) with respect to the function G(z ) =
λ+p zλ
z
Z
t λ−1 g (t )dt ,
0
provided that Gp,k (a, c ; z ) 6= 0 (0 < |z | < 1). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
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