Volume 10 (2009), Issue 1, Article 22, 12 pp.
CERTAIN CLASSES OF ANALYTIC FUNCTIONS INVOLVING SĂLĂGEAN
OPERATOR
SEVTAP SÜMER EKER AND SHIGEYOSHI OWA
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE AND L ETTERS
D ICLE U NIVERSITY
21280 - D IYARBAKIR
T URKEY
sevtaps@dicle.edu.tr
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577 - 8502
JAPAN
owa@math.kindai.ac.jp
Received 21 March, 2007; accepted 27 December, 2008
Communicated by G. Kohr
A BSTRACT. Using Sălăgean differential operator, we study new subclasses of analytic functions. Coefficient inequalities and distortion theorems and extreme points of these classes are
studied. Furthermore, integral means inequalities are obtained for the fractional derivatives of
these classes.
Key words and phrases: Sălăgean operator, coefficient inequalities, distortion inequalities, extreme points, integral means,
fractional derivative.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION AND D EFINITIONS
Let A denote the class of functions f (z) of the form
(1.1)
f (z) = z +
∞
X
aj z j
j=2
which are analytic in the open disc U = {z : |z| < 1}. Let S be the subclass of A consisting of
analytic and univalent functions f (z) in U. We denote by S ∗ (α) and K(α) the class of starlike
functions of order α and the class of convex functions of order α, respectively, that is,
0 zf (z)
∗
S (α) = f ∈ A : Re
> α, 0 5 α < 1, z ∈ U
f (z)
088-07
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S EVTAP S ÜMER E KER AND S HIGEYOSHI OWA
and
zf 00 (z)
K(α) = f ∈ A : Re 1 + 0
> α, 0 5 α < 1, z ∈ U .
f (z)
For f (z) ∈ A, Sălăgean [1] introduced the following operator which is called the Sălăgean
operator:
D0 f (z) = f (z)
D1 f (z) = Df (z) = zf 0 (z)
Dn f (z) = D(Dn−1 f (z))
We note that,
n
D f (z) = z +
∞
X
j n aj z j
(n ∈ N = 1, 2, 3, ...).
(n ∈ N0 = N ∪ {0}).
j=2
Let Nm,n (α, β) denote the subclass of A consisting of functions f (z) which satisfy the inequality
m
m
D f (z)
D f (z)
Re
>β n
− 1 + α
n
D f (z)
D f (z)
for some 0 5 α < 1, β ≥ 0, m ∈ N, n ∈ N0 and all z ∈ U. Also let Msm,n (α, β) (s =
0, 1, 2, . . . ) be the subclass of A consisting of functions f (z) which satisfy the condition:
f (z) ∈ Msm,n (α, β) ⇔ Ds f (z) ∈ Nm,n (α, β).
It is easy to see that if s = 0, then M0m,n (α, β) ≡ Nm,n (α, β). Furthermore, special cases of
our classes are the following:
(i) N1,0 (α, 0) = S ∗ (α) and N2,1 (α, 0) = K(α) which were studied by Silverman [2].
(ii) N1,0 (α, β) = SD(α, β) and M11,0 (α, β) = KD(α, β) which were studied by Shams at
all [3].
(iii) Nm,n (α, 0) = Km,n (α) and Msm,n (α, 0) = Msm,n (α) which were studied by Eker and
Owa [4].
Therefore, our present paper is a generalization of these papers. In view of the coefficient
inequalities for f (z) to be in the classes Nm,n (α, β) and Msm,n (α, β), we introduce two subem,n (α, β) and M
fs (α, β). Some distortion inequalities for f (z) and some integral
classes N
m,n
means inequalities for fractional calculus of f (z) in the above classes are discussed in this paper.
2. C OEFFICIENT I NEQUALITIES FOR
CLASSES
Nm,n (α, β) AND Msm,n (α, β)
Theorem 2.1. If f (z) ∈ A satisfies
∞
X
(2.1)
Ψ(m, n, j, α, β) |aj | 5 2(1 − α)
j=2
where
(2.2)
Ψ(m, n, j, α, β) = |j m − j n − αj n | + (j m + j n − αj n ) + 2β |j m − j n |
for some α(0 5 α < 1), β ≥ 0, m ∈ N and n ∈ N0 ,then f (z) ∈ Nm,n (α, β).
Proof. Suppose that (2.1) is true for α(0 5 α < 1), β ≥ 0, m ∈ N , n ∈ N0 . For f (z) ∈ A, let
us define the function F (z) by
m
D f (z)
Dm f (z)
F (z) = n
−β n
− 1 − α.
D f (z)
D f (z)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
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C ERTAIN C LASSES OF A NALYTIC F UNCTIONS I NVOLVING S ĂL ĂGEAN O PERATOR
3
It suffices to show that
F (z) − 1 F (z) + 1 < 1
(z ∈ U).
We note that
F (z) − 1 Dm f (z) − βeiθ |Dm f (z) − Dn f (z)| − αDn f (z) − Dn f (z) F (z) + 1 = Dm f (z) − βeiθ |Dm f (z) − Dn f (z)| − αDn f (z) + Dn f (z) −α + P∞ (j m − j n − αj n )aj z j−1 − βeiθ | P∞ (j m − j n )aj z j−1 | j=2
j=2
P∞ m
P∞ m
=
n
n
j−1
iθ
n
j−1
(2 − α) + j=2 (j + j − αj )aj z
− βe | j=2 (j − j )aj z | P
P
m
n
n
j−1
m
n
j−1
α+ ∞
+ β|e|iθ ∞
j=2 |j − j − αj | |aj ||z|
j=2 |j − j ||aj ||z|
P
P
5
∞
m
n
n
j−1 − β|e|iθ
m
n
j−1
(2 − α) − ∞
j=2 (j + j − αj ) |aj ||z|
j=2 |j − j ||aj ||z|
P∞ m
P
m
n
α + j=2 |j − j n − αj n | |aj | + β ∞
j=2 |j − j ||aj |
P
P∞ m
5
.
m
n
n
n
(2 − α) − ∞
j=2 (j + j − αj ) |aj | − β
j=2 |j − j ||aj |
The last expression is bounded above by 1, if
α+
∞
X
|j m − j n − αj n | |aj | + β
j=2
∞
X
|j m − j n ||aj |
j=2
5 (2 − α) −
∞
X
m
n
n
(j + j − αj ) |aj | − β
j=2
∞
X
|j m − j n ||aj |
j=2
which is equivalent to our condition (2.1). This completes the proof of our theorem.
By using Theorem 2.1, we have:
Theorem 2.2. If f (z) ∈ A satisfies
∞
X
j s Ψ(m, n, j, α, β) |aj | 5 2(1 − α),
j=2
where Ψ(m, n, j, α, β) is defined by (2.2) for some α(0 5 α < 1), β ≥ 0, m ∈ N and n ∈ N0 ,
then f (z) ∈ Msm,n (α, β).
Proof. From
f (z) ∈ Msm,n (α, β) ⇔ Ds f (z) ∈ Nm,n (α, β),
replacing aj by j s aj in Theorem 2.1, we have the theorem.
Example 2.1. The function f (z) given by
f (z) = z +
∞
X
j=2
∞
X
2(2 + δ)(1 − α)j
zj = z +
Aj z j
(j + δ)(j + 1 + δ)Ψ(m, n, j, α, β)
j=2
with
Aj =
2(2 + δ)(1 − α)j
(j + δ)(j + 1 + δ)Ψ(m, n, j, α, β)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
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4
S EVTAP S ÜMER E KER AND S HIGEYOSHI OWA
belongs to the class Nm,n (α, β) for δ > −2, 0 5 α < 1, β ≥ 0, j ∈ C and |j | = 1. Because,
we know that
∞
∞
X
X
2(2 + δ)(1 − α)
Ψ(m, n, j, α, β) |Aj | 5
(j + δ)(j + 1 + δ)
j=2
j=2
∞
∞ X
X
1
1
=
2(2 + δ)(1 − α)
−
j
+
δ
j+1+δ
j=2
j=2
= 2(1 − α).
Example 2.2. The function f (z) given by
∞
∞
X
X
2(2 + δ)(1 − α)j
j
f (z) = z +
z =z+
Bj z j
s (j + δ)(j + 1 + δ)Ψ(m, n, j, α, β)
j
j=2
j=2
with
2(2 + δ)(1 − α)j
j s (j + δ)(j + 1 + δ)Ψ(m, n, j, α, β)
s
belongs to the class Mm,n (α, β) for δ > −2, 0 5 α < 1, β ≥ 0, j ∈ C and |j | = 1. Because,
the function f (z) gives us that
∞
∞
X
X
2(2 + δ)(1 − α)
s
j Ψ(m, n, j, α, β) |Bj | 5
= 2(1 − α).
(j + δ)(j + 1 + δ)
j=2
j=2
Bj =
3. R ELATION
FOR
em,n (α, β) AND M
fsm,n (α, β)
N
In view of Theorem 2.1 and Theorem 2.2, we now introduce the subclasses
em,n (α, β) ⊂ Nm,n (α, β) and M
fs (α, β) ⊂ Ms (α, β)
N
m,n
m,n
which consist of functions
f (z) = z +
(3.1)
∞
X
aj z j
(aj ≥ 0)
j=2
whose Taylor-Maclaurin coefficients satisfy the inequalities (2.1) and (2.2), respectively. By the
em,n (α, β) and M
fs (α, β), we see:
coefficient inequalities for the classes N
m,n
Theorem 3.1.
em,n (α, β2 ) ⊂ N
em,n (α, β1 )
N
for some β1 and β2 , 0 5 β1 5 β2 .
Proof. For 0 5 β1 5 β2 we obtain
∞
∞
X
X
Ψ(m, n, j, α, β1 )aj 5
Ψ(m, n, j, α, β2 )aj .
j=2
j=2
em,n (α, β2 ), then f (z) ∈ N
em,n (α, β1 ). Hence we get the required result.
Therefore, if f (z) ∈ N
By using Theorem 3.1, we also have
Corollary 3.2.
fs (α, β2 ) ⊂ M
fs (α, β1 )
M
m,n
m,n
for some β1 and β2 , 0 5 β1 5 β2 .
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
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C ERTAIN C LASSES OF A NALYTIC F UNCTIONS I NVOLVING S ĂL ĂGEAN O PERATOR
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4. D ISTORTION I NEQUALITIES
em,n (α, β), then we have
Lemma 4.1. If f (z) ∈ N
P
∞
X
2(1 − α) − pj=2 Ψ(m, n, j, α, β)aj
.
aj 5
Ψ(m, n, p + 1, α, β)
j=p+1
Proof. In view of Theorem 2.1, we can write
∞
X
(4.1)
Ψ(m, n, j, α, β)aj 5 2(1 − α) −
j=p+1
p
X
Ψ(m, n, j, α, β)aj .
j=2
Clearly Ψ(m, n, j, α, β) is an increasing function for j. Then from (2.2) and (4.1), we have
Ψ(m, n, p + 1, α, β)
∞
X
aj 5 2(1 − α) −
j=p+1
p
X
Ψ(m, n, j, α, β)aj .
j=2
Thus, we obtain
P
2(1 − α) − pj=2 Ψ(m, n, j, α, β)aj
aj 5
= Aj .
Ψ(m,
n,
p
+
1,
α,
β)
j=p+1
∞
X
em,n (α, β), then
Lemma 4.2. If f (z) ∈ N
P
2(1 − α) − pj=2 Ψ(m, n, j, α, β)aj
jaj 5
= Bj .
Ψ(m − 1, n − 1, p + 1, α, β)
j=p+1
∞
X
fs (α), then
Corollary 4.3. If f (z) ∈ M
m,n
P
2(1 − α) − pj=2 j s Ψ(m, n, j, α, β)aj
aj 5
= Cj
(p + 1)s Ψ(m, n, p + 1, α, β)
j=p+1
∞
X
and
P
2(1 − α) − pj=2 j s Ψ(m, n, j, α, β)aj
jaj 5
= Dj .
s Ψ(m − 1, n − 1, p + 1, α, β)
(p
+
1)
j=p+1
∞
X
em,n (α, β). Then for |z| = r < 1
Theorem 4.4. Let f (z) ∈ N
r−
p
X
j
aj |z| − Aj r
p+1
5 |f (z)| 5 r +
j=2
p
X
aj |z|j + Aj rp+1
j=2
and
1−
p
X
jaj |z|j−1 − Bj rp 5 |f 0 (z)| 5 1 +
j=2
p
X
jaj |z|j + Bj rp
j=2
where Aj and Bj are given by Lemma 4.1 and Lemma 4.2.
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6
S EVTAP S ÜMER E KER AND S HIGEYOSHI OWA
Proof. Let f (z) given by (1.1). For |z| = r < 1,using Lemma 4.1, we have
|f (z)| 5 |z| +
p
X
aj |z| +
j=2
p
5 |z| +
X
∞
X
j
aj |z|j
j=p+1
j
∞
X
p+1
aj |z| + |z|
j=2
p
5r+
X
aj
j=p+1
aj |z|j + Aj rp+1
j=2
and
|f (z)| ≥ |z| −
p
X
aj |z| −
j=2
p
≥ |z| −
X
∞
X
j
aj |z|j
j=p+1
j
aj |z| − |z|
j=2
p
≥r−
X
∞
X
p+1
aj
j=p+1
aj |z|j − Aj rp+1 .
j=2
Furthermore, for |z| = r < 1 using Lemma 4.2, we obtain
0
|f (z)| 5 1 +
p
X
j−1
jaj |z|
+
j=2
51+
∞
X
j=p+1
p
X
j−1
jaj |z|
p
X
∞
X
p
+ |z|
j=2
51+
jaj |z|j−1
jaj
j=p+1
jaj |z|j−1 + Bj rp
j=2
and
0
|f (z)| ≥ 1 −
p
X
j−1
jaj |z|
j=2
p
≥1−
X
−
∞
X
j=p+1
j−1
jaj |z|
X
p
− |z|
∞
X
jaj
j=p+1
j=2
p
≥1−
jaj |z|j−1
jaj |z|j−1 − Bj rp .
j=2
This completes the assertion of Theorem 4.4.
fs (α, β). Then
Theorem 4.5. Let f (z) ∈ M
m,n
r−
p
X
j
aj |z| − Cj r
p+1
5 |f (z)| 5 r +
j=2
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
p
X
aj |z|j + Cj rp+1
j=2
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C ERTAIN C LASSES OF A NALYTIC F UNCTIONS I NVOLVING S ĂL ĂGEAN O PERATOR
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and
p
X
1−
jaj |z|j−1 − Dj rp 5 |f 0 (z)| 5 1 +
p
X
j=2
jaj |z|j + Dj rp
j=2
where Cj and Dj are given by Corollary 4.3.
Proof. Using a similar method to that in the proof of Theorem 4.4 and making use Corollary
4.3, we get our result.
Taking p = 1 in Theorem 4.4 and Theorem 4.5, we have:
em,n (α, β). Then for |z| = r < 1
Corollary 4.6. Let f (z) ∈ N
r−
2(1 − α)
2(1 − α)
r2 5 |f (z)| 5 r +
r2
Ψ(m, n, 2, α, β)
Ψ(m, n, 2, α, β)
and
1−
2(1 − α)
2(1 − α)
r 5 |f 0 (z)| 5 1 +
r.
Ψ(m − 1, n − 1, 2, α, β)
Ψ(m − 1, n − 1, 2, α, β)
fsm,n (α, β). Then for |z| = r < 1
Corollary 4.7. Let f (z) ∈ M
2(1 − α)
r−
2s Ψ(m, n, 2, α, β)
r2 5 |f (z)| 5 r +
2(1 − α)
2s Ψ(m, n, 2, α, β)
r2
and
1−
2s Ψ(m
2(1 − α)
2(1 − α)
r 5 |f 0 (z)| 5 1 + s
r.
− 1, n − 1, 2, α, β)
2 Ψ(m − 1, n − 1, 2, α, β)
5. E XTREME P OINTS
The determination of the extreme points of a family F of univalent functions enables us
to solve many extremal problems for F . Now, let us determine extreme points of the classes
em,n (α, β) and M
fs (α, β).
N
m,n
Theorem 5.1. Let f1 (z) = z and
fj (z) = z +
2(1 − α)
zj
Ψ(m, n, j, α, β)
(j = 2, 3, ...).
em,n (α, β) if and only if it can be expressed
where Ψ(m, n, j, α, β) is defined by (2.2). Then f ∈ N
in the form
∞
X
f (z) =
λj fj (z),
j=1
where λj > 0 and
P∞
j=1
λj = 1.
Proof. Suppose that
f (z) =
∞
X
λj fj (z) = z +
j=1
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
∞
X
j=2
λj
2(1 − α)
zj .
Ψ(m, n, j, α, β)
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S EVTAP S ÜMER E KER AND S HIGEYOSHI OWA
Then
∞
X
∞
Ψ(m, n, j, α, β)
j=2
X
2(1 − α)
λj =
2(1 − α)λj
Ψ(m, n, j, α, β)
j=2
= 2(1 − α)
∞
X
λj
j=2
= 2(1 − α)(1 − λ1 )
< 2(1 − α)
em,n (α, β) from the definition of the class of N
em,n (α, β).
Thus, f (z) ∈ N
em,n (α, β). Since
Conversely, suppose that f ∈ N
aj 5
2(1 − α)
Ψ(m, n, j, α, β)
(j = 2, 3, ...),
we may set
λj =
Ψ(m, n, j, α, β)
aj
2(1 − α)
and
λ1 = 1 −
∞
X
λj .
j=2
Then,
f (z) =
∞
X
λj fj (z).
j=1
This completes the proof of the theorem.
Corollary 5.2. Let g1 (z) = z and
gj (z) = z +
2(1 − α)
j s Ψ(m, n, j, α, β)
zj
(j = 2, 3, ...).
fs (α, β) if and only if it can be expressed in the form
Then g ∈ M
m,n
g(z) =
∞
X
λj gj (z),
j=1
where λj > 0 and
P∞
j=1
λj = 1.
em,n (α, β) are the functions f1 (z) = z and
Corollary 5.3. The extreme points of N
fj (z) = z +
2(1 − α)
zj
Ψ(m, n, j, α, β)
(j = 2, 3, ...).
fs (α, β) are given by g1 (z) = z and
Corollary 5.4. The extreme points of M
m,n
gj (z) = z +
2(1 − α)
j s Ψ(m, n, j, α, β)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
zj
(j = 2, 3, ...).
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C ERTAIN C LASSES OF A NALYTIC F UNCTIONS I NVOLVING S ĂL ĂGEAN O PERATOR
9
6. I NTEGRAL M EANS I NEQUALITIES
We shall use the following definitions for fractional derivatives by Owa [6] (also Srivastava
and Owa [7]).
Definition 6.1. The fractional derivative of order λ is defined, for a function f (z), by
Z z
d
f (ξ)
1
λ
(6.1)
Dz f (z) =
dξ (0 5 λ < 1),
Γ(1 − λ) dz 0 (z − ξ)λ
where the function f (z) is analytic in a simply-connected region of the complex z-plane containing the origin, and the multiplicity of (z − ξ)−λ is removed by requiring log(z − ξ) to be
real when (z − ξ) > 0.
Definition 6.2. Under the hypotheses of Definition 6.1, the fractional derivative of order (p+λ)
is defined, for a function f (z), by
Dzp+λ f (z) =
dp λ
D f (z)
dz p z
where 0 5 λ < 1 and p ∈ N0 = N ∪ {0}.
It readily follows from (6.1) in Definition 6.1 that
Dzλ z k =
(6.2)
Γ(k + 1) k−λ
z
Γ(k − λ + 1)
(0 5 λ < 1).
Further, we need the concept of subordination between analytic functions and a subordination
theorem by Littlewood [5] in our investigation.
Let us consider two functions f (z) and g(z), which are analytic in U. The function f (z) is
said to be subordinate to g(z) in U if there exists a function w(z) analytic in U with
w(0) = 0 and |w(z)| < 1
(z ∈ U),
such that
(z ∈ U).
f (z) = g(w(z))
We denote this subordination by
f (z) ≺ g(z).
Theorem 6.1 (Littlewood [5]). If f (z) and g (z) are analytic in U with f (z) ≺ g(z), then for
µ > 0 and z = reiθ (0 < r < 1)
Z 2π
Z 2π
µ
|f (z)| dθ 5
|g (z)|µ dθ.
0
0
em,n (α, β) and suppose that
Theorem 6.2. Let f (z) ∈ A given by (3.1) be in the class N
∞
X
j=2
(j − p)p+1 aj 5
2(1 − α)Γ(k + 1)Γ(3 − λ − p)
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)Γ(2 − p)
for some 0 5 p 5 2, 0 5 λ < 1 where (j − p)p+1 denotes the Pochhammer symbol defined by
(j − p)p+1 = (j − p)(j − p + 1) · · · j. Also given is the function fk (z) by
(6.3)
fk (z) = z +
2(1 − α)
zk
Ψ(m, n, k, α, β)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
(k ≥ 2).
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10
S EVTAP S ÜMER E KER AND S HIGEYOSHI OWA
If there exists an analytic function w(z) given by
{w(z)}k−1 =
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
2(1 − α)Γ(k + 1)
∞
X
×
(j − p)p+1
j=2
then for z = reiθ (0 < r < 1) and µ > 0,
Z 2π
Z
p+λ
µ
Dz f (z) dθ 5
0
2π
Γ(j − p)
aj z j−1 ,
Γ(j + 1 − λ − p)
p+λ
Dz fk (z)µ dθ.
0
Proof. By virtue of the fractional derivative formula (6.2) and Definition 6.2, we find from (1.1)
that
(
)
∞
1−λ−p
X
z
Γ(2
−
λ
−
p)Γ(j
+
1)
Dzp+λ f (z) =
1+
aj z j−1
Γ(2 − λ − p)
Γ(j
+
1
−
λ
−
p)
j=2
(
)
∞
X
z 1−λ−p
=
1+
Γ(2 − λ − p)(j − p)p+1 Φ(j)aj z j−1
Γ(2 − λ − p)
j=2
where
Φ(j) =
Γ(j − p)
.
Γ(j + 1 − λ − p)
Since Φ(j) is a decreasing function of j, we have
0 < Φ(j) 5 Φ(2) =
Γ(2 − p)
Γ(3 − λ − p)
(0 5 λ < 1
; 0 5 p 5 2 5 j).
Similarly, from (6.2), (6.3) and Definition 6.2, we obtain
z 1−λ−p
2(1 − α)Γ(2 − λ − p)Γ(k + 1) k−1
p+λ
Dz fk (z) =
1+
z
.
Γ(2 − λ − p)
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
For z = reiθ , 0 < r < 1, we must show that
µ
Z 2π ∞
X
j−1 Γ(2 − λ − p)(j − p)p+1 Φ(j)aj z dθ
1 +
0
j=2
µ
Z 2π 2(1
−
α)Γ(2
−
λ
−
p)Γ(k
+
1)
k−1
1 +
5
z dθ
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
0
(µ > 0).
Thus by applying Littlewood’s subordination theorem, it would suffice to show that
(6.4)
1+
∞
X
Γ(2 − λ − p)(j − p)p+1 Φ(j)aj z j−1
j=2
≺1+
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
2(1 − α)Γ(2 − λ − p)Γ(k + 1) k−1
z .
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
http://jipam.vu.edu.au/
C ERTAIN C LASSES OF A NALYTIC F UNCTIONS I NVOLVING S ĂL ĂGEAN O PERATOR
11
By setting
1+
∞
X
Γ(2 − λ − p)(j − p)p+1 Φ(j)aj z j−1
j=2
=1+
2(1 − α)Γ(2 − λ − p)Γ(k + 1)
{w(z)}k−1
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
we find that
∞
k−1
{w(z)}
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p) X
=
(j − p)p+1 Φ(j)aj z j−1
2(1 − α)Γ(k + 1)
j=2
which readily yields w(0) = 0.
Therefore, we have
∞
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p) X
(j − p)p+1 Φ(j)aj z j−1 |w(z)|k−1 = 2(1 − α)Γ(k + 1)
j=2
∞
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p) X
(j − p)p+1 Φ(j)aj |z|j−1
2(1 − α)Γ(k + 1)
j=2
5
5 |z|
∞
X
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
Φ(2)
(j − p)p+1 aj
2(1 − α)Γ(k + 1)
j=2
∞
Ψ(m, n, k, α, β)Γ(k + 1 − λ − p) Γ(2 − p) X
= |z|
(j − p)p+1 aj
2(1 − α)Γ(k + 1)
Γ(3 − λ − p) j=2
5 |z| < 1
by means of the hypothesis of Theorem 6.2.
For the special case p = 0, Theorem 6.2 readily yields the following result.
em,n (α, β) and suppose that
Corollary 6.3. Let f (z) ∈ A given by (3.1) be in the class N
∞
X
2(1 − α)Γ(k + 1)Γ(3 − λ)
Ψ(m, n, k, α, β)Γ(k + 1 − λ)
jaj 5
j=2
for 0 5 λ < 1. Also let the function fk (z) be given by (6.3). If there exists an analytic function
w(z) given by
∞
k−1
{w(z)}
Ψ(m, n, k, α, β)Γ(k + 1 − λ) X Γ(j + 1)
=
aj z j−1 ,
2(1 − α)Γ(k + 1)
Γ(j
+
1
−
λ)
j=2
then for z = reiθ and 0 < r < 1,
Z 2π
Z
λ
Dz f (z)µ dθ 5
0
2π
λ
Dz fk (z)µ dθ
(0 5 λ < 1, µ > 0).
0
fs (α, β) and suppose that
Corollary 6.4. Let f (z) ∈ A given by (3.1) be in the class M
m,n
∞
X
j=2
(j − p)p+1 aj 5
2(1 − α)Γ(k + 1)Γ(3 − λ − p)
+ 1 − λ − p)Γ(2 − p)
k s Ψ(m, n, k, α, β)Γ(k
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
http://jipam.vu.edu.au/
12
S EVTAP S ÜMER E KER AND S HIGEYOSHI OWA
for some 0 5 p 5 2, 0 5 λ < 1. Also let the function
2(1 − α)
(6.5)
gk (z) = z + s
zk ,
k Ψ(m, n, k, α, β)
If there exists an analytic function w(z) given by
{w(z)}k−1 =
(k ≥ 2).
k s Ψ(m, n, k, α, β)Γ(k + 1 − λ − p)
2(1 − α)Γ(k + 1)
∞
X
×
(j − p)p+1
j=2
Γ(j − p)
aj z j−1 ,
Γ(j + 1 − λ − p)
iθ
then for z = re (0 < r < 1) and µ > 0,
Z 2π
Z
p+λ
Dz f (z)µ dθ 5
0
2π
p+λ
Dz gk (z)µ dθ.
0
For the special case p = 0, Corollary 6.4 readily yields,
fsm,n (α, β) and suppose that
Corollary 6.5. Let f (z) ∈ A given by (3.1) be in the class M
∞
X
j=2
jaj 5
2(1 − α)Γ(k + 1)Γ(3 − λ)
+ 1 − λ)
k s Ψ(m, n, k, α, β)Γ(k
for 0 5 λ < 1. Also let the function gk (z) be given by (6.5). If there exists an analytic function
w(z) given by
∞
k s Ψ(m, n, k, α, β)Γ(k + 1 − λ) X Γ(j + 1)
{w(z)}k−1 =
aj z j−1 ,
2(1 − α)Γ(k + 1)
Γ(j + 1 − λ)
j=2
then for z = reiθ (0 < r < 1) and µ > 0,
Z 2π
Z
λ
µ
Dz f (z) dθ 5
0
2π
λ
Dz gk (z)µ dθ.
0
R EFERENCES
[1] G.S. SĂLĂGEAN , Subclasses of univalent functions, Complex analysis - Proc. 5th Rom.-Finn.
Semin., Bucharest 1981, Part 1, Lect. Notes Math., 1013 (1983), 362–372.
[2] H. SILVERMAN, Univalent functions with negative coefficients, Proc. Amer. Math. Soc., 51(1)
(1975), 109–116.
[3] S. SHAMS, S.R. KULKARNI AND J.M. JAHANGIRI, Classes of uniformly starlike and convex
functions Internat. J. Math. Math. Sci., 2004 (2004), Issue 55, 2959–2961.
[4] S. SÜMER EKER AND S. OWA, New applications of classes of analytic functions involving the
Sălăgean Operator, Proceedings of the International Symposium on Complex Function Theory and
Applications, Transilvania University of Braşov Printing House, Braşov, Romania, 2006, 21–34.
[5] J.E. LITTLEWOOD, On inequalities in the theory of functions, Proc. London Math. Soc., 23 (1925),
481–519.
[6] S. OWA, On the distortion theorems I, Kyungpook Math. J., 18 (1978), 53–59.
[7] H.M. SRIVASTAVA AND S. OWA (Eds.), Univalent Functions, Fractional Calculus, and Their
Applications, Halsted Press (Ellis Horwood Limited, Chichester) John Wiley and Sons, New York,
Chichester, Brisbane and Toronto, 1989
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 22, 12 pp.
http://jipam.vu.edu.au/