Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 50, 2005
INTEGRAL MEANS FOR STARLIKE AND CONVEX FUNCTIONS WITH
NEGATIVE COEFFICIENTS
SHIGEYOSHI OWA, MIHAI PASCU, DAISUKE YAGI, AND JUNICHI NISHIWAKI
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502
JAPAN
owa@math.kindai.ac.jp
D EPARTMENT OF M ATHEMATICS
T RANSILVANIA U NIVERSITY OF B RASOV
R-2200 B RASOV
ROMANIA
mihai.pascu@unitbv.ro
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502
JAPAN
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502
JAPAN
Received 17 February, 2005; accepted 06 April, 2005
Communicated by N.E. Cho
Memorial Paper for Professor Nicolae N. Pascu
A BSTRACT. Let T be the class of functions f (z) with negative coefficients which are analytic
and univalent in the open unit disk U with f (0) = 0 and f 0 (0) = 1. The classes T ∗ and C are
defined as the subclasses of T which are starlike and convex in U, respectively. In view of the
interesting results for integral means given by H. Silverman (Houston J. Math. 23(1977)), some
generalization theorems are discussed in this paper.
Key words and phrases: Univalent, Starlike, Convex, Integral mean.
2000 Mathematics Subject Classification. Primary 30C45.
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
041-05
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S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI , AND J UNICHI N ISHIWAKI
1. I NTRODUCTION
Let A denote the class of functions f (z) of the form
∞
X
(1.1)
f (z) = z +
an z n
n=2
that are analytic in the open unit disk U = {z ∈ C : |z| < 1}. Let S be the subclass of A
consisting of all univalent functions f (z) in U. A function f (z) ∈ A is said to be starlike with
respect to the origin in U if it satisfies
0 zf (z)
(1.2)
Re
>0
f (z)
(z ∈ U).
We denote by S ∗ the subclass of S consisting of all starlike functions f (z) with respect to the
origin in U. Further, a function f (z) ∈ A is said to be convex in U if it satisfies
zf 00 (z)
(1.3)
Re 1 + 0
>0
(z ∈ U).
f (z)
We also denote by K the subclass of S consisting of f (z) which are convex in U. By the above
definitions, we know that f (z) ∈ K if and only if zf 0 (z) ∈ S ∗ , and that K ⊂ S ∗ ⊂ S ⊂ A.
The class T is defined as the subclass of S consisting of all functions f (z) which are given
by
(1.4)
f (z) = z −
∞
X
an z n
(an ≥ 0).
n=2
Further, we denote by T ∗ = S ∗ ∩ T and C = K ∩ T . It is well-known by Silverman [6] that
Remark 1.1. A function f (z) ∈ T ∗ if and only if
∞
X
(1.5)
nan ≤ 1.
n=2
A function f (z) ∈ C if and only if
∞
X
(1.6)
n2 an ≤ 1.
n=2
For f (z) ∈ A and g(z) ∈ A, f (z) is said to be subordinate to g(z) in U if there exists an
analytic function ω(z) in U such that ω(0) = 0, |ω(z)| < 1 (z ∈ U), and f (z) = g(ω(z)). We
denote this subordination by
f (z) ≺ g(z). (cf. Duren [1]).
(1.7)
For subordinations, Littlewood [2] has given the following integral mean.
Theorem A. If f (z) and g(z) are analytic in U with f (z) ≺ g(z), then, for λ > 0 and |z| = r
(0 < r < 1),
Z 2π
Z 2π
iθ λ
(1.8)
|f (re )| dθ ≤
|g(reiθ )|λ dθ.
0
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I NTEGRAL M EANS FOR S TARLIKE AND C ONVEX F UNCTIONS WITH N EGATIVE C OEFFICIENTS
3
Furthermore, Silverman [6] has shown that
Remark 1.2. f1 (z) = z and fn (z) = z −
f1 (z) = z and fn (z) = z −
zn
n2
zn
n
(n ≥ 2) are extreme points of the class T ∗ (or T ).
(n ≥ 2) are extreme points of the class C.
Applying Theorem A with extreme points of T , Silverman [7] has proved the following
results.
Theorem B. Suppose that f (z) ∈ T ∗ , λ > 0 and f2 (z) = z −
z2
.
2
Then, for z = reiθ
(0 < r < 1),
2π
Z
λ
2π
Z
|f2 (z)|λ dθ.
|f (z)| dθ ≤
(1.9)
0
0
2
Theorem C. If f (z) ∈ T ∗ , λ > 0, and f2 (z) = z − z2 , then, for z = reiθ (0 < r < 1),
Z 2π
Z 2π
0
λ
(1.10)
|f (z)| dθ ≤
|f20 (z)|λ dθ.
0
0
In the present paper, we consider the generalization properties for Theorem B and Theorem
C with f (z) ∈ T ∗ and f (z) ∈ C.
Remark 1.3. More recently, applying Theorem A by Littlewood [2], Sekine, Tsurumi and
Srivastava [4]; and Sekine, Tsurumi, Owa and Srivastava [5] have discussed some interesting
properties of integral means inequalities for fractional derivatives of some general subclasses of
analytic functions f (z) in the open unit disk U. Further, Owa and Sekine [3] have considered
the integral means with some coefficient inequalities for certain analytic functions f (z) in U.
2. G ENERALIZATION P ROPERTIES
Our first result for the generalization properties is contained in
Theorem 2.1. Let f (z) ∈ T ∗ , λ > 0, and fk (z) = z −
(2.1)
k−3
X
j+1
k
j=0
zk
k
(k ≥ 2). If f (z) satisfies
(a2k+j−1 + ak+j+1 − ak−j−1 ) ≥ 0
for k ≥ 3, and if there exists an analytic function ω(z) in U given by
!
∞
X
(ω(z))k−1 = k
an z n−1 ,
n=2
then, for z = reiθ (0 < r < 1),
Z
(2.2)
2π
0
λ
Z
|f (z)| dθ ≤
2π
|fk (z)|λ dθ.
0
Proof. For f (z) ∈ T ∗ , we have to show that
λ
λ
Z 2π Z 2π ∞
X
z k−1 n−1 an z dθ ≤
1 −
1 − k dθ.
0
0
n=2
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
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S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI , AND J UNICHI N ISHIWAKI
By Theorem A, it suffices to prove that
1−
∞
X
an z n−1 ≺ 1 −
n=2
z k−1
.
k
Let us define the function ω(z) by
1−
(2.3)
∞
X
1
an z n−1 = 1 − (ω(z))k−1 .
k
n=2
It follows from (2.3) that
|ω(z)|k−1
!
∞
∞
X
X
= k
an z n−1 ≤ |z|
kan .
n=2
n=2
Thus, we only show that
∞
X
kan ≤
n=2
∞
X
nan ,
n=2
or
∞
X
1
an ≤
k
n=2
∞
X
!
nan .
n=2
Indeed, we see that
1
k
∞
X
n=2
!
nan
k−2
k−3
2
= 1−
a2 + 1 −
a3 + · · · + 1 −
ak−2
k
k
k
1
1
2
+ 1−
ak−1 + ak + 1 +
ak+1 + 1 +
ak+2
k
k
k
k+1
k+2
+ ··· + 1 +
a2k+1 + 1 +
a2k+2 + · · ·
k
k
k−2
k−3
=
(a2k−2 − a2 ) +
(a2k−3 − a3 ) + · · ·
k
k
2
1
k−1
+ (ak+2 − ak−2 ) + (ak+1 − ak−1 ) + 1 +
a2k−1
k
k
k
2k−2
X
k
k+1
a2k + 1 +
a2k+1 + · · · +
an .
+ 1+
k
k
n=2
Noting that
1+
k+j
2+j
≥1+
, (j = −1, 0, 1, . . . ),
k
k
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I NTEGRAL M EANS FOR S TARLIKE AND C ONVEX F UNCTIONS WITH N EGATIVE C OEFFICIENTS
5
we obtain
(2.4)
1
k
∞
X
!
nan
≥
n=2
k−2
k−3
(a2k−2 − a2 ) +
(a2k−3 − a3 ) + · · ·
k
k
2
1
+ (ak+2 − ak−2 ) + (ak+1 − ak−1 )
k
k
1
2
+ 1+
a2k−1 + 1 +
a2k + · · ·
k
k
2k−2
X
k−3
k−2
+ 1+
a3k−5 + 1 +
a3k−4 + · · · +
an
k
k
n=2
≥
=
1
2
(a2k−1 + ak+1 − ak−1 ) + (a2k + ak+2 − ak−2 ) + · · ·
k
k
∞
X
k−2
(a3k−4 + a2k−2 − a2 ) +
+
an
k
n=2
k−3
X
j+1
k
j=0
≥
∞
X
(a2k+j−1 + ak+j+1 − ak−j−1 ) +
∞
X
an
n=2
an
n=2
with the following condition
k−3
X
j+1
j=0
k
(a2k+j−1 + ak+j+1 − ak−j−1 ) ≥ 0.
Thus, we observe that the function ω(z) defined by (2.3) is analytic in U with ω(0) = 0,
|ω(z)| < 1 (z ∈ U). This completes the proof of the theorem.
Remark 2.2. Taking k = 2 in Theorem 2.1, we have Theorem B by Silverman [7].
Example 2.1. Let us define
(2.5)
f (z) = z −
37 2
1
1
1 5
z − z3 − z4 −
z
1200
18
48
100
and
1
f3 (z) = z − z 3
3
with k = 3 in Theorem 2.1. Since f (z) satisfies
(2.6)
∞
X
n=2
nan =
217
< 1,
600
we have f (z) ∈ T ∗ . Furthermore, f (z) satisfies,
1
1
1
1
37
(a5 + a4 − a2 ) =
+
−
= 0.
3
3 100 48 1200
Thus, f (z) satisfies the conditions in Theorem 2.1 with k = 3.
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
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S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI , AND J UNICHI N ISHIWAKI
If we take λ = 2, then we have
Z 2π
1 4
20
2
2
|f (z)| dθ ≤ 2πr 1 + r < π = 6.9813 . . . .
9
9
0
zk
k
Corollary 2.3. Let f (z) ∈ T ∗ , 0 < λ ≤ 2, and fk (z) = z −
(k ≥ 2). If f (z) satisfies the
iθ
conditions in Theorem 2.1, then, for z = re (0 < r < 1),
λ
λ
Z 2π
1 2(k−1) 2
1 2
λ
λ
(2.7)
|f (z)| dθ ≤ 2πr 1 + 2 r
< 2π 1 + 2
.
k
k
0
Proof. It follows that
Z
2π
Z
λ
2π
|fk (z)| dθ =
0
k−1 λ
z
dθ.
|z| 1 −
k λ
0
Applying Hölder’s inequality for 0 < λ < 2, we obtain that

λ
Z 2π
2−λ
Z 2π
Z
k−1
2
2
z λ
λ 2−λ

|z| 1 −
dθ ≤
(|z| ) dθ
k 0
0
2π
0
 λ2
λ ! λ2
k−1
1 − z dθ
k ! λ2
k−1 2
z
dθ
1 −
=
|z| dθ
k 0
0
λ2
2−λ
2λ
1
2
2(k−1)
= 2πr 2−λ
2π 1 + 2 r
k
λ2
1
= 2πrλ 1 + 2 r2(k−1)
k
λ
1 2
< 2π 1 + 2
.
k
2π
Z
2−λ
2
2λ
2−λ
Z
2π
Further, it is clear for λ = 2.
For the generalization of Theorem C by Silverman [7], we have
k
Theorem 2.4. Let f (z) ∈ T ∗ , λ > 0, and fk (z) = z − zk (k ≥ 2). If there exists an analytic
function ω(z) in U given by
∞
k−1 X
ω(z)
=
nan z n−1 ,
n=2
iθ
then, for z = re (0 < r < 1),
Z
(2.8)
2π
0
λ
Z
|f (z)| dθ ≤
0
2π
|fk0 (z)|λ dθ.
0
Proof. For f (z) ∈ T ∗ , it is sufficient to show that
(2.9)
1−
∞
X
nan z n−1 ≺ 1 − z k−1 .
n=2
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I NTEGRAL M EANS FOR S TARLIKE AND C ONVEX F UNCTIONS WITH N EGATIVE C OEFFICIENTS
7
Let us define the function ω(z) by
1−
(2.10)
∞
X
nan z n−1 = 1 − ω(z)k−1 ,
n=2
or, by
ω(z)k−1 =
∞
X
nan z n−1 .
n=2
Since f (z) satisfies
∞
X
nan ≤ 1,
n=2
the function ω(z) is analytic in U, ω(0) = 0, and |ω(z)| < 1 (z ∈ U).
Remark 2.5. If we take k = 2 in Theorem 2.4, then we have Theorem C by Silverman [7].
Using the Hölder inequality for Theorem 2.4, we have
Corollary 2.6. Let f (z) ∈ T ∗ , 0 < λ ≤ 2, and fk (z) = z −
zk
k
(k ≥ 2). If f (z) satisfies the
iθ
conditions in Theorem 2.4, then, for z = re (0 < r < 1),
Z 2π
λ
2+λ
|f 0 (z)|λ dθ ≤ 2π 1 + r2(k−1) 2 < 2 2 π.
0
3. I NTEGRAL M EANS FOR F UNCTIONS IN THE C LASS C
In this section, we discuss the integral means for functions f (z) in the class C.
Theorem 3.1. Let f (z) ∈ C, λ > 0, and fk (z) = z −
(3.1)
k−1
X
(k + j)(k − j)
k2
j=2
zk
k2
(k ≥ 2). If f (z) satisfies
(a2k−j − aj ) ≥ 0
for k ≥ 3, and if there exists an analytic function ω(z) in U given by
k−1
(ω(z))
=k
2
∞
X
an z n−1 ,
n=2
then, for z = reiθ (0 < r < 1),
Z
(3.2)
2π
λ
Z
2π
|f (z)| dθ ≤
0
|fk (z)|λ dθ.
0
Proof. For the proof, we need to show that
(3.3)
1−
∞
X
an z n−1 ≺ 1 −
n=2
z k−1
k2
by Theorem A. Define the function ω(z) by
(3.4)
1−
∞
X
n=2
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
an z n−1 = 1 −
1
ω(z)k−1 ,
k2
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S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI , AND J UNICHI N ISHIWAKI
or by
(3.5)
(ω(z))k−1 = k 2
∞
X
!
an z n−1
.
n=2
Therefore, we have to show that
∞
X
1
an ≤ 2
k
n=2
∞
X
!
n 2 an .
n=2
Using the same technique as in the proof of Theorem 2.1, we see that
! k−1
∞
∞
X (k + j)(k − j)
X
1 X 2
n an ≥
(a2k−j − aj ) +
an
k 2 n=2
k2
n=2
j=2
≥
∞
X
an .
n=2
Example 3.1. Consider the functions
(3.6)
f (z) = z −
1
1
1 2
z − z3 − z4
40
18
40
and
(3.7)
1
f3 (z) = z − z 3
9
with k = 3 in Theorem 3.1. Then we have that
∞
X
4
9
16
n 2 an =
+
+
= 1,
40 18 40
n=2
which implies f (z) ∈ C, and that
5
(a4 − a2 ) = 0.
9
Thus f (z) satisfies the conditions of Theorem 3.1. If we make λ = 2, then we see that
Z 2π
1 4
164
2
2
|f (z)| dθ ≤ 2πr 1 + r <
π = 6.3607 · · · .
81
81
0
k
Corollary 3.2. Let f (z) ∈ C, 0 < λ ≤ 2, and fk (z) = z − zk2 (k ≥ 2). If f (z) satisfies the
condition in Theorem 3.1, then, for k ≥ 3, then, for z = reiθ (0 < r < 1),
λ
Z 2π
1 2(k−1) 2
λ
λ
(3.8)
|f (z)| dθ ≤ 2πr 1 + 4 r
k
0
λ
1 2
< 2π 1 + 4
.
k
Further, we may have
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
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I NTEGRAL M EANS FOR S TARLIKE AND C ONVEX F UNCTIONS WITH N EGATIVE C OEFFICIENTS
Theorem 3.3. Let f (z) ∈ C, λ > 0, and fk (z) = z −
2k−2
X
(3.9)
zk
k2
9
(k ≥ 2). If f (z) satisfies
j(k − j)aj ≤ 0,
j=2
and if there exists an analytic function ω(z) in U given by
k−1
(ω(z))
=k
∞
X
nan z n−1 ,
n=2
iθ
then, for z = re (0 < r < 1),
Z
(3.10)
2π
0
λ
Z
|f (z)| dθ ≤
0
2π
|fk0 (z)|λ dθ.
0
Example 3.2. Take the functions
f (z) = z −
(3.11)
1 2
1
1
z − z3 − z4
24
18
48
and
1
f3 (z) = z − z 3
9
(3.12)
with k = 3 in Theorem 3.3. Since
∞
X
9
16
5
4
n 2 an =
+
+
= <1
24 18 48
6
n=2
and
1
1
−
= 0,
12 12
f (z) satisfies the conditions in Theorem 3.3. If we take λ = 2, then we have
Z 2π
1 4
20
0
2
|f (z)| dθ ≤ 2π 1 + r < π.
9
9
0
2(3 − 2)a2 + 3(3 − 3)a3 + 4(3 − 4)a4 =
Corollary 3.4. Let f (z) ∈ C, 0 < λ ≤ 2, and fk (z) = z −
zk
k2
(k ≥ 2). If f (z) satisfies the
condition in Theorem 3.3, then, for k ≥ 2, then, for z = reiθ (0 < r < 1),
λ
λ
Z 2π
1 2(k−1) 2
1 2
0
λ
|f (z)| dθ ≤ 2π 1 + r
< 2π 1 +
.
k
k
0
4. A PPLICATIONS FOR THE I NTEGRATED F UNCTIONS
For f (z) ∈ T , we define
I0 f (z) = f (z) = z −
∞
X
an z n
n=2
Z
If (z) = I1 f (z) =
0
z
∞
1 2 X an n+1
z
f (t)dt = z −
2
n
+
1
n=2
∞
X
1
n!
k+1
Ik f (z) = I(Ik−1 f (z)) =
z
−
an z n+k (k = 1, 2, 3, . . . ).
(k + 1)!
(n
+
k)!
n=2
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
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10
S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI ,
Theorem 4.1. Let f (z) ∈ T ∗ , λ > 0, and fj (z) = z −
AND J UNICHI
zj
j
N ISHIWAKI
(j = 2, 3, 4, . . . ).
If f (z) satisfies
j 2 +j−1
X j2 + j − k
(a2j 2 +2j−k − ak ) ≥ 0
j(j + 1)
k=2
(4.1)
for j = 2, 3, 4, . . . , and if there exists an analytic function ω(z) in U given by
!
∞
X
1
(ω(z))j−1 = j(j + 1)
an z n−1 ,
n
+
1
n=2
then
2π
Z
Z
λ
|If (z)| dθ ≤
(4.2)
0
2π
|Ifj (z)|λ dθ.
0
Proof. We have to prove
Z
0
2π
λ
λ
Z 2π ∞
X
2
2
j−1
n−1 1 −
an z dθ ≤
z dθ.
1 −
n+1
j(j + 1)
0
n=2
If
1−
∞
X
n=2
2
2
an z n−1 ≺ 1 −
z j−1 ,
n+1
j(j + 1)
then the proof is completed by Theorem A.
Let us define the function ω(z) by
1−
∞
X
n=2
2
2
an z n−1 = 1 −
(ω(z))j−1 .
n+1
j(j + 1)
Then
|ω(z)|j−1
∞
X
1
n−1 = j(j + 1)
an z n+1
n=2
!
∞
X
1
≤ |z| j(j + 1)
an .
n+1
n=2
Thus, we only show that
j(j + 1)
∞
X
n=2
∞
X
1
an ≤
nan
n+1
n=2
or
∞
X
∞
X
an ≤
n
n=2
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
n=2
1
1
+
j(j + 1) n + 1
an .
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I NTEGRAL M EANS FOR S TARLIKE AND C ONVEX F UNCTIONS WITH N EGATIVE C OEFFICIENTS
11
Indeed,
∞
X
n
n=2
1
1
+
an
j(j + 1) n + 1
1
1
1
1
=2
+
a2 + 3
+
a3 + · · ·
j(j + 1) 3
j(j + 1) 4
1
1
1
1
+ (j − 1)
+
aj−1 + j
+
aj
j(j + 1) j
j(j + 1) j + 1
1
1
+ (j + 1)
+
aj+1
j(j + 1) j + 2
1
1
2
+ · · · + (2j + 2j − 3)
+
a2j 2 +2j−3
j(j + 1) 2j 2 + 2j − 2
1
1
2
+ (2j + 2j − 2)
+
a2j 2 +2j−1 + · · ·
j(j + 1) 2j 2 + 2j − 1
j(j + 1) − 2
j(j + 1) − 3
≥ 1−
a2 + 1 −
a3 + · · ·
j(j + 1)
j(j + 1)
j(j + 1) − (j − 1)
j(j + 1) − j
+ 1−
aj−1 + 1 −
aj
j(j + 1)
j(j + 1)
j(j + 1) − (j + 1)
+ 1−
aj+1
j(j + 1)
j(j + 1) − (2j 2 + 2j − 3)
+ ··· + 1 −
a2j 2 +2j−3
j(j + 1)
j(j + 1) − (2j 2 + 2j − 2)
+ 1−
a2j 2 +2j−2 + · · ·
j(j + 1)
j2 + j − 3
j2 + j − 2
(a2j 2 +2j−2 − a2 ) +
(a2j 2 +2j−3 − a3 )
=
j(j + 1)
j(j + 1)
j2 + 1
j2
+ ··· +
(a2j 2 +j+1 − aj−1 ) +
(a2j 2 +j − aj )
j(j + 1)
j(j + 1)
j2 − 1
+
(a2j 2 +j−1 − aj+1 ) + · · · + a2 + a3 + · · · + a2j 2 +2j−2 + · · ·
j(j + 1)
j 2 +j−1
∞
X
X j2 + j − k
=
(a2j 2 +2j−k − ak ) +
an
j(j
+
1)
n=2
k=2
≥
∞
X
an
n=2
for
j 2 +j−1
X j2 + j − k
(a2j 2 +2j−k − ak ) ≥ 0.
j(j
+
1)
k=2
This completes the proof of Theorem 4.1.
Finally, we derive
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
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12
S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI ,
Theorem 4.2. Let f (z) ∈ T ∗ ,λ > 0,and fj (z) = z −
∞
X
6
an ≥
5
n=2
(4.3)
(j+k)!
−1
2(j−1)!
X
n=2
zj
j
2n(j − 1)!
1−
(j + k)!
AND J UNICHI
N ISHIWAKI
(j = 2, 3, 4, . . . ). If f (z) satisfies
an − a (j+k)! −n
(j−1)!
for k = 2, 3, 4, . . . , and if there exists an analytic function ω(z) in U given by
∞
(j + k)! X
n!
j−1
(ω(z))
=
an z n−1 ,
(j − 1)! n=2 (n + k)!
then
Z
2π
λ
Z
|Ik f (z)| dθ ≤
(4.4)
0
2π
|Ik fj (z)|λ dθ.
0
Proof. We have to show that
∞
X
n!(k + 1)!
(j − 1)!(k + 1)! j−1
1−
an z n−1 ≺ 1 −
z .
(n + k)!
(j + k)!
n=2
Define ω(z) by
1−
∞
X
n!(k + 1)!
n=2
(n + k)!
an z n−1 = 1 −
(j − 1)!(k + 1)!
(ω(z))j−1
(j + k)!
or by
∞
j−1
(ω(z))
n!
(j + k)! X
an z n−1 .
=
(j − 1)! n=2 (n + k)!
Then we have to show that
∞
∞
X
(j + k)! X
n!
an ≤
nan ,
(j − 1)! n=2 (n + k)!
n=2
that is, that
∞
X
n=2
∞
n!
(j − 1)! X
an ≤
nan .
(n + k)!
(j + k)! n=2
Since
∞
X
n=2
we obtain
∞
X
n!
1
an =
an
(n + k)!
(n
+
1)(n
+
2)
·
·
·
(n
+
k)
n=2
∞ X
1
1
1
1
=
−
−
· · · an
n+1 n+2
n+3 n+4
n=2
[ k2 ]
∞ X
1
1
≤
−
an
n
+
1
n
+
2
n=2
∞ X
1
1
≤
−
an ,
n
+
1
n
+
2
n=2
∞ ∞
X
1
1
(j − 1)! X
−
an ≤
nan .
n+1 n+2
(j + k)! n=2
n=2
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I NTEGRAL M EANS FOR S TARLIKE AND C ONVEX F UNCTIONS WITH N EGATIVE C OEFFICIENTS
13
Furthermore, we have
∞
X
an ≤
n=2
∞ X
2n(j − 1)!
n=2
2n
n
+
−
(j + k)!
n+1 n+2
an .
Let the function h(n) be given by
h(n) =
n
2
2n
−
=1− 2
.
n+1 n+2
n + 3n + 2
Since h(n) is increasing for n ≥ 2,
5
h(n) ≥ .
6
Thus, we only show that
∞
X
n=2
an ≤
∞ X
11
n=2
(j + k)! − 2n(j − 1)!
−
6
(j + k)!
an .
In fact,
∞ X
11
n=2
(j + k)! − 2n(j − 1)!
−
an
6
(j + k)!
11 (j + k)! − 4(j − 1)!
11 (j + k)! − 6(j − 1)!
=
−
a2 +
−
a3 + · · ·
6
(j + k)!
6
(j + k)!
11 4(j − 1)!
11 2(j − 1)!
+
−
a (j+k)! −2 +
−
a (j+k)! −1
2(j−1)!
2(j−1)!
6
(j + k)!
6
(j + k)!
11
11 2(j − 1)!
+
− 0 a (j+k)! +
−
a (j+k)! +1
2(j−1)!
2(j−1)!
6
6
(j + k)!
11 4(j − 1)!
11 (j + k)! − 6(j − 1)!
+
+
a (j+k)! +2 + · · · +
+
a (j+k)! −3
2(j−1)!
(j−1)!
6
(j + k)!
6
(j + k)!
11 (j + k)! − 4(j − 1)!
+
+
a (j+k)! −2 + · · ·
(j−1)!
6
(j + k)!
∞
11 X
(j + k)! − 4(j − 1)!
≥
an +
a (j+k)! −2 − a2
(j−1)!
6 n=2
(j + k)!
(j + k)! − 6(j − 1)!
4(j − 1)!
+
a (j+k)! −3 − a3 +
a (j+k)! +2 − a (j+k)! −2
(j−1)!
2(j−1)!
2(j−1)!
(j + k)!
(j + k)!
2(j − 1)!
+
a (j+k)! +1 − a (j+k)! −1
2(j−1)!
2(j−1)!
(j + k)!
∞
∞
X
5X
(j + k)! − 4(j − 1)!
=
an +
an +
a (j+k)! −2 − a2
2(j−1)!
6
(j
+
k)!
n=2
n=2
(j + k)! − 6(j − 1)!
+
a (j+k)! −3 − a3 + · · ·
2(j−1)!
(j + k)!
(j + k)! − {(j + k)! − 4(j − 1)!}
+
a (j+k)! +2 − a (j+k)! −2
2(j−1)!
2(j−1)!
(j + k)!
(j + k)! − {(j + k)! − 2(j − 1)!}
+
a (j+k)! +1 − a (j+k)! −1
2(j−1)!
2(j−1)!
(j + k)!
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
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14
S HIGEYOSHI OWA , M IHAI PASCU , DAISUKE YAGI ,
=
∞
X
≥
n=2
∞
X
an +
∞
5X
6
(j+k)!
−1
2(j−1)!
X
an +
n=2
n=2
AND J UNICHI
(j + k)! − 2n(j − 1)!
(j + k)!
N ISHIWAKI
a (j+k)! −n − an
(j−1)!
an
n=2
for
(j+k)!
−1
2(j−1)!
∞
X
6
an ≥
5
n=2
X
n=2
2n(j − 1)!
1−
(j + k)!
an − a (j+k)! −n .
(j−1)!
This completes the proof of Theorem 4.2.
Remark 4.3. Letting k = 2, if f (z) satisfies,
∞
X
6
an ≥
5
n=2
(4.5)
j(j+1)(j+2)
−1
2
X
n=2
2n
1−
j(j + 1)(j + 2)
(an − aj(j+1)(j+2)−n )
for j = 2, 3, 4, . . . , then
Z
2π
λ
Z
2π
|I2 fj (z)|λ dθ.
|I2 f (z)| dθ ≤
(4.6)
0
0
Remark 4.4. Letting k = 3, if f (z) satisfies,
(4.7)
∞
X
6
an ≥
5
n=2
j(j+1)(j+2)(j+3)
−1
2
X
n=2
2n
1−
j(j + 1)(j + 2)(j + 3)
(an − aj(j+1)(j+2)(j+3)−n )
for j = 2, 3, 4, . . . , then
Z
2π
λ
Z
|I3 f (z)| dθ ≤
(4.8)
0
2π
|I3 fj (z)|λ dθ.
0
R EFERENCES
[1] P.L. DUREN, Univalent Functions, Springer-Verlag, New York, 1983.
[2] J.E. LITTLEWOOD, On inequalities in the theory of functions, Proc. London Math. Soc., 23 (1925),
481–519.
[3] S. OWA AND T. SEKINE, Integral means of analytic functions, J. Math. Anal. Appl. (in press).
[4] T. SEKINE, K. TSURUMI AND H.M. SRIVASTAVA, Integral means for generalized subclasses of
analytic functions, Sci. Math. Jpon., 54 (2001), 489–501.
[5] T. SEKINE, K. TSURUMI, S. OWA AND H.M. SRIVASTAVA, Integral means inequalities for fractional derivatives of some general subclasses of analytic functions, J. Inequal. Pure Appl. Math., 3
(2002), Art. 66.
[6] H. SILVERMAN, Univalent functions with negative coefficients, Proc. Amer. Math. Soc., 51 (1975),
109–116.
[7] H. SILVERMAN, Integral means for univalent functions with negative coefficients, Houston J.
Math., 23 (1997), 169–174.
J. Inequal. Pure and Appl. Math., 6(2) Art. 50, 2005
http://jipam.vu.edu.au/