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volume 6, issue 2, article 50, 2005.

Received 17 February, 2005;

accepted 06 April, 2005.

Communicated by:N.E. Cho

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Journal of Inequalities in Pure and Applied Mathematics

INTEGRAL MEANS FOR STARLIKE AND CONVEX FUNCTIONS WITH NEGATIVE COEFFICIENTS

SHIGEYOSHI OWA¹, MIHAI PASCU², DAISUKE YAGI¹ AND JUNICHI NISHIWAKI¹

Department of Mathematics¹ Kinki University

Higashi-Osaka, Osaka 577-8502 Japan

EMail:owa@math.kindai.ac.jp Department of Mathematics² Transilvania University of Brasov R-2200 Brasov

Romania

EMail:mihai.pascu@unitbv.ro

c

2000Victoria University ISSN (electronic): 1443-5756 041-05

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Integral Means for Starlike and Convex Functions with Negative

Coefficients

Shigeyoshi Owa, Mihai Pascu, Daisuke Yagi and Junichi Nishiwaki

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Abstract

LetT be the class of functionsf(z)with negative coefficients which are analytic and univalent in the open unit diskUwithf(0) = 0andf⁰(0) = 1. The classes T^∗andCare defined as the subclasses ofT 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.

2000 Mathematics Subject Classification:Primary 30C45.

Key words: Univalent, Starlike, Convex, Integral mean.

Memorial Paper for Professor Nicolae N. Pascu

1. Introduction

LetAdenote the class of functionsf(z)of the form

(1.1) f(z) =z+

∞

X

n=2

a_nzⁿ

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)∈ Ais said to be starlike with respect to the origin inUif it satisfies

(1.2) Re

zf⁰(z) f(z)

>0 (z ∈U).

We denote byS^∗the subclass ofS consisting of all starlike functionsf(z)with respect to the origin inU. Further, a functionf(z)∈ Ais said to be convex in Uif it satisfies

(1.3) Re

1 + zf⁰⁰(z) f⁰(z)

>0 (z ∈U).

We also denote byKthe subclass ofSconsisting off(z)which are convex in U. By the above definitions, we know thatf(z)∈ Kif and only ifzf⁰(z)∈ S^∗, and thatK ⊂ S^∗ ⊂ S ⊂ A.

The classT is defined as the subclass ofS consisting of all functionsf(z) which are given by

(1.4) f(z) =z−

∞

X

n=2

anzⁿ (an ≥0).

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Further, we denote by T^∗ = S^∗ ∩ T and C = K ∩ T. It is well-known by Silverman [6] that

Remark 1. A functionf(z)∈ T^∗if and only if (1.5)

∞

X

n=2

nan ≤1.

A functionf(z)∈ C if and only if (1.6)

∞

X

n=2

n²a_n≤1.

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), andf(z) =g(ω(z)). We denote this subordination by

(1.7) f(z)≺g(z). (cf. Duren [1]).

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 >0and|z|=r(0< r <1),

(1.8)

Z 2π 0

|f(re^iθ)|^λdθ ≤ Z 2π

0

|g(re^iθ)|^λdθ.

Furthermore, Silverman [6] has shown that

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Remark 2. f₁(z) = z andf_n(z) = z − ^z_nⁿ (n ≥ 2)are extreme points of the classT^∗(orT). f₁(z) = z andf_n(z) =z − ^z_nⁿ2 (n ≥ 2)are extreme points of the classC.

Applying Theorem A with extreme points ofT, Silverman [7] has proved the following results.

Theorem B. Suppose thatf(z) ∈ T^∗, λ > 0 andf₂(z) = z− ^z₂². Then, for z =re^iθ(0< r <1),

(1.9)

Z 2π 0

|f(z)|^λdθ ≤ Z 2π

0

|f₂(z)|^λdθ.

Theorem C. If f(z) ∈ T^∗, λ > 0, and f₂(z) = z − ^z₂², then, for z = re^iθ (0< r <1),

(1.10)

Z 2π 0

|f⁰(z)|^λdθ ≤ Z 2π

0

|f₂⁰(z)|^λdθ.

In the present paper, we consider the generalization properties for Theorem Band TheoremCwithf(z)∈ T^∗andf(z)∈ C.

Remark 3. More recently, applying TheoremAby Littlewood [2], Sekine, Tsu- rumi and Srivastava [4]; and Sekine, Tsurumi, Owa and Srivastava [5] have discussed some interesting properties of integral means inequalities for frac- tional 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.

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2. Generalization Properties

Our first result for the generalization properties is contained in

Theorem 2.1. Let f(z) ∈ T^∗, λ > 0, and f_k(z) = z − ^z_k^k (k ≥ 2). Iff(z) satisfies

(2.1)

k−3

X

j=0

j+ 1

k (a2k+j−1+a_k+j+1−ak−j−1)≥0 fork ≥3, and if there exists an analytic functionω(z)inUgiven by

(ω(z))^k−1 =k

∞

X

n=2

a_nzⁿ⁻¹

! ,

then, forz =re^iθ (0< r <1), (2.2)

Z 2π 0

0

|fk(z)|^λdθ.

Proof. Forf(z)∈ T^∗, we have to show that Z 2π

0

1−

∞

X

n=2

a_nzⁿ⁻¹

λ

dθ≤ Z 2π

0

1− z^k−1 k

λ

dθ.

By TheoremA, it suffices to prove that 1−

∞

X

n=2

anzⁿ⁻¹ ≺1− z^k−1 k .

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Let us define the functionω(z)by

(2.3) 1−

∞

X

n=2

a_nzⁿ⁻¹ = 1− 1

k(ω(z))^k−1. It follows from (2.3) that

|ω(z)|^k−1 = k

∞

X

n=2

a_nzⁿ⁻¹

≤ |z|

∞

X

n=2

ka_n

! .

Thus, we only show that

∞

X

n=2

ka_n≤

∞

X

n=2

na_n,

or ∞

X

n=2

a_n≤ 1 k

∞

X

n=2

na_n

! .

Indeed, we see that 1

k

∞

X

n=2

nan

!

=

1− k−2 k

a2+

1− k−3 k

a3+· · ·+

1− 2 k

ak−2

+

1− 1 k

ak−1+a_k+

1 + 1 k

a_k+1+

1 + 2 k

a_k+2

+· · ·+

1 + k+ 1 k

a_2k+1+

1 + k+ 2 k

a_2k+2+· · ·

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= k−2

k (a2k−2−a2) + k−3

k (a2k−3−a3) +· · · + 2

k(a_k+2−ak−2) + 1

k(a_k+1−ak−1) +

1 + k−1 k

a2k−1

+

1 + k k

a_2k+

1 + k+ 1 k

a_2k+1+· · ·+

2k−2

X

n=2

a_n.

Noting that

1 + k+j

k ≥1 + 2 +j

k , (j =−1,0,1, . . .), we obtain

1 k

∞

X

n=2

na_n

! (2.4)

≥ k−2

k (a2k−2−a₂) + k−3

k (a2k−3−a₃) +· · · + 2

k(a_k+2−ak−2) + 1

k(a_k+1−ak−1) +

1 + 1

k

a2k−1+

1 + 2 k

a_2k+· · · +

1 + k−3 k

a3k−5+

1 + k−2 k

a3k−4+· · ·+

2k−2

X

n=2

a_n

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≥ 1

k(a2k−1 +ak+1−ak−1) + 2

k(a2k+ak+2−ak−2) +· · · +k−2

k (a3k−4+a2k−2−a₂) +

∞

X

n=2

a_n

=

k−3

X

j=0

j + 1

k (a2k+j−1+a_k+j+1−ak−j−1) +

∞

X

n=2

a_n

≥

∞

X

n=2

a_n

with the following condition

k−3

X

j=0

j+ 1

k (a2k+j−1+a_k+j+1−ak−j−1)≥0.

Thus, we observe that the function ω(z)defined by (2.3) is analytic inU with ω(0) = 0,|ω(z)|<1 (z ∈U). This completes the proof of the theorem.

Remark 4. 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

1200z²− 1

18z³− 1

48z⁴− 1 100z⁵ and

(2.6) f₃(z) =z− 1

3z³

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withk= 3in Theorem2.1. Sincef(z)satisfies

∞

X

n=2

na_n= 217 600 <1, we havef(z)∈ T^∗. Furthermore,f(z)satisfies,

1

3(a₅+a₄−a₂) = 1 3

1 100 + 1

48− 37 1200

= 0.

Thus,f(z)satisfies the conditions in Theorem2.1withk = 3.

If we takeλ= 2, then we have Z 2π

0

|f(z)|²dθ ≤2πr²

1 + 1 9r⁴

< 20

9 π = 6.9813. . . .

Corollary 2.2. Letf(z)∈ T^∗,0< λ≤2, andf_k(z) =z−^z_k^k (k ≥2). Iff(z) satisfies the conditions in Theorem2.1, then, forz =re^iθ (0< r <1),

(2.7)

Z 2π 0

|f(z)|^λdθ ≤2πr^λ

1 + 1

k²r^2(k−1) ^λ₂

<2π

1 + 1 k²

^λ₂ .

Proof. It follows that

Z 2π 0

|f_k(z)|^λdθ = Z 2π

0

|z|^λ

1− z^k−1 k

λ

dθ.

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Applying Hölder’s inequality for0< λ <2, we obtain that

Z 2π 0

|z|^λ

1− z^k−1 k

λ

dθ ≤ Z 2π

0

(|z|^λ)^2−λ² dθ ^2−λ₂



 Z 2π

0

1− z^k−1 k

λ!²_λ dθ





λ 2

= Z 2π

0

|z|^2−λ^2λ dθ

2−λ

2 Z 2π

0

1− z^k−1 k

2

dθ

!^λ₂

=

2πr^2−λ^2λ ^2−λ₂ 2π

1 + 1

k²r^2(k−1) ^λ₂

= 2πr^λ

1 + 1

k²r^2(k−1) ^λ₂

<2π

1 + 1 k²

^λ₂ .

Further, it is clear forλ= 2.

For the generalization of TheoremCby Silverman [7], we have

Theorem 2.3. Let f(z) ∈ T^∗, λ > 0, andf_k(z) = z − ^z_k^k (k ≥ 2). If there exists an analytic functionω(z)inUgiven by

ω(z)k−1

=

∞

X

n=2

nanzⁿ⁻¹,

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then, forz =re^iθ (0< r <1), (2.8)

Z 2π 0

|f⁰(z)|^λdθ ≤ Z 2π

0

|f_k⁰(z)|^λdθ.

Proof. Forf(z)∈ T^∗, it is sufficient to show that

(2.9) 1−

∞

X

n=2

nanzⁿ⁻¹ ≺1−z^k−1. Let us define the functionω(z)by

(2.10) 1−

∞

X

n=2

na_nzⁿ⁻¹ = 1−ω(z)^k−1, or, by

ω(z)^k−1 =

∞

X

n=2

na_nzⁿ⁻¹.

Sincef(z)satisfies

∞

X

n=2

na_n ≤1,

the functionω(z)is analytic inU,ω(0) = 0, and|ω(z)|<1 (z ∈U).

Remark 5. If we take k = 2 in Theorem 2.3, then we have Theorem C by Silverman [7].

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Using the Hölder inequality for Theorem2.3, we have

Corollary 2.4. Letf(z)∈ T^∗,0< λ≤2, andfk(z) =z−^z_k^k (k ≥2). Iff(z) satisfies the conditions in Theorem2.3, then, forz =re^iθ (0< r <1),

Z 2π 0

|f⁰(z)|^λdθ ≤2π 1 +r^2(k−1)^λ₂

<2^2+λ² π.

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3. Integral Means for Functions in the Class C

In this section, we discuss the integral means for functionsf(z)in the classC.

Theorem 3.1. Let f(z) ∈ C, λ > 0, and f_k(z) = z − ^z_k^k2 (k ≥ 2). If f(z) satisfies

(3.1)

k−1

X

j=2

(k+j)(k−j)

k² (a2k−j −a_j)≥0

fork ≥3, and if there exists an analytic functionω(z)inUgiven by (ω(z))^k−1 =k²

∞

X

n=2

a_nzⁿ⁻¹,

then, forz =re^iθ (0< r <1), (3.2)

Z 2π 0

0

|f_k(z)|^λdθ.

Proof. For the proof, we need to show that

(3.3) 1−

∞

X

n=2

a_nzⁿ⁻¹ ≺1− z^k−1 k² by TheoremA. Define the functionω(z)by

(3.4) 1−

∞

X

n=2

anzⁿ⁻¹ = 1− 1

k²ω(z)^k−1,

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or by

(3.5) (ω(z))^k−1 =k²

∞

X

n=2

a_nzⁿ⁻¹

! .

Therefore, we have to show that

∞

X

n=2

a_n≤ 1 k²

∞

X

n=2

n²a_n

! .

Using the same technique as in the proof of Theorem2.1, we see that 1

k²

∞

X

n=2

n²a_n

!

≥

k−1

X

j=2

(k+j)(k−j)

k² (a2k−j −a_j) +

∞

X

n=2

a_n

≥

∞

X

n=2

an.

Example 3.1. Consider the functions

(3.6) f(z) =z− 1

40z²− 1

18z³− 1 40z⁴ and

(3.7) f₃(z) =z− 1

9z³

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withk= 3in Theorem3.1. Then we have that

∞

X

n=2

n²a_n= 4 40+ 9

18 +16 40 = 1, which impliesf(z)∈ C, and that

5

9(a₄−a₂) = 0.

Thus f(z)satisfies the conditions of Theorem3.1. If we makeλ = 2, then we see that

Z 2π 0

|f(z)|²dθ ≤2πr²

1 + 1 81r⁴

< 164

81π = 6.3607· · · .

Corollary 3.2. Let f(z) ∈ C, 0 < λ ≤ 2, and fk(z) = z − ^z_k^k2 (k ≥ 2). If f(z)satisfies the condition in Theorem3.1, then, fork ≥ 3, then, forz = re^iθ (0< r <1),

Z 2π 0

|f(z)|^λdθ ≤2πr^λ

1 + 1

k⁴r^2(k−1) ^λ₂ (3.8)

<2π

1 + 1 k⁴

^λ₂ .

Further, we may have

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Theorem 3.3. Let f(z) ∈ C, λ > 0, and f_k(z) = z − ^z_k^k2 (k ≥ 2). If f(z) satisfies

(3.9)

2k−2

X

j=2

j(k−j)a_j ≤0,

and if there exists an analytic functionω(z)inUgiven by

(ω(z))^k−1 =k

∞

X

n=2

na_nzⁿ⁻¹,

then, forz =re^iθ (0< r <1), (3.10)

Z 2π 0

|f⁰(z)|^λdθ ≤ Z 2π

0

|f_k⁰(z)|^λdθ.

Example 3.2. Take the functions

(3.11) f(z) =z− 1

24z²− 1

18z³− 1 48z⁴ and

(3.12) f₃(z) =z− 1

9z³ withk= 3in Theorem3.3. Since

∞

X

n=2

n²an = 4 24+ 9

18+ 16 48 = 5

6 <1

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and

2(3−2)a₂+ 3(3−3)a₃+ 4(3−4)a₄ = 1 12− 1

12 = 0,

f(z)satisfies the conditions in Theorem3.3. If we takeλ = 2, then we have Z 2π

0

|f⁰(z)|²dθ ≤2π

1 + 1 9r⁴

< 20 9 π.

Corollary 3.4. Let f(z) ∈ C, 0 < λ ≤ 2, and fk(z) = z − ^z_k^k2 (k ≥ 2). If f(z)satisfies the condition in Theorem3.3, then, fork ≥ 2, then, forz = re^iθ (0< r <1),

Z 2π 0

|f⁰(z)|^λdθ ≤2π

1 + 1 kr^2(k−1)

^λ₂

<2π

1 + 1 k

^λ₂ .

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4. Applications for the Integrated Functions

Forf(z)∈ T, we define I₀f(z) = f(z) =z−

∞

X

n=2

a_nzⁿ

If(z) = I₁f(z) = Z z

0

f(t)dt= 1 2z²−

∞

X

n=2

a_n n+ 1zⁿ⁺¹ Ikf(z) = I(Ik−1f(z)) = 1

(k+ 1)!z^k+1−

∞

X

n=2

n!

(n+k)!anz^n+k (k = 1,2,3, . . .).

Theorem 4.1. Letf(z)∈ T^∗,λ >0, andf_j(z) =z− ^z_j^j (j = 2,3,4, . . .).

Iff(z)satisfies (4.1)

j²+j−1

X

k=2

j²+j−k

j(j+ 1) (a_2j²_+2j−k−a_k)≥0

forj = 2,3,4, . . . ,and if there exists an analytic functionω(z)inUgiven by (ω(z))^j−1 =j(j+ 1)

∞

X

n=2

1

n+ 1a_nzⁿ⁻¹

! ,

then (4.2)

Z 2π 0

|If(z)|^λdθ ≤ Z 2π

0

|If_j(z)|^λdθ.

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Proof. We have to prove

Z 2π 0

1−

∞

X

n=2

2

n+ 1a_nzⁿ⁻¹

λ

dθ ≤ Z 2π

0

1− 2

j(j+ 1)z^j−1

λ

dθ.

If

1−

∞

X

n=2

2

n+ 1a_nzⁿ⁻¹ ≺1− 2

j(j + 1)z^j−1, then the proof is completed by TheoremA.

Let us define the functionω(z)by 1−

∞

X

n=2

2

n+ 1a_nzⁿ⁻¹ = 1− 2

j(j+ 1)(ω(z))^j−1. Then

|ω(z)|^j−1 =

j(j+ 1)

∞

X

n=2

1

n+ 1a_nzⁿ⁻¹

≤ |z| j(j+ 1)

∞

X

n=2

1 n+ 1a_n

! .

Thus, we only show that

j(j+ 1)

∞

X

n=2

1

n+ 1an ≤

∞

X

n=2

nan

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or ∞

X

n=2

a_n ≤

∞

X

n=2

n

1

j(j + 1) + 1 n+ 1

a_n.

Indeed,

∞

X

n=2

n

1

j(j+ 1) + 1 n+ 1

a_n

= 2

1

j(j+ 1) +1 3

a₂+ 3

1

j(j+ 1) + 1 4

a₃+· · · + (j−1)

1

j(j + 1) +1 j

a_j−1+j

1

j(j+ 1) + 1 j+ 1

a_j

+ (j+ 1)

1

j(j+ 1) + 1 j+ 2

aj+1

+· · ·+ (2j²+ 2j−3)

1

j(j + 1) + 1 2j²+ 2j−2

a_2j²+2j−3

+ (2j²+ 2j−2)

1

j(j+ 1) + 1 2j²+ 2j−1

a_2j²+2j−1+· · ·

≥

1− j(j+ 1)−2 j(j+ 1)

a2+

1− j(j+ 1)−3 j(j+ 1)

a3+· · · +

1−j(j + 1)−(j−1) j(j+ 1)

aj−1 +

1− j(j+ 1)−j j(j+ 1)

aj

+

1−j(j + 1)−(j+ 1) j(j+ 1)

aj+1

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+· · ·+

1− j(j+ 1)−(2j²+ 2j−3) j(j+ 1)

a_2j²_+2j−3

+

1−j(j + 1)−(2j²+ 2j −2) j(j+ 1)

a_2j²+2j−2+· · ·

= j²+j−2

j(j + 1) (a_2j²+2j−2−a₂) + j²+j −3

j(j+ 1) (a_2j²+2j−3−a₃) +· · ·+ j²+ 1

j(j+ 1)(a_2j²_+j+1−aj−1) + j²

j(j+ 1)(a_2j²_+j −a_j) + j²−1

j(j+ 1)(a_2j²_+j−1−a_j+1) +· · ·+a₂+a₃+· · ·+a_2j²_+2j−2+· · ·

=

j²+j−1

X

k=2

j²+j−k

j(j+ 1) (a_2j²+2j−k−a_k) +

∞

X

n=2

a_n

≥

∞

X

n=2

a_n

for

j²+j−1

X

k=2

j²+j −k

j(j+ 1) (a_2j²+2j−k−a_k)≥0.

This completes the proof of Theorem4.1.

Finally, we derive

Theorem 4.2. Letf(z)∈ T^∗,λ >0,andf_j(z) = z− ^z_j^j (j = 2,3,4, . . .). If

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f(z)satisfies

(4.3)

∞

X

n=2

a_n≥ 6 5

(j+k)!

2(j−1)!−1

X

n=2

1− 2n(j−1)!

(j+k)! a_n−a(j+k)!

(j−1)!−n

fork = 2,3,4, . . . ,and if there exists an analytic functionω(z)inUgiven by (ω(z))^j−1 = (j+k)!

(j−1)!

∞

X

n=2

n!

(n+k)!a_nzⁿ⁻¹, then

(4.4)

Z 2π 0

|I_kf(z)|^λdθ ≤ Z 2π

0

|I_kf_j(z)|^λdθ.

Proof. We have to show that 1−

∞

X

n=2

n!(k+ 1)!

(n+k)! a_nzⁿ⁻¹ ≺1− (j−1)!(k+ 1)!

(j+k)! z^j−1. Defineω(z)by

1−

∞

X

n=2

n!(k+ 1)!

(n+k)! a_nzⁿ⁻¹ = 1−(j−1)!(k+ 1)!

(j +k)! (ω(z))^j−1 or by

(ω(z))^j−1 = (j+k)!

(j−1)!

∞

X

n=2

n!

(n+k)!anzⁿ⁻¹.

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Then we have to show that (j+k)!

(j −1)!

∞

X

n=2

n!

(n+k)!a_n ≤

∞

X

n=2

na_n, that is, that

∞

X

n=2

n!

(n+k)!a_n ≤ (j−1)!

(j +k)!

∞

X

n=2

na_n. Since

∞

X

n=2

n!

(n+k)!an =

∞

X

n=2

1

(n+ 1)(n+ 2)· · ·(n+k)an

=

∞

X

n=2

1

n+ 1 − 1 n+ 2

1

n+ 3 − 1 n+ 4

· · ·

a_n

≤

∞

X

n=2

1

n+ 1 − 1 n+ 2

[^k₂] a_n

≤

∞

X

n=2

1

n+ 1 − 1 n+ 2

a_n, we obtain

∞

X

n=2

1

n+ 1 − 1 n+ 2

a_n≤ (j−1)!

(j+k)!

∞

X

n=2

na_n. Furthermore, we have

∞

X

n=2

an≤

∞

X

n=2

2n(j −1)!

(j +k)! + 2n

n+ 1 − n n+ 2

an.

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Let the functionh(n)be given by h(n) = 2n

n+ 1 − n

n+ 2 = 1− 2 n²+ 3n+ 2. Sinceh(n)is increasing forn ≥2,

h(n)≥ 5 6. Thus, we only show that

∞

X

n=2

a_n≤

∞

X

n=2

11

6 − (j+k)!−2n(j−1)!

(j+k)!

a_n.

In fact,

∞

X

n=2

11

6 − (j +k)!−2n(j−1)!

(j+k)!

an

= 11

6 −(j+k)!−4(j−1)!

(j+k)!

a₂

+ 11

6 − (j+k)!−6(j−1)!

(j+k)!

a₃+· · · +

11

6 − 4(j−1)!

(j+k)!

a(j+k)!

2(j−1)!−2+ 11

6 −2(j−1)!

(j +k)!

a(j+k)!

2(j−1)!−1

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+ 11

6 −0

a(j+k)!

2(j−1)!

+ 11

6 − 2(j−1)!

(j+k)!

a(j+k)!

2(j−1)!+1

+ 11

6 + 4(j −1)!

(j+k)!

a(j+k)!

2(j−1)!+2+· · · +

11

6 + (j+k)!−6(j−1)!

(j+k)!

a(j+k)!

(j−1)!−3

+ 11

6 + (j+k)!−4(j−1)!

(j+k)!

a(j+k)!

(j−1)!−2+· · ·

≥ 11 6

∞

X

n=2

a_n+ (j +k)!−4(j −1)!

(j+k)!

a(j+k)!

(j−1)!−2−a₂

+ (j+k)!−6(j−1)!

(j+k)!

a(j+k)!

(j−1)!−3−a₃

+ 4(j−1)!

(j+k)!

a(j+k)!

2(j−1)!+2−a(j+k)!

2(j−1)!−2

+ 2(j−1)!

(j+k)!

a(j+k)!

2(j−1)!+1−a(j+k)!

2(j−1)!−1

=

∞

X

n=2

a_n+5 6

∞

X

n=2

a_n+ (j+k)!−4(j −1)!

(j+k)!

a(j+k)!

2(j−1)!−2−a₂

+ (j+k)!−6(j−1)!

(j+k)!

a^(j+k)!

2(j−1)!−3 −a3

+· · · + (j+k)!− {(j+k)!−4(j−1)!}

(j+k)!

a^(j+k)!

2(j−1)!+2−a^(j+k)!

2(j−1)!−2

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+ (j+k)!− {(j+k)!−2(j−1)!}

(j+k)!

a(j+k)!

2(j−1)!+1−a(j+k)!

2(j−1)!−1

=

∞

X

n=2

a_n+5 6

∞

X

n=2

a_n+

(j+k)!

2(j−1)!−1

X

n=2

(j+k)!−2n(j−1)!

(j+k)!

a(j+k)!

(j−1)!−n−a_n

≥

∞

X

n=2

a_n

for

∞

X

n=2

a_n≥ 6 5

(j+k)!

2(j−1)!−1

X

n=2

1− 2n(j−1)!

(j+k)! a_n−a(j+k)!

(j−1)!−n

.

This completes the proof of Theorem4.2.

Remark 6. Lettingk = 2, iff(z)satisfies,

(4.5)

∞

X

n=2

a_n ≥ 6 5

j(j+1)(j+2)

2 −1

X

n=2

1− 2n

j(j+ 1)(j+ 2)

(a_n−aj(j+1)(j+2)−n)

forj = 2,3,4, . . . ,then (4.6)

Z 2π 0

|I₂f(z)|^λdθ ≤ Z 2π

0

|I₂f_j(z)|^λdθ.

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Remark 7. Lettingk = 3, iff(z)satisfies,

(4.7)

∞

X

n=2

a_n≥ 6 5

j(j+1)(j+2)(j+3)

2 −1

X

n=2

1− 2n

j(j+ 1)(j + 2)(j+ 3)

×(a_n−aj(j+1)(j+2)(j+3)−n) forj = 2,3,4, . . . ,then

(4.8)

Z 2π 0

|I₃f(z)|^λdθ ≤ Z 2π

0

|I₃f_j(z)|^λdθ.

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References

[1] P.L. DUREN, Univalent Functions, Springer-Verlag, New York, 1983.

[2] J.E. LITTLEWOOD, On inequalities in the theory of functions, Proc. Lon- don 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 co- efficients, Houston J. Math., 23 (1997), 169–174.