ON A CERTAIN SUBCLASS OF STARLIKE FUNCTIONS WITH NEGATIVE COEFFICIENTS

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Subclass of Starlike Functions A. Y. Lashin vol. 10, iss. 2, art. 40, 2009

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ON A CERTAIN SUBCLASS OF STARLIKE FUNCTIONS WITH NEGATIVE COEFFICIENTS

A. Y. LASHIN

Department of Mathematics Faculty of Science

Mansoura University Mansoura, 35516, EGYPT.

EMail:aylashin@yahoo.com

Received: 11 December, 2007

Accepted: 29 May, 2009

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 30C45.

Key words: Univalent functions, Starlike functions, Integral means, Neighborhoods, Partial sums.

Abstract: We introduce the classH(α, β)of analytic functions with negative coefficients.

In this paper we give some properties of functions in the classH(α, β)and we obtain coefficient estimates, neighborhood and integral means inequalities for the functionf(z) belonging to the class H(α, β).We also establish some results concerning the partial sums for the functionf(z)belonging to the classH(α, β).

Acknowledgements: The author would like to thank the referee of the paper for his helpful suggestions.

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1. Introduction

Let Adenote the class of functions of the form

(1.1) f(z) = z+

∞

X

k=2

a_kz^k,

which are analytic in the unit discU ={z :|z|<1}.And letSdenote the subclass ofAconsisting of univalent functionsf(z)inU.

A functionf(z)inSis said to be starlike of orderαif and only if Re

zf⁰(z) f(z)

> α (z ∈U),

for someα(0≤ α <1). We denote by S^∗(α)the class of all functions inS which are starlike of orderα. It is well-known that

S^∗(α)⊆ S^∗(0)≡S^∗.

Further, a functionf(z)inSis said to be convex of orderαinU if and only if Re

1 + zf⁰⁰(z) f⁰(z)

> α (z ∈U),

for someα (0≤ α <1). We denote by K(α)the class of all functions inS which are convex of orderα.

The classes S^∗(α), andK(α) were first introduced by Robertson [8], and later were studied by Schild [10], MacGregor [4], and Pinchuk [7].

LetT denote the subclass ofS whose elements can be expressed in the form:

(1.2) f(z) =z−

∞

X

k=2

a_kz^k (a_k ≥0).

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We denote byT^∗(α)and C(α), respectively, the classes obtained by taking the intersections ofS^∗(α)andK(α)withT,

T^∗(α) =S^∗(α)∩T and C(α) = K(α)∩T.

The classesT^∗(α)andC(α)were introduced by Silverman [11].

LetH(α, β)denote the class of functionsf(z)∈Awhich satisfy the condition Re

αz²f⁰⁰(z)

f(z) + zf⁰(z) f(z)

> β

for someα≥0,0≤β <1, ^f(z)_z 6= 0 and z ∈U.

The classesH(α, β)andH(α,0)were introduced and studied by Obraddovic and Joshi [5], Padmanabhan [6], Li and Owa [2], Xu and Yang [14], Singh and Gupta [13], and others.

Further, we denote by H(α, β) the class obtained by taking intersections of the classH(α, β)withT, that is

H(α, β) = H(α, β) ∩T.

We note that

H(0, β) =T^∗(β) (Silverman [11]).

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2. Coefficient Estimates

Theorem 2.1. A functionf(z)∈T is in the classH(α, β)if and only if

(2.1)

∞

X

k=2

[(k−1)(αk+ 1) + (1−β)]a_k≤1−β.

The result is sharp.

Proof. Assume that the inequality (2.1) holds and let|z|<1.Then we have

αz²f⁰⁰(z)

f(z) +zf⁰(z) f(z) −1

=

−P∞

k=2(k−1)(αk+ 1)a_kz^k−1 1−P∞

k=2akz^k−1

≤ P∞

k=2(k−1)(αk+ 1)a_k 1−P∞

k=2a_k ≤1−β.

This shows that the values of ^αz²^f

00(z)

f(z) + ^zf_f_(z)⁰^(z) lie in the circle centered at w = 1 whose radius is1−β.Hencef(z)is in the classH(α, β).

To prove the converse, assume thatf(z)defined by (1.2) is in the class H(α, β).

Then

(2.2) Re

αz²f⁰⁰(z)

f(z) + zf⁰(z) f(z)

= Re

1−P∞

k=2[αk(k−1) +k)]akz^k−1 1−P∞

k=2a_kz^k−1

> β

forz ∈U.Choose values ofz on the real axis so that ^αz²^f

00(z)

f(z) +^zf_f_(z)⁰^(z) is real. Upon

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clearing the denominator in (2.2) and lettingz→1⁻through real values, we have β 1−

∞

X

k=2

a_k

!

≤1−

∞

X

k=2

[αk(k−1) +k]a_k,

which obviously is the required result (2.1).

Finally, we note that the assertion (2.1) of Theorem2.1is sharp, with the extremal function being

(2.3) f(z) = z− 1−β

[(k−1)(αk+ 1) + (1−β)]z^k (k ≥2).

Corollary 2.2. Letf(z)∈T be in the classH(α, β). Then we have

(2.4) a_k ≤ 1−β

[(k−1)(αk+ 1) + (1−β)] (k≥2).

Equality in (2.4) holds true for the functionf(z)given by (2.3).

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3. Some Properties of the Class H (α, β)

Theorem 3.1. Let0≤α₁ < α₂and0≤β <1.ThenH(α₂, β)⊂H(α₁, β).

Proof. It follows from Theorem2.1. That

∞

X

k=2

[(k−1)(α1k+ 1) + (1−β)]ak

<

∞

X

k=2

[(k−1)(α₂k+ 1) + (1−β)]a_k≤1−β

forf(z)∈H(α₂, β).Hencef(z)∈H(α₁, β).

Corollary 3.2. H(α, β)⊆T^∗(β).

The proof is now immediate becauseα≥0.

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4. Neighborhood Results

Following the earlier investigations of Goodman [1] and Ruscheweyh [9], we define theδ−neighborhood of function f(z)∈T by:

N_δ(f) = (

g ∈T :g(z) =z−

∞

X

k=2

b_kz^k,

∞

X

k=2

k|a_k−b_k| ≤δ )

.

In particular, for the identity function

e(z) = z, we immediately have

(4.1) N_δ(e) = (

g ∈T :g(z) =z−

∞

X

k=2

b_kz^k,

∞

X

k=2

k|b_k| ≤δ )

.

Theorem 4.1. H(α, β)⊆N_δ(e),whereδ = _(2α+2−β)^2(1−β) .

Proof. Letf(z)∈H(α, β).Then, in view of Theorem2.1, since[(k−1)(αk+ 1) + (1−β)]is an increasing function ofk(k ≥2), we have

(2α+ 2−β)

∞

X

k=2

ak ≤

∞

X

k=2

[(k−1)(αk+ 1) + (1−β)]ak ≤1−β,

which immediately yields (4.2)

∞

X

k=2

a_k≤ 1−β (2α+ 2−β).

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On the other hand, we also find from (2.1) (α+ 1)

∞

X

k=2

ka_k−β

∞

X

k=2

a_k ≤

∞

X

k=2

[(α(k−1) + 1)k−β)]a_k

=

∞

X

k=2

[(k−1)(αk+ 1) + (1−β)]a_k ≤1−β.

(4.3)

From (4.3) and (4.2), we have (α+ 1)

∞

X

k=2

ka_k ≤(1−β) +β

∞

X

k=2

a_k

≤(1−β) +β 1−β (2α+ 2−β)

≤ 2(α+ 1)(1−β) (2α+ 2−β) , that is,

(4.4)

∞

X

k=2

ka_k ≤ 2 (1−β)

(2α+ 2−β) =δ, which in view of the definition (4.1), prove Theorem4.1.

Lettingα= 0,in the above theorem, we have:

Corollary 4.2. T^∗(β)⊆N_δ(e),whereδ= ^2(1−β)_(2−β).

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5. Integral Means Inequalities

We need the following lemma.

Lemma 5.1 ([3]). Iff andgare analytic inU withf ≺g,then Z 2π

0

g(re^iθ)

δdθ ≤ Z 2π

0

f(re^iθ)

δdθ,

whereδ > 0, z =re^iθ and0< r <1.

Applying Lemma5.1, and (2.1), we prove the following theorem.

Theorem 5.2. Letδ >0.Iff(z)∈H(α, β),then forz =re^iθ,0< r <1,we have Z 2π

0

f(re^iθ)

δdθ ≤ Z 2π

0

f2(re^iθ)

δdθ,

where

(5.1) f₂(z) =z− (1−β)

(2α+ 2−β)z².

Proof. Letf(z)defined by (1.2) andf₂(z)be given by (5.1). We must show that Z 2π

0

1−

∞

X

k=2

a_kz^k−1

δ

dθ≤ Z 2π

0

1− (1−β) (2α+ 2−β)z

δ

dθ.

By Lemma5.1, it suffices to show that 1−

∞

X

k=2

a_kz^k−1 ≺1− (1−β) (2α+ 2−β)z.

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Setting

(5.2) 1−

∞

X

k=2

a_kz^k−1 = 1− (1−β)

(2α+ 2−β)w(z).

From (5.2) and (2.1), we obtain

|w(z)|=

∞

X

k=2

(2α+ 2−β) (1−β) akz^k−1

≤ |z|

∞

X

k=2

[(k−1)(αk+ 1) + (1−β)]

1−β a_k≤ |z|. This completes the proof of the theorem.

Lettingα= 0in the above theorem, we have:

Corollary 5.3. Letδ > 0.If f(z)∈T^∗(β),then forz =re^iθ,0< r <1,we have Z 2π

0

f(re^iθ)

δdθ ≤ Z 2π

0

f2(re^iθ)

δdθ,

where

f2(z) = z− (1−β) (2−β)z².

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6. Partial Sums

In this section we will examine the ratio of a function of the form (1.2) to its sequence of partial sums defined byf₁(z) = z andf_n(z) = z−Pn

k=2a_kz^k when the coefficients off are sufficiently small to satisfy the condition (2.1). We will determine sharp lower bounds forRe_f(z)

fn(z)

,Re_f

n(z) f(z)

,Re_f0(z) f_n⁰(z)

andRe_f0 n(z) f⁰(z)

. In what follows, we will use the well known result that

Re1−w(z)

1 +w(z) >0, z ∈U, if and only if

w(z) =

∞

X

k=1

c_kz^k

satisfies the inequality|w(z)| ≤ |z|. Theorem 6.1. Iff(z)∈H(α, β),then

(6.1) Re f(z)

f_n(z) ≥1− 1

c_n+1 (z ∈U, n∈N) and

(6.2) Re

f_n(z) f(z)

≥ c_n+1

1 +c_n+1 (z ∈U, n∈N), where

c_k=: [(k−1)(αk+1)+(1−β)]

1−β

.The estimates in (6.1) and (6.2) are sharp.

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Proof. We employ the same technique used by Silverman [12]. The functionf(z)∈ H(α, β),if and only ifP∞

k=2c_ka_k ≤1.It is easy to verify thatc_k+1 > c_k >1.Thus, (6.3)

n

X

k=2

ak+cn+1

∞

X

k=n+1

ak ≤

∞

X

k=2

ckak ≤1.

We may write c_n+1

f(z) fn(z)−

1− 1 cn+1

= 1−Pn

k=2a_kz^k−1−c_n+1P∞

k=n+1a_kz^k−1 1−Pn

k=2akz^k−1

= 1 +D(z) 1 +E(z).

Set 1 +D(z)

1 +E(z) = 1−w(z) 1 +w(z), so that

w(z) = E(z)−D(z) 2 +D(z) +E(z). Then

w(z) = c_n+1P∞

k=n+1a_kz^k−1 2−2Pn

k=2akz^k−1−cn+1P∞

k=n+1akz^k−1 and

|w(z)| ≤ c_n+1P∞ k=n+1a_k 2−2Pn

k=2ak−cn+1P∞ k=n+1ak

. Now|w(z)| ≤1if and only if

n

X

k=2

a_k+c_n+1

∞

X

k=n+1

a_k ≤1,

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which is true by (6.3). This readily yields the assertion (6.1) of Theorem6.1.

To see that

(6.4) f(z) = z−zⁿ⁺¹

c_n+1 gives sharp results, we observe that

f(z)

f_n(z) = 1− zⁿ c_n+1. Letting z →1⁻,we have

f(z)

f_n(z) = 1− 1 c_n+1,

which shows that the bounds in (6.1) are the best possible for eachn ∈N.Similarly, we take

(1 +c_n+1)

f_n(z)

f(z) − c_n+1 1 +cn+1

= 1−Pn

k=2a_kz^k−1+c_n+1P∞

k=n+1a_kz^k−1 1−P∞

k=2akz^k−1 := 1−w(z)

1 +w(z), where

|w(z)| ≤ (1 +c_n+1)P∞ k=n+1a_k 2−2Pn

k=2a_k+ (1−c_n+1)P∞

k=n+1a_k. Now|w(z)| ≤1if and only if

n

X

k=2

a_k+c_n+1

∞

X

k=n+1

a_k ≤1,

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which is true by (6.3). This immediately leads to the assertion (6.2) of Theorem6.1.

The estimate in (6.2) is sharp with the extremal functionf(z)given by (6.4). This completes the proof of Theorem6.1.

Corollary 6.2. Iff(z)∈T^∗(β),then

Re f(z)

fn(z) ≥ n

(n+ 1−β), (z ∈U) and

Ref_n(z)

f(z) ≥ n+ 1−β

(n+ 2−2β), (z ∈U). The result is sharp for everyn, with the extremal function

(6.5) f(z) =z− 1−β

(n+ 1−β)zⁿ⁺¹.

We now turn to the ratios involving derivatives. The proof of Theorem6.3below follows the pattern of that in Theorem6.1, and so the details may be omitted.

Theorem 6.3. Iff(z)∈H(α, β),then

(6.6) Ref⁰(z)

f_n⁰(z) ≥1− n+ 1 cn+1

(z ∈U),

and

(6.7) Re

f_n⁰(z) f⁰(z)

≥ c_n+1

n+ 1 +c_n+1 (z ∈U, n∈N).

The estimates in (6.6) and (6.7) are sharp with the extremal function given by (6.4).

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Corollary 6.4. Iff(z)∈T^∗(β),then

Re f⁰(z)

f_n⁰(z) ≥ βn

(n+ 1−β), (z ∈U), and

Ref_n⁰(z)

f⁰(z) ≥ n+ 1−β

n+ (1−β)(n+ 2), (z∈U). The result is sharp for everyn, with the extremal function given by (6.5).

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