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Partial Sums K. K. Dixit and Saurabh Porwal

vol. 10, iss. 4, art. 101, 2009

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A CONVOLUTION APPROACH ON PARTIAL SUMS OF CERTAIN ANALYTIC AND UNIVALENT

FUNCTIONS

K. K. DIXIT AND SAURABH PORWAL

Department of Mathematics Janta College, Bakewar, Etawah (U.P.) India-206124

EMail:kk.dixit@rediffmail.com saurabh.840@rediffmail.com

Received: 03 May, 2009

Accepted: 28 September, 2009

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

Key words: Analytic functions, Univalent functions, Convolution, Partial Sums.

Abstract: In this paper, we determine sharp lower bounds for Ren

f(z)∗ψ(z) fn(z)∗ψ(z)

o and

Re

nfn(z)∗ψ(z) f(z)∗ψ(z)

o

. We extend the results of ([1] – [5]) and correct the condi- tions for the results of Frasin [2, Theorem 2.7], [1, Theorem 2], Rosy et al. [4, Theorems 4.2 and 4.3], as well as Raina and Bansal [3, Theorem 6.2].

Acknowledgements: The authors are thankful to the referee for his valuable comments and sugges- tions.

The present investigation was supported by the University grant commission un- der grant No. F- 11-12/2006(SA-I).

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Partial Sums K. K. Dixit and Saurabh Porwal

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Contents

1 Introduction 3

2 Main Results 6

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1. Introduction

LetAdenote the class of functionsf of the form

(1.1) f(z) = z+

X

k=2

akzk,

which are analytic in the open unit discU = {z :|z|<1}. Further, byS we shall denote the class of all functions inAwhich are univalent inU. A function f(z)in Sis said to be starlike of orderα(0≤α <1), denoted byS(α), if it satisfies

Re

zf0(z) f(z)

> α (z ∈U),

and is said to be convex of orderα(0≤α <1), denoted byK(α), if it satisfies Re

1 + zf00(z) f0(z)

> α (z ∈U).

Let T(α) and C(α) be subclasses of S(α) and K(α), respectively, whose functions are of the form

(1.2) f(z) = z−

X

k=2

akzk, ak ≥0.

A sufficient condition for a function of the form (1.1) to be inS(α)is that (1.3)

X

k=2

(k−α)|ak| ≤1−α

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and to be inK(α)is that (1.4)

X

k=2

k(k−α)|ak| ≤1−α.

For functions of the form (1.2), Silverman [6] proved that the above sufficient conditions are also necessary.

Letφ(z)∈Sbe a fixed function of the form

(1.5) φ(z) =z+

X

k=2

ckzk, (ck≥c2 >0, k ≥2).

Very recently, Frasin [2] defined the classHφ(ck, δ)consisting of functionsf(z), of the form (1.1) which satisfy the inequality

(1.6)

X

k=2

ck|ak| ≤δ,

whereδ >0.

He shows that for suitable choices of ck and δ, Hφ(ck, δ) reduces to various known subclasses ofS studied by various authors (for a detailed study, see [2] and the references therein).

In the present paper, we determine sharp lower bounds forRenf(z)∗ψ(z)

fn(z)∗ψ(z)

o and Ren

fn(z)∗ψ(z) f(z)∗ψ(z)

o

, where

fn(z) =z+

n

X

k=2

akzk

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is a sequence of partial sums of a function f(z) =z+

X

k=2

akzk

belonging to the classHφ(ck, δ)and ψ(z) =z+

X

k=2

λkzk, (λk ≥0)

is analytic in open unit discU and the operator “*” stands for the Hadamard product or convolution of two power series, which is defined for two functions f, g ∈ A, wheref(z)andg(z)are of the form

f(z) =z+

X

k=2

akzk and g(z) =z+

X

k=2

bkzk

as

(f∗g) (z) =f(z)∗g(z) = z+

X

k=2

akbkzk.

In this paper, we extend the results of Silverman [5], Frasin ([1], [2]) Rosy et al. [4]

as well as Raina and Bansal [3] and we point out that some conditions on the results of Frasin ([2, Theorem 2.7], [1, Theorem 2]), Rosy et al. ([4, Theorem 4.2, 4.3]), Raina and Bansal ([3, Theorem 6.2]) are incorrect and we correct them. It is seen that this study not only gives a particular case of the results ([1] – [5]) but also gives rise to several new results.

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2. Main Results

Theorem 2.1. Iff ∈Hφ(ck, δ)andψ(z) =z+P

k=2λkzk,λk ≥0, then

(2.1) Re

f(z)∗ψ(z) fn(z)∗ψ(z)

≥ cn+1−λn+1δ

cn+1 (z ∈U) and

(2.2) Re

fn(z)∗ψ(z) f(z)∗ψ(z)

≥ cn+1

cn+1n+1δ (z∈U), where

ck

( λkδ if k = 2,3, . . . , n,

λkcn+1

λn+1 if k =n+ 1, n+ 2, . . . The results (2.1) and (2.2) are sharp with the function given by

(2.3) f(z) = z+ δ

cn+1

zn+1,

where0< δ ≤ λcn+1

n+1.

Proof. Define the functionω(z)by 1 +ω(z)

1−ω(z) = cn+1n+1

f(z)∗ψ(z) fn(z)∗ψ(z)−

cn+1−δλn+1 cn+1

(2.4)

= 1 +Pn

k=2λkakzk−1+ cn+1

n+1

P

k=n+1λkakzk−1 1 +Pn

k=2λkakzk−1 .

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It suffices to show that|ω(z)| ≤1. Now, from (2.4) we can write ω(z) =

cn+1

n+1

P

k=n+1λkakzk−1 2 + 2Pn

k=2λkakzk−1+cn+1

n+1

P

k=n+1λkakzk−1. Hence we obtain

|ω(z)| ≤

cn+1

n+1

P

k=n+1λk|ak| 2−2Pn

k=2λk|ak| − cn+1

n+1

P

k=n+1λk|ak|. Now|ω(z)| ≤1if

2 cn+1

n+1

X

k=n+1

λk|ak| ≤2−2

n

X

k=2

λk|ak|

or, equivalently, (2.5)

n

X

k=2

λk|ak|+ cn+1n+1

X

k=n+1

λk|ak| ≤1.

It suffices to show that the L.H.S. of (2.5) is bounded above byP k=2

ck

δ |ak|, which is equivalent to

(2.6)

n

X

k=2

ck−δλk δ

|ak|+

X

k=n+1

λn+1ck−cn+1λk λn+1δ

|ak| ≥0.

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To see that the function given by (2.3) gives a sharp result we observe that for z =reiπ/n

f(z)∗ψ(z)

fn(z)∗ψ(z) = 1 + δ cn+1

λn+1zn→1− δ cn+1

λn+1

= cn+1−δλn+1 cn+1 whenr →1.

To prove the second part of this theorem, we write 1 +ω(z)

1−ω(z) = cn+1n+1δ λn+1δ

fn(z)∗ψ(z) f(z)∗ψ(z) −

cn+1

cn+1n+1δ

=

1 +Pn

k=2λkakzk−1λcn+1

n+1δ

P

k=n+1λkakzk−1 1 +P

k=2λkakzk−1 , where

|ω(z)| ≤

cn+1n+1δ λn+1δ

P

k=n+1λk|ak| 2−2Pn

k=2λk|ak| − cn+1λ−λn+1δ

n+1δ

P

k=n+1λk|ak| ≤1.

This last inequality is equivalent to

n

X

k=2

λk|ak|+ cn+1

n+1

X

k=n+1

λk|ak| ≤1.

Making use of (1.6), we get (2.6). Finally, equality holds in (2.2) for the function f(z)given by (2.3).

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Taking ψ(z) = 1−zz in Theorem 2.1, we obtain the following result given by Frasin in [2].

Corollary 2.2. Iff ∈Hφ(ck, δ), then

(2.7) Re

f(z) fn(z)

≥ cn+1−δ

cn+1 (z ∈U) and

(2.8) Re

fn(z) f(z)

≥ cn+1

cn+1+δ (z ∈U), where

ck

( δ if k = 2,3, . . . , n, cn+1 if k =n+ 1, n+ 2, . . .

The results (2.7) and (2.8) are sharp with the function given by (2.3).

If we putψ(z) = (1−z)z 2 in Theorem2.1, we obtain:

Corollary 2.3. Iff ∈Hφ(ck, δ), then

(2.9) Re f0(z)

fn0 (z) ≥ cn+1−(n+ 1)δ cn+1

(z ∈U) and

(2.10) Refn0 (z)

f0(z) ≥ cn+1

cn+1+ (n+ 1)δ (z ∈U), where

(2.11) ck

( kδ if k = 2,3, . . . , n,

kcn+1

n+1 if k =n+ 1, n+ 2, . . .

The results (2.9) and (2.10) are sharp with the function given by (2.3).

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Remark 1. Frasin has shown in Theorem 2.7 of [2] that forf ∈Hφ(ck, δ), inequali- ties (2.9) and (2.10) hold with the condition

(2.12) ck

( kδ if k = 2,3, . . . , n, kδ 1 + cn+1n+1

if k =n+ 1, n+ 2, . . .

However, it can be easily seen that the condition (2.12) fork =n+ 1gives cn+1 ≥(n+ 1)δ

1 + cn+1 (n+ 1)δ

or, equivalentlyδ ≤ 0,which contradicts the initial assumptionδ > 0.So Theorem 2.7 of [2] does not seem suitable with the condition (2.12), but our condition (2.11) remedies this problem.

Taking ψ(z) = 1−zz , ck = [(1+β)k−(α+β)]

1−α

k+λ−1 k

, where λ ≥ 0, β ≥ 0,−1 ≤ α <1andδ= 1in Theorem2.1, we obtain the following result given by Rosy et al.

in [4].

Corollary 2.4. Iff is of the form (1.1) and satisfies the conditionP

k=2ck|ak| ≤1, whereck = [(1+β)k−(α+β)]

1−α

k+λ−1 k

, λ≥0, β ≥0,−1≤α <1, then

(2.13) Re

f(z) fn(z)

≥ cn+1−1 cn+1

(z ∈U) and

(2.14) Re

fn(z) f(z)

≥ cn+1

cn+1+ 1 (z ∈U). The results (2.13) and (2.14) are sharp with the function given by

(2.15) f(z) = z+ 1

cn+1zn+1.

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Taking

ψ(z) = z

(1−z)2, ck = [(1 +β)k−(α+β)]

1−α

k+λ−1 k

,

whereλ≥0, β ≥0,−1≤α <1andδ= 1in Theorem2.1, we obtain Corollary 2.5. Iff is of the form (1.1) and satisfies the condition

X

k=2

ck|ak| ≤1,

where

ck = [(1 +β)k−(α+β)]

1−α

k+λ−1 k

, (λ≥0, β ≥0, −1≤α <1),

then

(2.16) Re

f0(z) fn0 (z)

≥ cn+1−(n+ 1)

cn+1 (z ∈U)

and

(2.17) Re

fn0 (z) f0(z)

≥ cn+1

cn+1+ (n+ 1) (z ∈U), where

(2.18) ck

( k if k = 2,3, . . . , n,

kcn+1

n+1 if k =n+ 1, n+ 2, . . .

The results (2.16) and (2.17) are sharp with the function given by (2.15).

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Remark 2. Rosy et al. has obtained inequalities (2.16) & (2.17) in Theorem 4.2 &

4.3 of [4] without any restriction on ck. However, when we critically observe the proof of Theorem 4.2 we find that inequality (4.16) of [4, Theorem 4.2]

n

X

k=2

(ck−k)|ak|+

X

k=n+1

ck− cn+1k n+ 1

|ak| ≥0

cannot hold if condition (2.18) does not occur. So Theorems 4.2 & 4.3 of [4] are not proper and proper results are mentioned in Corollary2.5.

Taking ψ(z) = 1−zz , ck = λk −αµk, δ = 1−α, where 0 ≤ α < 1, λk ≥ 0, µk ≥0,andλk ≥ µk(k ≥2)in Theorem2.1, we obtain the following result given by Frasin in [1].

Corollary 2.6. Iff is of the form (1.1) and satisfies the condition

X

k=2

k−αµk)|ak| ≤1−α,

then

(2.19) Re

f(z) fn(z)

≥ λn+1−αµn+1−1 +α

λn+1−αµn+1 (z ∈U) and

(2.20) Re

fn(z) f(z)

≥ λn+1−αµn+1

λn+1−αµn+1+ 1−α (z ∈U), where

λk−αµk

( 1−α if k = 2,3, . . . , n, λn+1−αµn+1 if k =n+ 1, n+ 2, . . .

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The results (2.19) and (2.20) are sharp with the function given by

(2.21) f(z) =z+ 1−α

λn+1−αµn+1zn+1.

Taking ψ(z) = (1−z)z 2, ck = λk−αµk, δ = 1−αwhere0 ≤ α < 1, λk ≥ 0, µk ≥0,andλk≥µk(k ≥2)in Theorem2.1, we obtain:

Corollary 2.7. Iff is of the form (1.1) and satisfies the condition

X

k=2

k−αµk)|ak| ≤1−α,

then

(2.22) Re

f0(z) fn0 (z)

≥ λn+1−αµn+1−(n+ 1) (1−α)

λn+1−αµn+1 (z ∈U) and

(2.23) Re

fn0 (z) f0(z)

≥ λn+1−αµn+1

λn+1−αµn+1+ (n+ 1) (1−α) (z ∈U), where

(2.24) λk−αµk

( k(1−α) if k = 2,3, . . . , n,

k(λn+1−αµn+1)

n+1 if k =n+ 1, n+ 2, . . . The results (2.22) and (2.23) are sharp with the function given by (2.21).

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Remark 3. Frasin has obtained inequalities (2.22) & (2.23) in Theorem 2 of [1] under the condition

(2.25) λk+1−αµk+1

( k(1−α) if k = 2,3, . . . , n, k(1−α) + k(λn+1n+1−αµn+1) if k =n+ 1, n+ 2, . . . However, when we critically observe the proof of Theorem 2 of [1], we find that the last inequality of this theorem

(2.26)

n

X

k=2

λk−αµk 1−α −k

|ak|

+

X

k=n+1

λk−αµk

1−α −

1 + λn+1−αµn+1

(n+ 1) (1−α)

k

|ak| ≥0 cannot hold for the function given by (2.21) for supporting the sharpness of the results (2.22) & (2.23). So condition 2.25 of Theorem 2 in [1] is incorrect and the corrected results are mentioned in Corollary2.7.

Taking

ψ(z) = z

1−z, ck= {(1 +β)k−(α+β)}µk 1−α

andδ = 1, where−1≤ α <1,β ≥0, µk ≥0 (∀k∈N\ {1})in Theorem2.1, we obtain the following result given by Raina and Bansal in [3].

Corollary 2.8. Iff is of the form (1.2) and satisfies the conditionP

k=2ck|ak| ≤1, where

ck = {(1 +β)k−(α+β)}µk

1−α ,

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andkik=2 is a nondecreasing sequence such that µ2 ≥ 1−α

2 +β−α

0< 1−α

2 +β−α <1, −1≤α <1, β ≥0

,

then

(2.27) Re

f(z) fn(z)

≥ cn+1−1

cn+1 (z ∈U) and

(2.28) Re

fn(z) f(z)

≥ cn+1

cn+1+ 1 (z ∈U). The results (2.27) and (2.28) are sharp with the function given by

(2.29) f(z) = z− 1

cn+1zn+1. Taking ψ(z) = (1−z)z 2, ck = {(1+β)k−(α+β)}µk

1−α and δ = 1, where −1 ≤ α < 1, β ≥0, µk≥0 (∀k ∈N\ {1})in Theorem2.1, we obtain the following result given by Raina and Bansal in [3].

Corollary 2.9. Iff is of the form (1.2) and satisfies the condition

X

k=2

ck|ak| ≤1,

where

ck = {(1 +β)k−(α+β)}µk

1−α ,

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andkik=2 is a nondecreasing sequence such that µ2 ≥ 2 (1−α)

2 +β−α

0< 1−α

2 +β−α <1, −1≤α <1, β ≥0

.

Then

(2.30) Re

f0(z) fn0 (z)

≥ cn+1−(n+ 1) cn+1

(z ∈U)

and

(2.31) Re

fn0 (z) f0(z)

≥ cn+1

cn+1+ (n+ 1) (z ∈U), where

(2.32) ck

( k if k = 2,3, . . . , n,

kcn+1

n+1 if k =n+ 1, n+ 2, . . .

The results (2.30) and (2.31) are sharp with the function given by (2.29).

Remark 4. Raina and Bansal [3] have obtained inequalities (2.30) & (2.31) in Theo- rem 6.2 of [3] without any restriction onck. However, we easily see that condition (2.32) is must.

Remark 5. Takingψ(z) = 1−zz , ck= (k−α),ck =k(k−α),δ = 1−α,0≤α <1 in Theorem2.1, we obtain Theorems 1-3 given by Silverman in [5].

Remark 6. Taking ψ(z) = (1−z)z 2, ck = (k−α), ck = k(k−α), δ = 1− α, 0≤α <1in Theorem2.1, we obtain Theorems 4-5 given by Silverman in [5].

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References

[1] B.A. FRASIN, Partial sums of certain analytic and univalent functions, Acta Math. Acad. Paed. Nyir., 21 (2005), 135–145.

[2] B.A. FRASIN, Generalization of partial sums of certain analytic and univalent functions, Appl. Math. Lett., 21(7) (2008), 735–741.

[3] R.K. RAINA AND D. BANSAL, Some properties of a new class of analytic functions defined in terms of a Hadamard product, J. Inequal. Pure Appl. Math., 9(1) (2008), Art. 22. [ONLINE: http://jipam.vu.edu.au/article.

php?sid=957]

[4] T. ROSY, K.G. SUBRAMANIANAND G. MURUGUSUNDARAMOORTHY, Neighbourhoods and partial sums of starlike functions based on Ruscheweyh derivatives, J. Inequal. Pure Appl. Math., 4(4) (2003), Art. 64. [ONLINE:

http://jipam.vu.edu.au/article.php?sid=305]

[5] H. SILVERMAN, Partial sums of starlike and convex functions, J. Math. Anal.

Appl., 209 (1997), 221–227.

[6] H. SILVERMAN, Univalent functions with negative coefficients, Proc. Amer.

Math. Soc., 51 (1975), 109–116.

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