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International Journal of Mathematics and Mathematical Sciences
Volume 2012 (2012), Article ID 915625, 10 pages
Research Article

Subclass of Multivalent Harmonic Functions with Missing Coefficients

Department of Mathematics, Faculty of Science (Damietta Branch), Mansoura University, New Damietta 34517, Egypt

Received 20 March 2012; Accepted 8 July 2012

Academic Editor: Attila Gilányi

Copyright © 2012 R. M. El-Ashwah. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


We have studied subclass of multivalent harmonic functions with missing coefficients in the open unit disc and obtained the basic properties such as coefficient characterization and distortion theorem, extreme points, and convolution.

1. Introduction

A continuous function 𝑓=𝑢+𝑖𝑣 is a complex-valued harmonic function in a simply connected complex domain 𝐷 if both 𝑢 and 𝑣 are real harmonic in 𝐷. It was shown by Clunie and Sheil-Small [1] that such harmonic function can be represented by 𝑓=+𝑔, where and 𝑔 are analytic in 𝐷. Also, a necessary and sufficient condition for 𝑓 to be locally univalent and sense preserving in 𝐷 is that |(𝑧)|>|𝑔(𝑧)| (see also, [24]).

Denote by 𝐻 the family of functions 𝑓=+𝑔, which are harmonic univalent and sense-preserving in the open-unit disc 𝑈={𝑧|𝑧|<1} with normalization 𝑓(0)=(0)=𝑓𝑧(0)1=0.

For 𝑚1,0𝛽<1,and𝛾0,let 𝑅(𝑚,𝛽,𝛾)denote the class of all multivalent harmonic functions 𝑓=+𝑔 with missing coefficients that are sense-preserving in 𝑈, and ,𝑔 are of the form (𝑧)=𝑧𝑚+𝑛=𝑚+1𝑎𝑛+1𝑧𝑛+1,𝑔(𝑧)=𝑛=𝑚𝑏𝑛+1𝑧𝑛+1(𝑚1;𝑧𝑈)(1.1) and satisfying the following condition: Re1+𝛾𝑒𝑖𝜙𝑧𝑓(𝑧)𝑧𝑓(𝑧)𝛾𝑚𝑒𝑖𝜙𝑚𝛽(𝑚1;0𝛽<1;𝛾0;𝜙real),(1.2) where 𝑧=𝜕𝜕𝜃𝑧=𝑟𝑒𝑖𝜃,𝑓𝜕(𝑧)=𝑓𝜕𝜃𝑟𝑒𝑖𝜃.(1.3)

We note that:(i)𝑅(𝑚,𝛽,1)=𝑅(𝑚,𝛽) with 𝑎𝑚+1;𝑏𝑚0 (see Jahangiri et al. [5]);(ii)𝑅(1,𝛽,𝛾)=𝐽𝐻(𝛼,𝛽,𝛾) (see Kharinar and More [6]);(iii)𝑅(1,𝛽,1)=𝐺𝐻(𝛽)with 𝑎2;𝑏10 (see Rosy et al. [7] and Ahuja and Jahangiri [2]).

Also, the subclass denoted by 𝑇(𝑚,𝛽,𝛾) consists of harmonic functions 𝑓=+𝑔, so that and 𝑔 are of the form (𝑧)=𝑧𝑚𝑛=𝑚+1𝑎𝑛+1𝑧𝑛+1,𝑔(𝑧)=𝑛=𝑚𝑏𝑛+1𝑧𝑛+1𝑎𝑛+1;𝑏𝑛+1.0;𝑚1;𝑧𝑈(1.4)

We note that:(i)𝑇(𝑚,𝛽,1)=𝑇(𝑚,𝛽) with 𝑎𝑚+1;𝑏𝑚0(see Jahangiri et al. [5]);(ii)𝑇(1,𝛽,𝛾)=𝐽𝐻(𝛼,𝛽,𝛾)(see Kharinar and More [6]);(iii)𝑇(1,𝛽,1)=𝐺𝐻(𝛽) with 𝑎2;𝑏10(see Rosy et al. [7] and Ahuja and Jahangiri [2]).

From Ahuja and Jahangiri [2] with slight modification and among other things proved, if 𝑓=+𝑔is of the form (1.1) and satisfies the coefficient condition 𝑛=𝑚1(𝑛+1)𝑚𝛽||𝑎𝑚(1𝛽)𝑛+1||+(𝑛+1)+𝑚𝛽||𝑏𝑚(1𝛽)𝑛+1||𝑎2𝑚=1;𝑎𝑚+1=𝑏𝑚,=0(1.5) then the harmonic function 𝑓is sense-preserving, harmonic multivalent with missing coefficients and starlike of order 𝛽(0𝛽<1) in 𝑈.They also proved that the condition (1.5) is also necessary for the starlikeness of function 𝑓=+𝑔of the form (1.4).

In this paper, we obtain sufficient coefficient bounds for functions in the class 𝑅(𝑚,𝛽,𝛾). These sufficient coefficient conditions are shown to be also necessary for functions in the class 𝑇(𝑚,𝛽,𝛾). Basic properties such as distortion theorem, extreme points, and convolution for the class 𝑇(𝑚,𝛽,𝛾) are also obtained.

2. Coefficient Characterization and Distortion Theorem

Unless otherwise mentioned, we assume throughout this paper that 𝑚1,0𝛽<1,𝛾0,and𝜙is real. We begin with a sufficient condition for functions in the class 𝑅(𝑚,𝛽,𝛾).

Theorem 2.1. Let 𝑓=+𝑔 be such that and 𝑔 are given by (1.1). Furthermore, let 𝑛=𝑚1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)||𝑎𝑚(1𝛽)𝑛+1||+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)||𝑏𝑚(1𝛽)𝑛+1||2,(2.1) where 𝑎𝑚=1and𝑎𝑚+1=𝑏𝑚=0.Then 𝑓 is sense-preserving, harmonic multivalent in 𝑈 and 𝑓𝑅(𝑚,𝛽,𝛾).

Proof. To prove 𝑓𝑅(𝑚,𝛽,𝛾), by definition, we only need to show that the condition (2.1) holds for 𝑓. Substituting +𝑔for 𝑓 in (1.2), it suffices to show that Re1+𝛾𝑒𝑖𝜃𝑧(𝑧)𝑧𝑔(𝑧)𝑚𝛽+𝛾𝑒𝑖𝜃(𝑧)+𝑔(𝑧)(𝑧)+𝑔(𝑧)0,(2.2) where (𝑧)=(𝜕/𝜕𝑧)(𝑧) and 𝑔(𝑧)=(𝜕/𝜕𝑧)𝑔(𝑧). Substituting for ,𝑔,, and 𝑔 in (2.2), and dividing by 𝑚(1𝛽)𝑧𝑚, we obtain Re(𝐴(𝑧)/𝐵(𝑧))0, where 𝐴(𝑧)=1+𝑛=𝑚+1(𝑛+1)1+𝛾𝑒𝑖𝜃𝑚𝛽+𝛾𝑒𝑖𝜃𝑎𝑚(1𝛽)𝑛+1𝑧𝑛𝑚+1𝑧𝑧𝑚𝑛=𝑚+1(𝑛+1)1+𝛾𝑒𝑖𝜃+𝑚𝛽+𝛾𝑒𝑖𝜃𝑏𝑚(1𝛽)𝑛+1𝑧𝑛𝑚+1,𝐵(𝑧)=1+𝑛=𝑚+1𝑎𝑛+1𝑧𝑛𝑚+1+𝑧𝑧𝑚𝑛=𝑚+1𝑏𝑛+1𝑧𝑛𝑚+1.(2.3) Using the fact that Re(𝑤)0 if and only if |1+𝑤|>|1𝑤| in 𝑈, it suffices to show that|𝐴(𝑧)+𝐵(𝑧)||𝐴(𝑧)𝐵(𝑧)|0.Substituting for 𝐴(𝑧) and 𝐵(𝑧) gives ||||||||=|||||𝐴(𝑧)+𝐵(𝑧)𝐴(𝑧)𝐵(𝑧)2+𝑛=𝑚+1(𝑛+1)1+𝛾𝑒𝑖𝜃𝑚1+2𝛽+𝛾𝑒𝑖𝜃𝑎𝑚(1𝛽)𝑛+1𝑧𝑛𝑚+1𝑧𝑧𝑚𝑛=𝑚(𝑛+1)1+𝛾𝑒𝑖𝜃+𝑚1+2𝛽+𝛾𝑒𝑖𝜃𝑏𝑚(1𝛽)𝑛+1𝑧𝑛𝑚+1||||||||||𝑛=𝑚+1(𝑛+1)1+𝛾𝑒𝑖𝜃𝑚1+𝛾𝑒𝑖𝜃𝑚𝑎(1𝛽)𝑛+1𝑧𝑛𝑚+1𝑧𝑧𝑛=𝑚(𝑛+1)1+𝛾𝑒𝑖𝜃+𝑚1+𝛾𝑒𝑖𝜃𝑚𝑏(1𝛽)𝑛+1𝑧𝑛𝑚+1|||||2𝑛=𝑚+1(𝑛+1)(1+𝛾)𝑚(2𝛽+𝛾1)||𝑎𝑚(1𝛽)𝑛+1|||𝑧|𝑛𝑚+1𝑛=𝑚(𝑛+1)(1+𝛾)+𝑚(2𝛽+𝛾1)||𝑏𝑚(1𝛽)𝑛+1|||𝑧|𝑛𝑚+1𝑛=𝑚+1(𝑛+1)(1+𝛾)𝑚(1+𝛾)||𝑎𝑚(1𝛽)𝑛+1|||𝑧|𝑛𝑚+1𝑛=𝑚(𝑛+1)(1+𝛾)+𝑚(1+𝛾)𝑚||𝑏(1𝛽)𝑛+1|||𝑧|𝑛𝑚+121𝑛=𝑚+1(𝑛+1)(1+𝛾)𝑚(𝛽+𝛾)||𝑎𝑚(1𝛽)𝑛+1||𝑛=𝑚(𝑛+1)(1+𝛾)+𝑚(𝛽+𝛾)||𝑏𝑚(1𝛽)𝑛+1||0by(2.1)(2.4)
The harmonic functions 𝑓(𝑧)=𝑧𝑚+𝑛=𝑚+1𝑚(1𝛽)𝑥(𝑛+1)(1+𝛾)𝑚(𝛽+𝛾)𝑛𝑧𝑛+1+𝑛=𝑚𝑚(1𝛽)(𝑛+1)(1+𝛾)+𝑚(𝛽+𝛾)𝑦𝑛𝑧𝑛+1,(2.5) where 𝑛=𝑚+1|𝑥𝑛|+𝑛=𝑚|𝑦𝑛|=1, show that the coefficient boundary given by (2.1) is sharp. The functions of the form (2.5) are in the class 𝑅(𝑚,𝛽,𝛾) because 𝑛=𝑚+1(1+𝛾)(𝑛+1)𝑚(𝛽+𝛾)||𝑎𝑚(1𝛽)𝑛+1||+𝑛=𝑚(1+𝛾)(𝑛+1)+𝑚(𝛽+𝛾)||𝑏𝑚(1𝛽)𝑛+1||=𝑛=𝑚+1||𝑥𝑛||+𝑛=𝑚+1||𝑦𝑛||=1.(2.6) This completes the proof of Theorem 2.1.

In the following theorem, it is shown that the condition (2.1) is also necessary for functions 𝑓=+𝑔, where and 𝑔 are of the form (1.4).

Theorem 2.2. Let 𝑓=+𝑔 be such that and 𝑔 are given by (1.4). Then 𝑓𝑇(𝑚,𝛽,𝛾) if and only if 𝑛=𝑚1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑎𝑚(1𝛽)𝑛+1+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)𝑛+12.(2.7)

Proof. Since 𝑅(𝑚,𝛽,𝛾)𝑇(𝑚,𝛽,𝛾), we only need to prove the “only if” part of the theorem. To this end, for functions 𝑓 of the form (1.4), we notice that the condition Re{(1+𝛾𝑒𝑖𝜃)(𝑧𝑓(𝑧))/(𝑧𝑓(𝑧))𝛾𝑚𝑒𝑖𝜃}𝑚𝛽 is equivalent to Re1+𝛾𝑒𝑖𝜃𝑧(𝑧)𝑧𝑔(𝑧)𝑚𝛽+𝛾𝑒𝑖𝜃(𝑧)+𝑔(𝑧)(𝑧)+𝑔(𝑧)>0,(2.8) which implies that 𝑚Re1+𝛾𝑒𝑖𝜃𝑚𝛽+𝛾𝑒𝑖𝜃𝑧𝑚𝑛=𝑚+11+𝛾𝑒𝑖𝜃(𝑛+1)𝑚𝛽+𝛾𝑒𝑖𝜃𝑎𝑛+1𝑧𝑛+1𝑧𝑚𝑛=𝑚+1𝑎𝑛+1𝑧𝑛+1+𝑛=𝑚𝑏𝑛+1𝑧𝑛+1𝑛=𝑚1+𝛾𝑒𝑖𝜃(𝑛+1)+𝑚𝛽+𝛾𝑒𝑖𝜃𝑏𝑛+1𝑧𝑛+1𝑧𝑚𝑛=𝑚+1𝑎𝑛+1𝑧𝑛+1+𝑛=𝑚𝑏𝑛+1𝑧𝑛+1𝑚=Re(1𝛽)𝑛=𝑚+11+𝛾𝑒𝑖𝜃(𝑛+1)𝑚𝛽+𝛾𝑒𝑖𝜃𝑎𝑛+1𝑧𝑛𝑚+11𝑛=𝑚+1𝑎𝑛+1𝑧𝑛𝑚+1+𝑛=𝑚𝑏𝑛+1𝑧𝑛𝑚+1𝑛=𝑚1+𝛾𝑒𝑖𝜃(𝑛+1)+𝑚𝛽+𝛾𝑒𝑖𝜃𝑏𝑛+1𝑧𝑛𝑚+11𝑛=𝑚+1𝑎𝑛+1𝑧𝑛𝑚+1+𝑛=𝑚𝑏𝑛+1𝑧𝑛𝑚+1>0.(2.9) Since Re(𝑒𝑖𝜃)|𝑒𝑖𝜃|=1, the required condition is that (2.9) is equivalent to 1𝑛=𝑚+1([](1+𝛾)(𝑛+1)𝑚(𝛽+𝛾)/𝑚(1𝛽))𝑎𝑛+1𝑟𝑛𝑚+11𝑛=𝑚+1𝑎𝑛+1𝑟𝑛𝑚+1+𝑛=𝑚𝑏𝑛+1𝑟𝑛𝑚+1𝑛=𝑚([](1+𝛾)(𝑛+1)+𝑚(𝛽+𝛾)/𝑚(1𝛽))𝑏𝑛+1𝑟𝑛𝑚+11𝑛=𝑚+1𝑎𝑛+1𝑟𝑛𝑚+1+𝑛=𝑚𝑏𝑛+1𝑟𝑛𝑚+10.(2.10) If the condition (2.7) does not hold, then the numerator in (2.10) is negative for 𝑧=𝑟 sufficiently close to 1. Hence there exists 𝑧0=𝑟0 in (0,1) for which the quotient in (2.10) is negative. This contradicts the required condition for 𝑓𝑇(𝑚,𝛽,𝛾), and so the proof of Theorem 2.2 is completed.

Corollary 2.3. The functions in the class 𝑇(𝑚,𝛽,𝛾)are starlike of order (𝛾+𝛽)/(1+𝛾).

Proof. The proof follows from (1.5), by putting (2.7) in the form 𝑛=𝑚1(𝑛+1)𝑚((𝛾+𝛽)/(1+𝛾))𝑎𝑚(1((𝛾+𝛽)/(1+𝛾)))𝑛+1+(𝑛+1)+𝑚((𝛾+𝛽)/(1+𝛾))𝑏𝑚(1((𝛾+𝛽)/(1+𝛾)))𝑛+12.(2.11)

Theorem 2.4. Let 𝑓𝑇(𝑚,𝛽,𝛾). Then for |𝑧|=𝑟<1, we have ||||𝑓(𝑧)1+𝑏𝑚+1𝑟𝑟𝑚+𝑚(1𝛽)𝑚(1𝛽)+2(1+𝛾)𝑚(1+2𝛾+𝛽)+(1+𝛾)𝑏𝑚(1𝛽)+2(1+𝛾)𝑚+1𝑟𝑚+2,||𝑓𝑚||(𝑧)1𝑏𝑚+1𝑟𝑟𝑚𝑚(1𝛽)𝑚(1𝛽)+2(1+𝛾)𝑚(1+2𝛾+𝛽)+(1+𝛾)𝑏𝑚(1𝛽)+2(1+𝛾)𝑚+1𝑟𝑚+2.(2.12)

Proof. We prove the left-hand-side inequality for |𝑓|.The proof for the right-hand-side inequality can be done by using similar arguments.
Let 𝑓𝑇(𝑚,𝛽,𝛾), then we have ||||=|||||𝑧𝑓(𝑧)𝑚𝑛=𝑚+1𝑎𝑛+1𝑧𝑛+1+𝑛=𝑚𝑏𝑛+1𝑧𝑛+1|||||𝑟𝑚𝑏𝑚+1𝑟𝑚+1𝑛=𝑚+1𝑎𝑛+1+𝑏𝑛+1𝑟𝑚+2𝑟𝑚𝑏𝑚+1𝑟𝑚+1𝑚(1𝛽)(1+𝛾)(𝑚+2)𝑚(𝛾+𝛽)𝑛=𝑚+1(1+𝛾)(𝑚+2)𝑚(𝛾+𝛽)𝑎𝑚(1𝛽)𝑛+1+𝑏𝑛+1𝑟𝑛+1𝑟𝑚𝑏𝑚+1𝑟𝑚+1𝑚(1𝛽)(1+𝛾)(𝑚+2)𝑚(𝛾+𝛽)𝑛=𝑚+1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑎𝑚(1𝛽)𝑛+1+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)𝑛+1𝑟𝑛+11𝑏𝑚+1𝑟𝑟𝑚𝑚(1𝛽)(1+𝛾)(𝑚+2)𝑚(𝛾+𝛽)1(1+𝛾)(𝑚+1)+𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)𝑚+1𝑟𝑚+21𝑏𝑚+1𝑟𝑟𝑚𝑚(1𝛽)𝑚(1𝛽)+2(1+𝛾)𝑚(1+2𝛾+𝛽)+(1+𝛾)𝑏𝑚(1𝛽)+2(1+𝛾)𝑚+1𝑟𝑚+2.(2.13) This completes the proof of Theorem 2.4.

The following covering result follows from the left-side inequality in Theorem 2.4.

Corollary 2.5. Let 𝑓𝑇(𝑚,𝛽,𝛾),then the set 𝑤|𝑤|<2(1+𝛾)𝑚(1𝛽)+2(1+𝛾)(1+𝛾)2𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)+2(1+𝛾)𝑚+1(2.14) is included in 𝑓(𝑈).

3. Extreme Points

Our next theorem is on the extreme points of convex hulls of the class 𝑇(𝑚,𝛽,𝛾), denoted by clco𝑇(𝑚,𝛽,𝛾).

Theorem 3.1. Let 𝑓=+𝑔 be such that and 𝑔 are given by (1.4). Then 𝑓clco𝑇(𝑚,𝛽,𝛾) if and only if 𝑓 can be expressed as 𝑓(𝑧)=𝑛=𝑚𝑋𝑛+1𝑛+1(𝑧)+𝑌𝑛+1𝑔𝑛+1(,𝑧)(3.1) where 𝑚(𝑧)=𝑧𝑚,𝑛+1(𝑧)=𝑧𝑚𝑚(1𝛽)(𝑧1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑛+1𝑔(𝑛=𝑚+1,𝑚+2,...),𝑛+1(𝑧)=𝑧𝑚+𝑚(1𝛽)(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑧𝑛+1𝑋(𝑛=𝑚,𝑚+1,𝑚+2,...),𝑛+10,𝑌𝑛+10,𝑛=𝑚𝑋𝑛+1+𝑌𝑛+1=1.(3.2) In particular, the extreme points of the class 𝑇(𝑚,𝛽,𝛾) are {𝑛+1} and {𝑔𝑛+1},respectively.

Proof. For functions 𝑓(𝑧) of the form (3.1), we have 𝑓(𝑧)=𝑛=𝑚𝑋𝑛+1+𝑌𝑛+1𝑧𝑚𝑛=𝑚𝑚(1𝛽)𝑋(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑛+1𝑧𝑛+1+𝑛=𝑚𝑚(1𝛽)𝑌(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑛+1𝑧𝑛+1.(3.3) Then 𝑛=𝑚+1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑚(1𝛽)𝑚(1𝛽)𝑋(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑛+1+𝑛=𝑚(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑚(1𝛽)𝑚(1𝛽)𝑌(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑛+1=𝑛=𝑚+1𝑋𝑛+1+𝑛=𝑚𝑌𝑛+1=1𝑋𝑚1,(3.4) and so 𝑓(𝑧)clco𝑇(𝑚,𝛽,𝛾).Conversely, suppose that 𝑓(𝑧)clco𝑇(𝑚,𝛽,𝛾). Set 𝑋𝑛+1=(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑎𝑚(1𝛽)𝑛+1(𝑌𝑛=𝑚+1,...),𝑛+1=(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)𝑛+1(𝑛=𝑚,𝑚+1,...),(3.5) then note that by Theorem 2.2,0𝑋𝑛+11(𝑛=𝑚+1,...) and 0𝑌𝑛+11(𝑛=𝑚,𝑚+1,...).
Consequently, we obtain 𝑓(𝑧)=𝑛=𝑚𝑋𝑛+1𝑛+1(𝑧)+𝑌𝑛+1𝑔𝑛+1(.𝑧)(3.6) Using Theorem 2.2 it is easily seen that the class 𝑇(𝑚,𝛽,𝛾)is convex and closed, and so clco𝑇(𝑚,𝛽,𝛾)=𝑇(𝑚,𝛽,𝛾).

4. Convolution Result

For harmonic functions of the form 𝑓(𝑧)=𝑧𝑚𝑛=𝑚+1𝑎𝑛+1𝑧𝑛+1+𝑛=𝑚𝑏𝑛+1𝑧𝑛+1,(4.1)𝐺(𝑧)=𝑧𝑚𝑛=𝑚+1𝐴𝑛+1𝑧𝑛+1+𝑛=𝑚𝐵𝑛+1𝑧𝑛+1,(4.2) we define the convolution of two harmonic functions 𝑓 and 𝐺 as (𝑓𝐺)(𝑧)=𝑓(𝑧)𝐺(𝑧)=𝑧𝑚𝑛=𝑚+1𝑎𝑛+1𝐴𝑛+1𝑧𝑛+1+𝑛=𝑚𝑏𝑛+1𝐵𝑛+1𝑧𝑛+1.(4.3) Using this definition, we show that the class 𝑇(𝑚,𝛽,𝛾) is closed under convolution.

Theorem 4.1. For 0𝛽<1, let 𝑓(𝑧)𝑇(𝑚,𝛽,𝛾) and 𝐺(𝑧)𝑇(𝑚,𝛽,𝛾). Then 𝑓(𝑧)𝐺(𝑧)𝑇(𝑚,𝛽,𝛾).

Proof. Let the functions 𝑓(𝑧) defined by (4.1) be in the class𝑇(𝑚,𝛽,𝛾), and let the functions 𝐺(𝑧) defined by (4.2) be in the class 𝑇(𝑚,𝛽,𝛾). Obviously, the coefficients of 𝑓 and 𝐺 must satisfy a condition similar to the inequality (2.7). So for the coefficients of 𝑓𝐺we can write 𝑛=𝑚1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑎𝑚(1𝛽)𝑛+1𝐴𝑛+1+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)𝑛+1𝐵𝑛+1𝑛=𝑚1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑚𝑎(1𝛽)𝑛+1+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑚𝑏(1𝛽)𝑛+1,(4.4) where the right hand side of this inequality is bounded by 2 because 𝑓𝑇(𝑚,𝛽,𝛾). Then, 𝑓(𝑧)𝐺(𝑧)𝑇(𝑚,𝛽,𝛾).

Finally, we show that 𝑇(𝑚,𝛽,𝛾) is closed under convex combinations of its members.

Theorem 4.2. The class 𝑇(𝑚,𝛽,𝛾)is closed under convex combination.

Proof. For 𝑖=1,2,3,.... let 𝑓𝑖𝑇(𝑚,𝛽,𝛾),where the functions 𝑓𝑖 are given by 𝑓𝑖(𝑧)=𝑧𝑚𝑛=𝑚+1𝑎𝑛+1,𝑖𝑧𝑛+1+𝑛=𝑚𝑏𝑛+1,𝑖𝑧𝑛+1𝑎𝑛+1,𝑖;𝑏𝑛+1,𝑖.0;𝑚1(4.5) For 𝑖=1𝑡𝑖=1;0𝑡𝑖1,the convex combination of 𝑓𝑖 may be written as 𝑖=1𝑡𝑖𝑓𝑖(𝑧)=𝑧𝑚𝑛=𝑚+1𝑖=1𝑡𝑖𝑎𝑛+1,𝑖𝑧𝑛+1+𝑛=𝑚𝑖=1𝑡𝑖𝑏𝑛+1,𝑖𝑧𝑛+1(4.6) Then by (2.7), we have 𝑛=𝑚1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑚(1𝛽)𝑖=1𝑡𝑖𝑎𝑛+1,𝑖+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑚(1𝛽)𝑖=1𝑡𝑖𝑏𝑛+1,𝑖=𝑖=1𝑡𝑖𝑛=𝑚1(1+𝛾)(𝑛+1)𝑚(𝛾+𝛽)𝑎𝑚(1𝛽)𝑛+1,𝑖+(1+𝛾)(𝑛+1)+𝑚(𝛾+𝛽)𝑏𝑚(1𝛽)𝑛+1,𝑖2𝑖=1𝑡𝑖=2.(4.7) This is the condition required by (2.7), and so 𝑖=1𝑡𝑖𝑓𝑖(𝑧)𝑇(𝑚,𝛽,𝛾).This completes the proof of Theorem 4.2.

Remark 4.3. Our results for 𝑚=1 correct the results obtained by Kharinar and More [6].


The author thanks the referees for their valuable suggestions which led to the improvement of this study.


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