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Neighborhoods and partial sums of certain subclass of starlike functions

Abstract

The main purpose of the present paper is to derive the neighborhoods and partial sums of a certain subclass of starlike functions.

MSC: 30C45.

1 Introduction

Let A m denote the class of functions f of the form

f(z)=z+ k = m + 1 a k z k ( m N : = { 1 , 2 , 3 , } ) ,
(1.1)

which are analytic in the open unit disk

U:= { z : z C  and  | z | < 1 } .

A function f A m is said to be in the class S m (β) of starlike functions of order β if it satisfies the inequality

( z f ( z ) f ( z ) ) >β(zU;0β<1).
(1.2)

Assuming that α0, 0β<1 and f A m , we say that a function f H m (α,β) if it satisfies the condition

( z f ( z ) f ( z ) + α z 2 f ( z ) f ( z ) ) >αβ ( β + m 2 1 ) +β m α 2 (zU).
(1.3)

For convenience, throughout this paper, we write

γ m :=αβ ( β + m 2 1 ) +β m α 2 .
(1.4)

Recently, Ravichandran et al. [1] proved that H m (α,β) S m (β). Subsequently, Liu et al. [2] derived various properties and characteristics such as inclusion relationships, Hadamard products, coefficient estimates, distortion theorems and cover theorems for the class H m (α,β) and a subclass of H m (α,β) with negative coefficients. Furthermore, Singh et al. [3] generalized the class H m (α,β) and found several sufficient conditions for starlikeness. In the present paper, we aim at proving the neighborhoods and partial sums of the class H m (α,β).

2 Main results

Following the earlier works (based upon the familiar concept of a neighborhood of analytic functions) by Goodman [4] and Ruscheweyh [5], and (more recently) by Altintaş et al. [69], Cǎtaş [10], Frasin [11], Keerthi et al. [12] and Srivastava et al. [13], we begin by introducing here the δ-neighborhood of a function f A m of the form (1.1) by means of the definition

N δ ( f ) : = { g A m : g ( z ) = z + k = m + 1 b k z k  and k = m + 1 k ( 1 + k α α ) γ m 1 γ m | a k b k | δ ( δ , α 0 ; 0 β < 1 ; γ m < 1 ) } .
(2.1)

By making use of the definition (2.1), we now derive the following result.

Theorem 1 If f A m satisfies the condition

f ( z ) + ε z 1 + ε H m (α,β) ( ε C ; | ε | < δ ; δ > 0 ) ,
(2.2)

then

N δ (f) H m (α,β).
(2.3)

Proof By noting that the condition (1.3) can be rewritten as follows:

| z f ( z ) f ( z ) + α z 2 f ( z ) f ( z ) 1 z f ( z ) f ( z ) + α z 2 f ( z ) f ( z ) ( 2 γ m 1 ) |<1(zU),
(2.4)

we easily find from (2.4) that a function g H m (α,β) if and only if

z g ( z ) + α z 2 g ( z ) g ( z ) z g ( z ) + α z 2 g ( z ) ( 2 γ m 1 ) g ( z ) σ ( z U ; σ C ; | σ | = 1 ) ,

which is equivalent to

( g h ) ( z ) z 0(zU),
(2.5)

where

h(z)=z+ k = m + 1 c k z k ( c k : = k + α k ( k 1 ) 1 [ k + α k ( k 1 ) ( 2 γ m 1 ) ] σ 2 ( γ m 1 ) σ ) .
(2.6)

It follows from (2.6) that

| c k | = | k + α k ( k 1 ) 1 [ k + α k ( k 1 ) ( 2 γ m 1 ) ] σ 2 ( γ m 1 ) σ | k + α k ( k 1 ) 1 + [ k + α k ( k 1 ) ( 2 γ m 1 ) ] | σ | 2 ( 1 γ m ) | σ | = k ( 1 + k α α ) γ m 1 γ m ( | σ | = 1 ) .

If f A m satisfies the condition (2.2), we deduce from (2.5) that

( f h ) ( z ) z ε ( | ε | < δ ; δ > 0 ) ,

or, equivalently,

| ( f h ) ( z ) z |δ(zU;δ>0).
(2.7)

We now suppose that

q(z)=z+ k = m + 1 d k z k N δ (f).

It follows from (2.1) that

| ( ( q f ) h ) ( z ) z | = | k = m + 1 ( d k a k ) c k z k 1 | | z | k = m + 1 k ( 1 + k α α ) γ m 1 γ m | d k a k | < δ .
(2.8)

Combining (2.7) and (2.8), we easily find that

| ( q h ) ( z ) z |=| ( [ f + ( q f ) ] h ) ( z ) z || ( f h ) ( z ) z || ( ( q f ) h ) ( z ) z |>0,

which implies that

( q h ) ( z ) z 0(zU).

Therefore, we conclude that

q(z) N δ (f) H m (α,β).

We thus complete the proof of Theorem 1. □

Next, we derive the partial sums of the class H m (α,β). For some recent investigations involving the partial sums in analytic function theory, one can refer to [1416] and the references cited therein.

Theorem 2 Let f A m be given by (1.1) and define the partial sums f n (z) of f by

f n (z)=z+ k = m + 1 n a k z k (nN;nm+1).
(2.9)

If

k = m + 1 k ( 1 + k α α ) γ m 1 γ m | a k |1(α0;0β<1; γ m <1),
(2.10)

then

(1) f H m (α,β);

(2)

( f ( z ) f n ( z ) ) n ( 1 + α + n α ) ( n + 1 ) ( 1 + n α ) γ m (nN;nm+1;zU)
(2.11)

and

( f n ( z ) f ( z ) ) ( n + 1 ) ( 1 + n α ) γ m ( n + 1 ) ( 1 + n α ) + 1 2 γ m (nN;nm+1;zU).
(2.12)

The bounds in (2.11) and (2.12) are sharp.

Proof (1) Suppose that f 1 (z)=z. We know that z H m (α,β), which implies that

f 1 ( z ) + ε z 1 + ε =z H m (α,β).

From (2.10), we easily find that

k = m + 1 k ( 1 + k α α ) γ m 1 γ m | a k 0|1,

which implies that f N 1 (z). In view of Theorem 1, we deduce that

f N 1 (z) H m (α,β).

(2) It is easy to verify that

( n + 1 ) [ 1 + ( n + 1 ) α α ] γ m 1 γ m = ( n + 1 ) ( 1 + n α ) γ m 1 γ m > n ( 1 + n α α ) γ m 1 γ m > 1 ( n N ) .

Therefore, we have

k = m + 1 n | a k |+ ( n + 1 ) ( 1 + n α ) γ m 1 γ m k = n + 1 | a k | k = m + 1 k ( 1 + k α α ) γ m 1 γ m | a k |1.
(2.13)

We now suppose that

ψ ( z ) = ( n + 1 ) ( 1 + n α ) γ m 1 γ m ( f ( z ) f n ( z ) n ( 1 + α + n α ) ( n + 1 ) ( 1 + n α ) γ m ) = 1 + ( n + 1 ) ( 1 + n α ) γ m 1 γ m k = n + 1 a k z k 1 1 + k = m + 1 n a k z k 1 .
(2.14)

It follows from (2.13) and (2.14) that

| ψ ( z ) 1 ψ ( z ) + 1 | ( n + 1 ) ( 1 + n α ) γ m 1 γ m k = n + 1 | a k | 2 2 k = m + 1 n | a k | ( n + 1 ) ( 1 + n α ) γ m 1 γ m k = n + 1 | a k | 1(zU),

which shows that

( ψ ( z ) ) 0(zU).
(2.15)

Combining (2.14) and (2.15), we deduce that the assertion (2.11) holds true.

Moreover, if we put

f(z)=z+ 1 γ m ( n + 1 ) ( 1 + n α ) γ m z n + 1 ( n N { 1 , 2 , , m 1 } ; m N ) ,
(2.16)

then for z=r e i π / n , we have

f ( z ) f n ( z ) =1+ 1 γ m ( n + 1 ) ( 1 + n α ) γ m z n n ( 1 + α + n α ) ( n + 1 ) ( 1 + n α ) γ m ( r 1 ) ,

which implies that the bound in (2.11) is the best possible for each nN{1,2,,m1}.

Similarly, we suppose that

φ ( z ) = ( n + 1 ) ( 1 + n α ) + 1 2 γ m 1 γ m ( f n ( z ) f ( z ) ( n + 1 ) ( 1 + n α ) γ m ( n + 1 ) ( 1 + n α ) + 1 2 γ m ) = 1 ( n + 1 ) ( 1 + n α ) + 1 2 γ m 1 γ m k = n + 1 a k z k 1 1 + k = m + 1 a k z k 1 .
(2.17)

In view of (2.13) and (2.17), we conclude that

| φ ( z ) 1 φ ( z ) + 1 | ( n + 1 ) ( 1 + n α ) + 1 2 γ m 1 γ m k = n + 1 | a k | 2 2 k = m + 1 n | a k | n ( 1 + α + n α ) 1 γ m k = n + 1 | a k | 1(zU),

which implies that

( φ ( z ) ) 0(zU).
(2.18)

Combining (2.17) and (2.18), we readily get the assertion (2.12) of Theorem 2. The bound in (2.12) is sharp with the extremal function f given by (2.16).

The proof of Theorem 2 is thus completed. □

Finally, we turn to ratios involving derivatives. The proof of Theorem 3 below is much akin to that of Theorem 2, we here choose to omit the analogous details.

Theorem 3 Let f A m be given by (1.1) and define the partial sums f n (z) of f by (2.9). If the condition (2.10) holds, then

( f ( z ) f n ( z ) ) ( n + 1 ) ( n α + γ m ) γ m ( n + 1 ) ( 1 + n α ) γ m (nN;nm+1;zU)
(2.19)

and

( f n ( z ) f ( z ) ) ( n + 1 ) ( 1 + n α ) γ m ( n + 1 ) ( 2 + n α γ m ) γ m (nN;nm+1;zU).
(2.20)

The bounds in (2.19) and (2.20) are sharp with the extremal function given by (2.16).

Remark By setting α=0 and m=1 in Theorems 2 and 3, we get the corresponding results obtained by Silverman [16].

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

The present investigation was supported by the National Natural Science Foundation under Grants 11226088 and 11101053, the Key Project of Chinese Ministry of Education under Grant 211118, the Excellent Youth Foundation of Educational Committee of Hunan Province under Grant 10B002, the Open Fund Project of Key Research Institute of Philosophies and Social Sciences in Hunan Universities under Grants 11FEFM02 and 12FEFM02, and the Key Project of Natural Science Foundation of Educational Committee of Henan Province under Grant 12A110002 of the People’s Republic of China.

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Correspondence to Zhi-Gang Wang.

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Wang, ZG., Yuan, XS. & Shi, L. Neighborhoods and partial sums of certain subclass of starlike functions. J Inequal Appl 2013, 163 (2013). https://doi.org/10.1186/1029-242X-2013-163

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Keywords

  • analytic functions
  • starlike functions
  • neighborhoods
  • partial sums