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Properties for certain subclasses of meromorphic functions defined by a multiplier transformation

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Abstract

Some inclusion and convolution properties of certain subclasses of meromorphic functions associated with a family of multiplier transformations, which are defined by means of the Hadamard product (or convolution), are investigated. We also obtain closure properties for certain integral operators.

MSC:30C45, 30C80.

1 Introduction

Let A denote the class of analytic functions f in the open unit disk U={zC:|z|<1} with the usual normalization f(0)= f (0)1=0. Let S (α) and K(α) denote the subclasses of A consisting of starlike and convex functions of order α (0α<1) and let S (0)= S and K(0)=K. If f and g are analytic in U, we say that f is subordinate to g in U, written as fg or f(z)g(z), if there exists a Schwarz function w such that f(z)=g(w(z)) (zU).

A function fA is said to be prestarlike of order α in U if

z ( 1 z ) 2 ( 1 α ) f(z) S (α)(0α<1),

where fg denotes the familiar Hadamard product (or convolution) of two analytic functions f and g in U. We denote this class by R(α) (see, for details, [1]). We note that R(0)=K and R(1/2)= S (1/2).

Let N be the class of all functions h which are analytic and univalent in U and for which h(U) is convex with h(0)=1.

Let denote the class of functions of the form

f(z)= 1 z + k = 0 a k z k

which are analytic in the punctured open unit disk D=U{0}.

For any n N 0 ={0,1,2,}, we denote the multiplier transformations D λ n of functions fM by

D λ n f(z)= 1 z + k = 0 ( k + 1 + l λ ) n a k z k (λ>0;zD).

Obviously, we have

D λ s ( D λ t f ( z ) ) = D λ s + t f(z)

for all nonnegative integers s and t. The operators D λ n and D 1 n are the multiplier transformations introduced and studied by Sarangi and Uraligaddi [2] and Uralegaddi and Somanatha [3, 4], respectively. Analogous to D λ n , we here define a new multiplier transformation I λ , μ n as follows.

Let f n (z)=1/z+ k = 0 ( ( k + 1 + λ ) / λ ) n z k , n N 0 , and let f n , μ be such that

f n (z) f n , μ (z)= 1 z + k = 0 ( μ ) k + 1 ( 1 ) k + 1 z k (μ>0;zD),

where ( ν ) k is the Pochhammer symbol (or the shifted factorial) defined (in terms of the gamma function) by

( ν ) k := Γ ( ν + k ) Γ ( ν ) ={ 1 if  k = 0  and  ν C { 0 } , ν ( ν + 1 ) ( ν + k 1 ) if  k N : = { 1 , 2 , }  and  ν C .

Then

I λ , μ n f(z)= f n , μ (z)f(z).
(1.1)

We note that I 1 , 2 0 f(z)=z f (z)+2f(z) and I 1 , 2 1 f(z)=f(z). It is easily verified from (1.1) that

z ( I λ , μ n + 1 f ( z ) ) =λ I λ , μ n f(z)(λ+1) I λ , μ n + 1 f(z)
(1.2)

and

z ( I λ , μ n f ( z ) ) =μ I λ , μ + 1 n f(z)(μ+1) I λ , μ n f(z).
(1.3)

The definition (1.1) of the multiplier transformation I λ , μ n is motivated essentially by the Choi-Saigo-Srivastava operator [5] for analytic functions, which includes the Noor integral operator studied by Liu [6] (also, see [79]).

We also define the function ϕ(a,c;z) by

ϕ(a,c;z):= 1 z + k = 0 ( a ) k + 1 ( c ) k + 1 z k ( z U ; a R ; c R Z 0 ; Z 0 : = { 1 , 2 , } ) .
(1.4)

By using the operator I λ , μ n , we introduce the following class of analytic functions for γ>0, λ>0, sR, μ>0 and hN:

M λ , μ n (γ;h):= { f M : ( 1 γ ) z I λ , μ n f ( z ) + γ z 2 ( I λ , μ n f ( z ) ) h ( z ) } .

In the present paper, we derive some inclusion relations, convolution properties and integral preserving properties for the class M λ , μ n (γ;h).

The following lemmas will be required in our investigation.

Lemma 1.1 [[10], Lemma 2, p.192]

Let g be analytic in U and h be analytic and convex univalent in U with h(0)=g(0). If

g(z)+ 1 γ z g (z)h(z) ( Re { γ } 0 ; γ 0 ) ,
(1.5)

then

g(z) h ˜ (z)=γ z γ 0 z t γ 1 h(t)dth(z)

and h ˜ is the best dominant of (1.5).

Lemma 1.2 [[1], Theorem 2.4, p.54]

Let f S (α) and gR(α). Then for any analytic function F in U,

g ( f F ) g f (U) co ¯ ( F ( U ) ) ,

where co ¯ (F(U)) denotes the convex hull of F(U).

Lemma 1.3 [[11], Lemma 5, p.656]

Let 0<ac. Then

Re { z ϕ ( a , c ; z ) } > 1 2 (zU),

where ϕ is given by (1.4).

2 Inclusion relations

Theorem 2.1 If 0 γ 1 < γ 2 , then

M λ , μ n ( γ 2 ;h) M λ , μ n ( γ 1 ;h).

Proof

Let

g(z)=z I λ , μ n f(z) ( f M λ , μ n ( γ 2 ; h ) : z U ) .
(2.1)

Then the function g is analytic in U with g(0)=1. Differentiating both sides of (2.1), we have

(1+ γ 2 )z I λ , μ n f(z)+ γ 2 z 2 ( I λ , μ n f ( z ) ) =g(z)+ γ 2 z g (z)h(z).
(2.2)

Hence an application of Lemma 1.1 with μ=1/ γ 2 yields

g(z)h(z).
(2.3)

Since 0 γ 1 / γ 2 <1 and h is convex univalent in U, it follows from (2.1), (2.2) and (2.3) that

( 1 + γ 1 ) z I λ , μ n f ( z ) + γ 1 z 2 ( I λ , μ n f ( z ) ) = γ 1 γ 2 [ ( 1 γ 2 ) z I λ , μ n f ( z ) + γ 2 z 2 ( I λ , μ n f ( z ) ) ] + ( 1 γ 1 γ 2 ) g ( z ) h ( z ) .

Therefore f M λ , μ n ( γ 1 ;h), and so we complete the proof of Theorem 2.1. □

Theorem 2.2 If 0< μ 1 μ 2 , then

M λ , μ 2 n (γ;h) M λ , μ 1 n (γ;h).

Proof Let f M λ , μ 2 n (γ;h). Then

(2.4)

In view of Lemma 1.3, we see that the function zϕ( μ 1 , μ 2 ;z) has the Herglotz representation

zϕ( μ 1 , μ 2 ;z)= | x | = 1 d μ ( x ) 1 x z (zU),
(2.5)

where μ(x) is a probability measure defined on the unit circle |x|<1 and

| x | = 1 dμ(x)=1.

Since h is convex univalent in U, it follows from (2.4) and (2.5) that

(1+γ)z I λ , μ 1 n f(z)+γ z 2 ( I λ , μ 1 n f ( z ) ) = | x | = 1 h(xz)dμ(x)h(z),

which completes the proof of Theorem 2.2. □

Theorem 2.3 If μ>0, then

M λ , μ + 1 n (γ; h ˜ ) M λ , μ n (γ;h),

where

h ˜ (z)=μ z μ 0 z t μ 1 h(t)dth(z).

Proof

Let

g(z)=(1+γ)z I λ , μ n f(z)+γ z 2 ( I λ , μ n f ( z ) ) (fM;zU).
(2.6)

Then from (1.4) and (2.6), we have

z 1 g(z)=γμ I λ , μ + 1 n f(z)+(1γμ) I λ , μ n f(z).
(2.7)

Differentiating both sides of (2.6) and using (1.4), we obtain

(2.8)

By a simple calculation with (2.7) and (2.8), we get

g(z)+ z g ( z ) μ =(1+γ) I λ , μ + 1 n f ( z ) z +γ ( I λ , μ + 1 n f ( z ) ) .
(2.9)

If f M λ , μ + 1 n (γ;h), then it follows from (2.9) that

g(z)+ z g ( z ) μ h(z)(μ>0).

Hence an application of Lemma 1.1 yields

g(z) h ˜ (z)=μ z μ 0 z t μ 1 h(t)dth(z),

which shows that

f M λ , μ + 1 n (γ; h ˜ ) M λ , μ n (γ;h).

 □

Theorem 2.4 If sR and λ>0, then

M λ , μ n (γ; h ˜ ) M λ , μ n + 1 (γ;h),

where

h ˜ (z)=λ z λ 0 z t λ 1 h(t)dth(z).

Proof By using the same techniques as in the proof of Theorem 2.3 and (1.5), we have Theorem 2.4 and so we omit the detailed proof involved. □

Theorem 2.5 Let γ>0, β>0 and f M λ , μ n (γ;βh+1β). If β β 0 , where

β 0 = 1 2 ( 1 1 γ 0 1 u 1 γ 1 1 + u d u ) 1 ,
(2.10)

then f M λ , μ n (0;h). The bound β 0 is sharp for the function

h(z)= 1 1 z (zU).

Proof

Let

g(z)=z I λ , μ n f(z) ( f M λ , μ n ( γ ; β h + 1 β ) ; γ > 0 ; β > 0 ) .
(2.11)

Then we have

g ( z ) + γ z g ( z ) = ( 1 + γ ) z I λ , μ n f ( z ) + γ z 2 ( I λ , μ n f ( z ) ) β h ( z ) + 1 β .

Hence an application of Lemma 1.1 yields

g(z) β γ z 1 γ 0 z t 1 γ 1 h(t)dt+1β=(hψ)(z),
(2.12)

where

ψ(z)= β γ z 1 γ 0 z t 1 γ 1 1 t dt+1β.
(2.13)

If 0<β β 0 , where β 0 is given by (2.10), then from (2.13), we have

Re { ψ ( z ) } = β γ 0 1 u 1 γ 1 Re { 1 1 u z d u } + 1 β > β γ 0 1 u 1 γ 1 1 + u d u + 1 β 1 2 .

By using the Herglotz representation for ψ, it follows from (2.11) and (2.12) that

z I λ , μ n f(z)(hψ)(z)h(z)

since h is convex univalent in U. This shows that f M λ , μ n (0;h).

For h(z)=1/(1z) and fM defined by

z I λ , μ n f(z)= β γ z 1 γ 0 z t 1 γ 1 1 t dt+1β,

it is easy to verify that

(1+γ)z I λ , μ n f(z)+γ z 2 ( I λ , μ n f ( z ) ) =βh(z)+1β.

Thus f M λ , μ n (γ;βh+1β). Furthermore, for β> β 0 , we have

Re { z I λ , μ n f ( z ) } β γ 0 1 u 1 γ 1 1 + u du+1β< 1 2 (z1),

which implies that f M λ , μ n (0;h). Hence the bound β 0 cannot be increased when h(z)=1/(1z) (zU). □

3 Convolution properties

Theorem 3.1 If f M λ , μ n (γ;h) and

Re { z g ( z ) } > 1 2 (gM;zU),

then

fg M λ , μ n (γ;h).

Proof Let f M λ , μ n (γ;h) and gM. Then we have

(1+γ)z I λ , μ n (fg)(z)+γ z 2 ( I λ , μ n ( f g ) ( z ) ) =zg(z)ψ(z),

where

ψ(z)=(1+γ)z I λ , μ n f ( z ) + γ z 2 ( I λ , μ n f ( z ) ) h(z).

The remaining part of the proof of Theorem 3.1 is similar to that of Theorem 2.2, and so we omit the details involved. □

Corollary 3.1 Let f M λ , μ n (γ;h) be given by (1.1). Then the function

σ m (z)= 0 1 t S m (tz)dt(zU),

where

S m (z)= 1 z + n = 1 m 1 a n z n 1 ( m N { 1 } ; z U ) ,

is also in the class M λ , μ n (γ;h).

Proof

We have

σ m (z)= 1 z + n = 1 m 1 a n n 1 z n + 1 =(f g m )(z) ( m N { 1 } ) ,
(3.1)

where

f(z)= 1 z + n = 1 a n z n 1 M λ , μ n (γ;h)

and

g m (z)= 1 z + n = 1 m 1 z n n 1 M,

while, it is known [4] that

Re { z g m ( z ) } =Re { 1 + n = 1 m 1 z n n + 1 } > 1 2 ( m N { 1 } ; z U ) .
(3.2)

In view of (3.1) and (3.2), an application of Theorem 3.1 leads to σ m M λ , μ n (γ;h). □

Theorem 3.2 If f M λ , μ n (γ;h) and

z 2 g(z)R(α)(gM;zU),

then

(fg) M λ , μ n (γ;h).

Proof

By using a similar method as in the proof of Theorem 3.1, we have

(1+γ)z I λ , μ n (fg)(z)+γ z 2 ( I λ , μ n ( f g ) ( z ) ) = z 2 g ( z ) ( z ψ ( z ) ) z 2 g ( z ) z (zU),
(3.3)

where

ψ(z)=(1+γ)z I λ , μ n f(z)+γ z 2 ( I λ , μ n f ( z ) ) h(z).

Since h is convex univalent in U, it follows from (3.3) and Lemma 1.2 that Theorem 3.2 holds true. □

If we take α=0 and α=1/2 in Theorem 3.2, we have the following corollary.

Corollary 3.2 If f M λ , μ n (γ;h) and gM satisfies one of the following conditions:

  1. (i)

    z 2 g(z) is convex univalent in U

or

  1. (ii)

    z 2 g(z) S ( 1 2 ),

then (fg) M λ , μ n (γ;h).

4 Integral operators

Theorem 4.1 If f M λ , μ n (γ;h), then the function F defined by

F(z)= c 1 z c 0 z t c 1 f(t)dt ( Re { c } > 1 )
(4.1)

is in the class M λ , μ n (γ; h ˜ ), where

h ˜ (z)=(c1) z ( c 1 ) 0 z t c h(t)dth(z).

Proof Let f M λ , μ n (γ;h). Then from (4.1), we obtain

(c1)f(z)=z F (z)+cF(z).
(4.2)

Define the function G by

z 1 G(z)=(1+γ) I λ , μ n F(z)+γz ( I λ , μ n F ( z ) ) (zD).
(4.3)

Differentiating both sides of (4.3) with respect to z, we get

z G (z)G(z)=(1+γ)z I λ , μ n ( z F ( z ) ) +γ z 2 ( I λ , μ n ( z F ( z ) ) ) .
(4.4)

Furthermore, it follows from (4.2), (4.3) and (4.4) that

(4.5)

Since f M λ , μ n (γ;h), from (4.5), we have

G(z)+ 1 c 1 z G (z)h(z) ( Re { c } > 1 ) ,

and so an application of Lemma 1.1 yields

G(z) h ˜ (z)= c 1 z c 1 0 z t c h(t)dth(z).

Therefore we conclude that

F M λ , μ n (γ; h ˜ ) M λ , μ n (γ;h).

 □

Theorem 4.2 If fM and F are defined as in Theorem  4.1, if

(1α)z I λ , μ n F(z)+αz I λ , μ n f(z)h(z)(α>0),
(4.6)

then F M λ , μ n (0; h ˜ ), where

h ˜ (z)= c 1 α z α c 1 0 z t c 1 α 1 h(t)h(z) ( Re { c } > 1 ) .

Proof

Let

G(z)=z I λ , μ n F(z)(zD).
(4.7)

Then G is analytic in U with G(0)=1 and

z G (z)= z 2 ( I λ , μ n F ( z ) ) +G(z).
(4.8)

It follows from (4.2), (4.6), (4.7) and (4.8) that

( 1 α ) z I λ , μ n F ( z ) + α z I λ , μ n f ( z ) = ( 1 α ) z I λ , μ n F ( z ) + α c 1 [ c z I λ , μ n F ( z ) + z 2 ( I λ , μ n F ( z ) ) ] = G ( z ) + α c 1 z G ( z ) h ( z ) ( Re { c } > 1 ; α > 0 ) .

Therefore, by Lemma 1.1, we conclude that Theorem 4.2 holds true as stated. □

Theorem 4.3 Let F M λ , μ n (γ;h). If the function f is defined by

F(z)= c 1 z c 0 z t c 1 f(t)dt(c>1),
(4.9)

then

σf(σz) M λ , μ n (γ;h),

where

σ=σ(c)= 1 + ( c 1 ) 2 1 c 1 .
(4.10)

The bound σ is sharp for the function

h(z)=β+(1β) 1 + z 1 z (β1;zU).
(4.11)

Proof We note that for FM,

F(z)=F(z) 1 z ( 1 z ) andz F (z)=F(z) ( 1 z ( 1 z ) 2 2 z 2 ( 1 z ) ) .

Then from (4.9), we have

f(z)= c F ( z ) + z F ( z ) c 1 =(Fg)(z)(c>1;zD),
(4.12)

where

g(z)= 1 c 1 ( ( c 2 ) 1 z ( 1 z ) + 1 z ( 1 z ) 2 ) M.
(4.13)

Next, we show that

Re { z g ( z ) } > 1 2 ( | z | < σ ) ,
(4.14)

where σ=σ(c) is given by (4.10). Letting

1 1 z =R e i θ ( | z | = r < 1 ; R > 0 ) ,

we see that

cosθ= 1 + R 2 ( 1 r 2 ) 2 R andR 1 1 + r .
(4.15)

Then for (4.13) and (4.15), we have

2 Re { z g ( z ) } = 2 c 1 [ ( c 2 ) R cos θ + R 2 ( 2 cos 2 θ 1 ) ] = R 2 c 1 [ c ( 1 r 2 ) + R 2 ( 1 r 2 ) 2 2 ] + 1 R 2 c 1 [ c 1 2 r ( c 1 ) r 2 ] + 1 .

This evidently gives (4.14), which is equivalent to

Re { σ z g ( σ z ) } > 1 2 (zU).
(4.16)

Let F M λ , μ n (γ;h). Then, by using (4.12) and (4.16), an application of Theorem 3.1 yields

σf(σz)=F(z)σg(σz) M λ , μ n (γ;h).

For h given by (4.11), we consider the function FM defined by

(1+γ)z I λ , μ n F(z)+γ z 2 ( I λ , μ n F ( z ) ) =β+(1β) 1 + z 1 z (β1;zU).
(4.17)

Then from (4.3), (4.5) and (4.17), we find that

( 1 + γ ) z I λ , μ n f ( z ) + γ z 2 ( I λ , μ n f ( z ) ) = β + ( 1 β ) 1 + z 1 z + z c 1 ( β + ( 1 β ) 1 + z 1 z ) = β + ( 1 β ) ( c 1 + 2 z ( c 1 ) z 2 ) ( c 1 ) ( 1 z ) 2 = β ( z = σ ) .

Therefore we conclude that the bound σ=σ(c) cannot be increased for each c (c>1). □

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Acknowledgements

The authors would like to express their thanks to the referees for valuable advice regarding a previous version of this paper. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2012-0002619).

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Correspondence to Min Yoon.

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All authors jointly worked on the results and they read and approved the final manuscript.

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Cho, N.E., Yoon, M. Properties for certain subclasses of meromorphic functions defined by a multiplier transformation. J Inequal Appl 2013, 46 (2013) doi:10.1186/1029-242X-2013-46

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Keywords

  • meromorphic function
  • subordination
  • starlike of order α
  • convex of order α
  • prestarlike of order α
  • convolution
  • integral operator