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APPLICATIONS OF SUBORDINATION ON SUBCLASSES OF MEROMORPHICALLY UNIVALENT FUNCTIONS WITH

INTEGRAL OPERATOR

Indu Bala Bapna and Nisha Mathur

Abstract. In this paper we are concerned with applications of differential sub-ordination for class of meromorphic univalent functions defined by integral operator,

Pβαf(z) = β α

Γ(α). 1

zβ+1

z Z

0

tβlogz

t

α−1 f(t)dt

In the present paper, our aim is to study the Coefficient Bounds, Integral Represen-tation, Linear Combinations, Weighted and Arithmetic Mean.

Keywords: Meromorphic Functions, Differential Subordination, Integral Oper-ator, Coefficient Bounds, Integral Representation, Linear Combination, Weighted Means and Arithmetic Means.

2000Mathematics Subject Classification: 30c45. 1. Introduction

Let Σ be a class of all Meromorphic functions f(z) of the form by [4]

f(z) = 1

z +

∞ X k=1

akzk, ak≥0 (1.1)

and

g(z) = 1

z+

∞ X k=1

bkzk, bk≥0 (1.2)

Which are univalent in the punctured unit diskU ={z:z∈C ,0<|z|<1}=

U\ {0} with a simple pole at the origin. If f(z) and g(z) are analytic inU, we say that f(z) is subordinate to g(z), written f ≺ g if there exists a Schwarz function

w(z) in U withw(0) = 0 and|w(z)|<1(z∈U), such that f(z) =g(w(z)),(z∈U). In particular, if the function g(z) is univalent in U, we have the following[5],

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Definition 1. Analogous to the operators defined by Jung, Kim, and Srivas-tava[3] on the normalized Analytic functions, by [1] define the following integral operator

Pβα:X→X

Pβα =Pβαf(z) = β α

Γ(α). 1

zβ+1

z Z

0

tβlogz

t

α−1

f(t)dt (1.3)

(α >0, β >0; z∈U) Where Γ(α) is the familiar Gamma Function.

Using the integral representation of the Gamma and Beta function, it can be shown that

For f(z)∈Σ, given by (1.1) we have

Pβαf(z) = 1

z+

∞ X k=1

β

k+β+ 1 α

akzk, (α >0, β >0) (1.4)

z(Pβαf(z))′ =βPα−1

β f(z)−(β+ 1)P α

βf(z), (α >1, β >0) (1.5) Definition 2. Let A and B (−1 ≤ B < A ≤ 1) be defined parameters. We say that a function f(z) ∈ Σ is in the class Σ(A, B); if it satisfies the following subordination condition by [5]

−z2 pαβf(z)′

≺ 1 +Az

1 +Bz (z∈U) (1.6)

By the definition of differential subordinate, (1.6) is equivalent to the following condition

1 +z2pα βf(z)

A+Bz2pα βf(z)

<1 (z∈U) (1.7)

In particular, we can write X

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Where Σ(β) denotes the class of function in Σ satisfying following form

Re−z2 pαβf(z)′

> β (0≤β <1;z∈U) Here are some applications of differential subordination {[2],[6]}.

2. Coefficient Bounds

Theorem 2.1Let the functionf(z) of the form (1.1) be inΣ Then the function

f(z) belongs to the class Σ(A, B) if and only if

(1−B) ∞ X k=1 k β k+β+ 1

α

ak<(A−B) (2.1) Where −1≤B < A≤1. The result is sharp for the functionf(z) given by

f(z) = 1

z +

(A−B) (1−B)k

β k+β+1

αzk

Proof: Assume that the condition (2.1) is true. We must show thatf ∈Σ(A, B) or equivalently prove that

1 +z2pα βf(z)

A+Bz2pα βf(z)

<1 (2.2)

1 +z2pα βf(z)

A+Bz2pα βf(z)

′ = 1 +

−1 + ∞ P k=1

kk+ββ+1αakzk+1

A+B

−1 + ∞ P k=1

kk+ββ+1αakzk+1 = ∞ P k=1

kk+ββ+1αakzk+1

A−B+B

∞ P k=1

kk+ββ+1αakzk+1 ≤ ∞ P k=1

kk+ββ+1αak

A−B+B

∞ P

kk+ββ+1αak

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The last inequality is true by (2.1).

Conversely, suppose that f ∈ Σ(A, B). We must show that the condition (2.1) holds true. We have

1 +z2

βf(z) ′

A+Bz2pα βf(z)

′ <1

Hence we get

∞ P k=1

kk+ββ+1αak

A−B+B

∞ P k=1

kk+ββ+1αak

<1,

Since Re(z)<|z|, so we have

Re        ∞ P k=1

kk+ββ+1αak

A−B+B

∞ P k=1

kk+ββ+1αak        <1

We Choose the value of zon the real axis and letting z→1−

, then we obtain        ∞ P k=1

kk+ββ+1αak

A−B+B

∞ P k=1

kk+ββ+1αak        <1 Then

(1−B) ∞ X k=1 k β k+β+ 1

α

ak<(A−B) The result is sharp for the function

f(z) = 1

z +

∞ X k=1

(A−B) (k+β+ 1)α (1−B)kβα z

k

Corollary 2.2 Let f ∈Σ(A, B), then we have

ak ≤

(A−B) (k+β+ 1)α

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3. Integral Representation

In the next theorem we obtain an integral representation forpαβf(z) Theorem 3.1 Let f ∈Σ(A, B), then

βf(z) = z Z

0

(Aφ(t)−1)

t2(1(t))dt , where|φ(z)|<1, z∈U (3.1)

Proof: Let f(z)∈Σ(A, B) letting

−z2pα βf(z)

=y(z) We have

y(z)≺ 1 +Az

1 +Bz (3.2)

Or we can write

y(z)−1

By(z)−A

<1,

so that consequently, we have

y(z)−1

By(z)−A =φ(z), |φ(z)|<1 (z∈U)

We can write

−z2

βf(z)′

= 1−Aφ(z) 1−Bφ(z), Which gives

− pαβf(z)′ = 1

z2

1−Aφ(z) 1−Bφ(z),

βf(z) = z Z

0

1

t2

Aφ(t)−1

1−Bφ(t)dt (3.3)

And this gives the required result.

4. Linear Combination

In the theorem below, we prove a linear combination for the class Σ(A, B). Theorem 4.1 Let fi(z) =z−1

+ ∞ P

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Belong to Σ(A, B) then

F(z) = l P i=1

cifi(z)∈Σ(A, B) Where l P i=1

ci = 1 Proof: By theorem 2.1, We can write for everyi∈ {1,2. . . , l}

∞ X k=1

k(1−B) (A−B)

β k+β+ 1

α

ak,i<1, (4.1)

Therefore

F(z) = l X

i=1 ci z

−1

+ ∞ X k=1

ak,izk !!

=z−1

+ l X i=1

∞ X k=1

ciak,izk

=z−1

+ ∞ X k=1

l X i=1

ciak,i !

zk (4.2)

however

∞ X k=1

k(1−B) (A−B)

β k+β+ 1

α l

X i=1

ak,ici !

= l X i=1

"∞ X k=1

k(1−B) (A−B)

β k+β+ 1

α

ak,i #

ci

≤1 (4.3)

then F(z)∈Σ(A, B)

hence the proof is complete.

5. Weighted Mean

Definition 3. f(z) and g(z) belong to Σ,then the weighted mean hj(z) of f(z) and g(z) is given by

hj(z) = 1

2[(1−j)f(z) + (1 +j)g(z)]

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f(z) and g(z) is also in Σ(A, B) Proof: We have forhj(z) by definition

h(z) = 1 2

"

(1−j) z−1

+ ∞ X k=1

akzk !

+ (1 +j) z−1

+ ∞ X k=1

bkzk !#

=z−1

+1 2

∞ X k=1

[(1−j)ak+ (1 +j)bk]zk (5.1) Since f(z) andg(z) are in the class Σ(A, B) so by theorem 2.1 we must prove that

∞ X k=1

k(1−B)

β k+β+ 1

α 1

2(1−j)ak+ 1

2(1 +j)bk

= 1

2(1−j)(1−B) ∞ X k=1 k

β k+β+ 1

α

ak+ 1

2(1 +j)(1−B) ∞ X k=1 k

β k+β+ 1

α

bk

≤ 1

2(1−j)(A−B) + 1

2(1 +J)(A−B)

≤(A−B) (5.2)

hence proved

6. Arithmetic Mean

Definition 4. Let f1(z),f2(z). . . fl(z) belong to Σ(A, B), then the arithmetic mean h(z) of fi(z) is given by

h(z) = 1

l

l X k=1

fi(z)

In the theorem below we will prove the arithmetic mean for this class Σ(A, B). Theorem 6.1 If f1(z),f2(z). . . fl(z) are in the class Σ(A, B), then the arithmetic mean h(z) of fi(z) is given by

h(z) = 1

l

l X

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is also in the class Σ(A, B). Proof: We have forh(z) by def. 4

h(z) = 1

l

l X k=1

z−1

+ ∞ X k=1

ak,izk !

=z−1

+ ∞ X k=1

1

l

l X k=1

ak,i !

zk (6.2)

Sincefi(z)∈Σ(A, B) for everyi= 1,2, . . . , l so by using theorem 2.1, we prove that

(1−B) ∞ X k=1 k

β k+β+ 1

α 1

l

l X i=1

ak,i !

= 1

l

l X i=1

∞ X k=1

k(1−B) !

β k+β+ 1

α

ak,i

≤ 1 l

l X i=1

(A−B) (6.3)

The proof is complete.

References

[1] A. Y. Lashin, On Certain Subclasses of Meromorphic Functions Associated with Certain Integral Operators, Computers and Mathematics with Application, 59, (2009), 524-531.

[2] H. M. Srivastava and J. Patel, Applications of Differential Subordination to Cer-tain Subclasses of Meromorphically Multivalent Functions, J. Ineq. Pure and Appl. Math., 6 (3), (2005), Art. 88.

[3] I. B. Jung, Y. C. Kim, H. M. Srivastava,The Hardy Space of Analytic Functions Asociated with Certain One-parameter of Families of Integral Operator, J. Math. Anal. Appl., 176 (1), (1993), 138-147. [4] J. Patel, P. Sahoo,On Certain Subclass of Meromorphically Univalent Functions, Kyungpook Math. J., 42, (2002), 121-132. [5] S. S. Miller and P. T. Mocanu, Differential Subordinations and Univalent Func-tions, Michigan Math. J., 28, (1981), 157-171.

[6] W. G. Atshan and S. R. Kulkarni, On Application of Differential Subordination for Certain Subclass pf Meromorphically P-Valent Functions With Positive Coeffi-cients Defined by Linear Operator, jipam, 10, (2009), Issue 2, Art. 53, pg. 11.

Indu Bala Bapna,

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Bhilwara, Rajasthan, India email:[email protected] Nisha Mathur,

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