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Volume 2007, Article ID 71616,8pages doi:10.1155/2007/71616

Research Article

Some Subordination Results of Multivalent Functions

Defined by Integral Operator

¨

Oznur ¨

Ozkan

Received 29 August 2006; Accepted 1 March 2007

Recommended by Andrei Ronto

The object of the present paper is to give some subordination properties of integral oper-atorᏼαwhich was studied by Jung in 1993.

Copyright © 2007 ¨Oznur ¨Ozkan. This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction and definitions

LetU= {z:z∈Cand|z|<1}, letᏴ(U) be the set of all functionsanalytic inU, and let

Ꮽp=f Ᏼ(U) : f(z)=zp+ap+1zp+1+···

(1.1)

for allz∈Uandp∈N={1, 2, 3,. . .}withᏭ:=Ꮽ1. Forp∈N, let

Ᏼp=f Ᏼ(U) : f(z)=p+bpzp+··· (1.2)

withᏴ:=Ᏼ1.

Given two functions f andg, which are analytic inU, then we say that the function f issubordinatetogand write f ≺gor (more precisely)

f(z)≺g(z) (z∈U), (1.3)

if there exists a Schwarz functionw(z), analytic inUwith

w(0)=0, w(z)<1 (z∈U) (1.4)

such that

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In particular, ifgis univalent inU, then we write the following equivalence:

f(z)≺g(z)⇐⇒f(0)=g(0), f(U)⊂g(U). (1.6)

For analytic functions fj(j=1, 2) given by

fj(z)=z+

k=2

ak,jzk (j=1, 2), (1.7)

let f1∗f2denote theHadamardproduct(or convolution)of f1and f2, defined by

f1∗f2(z) :=z+

k=2

ak,1ak,2zk=:

f2∗f1(z). (1.8)

Next, for real parametersAandBsuch that1≤B < A≤1, we define the function

ϕA,B(z) :=1 +1 +AzBz (z∈U). (1.9)

It is obvious thatϕA,B(z) for1≤B≤1 is the conformal map of the unit diskUonto the

disk symmetrical with respect to the real axis having the center

1−AB

1−B2 (B= ∓1) (1.10)

and the radius

A−B

1−B2 (B= ∓1). (1.11)

Furthermore, the boundary circle of this disk intersects the real axis at the points

1−A 1−B,

1 +A

1 +B (B= ∓1). (1.12)

Let (a)vdenote thePochhammersymbol(or the shifted factorial), since

(1)n=n! forn∈N0:=N∪ {0}, (1.13)

defined (fora,v∈Cand in terms of the Gamma function) by

(a)v:=Γ(Γa+v)

(a) =

⎧ ⎪ ⎨ ⎪ ⎩

1 v=0,a∈C\{0},

a(a+ 1)···(a+n−1) v=n∈N;a∈C. (1.14)

Recently, the following integral operator was studied by Jung et al. [1] for p=1 and Shams et al. [2]:

αf =αf(z) :=(p+ 1)α

zΓ(α)

z

0

logz t

α−1

f(t)dt=zp+

n=p+1

p+ 1

n+ 1

α

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for f Ꮽpandα >0. Moreover, Jung et al. [1] have shown that

αf(z)=z+ n=2

2

n+ 1

α

anzn (1.16)

for f Ꮽandα >0.

Therefore, Shams et al. [2] showed the following equality:

zαf(z)=(p+ 1)α−1f(z)αf(z) (1.17)

using (1.15).

Letᏼ(γ) denote the subclass ofᏴconsisting of functions f with the following condi-tion:

Ref(z)> γ (1.18)

for 0≤γ <1 and for allz∈U.

2. Main results

In proving our main results, we need the following lemmas.

Lemma2.1 [3, page 71]. Lethbe analytic, univalent, convex inUwithh(0)=1. Also letp be analytic inUwithp(0)=h(0). If

p(z) +z p (z)

γ ≺h(z) (z∈U;γ=0), (2.1)

then

p(z)≺q(z)≺h(z), (2.2)

where

q(z)= γ

z

0t

γ−1h(t)dt zU;Re(γ)0;γ=0. (2.3)

Lemma2.2 [4]. If f ᏼ(γ)for0≤γ <1, then

Ref(z)2γ−1 +2(1−γ)

1 +|z| (z∈U). (2.4)

Lemma2.3 [3, page 132]. Letqbe univalent inUand letθandφbe analytic in a domain

containingq(U)withφ(w)=0, whenw∈q(U). Set

Q(z)=zq(z)·φq(z), h(z)=θq(z)+Q(z), (2.5)

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In addition, assume that (iii)

Rezh(z)

Q(z) =Re

θq(z)

φq(z) + zQ(z)

Q(z)

>0. (2.6)

IfPis analytic inU, withP(0)=q(0),P(U), and

θP(z)+zP(z)·φP[z]≺θq(z)+zq(z)·φq[z]=h(z), (2.7)

thenP≺q, andqis the best dominant.

Theorem2.4. Let α >0,λ >0, and−1≤Bj< Aj≤1 (j=1, 2). If each of the functions

fj(z)Ꮽp(j=1, 2)satisfies the following subordination condition:

λ

α−1f

j(z)

zp + (1−λ) ᏼαf

j(z)

zp ≺ϕAj,Bj(z), (2.8)

then

λ

α−1Ω(z)

zp + (1−λ) ᏼαΩ(z)

zp ≺ϕ1−2β,1(z), (2.9)

where

Ω(z) :=αf1f2(z), (2.10)

β:=14

A1−B1A2−B2

1−B11−B2

1 p+ 1

λ

1

0

u(p+1)/λ−1 1 +u du

. (2.11)

The result is sharp forB1=B2= −1.

Proof. We assume that each of the functions fj(z)Ꮽp(j=1, 2) satisfies the

subordina-tion condisubordina-tion (2.8). If we take

ϕj(z) :α−1f

j(z)

zp + (1−λ) ᏼαf

j(z)

zp , (2.12)

then we write

ϕj(z)ᏼ(γj) (2.13)

forγj=(1−Aj)/(1−Bj) andj=1, 2. Using (1.17) and (2.12), we obtain the equality

αf

j(z)= p+ 1

λ z

p−(p+1) z

0t

(p+1)/λ−1ϕ

j(t)dt (2.14)

forj=1, 2. Then, from definition (2.10) we write

αΩ(z)= p+ 1

λ z

p−(p+1) z

0t

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where, for convenience,

ϕ0(z)α− 1Ω(z)

zp + (1−λ) ᏼαΩ(z)

zp =

p+ 1 λ z

(p+1) z

0t

(p+1)/λ−1ϕ1ϕ2(t)dt.

(2.16)

Sinceϕ1(z)ᏼ(γ1) and

ϕ3(z) :=ϕ2(z)−γ2 21−γ2 +

1 2

1

2

, (2.17)

we see that (ϕ1∗ϕ3)(z)ᏼ(γ1) by applying the well-known Herglotz formula. Thus,

h1∗h2(z)γ3 forγ3:=121−γ11−γ2. (2.18)

If we change variablet=uzand take real part in (2.16), then we obtain the following inequality:

Reh0(z)= p+ 1

λ

z

0u

(p+1)/λ−1Reϕ1ϕ2(uz)du. (2.19)

UsingLemma 2.2in last equality, we obtain

Reϕ0(z) p+ 1

λ

z

0u

(p+1)/λ−12γ31 +2

1−γ3 1 +u|z|

du

> p+ 1 λ

z

0u (p+1)/λ−1

2γ3−1 +2

1−γ3 1 +u

du

=14

A1−B1A2−B2

1−B11−B2

1−p+ 1

λ

1

0

u(p+1)/λ−1 1 +u du

=β.

(2.20)

Thus, we have obtained (2.11).

Now, we must prove that this result is sharp whenB1=B2= −1. LetB1=B2= −1. From (2.14), we can write

αf j(z)=

p+ 1 λ z

p−(p+1) z

0t

(p+1)/λ−11 +Ajt

1−t dt (2.21)

forj=1, 2. Since

1 +A1z 1−z

1 +A2z 1−z

=11 +A11 +A2+

1 +A11 +A2

1−z (2.22)

and with change of variablet=uzin (2.16),

ϕ0(z)= p+ 1

λ z (p+1)

z

0t (p+1)/λ−1

11 +A11 +A2+

1 +A11 +A2 1−uz

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If we choosezon the real axis and letz→ −1, we obtain

h0(z)−→11 +A11 +A21−p+ 1

λ

1

0

u(p+1)/λ−1 1 +u du

. (2.24)

Thus, we complete the proof ofTheorem 2.4.

Corollary2.5. Letα >0and the functionfj(z)Ꮽp(j=1, 2). If the condition

α−1f

j(z)

z

1 +z

1−z (2.25)

is satisfied, then

α−1Ω(z)

z

1 + (16 ln 29)z

1−z , (2.26)

where

Ω:=αf1f2(z), β=58 ln 2. (2.27)

Proof. By puttingp=1,λ=1,A=1,B= −1 inTheorem 2.4, we obtainCorollary 2.5.

Theorem2.6. Let0≤ρ <1. For f Ꮽp, if

Reᏼα−α1f(z)

f(z)

> ρ, (2.28)

then

αf(z)

zp γ

1

(1−z)2γ(1−ρ2)=q(z), (2.29)

whereq(z)is the best dominant.

Proof. Let f Ꮽpand

p(z) :=

αf(z)

zp γ

. (2.30)

Taking logarithmic derivative and multiplying byz, we find that

z p(z)

p(z) =(p+ 1)γ

α−1f(z)

αf(z) 1

, (2.31)

1 + 1

(p+ 1)γ z p(z)

p(z) =

α−1f(z)

αf(z) . (2.32)

Therefore, from (2.28) we write

α−1f(z)

αf(z)

1 + (12ρ)z

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Thus, using (2.33) in (2.32), we obtain

1 + 1

(p+ 1)γ z p(z)

p(z)

1 + (12ρ)z

1−z . (2.34)

Now, define the functionsθandφby

θ(w) :=1, φ(w) := 1

(p+ 1)γw, Ᏸ={w:w=0} (2.35)

inLemma 2.3. If we take

q(z) := 1

(1−z)2γ(1−ρ2), (2.36)

thenqsatisfies the conditions ofLemma 2.3. Thus, the following functions:

Q(z)=zq(z)·φq(z)= 1

(p+ 1)γ zq(z)

q(z) =

2(1−ρ)z 1−z ,

h(z)=θq(z)+Q(z)=1 + 1 (p+ 1)γ

zq(z) q(z) =

1 + (12ρ)z 1−z .

(2.37)

Since h is convex, preconditions of Lemma 2.3 are satisfied. Consequently, from

Lemma 2.3we writep≺q, andqis the best dominant.

Corollary2.7. Let0≤ρ <1. If f Ꮽpsatisfies (2.28), then

Reᏼα−1fj(z)

zp

γ/21−ρ2

>21 (2.38)

and21is the best possible.

Proof. From (2.29), there exists a Schwarz functionw(z) analytic inUwith

w(0)=0, w(z)<1 (z∈U) (2.39)

such that

αf(z)

zp γ

= 1

1−w(z)2γ

1−ρ2, (2.40)

that is,

αf(z)

zp

γ/2(1−ρ2)

=1−w(z)−γ. (2.41)

In the last equality, if we take a real part and use the following inequality:

Rew1/mRe(w)1/m Re(w)>0, (2.42)

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Acknowledgment

The present investigation was supported by Baskent University, Ankara, Turkey.

References

[1] I. B. Jung, Y. C. Kim, and H. M. Srivastava, “The Hardy space of analytic functions associated

with certain one-parameter families of integral operators,”Journal of Mathematical Analysis and

Applications, vol. 176, no. 1, pp. 138–147, 1993.

[2] S. Shams, S. R. Kulkarni, and J. M. Jahangiri, “Subordination properties of p-valent functions

defined by integral operators,”International Journal of Mathematics and Mathematical Sciences,

vol. 2006, Article ID 94572, 3 pages, 2006.

[3] S. S. Miller and P. T. Mocanu,Differential Subordinations. Theory and Applications, vol. 225 of

Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 2000.

[4] J.-L. Liu and H. M. Srivastava, “Certain properties of the Dziok-Srivastava operator,”Applied

Mathematics and Computation, vol. 159, no. 2, pp. 485–493, 2004.

¨

Oznur ¨Ozkan: Department of Statistics and Computer Sciences, Bas¸kent University, Baglıca,

06530 Ankara, Turkey

References

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