actamathscientia32b(3)(2012)929-941cho

Upload: kumarn38

Post on 05-Apr-2018

214 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    1/13

    Acta Mathematica Scientia 2012,32B(3):929941

    http://actams.wipm.ac.cn

    SANDWICH-TYPE THEOREMS FOR

    MEROMORPHIC MULTIVALENT FUNCTIONS

    ASSOCIATED WITH THE LIU-SRIVASTAVA

    OPERATOR

    Nak Eun Cho

    Department of Applied Mathematics, Pukyong National University, Busan 608-737, Korea

    E-mail: [email protected]

    Abstract The purpose of this article is to obtain some subordination and superordi-

    nation preserving properties of meromorphic multivalent functions in the punctured open

    unit disk associated with the Liu-Srivastava operator. The sandwich-type results for these

    meromorphic multivalent functions are also considered.

    Key words Subordination; superordination; meromorphic multivalent function; Liu-

    Srivastava operator; sandwich-type result

    2000 MR Subject Classification 30C80; 30C45; 30D30

    1 Introduction

    Let H = H(U) denote the class of analytic functions in the open unit disk

    U = {z C : |z| < 1}.

    For n N = {1, 2, } and a C, let

    H[a, n] = {f H : f(z) = a + anzn + an+1z

    n+1 + }.

    Let f and F be members of H. The function f is said to be subordinate to F, or F is

    said to be superordinate to f, if there exists a function w analytic in U, with w(0) = 0 and

    |w(z)| < 1 (z U), such that f(z) = F(w(z)) (z U).

    In such a case, we write

    f F (z U) or f(z) F(z) (z U).

    Received April 15, 2010; revised November 23, 2010. This research was supported by the Basic Science Re-

    search Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education,

    Science and Technology (2010-0017111).

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    2/13

    930 ACTA MATHEMATICA SCIENTIA Vol.32 Ser.B

    If the function F is univalent in U, then we have (cf. [1])

    f F (z U) f(0) = F(0) and f(U) F(U).

    Definition 1 [1] Let : C2 C and let h be univalent in U. If p is analytic in U and

    satisfies the differential subordination:

    (p(z), zp(z)) h(z) (z U), (1.1)

    then p is called a solution of the differential subordination. The univalent function q is called a

    dominant of the solutions of the differential subordination, or more simply a dominant, if p q

    for all p satisfying (1.1). A dominant q that satisfies q q for all dominants q of (1.1) is said

    to be the best dominant.

    Definition 2 [2] Let : C2 C and let h be analytic in U. If p and (p(z), zp(z)) are

    univalent in U and satisfy the differential superordination:

    h(z) (p(z), zp(z)) (z U), (1.2)

    then p is called a solution of the differential superordination. An analytic function q is called asubordinant of the solutions of the differential superordination, or more simply a subordinant if

    q p for all p satisfying (1.2). A univalent subordinant q that satisfies q q for all subordinants

    q of (1.2) is said to be the best subordinant.

    Definition 3 [2] Denote by Q the class of functions f that are analytic and injective on

    U\E(f), where

    E(f) =

    U : limz

    f(z) =

    ,

    and are such that

    f() = 0 ( U\E(f)).

    Let p denote the class of functions of the form

    f(z) = zp +

    k=1

    akzkp (p N = {1, 2, }), (1.3)

    which are analytic and p-valent in the punctured unit disk D = U\{0} with the following

    additional condition:

    limz0

    (zpf(z)) = 0 (z D). (1.4)

    For functions f p given by (1.3), and g p given by

    g(z) = zp +

    k=1

    bkzkp (p N),

    define the Hadamard product (or convolution) of f and g by

    (f g)(z) := zp +k=1

    akbkzkp.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    3/13

    No.3 Nak Eun Cho: SANDWICH-TYPE THEOREMS 931

    For parameters j C (j = 1, , l) and j C\Z

    0 (Z

    0 := 0, 1, 2, ; j = 1, , m),

    the generalized hypergeometric function lFm(1, , l; 1, , m; z) is defined by the follow-

    ing infinite series (cf. [35]):

    lFm(1, , l; 1, , m; z) :=k=0

    (1)k (l)k(1)k (m)k

    zk

    k!

    (l m + 1; l, m N0 := N {0}; z U),

    where ()k is the Pochhammer symbol (or the shifted factorial) defined (in terms of the Gamma

    function) by

    ()k :=(+ k)

    ()=

    1 if k = 0 and C\{0},(+ 1) (+ k 1) ifk N and C.

    Corresponding to a function Fp(1, , l; 1, , m; z) defined by

    Fp(1, , l; 1, , m; z) := z

    p

    lFm(1, , l; 1, , m; z),

    the Liu-Srivastava operator Hl,mp (1, , lq; 1, , m) : p p is defined by the fol-

    lowing Hadamard product (or convolution):

    Hl,mp (1, , l; 1, , m)f(z) := Fp(1, , l; 1, , m; z) f(z).

    We observe that, for a function f given by (1.3), we have

    Hl,mp (1, , l; 1, , m)f(z) =1

    zp+

    k=1

    (1)k (l)k(1)k (m)k

    akk!

    zkp.

    To make the notation simple, we write

    Hl,mp (1) := Hl,mp (1, , l; 1, , m) (l m + 1; l, m N0). (1.5)

    The operator Hl,mp (1) was defined and studied by Liu and Srivastava [6]. We also note that

    the definition of operator Hl,mp (1) was motivated essentially by Dziok and Srivastava [3].

    Some interesting developments involving the Dziok-Srivastava operator were considered by (for

    example) Dziok and Srivastava ([7], also see [812]) and Liu and Srivastava [5, 13, 14].

    Using the principle of subordination, various subordination theorems involving certain in-

    tegral operators for analytic functions in U were established in Bulboaca [15], Miller et al [16]

    and Owa and Srivastava [17]. Recently, Miller and Mocanu [2] also considered differential su-

    perordinations, as the dual problem of differential subordinations (see, also [18]). It should be

    also remarked that, in recent years, several authors obtained many interesting results involving

    various linear and nonlinear operators associated with differential subordination and superordi-

    nation [1925]. In this article, we investigate the subordination and superordination preserving

    properties of the linear operator Hl,mp (1) defined by (1.5) with the sandwich-type theorems.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    4/13

    932 ACTA MATHEMATICA SCIENTIA Vol.32 Ser.B

    2 A Set of Lemmas

    The following lemmas will be required in our present investigation.

    Lemma 1 [1] Let p Q with p(0) = a and let

    q(z) = a + anzn +

    be analytic in U withq(z) a and n 1.

    If q is not subordinate to p, then there exist points

    z0 = r0ei U and 0 U\E(f),

    for which

    q(Ur0) p(U), q(z0) = p(0) and z0q(z0) = m0p

    (0) (m n).

    A function L(z, t) defined on U [0, ) is the subordination chain (or Lowner chain) if

    L(, t) is analytic and univalent in U for all t [0, ), L(z, ) is continuously differentiable on

    [0, ) for all z U and L(z, s) L(z, t) (z U; 0 s < t).

    Lemma 2 [2] Let q H[a, 1] and : C2 C. Also set

    (q(z), zq (z)) h(z) (z U).

    If L(z, t) = (q(z),tzq(z)) is a subordination chain and p H[a, 1] Q, then,

    h(z) (p(z), zp(z)) (z U).

    implies that

    q(z) p(z) (z U).

    Furthermore, if (q(z), zp

    (z)) = h(z) has a univalent solution q Q, then q is the bestsubordinant.

    Lemma 3 [26] Suppose that the function H : C2 C satisfies the following condition:

    R{H(is, t)} 0

    for all real s and

    t n(1 + s2)/2 (n N).

    If the function p(z) = 1 + pnzn + is analytic in U and

    R{H(p(z), zp(z))} > 0 (z U),

    then,

    R{p(z)} > 0 (z U).

    Lemma 4 [27] Let , C with = 0 and let h H(U) with h(0) = c. If

    R{h(z) + } > 0 (z U),

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    5/13

    No.3 Nak Eun Cho: SANDWICH-TYPE THEOREMS 933

    then, the solution of the differential equation

    q(z) +zq (z)

    q(z) + = h(z) (z U; q(0) = c)

    is analytic in U and satisfies the inequality

    R{q(z) + } > 0 (z U).

    Lemma 5 [28] The function L(z, t) = a1(t)z + with a1(t) = 0 and limt

    |a1(t)| =

    is a subordination chain if and only if

    R

    zL(z,t)

    z

    L(z,t)t

    > 0 (z U; 0 t < ).

    3 Main Results

    We begin with proving the following subordination theorem involving the operator Hl,mp (1)

    defined by (1.5).

    Theorem 1 Let f, g p. Suppose that

    R

    1 +

    z(z)

    (z)

    > (3.1)

    (z) :=

    p

    pzpHl,mp (1 + 1)g(z) +

    pzpHl,mp (1)g(z); 1 > 0; 0 < p; z U

    ,

    where

    =(p )2 + p221 |(p )

    2 p221|

    4p(p )1(0 < 1/2). (3.2)

    Then, the following subordination relation

    p

    pzpHl,mp (1 + 1)g(z) +

    pzpHl,mp (1)f(z) (z) (z U) (3.3)

    implies thatzpHl,mp (1)f(z) z

    pHl,mp (1)g(z) (z U). (3.4)

    Moreover, the function zpHl,mp (1)g(z) is the best dominant.

    Proof Let us define the functions F and G, respectively, by

    F(z) := zpHl,mp (1)f(z) and G(z) := zpHl,mp (1)g(z). (3.5)

    We first show that, if the function q is defined by

    q(z) := 1 +zG(z)

    G(z)(z U), (3.6)

    then,

    R{q(z)} > 0 (z U).

    Taking the logarithmic differentiation on both sides of the second equation in (3.5) and

    using the equation

    z(Hl,mp (1)g(z)) = 1H

    l,mp (1 + 1)g(z) (1 + p)H

    l,mp (1)g(z),

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    6/13

    934 ACTA MATHEMATICA SCIENTIA Vol.32 Ser.B

    we obtain

    p1(z) = p1G(z) + (p )zG(z). (3.7)

    Now, by differentiating both sides of (3.7), we obtain the relationship:

    1 +z(z)

    (z)= 1 +

    zG(z)

    G(z)+

    zq (z)

    q(z) + p1/(p )

    = q(z) + zq

    (z)q(z) + p1/(p )

    h(z). (3.8)

    We also note from (3.1) that

    R

    h(z) +

    p1p

    > 0 (z U),

    and so by Lemma 4, we conclude that the differential equation (3.8) has a solution q H(U)

    with

    q(0) = h(0) = 1.

    Let us put

    H(u, v) = u +v

    u + p1/(p ) + , (3.9)

    where is given by (3.2). From (3.1), (3.8), and (3.9), we obtain

    R{H(q(z), zq (z))} > 0 (z U).

    Now, we proceed to show that

    R{H(is, t)} 0

    s R; t

    1

    2(1 + s2)

    . (3.10)

    Indeed, From (3.9), we have

    R{H(is, t)} = R

    is + tis + p1/(p )

    +

    =tp1/p

    |p1/(p ) + is|2+

    E(s)

    2|p1/(p ) + is|2, (3.11)

    where

    E(s) :=

    p1

    p 2

    s2

    p1p

    2

    p1p

    1

    . (3.12)

    For given by (3.2), we can prove easily that the expression E(s) given by (3.12) is greater

    than or equal to zero. Hence, from (3.9), we see that (3.10) holds true. Thus, using Lemma 3,

    we conclude that

    R{q(z)} > 0 (z U).

    Moreover, we see that the condition:

    G(0) = 0

    is satisfied. Hence, the function G defined by (3.5) is convex in U.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    7/13

    No.3 Nak Eun Cho: SANDWICH-TYPE THEOREMS 935

    Next, we prove that the subordination condition (3.3) implies that

    F(z) G(z) (z U) (3.13)

    for the functions F and G defined by (3.5). Without loss of generality, we can assume that G

    is analytic and univalent on U and

    G

    () = 0 ( U).

    For this purpose, we consider the function L(z, t) given by

    L(z, t) := G(z) +(p )(1 + t)

    p1zG(z) (z U; 0 t < ; 0 < 1).

    We note that

    L(z, t)

    z

    z=0

    = G(0)

    p1 + (p )(1 + t)

    1

    = 0 (0 t < ; 1 > 0; 0 < 1).

    This shows that the function

    L(z, t) = a1(t)z +

    satisfies the condition

    a1(t) = 0 (0 t < ).

    Furthermore, we have

    R

    z L(z, t)/z

    L(z, t)/t

    = R

    p1

    p + (1 + t)

    1 +

    zG(z)

    G(z)

    > 0.

    Therefore, by virtue of Lemma 5, L(z, t) is a subordination chain. We observe from the

    definition of subordination chain that

    L(, t) L(U, 0) = (U) ( U; 0 t < ).

    Now, suppose that F is not subordinate to G, then by Lemma 1, there exists points z0 U

    and 0 U, such that

    F(z0) = G(0) and z0F(z0) = (1 + t)0G(0) (0 t < ).

    Hence, we have

    L(0, t) = G(0) +(p )(1 + t)

    p10G

    (0) = F(z0) +p

    p1z0F

    (z0)

    =p

    pz0(H

    l,mp (1 + 1)f(z0)) +

    pz0(H

    l,mp (1)f(z0)) (U),

    by virtue of the subordination condition (3.3). This contradicts the above observation L(0, t)

    (U). Therefore, the subordination condition (3.3) must imply the subordination given by

    (3.13). Considering F(z) = G(z), we see that the function G is the best dominant. This

    evidently completes the proof of Theorem 1.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    8/13

    936 ACTA MATHEMATICA SCIENTIA Vol.32 Ser.B

    We next provide a dual problem of Theorem 1, in the sense that the subordinations are

    replaced by superordinations.

    Theorem 2 Let f, g p. Suppose that

    R

    1 +

    z(z)

    (z)

    >

    (z) := p

    zpHl,mp (1 + 1)g(z) + p

    zpHl,mp (1)g(z); 1 > 0; 0 < p; z U

    ,

    where is given by (3.2), and

    p

    zpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z)

    is univalent in U and zpHl,mp (1)f(z) H[1, 1]Q. Then, the following superordination relation

    (z) p

    zpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z) (z U) (3.14)

    implies that

    zp

    Hl,mp (1)g(z) z

    p

    Hl,mp (1)f(z) (z U).

    Moreover, the function zpHl,mp (1)g(z) is the best subordinant.

    Proof The first part of the proof is similar to that of Theorem 1 and so we will use the

    same notation as in the proof of Theorem 1.

    Now, let us define the functions F and G by (3.5). We first note that, if the function q is

    defined by (3.6), using (3.7), then we obtain

    (z) = G(z) +p

    p1zG(z) =: (G(z), zG(z)). (3.15)

    Then using the same method as in the proof of Theorem 1, we can prove that

    R{q(z)} > 0 (z U),

    that is, G defined by (3.5) is convex(univalent) in U.

    Next, we prove that the subordination condition (3.14) implies that

    G(z) F(z) (z U) (3.16)

    for the functions F and G defined by (3.5). Now, consider the function L(z, t) defined by

    L(z, t) := G(z) +(p )t

    p1zG(z) (z U; 0 t < ).

    As G is convex and p1/(p ) > 0, we can prove easily that L(z, t) is a subordination

    chain as in the proof of Theorem 1. Therefore, according to Lemma 2, we conclude that the

    superordination condition (3.14) must imply the superordination given by (3.16). Furthermore,

    as the differential equation (3.15) has the univalent solution G, it is the best subordinant of the

    given differential superordination. Therefore, we complete the proof of Theorem 2.

    If we combine Theorems 1 and 2, then, we obtain the following sandwich-type theorem.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    9/13

    No.3 Nak Eun Cho: SANDWICH-TYPE THEOREMS 937

    Theorem 3 Let f, gk p (k = 1, 2). Suppose that

    R

    1 +

    zk(z)

    k(z)

    > (3.17)

    k(z) : =p

    pzpHl,mp (1 + 1)gk(z) +

    pzpHl,mp (1)gk(z);

    k = 1, 2; 1 > 0; 0 < p; z U

    ,

    where is given by (3.2), and

    p

    pzpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z)

    is univalent in U and zpHl,mp (1)f(z) H[1, 1] Q. Then, the following relation

    1(z) p

    pzpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z) 2(z) (z U)

    implies that

    zpHl,mp (1)g1(z) zpHl,mp (1)f(z) z

    pHl,mp (1)g2(z) (z U).

    Moreover, the functions zpHl,mp (1)g1(z) and zpHl,mp (1)g2(z) are the best subordinant

    and the best dominant, respectively.

    The assumption of Theorem 3, that the functions

    p

    pzpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z) and z

    pHl,mp (1)f(z)

    need to be univalent in U, may be replaced by another condition in the following result.

    Corollary 1 Let f, gk p (k = 1, 2). Suppose that condition (3.17) is satisfied and

    R

    1 + z

    (z)(z)

    > (3.18)

    (z) :=

    p

    pzpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z); 1 > 0; 0 < p; z U

    ,

    where is given by (3.2). Then, the following relation

    1(z) p

    pzpHl,mp (1 + 1)f(z) +

    pzpHl,mp (1)f(z) 2(z) (z U)

    implies that

    zpHl,mp (1)g1(z) zpHl,mp (1)f(z) z

    pHl,mp (1)g2(z) (z U).

    Moreover, the functions zpHl,mp (1)g1(z) and zpHl,mp (1)g2(z) are the best subordinant

    and the best dominant, respectively.

    Proof To prove Corollary 1, we have to show that condition (3.18) implies the univalence

    of (z) and

    F(z) := zpHl,mp (1)f(z).

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    10/13

    938 ACTA MATHEMATICA SCIENTIA Vol.32 Ser.B

    As 0 < 1/2 from Theorem 1, condition (3.18) means that is a close-to-convex function

    in U (see [29]) and hence is univalent in U. Furthermore, using the same techniques as in

    the proof of Theorem 1, we can prove the convexity(univalence) of F and so the details may be

    omitted here. Therefore, by applying Theorem 3, we obtain Corollary 1.

    Taking q = s + 1, 1 = = p, i = i (i = 2, 3, , s), s+1 = 1, and = 0 in Theorem

    3, we have the following result.

    Corollary 2 Let f, g p. Suppose that

    R

    1 +

    zk(z)

    k(z)

    >

    1

    2p(z U; k(z) := z

    p(zg k(z) + (1 + p)gk(z)); k = 1, 2) ,

    and zp(zf(z) + (1 + p)f(z)) is univalent in U, and zpf(z) H[1, 1] Q. Then,

    zp(zg 1(z) + (1 + p)g1(z)) zp(zf(z) + (1 + p)f(z)) zp(zg 2(z) + (1 + p)g2(z)) (z U)

    implies that

    zpg1(z) zpf(z) zpg2(z) (z U).

    Moreover, the functions zpg1(z) and zpg2(z) are the best subordinant and the best domi-

    nant, respectively.The proof of Theorem 4 below is similar to that of Theorem 3 using (1.6), and so it is

    omitted.

    Theorem 4 Let f, gk p (k = 1, 2). Suppose that

    R

    1 +

    zk(z)

    k(z)

    >

    k(z) : =

    p

    pzp+1Hl,mp (1 + 1)gk(z) +

    pzp+1Hl,mp (1)gk(z);

    k = 1, 2; 1 > (p )/p; 0 < p; z U,where

    =(p )2 + (p(1 1) + )2 |(p )2 (p(1 1) + )2|

    4(p )(p(1 1) + ),

    andp

    pzp+1Hl,mp (1 + 1)f(z) +

    pzp+1Hl,mp (1)f(z)

    is univalent in U and zp+1(Hl,mp (1)f(z)) H[0, 1] Q. Then, the following relation

    1(z) p

    pzp+1Hl,mp (1 + 1)f(z) +

    pzp+1Hl,mp (1)f(z) 2(z) (z U)

    implies that

    zpHl,mp (1)g1(z) zpHl,mp (1)f(z) z

    pHl,mp (1)g2(z) (z U).

    Moreover, the functions zpHl,mp (1)g1(z) and zpHl,mp (1)g2(z) are the best subordinant

    and the best dominant, respectively.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    11/13

    No.3 Nak Eun Cho: SANDWICH-TYPE THEOREMS 939

    Next, we consider the integral operator F ( > 0) defined by (cf. [3032])

    Fu(f)(z) :=

    z+p

    z0

    t+p1f(t)dt (f p; > 0) (3.19)

    Now, we obtain the following result involving the integral operator defined by (3.19).

    Theorem 5 Let f, gk p(k = 1, 2). Suppose also that

    R

    1 +

    zk(z)

    k(z)

    >

    k(z) := z

    pHl,mp (1)gk(z); k = 1, 2; z U

    , (3.20)

    where

    =1 + 2 |1 2|

    4( > 0), (3.21)

    and zpHl,mp (1)f(z) is univalent in U and zpHl,mp (1)F(f)(z) H[1, 1] Q. Then, the follow-

    ing relation

    1(z) zpHl,mp (1)f(z) 2(z) (z U)

    implies that

    zp

    Hl,m

    p (1)F(g1)(z) zp

    Hl,m

    p (1)F(f)(z) zp

    Hl,m

    p (1)F(g2)(z) (z U).

    Moreover, the functions zpHl,mp (1)F(g1)(z) and zpHl,mp (1)F(g2)(z) are the best sub-

    ordinant and the best dominant, respectively.

    Proof Let us define the functions F and Gk (k = 1, 2) by

    F(z) := zpHl,mp (1)F(f)(z) and Gk(z) := zpHl,mp (1)F(gk)(z),

    respectively. Without loss of generality, as in the proof of Theorem 1, we can assume that Gk

    is analytic and univalent on U and

    Gk() = 0 ( U).

    From the definition of the integral operator F defined by (3.19), we obtain

    z(Hl,mp (1)F(f)(z)) = Hl,mp (1)f(z) ( + p)H

    l,mp (1)F(f)(z). (3.22)

    Then, from (3.20) and (3.22), we have

    k(z) = Gk(z) + zG

    k(z). (3.23)

    Setting

    qk(z) = 1 +zGk(z)

    Gk(z)(k = 1, 2; z U)

    and differentiating both sides of (3.23), we obtain

    1 +zk(z)

    k(z)= qk(z) +

    zq k(z)

    qk(z) + .

    The remaining part of the proof is similar to that of Theorem 1 and so is omitted the proof

    involved.

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    12/13

    940 ACTA MATHEMATICA SCIENTIA Vol.32 Ser.B

    Using the same methods as in the proof of Corollary 1, we have the following result.

    Corollary 3 Let f, gk p(k = 1, 2). Suppose that condition (3.20) is satisfied and

    R

    1 +

    z (z)

    (z)

    >

    (z) := zpHl,mp (1)f(z) : z U

    ,

    where is given by (3.21). Then,

    1(z) zpHl,mp (1)f(z) 2(z) (z U)

    implies that

    zpHl,mp (1)F(g1)(z) zpHl,mp (1)F(f)(z) z

    pHl,mp (1)F(g2)(z) (z U).

    Moreover, the functions zpHl,mp (1)F(g1)(z) and zpHl,mp (1)F(g2)(z) are the best sub-

    ordinant and the best dominant, respectively.

    Taking q = s + 1, 1 = 1 = p, i = i (i = 2, 3, , s), and s+1 = 1 in Theorem 5, we

    have the following result.

    Corollary 4 Let f, gk p (k = 1, 2). Suppose also that

    R

    1 + z

    k(z)k(z)

    > (z U; k(z) := zpgk(z); k = 1, 2) ,

    where is given by (3.21), and zpf(z) is univalent in U and zp(F(f)(z)) H[1, 1] Q, then,

    zpg1(z) zpf(z) zpg2(z) (z U)

    implies that

    zp(F(g1)(z)) zp(F(f)(z)) z

    p(F(g2)(z)) (z U).

    Moreover, the functions zp(F(g1)(z)) and zp(F(g2)(z)) are the best subordinant and the

    best dominant, respectively.

    References

    [1] Miller S S, Mocanu P T. Differential Subordination, Theory and Applications. New York, Basel: Marcel

    Dekker Inc, 2000

    [2] Miller S S, Mocanu P T. Subordinants of differential superordinations. Complex Var Theory Appl, 2003,

    48: 815826

    [3] Dziok J, Srivastava H M. Classes of analytic functions associated with the generalized hypergeometric

    function. Appl Math Comput, 1999, 103: 113

    [4] Dziok J, Srivastava H M. Certain subclasses of analytic functions associated with the generalized hyper-

    geometric function. Integral Transforms Spec Funct, 2003, 14: 718

    [5] Liu J L, Srivastava H M. Certain properties of the Dziok-Srivastava operator. Appl Math Comput, 2004,

    159: 485493

    [6] Liu J L, Srivastava H M. Classes of meromorphically multivalent functions associated with the generalized

    hypergeometric function. Math Comput Modelling, 2004, 39: 2134

    [7] Dziok J, Srivastava H M. Some subclasses of analytic functions with fixed argument of coefficients associ-

    ated with the generalized hypergeometric function. Adv Stud Contemp Math, 2002, 5: 115125

    [8] Liu J L. On subordinations for certain analytic functions associated with the Dziok-Srivastava linear

    operator. Taiwanese J Math, 2009, 13: 349357

    [9] Patel J, Mishra A K, Srivastava H M. Classes of multivalent analytic functions involving the Dziok-

    Srivastava operator. Comput Math Appl, 2007, 54: 599616

  • 7/31/2019 ActaMathScientia32B(3)(2012)929-941Cho

    13/13

    No.3 Nak Eun Cho: SANDWICH-TYPE THEOREMS 941

    [10] Ramachandran C, Shanmugam T N, Srivastava H M, Swaminathan A. A unified class of k-uniformly

    convex functions defined by the Dziok-Srivastava operator. Appl Math Comput, 2007, 190: 16271636

    [11] Srivastava H M, Owa S. Some characterizations and distortions theorems involving fractional calculus,

    generalized hypergeometric functions, Hadamard products, linear operators, and certain subclasses of

    analytic functions. Nagoya Math J, 1987, 106: 128

    [12] Owa S, Srivastava H M. Univalent and starlike generalized hypergeometric functions. Canad J Math, 1987,

    39: 10571077

    [13] Liu J L, Srivastava H M. A linear operator and associated families of meromorphically multivalent func-

    tions. J Math Anal Appl, 2001, 259: 566581

    [14] Liu J L, Srivastava H M. Certain properties of the Dziok-Srivastava operator. Appl Math Comput, 2004,

    159: 485493

    [15] Bulboaca T. Integral operators that preserve the subordination. Bull Korean Math Soc, 1997, 32: 627636

    [16] Miller S S, Mocanu P T, Reade M O. Subordination-preserving integral operators. Trans Amer Math Soc,

    1984, 283: 605615

    [17] Owa S, Srivastava H M. Some subordination theorems involving a certain family of integral operators.

    Integral Transforms Spec Funct, 2004, 15: 445454

    [18] Bulboaca T. A class of superordination-preserving integral operators. Indag Math N S, 2002, 13: 301311

    [19] Ali R M, Ravichandran V, Seenivasagan N. Subordination and superordination of the Liu-Srivastava linear

    operator on meromorphic functions. Bull Malays Math Sci Soc (2), 2008, 31(2): 192207

    [20] Ali R M, Ravichandran V, Seenivasagan N. Subordination and superordination on Schwarzian derivatives.

    J Inequal Appl, 2008, Art. ID 712328, 18 pp

    [21] Ali R M, Ravichandran V, Seenivasagan N. Differential subordination and superordination of analyticfunctions defined by the multiplier transformation. Math Inequal Appl, 2009, 12: 123139

    [22] Ali R M, Ravichandran V, Seenivasagan N. Differential subordination and superordination for meromorphic

    functions defined by certain multiplier transformation. Bull Malays Math Sci Soc (2), 2010, 33(2): 311324

    [23] Cho N E, Kwon O S, Owa S, Srivastava H M. A class of integral operators preserving subordination and

    superordination for meromorphic functions. Appl Math Comput, 2007, 193: 463474

    [24] Cho N E, Kwon O S. A class of integral operators preserving subordination and superordination. Bull

    Malays Math Sci Soc (2), 2010, 33(3): 429437

    [25] Wang Z G, Xiang R G, Darus M. A family of integral operators preserving subordination and superordi-

    nation. Bull Malays Math Sci Soc (2), 2010, 33(1): 121131

    [26] Miller S S, Mocanu P T. Differential subordinations and univalent functions. Michigan Math J, 1981, 28:

    157171

    [27] Miller S S, Mocanu P T. Univalent solutions of Briot-Bouquet differential equations. J Different Eq, 1985,

    567: 297309

    [28] Pommerenke C. Univalent Functions. Vanderhoeck and Ruprecht: Gottingen, 1975

    [29] Kaplan W. Close-to-convex schlicht functions. Michigan Math J, 1952, 2: 169185

    [30] Goel R M, Sohi N S. On a class of meromorphic functions. Glas Mat, 1982, 17(37): 1928

    [31] Kumar V, Shukla S L. Certain integrals for classes of p-valent functions. Bull Austral Math Soc, 1982,

    25: 8597

    [32] Srivastava H M, Owa S. Current Topics in Analytic Function Theory. Singapore, New Jersey, London,

    and Hong Kong: World Scientific Publishing Company, 1992