Applications Of Differential Subordination To Certain ... - EMIS

Journal of Inequalities in Pure and Applied Mathematics APPLICATIONS OF DIFFERENTIAL SUBORDINATION TO CERTAIN SUBCLASSES OF MEROMORPHICALLY MULTIVALENT FUNCTIONS volume 6, issue 3, article 88, 2005.

H.M. SRIVASTAVA AND J. PATEL Department of Mathematics and Statistics University of Victoria Victoria, British Columbia V8W 3P4 Canada. EMail: [email protected] Department of Mathematics Utkal University Vani Vihar, Bhubaneswar 751004 Orissa, India. EMail: [email protected]

Received 18 April, 2005; accepted 05 July, 2005. Communicated by: Th.M. Rassias

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2000 Victoria University ISSN (electronic): 1443-5756 210-05

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Abstract By making use of the principle of differential subordination, the authors investigate several inclusion relationships and other interesting properties of certain subclasses of meromorphically multivalent functions which are defined here by means of a linear operator. They also indicate relevant connections of the various results presented in this paper with those obtained in earlier works. 2000 Mathematics Subject Classification: Primary 30C45; Secondary 30D30, 33C20. Key words: Meromorphic functions, Differential subordination, Hadamard product (or convolution), Multivalent functions, Linear operator, Hypergeometric function.

The present investigation was supported, in part, by the Natural Sciences and Engineering Research Council of Canada under Grant OGP0007353 and, in part, by the University Grants Commission of India under its DRS Financial Assistance Program.

Contents 1 2 3

Introduction and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Preliminary Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 The Main Subordination Theorems and The Associated Functional Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 References

An Inequality of Ostrowski Type via Pompeiu’s Mean Value Theorem H.M. Srivastava and J. Patel

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1.

Introduction and Definitions

For any integer m > −p, let Σp,m denote the class of all meromorphic functions f (z) normalized by f (z) = z

(1.1)

−p

+

∞ X

ak z k

(p ∈ N := {1, 2, 3, · · · }),

k=m

which are analytic and p-valent in the punctured unit disk ∗

U = {z : z ∈ C

and 0 < |z| < 1} = U \ {0}.

For convenience, we write


Σp,−p+1 = Σp . If f (z) and g(z) are analytic in U, we say that f (z) is subordinate to g(z), written symbolically as follows: f ≺g

in

U

or

f (z) ≺ g(z)

(z ∈ U),

if there exists a Schwarz function w(z), which (by definition) is analytic in U with w(0) = 0 and |w(z)| < 1 (z ∈ U) such that

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Indeed it is known that (z ∈ U) =⇒ f (0) = g(0)

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f (z) = g w(z) (z ∈ U).

f (z) ≺ g(z)

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and

f (U) ⊂ g(U).

J. Ineq. Pure and Appl. Math. 6(3) Art. 88, 2005


In particular, if the function g(z) is univalent in U, we have the following equivalence (cf., e.g., [5]; see also [6, p. 4]): f (z) ≺ g(z) (z ∈ U)

⇐⇒

f (0) = g(0)

and f (U) ⊂ g(U).

For functions f (z) ∈ Σp,m , given by (1.1), and g(z) ∈ Σp,m defined by g(z) = z

(1.2)

−p

+

∞ X

bk z k

(m > −p; p ∈ N),

k=m

we define the Hadamard product (or convolution) of f (z) and g(z) by (f ? g)(z) := z −p +

(1.3)

∞ X

ak bk z k =: (g ? f )(z)


k=m

(m > −p; p ∈ N; z ∈ U). Following the recent work of Liu and Srivastava [3], for a function f (z) in the class Σp,m , given by (1.1), we now define a linear operator Dn by D0 f (z) = f (z), 0 p+1 z f (z) D1 f (z) = z −p + (p + k + 1)ak z k = , zp k=m ∞ X

and (in general)

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Dn f (z) = D Dn−1 f (z) = z −p +

∞ X

(p + k + 1)n ak z k

=

z

p+1

n−1

D zp

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k=m

(1.4)

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0 f (z)

(n ∈ N).



It is easily verified from (1.4) that 0 (1.5) z Dn f (z) = Dn+1 f (z) − (p + 1)Dn f (z) f ∈ Σp,m ; n ∈ N0 := N ∪ {0} . The case m = 0 of the linear operator Dn was introduced recently by Liu and Srivastava [3], who investigated (among other things) several inclusion relationships involving various subclasses of meromorphically p-valent functions, which they defined by means of the linear operator Dn (see also [2]). A special case of the linear operator Dn for p = 1 and m = 0 was considered earlier by Uralegaddi and Somanatha [13]. Aouf and Hossen [1] also obtained several results involving the operator Dn for m = 0 and p ∈ N. Making use of the principle of differential subordination as well as the linear operator Dn , we now introduce a subclass of the function class Σp,m as follows. Definition. For fixed parameters A and B (−1 5 B < A 5 1), we say that a function f (z) ∈ Σp,m is in the class Σnp,m (A, B), if it satisfies the following subordination condition: 0 z p+1 Dn f (z) 1 + Az (1.6) − ≺ (n ∈ N0 ; z ∈ U). p 1 + Bz


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In view of the definition of differential subordination, (1.6) is equivalent to the following condition: z p+1 Dn f (z)0 + p 0 < 1 (z ∈ U). p+1 n Bz D f (z) + pA

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For convenience, we write Σnp,m

2α 1− , −1 = Σnp,m (α) , p

where Σnp,m (α) denotes the class of functions in Σp,m satisfying the following inequality: 0 < − z p+1 Dn f (z) > α (0 5 α < p; z ∈ U). An Inequality of Ostrowski Type via Pompeiu’s Mean Value Theorem

In particular, we have Σnp,0 (A, B) = Rn,p (A, B), where Rn,p (A, B) is the function class introduced and studied by Liu and Srivastava [3]. The function class H(p; A, B), considered by Mogra [7], happens to be a further special case of the Liu-Srivastava class Rn,p (A, B) when n = 0. In the present paper, we derive several inclusion relationships for the function class Σnp,m (A, B) and investigate various other properties of functions belonging to the class Σnp,m (A, B), which we have defined here by means of the linear operator Dn . These include (for example) some mapping properties involving the linear operator Dn . Relevant connections of the results presented in this paper with those obtained in earlier works are also pointed out.

H.M. Srivastava and J. Patel

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2.

Preliminary Lemmas

In proving our main results, we need each of the following lemmas. Lemma 1 (Miller and Mocanu [5]; see also [6]). Let the function h(z) be analytic and convex (univalent) in U with h(0) = 1. Suppose also that the function φ(z) given by (2.1)

φ(z) = 1 + cp+m z p+m + cp+m+1 z p+m+1 + · · · An Inequality of Ostrowski Type via Pompeiu’s Mean Value Theorem

is analytic in U. If (2.2)

φ(z) +

zφ0 (z) ≺ h(z) γ

γ (0 5 γ < 1; z ∈ U).

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Lemma 2 (cf., e.g., Pashkouleva [9]). Let the function ϕ(z), given by (2.3), be in the class P(γ). Then 2


(z ∈ U).

Then, for any function F (z) analytic in U, (Φ?F )(U) is contained in the convex hull of F (U).

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3.

The Main Subordination Theorems and The Associated Functional Inequalities

Unless otherwise mentioned, we shall assume throughout the sequel that m is an integer greater than −p, and that −1 5 B < A 5 1, λ > 0, n ∈ N0 , and p ∈ N. Theorem 1. Let the function f (z) defined by (1.1) satisfy the following subordination condition: 0 0 (1 − λ)z p+1 Dn f (z) + λz p+1 Dn+1 f (z) 1 + Az − ≺ (z ∈ U). p 1 + Bz Then (3.1)

0 z p+1 Dn f (z) 1 + Az − ≺ Q(z) ≺ p 1 + Bz


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(z ∈ U),

where the function Q(z) given by  A A 1 Bz −1   B + 1 − B (1 + Bz) 2 F1 1, 1; λ(p+m) + 1; 1+Bz Q(z) =  A 1 + z λ(p+m)+1 is the best dominant of (3.1). Furthermore, 0 ! z p+1 Dn f (z) >ρ (3.2) < − p

An Inequality of Ostrowski Type via Pompeiu’s Mean Value Theorem

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(z ∈ U),



where

ρ=

 A A 1 B  (1 − B)−1 2 F1 1, 1; λ(p+m) + 1− B + 1; B−1 B  1 −

A λ(p+m)+1

(B 6= 0) (B = 0).

The inequality in (3.2) is the best possible. Proof. Consider the function φ(z) defined by 0 z p+1 Dn f (z) (3.3) φ(z) = − p


(z ∈ U).


Then φ(z) is of the form (2.1) and is analytic in U. Applying the identity (1.5) in (3.3) and differentiating the resulting equation with respect to z, we get

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0 0 (1 − λ)z p+1 Dn f (z) + λz p+1 Dn+1 f (z) − p = φ(z) + λ zφ0 (z) ≺

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1 + Az 1 + Bz

Now, by using Lemma 1 for γ = 1/λ, we deduce that 0 z p+1 Dn f (z) − ≺ Q(z) p Z z 1 1 + At − 1/λ(p+m) 1/λ(p+m) −1 = z t dt λ(p + m) 1 + Bt 0

(z ∈ U).

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=

 A 1 Bz A  + 1− B (1 + Bz)−1 2 F1 1, 1; λ(p+m) + 1; 1+Bz B  1 +

A λ(p+m)+1

z

(B 6= 0) (B = 0),

by change of variables followed by the use of the identities (2.4), (2.5), and (2.6) (with b = 1 and c = a + 1). This proves the assertion (3.1) of Theorem 1. Next, in order to prove the assertion (3.2) of Theorem 1, it suffices to show that (3.4) inf < Q(z) = Q(−1). |z| α (0 5 α < p; z ∈ U), < −z then

π < − z p+1 f 0 (z) > α + (p − α) −1 2 The result is the best possible.

(z ∈ U).

Remark 1. From Corollary 2, we note that, if f (z) ∈ Σp,−p+2 satisfies the following inequality: p(π − 2) < − z p+1 {(p + 2)f 0 (z) + zf 00 (z)} > − (z ∈ U), 4−π


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then < −z

p+1 0

f (z) > 0

(z ∈ U).

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This result is the best possible. The result (asserted by Remark 1 above) was also obtained by Pap [8]. Theorem 2. If f (z) ∈ Σnp,m (α) (0 5 α < p), then n 0 o 0 (3.5) < − z p+1 (1 − λ) Dn f (z) + λ Dn+1 f (z) >α

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(|z| < R),

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where

1 p+m p 2 2 R= 1 + λ (p + m) − λ(p + m) .

The result is the best possible.

Page 14 of 31 J. Ineq. Pure and Appl. Math. 6(3) Art. 88, 2005


Proof. We begin by writing 0 −z p+1 Dn f (z) = α + (p − α)u(z)

(3.6)

(z ∈ U).

Then, clearly, u(z) is of the form (2.1), is analytic in U, and has a positive real part in U. Making use of the identity (1.5) in (3.6) and differentiating the resulting equation with respect to z, we observe that 0 0 p+1 n n+1 z (1 − λ) D f (z) + λ D f (z) +α (3.7) − = u(z) + λzu0 (z). p−α Now, by applying the following estimate [4]: |zu0 (z)| 2(p + m)rp+m 5 0; z ∈ U∗ ). z 0 Then (3.10)

0 z p+1 Dn Fδ,p (f )(z) 1 + Az − ≺ Θ(z) ≺ p 1 + Bz

(z ∈ U),

where the function Θ(z) given by  A A Bz δ −1   B + 1 − B (1 + Bz) 2 F1 1, 1; p+m + 1; 1+Bz Θ(z) =  1 + Aδ z δ+p+m is the best dominant of (3.10). Furthermore, 0 ! z p+1 Dn Fδ,p (f )(z) >κ (3.11) < − p


(B 6= 0) (B = 0)

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(z ∈ U),

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where κ=


 A A δ B −1   B + 1 − B (1 − B) 2 F1 1, 1; p+m + 1; B−1  1 −

Aδ δ+p+m


Close

(B 6= 0) (B = 0).



Proof. Setting 0 z p+1 Dn Fδ,p (f )(z) φ(z) = − p

(3.12)

(z ∈ U),

we note that φ(z) is of the form (2.1) and is analytic in U. Using the following operator identity: 0 (3.13) z Dn Fδ,p (f )(z) = δ Dn f (z) − (δ + p) Dn Fδ,p (f )(z) in (3.12), and differentiating the resulting equation with respect to z, we find that 0 z p+1 Dn f (z) zφ0 (z) 1 + Az − = φ(z) + ≺ (z ∈ U). p δ 1 + Bz Now the remaining part of Theorem 3 follows by employing the techniques that we used in proving Theorem 1 above.


Setting m = 0 in Theorem 3, we obtain the following result which improves the corresponding work of Liu and Srivastava [3, Theorem 2].

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Corollary 4. If δ > 0 and f (z) ∈ Rn,p (A, B), then Fδ,p (f )(z) ∈ Rn,p (1 − 2ξ, −1) ⊂ Rn,p (A, B),


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where

Close

ξ=

 A A B  B + 1− B (1 − B)−1 2 F1 1, 1; pδ + 1; B−1  1 −

Aδ δ+p


(B 6= 0) (B = 0).



Remark 2. By observing that z

(3.14)

p+1

Z z 0 0 δ D Fδ,p (f )(z) = δ tδ+p Dn f (t) dt z 0 (f ∈ Σp,m ; z ∈ U), n

Corollary 4 can be restated as follows. If δ > 0 and f (z) ∈Rn,p (A, B), then Z z 0 δ δ+p n < − δ t D f (t) dt > ξ pz 0

(z ∈ U),


where ξ is given as in Corollary 4. In view of (3.14), Theorem 3 for 2α A=1− , B = −1, and n = 0 p yields


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Corollary 5. If δ > 0 and if f (z) ∈ Σp,m satisfies the following inequality: p+1 0 < − z f (z) > α (0 5 α < p; z ∈ U),

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then

δ < − δ z

Z

z

Close

f (t)dt

δ+p 0

t

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0

> α + (p − α) The result is the best possible.

2 F1

1, 1;

1 δ + 1; p+m 2

−1

(z ∈ U).



Theorem 4. Let f (z) ∈ Σp,m . Suppose also that g(z) ∈ Σp,m satisfies the following inequality: < z p Dn g(z) > 0 (z ∈ U). If n D f (z) Dn g(z) − 1 < 1 then

(n ∈ N0 ; z ∈ U),

0 ! z Dn f (z) < − >0 Dn f (z)

(|z| < R0 ),


where

p 9(p + m)2 + 4p(2p + m) − 3(p + m) R0 = . 2(2p + m)

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Proof. Letting (3.15)

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Dn f (z) w(z) = n − 1 = κp+m z p+m + κp+m+1 z p+m+1 + · · · , D g(z)

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we note that w(z) is analytic in U, with w(0) = 0 and w(z) 5 |z|p+m

Close

(z ∈ U).

Then, by applying the familiar Schwarz lemma, we get w(z) = z p+m Ψ(z),



where the function Ψ(z) is analytic in U and Ψ(z) 5 1 (z ∈ U). Therefore, (3.15) leads us to (3.16)

Dn f (z) = Dn g(z) 1 + z p+m Ψ(z)

(z ∈ U).

Making use of logarithmic differentiation in (3.16), we obtain 0 0 z Dn f (z) z Dn g(z) z p+m (p + m)Ψ(z) + zΨ0 (z) (3.17) = + . Dn f (z) Dn g(z) 1 + z p+m Ψ(z) p


n

Setting φ(z) = z D g(z), we see that the function φ(z) is of the form (2.1), is analytic in U, 0 (z ∈ U), and


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0 z Dn g(z) zφ0 (z) = − p, Dn g(z) φ(z)

so that we find from (3.17) that 0 ! z Dn f (z) (3.18) < − Dn f (z) 0 p+m 0 zφ (z) z (p + m)Ψ(z) + zΨ (z) − = p − φ(z) 1 + z p+m Ψ(z)

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(z ∈ U). J. Ineq. Pure and Appl. Math. 6(3) Art. 88, 2005


Now, by using the following known estimates [10] (see also [4]): 0 φ (z) 2(p + m)rp+m−1 and φ(z) 5 1 − r2(p+m) (p + m)Ψ(z) + zΨ0 (z) 5 (p + m) (|z| = r < 1) 1 − rp+m 1 + z p+m Ψ(z) in (3.18), we obtain 0 ! z Dn f (z) p − 3(p + m)rp+m − (2p + m)r2(p+m) < − = Dn f (z) 1 − r2(p+m)

(|z| = r < 1),


which is certainly positive, provided that r < R0 , R0 being given as in Theorem 4. Theorem 5. Let −1 5 Bj < Aj 5 1 (j = 1, 2). If each of the functions fj (z) ∈ Σp satisfies the following subordination condition: (3.19) (1 − λ)z p Dn fj (z) + λz p Dn+1 fj (z) ≺

1 + Aj z 1 + Bj z

(j = 1, 2; z ∈ U),

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then (3.20)

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(1 − λ)z p Dn H(z) + λz p Dn+1 H(z) ≺

1 + (1 − 2η)z 1−z

(z ∈ U),

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where H(z) = Dn (f1 ? f2 )(z)



and 4(A1 − B1 )(A2 − B2 ) 1 1 1 η =1− 1 − 2 F1 1, 1; + 1; . (1 − B1 )(1 − B2 ) 2 λ 2 The result is the best possible when B1 = B2 = −1. Proof. Suppose that each of the functions fj (z) ∈ Σp (j = 1, 2) satisfies the condition (3.19). Then, by letting (3.21)

ϕj (z) = (1 − λ)z p Dn fj (z) + λz p Dn+1 fj (z)

we have

(j = 1, 2),

1 − Aj ϕj (z) ∈ P(γj ) γj = ; j = 1, 2 . 1 − Bj By making use of the operator identity (1.5) in (3.21), we observe that Z 1 −p−(1/λ) z (1/λ)−1 n D fj (z) = z t ϕj (t)dt (j = 1, 2), λ 0

which, in view of the definition of H(z) given already with (3.20), yields Z 1 −p−(1/λ) z (1/λ)−1 n t ϕ0 (t)dt, (3.22) D H(z) = z λ 0 where, for convenience,


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p

(3.23)


n

p

n+1

ϕ0 (z) = (1 − λ)z D H(z) + λz D H(z) Z 1 −(1/λ) z (1/λ)−1 t (ϕ1 ? ϕ2 )(t)dt. = z λ 0



Since ϕ1 (z) ∈ P(γ1 ) and ϕ2 (z) ∈ P(γ2 ), it follows from Lemma 3 that (3.24) (ϕ1 ? ϕ2 )(z) ∈ P(γ3 ) γ3 = 1 − 2(1 − γ1 )(1 − γ2 ) . Now, by using (3.24) in (3.23) and then appealing to Lemma 2 and Lemma 4, we get Z 1 1 (1/λ)−1 < ϕ0 (z) = u < (ϕ1 ? ϕ2 ) (uz)du λ 0 Z 1 1 (1/λ)−1 2(1 − γ3 ) = u 2γ3 − 1 + du λ 0 1 + u|z| Z 2(1 − γ3 ) 1 1 (1/λ)−1 u 2γ3 − 1 + du > λ 0 1+u Z 4(A1 − B1 )(A2 − B2 ) 1 1 (1/λ)−1 −1 =1− 1− u (1 + u) du (1 − B1 )(1 − B2 ) λ 0 4(A1 − B1 )(A2 − B2 ) 1 1 1 =1− 1 − 2 F1 1, 1; + 1; (1 − B1 )(1 − B2 ) 2 λ 2 = η (z ∈ U).


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When B1 = B2 = −1, we consider the functions fj (z) ∈ Σp (j = 1, 2), which satisfy the hypothesis (3.19) of Theorem 5 and are defined by Z 1 −(1/λ) z (1/λ)−1 1 + Aj t n D fj (z) = z t dt (j = 1, 2). λ 1−z 0

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Thus it follows from (3.23) and Lemma 4 that Z 1 1 (1/λ)−1 (1 + A1 )(1 + A2 ) ϕ0 (z) = u 1 − (1 + A1 )(1 + A2 ) + du λ 0 1 − uz = 1 − (1 + A1 )(1 + A2 ) 1 z −1 + (1 + A1 )(1 + A2 )(1 − z) 2 F1 1, 1; + 1; λ z−1 −→ 1 − (1 + A1 )(1 + A2 ) 1 1 1 + (1 + A1 )(1 + A2 ) 2 F1 1, 1; + 1; as z → −1, 2 λ 2


which evidently completes the proof of Theorem 5. Title Page

By setting Aj = 1 − 2αj , Bj = −1 (j = 1, 2),

and

n=0

in Theorem 5, we obtain the following result which refines the work of Yang [15, Theorem 4]. Corollary 6. If the functions fj (z) ∈ Σp (j = 1, 2) satisfy the following inequality: (3.25) < (1 + λp)z p fj (z) + λz p+1 fj0 (z) > αj

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(0 5 αj < 1; j = 1, 2; z ∈ U), J. Ineq. Pure and Appl. Math. 6(3) Art. 88, 2005


then < (1 + λp)z p (f1 ? f2 )(z) + λz p+1 (f1 ? f2 )0 (z) > η0

(z ∈ U),

where 1 1 1 . η0 = 1 − 4(1 − α1 )(1 − α2 ) 1 − 2 F1 1, 1; + 1; 2 λ 2 The result is the best possible. Theorem 6. If f (z) ∈ Σp,m satisfies the following subordination condition: (1 − λ)z p Dn f (z) + λz p Dn+1 f (z) ≺

1 + Az 1 + Bz


(z ∈ U),

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then
ρ1/q

(q ∈ N; z ∈ U),

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where ρ is given as in Theorem 1. The result is the best possible. Proof. Defining the function φ(z) by (3.26)


φ(z) = z p Dn f (z)

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(f ∈ Σp,m ; z ∈ U),

we see that the function φ(z) is of the form (2.1) and is analytic in U. Using the identity (1.5) in (3.26) and differentiating the resulting equation with respect to z, we find that (1 − λ)z p Dn f (z) + λz p Dn+1 f (z) = φ(z) + λ zφ0 (z) ≺

1 + Az 1 + Bz

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(z ∈ U).

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Now, by following the lines of proof of Theorem 1 mutatis mutandis, and using the elementary inequality: 1/q < w1/q = h 1 2 2 − 2 F1 1, 1; λ(p+m) + 1; 12

then

1 < z p f (z) > 2 The result is the best possible.


and q = 1

,

(z ∈ U),

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(z ∈ U).



From Corollary 6 and Theorem 6 (for m = −p + 1, A = 1 − 2η0 , B = −1, and q = 1), we deduce the following result. Corollary 8. If the functions fj (z) ∈ Σp (j = 1, 2) satisfy the inequality (3.25), then 1 1 p < z (f1 ? f2 )(z) > η0 + (1 − η0 ) 2 F1 1, 1; + 1; −1 (z ∈ U), λ 2 where η0 is given as in Corollary 6. The result is the best possible. Theorem 7. Let f (z) ∈ Σnp,m (A, B) and let g(z) ∈ Σp,m satisfy the following inequality: 1 < z p g(z) > (z ∈ U). 2 Then (f ? g)(z) ∈ Σnp,m (A, B).

Since

and the function


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Proof. We have 0 0 z p+1 Dn (f ? g)(z) z p+1 Dn f (z) − =− ? z p g(z) p p


(z ∈ U).

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1 < z p g(z) > 2 1 + Az 1 + Bz

(z ∈ U)



is convex (univalent) in U, it follows from (1.6) and Lemma 5 that (f ? g)(z) ∈ Σnp,m (A, B). This completes the proof of Theorem 7. In view of Corollary 7 and Theorem 7, we have Corollary 9 below. Corollary 9. If f (z) ∈ Σnp,m (A, B) and the function g(z) ∈ Σp,m satisfies the inequality (3.27), then (f ? g)(z) ∈ Σnp,m (A, B).


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References [1] M.K. AOUF AND H.M. HOSSEN, New criteria for meromorphic p-valent starlike functions, Tsukuba J. Math., 17 (1993), 481–486. [2] J.-L. LIU AND H.M. SRIVASTAVA, Classes of meromorphically multivalent functions associated with the generalized hypergeometric function, Math. Comput. Modelling, 39 (2004), 21–34. [3] J.-L. LIU AND H.M. SRIVASTAVA, Subclasses of meromorphically multivalent functions associated with a certain linear operator, Math. Comput. Modelling, 39 (2004), 35–44. [4] T.H. MacGREGOR, Radius of univalence of certain analytic functions, Proc. Amer. Math. Soc., 14 (1963), 514–520.


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[5] S.S. MILLER AND P.T. MOCANU, Differential subordinations and univalent functions, Michigan Math. J., 28 (1981), 157–171. [6] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and Applications, Series on Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker, New York and Basel, 2000.

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[7] M.L. MOGRA, Meromorphic multivalent functions with positive coefficients. I and II, Math. Japon., 35 (1990), 1–11 and 1089–1098.

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[8] M. PAP, On certain subclasses of meromorphic m-valent close-to-convex functions, Pure Math. Appl., 9 (1998), 155–163.

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[9] D.Ž. PASHKOULEVA, The starlikeness and spiral-convexity of certain subclasses of analytic functions, in : H.M. Srivastava and S. Owa (Editors), Current Topics in Analytic Function Theory, pp. 266–273, World Scientific Publishing Company, Singapore, New Jersey, London and Hong Kong, 1992. [10] J. PATEL, Radii of γ-spirallikeness of certain analytic functions, J. Math. Phys. Sci., 27 (1993), 321–334. [11] R. SINGH AND S. SINGH, Convolution properties of a class of starlike functions, Proc. Amer. Math. Soc., 106 (1989), 145–152. [12] J. STANKIEWICZ AND Z. STANKIEWICZ, Some applications of the Hadamard convolution in the theory of functions, Ann. Univ. Mariae Curie-Skłodowska Sect. A, 40 (1986), 251–265. [13] B.A. URALEGADDI AND C. SOMANATHA, New criteria for meromorphic starlike univalent functions, Bull. Austral. Math. Soc., 43 (1991), 137– 140. [14] E.T. WHITTAKER AND G.N. WATSON, A Course on Modern Analysis: An Introduction to the General Theory of Infinite Processes and of Analytic Functions; With an Account of the Principal Transcendental Functions, Fourth Edition (Reprinted), Cambridge University Press, Cambridge, 1927. [15] D.-G. YANG, Certain convolution operators for meromorphic functions, Southeast Asian Bull. Math., 25 (2001), 175–186.


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