• 検索結果がありません。

A New Criterion for Meromorphically Convex Functions of Order α

N/A
N/A
Protected

Academic year: 2022

シェア "A New Criterion for Meromorphically Convex Functions of Order α"

Copied!
10
0
0

読み込み中.... (全文を見る)

全文

(1)

A New Criterion for Meromorphically Convex Functions of Order α

H. E. Darwish, M. K. Aouf

and

G. S. Sˇ alˇ agean

Dedicated to Professor Dumitru Acu on his 60th anniversary

Abstract

Let Jn(α) denote the classes of functions of the form f(z) = 1

z + X k=0

akzk,

which are regular in the punctured disc U={z : 0<|z|<1} and satisfy

Re

(Dn+1f(z))0 (Dnf(z))0 2

<−n+α

n+ 1, z∈U, n∈N={0,1, ...} and α∈[0,1), where

Dnf(z) = 1 z

zn+1f(z) n!

(n) .

In this paper it is proved that Jn+1(α)⊂Jn(α) (nN and α∈ [0,1)). Since J0(α) is the class of meromorphically convex functions of order α, α [0,1), all functions in Jn(α) are meromorphically convex functions of order α. Further we consider the integrals of the functions in Jn(α).

2000 Mathematics Subject Classification: 30C45 and 30C50.

Keywords: Meromorphically convex, Hadamard product.

71

(2)

1 Introduction

Let Σ denote the class of functions of the form

(1) f(z) = 1

z + X

k=0

akzk,

which are regular in the punctured disc U ={z : 0<|z|<1}.

The Hadamard product of two functions f, g Σ will be denoted by f∗g. Let

Dnf(z) = 1

z(1−z)n+1 ∗f(z) = 1 z

zn+1f(z) n!

(n)

=

= 1

z + (n+ 1)a0+ (n+ 1)(n+ 2)

2! a1z+...

In this paper among other things we shall show that a function f in Σ, which satisfies one of the conditions

(2) Re

(Dn+1f(z))0 (Dnf(z))0 2

<−n+α

n+ 1, z ∈U, n∈N,

for α∈[0,1), is meromorphically convex of order α in U. More precisely, it is proved that for the classes Jn(α) of functions in Σ satisfying (2) (3) Jn+1(α)⊂Jn(α), (n N, α[0,1))

holds. Since J0(α) = Σk(α) (the class of meromorphically convex functions

of order α,

α [0,1) ) the convexity of the members of Jn(α) is a consequence of (3).

We note that in [1] Aouf and Hossen obtained a new criterion for mero- morphic univalent functions via the basic inclusion relationship Mn+1(α)

(3)

Mn(α), n N and α [0,1), where Mn(α) is the class consisting of functions in Σ satisfying

Re

Dn+1f(z) Dnf(z) 2

<−n+α

n+ 1, z ∈U, n∈N and α [0,1).

Futher, for c∈(0,∞) let F(z) = c

zc+1 Z z

0

tcf(t)dt.

In this paper it is shown that F Jn(α) whenever f ∈Jn(α). Also it is shown that if f ∈Jn(α), then

(4) F(z) = n+ 1

zn+2 Z z

0

tn+1f(t)dt.

belongs to Jn+1(α). Some known results of Bajpai [2], Goel and Sohi [3], Uralegaddi and Ganigi [10] and Sˇalˇagean [8] are extended. Techniques are similar to those in [6] (see also [4], [5] and [8]).

2 The classes J

n

(α)

In order to prove our main results (Theorem 1 and Theorem 2 below) we shall need the following lemma due to S. S. Miller and P. T. Mocanu [4], [5]

(see also [6] or [8])

Lemma A. Let the function Ψ : C2 C satisfy Re Ψ(ix, y)0

for all real x and for all real y, y ≤ −(1 +x2)/2. If p(z) = 1 +p1z +...

is analytic in the unit disc U ={z : |z|<1} and Re Φ(p(z), zp0(z))>0, z ∈U,

(4)

then Rep(z)>0 for z ∈U.

Theorem 1. Jn+1(α)⊂Jn(α) for each integer n∈N and α [0,1).

Proof. Let f be in Jn+1(α). Then

(1) Re

(Dn+2f(z))0 (Dn+1f(z))0 2

<−n+ 1 +α

n+ 2 , z ∈U. We have to show that (1) implies the inequality

(2) Re

(Dn+1f(z))0 (Dnf(z))0 2

<−n+α

n+ 1, z ∈U. We define the regular in U ={z :|z|<1} function p by

p(z) = n+ 2−α

1−α n+ 1 1−α

(Dn+1f(z))0

(Dnf(z))0 , z∈U and p(0) = 1. We have

(3) (Dn+1f(z))0

(Dnf(z))0 = 1

n+ 1(n+ 2−α−(1−α)p(z)).

Differentiating (3) logarithmically, we obtain (4) (Dn+1f(z))00

(Dn+1f(z))0 (Dnf(z))00

(Dnf(z))0 = −(1−α)zp0(z) n+ 2−α−(1−α)p(z). From the identity

(5) z(Dnf(z))0 = (n+ 1)Dn+1f(z)(n+ 2)Dnf(z) we have

(6) z(Dnf(z))00 = (n+ 1)(Dn+1f(z))0(n+ 3)(Dnf(z))0. Using the identity (6) the equation (4) may be written

(n+2)

(Dn+2f(z))0 (Dn+1f(z))0 2

+n+1+α+(1−α)p(z) = −(1−α)zp0(z) n+ 2−α−(1−α)p(z)

(5)

or (7)

p(z) + zp0(z)

n+ 2−α−(1−α)p(z) = n+ 2 1−α

2 (Dn+2f(z))0

(Dn+1f(z))0 −n+ 1 +α n+ 2

Since f ∈Jn+1(α), from (1) and (7) we have

(8) Re

p(z) + zp0(z)

n+ 2−α−(1−α)p(z)

>0, z ∈U.

We define the function Ψ by

(9) Ψ(u, v) =u+v/(n+ 2−α−(1−α)u).

In order to use Lemma A we must verify that Re Ψ(ix, y)0 whenever x and y are real numbers such that y≤ −(1 +x2)/2. We have

Re Ψ(ix, y) = yRe 1

n+ 2−α−(1−α)xi =y n+ 2−α

(n+ 2−α)2+ (1−α)2x2

≤ − (1 +x)2(n+ 2−α)

(n+ 2−α)2+ (1−α)2x2 <0.

From (8) and (9) we obtain

Re Ψ(p(z), zp0(z))>0, z ∈U.

By Lemma A we conclude that Rep(z) > 0 in U and (see (5)) this implies the inequality (2).

3 Integral operators

Theorem 2. Let f be a function in Σ; if for a given n N and c∈(0,∞) f satisfies the condition

(10) Re

(Dn+1f(z))0 (Dnf(z))0 2

< 1−α−2(n+α)(c+ 1−α)

2(n+ 1)(c+ 1−α) , z ∈U,

(6)

then

(11) F(z) = c

zc+1 Z z

0

tcf(t)dt belongs to Jn(α).

Proof. Using the identity

(12) z(DnF(z))0 =c(Dnf(z))(c+ 1)(DnF(z))

obtained from (11) and the identity (5) for the function F the condition (10) may be written

(13) Re







(n+ 2)(Dn+2F(z))0

(Dn+1F(z))0 −n−2 +c (n+ 1)(n+ 1−c) (DnF(z))0

(Dn+1F(z))0







−2< 1−α−2(n+α)(c+ 1−α) 2(n+ 1)(c+ 1−α) .

We have to prove that (13) implies the inequality

(14) Re

(Dn+1F(z))0 (DnF(z))0 2

<−n+α

n+ 1, z ∈U. We define the regular in U function p by

(15) p(z) = n+ 2−α

1−α n+ 1 1−α

(Dn+1F(z))0

(DnF(z))0 , z ∈U and p(0) = 1.

From (6) we have

(16) (Dn+1F(z))0

(DnF(z))0 = n+ 2−α−(1−α)p(z) n+ 1

and for the function F instead of f the identity (15) becomes (17) z(DnF(z))00= (n+ 1)(Dn+1F(z))0(n+ 3)(DnF(z))0.

(7)

Differentiating (16) logarithmically and using (17) we obtain (n+ 2)(Dn+2F(z))0

(Dn+1F(z))0 −n−2 +c (n+ 1)(n+ 1−c) (DnF(z))0

(Dn+1F(z))0

2 =

= 1

n+ 1

−n−α−(1−α)p(z)− (1−α)zp0(z) 1−α+c−(1−α)p(z)

and (13) is equivalent to (18)

1−α n+ 1 Re

p(z) + zp0(z)

1−α+c−(1−α)p(z) + 1 2(c+ 1−α)

>0, z ∈U.

We define the function Ψ by Ψ(u, v) = 1−α

n+ 1

u+ v

1−α+c−(1−α)u + 1 2(c+ 1−α)

For the real numbers x, y with y ≤ −(1 +x2)/2, we have Re Ψ(ix, y) = 1−α

n+ 1

1

2(c+ 1−α) + Re y

c+ 1−α−(1−α)x i

=

= 1−α n+ 1

1

2(c+ 1−α) (1 +x2)(c+ 1−α) 2 [(c+ 1−α)2+ (1−α)2x2]

=

= 1−α n+ 1

−2(1−α)c−c2

2 [(c+ 1−α)2+ (1−α)2x2] (c+ 1−α) <0.

We obtained that Ψ satisfies the conditions of Lemma A. This implies Rep(z) > 0, and from (15) it follows that (12) implies (14), that is F Jn(α).

Putting n = 0 in the statement of Theorem 2 we obtain the following result

Corollary 1. If f Σ and satisfies Re

1 + zf00(z) f0(z)

< 1−α−2α(c+ 1−α) 2(c+ 1−α) ,

(8)

then F given by (11) belongs to Σk(α).

Putting α= 0 and c= 1 in Corollary 1 we obtain the following result Corollary 2. If f Σ and satisfies

Re

1 + zf00(z) f0(z)

< 1 4, then the function F defined by

F(z) = 1 z2

Z z

0

tf(t)dt belongs to Σk(0).

Remark 1. Corollary 1 was obtained by Goel and Sohi [3]. In [8] there is another extention of this result.

Remark 2. Corollary 2 extends a result of Bajpai [2]

Theorem 3. If f ∈Jn(α), then F(z) = n+ 1

zn+2 Z z

0

tn+1f(t)dt belongs to Jn+1(α).

Proof. From (11) we have

cDnf(z) = (n+ 1)Dn+1F(z)(n+ 1−c)DnF(z) and

cDn+1f(z) = (n+ 2)Dn+2F(z)(n+ 2−c)Dn+1F(z).

Taking c=n+ 1 in the above relations we obtain (n+ 2)(Dn+2F(z))0(Dn+1F(z))0

(n+ 1)(Dn+1F(z))0 = (Dn+1f(z))0 (Dnf(z))0 which reduces to

(n+ 2)(Dn+2F(z))0

(n+ 1)(Dn+1F(z))0 1

n+ 1 = (Dn+1f(z))0 (Dnf(z))0

(9)

Thus Re

(n+ 2)(Dn+2F(z))0

(n+ 1)(Dn+1F(z))0 1 n+ 1 2

= Re

(Dn+1f(z))0 (Dnf(z))0 2

<−n+α n+ 1 from which it follows that

Re

(Dn+2F(z))0 (Dn+1F(z))0 2

<−n+α+ 1 n+ 2 . This completes the proof of Theorem 3.

Remark 3. Taking α = 0 in the above theorems we get the results obtained by Uralegaddi and Ganigi [10].

References

[1] M. K. Aouf and H. M. Hossen,New criteria for meromorphic univalent functions of order α, Nihonkai Math. J., 5 (1994), no. 1, 1-11.

[2] S. K. Bajpai, A note on a class of meromorphic univalent functions, Rev. Roum. Math. Pures Appl., 22 (1977), 295-297.

[3] R. M. Goel and N. S. Sohi,On a class of meromorphic functions, Glas.

Mat., 17 (1981), 19-28.

[4] S. S. Miller and P. T. Mocanu,Second order differential inequalities in the complex plane, J. Math. Anal. Appl., 65 (1978), 289-305.

[5] S. S. Miller and P. T. Mocanu,Differential subordination and univalent functions, Michigan Math. J., 28 (1981), 157-171.

[6] P. T. Mocanu and G. St. Sˇalˇagean,Integral operators and meromorphic starlike functions, Mathematica (Cluj), 32 (55) (1990), no. 2, 147-152.

(10)

[7] St. Ruscheweyh, New criteria for univalent functions, Proc. Amer.

Math. Soc.,49 (1975), 109-115.

[8] G. St. Sˇalˇagean, Integral operators and meromorphic functions, Rev.

Roumaine Math. Pures Appl.,33 (1988), no. 1-2, 135-140.

[9] R. Singh and S. Singh, Integrals of certain univalent functions, Proc.

Amer. Math. Soc., 77 (1973), 336-349.

[10] B. A. Uralegaddi and M. D. Ganigi,A new criterion for meromorphic convex functions, Tamkang J. Math., 19 (1988), no. 1, 43-48.

H. E. Darwish

Department of Mathematics, Faculty of Science, University of Mansoura,

Mansoura, Egypt.

E-mail: sinfac@mum.mans.eun.eg

M. K. Aouf

Department of Mathematics, Faculty of Science, University of Mansoura,

Mansoura, Egypt.

E-mail: sinfac@mum.mans.eun.eg

G. S. Sˇalˇagean

Babes-Bolyai University, Faculty of Mathematics and Computer Science, str. M. Kogalniceanu nr. 1, 400084 Cluj-Napoca, Romania.

E-mail:salagean@math.ubbcluj.ro

参照

関連したドキュメント

We construct an example that the class of asymptotically nonexpansive mappings include properly the class of nonexpansive mappings in locally convex spaces, prove a theorem on

Keywords: Convex order ; Fréchet distribution ; Median ; Mittag-Leffler distribution ; Mittag- Leffler function ; Stable distribution ; Stochastic order.. AMS MSC 2010: Primary 60E05

SAS ∗ with some invertible bounded linear or conjugate-linear operator S on H pre- serves Lebesgue decompositions in both directions, we see that the transformation in (2.5) is

Making use of certain linear operator, we define a new subclass of meromorphically uniformly convex functions with positive coefficients and obtain coefficient estimates, growth

The line graph L(G) of a graph G is defined to have as its vertices the edges of G, with two being adjacent if the corresponding edges share a vertex in G.. Line graphs have a

We also obtain the above mentioned result of Goel and Sohi [3, Theorem 4], by using a different technique, for complex y The results are shown to be sharp..

The sharp function given in the proof of Theorem 4 is independent of the point under consideration whereas the sharp function used for inequality. (19) is dependent on

Srivastava, Inclusion relationships and argument properties for certain subclasses of multivalent functions associated with a family of linear operators, J.. Sohi, A new criterion