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Integral Operator Pravati Sahoo and Saumya Singh

vol. 10, iss. 3, art. 77, 2009

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STARLIKENESS CONDITIONS FOR AN INTEGRAL OPERATOR

PRAVATI SAHOO AND SAUMYA SINGH

Department of Mathematics Banaras Hindu University Banaras 221 005, India

EMail:pravatis@yahoo.co.in bhu.saumya@gmail.com

Received: 30 June, 2009

Accepted: 26 August, 2009

Communicated by: R.N. Mohapatra 2000 AMS Sub. Class.: 30C45, 30D30.

Key words: Meromorphic functions; Differential subordination; Starlike functions, Convex functions

Abstract: Let for fixednN,Σndenotes the class of function of the following form

f(z) = 1 z +

X

k=n

akzk,

which are analytic in the punctured open unit disk={zC: 0<|z|<

1}. In the present paper we defined and studied an operator in F(z) =

c+ 1µ zc+1

Z z

0

f(t) t

µ

tc+µdt 1µ

, forfΣn and c+1−µ >0.

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Integral Operator Pravati Sahoo and Saumya Singh

vol. 10, iss. 3, art. 77, 2009

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Contents

1 Introduction 3

2 Main Results 5

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Integral Operator Pravati Sahoo and Saumya Singh

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1. Introduction

LetH(∆) = H denote the class of analytic functions in∆, where∆ = {z ∈ C :

|z|<1}. For a fixed positive integernanda∈C, let

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

withH0 =H[0,1]. LetAnbe the class of analytic functions defined on the unit disc with the normalized conditionsf(0) = 0 =f0(0)−1, that isf ∈ Anhas the form

(1.1) f(z) = z+

X

k=n+1

akzk, (z ∈∆andn∈N).

LetA1 =Aand letS be the class of all functionsf ∈ Awhich are univalent in∆.

A functionf ∈ Ais said to be inS ifff(∆)is a starlike domain with respect to the origin. Let for0≤α <1,

S(α) =

f ∈ A: Rezf0(z)

f(z) > α, z∈∆

be the class of all starlike functions of orderα. SoS(0)≡ S. We denoteSn(α)≡ S(α)T

Anforn∈N.

A functionf ∈ Ais said to be inC ifff(∆)is a convex domain. Let for0≤α <

1,

C(α) =

f ∈ A : Re

1 + zf00(z) f(z)

> α, z ∈∆

be the class of convex functions of orderα. SoC(0)≡ C.

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Integral Operator Pravati Sahoo and Saumya Singh

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Let for fixedn ∈N,Σndenote the class of meromorphic functions of the follow- ing form

(1.2) f(z) = 1

z +

X

k=n

akzk,

which are analytic in the punctured open unit disk∆ ={z :z ∈C and 0<|z|<

1}= ∆− {0}. LetΣ0 = Σ.

A function f ∈ Σis said to be meromorphically starlike of order α in∆ if it satisfies the condition

−Re

zf0(z) f(z)

> α, (0≤α <1;z ∈∆).

We denote by Σ(α), the subclass of Σconsisting of all meromorphically starlike functions of orderαin∆andΣn(α)≡Σ(α)T

Σnforn ∈N.

We say thatf(z)is subordinate tog(z)andf ≺g in∆orf(z)≺ g(z) (z ∈∆) if there exists a Schwarz functionw(z), which (by definition) is analytic in∆with w(0) = 0and|w(z)| < 1, such that f(z) = g(w(z)), z ∈ ∆.Furthermore, if the functiong is univalent in∆, f(z) ≺ g(z) (z ∈ ∆) ⇔ f(0) = g(0) andf(∆) ⊂ g(∆).

In the present paper, forf(z) ∈ Σn, we define and study a generalized operator I[f]

(1.3) I[f] =F(z) =

c+ 1−µ zc+1

Z z

0

f(t) t

µ

tc+µdt 1µ

, (c+1−µ >0, z∈∆), which is similar to the Alexander transform when c = µ = 1 and is similar to Bernardi transformation whenµ= 1andc >0.

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Integral Operator Pravati Sahoo and Saumya Singh

vol. 10, iss. 3, art. 77, 2009

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2. Main Results

For our main results we need the following lemmas.

Lemma 2.1 (Goluzin [5]). Iff ∈ AnT

S,then

Re f(z)

z n2

> 1 2.

This inequality is sharp with extremal functionf(z) = z

(1−zn)2n.

Lemma 2.2 ([9]). Let uandv denote complex variables,u = α+iρ, v = σ+iδ and letΨ(u, v)be a complex valued function that satisfies the following conditions:

(i) Ψ(u, v)is continuous in a domainΩ⊂C2; (ii) (1,0)∈ΩandRe(Ψ(1,0)) >0;

(iii) Re(Ψ(iρ, σ))≤0whenever(iρ, σ)∈Ω,σ ≤ −1+ρ22 andρ,σare real.

Ifp(z)∈ H[a, n]is a function that is analytic in∆, such that(p(z), zp0(z))∈Ωand Re(Ψ(p(z), zp0(z)))>0hold for allz ∈∆, thenRep(z)>0,whenz∈∆.

Lemma 2.3 ([9, p. 34], [8]). Letp∈ H[a, n]

(i) IfΨ∈Ψn[Ω, M, a],then

Ψ(p(z), zp02p00(z);z)∈Ω⇒ |p(z)|< M.

(ii) IfΨ∈Ψn[M, a],then

|Ψ(p(z), zp02p00(z);z)|< M ⇒ |p(z)|< M.

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vol. 10, iss. 3, art. 77, 2009

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Lemma 2.4 ([6]). Leth(z)be an analytic and convex univalent function in∆,with h(0) =a, c6= 0andRec≥0.Ifp∈ H[a, n]and

p(z) + zp0(z)

c ≺h(z), then

p(z)≺q(z)≺h(z), where

q(z)≺ c nznc

Z z

0

tcn−1f(t)dt, z ∈∆.

The functionqis convex and the best dominant.

Theorem 2.5. Let c > 0and 0 < µ < 1. If f ∈ Σn(α)for 0 < α < 1, then I(f) =F(z)∈Σn(β), where

β =β(α, c, µ) (2.1)

= 1 4µ

h

2c+ 2αµ+n+ 2

−p

[4(c−αµ)]2+ (n+ 2)(n+ 2 + 4c+ 4µα)−16α−8µni .

Proof. Here we have the conditions

(2.2) 0< α <1, 0< µ <1 and c > 0, which will imply thatβ <1.

Letf(z)∈Σn(α). We first show thatF(z)defined by (1.3) will become nonzero forz ∈∆. Again sincef ∈Σn(α),we havef(z)6= 0, forz ∈∆.

Letg(z) = (f(z))1 µ, then a simple computation shows thatg(z)∈Sn(αµ).

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If we define

Ig = g(z)

z

{1−αµ1 } ,

thenI(g)∈Sn and by Goluzin’s subordination result (by Lemma2.1), we obtain Ig

z n2

≺ 1 1 +z. From the relation betweenIg,g andf we get that

g(z)

z ≺(1 +z)n2(αµ−1), which implies

z(f(z))µ≺(1 +z)n2(1−αµ)

and since0< αµ <1, we havez(f(z))µ≺(1 +z)n2. Combining this with min|z|=1Re(1 +z)n2 = 0,

we deduce that

(2.3) Re[z(f(z))µ]>0.

By differentiating (1.3), we obtain (2.4) (c+ 1)(F(z))µ+z d

dz(F(z))µ = (c+ 1−µ)(f(z))µ. If we let

(2.5) P(z)

z = (F(z))µ,

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then (2.4) becomes

P(z) + 1

czP0(z) = c+ 1−µ

c z(f(z))µ. Hence from (2.3) we have

(2.6) Re Ψ(P(z), zP0(z)) = Re

P(z) + zP0(z) c

,

whereΨ(r, s) = r + sc.To show that ReP(z) > 0,condition (iii) of Lemma 2.2 must be satisfied. Sincec >0, (2.6) implies that

Re Ψ(iρ, σ) = Re iρ+σ

c

≤ −n(1 +ρ2) 2c ≤0,

whenσ ≤ −n(1+ρ2 2),for allρ ∈ R. Hence from (2.6) we deduce thatReP(z) >0, which implies thatF(z)6= 0forz ∈∆.

We next determineβ such thatF ∈Σn(β).Let us definep(z)∈ H[1, n]by

(2.7) −zF0(z)

F(z) = (1−β)p(z) +β.

By applying (part iii) of Lemma2.2again with differentΨwe finish the proof of the theorem. Sincef ∈Σn(α),by differentiating (2.4) we easily get

Re Ψ(p(z), zp0(z))>0, where

Ψ(r, s) = (1−β)r+β+ (1−β)σ

c+ 1−µβ−µ(1−β)p(z)−α.

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Forβ ≤ β(α, c, µ),where β(α, c, µ)is given by (2.1), a simple calculation shows that the admissibility condition (iii) of Lemma 2.2 is satisfied. Hence by Lemma 2.2, we get Rep(z) > 0. Using this result in (2.7) together with β < 1shows that F(z)∈Σn(β).

Theorem 2.6. Let 0 < c + 1 − µ < 1. If, for 0 < α < 1, f ∈ Σ(α), then I(f)∈Σ(β), where

β =β(α, µ, c) (2.8)

= 1 2µ

h

2c+ 2αµ+ 3−p

[2(c−αµ)]2+ 3(3 + 4c)−4µ(2 +α)i . The proof is very similar to that of Theorem2.5.

In the special case when the meromorphic function given in (1.2) has a coefficient a0 = 0, it is possible to obtain a stronger result than (2.8).

Theorem 2.7. Letc > 0,0 < µ < 1, 0 < α < 1, f ∈ Σ1(α),thenI(f) ∈ Σ1(β), where

(2.9) β =β(α, µ, c) = 1 2µ

h

c+αµ+ 1−p

(c−αµ)2+ 4(c+ 1−µ)i . The proof is similar to that of Theorem2.5.

Corollary 2.8. Letn ≥ 1, c+n+ 1 >0andg(z)∈ H[0, n]. If|((g(z))µ)0| ≤λ and

(2.10) F(z) =

1 zc+1

Z z

0

(g(t))µtcdt 1µ

,

then

|((F(z))µ)0| ≤ λ c+n+ 1.

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Integral Operator Pravati Sahoo and Saumya Singh

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Proof. From (2.10) we deduce(c+ 1)(F(z))µ+z((F(z))µ)0 = gµ(z). If we set z((F(z))µ)0 =P(z), thenP ∈ H[0, n]and

(c+ 1)P(z) +zP0(z) =z(gµ(z))0 ≺λz.

From part(i) of Lemma2.3, it follows that this differential subordination has the best dominant

P(z)≺Q(z) = λz c+n+ 1. Hence we have

|((F(z))µ)0| ≤ λ c+n+ 1.

Corollary 2.9. Letc+n+ 1>0andf ∈Σnbe given as f(z) = 1

z +g(z), wheren ≥1andg(z)∈ H[0, n]. LetF be defined by

(2.11) F(z)≡ 1

z +G(z) = 1 z +

1 zc+1

Z z

0

(g(t))µtcdt µ1

.

Then

|((g(z))µ)0| ≤ n(c+n+ 1)

√n2 + 1 . Proof. From Corollary2.8we obtain

|((G(z))µ)0| ≤ n

√n2+ 1,

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since from (2.11), we have

|z2((F(z))µ)0 + 1|=|G0(z)|.

Hence from [2], we conclude thatF ∈Σn.

Corollary 2.10. Let n be a fixed positive integer and c > 0. Let q be a convex function in∆, withq(0) = 1and lethbe defined by

(2.12) h(z) =q(z) + n+ 1

c zq0(z).

Iff ∈ΣnandF(z)is given by(1.3), then

−c+ 1−µ

c z2((f(z))µ)0 ≺h(z)⇒ −z2((F(z))µ)0 ≺q(z), and this result is sharp.

Proof. From the definition ofh(z),it is a convex function. If we obtain p(z) =−z2(Fµ(z))0,

thenp∈ H[1, n+ 1]and from (2.3), we get p(z) + 1

czp0(z) =−c+ 1−µ

c z2((f(z))µ)0 ≺h(z).

The conclusion of the corollary follows by Lemma2.4.

Corollary 2.11. Letn ≥ 1andc > 0. Letf ∈ Σn and letF(z)given by(1.3). If λ >0, then

|z2((f(z))µ)0+ 1|< λ⇒ |z2((F(z))µ)0+ 1|< λc c+n+ 1.

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In particular,

|z2((f(z))µ)0+ 1|< c+n+ 1

c ⇒ |z2((F(z))µ)0+ 1|<1.

Hence(F(z))µis univalent.

Proof. If we take

q(z) = 1 + λcz c+n+ 1, then (2.12) becomes

h(z) = 1 +λz.

The conclusion of the corollary follows by Corollary2.10.

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References

[1] H. AL-AMIRI AND P.T. MOCANU, Some simple criteria of starlikeness and convexity for meromorphic functions, Mathematica(Cluj), 37(60) (1995), 11–

21.

[2] S.K. BAJPAI, A note on a class of meromorphic univalent functions, Rev.

Roumine Math. Pures Appl., 22 (1977), 295–297.

[3] P.L. DUREN, Univalent Functions, Springer-Verlag, Berlin-New York, 1983.

[4] G.M. GOEL AND N.S. SOHI, On a class of meromorphic functions, Glasnik Mat. Ser. III, 17(37) (1981), 19–28.

[5] G. GOLUZIN, Some estimates for coefficients of univalent functions (Rus- sian), Mat. Sb., 3(45), 2 (1938), 321–330.

[6] D. J. HALLENBECKANDSt. RUSCHWEYH, Subordination by convex func- tions, Proc. Amer. Math. Soc., 52 (1975), 191–195.

[7] P. T. MOCANU, Starlikeness conditions for meromorphic functions, Proc.

Mem. Sect. Sci., Academia Romania, 4(19) (1996), 7–12.

[8] S.S. MILLERANDP.T. MOCANU, Differential Subordinations and univalent functions, Michigan Math. J., 28 (1981), 157–171.

[9] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and Applications, Monographs and Textbooks in Pure and Appl. Math., Vol. 225, Marcel Dekker, New York, 2000.

[10] P. T. MOCANUANDGr. St. S ˇAL ˇAGEAN, Integral operators and meromorphic starlike functions, Mathematica(Cluj)., 32(55), 2 (1990), 147–152.

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[11] Gr. St. S ˇAL ˇAGEAN, Meromorphic starlike univalent functions, Babe¸s-Bolyai Uni. Fac. Math. Res. Sem., 7 (1986), 261–266.

[12] Gr. St. S ˇAL ˇAGEAN, Integral operators and meromorphic functions, Rev.

Roumine Math. Pures Appl., 33(1-2) (1988), 135–140.

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