Subordinations for analytic functions defined by the Dziok–Srivastava linear operator

Subordinations for analytic functions defined by the Dziok–Srivastava linear operator

Applied Mathematics and Computation 187 (2007) 13–19 www.elsevier.com/locate/amc Subordinations for analytic functions defined by the Dziok–Srivastava...

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Applied Mathematics and Computation 187 (2007) 13–19 www.elsevier.com/locate/amc

Subordinations for analytic functions defined by the Dziok–Srivastava linear operator R. Aghalary

a,*

, S.B. Joshi b, R.N. Mohapatra c, V. Ravichandran

d

a

b

Department of Mathematics, University of Urmia, Urmia, Iran Department of Mathematics, Walchand College of Engineering, Sangli 416415, Maharashtra, India c Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA d School of Mathematical Sciences, Universiti Sains Malaysia, 11800 USM Penang, Malaysia

Dedicated to Professor H.M. Srivastava on the occasion of his 65th birthday

Abstract In the present investigation, we obtain certain sufficient conditions for a normalized analytic function f(z) defined by the Dziok–Srivastava linear operator H lm ½a1  to satisfy the certain subordination. Our results extend corresponding previously known results on starlikeness, convexity, and close to convexity. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Univalent functions; Starlike functions; Convex functions; Differential subordination; Convolution; Dziok–Srivastava linear operator

1. Introduction Let H denote the class of analytic functions defined on the open unit disc D ¼ fz 2 C :j z j< 1g. Let A denote the subclass of H consisting of functions f(z) normalized by f(0) = f 0 (0)  1 = 0. For the functions f and g in H, we say that f is subordinate to g in D, and write f  g, if there exists a Schwarz P function x in 1 k H with jx(z)j < 1 and x(0) = 0 such that f(z) = g(x(z)) in D. For two functions f ðzÞ ¼ z þ k¼2 ak z and P1 gðzÞ ¼ z þ k¼2 bk zk , the Hadamard product (or convolution) of f and g is defined by 1 X ak bk zk ¼: ðg  f ÞðzÞ: ðf  gÞðzÞ :¼ z þ k¼2

*

Corresponding author. E-mail addresses: [email protected] (R. Aghalary), [email protected] (S.B. Joshi), [email protected] (R.N. Mohapatra), [email protected] (V. Ravichandran). URL: http://cs.usm.my/~vravi (V. Ravichandran). 0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.08.097

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For aj 2 Cðj ¼ 1; 2; . . . ; lÞ and bj 2 C n f0; 1; 2; . . .gðj ¼ 1; 2; . . . ; mÞ; the generalized hypergeometric function lFm(a1, . . . , al;b1, . . . , bm;z) is defined by the infinite series 1 X ða1 Þn    ðal Þn zn ðl 6 m þ 1; l; m 2 N0 :¼ f0; 1; 2; . . .gÞ; l F m ða1 ; . . . ; al ; b1 ; . . . ; bm ; zÞ :¼ ðb1 Þn    ðbm Þn n! n¼0 where (a)n is the Pochhammer symbol defined by  1; ðn ¼ 0Þ; Cða þ nÞ ðaÞn :¼ ¼ CðaÞ aða þ 1Þða þ 2Þ    ða þ n  1Þ; ðn 2 N :¼ f1; 2; 3; . . . gÞ: Corresponding to the function hða1 ; . . . ; al ; b1 ; . . . ; bm ; zÞ :¼ zl F m ða1 ; . . . ; al ; b1 ; . . . ; bm ; zÞ; the Dziok–Srivastava operator [3] (see also [4,13]) H(l,m)(a1, . . . , al; b1, . . . , bm) is defined by the Hadamard product 

H ðl;mÞ ða1 ; . . . ; al ; b1 ; . . . ; bm Þf ðzÞ :¼ hða1 ; . . . ; al ; b1 ; . . . ; bm ; zÞ f ðzÞ 1 X ða1 Þn1 . . . ðal Þn1 an zn ¼zþ : ðb1 Þn1 . . . ðbm Þn1 ðn  1Þ! n¼2

ð1:1Þ

For brevity, we write H lm ½a1 f ðzÞ :¼ H ðl;mÞ ða1 ; . . . ; al ; b1 ; . . . ; bm Þf ðzÞ: Special cases of the Dziok–Srivastava linear operator includes the Hohlov linear operator [5], the Carlson– Shaffer linear operator L(a, c) [2], the Ruscheweyh derivative operator Dn [12], the generalized Bernardi–Libera–Livingston linear integral operator (cf. [1,6,7]), and the Srivastava–Owa fractional derivative operators (cf. [9,10]). In the present paper, we obtain certain sufficient conditions for a function f 2 A to satisfy either of the following subordinations: H lm ½a1 þ 1f ðzÞ kð1  zÞ ;  kz H lm ½a1 f ðzÞ

H lm ½a1 f ðzÞ 1 þ Az  ; z 1z

H lm ½a1 f ðzÞ kð1  zÞ  : z kz

Our results extend corresponding previously known results on starlikeness, convexity, and close to convexity. To prove our main results, we need the following: Lemma 1.1 (cf. Miller and Mocanu [8, Theorem 3.4h, p.132]). Let q(z) be univalent in the unit disk D and let # and u be analytic in a domain D  q(D), with u(w) 5 0 when w 2 q(D). Set QðzÞ :¼ zq0 ðzÞuðqðzÞÞ;

hðzÞ :¼ #ðqðzÞÞ þ QðzÞ:

Suppose that (1) Q(z) is starlike univalent in D, and (2) R zh0ðzÞ > 0 for z 2 D. QðzÞ If p(z) is analytic in D with p(0) = q(0), p(D)  D and #ðpðzÞÞ þ zp0 ðzÞuðpðzÞÞ  #ðqðzÞÞ þ zq0 ðzÞuðqðzÞÞ; then p(z)  q(z) and q(z) is the best dominant. 2. Main results We begin with the following:

ð1:2Þ

R. Aghalary et al. / Applied Mathematics and Computation 187 (2007) 13–19

Theorem 2.1. Let a1 P 0, a 2 R satisfy jaj 6 1 and k > 1. If f 2 A satisfies H lm ½a1 f ðzÞ=z 6¼ 0 in D and  l a   H m ½a1 þ 1f ðzÞ H lm ½a1 þ 2f ðzÞ  1  hðzÞ; ða þ 1Þ 1 H lm ½a1 f ðzÞ H lm ½a1 þ 1f ðzÞ

15

ð2:1Þ

where

!  1þa kð1  zÞ ðk  1Þz hðzÞ ¼ a1  ; 2 kz kð1  zÞ

then H lm ½a1 þ 1f ðzÞ kð1  zÞ :  kz H lm ½a1 f ðzÞ Proof. The condition (2.1) and H lm ½a1 f ðzÞ=z 6¼ 0 in D implies that H lm ½a1 þ 1f ðzÞ=z 6¼ 0 in D. Define the function p(z) by pðzÞ :¼

H lm ½a1 þ 1f ðzÞ : H lm ½a1 f ðzÞ

Clearly p(z) is analytic in D. A computation shows that 0

0

zp0 ðzÞ z½H lm ½a1 þ 1f ðzÞ z½H lm ½a1 f ðzÞ ¼  : pðzÞ H lm ½a1 þ 1f ðzÞ H lm ½a1 f ðzÞ

ð2:2Þ

By using the identity z½H lm ½a1 f ðzÞ0 ¼ a1 H lm ½a1 þ 1f ðzÞ  ða1  1ÞH lm ½a1 f ðzÞ;

ð2:3Þ

we get, from (2.2), ða1 þ 1Þ

H lm ½a1 þ 2f ðzÞ zp0 ðzÞ : ¼ 1 þ a1 pðzÞ þ l pðzÞ H m ½a1 þ 1f ðzÞ

ð2:4Þ

Using (2.4) in (2.1), we get a1 ðpðzÞÞ

aþ1

a1

þ zp0 ðzÞðpðzÞÞ

 hðzÞ:

ð2:5Þ

Let q(z) be the function defined by qðzÞ :¼

kð1  zÞ : kz

It is clear that q is convex univalent in D. Since hðzÞ ¼ a1 ðqðzÞÞaþ1 þ zq0 ðzÞðqðzÞÞa1 ; we see that (2.5) can be written as (1.2) when # and u are given by #ðwÞ ¼ a1 waþ1

and

uðwÞ ¼ wa1 :

Clearly u and # are analytic in C n f0g. Now a1

QðzÞ ¼ zq0 ðzÞuðqðzÞÞ ¼ zq0 ðzÞðqðzÞÞ  hðzÞ ¼ #ðqðzÞÞ þ QðzÞ ¼

kð1  zÞ kz

¼

ð1  kÞzka ð1  zÞa1 1þa

ðk  zÞ

1þa a1 

ðk  1Þz kð1  zÞ

2

! :

;

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By our assumptions on the parameters a and k, we see that  0    zQ ðzÞ zð1  aÞ z þ ð1 þ aÞ R ¼R 1þ QðzÞ 1z kz 1 ð1 þ aÞk > 1 þ ð1  aÞ þ 2 1þk ð1 þ aÞðk  1Þ ¼ > 0; 2ð1 þ kÞ and therefore Q(z) is starlike. Also we have R

zh0 ðzÞ kð1  zÞ zQ0 ðzÞ ¼ a1 ð1 þ aÞR þR P 0: QðzÞ kz QðzÞ

By an application of Lemma 1.1, we have p(z)  q(z) or H lm ½a1 þ 1f ðzÞ kð1  zÞ :  kz H lm ½a1 f ðzÞ



Since 1 H 10 ½3f ðzÞ ¼ z2 f 00 ðzÞ þ zf 0 ðzÞ; 2 by taking a = 0, l = 1, m = 0 and a1 = 1 in Theorem 2.1, we get the following corollary: H 10 ½1f ðzÞ ¼ f ðzÞ; H 10 ½2f ðzÞ ¼ zf 0 ðzÞ;

Corollary 2.1. Let f 2 A be so that f(z)/z 5 0 in D. If k > 1 and 1þ

zf 00 ðzÞ kð1  zÞ ðk  1Þz  ;  f 0 ðzÞ kz ð1  zÞðk  zÞ

then zf 0 ðzÞ kð1  zÞ  : f ðzÞ kz Remark 2.1. The function hðzÞ ¼

kð1  zÞ ðk  1Þz z 1  ¼ þ ðk  zÞ ð1  zÞðk  zÞ kz 1z

ðkþ1Þ ð5k1Þ takes real value for real value of z,h(0) = 1 and h(D) is the region RhðzÞ < 2ðk1Þ for 1 < k 6 2 and RhðzÞ < 2ðkþ1Þ for 2 < k. Hence this result generalizes the result obtained by Owa et al. [11].

We note that the image of the function hðzÞ ¼ 1 

ðk  1Þz kð1  zÞ

is

 hðDÞ ¼ C 

2

 5k  1 ;1 : 4k

Hence by taking a = 1, a1 = 1, l = 1, and m = 0 in Theorem 2.1, we get the following corollary: Corollary 2.2. Let k > 1, f 2 A and f(z)/z 5 0 in D. If f satisfies 0 1 00 ðzÞ 1 þ zff 0 ðzÞ 5k  1 ; R@ zf 0 ðzÞ A < 4k f ðzÞ

R. Aghalary et al. / Applied Mathematics and Computation 187 (2007) 13–19

17

then zf 0 ðzÞ kð1  zÞ  : f ðzÞ kz Theorem 2.2. Let a1 > 0, 1 6 a < 0, 1 < A < 1. If f 2 A satisfy the condition H lm ½a1 f ðzÞ=z 6¼ 0 in D and  l a   H m ½a1 f ðzÞ H l ½a1 þ 1f ðzÞ a1 m  hðzÞ; z z

ð2:6Þ

where !  a 1 þ Az 1 þ Az ð1 þ AÞz hðzÞ ¼ þ a1 ; 2 1z 1z ð1  zÞ then H lm ½a1 f ðzÞ 1 þ Az  : z 1z Proof. Define the function p(z) by pðzÞ :¼

H lm ½a1 f ðzÞ : z

ð2:7Þ

It is clear that p(0) = 1 and p is analytic in D. By using the identity (2.3), we get, from (2.7), a1 H lm ½a1 þ 1f ðzÞ ¼ zp0 ðzÞ þ a1 pðzÞ:

ð2:8Þ

Using (2.8) in (2.6), we see that the subordination becomes a1 pðzÞ

1þa

a

þ pðzÞ zp0 ðzÞ  hðzÞ:

Define the function q(z) by qðzÞ :¼

1 þ Az : 1z

It is clear that q(z) is univalent in D and q(D) is the region RqðzÞ > ð1  AÞ=2. By defining the functions # and u by #ðwÞ ¼ a1 w1þa

and

uðwÞ ¼ wa ;

we observe that (2.6) can be written as (1.2). Note that u and # are analytic in C n f0g. Also we see that QðzÞ :¼ zq0 ðzÞuðqðzÞÞ ¼ and

ð1 þ AÞzð1 þ AzÞ ð1  zÞ

2þa

a

;

!  a 1 þ Az 1 þ Az ð1 þ AÞz þ hðzÞ :¼ #ðqðzÞÞ þ QðzÞ ¼ a1 : 2 1z 1z ð1  zÞ

By our assumptions, we have   zQ0 ðzÞ Az z ajAj 2 þ a að1  jAjÞ ¼R 1þa þ ð2 þ aÞ  ¼ > 0; R >1 QðzÞ 1 þ Az 1z 1 þ jAj 2 2ð1 þ jAjÞ

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and

R. Aghalary et al. / Applied Mathematics and Computation 187 (2007) 13–19

 0  zh0 ðzÞ # ðqðzÞÞ zQ0 ðzÞ zQ0 ðzÞ ¼R þ P 0: R ¼ a1 ð1 þ aÞ þ R QðzÞ uðqðzÞÞ QðzÞ QðzÞ

The results now follows by an application of Lemma 1.1. ð1þAÞz ð1zÞð1þAzÞ

The function hðzÞ ¼ a1 þ with respect to the real axis and 1 1 ; RhðzÞ > a1 þ  2 1  jAj

h

takes real values for real values of z with h(0) = a1 and h(D) is symmetric

z 2 D:

Consequently, by letting a = 1 in Theorem 2.2, we obtain the following corollary, which is extension to [11, Theorem 1, p. 64] which does not extend as for the sharpness. Corollary 2.3. Let 1 < A < 1, a1 > 0 and f 2 A be so that H lm ½a1 f ðzÞ=z 6¼ 0 in D and  R

 H lm ½a1 þ 1f ðzÞ 1 1 ;  >1þ 2a1 a1 ð1  jAjÞ H lm ½a1 f ðzÞ

then H lm ½a1 f ðzÞ 1 þ Az  : z 1z Theorem 2.3. Let a P 1, k > 1, f 2 A and H lm ½a1 f ðzÞ 6¼ 0 for 0 < jzj < 1. If f satisfies 

H lm ½a1 f ðzÞ z

a    a H lm ½a1 þ 1f ðzÞ k1þa ð1  zÞ ðk  1Þz a1 a1 ð1  zÞ   ; z kz ðk  zÞ1þa

then H lm ½a1 f ðzÞ kð1  zÞ  : z kz Proof. The proof of Theorem 2.3, also based upon Lemma 1.1 is similar to that of Theorem 2.1. Indeed, in this case, the results follows from Lemma 1.1 when we define the functions u and # by #(w) = a1w(1+a) and u(w) = w(2+a). h   ðk1Þz 3k1 Finally we note that R 1  ðkzÞð1zÞ for z 2 D and so from above Theorem by choosing l = 1, < 2ðk1Þ m = 0, and a1 = 1 we can get the following corollary: Corollary 2.4. Let k > 1, f 2 A and f 0 (z) 5 0 in D. If f satisfies 

zf 00 ðzÞ R 1þ 0 f ðzÞ

 <

3k  1 ; 2ðk  1Þ

then f 0 ðzÞ 

kð1  zÞ : kz

Acknowledgement The research of V. Ravichandran is supported by a post-doctoral research fellowship from Universiti Sains Malaysia.

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