ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS INVOLVING ...

Bull. Korean Math. Soc. 46 (2009), No. 5, pp. 905–915 DOI 10.4134/BKMS.2009.46.5.905

ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS INVOLVING A GENERALIZED DIFFERENTIAL OPERATOR Chellian Selvaraj and Kuppathai A. Selvakumaran Abstract. Making use of a generalized differential operator we introduce some new subclasses of multivalent analytic functions in the open unit disk and investigate their inclusion relationships. Some integral preserving properties of these subclasses are also discussed.

1. Introduction and preliminaries Let Ap denote the class of functions f (z) of the form (1)

f (z) = z p +

∞ X

an z p+n ,

(p ∈ N = {1, 2, 3, . . .}),

n=1

which are analytic and p-valent in the open unit disk U = {z : z ∈ C and |z| < 1}. For functions f given by (1) and g given by g(z) = z p +

∞ X

bn z p+n ,

n=1

the Hadamard product (or convolution) of f and g is defined by p

(f ∗ g)(z) = z +

∞ X

an bn z p+n .

n=1

Given two functions f and g, which are analytic in U, the function f is said to be subordinate to g in U if there exists a function w analytic in U with w(0) = 0,

|w(z)| < 1

(z ∈ U),

such that f (z) = g(w(z))

(z ∈ U).

Received July 23, 2008; Revised January 21, 2009. 2000 Mathematics Subject Classification. 30C45, 30C50, 30C75. Key words and phrases. multivalent function, convex univalent function, starlike with respect to symmetric points, Hadamard product, subordination, integral operator. c °2009 The Korean Mathematical Society

905

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C. SELVARAJ AND K. A. SELVAKUMARAN

We denote this subordination by f (z) ≺ g(z). Furthermore, if the function g is univalent in U, then f (z) ≺ g(z) (z ∈ U) ⇐⇒ f (0) = g(0) and f (U) ⊂ g(U). Let P denote the class of analytic functions h(z) with h(0) = 1, which are convex and univalent in U and for which R{h(z)} > 0 (z ∈ U). Analogous to the operator defined recently by Selvaraj and Santhosh Moni δ [6], we define an operator Dλ,g f on Ap as follows: For a fixed function g ∈ Ap given by (2)

g(z) = z p +

∞ X

bn z p+n ,

(bn ≥ 0; p ∈ N = {1, 2, 3, . . .}),

n=1 δ Dλ,g f (z) : Ap −→ Ap is defined by 0 Dλ,g f (z) = (f ∗ g)(z), 1 f (z) = (1 − λ)(f ∗ g)(z) + Dλ,g

(3)

λ z((f ∗ g)(z))0 , p

δ−1 δ 1 Dλ,g f (z) = Dλ,g (Dλ,g f (z)).

If f (z) ∈ Ap , then we have (4)

δ Dλ,g f (z) = z p +

¶δ ∞ µ X λn 1+ an bn z p+n , p n=1

where δ ∈ N0 = N ∪ {0} and λ ≥ 0. It easily follows from (3) that (5)

λz δ δ+1 δ f (z). f (z) − (1 − λ)Dλ,g (Dλ,g f (z))0 = Dλ,g p

Throughout this paper, we assume that p, k ∈ N, εk = exp( 2πi k ), and (6)

δ fp,k (λ; g; z) =

k−1 1 X −jp δ ε (Dλ,g f (εjk z)) = z p + · · · , k j=0 k

(f ∈ Ap ).

Clearly, for k = 1, we have δ δ fp,1 (λ; g; z) = Dλ,g f (z). δ Making use of the operator Dλ,g f (z), we now introduce and study the following subclasses of Ap of p-valent analytic functions. δ Definition. A function f ∈ Ap is said to be in the class Sp,k (λ; g; h), if it satisfies

(7)

δ z(Dλ,g f (z))0 δ (λ; g; z) pfp,k

≺ h(z) (z ∈ U),

δ (λ; g; z) 6= 0 (z ∈ U). where h ∈ P and fp,k

ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS

907

Remark 1.1. If we let δ=0

g(z) = z p l Fm (α1 , . . . , αl ; β1 , . . . , βm ; z),

and

l,m δ then Sp,k (λ; g; h) reduces to the function class Sp,k (α1 ; h) introduced and investigated by Zhi-Gang, Wang Yue-Ping Jiang, and H. M. Srivastava [10].

Remark 1.2. If we let δ=0

and g(z) = z p +

∞ X (a)n p+n z , (c)n n=1

δ then Sp,k (λ; g; h) reduces to the function class Tp,k (a, c; h) introduced and investigated by N-Eng Xu and Ding-Gong Yang [7].

Remark 1.3. Let

1+z . 1−z 0 Then S1,2 (λ; g; h) = Ss∗ . The class Ss∗ of functions starlike with respect to symmetric points has been studied by several authors (see [3], [5], [9]). g(z) = h(z) =

δ (λ; g; h), if it Definition. A function f ∈ Ap is said to be in the class Kp,k satisfies δ z(Dλ,g f (z))0

(8)

pϕδp,k (λ; g; z)

≺ h(z) (z ∈ U)

δ for some ϕ(z) ∈ Sp,k (λ; g; h), where h ∈ P and ϕδp,k (λ; g; z) 6= 0 is defined as in (6). δ Definition. A function f ∈ Ap is said to be in the class Cp,k (α, λ; g; h), if it satisfies

(9)

(1 − α)

δ z(Dλ,g f (z))0

pϕδp,k (λ; g; z)

+α

δ (z(Dλ,g f (z))0 )0

p(ϕδp,k (λ; g; z))0

≺ h(z)

(z ∈ U)

δ for some α (α ≥ 0) and ϕ(z) ∈ Sp,k (λ; g; h), where h ∈ P and (ϕδp,k (λ; g; z))0 6= 0.

We need the following lemmas to derive our results. Lemma 1.4 ([1]). Let β (β 6= 0) and γ be complex numbers and let h(z) be analytic and convex univalent in U with R{βh(z) + γ} > 0 (z ∈ U). If q(z) is analytic in U with q(0) = h(0), then the subordination q(z) +

zq 0 (z) ≺ h(z) βq(z) + γ

(z ∈ U)

implies that q(z) ≺ h(z)

(z ∈ U).

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C. SELVARAJ AND K. A. SELVAKUMARAN

Lemma 1.5 ([2]). Let h(z) be analytic and convex univalent in U and let w(z) be analytic in U with R{w(z)} ≥ 0 (z ∈ U). If q(z) is analytic in U with q(0) = h(0), then the subordination q(z) + w(z)zq 0 (z) ≺ h(z)

(z ∈ U)

implies that q(z) ≺ h(z)

(z ∈ U).

δ Lemma 1.6. Let f (z) ∈ Sp,k (λ; g; h). Then

(10)

δ z(fp,k (λ; g; z))0 δ (λ; g; z) pfp,k

≺ h(z)

(z ∈ U).

Proof. For f (z) ∈ Ap , we have from (6) that δ fp,k (λ; g; εjk z) =

=

k−1 1 X −mp δ ε Dλ,g f (εm+j z) k k m=0 k k−1 X −(m+j)p εjp δ k ε Dλ,g f (εm+j z) k k m=0 k

δ = εjp k fp,k (λ; g; z),

(j ∈ {0, 1, . . . , k − 1})

and δ (λ; g; z))0 = (fp,k

k−1 1 X j(1−p) δ ε (Dλ,g f (εjk z))0 . k j=0 k

Hence δ z(fp,k (λ; g; z))0 δ (λ; g; z) pfp,k

(11)

=

k−1 j(1−p) δ z(Dλ,g f (εjk z))0 1 X εk δ (λ; g; z) k j=0 pfp,k

=

k−1 j δ f (εjk z))0 1 X εk z(Dλ,g δ (λ; g; εj z) k j=0 pfp,k k

(z ∈ U).

δ Since f (z) ∈ Sp,k (λ; g; h), we have

(12)

δ f (εjk z))0 εjk z(Dλ,g δ (λ; g; εj z) pfp,k k

≺ h(z)

for j ∈ {0, 1, . . . , k − 1}.

Noting that h(z) is convex univalent in U, from(11) and (12) we conclude that (10) holds true. ¤

ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS

909

2. A set of inclusion relationships Theorem 2.1. Let h(z) ∈ P with (13)

R{h(z)} > 1 −

1 λ

(z ∈ U; λ > 1).

δ+1 δ δ If f (z) ∈ Sp,k (λ; g; h), then f (z) ∈ Sp,k (λ; g; h) provided fp,k (λ; g; z) 6= 0 (z ∈ U).

Proof. By using (5) and (6), we have (14) δ (λ; g; z)+ (1−λ)fp,k

k−1 λz δ 1 X −jp δ+1 δ+1 (fp,k (λ; g; z))0 = ε (Dλ,g f (εjk z)) = fp,k (λ; g; z). p k j=0 k

δ+1 Let f (z) ∈ Sp,k (λ; g; h) and

(15)

w(z) =

δ z(fp,k (λ; g; z))0 δ (λ; g; z) pfp,k

.

Then w(z) is analytic in U, with w(0) = 1, and from (14) and (15) we have (16)

1 − λ + λw(z) =

δ+1 fp,k (λ; g; z) δ (λ; g; z) fp,k

.

Differentiating (16) with respect to z and using (15), we get (17)

δ+1 z(fp,k (λ; g; z))0 zw0 (z) . w(z) + p = δ+1 pfp,k (λ; g; z) λ (1 − λ) + pw(z)

From (17) and Lemma 1.6 we note that (18)

w(z) +

p λ (1

zw0 (z) ≺ h(z) (z ∈ U). − λ) + pw(z)

In view of (13) and (18), we deduce from Lemma 1.4 that (19)

w(z) ≺ h(z) (z ∈ U).

Suppose that q(z) =

δ z(Dλ,g f (z))0 δ (λ; g; z) pfp,k

.

Then q(z) is analytic in U, with q(0) = 1, and we obtain from (5) that µ ¶ 1 δ+1 1 δ δ (20) fp,k (λ; g; z)q(z) = Dλ,g f (z) + 1 − Dλ,g f (z). λ λ Differentiating both sides of (20) with respect to z, we get ¶ µ δ+1 δ z(Dλ,g f (z))0 (λ; g; z))0 ¡1 ¢ z(fp,k q(z) = (21) zq 0 (z) + p −1 + . δ (λ; g; z) δ (λ; g; z) λ fp,k λfp,k

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C. SELVARAJ AND K. A. SELVAKUMARAN

Now, we find from (14), (15) and (21) that (22)

q(z) +

δ+1 z(Dλ,g f (z))0 zq 0 (z) = ≺ h(z) (z ∈ U), p δ+1 pfp,k (λ; g; z) λ (1 − λ) + pw(z)

δ+1 since f (z) ∈ Sp,k (λ; g; h). From (13) and (19) we observe that ½ ¾ p R (1 − λ) + pw(z) > 0. λ

Therefore, from (22) and Lemma 1.5 we conclude that q(z) ≺ h(z) which shows that f (z) ∈

(z ∈ U)

δ Sp,k (λ; g; h).

¤

Theorem 2.2. Let h(z) ∈ P with (23)

R{h(z)} > 1 −

1 λ

(z ∈ U; λ > 1).

δ+1 δ+1 If f (z) ∈ Kp,k (λ; g; h) with respect to ϕ(z) ∈ Sp,k (λ; g; h), then f (z) ∈ δ δ Kp,k (λ; g; h) provided ϕp,k (λ; g; z) 6= 0 (z ∈ U). δ+1 δ+1 Proof. Let f (z) ∈ Kp,k (λ; g; h). Then there exists a function ϕ(z) ∈ Sp,k (λ; g; h) such that

(24)

δ+1 z(Dλ,g f (z))0

pϕδ+1 p,k (λ; g; z)

≺ h(z) (z ∈ U).

δ An application of Theorem 2.1 yields ϕ(z) ∈ Sp,k (λ; g; h) and Lemma 1.6 leads to

(25)

ψ(z) =

z(ϕδp,k (λ; g; z))0 pϕδp,k (λ; g; z)

≺ h(z) (z ∈ U).

Let q(z) =

δ z(Dλ,g f (z))0

pϕδp,k (λ; g; z)

.

By using (5), q(z) can be written as follows (26)

ϕδp,k (λ; g; z)q(z)

µ ¶ 1 1 δ+1 δ = Dλ,g f (z) + 1 − Dλ,g f (z). λ λ

Differentiating both sides of (26) with respect to z and using (14) (with f replaced by ϕ), we get (27)

q(z) +

p λ (1

δ+1 z(Dλ,g f (z))0 zq 0 (z) . = − λ) + pψ(z) pϕδ+1 p,k (λ; g; z)

ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS

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Now, from (24) and (27) we find that (28)

q(z) +

zq 0 (z) ≺ h(z) (z ∈ U). p λ (1 − λ) + pψ(z)

Combining (23), (25) and (28), we deduce from Lemma 1.5 that q(z) ≺ h(z) which shows that f (z) ∈

δ (λ; g; h) Kp,k

(z ∈ U)

δ (λ; g; h). with respect to ϕ(z) ∈ Sp,k

¤

Corollary 2.3. Let 0 < α ≤ 1, −1 ≤ B < A ≤ 1 and µ ¶α 1 + Az (29) h(z) = (z ∈ U). 1 + Bz · ¸ ¡ 1−A ¢α −1 δ+1 δ+1 δ If λ ≤ 1 − 1−B , then Sp,k (λ; g; h) ⊂ Sp,k (λ; g; h) and Kp,k (λ; g; h) ⊂ δ Kp,k (λ; g; h).

Proof. The analytic function h(z) defined by (29) is convex univalent in U (see [8]), h(0) = 1 and h(U) is symmetric with respect to real axis. Thus h(z) ∈ P and µ ¶α µ ¶α 1−A 1+A 0≤ < R{h(z)} < (z ∈ U; 0 < α ≤ 1; −1 < B < A ≤ 1). 1−B 1+B Hence, by using Theorems 2.1 and 2.2 we have the corollary.

¤

Corollary 2.4. Let 0 < α ≤ 1 and µ µ √ ¶¶2 2 1 + αz √ (30) h(z) = 1 + 2 log (z ∈ U). π 1 − αz √ ¢−2 2¡ δ+1 δ+1 δ If λ ≤ π8 arctan α , then Sp,k (λ; g; h) ⊂ Sp,k (λ; g; h) and Kp,k (λ; g; h) ⊂ δ Kp,k (λ; g; h) (z ∈ U). Proof. The function h(z) defined by (30) is in the class P (cf. [4]) and satisfies h(z) = h(z). Therefore, √ ¢2 8¡ 1 R{h(z)} > h(−1) = 1 − 2 arctan α ≥ (z ∈ U; 0 < α ≤ 1). π 2 Hence, by Theorems 2.1 and 2.2 we have the desired result. ¤ Theorem 2.5. Let 0 ≤ α1 < α2 . Then δ δ Cp,k (α2 , λ; g; h) ⊂ Cp,k (α1 , λ; g; h). δ δ (λ; g; h) Proof. Let f (z)∈ Cp,k (α2 , λ; g; h).Then there exists a function ϕ(z)∈ Sp,k such that δ δ f (z))0 z(Dλ,g f (z))0 )0 (z(Dλ,g (31) (1 − α2 ) δ + α2 ≺ h(z) (z ∈ U). pϕp,k (λ; g; z) p(ϕδp,k (λ; g; z))0

912

C. SELVARAJ AND K. A. SELVAKUMARAN

Suppose that (32)

q(z) =

δ z(Dλ,g f (z))0

pϕδp,k (λ; g; z)

.

Then q(z) is analytic in U, with q(0) = 1. Differentiating both sides of (32) we get (33)

q(z) +

ϕδp,k (λ; g; z) (ϕδp,k (λ; g; z))

q 0 (z) = 0

δ (z(Dλ,g f (z))0 )0

p(ϕδp,k (λ; g; z))0

.

Now, using (31), (32) and (33) we deduce that q(z) + w(z)zq 0 (z) ≺ h(z),

(34) where

µ w(z) = α2

z(ϕδp,k (λ; g; z))0

¶−1 .

ϕδp,k (λ; g; z)

In view of Lemma 1.6 and α2 > 0, we observe that w(z) is analytic in U and R{w(z)} > 0. Consequently, in view of (34), we deduce from Lemma 1.5 that (35)

q(z) ≺ h(z). α1 α2

Since 0 ≤ < 1 and since h(z) is convex univalent in U, we deduce from (31) and (35) that (1 − α1 ) µ

δ z(Dλ,g f (z))0

pϕδp,k (λ; g; z)

+ α1

δ (z(Dλ,g f (z))0 )0

p(ϕδp,k (λ; g; z))0

δ δ z(Dλ,g f (z))0 (z(Dλ,g f (z))0 )0 α1 = (1 − α2 ) δ + α2 α2 pϕp,k (λ; g; z) p(ϕδp,k (λ; g; z))0

¶

µ

¶ α1 + 1− q(z) α2

≺ h(z). δ Thus f (z) ∈ Cp,k (α1 , λ; g; h) which completes the proof of Theorem 2.5.

¤

3. Integral operator Theorem 3.1. Let h(z) ∈ P and

½

R(c) R{h(z)} > max 0, − p

¾ (z ∈ U),

δ where c is a complex number such that R(c) > −p. If f (z) ∈ Sp,k (λ; g; h), then the function Z c + p z c−1 t f (t)dt (36) F (z) = zc 0 δ δ (λ; g; z) 6= 0 (0 < |z| < 1) (λ; g; h), provided that Fp,k is also in the class Sp,k δ where Fp,k (λ; g; z) is defined as in (6).

ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS

913

δ Proof. Let f (z) ∈ Sp,k (λ; g; h). Then from (36) and R(c) > −p, we note that F (z) ∈ Ap and δ δ δ (c + p)Dλ,g f (z) = cDλ,g F (z) + z(Dλ,g F (z))0 .

(37)

Also, from the above, we have (c +

δ p)fp,k (λ; g; z)

¶ µ k−1 1 X −jp j j j δ δ 0 = ε cDλ,g F (εk z) + εk z(Dλ,g F (εk z)) k j=0 k δ δ (λ; g; z) + z(Fp,k (λ; g; z))0 . = cFp,k

(38) Let

δ z(Fp,k (λ; g; z))0

w(z) =

δ (λ; g; z) pFp,k

.

Then w(z) is analytic in U, with w(0) = 1, and from (38) we observe that (39)

pw(z) + c = (c + p)

δ fp,k (λ; g; z) δ (λ; g; z) Fp,k

.

Differentiating both sides of (39) with respect to z and using Lemma 1.6, we obtain (40)

w(z) +

δ z(fp,k (λ; g; z))0 zw0 (z) = ≺ h(z). δ (λ; g; z) pw(z) + c pfp,k

In view of (40), Lemma 1.5 leads to w(z) ≺ h(z). If we let q(z) =

δ z(Dλ,g F (z))0 δ (λ; g; z) pFp,k

,

then q(z) is analytic in U, with q(0) = 1, and it follows from (37) that δ Fp,k (λ; g; z)q(z) =

(41)

c+p δ c δ Dλ,g f (z) − Dλ,g F (z). p p

Differentiating both sides of (41), we get zq 0 (z) +

δ z(Fp,k (λ; g; z))0 δ (λ; g; z) Fp,k

q(z) = (c + p)

δ z(Dλ,g f (z))0 δ (λ; g; z) pFp,k

−c

δ z(Dλ,g F (z))0 δ (λ; g; z) pFp,k

or equivalently, (42)

δ f (z))0 ¡ ¢ z(Dλ,g zq 0 (z) + pw(z) + c q(z) = (c + p) δ . pFp,k (λ; g; z)

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C. SELVARAJ AND K. A. SELVAKUMARAN

Now, from (39) and (42) we deduce that q(z) +

δ z(Dλ,g f (z))0 zq 0 (z) c+p = δ pw(z) + c pw(z) + c pFp,k (λ; g; z)

= (43)

δ z(Dλ,g f (z))0 δ (λ; g; z) pfp,k

≺ h(z),

δ because f (z) ∈ Sp,k (λ; g; h).

Combining, R{h(z)} > max{0, − R(c) p } and w(z) ≺ h(z) we have R{pw(z) + c} > 0 (z ∈ U). Therefore, from (43) and Lemma 1.5 we find that q(z) ≺ h(z), which shows δ that F (z) ∈ Sp,k (λ; g; h). ¤ By applying similar method as in Theorem 3.1, we have: Theorem 3.2. Let h(z) ∈ P and ½ ¾ R(c) R{h(z)} > max 0, − p

(z ∈ U;

R(c) > −p).

δ δ If f (z) ∈ Kp,k (λ; g; h) with respect to ϕ(z) ∈ Sp,k (λ; g; h), then the function Z z c+p F (z) = tc−1 f (t)dt zc 0 δ belongs to the class Kp,k (λ; g; h) with respect to Z c + p z c−1 G(z) = t g(t)dt, zc 0

provided that Gδp,k (λ; g; z) 6= 0

(0 < |z| < 1).

Acknowledgement. The authors thank the referees for their valuable comments and helpful suggestions. References [1] S. S. Miller and P. T. Mocanu, On some classes of first-order differential subordinations, Michigan Math. J. 32 (1985), no. 2, 185–195. [2] , Differential subordinations and inequalities in the complex plane, J. Differential Equations 67 (1987), no. 2, 199–211. [3] S. Owa, Z. Wu, and F. Y. Ren, A note on certain subclass of Sakaguchi functions, Bull. Soc. Roy. Sci. Liege 57 (1988), no. 3, 143–149. [4] F. Rønning, Uniformly convex functions and a corresponding class of starlike functions, Proc. Amer. Math. Soc. 118 (1993), no. 1, 189–196. [5] K. Sakaguchi, On a certain univalent mapping, J. Math. Soc. Japan 11 (1959), 72–75. [6] C. Selvaraj and C. Santhosh Moni, Classes of analytic functions of complex order involving a family of generalised linear operators, Pre-print. [7] N.-E. Xu and D.-G. Yang, Some classes of analytic and multivalent functions involving a liner operator, Mathematical and Computer Modelling (2008), doi: 10.1016/ j.mcm.2008.04.005 - In press.

ON CERTAIN CLASSES OF MULTIVALENT FUNCTIONS

[8]

915

, An application of differential subordinations and some criteria for starlikeness, Indian J. Pure Appl. Math. 36 (2005), no. 10, 541–556. [9] D.-G. Yang and J.-L. Liu, On Sakaguchi functions, Int. J. Math. Math. Sci. 2003, no. 30, 1923–1931. [10] Z.-G. Wang, Y.-P. Jiang, and H. M. Srivastava, Some subclasses of multivalent analytic functions involving the Dziok-Srivastava operator, Integral Transforms Spec. Funct. 19 (2008), no. 1-2, 129–146. Chellian Selvaraj Department of Mathematics Presidency College (Autonomous) Chennai-600 005, India E-mail address: [email protected] Kuppathai A. Selvakumaran Department of Mathematics R. M. K. Engg. College Kavaraipettai-601 206, India E-mail address: [email protected]