Some convolution properties of certain classes of analytic functions

Applied Mathematics Letters 21 (2008) 1155–1160 www.elsevier.com/locate/aml

Some convolution properties of certain classes of analytic functions Khalida Inayat Noor a , Nak Eun Cho b,∗ a Mathematics Department, COMSATS Institute of Information Technology, Islamabad, Pakistan b Department of Applied Mathematics, Pukyong National University, Busan 608-737, South Korea

Received 8 May 2006; received in revised form 6 August 2007; accepted 11 December 2007

Abstract r

In this work, we discuss certain classes Tα 0 (β, k) (α ≥ 0, 0 ≤ β < 1, k ≥ 2, 0 < r0 ≤ 1) of analytic functions in the unit disk. Some convolution properties and radii problems are investigated. c 2008 Elsevier Ltd. All rights reserved.

Keywords: Convolution; Convex function; Starlike function; Univalent function; Function with positive real part

1. Introduction Let A denote the class of functions of the form f (z) = z +

∞ X

an z n ,

n=2

that are analytic in the unit disk E = {z : |z| < 1}. For r0 ∈ (0, 1], we let Er0 = {z ∈ C : |z| < r0 } and we denote by r Pk 0 (β) the class of functions p : E −→ C analytic in E satisfying the properties p(0) = 1 and 2π

Z 0

Re p(z) − β dθ ≤ kπ, 1−β

for all z = r eiθ ∈ Er0 , r

(1.1)

where k ≥ 2, 0 ≤ β < 1 and Pk (β) ≡ Pk1 (β). The class Pk 0 (β) has been introduced in [5]. We note that, for β = 0, we obtain the class Pk defined in [6] and for β = 0, k = 2, we have the class P of functions with positive real part (see also [1–4,7–10]). The case k = 2 gives us the class P(β) of functions with positive real part greater than β. Also we can write Z 1 2π 1 + (1 − 2β)ze−it p(z) = dµ(t), (1.2) 2 0 1 − ze−it ∗ Corresponding author.

E-mail addresses: [email protected], [email protected] (K. Inayat Noor), [email protected] (N.E. Cho). c 2008 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2007.12.014

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K. Inayat Noor, N.E. Cho / Applied Mathematics Letters 21 (2008) 1155–1160

where µ is a function with bounded variation on [0, 2π ] such that 2π

Z

2π

Z dµ(t) = 2

and

(1.3)

|dµ(t)| ≤ k. 0

0

r

From (1.1), we can write, for p ∈ Pk 0 (β),  p(z) =

k 1 + 4 2



p1 (z) −



k 1 − 4 2



p2 (z),

(1.4)

where p1 , p2 ∈ P(β). The convolution or Hadamard product of two analytic functions f (z) =

∞ X

am z m+1

and

g(z) =

m=0

∞ X

bm z m+1

m=0

in E is the function f ? g defined by ( f ? g)(z) =

∞ X

am bm z m+1 ,

z ∈ E.

m=0

r

Definition 1.1. For α ≥ 0, 0 ≤ β < 1, k ≥ 2 and r0 ∈ (0, 1], we denote by Tα 0 (β, k) the class of functions r f : E −→ C analytic in E with the property that f 0 (z) + αz f 00 (z) ∈ Pk 0 (β). 2. Preliminary results Using relation (1.4) and some known results [8,9], the following lemmas can be derived easily. Lemma 2.1. For α, β ∈ [0, 1) with (1 − α)(1 − β) ≤ r

r

1 2

and k ≥ 2,

r

Pk 0 (α) ? Pk 0 (β) ⊂ Pk 0 (δ),

(2.1)

where δ = 1 − 2(1 − α)(1 − β). The result is sharp. Lemma 2.2. If p is an analytic function in E with p(0) = 1 and if λ1 is a complex number satisfying Re λ1 ≥ 0 (λ1 6= 0), then r

r

p(z) + λ1 zp 0 (z) ∈ Pk 0 (β) H⇒ p(z) ∈ Pk 0 (δ1 ), where δ1 = β + (1 − β)(2γ − 1)

(2.2)

and γ = γ (Re λ1 ) =

1

Z

1+t

Re λ1

−1

dt,

0

which is an increasing function of Re λ1 and

1 2

≤ γ < 1. The result is sharp.

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3. Main results r

r

Theorem 3.1. Let f, g ∈ Tα 0 (β, k) and let F(z) = ( f ? g)(z) for z ∈ E. Then F ∈ Tα 0 (σ, k), where σ = min(σ1 , σ2 ) with σ1 = 1 − 2(1 − δ1 )2 ,

σ2 = (2σ1 − 1) + 2(1 − σ1 ) ln 2,

and δ1 is given by (2.2). r

Proof. Since f ∈ Tα 0 (β, k), we have r

f 0 (z) + αz f 00 (z) ∈ Pk 0 (β). r

r

Let f 0 (z) = p(z), where p is analytic in E with p(0) = 1. Then p(z)+αzp 0 (z) ∈ Pk 0 (β) implies that p(z) ∈ Pk 0 (δ1 ) r by using Lemma 2.2 with λ1 = α, where δ1 is given by (2.2) with λ1 replaced by α. Similarly g 0 ∈ Pk 0 (δ1 ). So for F = f ? g, we have, from Lemma 2.1, r

(z F 0 (z))0 = ( f 0 ? g 0 )(z) ∈ Pk 0 (σ1 ), where σ1 = 1 − 2(1 − δ1 )2 and δ1 is given by (2.2) with λ1 replaced by α. Since r

(z F 0 (z))0 = F 0 (z) + z F 00 (z) ∈ Pk 0 (σ1 ), r

by using Lemma 2.2 with α = 1, we obtain F 0 (z) ∈ Pk 0 (σ2 ), where σ2 = (2σ1 − 1) + 2(1 − σ1 )ln2. Thus F 0 (z) + αz F 00 (z) = (1 − α)F 0 (z) + α(z F 0 (z))0 = (1 − α)H1 (z) + α H2 (z),
0 r r r where H1 (z) = F 0 (z) ∈ Pk 0 (σ2 ) and H2 (z) = z F 0 (z) ∈ Pk 0 (σ1 ). Since Pk 0 (σ ) is a convex set [4] with r0 σ = min(σ1 , σ2 ), we conclude that F = f ? g ∈ Tα (σ, k).  r

r

Theorem 3.2. Let φ be convex univalent in E and let f ∈ Tα 0 (β, k). Then φ ? f ∈ Tα 0 (β, k). 1 Proof. Since φ is convex, it is well known that φ(z) z ∈ P( 2 ). Now     φ(z) φ(z) ? f 0 (z) + ? αz f 00 (z) (φ ? f )0 (z) + αz(φ ? f )00 (z) = z z φ(z) = ? [ f 0 (z) + αz f 00 (z)] z φ(z) ? p(z), = z       φ(z) φ(z) k 1 k 1 = + ? p1 (z) − − ? p2 (z) , 4 2 z 4 2 z r

r

where p ∈ Pk 0 (β) and pi ∈ P(β). Using (1.4) and Lemma 2.1 with k = 2, we see that φ ? f ∈ Tα 0 (β, 2).



Remark 3.1. Let, for z ∈ E, Ψ1 (z) =

∞ X 1+c n z , n +c n=1

c > −1,

∞ n X z = − log(1 − z), n n=1   1 1 + xz Ψ3 (z) = log , log 1 = 0, |x| ≤ 1, x 6= 1. 1−x 1−z

Ψ2 (z) =

r

It is known [9] that Ψi (i = 1, 2, 3) are convex univalent in E. So, using Theorem 3.2, we see that Tα 0 (β, k) is invariant under the following integral operators:

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R z c−1 (i) ( f ? Ψ1 )(z) = 1+c t f (t)dt, c > −1, Rzzc f (t)0 (ii) ( f ? Ψ2 )(z) = 0 t dt, Rz f (xt) (iii) ( f ? Ψ3 )(z) = 0 f (t)− dt, |x| ≤ 1, x 6= 1. t−xt Remark 3.2. Let z

Z

1 1− α1 k(z) = z α

0

1 −1

tα dt, 1−t

0 < α ≤ 1, z ∈ E.

(3.1) r

The function k(z) is convex [9]. From a result [1, p. 42], we note that the functions in Tα 0 (β, k) for 0 < α ≤ 1 are obtained by taking the convolution (or Hadamard product) of the convex function k(z) with the function J (z) defined by Z z J (z) = p(t)dt, 0 r

where p ∈ Pk 0 (β). If f has the form f (z) = z +

∞ X

an z n ,

(3.2)

n=2

then we define the sections of f as f n (z) = z +

n X

ak z k .

k=2

Then we can write f n (z) = ( f ? gn )(z),

(3.3)

where gn (z) = z +

n X

zk =

k=2

z − z n+1 . 1−z

(3.4)

It is known [9] that gn is starlike for z ∈ Ern , where  rn =

1 2n

1 n

.

(3.5) r

Theorem 3.3. Let f ∈ Tα 0 (β, k) and let the function Fn : E −→ C be defined by Z z f n (t) Fn (z) = dt, t 0 where f n is given by (3.3). Then Fn ∈ Tαrn (β, k), where rn is given by (3.5). r

Proof. We can write, for f ∈ Tα 0 (β, k), Z z f (z) = (k ? J )(z) = k(z) ? p(t)dt, 0

where p ∈

r Pk 0 (β)

and k is a convex function given by (3.1). Now we have

f n (z) = (gn ? f )(z) = zG 0n (z) ? f (z), = zG 0n (z) ? k(z) ?

z

Z

p(t)dt, 0

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K. Inayat Noor, N.E. Cho / Applied Mathematics Letters 21 (2008) 1155–1160

where G n (z) =

Rz 0

gn (ξ ) ξ dξ

r

and p ∈ Pk 0 (β). Therefore, using (1.4), we have

f n (z) z         k(z) G n (z) k G n (z) k 1 1 = ? + ? p1 (z) − − ? p2 (z) , z 4 2 z 4 2 z

Fn0 (z) =

where pi ∈ P(β) (i = 1, 2). Since G n is convex in Ern , where rn is given by (3.5), we conclude that P(β) in Ern (i = 1, 2) by using Lemma 2.1 and thus Fn0 (z) =

G n (z) z

? pi (z) ∈

k(z) ? h(z), z

where h ∈ Pkrn (β). This implies Z z h(t)dt ∈ Tαrn (β, k), Fn (z) = k(z) ? 0

where rn is given by (3.5).

 r

Remark 3.3. (i) It is known that f with f 0 ∈ P is close-to-convex and so univalent in E [2]. If p ∈ Pk 0 (β), then √ r p ∈ P(β) ⊂ P in Erk , where rk = 21 [k − k 2 − 4] [6]. Therefore by Lemma 2.2, f ∈ Tα 0 (β, k) is univalent in r Erk . In particular, Tα 0 (β, 2) consists entirely of univalent functions in E. r (ii) It can easily be shown that the class Tα 0 (β, k) is a convex set. r

Theorem 3.4. Let f 0 ∈ Pk 0 (β). Then f ∈ Tαrα (β, k), where rα =

√ 2α +

1 4α 2 − 2α + 1

.

(3.6)

Proof. We can write φα (z) ? f 0 (z) z      φα (z) k k 1 1 = ? + h 1 (z) − − h 2 (z) , z 4 2 4 2

(1 − α) f 0 (z) + α(z f 0 (z))0 =

where h i ∈ P(β) (i = 1, 2) and z z φα (z) = (1 − α) +α 1−z (1 − z)2 ∞ X = z+ (1 + (n − 1)α) z n . n=2

It is known [3] that φα is convex in Erα , where rα is given by (3.6), and consequently φαz(z) ∈ P( 12 ). Now using r Lemma 2.1 (with k = 2) along with relation (1.4) and the fact that f 0 ∈ Pk 0 (β), we obtain the required result.  Alternative Proof of Theorem 3.4. We can write φα (z) f 0 (z) + α f 00 (z) = ? f 0 (z). z Since φαz(z) ∈ P( 21 ) = P2 0 ( 11 ) ⊂ Pk 0 ( 21 ) in the corresponding disk and f 0 ∈ Pk 0 (β), the result follows from Lemma 2.1.  r

r

Theorem 3.5. Let f : E −→ C be analytic in E such that (1 − α)

f (z) r + α f 0 (z) ∈ Pk 0 (β), z

r

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for some 0 < α < 1. Then f 0 ∈ PkRα (β), where Rα is the smallest positive root of the equation 2 α

−1−r 2 − 1+r α



1 −1 α

1

Z 0

1 −1

tα dt = 0. 1 + tr

(3.7)

Proof. We can write f (z) = p(z) ? k(z), r

where p ∈ Pk 0 (β) and k is defined by (3.1). That is, from Remark 3.2, f 0 (z) = p(z) ? k 0 (z) and 1 1 k (z) = − α(1 − z) α 0



 1 Z z α1 −1 −α 1 t −1 z dt. α 1 −t 0

(3.8)

Powers in (3.8) are the principal values. The function k 0 is analytic, k 0 (0) = 1 and Z 1 1    1+z 2 1 1 1 α −1 0 2k (z) − 1 = + −1 − dt. t 1−z α α 1−z 1 − tz 0 So  Z 1 α1 −1 1 t (1 − t)r −1 dt α (1 − r )(1 − tr ) 0  Z 1 α1 −1  −1−r t 2 1 − −1 dt, 1+r α α 0 1 − tr

1−r 2 Re{2k (z) − 1} ≥ − 1+r α 0

=

2 α



where r = |z|, z ∈ E. Therefore Re k 0 (z) > 12 in E Rα , where Rα is the smallest positive root of (3.7). We now apply Lemma 2.1 with relation (1.4) and this gives us the required result.  Acknowledgements The authors would like to thank the referees for their valuable suggestions and comments. The alternative proof of Theorem 3.4 is due to a referee. The research of Prof. Dr. Khalida Noor is supported by the Higher Education Commission, Pakistan, through research grant No: 1-28/HEC/HRD/2005/90. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

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