1 + Bz. ,. } where 0
International Journal of Analysis and Applications ISSN 2291-8639 Volume 1, Number 2 (2013), 100-105 http://www.etamaths.com
A SUBORDINATION THEOREM INVOLVING A MULTIPLIER TRANSFORMATION SUKHWINDER SINGH BILLING
Abstract. We, here, study a certain differential subordination involving a multiplier transformation which unifies some known differential operators. As a special case to our main result, we find some new results providing the best dominant for z p /f (z), z/f (z) and z p−1 /f 0 (z), 1/f 0 (z).
1. Introduction Let A be the class of all functions f analytic in the open unit disk E = {z ∈ C : |z| < 1} and normalized by the conditions that f (0) = f 0 (0) − 1 = 0. Thus, f ∈ A has the Taylor series expansion f (z) = z +
∞ X
ak z k .
k=2
Let Ap denote the class of functions of the form f (z) = z p +
∞ X
ak z k , p ∈ N =
k=p+1
{1, 2, 3, · · · }, which are analytic and multivalent in the open unit disk E. Note A1 = A. For f ∈ Ap , define the multiplier transformation Ip (n, λ) as n ∞ X k+λ Ip (n, λ)f (z) = z p + ak z k , (λ ≥ 0, n ∈ N0 = N ∪ {0}). p+λ k=p+1
The operator Ip (n, λ) has been recently studied by Aghalary et al. [3]. I1 (n, 0) is the well-known S˘ al˘ agean [1] derivative operator Dn , defined for f ∈ A as under: Dn f (z) = z +
∞ X
k n ak z k .
k=2
For two analytic functions f and g in the unit disk E, we say that f is subordinate to g in E and write as f ≺ g if there exists a Schwarz function w analytic in E with w(0) = 0 and |w(z)| < 1, z ∈ E such that f (z) = g(w(z)), z ∈ E. In case the function g is univalent, the above subordination is equivalent to: f (0) = g(0) and f (E) ⊂ g(E). 2010 Mathematics Subject Classification. 30C80, 30C45. Key words and phrases. Differential subordination, Multiplier transformation, Analytic function; Univalent function, p-valent function. c
2013 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.
100
A SUBORDINATION THEOREM
101
Let Φ : C2 × E → C be an analytic function, p be an analytic function in E such that (p(z), zp0 (z); z) ∈ C2 × E for all z ∈ E and h be univalent in E. Then the function p is said to satisfy first order differential subordination if (1)
Φ(p(z), zp0 (z); z) ≺ h(z), Φ(p(0), 0; 0) = h(0).
A univalent function q is called a dominant of the differential subordination (1) if p(0) = q(0) and p ≺ q for all p satisfying (1). A dominant q˜ that satisfies q˜ ≺ q for each dominant q of (1), is said to be the best dominant of (1). Obradovi˘c [2], introduced and studied the class N (α), 0 < α < 1 of functions f ∈ A satisfying the following inequality ( 1+α ) z > 0, z ∈ E. < f 0 (z) f (z) He called it, the class of non-Bazilevi˘c functions. In 2005, Wang et al. [6] introduced the generalized class N (λ, α, A, B) of nonBazilevi˘c functions which is analytically defined as: α α z z zf 0 (z) 1 + Az , N (λ, α, A, B) = f ∈ A : (1 + λ) −λ ≺ f (z) f (z) f (z) 1 + Bz where 0 < α < 1, λ ∈ C, −1 ≤ B ≤ 1, A 6= B, A ∈ R. Wang αet al. [6] studied the class N (λ, α, A, B) and made some estimates on z . f (z) Using the concept of differential subordination, et al. [5] studied α Shanmugam α z zf 0 (z) z the differential operator (1+λ) −λ and obtained the best f (z) f (z) f (z) α z dominant for . f (z) The main objective of this paper is to unify the above mentioned differential operators. For this, we establish a differential subordination involving the multiplier transformation Ip (n, λ), defined above. As special cases of main theorem, we obtain best dominant for z p /f (z), z/f (z) and z p−1 /f 0 (z), 1/f 0 (z) and some known results also appear as special cases to our main result. To prove our main result, we shall make use of the following lemma of Miller and Macanu [4]. Lemma 1.1. Let q be univalent in E and let θ and φ be analytic in a domain D containing q(E), with φ(w) 6= 0, when w ∈ q(E). Set Q(z) = zq 0 (z)φ[q(z)], h(z) = θ[q(z)] + Q(z) and suppose that either (i) h is convex, or (ii) Q is starlike. In addition, assume that 0 (z) (iii) < zh Q(z) > 0, z ∈ E. If p is analytic in E, with p(0) = q(0), p(E) ⊂ D and θ[p(z)] + zp0 (z)φ[p(z)] ≺ θ[q(z)] + zq 0 (z)φ[q(z)], then p(z) ≺ q(z) and q is the best dominant.
102
BILLING
2. Main Results In what follows, all the powers taken are the principal ones. Theorem 2.1. Let α and β be non-zero complex numbers such that < (β/α) > 0 β zp and let f ∈ Ap , 6= 0, z ∈ E, satisfy the differential subordination Ip (n, λ)f (z) (2) β Ip (n + 1, λ)f (z) 1 + Az α (A − B)z zp , 1+α−α ≺ + Ip (n, λ)f (z) Ip (n, λ)f (z) 1 + Bz β(p + λ) (1 + Bz)2 then
zp Ip (n, λ)f (z)
β ≺
1 + Az , −1 ≤ B < A ≤ 1, z ∈ E, 1 + Bz
1 + Az is the best dominant. 1 + Bz β zp , a little calculation yields that Proof: On writing u(z) = Ip (n, λ)f (z) β zp Ip (n + 1, λ)f (z) α (3) 1+α−α = u(z) + zu0 (z), Ip (n, λ)f (z) Ip (n, λ)f (z) β(p + λ) and
Define the functions θ and φ as follows: θ(w) = w and φ(w) =
α . β(p + λ)
Clearly, the functions θ and φ are analytic in domain D = C and φ(w) 6= 0, w ∈ D. 1 + Az Select q(z) = , −1 ≤ B < A ≤ 1, z ∈ E and define the functions Q and h 1 + Bz as follows: α (A − B)z α zq 0 (z) = , Q(z) = zq 0 (z)φ(q(z)) = β(p + λ) β(p + λ) (1 + Bz)2 and (4) h(z) = θ(q(z))+Q(z) = q(z)+
α 1 + Az α (A − B)z zq 0 (z) = + . β(p + λ) 1 + Bz β(p + λ) (1 + Bz)2
A little calculation yields 0 zQ (z) zq 00 (z) 1 − Bz < =< 1+ 0 =< > 0, z ∈ E, Q(z) q (z) 1 + Bz i.e. Q is starlike in E and 0 zh (z) zq 00 (z) β 1 − Bz β < =< 1+ 0 + (p + λ) =< +(p+λ)< > 0, z ∈ E. Q(z) q (z) α 1 + Bz α Thus conditions (ii) and (iii) of Lemma 1.1, are satisfied. In view of (2), (3) and (4), we have θ[u(z)] + zu0 (z)φ[u(z)] ≺ θ[q(z)] + zq 0 (z)φ[q(z)]. Therefore, the proof follows from Lemma 1.1. For p = 1 and λ = 0 in above theorem, we get the following result involving S˘ al˘ agean operator.
A SUBORDINATION THEOREM
103
Theorem 2.2. If α, β are non-zero complex numbers such that < (β/α) > 0. If β z 6= 0, z ∈ E, satisfies f ∈ A, Dn f (z) β z Dn+1 f (z) 1 + Az α (A − B)z , −1 ≤ B < A ≤ 1, z ∈ E, 1 + α − α ≺ + n n D f (z) D f (z) 1 + Bz β (1 + Bz)2 then
z Dn f (z)
β ≺
1 + Az , z ∈ E. 1 + Bz
3. Dominant for z p /f (z), z/f (z) This section is concerned with the results giving the best dominant for z p /f (z) and z/f (z). Select λ = n = 0 in Theorem 2.1, we obtain the following result. Corollary 3.1. Let α, β be non-zero complex numbers such that < (β/α) > 0 p β z and let f ∈ Ap , 6= 0, z ∈ E, satisfy f (z) p β p β z z α (A − B)z zf 0 (z) 1 + Az + , z ∈ E, (1 + α) −α ≺ f (z) pf (z) f (z) 1 + Bz pβ (1 + Bz)2 then
zp f (z)
β ≺
1 + Az , −1 ≤ B < A ≤ 1, z ∈ E. 1 + Bz
Taking β = 1 in above theorem, we obtain: Corollary 3.2. Suppose that α is a non-zero complex number such that < (1/α) > zp 0 and suppose that f ∈ Ap , 6= 0, z ∈ E, satisfies f (z) (1 + α) then
zp z p+1 f 0 (z) 1 + Az α (A − B)z −α ≺ + , z ∈ E, 2 f (z) p(f (z)) 1 + Bz p (1 + Bz)2 zp 1 + Az ≺ , −1 ≤ B < A ≤ 1, z ∈ E. f (z) 1 + Bz
On writing α = −1 in Corollary 3.1, we get: Corollary 3.3. Let β be a complex number with < (β) < 0 and let f ∈ Ap , 0, z ∈ E, satisfy p β zf 0 (z) z 1 + Az 1 (A − B)z ≺ − , −1 ≤ B < A ≤ 1, z ∈ E, pf (z) f (z) 1 + Bz pβ (1 + Bz)2 then
zp f (z)
β ≺
1 + Az , z ∈ E. 1 + Bz
Selecting α = β = 1/2 in Corollary 3.1, we get:
zp f (z)
β 6=
104
BILLING
r
zp 6= 0, z ∈ E, satisfies f (z) r p z zf 0 (z) 2(1 + Az) 2 (A − B)z , z ∈ E, 3− ≺ + f (z) pf (z) 1 + Bz p (1 + Bz)2
Corollary 3.4. If f ∈ Ap ,
then r
zp 1 + Az ≺ , −1 ≤ B < A ≤ 1, z ∈ E. f (z) 1 + Bz
Taking p = 1 in Corollary 3.2, we have the following result. Corollary 3.5. If α is a non-zero complex number such that < (1/α) > 0 and z if f ∈ A, 6= 0, z ∈ E, satisfies f (z) (1 + α)
1 + Az z z 2 f 0 (z) (A − B)z ≺ , −1 ≤ B < A ≤ 1, z ∈ E, −α +α f (z) (f (z))2 1 + Bz (1 + Bz)2
then z 1 + Az ≺ , z ∈ E. f (z) 1 + Bz Setting p = 1 in Corollary 3.3, we have the following result. Corollary 3.6. If β is a complex number with < (β) < 0 and if f ∈ A,
z f (z)
β 6=
0, z ∈ E, satisfies z β+1 f 0 (z) 1 + Az 1 (A − B)z ≺ − , −1 ≤ B < A ≤ 1, z ∈ E, (f (z))β+1 1 + Bz β (1 + Bz)2 then
z f (z)
β ≺
1 + Az , z ∈ E. 1 + Bz
Setting p = 1 in Corollary 3.1, we obtain, below, the result of Shanmugam et al. [5]. Corollary 3.7. If α, β are non-zero complex numbers such that < (β/α) > 0. β z If f ∈ A, 6= 0, z ∈ E, satisfies f (z) β 1+β z z 1 + Az α (A − B)z (1 + α) − αf 0 (z) ≺ + , z ∈ E, f (z) f (z) 1 + Bz β (1 + Bz)2 then
z f (z)
β ≺
1 + Az , −1 ≤ B < A ≤ 1, z ∈ E. 1 + Bz
4. Dominant for z p−1 /f 0 (z), 1/f 0 (z) We obtain here, the best dominant for z p−1 /f 0 (z) and 1/f 0 (z) as special cases to our main result. Select λ = 0 and n = 1 in Theorem 2.1, we obtain:
A SUBORDINATION THEOREM
105
Corollary 4.1. Let α, β be non-zero complex numbers such that < (β/α) > 0 p−1 β pz and let f ∈ Ap , 6= 0, z ∈ E, satisfy f 0 (z) p−1 β p−1 β α 1 + Az pz zf 00 (z) pz α (A − B)z − ≺ , z ∈ E, (1 + α) 1 + + 0 0 0 f (z) p f (z) f (z) 1 + Bz pβ (1 + Bz)2 then p−1 β pz 1 + Az ≺ , −1 ≤ B < A ≤ 1, z ∈ E. 0 f (z) 1 + Bz Taking β = 1 in above theorem, we obtain: Corollary 4.2. Suppose that α is a non-zero complex number such that < (1/α) > pz p−1 0 and suppose that f ∈ Ap , 0 6= 0, z ∈ E, satisfies f (z) pz p−1 z p−1 zf 00 (z) 1 + Az α (A − B)z , z ∈ E, (1 + α) 0 −α 0 1+ 0 ≺ + f (z) f (z) f (z) 1 + Bz p (1 + Bz)2 then z p−1 1 + Az ≺ , −1 ≤ B < A ≤ 1, z ∈ E. 0 f (z) p(1 + Bz) Taking p = 1 in Corollary 4.2, we have the following result. Corollary 4.3. If α is a non-zero complex number such that < (1/α) > 0 and 1 if f ∈ A, 0 6= 0, z ∈ E, satisfies f (z) zf 00 (z) 1 + Az (A − B)z 1 1−α 0 ≺ +α , −1 ≤ B < A ≤ 1, z ∈ E, f 0 (z) f (z) 1 + Bz (1 + Bz)2 then 1 1 + Az ≺ , z ∈ E. f 0 (z) 1 + Bz References [1] G. S. S˘ al˘ agean, Subclasses of univalent functions, Lecture Notes in Math., 1013 362–372, Springer-Verlag, Heideberg,1983. [2] M. Obradovi˘ c, A class of univalent functions, Hokkaido Mathematical Journal, 27(2)(1998) 329–335. [3] R. Aghalary, R. M. Ali, S. B. Joshi and V. Ravichandran, Inequalities for analytic functions defined by certain linear operators, Int. J. Math. Sci., 4(2005) 267–274. [4] S. S. Miller and P. T. Mocanu, Differential Suordinations : Theory and Applications, (No. 225), Marcel Dekker, New York and Basel, 2000. [5] T. N. Shanmugam, S. Sivasubramanian and H. Silverman, On sandwich theorems for some classes of analytic functions, International J. Math. and Math. Sci., Article ID 29684(2006) pp.1–13. [6] Z. Wang, C. Gao and M. Liao, On certain generalized class of non-Bazilevi˘ c functions, Acta Mathematica Academiae Paedagogicae Ny´ıregyh´ aziensis, New Series, 21(2)(2005) 147–154. Department of Applied Sciences, Baba Banda Singh Bahadur Engineering College, Fatehgarh Sahib-140 407, Punjab, India
