Hindawi Publishing Corporation Geometry Volume 2013, Article ID 671826, 4 pages http://dx.doi.org/10.1155/2013/671826
Research Article On a Subclass of Meromorphic Functions Defined by Hilbert Space Operator Thomas Rosy and S. Sunil Varma Department of Mathematics, Madras Christian College, Tambaram, Chennai, Tamil Nadu 600 059, India Correspondence should be addressed to Thomas Rosy;
[email protected] Received 26 April 2013; Accepted 30 May 2013 Academic Editor: JinLin Liu Copyright Β© 2013 T. Rosy and S. S. Varma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this paper, we define a new operator on the class of meromorphic functions and define a subclass using Hilbert space operator. Coefficient estimate, distortion bounds, extreme points, radii of starlikeness, and convexity are obtained.
let π(π) denote the operator on π» defined by the RieszDunford integral [4]
1. Introduction Let Ξ£ denote the class of meromorphic functions
π (π) =
β
1 π (π§) = + β ππ π§π π§ π=1
(1)
defined on π = {π§ β πΆ : 0 < |π§| < 1}. For π β Ξ£ given by (1) and π β Ξ£ given by
where πΌ is the identity operator on π» and πΆ is a positively oriented simple closed rectifiable closed contour containing the spectrum π(π) in the interior domain [5]. The operator π(π) can also be defined by the following series: π(π) (0) π π π! π=0
π (π) = β
(2)
the Hadamard product (or convolution) [1] of π and π is defined by 1 β + β π π π§π , π§ π=1 π π
(4)
β
1 β π (π§) = + β ππ π§π , π§ π=1
(π β π) (π§) =
1 β« (π§πΌ β π)β1 π (π§) ππ§, 2ππ πΆ
π§ β π.
(3)
Many subclasses of meromorphic functions have been defined and studied in the past. In particular, the subclasses πΌ ππ,π1 (π, πΌ), 0 β€ πΌ < 1 and 0 β€ π < 1 [2], and π(πΌ, π½), 0 β€ πΌ < 1 and 0 < π½ β€ 1 [3], are considered by researchers. Let π» be a complex Hilbert space and πΏ(π») denote the algebra of all bounded linear operators on π». For a complex-valued function π analytic in a domain πΈ of the complex plane containing the spectrum π(π) of the bounded linear operator π,
(5)
which converges in the norm topology. In this paper, we introduce a subclass ππ (πΌ, π, π) of Ξ£ defined using Hilbert space operator and prove a necessary and sufficient condition for the function to belong to this class, the distortion theorem, radius of starlikeness, and convexity. In [6], Atshan and Buti had defined an operator acting on analytic functions in terms of a definite integral. We modify their operator for meromorphic functions as follows. Lemma 1. For π β Ξ£ given by (1), 0 β€ π β€ 1, and 0 β€ π β€ 1, if the operator πππ : Ξ£ β Ξ£ is defined by πππ π (π§) =
β
1 (1 β π)
π+1
Ξ (π + 1)
β« π‘π+1 πβ(π‘/(1βπ)) π (π§π‘) ππ‘, 0
(6)
2
Geometry
then
Conversely, let πππ π (π§) =
1 β + β πΏ (π, π, π) ππ π§π , π§ π=1
(7)
where πΏ(π, π, π) = (1 β π)π+1 Ξ(π + π + 2)/Ξ(π + 1). Denote by Ξ£π the class of all functions π β Ξ£ with ππ β₯ 0. Definition 2. For 0 β€ π < 1, 0 β€ πΌ < 1, a function π β Ξ£π is in the class ππ (πΌ, π, π) if σ΅©σ΅© σ΅© σ΅©σ΅©π(ππ π (π))σΈ β {(π β 1) ππ π (π) + ππ(ππ π (π))σΈ }σ΅©σ΅©σ΅© π π π σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σΈ < σ΅©σ΅©σ΅©π(πππ π (π)) σ΅© σΈ σ΅© σ΅© + (1 β 2πΌ) {(π β 1) πππ π (π) + ππ(πππ π (π)) }σ΅©σ΅©σ΅© σ΅©
(8)
for all operators π with βπβ < 1 and π =ΜΈ 0, 0 being the zero operator on π».
2. Coefficient Bounds Theorem 3. A meromorphic function π β Ξ£π given by (1) is in the class ππ (πΌ, π, π) if and only if β
β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1 β πΌ.
(9)
π=1
σ΅© σ΅©σ΅© σ΅©σ΅©π(ππ π (π))σΈ β {(π β 1) ππ π (π) + ππ(ππ π (π))σΈ }σ΅©σ΅©σ΅© π π π σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σΈ < σ΅©σ΅©σ΅©π(πππ π (π)) σ΅© σΈ σ΅© σ΅© + (1 β 2πΌ) {(π β 1) πππ π (π) + ππ(πππ π (π)) }σ΅©σ΅©σ΅© . σ΅©
(11)
This implies that σ΅©σ΅© β σ΅©σ΅© σ΅©σ΅© σ΅© σ΅©σ΅© β [(π + 1) (1 β π)] πΏ (π, π, π) ππ ππ+1 σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π=1 σ΅©σ΅© σ΅©σ΅© β σ΅©σ΅© β€ σ΅©σ΅©σ΅©2 (1 β πΌ) β β [π + (1 β 2πΌ) (π β 1) + π (1 β 2πΌ) π] σ΅©σ΅© π=1 σ΅© σ΅©σ΅© σ΅©σ΅© Γ πΏ (π, π, π) ππ ππ+1 σ΅©σ΅©σ΅© . σ΅©σ΅© σ΅© (12) Choose π = ππΌ, 0 < π < 1. Then, ββ π=1 [(π + 1)(1 β [π + (1 β 2πΌ)(π β 1) + π(1 β π)]πΏ(π, π, π)ππ ππ /(2π(1 β πΌ) β ββ π=1 2πΌ)π]πΏ(π, π β
π)ππ ππ ) β€ 1. As π β 1, we obtain (9). Corollary 4. If π of the form (1) is in ππ (πΌ, π, π), then ππ β€
1βπΌ . [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π)
(13)
The result is sharp for π(π§) = 1/π§ + ((1 β πΌ)/[π + πΌ β πΌπ(π + 1)]πΏ(π, π, π))π§π .
The result is sharp for the function π(π§) = 1/π§ + ((1 β πΌ)/[π + πΌ β πΌπ(π + 1)]πΏ(π, π, π))π§π .
π Proof. Let π(π) = πβ1 + ββ π=1 ππ π . Assume that (9) holds. Then,
Theorem 5. Let π0 (π§) = 1/π§ and ππ (π§) = 1/π§ + ((1 β πΌ)/[π + πΌβπΌπ(π+1)]πΏ(π, π, π))π§π , π β₯ 1, 0 β€ πΌ < 1, 0 β€ π < 1. Then, π is in the class ππ (πΌ, π, π) if and only if it can be expressed in the form π(π§) = ββ π=0 ππ ππ (π§), where ππ β₯ 0 (π = 0, 1, 2, . . .) and ββ π=0 ππ = 1.
σ΅©σ΅© σ΅© σ΅©σ΅©π(ππ π (π))σΈ β {(π β 1) ππ π (π) + ππ(ππ π (π))σΈ }σ΅©σ΅©σ΅© π π π σ΅©σ΅© σ΅©σ΅© σΈ σ΅©σ΅© β σ΅©σ΅©σ΅©π(πππ π (π)) σ΅©
σΈ σ΅© σ΅© + (1 β 2πΌ) {(π β 1) πππ π (π) + ππ(πππ π (π)) }σ΅©σ΅©σ΅© σ΅© σ΅©σ΅© β σ΅© σ΅© σ΅©σ΅© σ΅©σ΅© = σ΅©σ΅©σ΅© β [(π + 1) β π (π + 1)] πΏ (π, π, π) ππ ππ σ΅©σ΅©σ΅© σ΅©σ΅©π=1 σ΅©σ΅© σ΅© σ΅© σ΅©σ΅© β1 (10) β σ΅©σ΅©σ΅©2π (1 β πΌ) β
β β [π + (1 β 2πΌ) (π β 1) + π (1 β 2πΌ) π] π=1
σ΅© Γ πΏ (π, π, π) ππ ππ σ΅©σ΅©σ΅©σ΅© β
β€ β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β (1 β πΌ) π=1
β€ 0, by (9) , and hence π β ππ (πΌ, π, π).
β Proof. Suppose that π(π§) = ββ π=0 ππ ππ (π§) = 1/π§+βπ=1 ππ ((1β π πΌ)/[π + πΌ β πΌπ(π + 1)]πΏ(π, π, π))π§ ; then, we have β
β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π)
π=1
Γ
ππ (1 β πΌ) [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π)
(14)
β
= β ππ = 1 β π0 < 1. π=1
By Theorem 3, π β ππ (πΌ, π, π). Conversely, assume that π is in the class ππ (πΌ, π, π); then, by Corollary 4, ππ β€
1βπΌ . [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π)
(15)
Set ππ = ([π+πΌβπΌπ(π+1)]πΏ(π, π, π)/(1βπΌ))ππ , π = 1, 2, . . . and β π0 = 1 β ββ π=1 ππ . Then, π(π§) = βπ=0 ππ ππ (π§) β ππ (πΌ, π, π).
Geometry
3
3. Distortion Bounds
4. Radii Results
In this section, we obtain growth and distortion bounds for the class ππ (πΌ, π, π).
We now find the radius of meromorphically starlikeness and convexity for functions π in the class ππ (πΌ, π, π).
Theorem 6. If π β ππ (πΌ, π, π), then
Theorem 8. Let π β ππ (πΌ, π, π). Then, π is meromorphically starlike of order πΏ, (0 β€ πΏ < 1) in |π§| < π1 , where
1 1βπΌ σ΅© σ΅©σ΅© β βπβ , σ΅©σ΅©π (π)σ΅©σ΅©σ΅© β₯ βπβ (1 + πΌ β 2πΌπ) (1 β π)2 (π + 1) (π + 2) 1 1βπΌ σ΅© σ΅©σ΅© + βπβ . σ΅©σ΅©π (π)σ΅©σ΅©σ΅© β€ βπβ (1 + πΌ β 2πΌπ) (1 β π)2 (π + 1) (π + 2) (16) The result is sharp for π(π§) = 1/π§ + ((1 β πΌ)/(1 + πΌ β 2πΌπ)(1 β π)2 (π + 1)(π + 2))π§.
π1 = inf[ πβ₯1
(1 β πΏ) (π+πΌ β πΌπ (π+1)) πΏ (π, π, π) ] (1 β πΌ) (π + 2 β πΏ)
1/(π+1)
,
(22)
π β₯ 1. π Proof. Let π(π) = πβ1 + ββ π=1 ππ π . Since π β ππ (πΌ, π, π) is meromorphically starlike of order πΏ,
σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© ππσΈ (π) σ΅©σ΅© β€ 1 β πΏ. σ΅©σ΅©β β 1 σ΅©σ΅© σ΅©σ΅© π (π) σ΅© σ΅©
Proof. By Theorem 3,
(23)
Substituting for π, the above inequality becomes
β
(1 β πΌ β 2πΌπ) πΏ (1, π, π) β ππ
β
π=2
β(
(17)
β
β€ β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1 β πΌ.
π=1
π+2βπΏ ) βπβπ+1 ππ β€ 1. 1βπΏ
(24)
By Theorem 3,
π=1
β
[π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1. 1βπΌ π=1
Therefore,
β
β
β ππ β€
π=1
=
1βπΌ (1 + πΌ β 2πΌπ) πΏ (1, π, π)
Thus, (24) will be true if (18)
1βπΌ 2
(1 + πΌ β 2πΌπ) (1 β π) (π + 1) (π + 2)
(
.
β
1 1 σ΅© σ΅© β β ππ βπβπ β€ σ΅©σ΅©σ΅©π (π)σ΅©σ΅©σ΅© β€ + β π βπβπ . βπβ π=1 βπβ π=1 π
(19)
Since βπβ < 1, the above inequality becomes β
[π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) π+2βπΏ ) βπβπ+1 β€ , (26) 1βπΏ (1 β πΌ)
that is,
π Also, if π(π) = πβ1 + ββ π=1 ππ π , then β
(25)
1/(π+1)
. (27)
Theorem 9. Let π β ππ (πΌ, π, π). Then, π is meromorphically convex of order πΏ, (0 β€ πΏ < 1) in |π§| < π2 , where
β
1 1 σ΅© σ΅© β βπβ β ππ β€ σ΅©σ΅©σ΅©π (π)σ΅©σ΅©σ΅© β€ + βπβ β ππ . βπβ βπβ π=1 π=1
(1 β πΏ) (π + πΌ β πΌπ (π + 1)) πΏ (π, π, π) ] βπβ β€ [ (1 β πΌ) (π + 2 β πΏ)
(20)
Using (18), we get the result. Theorem 7. If π β ππ (πΌ, π, π), then 1 1βπΌ σ΅© σ΅©σ΅© σΈ σ΅©σ΅©π (π)σ΅©σ΅©σ΅© β₯ σ΅© βπβ2 β (1 + πΌ β 2πΌπ) (1 β π)2 (π + 1) (π + 2) , σ΅© 1 1βπΌ σ΅© σ΅©σ΅© σΈ σ΅©σ΅©π (π)σ΅©σ΅©σ΅© β€ σ΅© βπβ2 + (1 + πΌ β 2πΌπ) (1 β π)2 (π + 1) (π + 2) . σ΅© (21) The result is sharp for π(π§) = 1/π§ + ((1 β πΌ)/(1 + πΌ β 2πΌπ)(1 β π)2 (π + 1)(π + 2))π§.
π2 = inf [ πβ₯ 1
(1 β πΏ) (π + πΌ β πΌπ (π + 1)) πΏ (π, π, π) ] (1 β πΌ) π (π + 2 β πΏ)
1/(π+1)
. (28)
Theorem 10. The class ππ (πΌ, π, π) is closed under convex combination. π β1 + Proof. Let π(π§) = πβ1 + ββ π=0 ππ π and π(π§) = π π π π β π (πΌ, π, π). Then, by Theorem 3, ββ π π=0 π β
β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1 β πΌ,
π=1 β
β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1 β πΌ.
π=1
(29)
4
Geometry
Define β(π) = ππ(π) + (1 β π)π(π). Then, β(π) = πβ1 + π ββ π=1 [πππ + (1 β π)ππ ]π . Now,
that is, βππ ππ β€
β
β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) [πππ + (1 β π) ππ ]
(36)
It is enough to find the largest π such that
π=1
β
= π β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ π=1
(30)
β
1βπΌ . [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π)
+ (1 β π) β [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ
1βπΌ (1 β π) [π + πΌ β πΌπ (π + 1)] β€ [π+πΌ β πΌπ (π +1)] πΏ (π β
π, π) (1 β πΌ) [π + π β ππ (π + 1)] (37) which yields
π=1
π β€ ((π + πΌ β πΌπ (π + 1))2 πΏ (π, π, π) β (1 β πΌ)2 π)
β€ π (1 β πΌ) + (1 β π) (1 β πΌ) (using (30)) = 1 β πΌ.
Γ ((1 β πΌ)2 (1 β π (π + 1))
Thus, β β ππ (πΌ, π, π).
(38) β1
+ (π + πΌ β πΌπ (π + 1))2 πΏ (π, π, π)) .
5. Hadamard Product
It follows from (38) that
Theorem 11. The Hadamard product π β π of two functions π and π in ππ (πΌ, π, π) belongs to the class ππ (π, π, π), where π = 1 β 2(1 β πΌ)2 (1 β π)/(1 β πΌ)2 (1 β 2π) + (1 β πΌ β 2πΌπ)2 (1 β π)2 (π + 1)(π + 2).
π β€ 1 β (1 β πΌ)2 (π + 1) (1 β π) Γ ((1 β πΌ)2 (1 β π (π + 1)) β1
+ (π + πΌ β πΌπ (π + 1))2 πΏ (π, π, π)) .
Proof. It suffices to prove that Let
β
[π + π β ππ (π + 1)] πΏ (π, π, π) ππ ππ β€ 1, β 1βπ π=1
(39)
(31)
πΏ (π) = (1 β πΌ)2 (π + 1) (1 β π) Γ ((1 β πΌ)2 (1 β π (π + 1))
where
β1
+ (π + πΌ β πΌπ (π + 1))2 πΏ (π, π, π)) .
π = 1 β (2(1 β πΌ)2 (1 β π)) Γ ((1 β πΌ)2 (1 β 2π) + (1 β πΌ β 2πΌπ)2 2
(32)
β1
Γ (1 β π) (π + 1) (π + 2)) .
[π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1, 1βπΌ
ββ π=1 [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) ππ β€ 1. 1βπΌ
(33)
By Cauchy-Schwartz inequality, ββ π=1 [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) βππ ππ β€ 1. 1βπΌ
(34)
We have to find the largest π such that [π + π β ππ (π + 1)] πΏ (π, π, π) ππ ππ 1βπ [π + πΌ β πΌπ (π + 1)] πΏ (π, π, π) βππ ππ , β€ 1βπΌ
Clearly πΏ(π) is an increasing function of π (π β₯ 1). Letting π = 1, we prove the assertion.
References
Since π, π β ππ (πΌ, π, π) ββ π=1
(40)
(35)
[1] P. L. Duren, Univalent Functions, Springer, New York, NY, USA, 1983. [2] T. Rosy, G. Murugusundaramoorthy, and K. Muthunagai, βCertain subclasses of meromorphic functions associated with hypergeometric functions,β International Journal of Nonlinear Science, vol. 12, no. 2, pp. 1749β1759, 2011. [3] G. Murugusundaramoorthy and T. Rosy, βSubclasses of analytic functions associated with Fox-Wrightβs generalized hypergeometric functions based on Hilbert space operator,β Studia Universitatis BabesΒΈ-Bolyai, vol. 56, no. 3, pp. 61β72, 2011. [4] N. Dunford and J. T. Schwartz, Linear Operators, Part I: General Theory, Wiley-Interscience, New York, NY, USA, 1958. [5] K. Fan, βAnalytic functions of a proper contraction,β Mathematische Zeitschrift, vol. 160, no. 3, pp. 275β290, 1978. [6] W. G. Atshan and R. H. Buti, βFractional calculus of a class of univalent functions with negative coefficients defined by Hadamard product with Rafid-Operator,β European Journal of Pure and Applied Mathematics, vol. 4, no. 2, pp. 162β173, 2011.
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