Acta Universitatis Apulensis ISSN: 1582-5329 No. 29/212 pp. 197-215 SOME SUBCLASSES OF ANALYTIC FUNCTIONS DEFINED BY GENERALIZED DIFFERENTIAL OPERATOR Maslina Darus and Imran Faisal Abstract. Let A denote the class of analytic functions in the open unit disk U = {z : z < 1} normalized by f) = f ) 1 =. In this paper, we introduce and study certain subclasses of analytic functions. Several inclusion relations of various subclasses of analytic functions, coefficient estimates, growth and distortion theorems, Hadamard Product, extreme points, integral means and inclusion properties are given. Moreover, some sufficient conditions for subordination and superordination of analytic functions, and application of Φ like function are also discussed. 2 Mathematics Subject Classification: 3C45. Keywords and phrases: Analytic function, differential operator, subordination, Φ Like function, extreme points, coefficient estimates. 1. Introduction and Preliminaries Let A denote the class of analytic functions of the form fz) = z + a k z k 1) in the open unit disk U = {z : z < 1} normalized by f) = f ) 1 =. For a function f A, α, β, µ, λ and n N o we define the differential operator, as follow: D fz) = fz), Dλ 1 µ + β λ α, β, µ)fz) = α )fz) + µ + λ )zf z), Dλ 2 α, β, µ)fz) = DD1 λ α, β, µ)fz)),. Dλ n α, β, µ)fz) = DDn 1 λ α, β, µ)fz)). 2) 197
If f is given by 1), then from 2), we see that Dλ n α, β, µ)fz) = z + α + µ + λ)k 1) + β ) nak z k 3) which generalizes many operators. Indeed, if in the definition of Dλ n α, β, µ)f we substitute the following: β = 1, µ =, we get Dλ nα, 1, )fz) = Dn fz) = z + and El-Deeb differential operator[1]. α+λk 1)+1 α+1 ) n a k z k of Aouf, El-Ashwah α = 1, β = o and µ =, we get Dλ n1,, )fz) = Dn fz) = z + 1 + λk 1))n a k z k of Al-Oboudi differential operator[9]. α = 1, β = o, µ = and λ = 1, we get D1 n1,, )fz) = Dn fz) = z + k)n a k z k of Sălăgean s differential operator [5]. 2 )n a k z k of Uralegaddi and So- α = 1, β = 1, λ = 1 and µ =, we get D1 n1, 1, )fz) = Dn fz) = z + manatha differential operator [2]. α+1 )n a k z k of Cho and Srivastava dif- β = 1, λ = 1 and µ =, we get D1 nα, 1, )fz) = Dn fz) = z + ferential operator [3,4]. k+1 k+α Now we define some new subclasses of analytic functions by using extended multiplier transformations operator. Definition 1. Let f A. Then f M α,β,µ δ) if and only if zd n R λ α, β, µ)fz)) ) Dλ n > δ, δ < 1, z U. α, β, µ)fz) Definition 2. Let f A. Then f N α,β,µ δ) if and only if zd n R λ α, β, µ)fz)) ) ) Dλ n > δ, δ < 1, z U. α, β, µ)fz)) 198
Definition 3. Φ Like Function)[6,11] Let Φ be an analytic function in a domain containing fu), Φ) = and Φ ) >. The function f A is called Φ-like if zf ) z) R >, z U. Φfz)) This concept was introduced by Brickman [6] and established that a function f A is univalent if and only if f is Φ Like for some Φ. Definition 4.[1] Let Φ be an analytic function in a domain containing fu), Φ) = and Φ ) = 1, and Φω) for ω fu) \ {}. Let q be a fixed analytic function in U, q) = 1. The function f A is called Φ-Like with respect to q if zf z) qz), z U. Φfz)) Definition 5.[7] Denote by Q the set of all functions f that are analytic and injective on U Ef) where Ef) = {ζ U : lim z ζ fz) = } and are such that f ζ) for ζ U Ef). Definition 6. Subordination Principal) [8] For two functions f and g, analytic in U, we say that the function f is subordinate to g in U, and write fz) gz) z U), if there exists a Schwarz function w, which is analytic in U with w) = and wz) < 1, such that fz) = gwz)). Lemma 1.[8] Let qz) be univalent in the unit disk U and θ and φ be analytic in a domain D containing qu) with φw) when w qu), Qz) = zqz)φq z)) hz) = θqz)) + Qz), suppose that Qz) is starlike univalent in U and R zh z) Qz) ) > for z U if p is analytic in U with pu) D and θpz)) + zp z)φpz)) θqz)) + zq z)φqz)), then and q is the best dominant pz) qz), 199
2.Characterization properties In this section, we study the coefficient estimates and intend to prove the distortion theorems via two different techniques for the functions contained in the classes M α,β,µ δ) and N α,β,µ δ). Theorem 1.If an analytic function f belongs to A satisfies the following condition i.e α + µ + λ)k 1) + β 1 δ, δ < 1. 4) then f M α,β,µ δ). Proof. Let f M α,β,µ δ), then from Definition 1, we have We also know that this implies that such that R zdn λ α, β, µ)fz)) Dλ nα, β, µ)fz) ) > δ, δ < 1, z U. Rw) > α if and only if 1 α + w > 1 + α w, 2 δ)z + δz + 1 δ)d n λ α, β, µ)fz) + zdn λ α, β, µ)fz)) > 1 + δ)d n λ α, β, µ)fz) zdn λ α, β, µ)fz)) α + µ + λ)k 1) + β ) nak 1 + k δ) z k α + µ + λ)k 1) + β ) nak 1 k + δ) z k > So after simplification, we have or 2 2δ) α + µ + λ)k 1) + β 2k 2δ) α + µ + λ)k 1) + β 1 δ) 2
is our required result. Finally the result is sharp with extremal function f given by fz) = z + 1 δ α+µ+λ)k 1)+β ) n z k. Taking suitable values of parameters and function f in the class M α,β,µ δ), Theorem 1 gives us the following immediate results. Corollary 1.Let the function f defined by 1) be in the class M α,β,µ δ). Then we have a k 1 δ) α+µ+λ)k 1)+β ) n k 2. Corollary 2.Let the hypotheses of Theorem 1, be satisfied. Then for δ = µ = λ =, we have a k 1 k, k 2. Theorem 2.If an analytic function f belongs to A satisfies the following condition i.e α + µ + λ)k 1) + β k 1 δ, δ < 1 5) then f N α,β,µ δ). f A, α, β, µ, λ,, n N o ) Proof. The proof is similar to that of the proof of Theorem 2. Similarly, for a function f to be in the class N α,β,µ δ), we deduced a result from Theorem 2, as follow: Corollary 3.Let the function f defined by 1) be in the class N α,β,µ δ). Then we have 1 δ) a k ) n k 2. k α+µ+λ)k 1)+β 21
3. Growth and distortion theorems Theorem 3.If an analytic function f belongs to A satisfies the following condition i.e α + µ + λ)k 1) + β 1 δ, δ < 1, 6) then f M α,β,µ δ) and fz) z 1 δ 2 δ α + µ + λ + β )n z 2 and fz) z + 1 δ 2 δ α + µ + λ + β )n z 2. But Proof. Let f M α,β,µ δ), this implies that implies that α + µ + λ)k 1) + β 1 δ, δ < 1. α + µ + λ) + β ) n 2 δ) a k α + µ + λ)k 1) + β 1 δ, δ < 1, α + µ + λ) + β ) n 2 δ) a k 1 δ, δ < 1 Since a k 2 δ) 1 δ α+µ+λ)+β fz) = z + a k z k ) n δ < 1. 22
Similarly fz) z + a k z 2. fz) z + 1 δ 2 δ α + µ + λ + β )n z 2. fz) = z + fz) z a k z k a k z 2. fz) z 1 δ 2 δ α + µ + λ + β )n z 2. Theorem 4.Let the hypotheses of Theorem 2, be satisfied,then fz) z 1 δ 22 δ) α + µ + λ + β )n z 2, and fz) z + 1 δ 22 δ) α + µ + λ + β )n z 2. Proof. Working on the same lines as in the proof of Theorem 3, one can easily prove this theorem as well. Theorem 5.If an analytic function f belongs to A satisfies the following condition i.e α + µ + λ)k 1) + β 1 δ, δ < 1, 7) then f M α,β,µ δ) and Dλ n 1 δ α, β, µ)fz) z 2 δ z 2, and Dλ n 1 δ α, β, µ)fz) z + 2 δ z 2. 23
But Proof. Let f M α,β,µ δ), this implies that implies that Because α + µ + λ)k 1) + β 1 δ, δ < 1. 2 δ) this implies that similarly this implies that α + µ + λ)k 1) + β ) n 2 δ) a k α + µ + λ)k 1) + β 1 δ α + µ + λ)k 1) + β 1 δ, δ < 1 α + µ + λ)k 1) + β 1 δ 2 δ) Dλ n α, β, µ)fz) = z + α + µ + λ)k 1) + β ) nak z k Dλ n α, β, µ)fz) z + α + µ + λ)k 1) + β z k, Dλ n 1 δ α, β, µ)fz) z + 2 δ) zk, Dλ n α, β, µ)fz) = z + α + µ + λ)k 1) + β ) nak z k Dλ n α, β, µ)fz) z α + µ + λ)k 1) + β z k, Dλ n 1 δ α, β, µ)fz) z 2 δ) zk. 24
and and Theorem 6.Let the hypotheses of Theorem 5, be satisfied,then Dλ n 1 δ α, β, µ)fz) z 22 δ) z 2, Dλ n 1 δ α, β, µ)fz) z + 22 δ) z 2. Proof. Use the same method of the proof of Theorem 5 to get the result. Theorem 7.Let the hypotheses of Theorem 3, be satisfied, then f z) 1 2 1 δ 2 δ α + µ + λ + β )n z f z) 1 + 2 1 δ 2 δ α + µ + λ + β )n z. Proof. The proof is similar to that of the proof of Theorem 3. Theorem 8.a). If f 1 z) = z, and f i z) = z + 4.Extreme Points 1 δ) α+µ+λ)k 1)+β ) n z i i = 2, 3, 4 Then f M α,β,µ δ), if and only if it can be expressed in the form fz) = i=1 λ if i z), where λ i and i=1 λ i = 1. b). If f 1 z) = z, f i z) = z + k 1 δ) α+µ+λ)k 1)+β ) n z i i = 2, 3, 4 Then f N α,β,µ δ), if and only if it can be expressed in the form fz) = i=1 λ if i z), where λ i and i=1 λ i = 1. Proof. a) Let fz) = i=1 λ if i z), i = 1, 2, 3, λ i with i=1 λ i = 1 This implies that fz) = λ i f i z), i=1 25
or Since fz) = 1 δ) λ 1 z) + λ i z + ) n z i ), i=2 α+µ+λ)k 1)+β fz) = 1 δ) λ 1 z) + λ i z) + λ i ) n z i ), i=2 i=2 α+µ+λ)k 1)+β fz) = 1 δ) λ i z) + λ i ) n z i ), i=1 i=2 α+µ+λ)k 1)+β fz) = 1 δ) z + λ i ) n z i, α+µ+λ)k 1)+β 8) i=2 1 δ) λ i i=2 α+µ+λ)k 1)+β = α + µ + λ)k 1) + β ) n λ i 1 δ) i=2 = 1 λ 1 )1 δ) < 1 δ) The condition 6) for fz) = i=1 λ if i z), is satisfied Thus f M α,β,µ δ). Conversely, we suppose that f M α,β,µ δ), since We put and then a k λ i = 1 δ) α+µ+λ)k 1)+β ) n α+µ+λ)k 1)+β 1 δ) λ 1 = 1 fz) = λ i, i=2 λ i f i z). i=1 ) n k 2. a i k 2, The proof of the second part of the Theorem 8 is similar to 1st part. ) n 26
5.Integral means inequalities For any two functions f and g analytic in U, f is said to be subordinate to g in U, denoted by f g if there exists a Schwarz function w analytic in U satisfying w) = and wz) < 1 such that fz) = gwz)), z U. In particular, if the function g is univalent in U, the above subordination is equivalent to f) = g) and fu) gu). In 1925, Littlewood [12] proved the following Subordination Theorem. Theorem 9.If f and g are any two functions, analytic in U, with f g, then for µ > and z = re iθ, < r < 1), fz) µ dθ gz) µ dθ. Theorem 1.a). Let f M α,β,µ δ), and f k be defined by f k z) = z + ) n α+µ+λ)k 1)+β 1 δ) If there exists an analytic function wz) given by z k k 2. [wz)] k 1 = 1 δ) α+µ+λ)k 1)+β ) n a k z k 1, then for z = re iθ, and < r < 1), fre iθ ) µ dθ b). Let f N α,β,µ δ), and f k be defined by k f k z) = z + f k re iθ ) µ dθ. ) n α+µ+λ)k 1)+β 1 δ) If there exists an analytic function wz) given by z k k 2. [wz)] k 1 = k 1 δ) α+µ+λ)k 1)+β ) n a k z k 1, 27
then for z = re iθ, and < r < 1), fre iθ ) µ dθ f k re iθ ) µ dθ. Proof. b). We have to show that or or 1 + z + a k z k µ dθ a k z k 1 µ dθ fre iθ ) µ dθ k z + k 1 + f k re iθ ) µ dθ, ) n α+µ+λ)k 1)+β 1 δ) ) n α+µ+λ)k 1)+β 1 δ) z k µ dθ, z k 1 µ dθ, By using Theorem 9, it is enough to show that k 1 + a k z k 1 1 + Now by taking k 1 + a k z k 1 = 1 + After simplification we get ) n α+µ+λ)k 1)+β 1 δ) ) n α+µ+λ)k 1)+β 1 δ) z k 1, wz)) k 1, [wz)] k = k 1 δ) α+µ+λ)k 1)+β ) n a k z k 1, this implies that w) =, and [wz)] k = k 1 δ) α+µ+λ)k 1)+β ) n a k z k 1, 28
or By using 4) we get [wz)] k = k 1 δ) α+µ+λ)k 1)+β [wz)] k 1 z < 1. ) n a k z k 1, The proof of the 1st part of the Theorem 1 is similar to second part. 6.Hadamard Product Let f, g A, where fz) is given in 1) and gz) = z+ b k z k, then the modified Hadamard productf g is defined by f g) = z + a k b k z k. M α,β,µ Theorem 11.a). If fz) = z + δ) then prove that f g)z) M α,β,µ b). If fz) = z + prove that f g)z) N α,β,µ that a k z k M α,β,µ δ) and gz) = z + b k z k δ). a k z k N α,β,µ δ) and gz) = z + δ). Proof. a). Since it is given that fz) = z + b k z k N α,β,µ δ) then a k z k M α,β,µ δ), this implies α + µ + λ)k 1) + β 1 δ, δ < 1. Similarly gz) = z + b k z k M α,β,µ δ), implies that α + µ + λ)k 1) + β ) n bk 1 δ, δ < 1. Since α + µ + λ)k 1) + β b k α + µ + λ)k 1) + β 1 δ, δ < 1. 29
Implies that α + µ + λ)k 1) + β b k 1 δ, δ < 1. Orf g)z) M α,β,µ δ). The proof of the second part of the Theorem 11 is similar to 1st part. 7.Subordination and inclusion properties Here, we investigate the inclusion properties of the new classes defined above. We also find out some subordination results by using the Φ like function. Theorem 12.Let α 1 α 2 < 1, β 1 β 2 < 1, µ 1 µ 2 < 1, λ 1 λ 2 < 1, and δ 1 δ 2 < 1, then M α,β,µ δ 2 ) M α,β,µ δ 1 ) M α,β,µ λ 2,k δ) Mα,β,µ λ 2,k δ) M α,β,µ 2 δ) M α,β,µ 1 δ) M α 1,β,µ δ) M α 2,β,µ δ) M α,β 1,µ δ) M α,β 2,µ δ). Proof. To prove M α,β,µ δ 2 ) M α,β,µ δ 1 ), since we have δ 1 δ 2 < 1, this implies that δ 1 δ 2, or 1 δ 2 ) 1 δ 1 ), so let then from Theorem 1, we have f M α,β,µ δ 2 ), α + µ + λ)k 1) + β 1 δ 2, δ 2 < 1 This implies that α + µ + λ)k 1) + β 1 δ 2 1 δ 1, 21
and therefore α + µ + λ)k 1) + β 1 δ 1, implies that and hence fz) M α,β,µ δ 1 ), M α,β,µ δ 2 ) M α,β,µ δ 1 ). The proof of the remaining parts of the Theorem 12 is similar to 1st part. Theorem 13.Let α 1 α 2 < 1, β 1 β 2 < 1, µ 1 µ 2 < 1, λ 1 λ 2 < 1, and δ 1 δ 2 < 1, then N α,β,µ δ 2 ) N α,β,µ δ 1 ) N α,β,µ λ 2,k δ) N α,β,µ δ) λ 2,k N α,β,µ 2 δ) N α,β,µ 1δ) N α 1,β,µ δ) N α 2,β,µ δ) N α,β 1,µ δ) N α,β 2,µ δ) implies Proof. To prove N α 1,β,µ δ) N α 2,β,µ δ), we have α 1 α 2 < 1 α 1 α 2, 1/α 2 + β 1/α 1 + β 1 + µ + λ)k 1) α 2 + β 1 + µ + λ)k 1), α 1 + β therefore so let α2 + µ + λ)k 1) + β k α 2 + β α1 + µ + λ)k 1) + β k, α 1 + β fz) N α 1,β,µ δ), 211
then from Theorem 2, we have implies that This show that and implies α + µ + λ)k 1) + β k 1 δ, α 1 + β α + µ + λ)k 1) + β k α 2 + β α + µ + λ)k 1) + β k 1 δ α 1 + β α + µ + λ)k 1) + β k 1 δ, α 2 + β fz) N α 2,β,µ δ) N α 1,β,µ δ) N α 2,β,µ δ) The proof of the remaining parts of the Theorem 13 is similar to 4th part. Theorem 14.Let q be univalent in U such that zq z)/qz) is starlike univalent in U and R 1 + α ) γ qz) + zq z) q z) zq z) >, α, γ C, γ. 9) qz) If f A satisfies the subordination then α zdn λ α, β, µ)fz)) ΦDλ nα, β, µ)fz)) ) + γ1 + zdn λ α, β, µ)fz)) Dλ nα, β, zφ Dλ n α, β, µ)fz)) µ)fz)) ΦDλ n ) α, β, µ)fz)) and q is the best dominant. Proof. Let αqz) + γzq z)/qz), zdλ n α, β, µ)fz)) ΦDλ n qz), α, β, µ)fz)) pz) = zdn λ α, β, µ)fz)) ΦDλ n, α, β, µ)fz)) 212
then after computation, we have zp z)/pz) = 1 + zdn λ α, β, µ)fz)) Dλ nα, β, zφ Dλ n α, β, µ)fz)) µ)fz)) ΦDλ n, α, β, µ)fz)) which yields the following subordination By setting, αpz) + γzp z)/pz) αqz) + γzq z)/qz), α, γ C. θω) = αω φω) = γ/ω, γ, it can be easily observed that θω) is analytic in C and φω) is analytic in C {} and that φω) when ω C {} Also, by letting and Qz) = zq z)φqz)) = γzq z)/qz), hz) = θqz)) + Qz) = αqz) + γzq z)/qz), we find that Qz) is starlike univalent in U and that So by Lemma 1, we have R zh z) Qz) = R1 + α γ qz) + zq z) q z) zq z) qz) ) >. zdλ n α, β, µ)fz)) ΦDλ n qz). α, β, µ)fz)) We deduce the following result by taking Φ an identity function i.e. Φω) = ω in Theorem 14, as follows: Corollary 4.Let qz) be univalent in U If q satisfies 9) and α zdn λ α, β, µ)fz)) Dλ nα, β, µ)fz)) ) + γ1 + zdn λ α, β, µ)fz)) Dλ nα, β, zdn λ α, β, µ)fz)) ) µ)fz)) α, β, µ)fz)) αqz) + γzq z)/qz), D n λ then and q is the best dominant. zdλ n α, β, µ)fz)) qz), α, β, µ)fz)) D n λ 213
Similarly, by taking suitable values of a function q in Theorem 14, we deduce Corollary 5 and Corollary 6 as follows: implies Corollary 5.If f A and assume that 9) holds then and 1+Az 1+Bz 1 + zdn λ α, β, µ)fz)) Dλ nα, β, 1 + Az µ)fz)) 1 + Bz + A B)z 1 + Az)1 + Bz) zdλ n α, β, µ)fz)) 1 + Az, 1 B A 1, α, β, µ)fz)) 1 + Bz D n λ is the best dominant. Corollary 6.If f A and assume that 9) holds then implies, and 1+z 1 z 1 + zdn λ α, β, µ)fz)) Dλ nα, β, 1 + z µ)fz)) 1 z + 2z 1 + z)1 z), is the best dominant. zdλ n α, β, µ)fz)) 1 + z α, β, µ)fz)) 1 z, D n λ Acknowledgements. The work presented here was supported by UKM-ST-6- FRGS17-29. References [1] M.K. Aouf, R.M. El-Ashwah and S.M. El-Deeb, Some inequalities for certain p-valent functions involving extended multiplier transformations, Proc. Pakistan Acad. Sci., 46, 29), 217 221. [2] B.A. Uralegaddi and C. Somanatha, Certain classes of univalent functions, In: Current Topics in Analytic Function Theory, Eds. Srivastava, H.M. and Owa, S., World Scientific Publishing Company, Singapore, 1992, 371 374. [3] N.E. Cho and H.M. Srivastava, Argument estimates of certain analytic functions defined by a class of multiplier transformations, Math. Comp. Mode., 37, 23), 39 49. [4] N.E. Cho and T.H. Kim, Multiplier transformations and strongly close-toconvex functions, Bull. Korean Math. Soc., 4, 23, 399 41. [5] G.S. Salagean, Subclasses of univalent functions, Lecture Notes in Mathematics 113, Springer-Verlag, 1983), 362 372. 214
[6] L. Brickman, Φ like analytic functions I, Bull. Amer. Math. Soc., 79, 1973), 555 558. [7] S.S. Miller and P.T. Mocanu, Subordinants of differential superordinations, Complex Variables, 48, 23), 815 826. [8] S.S. Miller and P.T. Mocanu, Differential Subordinations: Theory and Applications, Series on Monographs and Textbooks in Pure and Appl. Math. vol. 255, Marcel Dekker, Inc, new york, 2. [9] F.M. Al-Oboudi, On univalent functions defined by a generalized Salagean operator, Int. J. Math. Sci., 24), 1429 1436. [1] R.W. Ibrahim and M. Darus, New classes of analytic functions involving generalized noor integral operator, Journal of Ineq. and App., 28), Article ID 39435. [11] St. Ruscheweyh, A subordination theorem for φ like functions, J. London Math. Soc., 13, 1976), 275 28. [12] J.E. Littlewood, On inequalities in the theory of functions, Proc. London Math. Soc., 23, 1925), 481 519. Maslina Darus School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 436 Selangor D. Ehsan, Malaysia email:maslina@ukm.mycorresponding author) Imran Faisal School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 436 Selangor D. Ehsan, Malaysia email:faisalmath@gmail.com 215