Journal of Inequalities in Pure and Applied Mathematics

Journal of Inequalities in Pure and Applied Mathematics APPLICATIONS OF DIFFERENTIAL SUBORDINATION TO CERTAIN SUBCLASSES OF MEROMORPHICALLY MULTIVALENT FUNCTIONS H.M. SRIVASTAVA AND J. PATEL Department of Mathematics and Statistics University of Victoria Victoria, British Columbia V8W 3P4 Canada. EMail: harimsri@math.uvic.ca Department of Mathematics Utkal University Vani Vihar, Bhubaneswar 7514 Orissa, India. EMail: jpatelmath@sify.com volume 6, issue 3, article 88, 25. Received 18 April, 25; accepted 5 July, 25. Communicated by: Th.M. Rassias Abstract Home Page c 2 Victoria University ISSN electronic: 1443-5756 21-5

Abstract By making use of the principle of differential subordination, the authors investigate several inclusion relationships and other interesting properties of certain subclasses of meromorphically multivalent functions which are defined here by means of a linear operator. They also indicate relevant connections of the various results presented in this paper with those obtained in earlier works. 2 Mathematics Subject Classification: Primary 3C45; Secondary 3D3, 33C2. Key words: Meromorphic functions, Differential subordination, Hadamard product or convolution, Multivalent functions, Linear operator, Hypergeometric function. The present investigation was supported, in part, by the Natural Sciences and Engineering Research Council of Canada under Grant OGP7353 and, in part, by the University Grants Commission of India under its DRS Financial Assistance Program. 1 Introduction and Definitions........................... 3 2 Preliminary Lemmas.................................. 7 3 The Main Subordination s and The Associated Functional Inequalities.................................... 1 References Page 2 of 31

1. Introduction and Definitions For any integer m > p, let Σ p,m denote the class of all meromorphic functions fz normalized by 1.1 fz = z p + a k z k p N := {1, 2, 3, }, k=m which are analytic and p-valent in the punctured unit disk U = {z : z C and < z < 1} = U \ {}. For convenience, we write Σ p, p+1 = Σ p. If fz and gz are analytic in U, we say that fz is subordinate to gz, written symbolically as follows: f g in U or fz gz z U, if there exists a Schwarz function wz, which by definition is analytic in U with w = and wz < 1 z U such that Indeed it is known that fz = g wz z U. Page 3 of 31 fz gz z U = f = g and fu gu.

In particular, if the function gz is univalent in U, we have the following equivalence cf., e.g., [5]; see also [6, p. 4]: fz gz z U f = g and fu gu. For functions fz Σ p,m, given by 1.1, and gz Σ p,m defined by 1.2 gz = z p + b k z k m > p; p N, k=m we define the Hadamard product or convolution of fz and gz by 1.3 f gz := z p + a k b k z k =: g fz k=m m > p; p N; z U. Following the recent work of Liu and Srivastava [3], for a function fz in the class Σ p,m, given by 1.1, we now define a linear operator D n by and in general 1.4 D 1 fz = z p + k=m D n fz = D D n 1 fz = z p + D fz = fz, z p+1 p + k + 1a k z k fz =, z p = p + k + 1 n a k z k k=m z p+1 D n 1 fz z p n N. Page 4 of 31

It is easily verified from 1.4 that 1.5 z D n fz = D n+1 fz p + 1D n fz f Σ p,m ; n N := N {}. The case m = of the linear operator D n was introduced recently by Liu and Srivastava [3], who investigated among other things several inclusion relationships involving various subclasses of meromorphically p-valent functions, which they defined by means of the linear operator D n see also [2]. A special case of the linear operator D n for p = 1 and m = was considered earlier by Uralegaddi and Somanatha [13]. Aouf and Hossen [1] also obtained several results involving the operator D n for m = and p N. Making use of the principle of differential subordination as well as the linear operator D n, we now introduce a subclass of the function class Σ p,m as follows. Definition. For fixed parameters A and B 1 B < A 1, we say that a function fz Σ p,m is in the class Σ n p,ma, B, if it satisfies the following subordination condition: 1.6 zp+1 D n fz p 1 + Az 1 + Bz n N ; z U. In view of the definition of differential subordination, 1.6 is equivalent to the following condition: z p+1 D n fz + p Bz p+1 D n fz + pa < 1 z U. Page 5 of 31

For convenience, we write 1 2αp, 1 = Σ n p,m α, Σ n p,m where Σ n p,mα denotes the class of functions in Σ p,m satisfying the following inequality: In particular, we have R z p+1 D n fz > α α < p; z U. Σ n p, A, B = R n,p A, B, where R n,p A, B is the function class introduced and studied by Liu and Srivastava [3]. The function class Hp; A, B, considered by Mogra [7], happens to be a further special case of the Liu-Srivastava class R n,p A, B when n =. In the present paper, we derive several inclusion relationships for the function class Σ n p,ma, B and investigate various other properties of functions belonging to the class Σ n p,ma, B, which we have defined here by means of the linear operator D n. These include for example some mapping properties involving the linear operator D n. Relevant connections of the results presented in this paper with those obtained in earlier works are also pointed out. Page 6 of 31

2. Preliminary Lemmas In proving our main results, we need each of the following lemmas. Lemma 1 Miller and Mocanu [5]; see also [6]. Let the function hz be analytic and convex univalent in U with h = 1. Suppose also that the function φz given by 2.1 φz = 1 + c p+m z p+m + c p+m+1 z p+m+1 + is analytic in U. If 2.2 φz + zφ z γ then φz ψz = hz γ p + m z γ z p+m and ψz is the best dominant of 2.2. Rγ ; γ ; z U, t γ p+m 1 htdt hz z U, With a view to stating a well-known result Lemma 2 below, we denote by Pγ the class of functions ϕz given by 2.3 ϕz = 1 + b 1 z + b 2 z 2 +, which are analytic in U and satisfy the following inequality: Page 7 of 31 R ϕz > γ γ < 1; z U.

Lemma 2 cf., e.g., Pashkouleva [9]. Let the function ϕz, given by 2.3, be in the class Pγ. Then R{ϕz} 2γ 1 + 21 γ 1 + z Lemma 3 see [12]. For γ 1, γ 2 < 1, γ < 1; z U. Pγ 1 Pγ 2 Pγ 3 The result is the best possible. For real or complex numbers a, b, and c c / Z Gauss hypergeometric function is defined by γ3 := 1 21 γ 1 1 γ 2. := {, 1, 2, }, the 2F 1 a, b; c; z = 1 + ab c z aa + 1bb + 1 + z2 1! cc + 1 2! +. We note that the above series converges absolutely for z U and hence represents an analytic function in U see, for details, [14, Chapter 14]. Each of the identities asserted by Lemma 4 below is well-known cf., e.g., [14, Chapter 14]. Lemma 4. For real or complex parameters a, b, and c c / Z, 2.4 1 t b 1 1 t c b 1 1 zt a dt = ΓbΓc b Γc 2F 1 a, b; c; z Rc > Rb > ; Page 8 of 31

2.5 2.6 2.7 2F 1 a, b; c; z = 1 z a z 2F 1 a, c b; c; ; z 1 2F 1 a, b; c; z = 2 F 1 a, b 1; c; z + az c 2 F 1 a + 1, b; c + 1; z; a + b + 1 2F 1 a, b; a + b + 1 ; 1 π Γ 2 =. 2 2 a + 1 b + 1 Γ Γ 2 2 We now recall a result due to Singh and Singh [11] as Lemma 5 below. Lemma 5. Let Φz be analytic in U with Φ = 1 and R Φz > 1 2 z U. Then, for any function F z analytic in U, Φ F U is contained in the convex hull of F U. Page 9 of 31

3. The Main Subordination s and The Associated Functional Inequalities Unless otherwise mentioned, we shall assume throughout the sequel that m is an integer greater than p, and that 1 B < A 1, λ >, n N, and p N. 1. Let the function fz defined by 1.1 satisfy the following subordination condition: Then 1 λzp+1 D n fz + λz p+1 D n+1 fz p 3.1 zp+1 D n fz p Qz 1 + Az 1 + Bz 1 + Az 1 + Bz z U, z U. where the function Qz given by A + 1 A B B 1 + Bz 1 1 2 F 1 1, 1; + 1; Bz B λp+m 1+Bz Qz = A 1 + z B = λp+m+1 is the best dominant of 3.1. Furthermore, 3.2 R zp+1 D n fz > ρ z U, p Page 1 of 31

where A + 1 A B B 1 B 1 1 2 F 1 1, 1; + 1; B B λp+m B 1 ρ = A 1 B =. λp+m+1 The inequality in 3.2 is the best possible. Proof. Consider the function φz defined by 3.3 φz = zp+1 D n fz p z U. Then φz is of the form 2.1 and is analytic in U. Applying the identity 1.5 in 3.3 and differentiating the resulting equation with respect to z, we get 1 λzp+1 D n fz + λz p+1 D n+1 fz p = φz + λ zφ z 1 + Az 1 + Bz Now, by using Lemma 1 for γ = 1/λ, we deduce that zp+1 D n fz Qz p { } 1 z = λp + m z 1/λp+m { } 1/λp+m t 1 1 + At 1 + Bt dt z U. Page 11 of 31

A + 1 A B B 1 + Bz 1 1 2 F 1 1, 1; + 1; Bz B λp+m 1+Bz = A 1 + z B =, λp+m+1 by change of variables followed by the use of the identities 2.4, 2.5, and 2.6 with b = 1 and c = a + 1. This proves the assertion 3.1 of 1. Next, in order to prove the assertion 3.2 of 1, it suffices to show that { } 3.4 inf R Qz = Q 1. z <1 Indeed, for z r < 1, 1 + Az R 1 + Bz Upon setting 1 Ar 1 Br z r < 1. Gs, z = 1 + Asz and 1 + Bsz 1 dνs = λp + m s{1/λp+m} 1 ds s 1, which is a positive measure on the closed interval [, 1], we get Qz = 1 Gs, z dνs, Page 12 of 31

so that R Qz 1 1 Asr dνs = Q r z r < 1. 1 Bsr Letting r 1 in the above inequality, we obtain the assertion 3.2 of 1. Finally, the estimate in 3.2 is the best possible as the function Qz is the best dominant of 3.1. For λ = 1 and m =, 1 yields the following result which improves the corresponding work of Liu and Srivastava [3, 1]. Corollary 1. The following inclusion property holds true for the function class R n,p A, B: R n+1,p A, B R n,p 1 2ϱ, 1 R n,p A, B, where A + 1 A B B 1 B 1 2 F 1 1, 1; 1 + 1; B B p B 1 ϱ = 1 A B =. p+1 The result is the best possible. Putting A = 1 2α p, B = 1, λ = 1, n =, and m = p + 2 Page 13 of 31 in 1, we get Corollary 2 below.

Corollary 2. If fz Σ p, p+2 satisfies the following inequality: { R z p+1 p + 2f z + zf z} > α α < p; z U, then R z p+1 f z π > α + p α 2 1 The result is the best possible. z U. Remark 1. From Corollary 2, we note that, if fz Σ p, p+2 satisfies the following inequality: R z p+1 {p + 2f z + zf pπ 2 z} > z U, 4 π then This result is the best possible. R z p+1 f z > z U. The result asserted by Remark 1 above was also obtained by Pap [8]. 2. If fz Σ n p,mα α < p, then { 3.5 R z p+1 1 λ D n fz + λ D n+1 fz } > α z < R, where The result is the best possible. 1 R = 1 + λ2 p + m 2 p+m λp + m. Page 14 of 31

Proof. We begin by writing 3.6 z p+1 D n fz = α + p αuz z U. Then, clearly, uz is of the form 2.1, is analytic in U, and has a positive real part in U. Making use of the identity 1.5 in 3.6 and differentiating the resulting equation with respect to z, we observe that z 1 p+1 λ D n fz + λ D n+1 fz + α 3.7 p α Now, by applying the following estimate [4]: in 3.7, we get zu z R{uz} 2p + mrp+m 1 r 2p+m z = r < 1 = uz + λzu z. z 1 p+1 λ D n fz + λ D n+1 fz + α 3.8 R p α R uz 2λp + mrp+m 1. 1 r 2p+m It is easily seen that the right-hand side of 3.8 is positive, provided that r < R, where R is given as in 2. This proves the assertion 3.5 of 2. Page 15 of 31

In order to show that the bound R is the best possible, we consider the function fz Σ p,m defined by Noting that z p+1 D n fz = α + p α 1 + z p+m α < p; z U. 1 zp+m for z 1 p+1 λ D n fz + λ D n+1 fz + α p α we complete the proof of 2. = 1 z2p+m + 2λp + mz p+m 1 z 2p+m = iπ z = R exp, p + m Putting λ = 1 in 2, we deduce the following result. Corollary 3. If fz Σ n p,mα α < p, then fz Σ n+1 p,m α for z < R, where 1 p+m R = 1 + p + m2 p + m. The result is the best possible. Page 16 of 31

3. Let fz Σ n p,ma, B and let 3.9 F δ,p fz = δ z δ+p Then z 3.1 zp+1 D n F δ,p fz p t δ+p 1 ftdt δ > ; z U. Θz 1 + Az 1 + Bz z U, where the function Θz given by A + 1 A B B 1 + Bz 1 δ 2 F 1 1, 1; + 1; Bz B p+m 1+Bz Θz = 1 + Aδ z B = δ+p+m is the best dominant of 3.1. Furthermore, 3.11 R zp+1 D n F δ,p fz > κ z U, p where A + 1 A B B 1 B 1 δ 2 F 1 1, 1; + 1; B B p+m B 1 κ = 1 Aδ B =. δ+p+m The result is the best possible. Page 17 of 31

Proof. Setting 3.12 φz = zp+1 D n F δ,p fz p z U, we note that φz is of the form 2.1 and is analytic in U. Using the following operator identity: 3.13 z D n F δ,p fz = δ D n fz δ + p D n F δ,p fz in 3.12, and differentiating the resulting equation with respect to z, we find that zp+1 D n fz = φz + zφ z 1 + Az z U. p δ 1 + Bz Now the remaining part of 3 follows by employing the techniques that we used in proving 1 above. Setting m = in 3, we obtain the following result which improves the corresponding work of Liu and Srivastava [3, 2]. Corollary 4. If δ > and fz R n,p A, B, then F δ,p fz R n,p 1 2ξ, 1 R n,p A, B, where A + 1 A B B 1 B 1 2 F 1 1, 1; δ + 1; B B p B 1 ξ = 1 Aδ B =. δ+p The result is the best possible. Page 18 of 31

Remark 2. By observing that 3.14 z p+1 D n F δ,p fz = δ z δ Corollary 4 can be restated as follows. z f Σ p,m ; z U, t δ+p D n ft dt If δ > and fz R n,p A, B, then R δ z t δ+p D n ft dt > ξ z U, pz δ where ξ is given as in Corollary 4. In view of 3.14, 3 for A = 1 2α p, B = 1, and n = yields Corollary 5. If δ > and if fz Σ p,m satisfies the following inequality: R z p+1 f z > α α < p; z U, then R δz z δ t δ+p f tdt > α + p α The result is the best possible. δ [2F 1 1, 1; p + m + 1; 1 ] 1 2 z U. Page 19 of 31

4. Let fz Σ p,m. Suppose also that gz Σ p,m satisfies the following inequality: R z p D n gz > z U. If then where Proof. Letting R = D n fz D n gz 1 < 1 n N ; z U, R z D n fz > z < R D n, fz 9p + m2 + 4p2p + m 3p + m. 22p + m 3.15 wz = Dn fz D n gz 1 = κ p+mz p+m + κ p+m+1 z p+m+1 +, we note that wz is analytic in U, with w = and wz z p+m Then, by applying the familiar Schwarz lemma, we get wz = z p+m Ψz, z U. Page 2 of 31

where the function Ψz is analytic in U and Ψz 1 z U. Therefore, 3.15 leads us to 3.16 D n fz = D n gz 1 + z p+m Ψz z U. Making use of logarithmic differentiation in 3.16, we obtain 3.17 z D n fz D n fz = z D n gz D n gz + zp+m{ p + mψz + zψ z }. 1 + z p+m Ψz Setting φz = z p D n gz, we see that the function φz is of the form 2.1, is analytic in U, Rφz > z U, and so that we find from 3.17 that 3.18 R z D n fz z D n gz D n gz = zφ z φz p, D n fz p zφ z φz z p+m{ p + mψz + zψ z } 1 + z p+m Ψz z U. Page 21 of 31

Now, by using the following known estimates [1] see also [4]: φ z 2p + mrp+m 1 φz and 1 r 2p+m p + mψz + zψ z p + m 1 + z p+m Ψz z = r < 1 1 rp+m in 3.18, we obtain z D n fz R D n fz p 3p + mrp+m 2p + mr 2p+m 1 r 2p+m z = r < 1, which is certainly positive, provided that r < R, R being given as in 4. 5. Let 1 B j < A j 1 j = 1, 2. If each of the functions f j z Σ p satisfies the following subordination condition: 3.19 1 λz p D n f j z + λz p D n+1 f j z 1 + A jz 1 + B j z then 3.2 1 λz p D n Hz + λz p D n+1 Hz where Hz = D n f 1 f 2 z 1 + 1 2ηz 1 z j = 1, 2; z U, z U, Page 22 of 31

and η = 1 4A 1 B 1 A 2 B 2 1 B 1 1 B 2 The result is the best possible when B 1 = B 2 = 1. [1 12 2 F 1 1, 1; 1 λ + 1; 1 ]. 2 Proof. Suppose that each of the functions f j z Σ p j = 1, 2 satisfies the condition 3.19. Then, by letting 3.21 ϕ j z = 1 λz p D n f j z + λz p D n+1 f j z j = 1, 2, we have ϕ j z Pγ j γ j = 1 A j ; j = 1, 2. 1 B j By making use of the operator identity 1.5 in 3.21, we observe that D n f j z = 1 z λ z p 1/λ t 1/λ 1 ϕ j tdt j = 1, 2, which, in view of the definition of Hz given already with 3.2, yields 3.22 D n Hz = 1 z λ z p 1/λ t 1/λ 1 ϕ tdt, where, for convenience, 3.23 ϕ z = 1 λz p D n Hz + λz p D n+1 Hz = 1 z λ z 1/λ t 1/λ 1 ϕ 1 ϕ 2 tdt. Page 23 of 31

Since ϕ 1 z Pγ 1 and ϕ 2 z Pγ 2, it follows from Lemma 3 that 3.24 ϕ 1 ϕ 2 z Pγ 3 γ3 = 1 21 γ 1 1 γ 2. Now, by using 3.24 in 3.23 and then appealing to Lemma 2 and Lemma 4, we get R { ϕ z } = 1 λ 1 λ > 1 λ 1 1 1 u 1/λ 1 R { ϕ 1 ϕ 2 } uzdu u 1/λ 1 2γ 3 1 + 21 γ 3 1 + u z u 1/λ 1 2γ 3 1 + 21 γ 3 1 + u = 1 4A 1 B 1 A 2 B 2 1 B 1 1 B 2 = 1 4A 1 B 1 A 2 B 2 1 B 1 1 B 2 = η z U. du du 1 1 1 λ [1 12 2 F 1 1, 1; 1 λ + 1; 1 2 u 1/λ 1 1 + u 1 du ] When B 1 = B 2 = 1, we consider the functions f j z Σ p j = 1, 2, which satisfy the hypothesis 3.19 of 5 and are defined by D n f j z = 1 z 1 + λ z 1/λ t 1/λ 1 Aj t dt j = 1, 2. 1 z Page 24 of 31

Thus it follows from 3.23 and Lemma 4 that ϕ z = 1 1 u 1/λ 1 1 1 + A 1 1 + A 2 + 1 + A 11 + A 2 du λ 1 uz = 1 1 + A 1 1 + A 2 + 1 + A 1 1 + A 2 1 z 1 2F 1 1, 1; 1 λ + 1; z z 1 1 1 + A 1 1 + A 2 + 1 2 1 + A 11 + A 2 2 F 1 1, 1; 1 λ + 1; 1 2 which evidently completes the proof of 5. By setting A j = 1 2α j, B j = 1 j = 1, 2, and n = as z 1, in 5, we obtain the following result which refines the work of Yang [15, 4]. Corollary 6. If the functions f j z Σ p j = 1, 2 satisfy the following inequality: 3.25 R 1 + λpz p f j z + λz p+1 f jz > α j α j < 1; j = 1, 2; z U, Page 25 of 31

then where R 1 + λpz p f 1 f 2 z + λz p+1 f 1 f 2 z > η z U, η = 1 41 α 1 1 α 2 [1 12 2 F 1 1, 1; 1 λ + 1; 1 ]. 2 The result is the best possible. 6. If fz Σ p,m satisfies the following subordination condition: 1 λz p D n fz + λz p D n+1 fz 1 + Az 1 + Bz z U, then R z p D n fz 1/q > ρ 1/q q N; z U, where ρ is given as in 1. The result is the best possible. Proof. Defining the function φz by 3.26 φz = z p D n fz f Σ p,m ; z U, we see that the function φz is of the form 2.1 and is analytic in U. Using the identity 1.5 in 3.26 and differentiating the resulting equation with respect to z, we find that Page 26 of 31 1 λz p D n fz + λz p D n+1 fz = φz + λ zφ z 1 + Az 1 + Bz z U.

Now, by following the lines of proof of 1 mutatis mutandis, and using the elementary inequality: R w 1/q Rw 1/q Rw > ; q N, we arrive at the result asserted by 6. A = Upon setting [2F 1 1, 1; 1 λp + m + 1; 1 ] 1 [2 2 F 1 1, 1; 2 B = 1, n =, and q = 1 in 6, we deduce Corollary 7 below. Corollary 7. If fz Σ p,m satisfies the following inequality: 1 λp + m + 1; 1 ] 1, 2 3.27 R 1 + λpz p fz + λz p+1 f z 1 3 2 2 F 1 1, 1; + 1; 1 λp+m 2 > ] z U, 1 2 [2 2 F 1 1, 1; + 1; 1 λp+m 2 then The result is the best possible. R z p fz > 1 2 z U. Page 27 of 31

From Corollary 6 and 6 for m = p + 1, A = 1 2η, B = 1, and q = 1, we deduce the following result. Corollary 8. If the functions f j z Σ p j = 1, 2 satisfy the inequality 3.25, then R z p f 1 f 2 z > η + 1 η [2F 1 1, 1; 1 λ + 1; 1 ] 1 z U, 2 where η is given as in Corollary 6. The result is the best possible. 7. Let fz Σ n p,ma, B and let gz Σ p,m satisfy the following inequality: Then Proof. We have Since zp+1 D n f gz p and the function R z p gz > 1 2 z U. f gz Σ n p,ma, B. = zp+1 D n fz z p gz p R z p gz > 1 2 1 + Az 1 + Bz z U z U. Page 28 of 31

is convex univalent in U, it follows from 1.6 and Lemma 5 that f gz Σ n p,ma, B. This completes the proof of 7. In view of Corollary 7 and 7, we have Corollary 9 below. Corollary 9. If fz Σ n p,ma, B and the function gz Σ p,m satisfies the inequality 3.27, then f gz Σ n p,ma, B. Page 29 of 31

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