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On Differential Subordination, Superordination and Sandwich Theorems of Symmetric Analytic Functions

On Differential Subordination, Superordination and Sandwich Theorems of Symmetric Analytic Functions In this paper, we study the interesting properties of differential subordination superordination for the classes of symmetric functions analytic in the unit disc. We derive swich results on the basis of this theory. Mathematics Subject Classification 2010: 30C45, 30C50. Key words: analytic functions, convex functions, symmetric functions, differential subordination superordination. 1. Introduction Let H(E) denote the class of analytic functions in the open unit disc E = {z : z C |z| < 1} let H[a, 1] denote the subclass of the functions f H(E) of the formf (z) = a + a1 z + a2 z 2 + ..., a C. Also, let A be the subclass of functions f H(E) of the form (1.1) f (z) = z + k=2 ak z k . If f g are analytic in E, we say that f is subordinate to g, written f g or f (z) g(z), if there exists a Schwarz function with w(0) = 0 |w(z)| < 1 in E such that f (z) = g(w(z)). Suppose that h k are two analytic functions in E let (r, s, t; z) : C3 ×EC. If h (h(z), zh (z), z 2 h (z); z) are univalent functions in E if h satisfies the second order superordination (1.2) k(z) (h(z), zh (z), z 2 h (z); z), then k is said to be a solution of the differential superordination (1.2). A function q H(E) is called a subordinant to (1.2), if q(z) h(z) for all the functions h satisfying (1.2). A univalent subordinant q that satisfies q(z) q(z) for all of the subordinants q of (1.2), is said to be the best subordinant. Recently, Miller Mocanu [7] obtained sufficient conditions on the functions k, q for which the following implications hold k(z) (h(z), zh (z), z 2 h (z); z) = q(z) h(z). Using these results, the authors in [1] considered certain classes of first-order differential superordinations, see also [4], as well as superordination-preserving integral operators [3]. Aouf et al. [1, 2], obtained sufficient conditions for certain normalized (z) analytic functions f to satisfyq1 (z) zf(z) q2 (z), where q1 q2 are f given univalent normalized functions in E. Very recently, Shanmugam et al. [12, 13] obtained the such called swich results for certain classes of analytic functions. For interested readers we refer to the work done by the authors [1, 8, 9, 11]. In [14], Sakaguchi defined the class of starlike functions with respect to symmetrical points as follows: Let f A. Then f is said to be starlike with respect to symmetrical zf (z) points in E if, only if,Re f (z)-f (-z) > 0, z E. Obviously, it forms a subclass of close-to-convex functions hence univalent. Moreover, this class includes the class of convex functions odd starlike functions with respect to the origin, see [14]. In this paper, we study several applications of the theory of differential subordination superordination for normalized univalent in the unit disc. On the basis of this theory we also investigate some important swhich results of which will be useful in obtaining applications of geometric function theory. 2. Preliminary results Definition 2.1 ([7]). Let Q be the set of all functions f that are analytic injective on E\U (f ), where U (f ) = { E : limz f (z) = } , are such that f () = 0 for E\U (f ). To establish our main results we need the following Lemmas. Lemma 2.2 (Miller Mocanu [6]). Let q be univalent in the unit disc E, let be analytic in a domain D containing q(E), with (w) = 0 when w q(E). Set Q(z) = zq (z)(q(z)), h(z) = (q(z) + Q(z) suppose that: (i) Q is a starlike function in E, (ii) Re zh (z) > 0, z E. Q(z) If p is analytic in E with p(0) = q(0), p(E) D (2.1) (p(z)) + zp (z)(p(z)) (q(z)) + zq (z)(q(z)), then p(z) q(z), q is the best dominant of (2.1). Lemma 2.3 (Shanmugam et al. [13]). Let µ, C with = 0, let q be a convex function in E with Re 1 + zq (z) q (z) > max 0; -Re µ , z E. If p is analytic in E (2.2) µp(z) + zp (z) µq(z) + zq (z), then p(z) q(z), q is the best dominant of (2.2). Lemma 2.4 (Bulboaca [5]). Let q be a univalent function in the unit disc E, let be analytic in a domain D containing q(E). Suppose that: (q(z)) (i) Re (q(z) > 0, for z E, (ii) h(z) = zq (z)(q(z)) is starlike in E. If p H[q(0), 1] Q with p(E) D, (p(z) + zp (z))(p(z)) is univalent in E, (2.3) (q(z)) + zq (z)(q(z)) (p(z)) + zp (z)(p(z)), then q(z) p(z), q is the best subordinant of (2.3). Note that this result generalize a similar one obtained in [4]. Lemma 2.5 (Miller Mocanu [7]). Let q be convex in E let C, with Re > 0. If p H[q(0), 1] Q p(z) + zp (z) is univalent in E, then (2.4) q(z) + zq (z) p(z) + zp (z), 1 (1-z)2ab implies q(z) p(z), q is the best subordinant of (2.4). Lemma 2.6 (Royster [10]). The function q(z) = in E if only if |2ab - 1| 1 or |2ab + 1| 1. is univalent 3. Main results Theorem 3.1. Let q be univalent in E, with q(0) = 1, suppose that (3.1) Re 1 + zq (z) q(z) > max 0; - Re 1 , z E, where C = C\{0}. If f A satisfies the subordination (3.2) (1 + ) - 2z 2 (f (z) + f (-z)) q(z) + zq (z), ((f (z) - f (-z)))2 2z then ( (f (z)-f (-z)) ) q(z), q is the best dominant of Eq. (3.2). Proof. Set (3.3) = h(z), where h(z) is analytic in E with h(0) = 1. A simple computation shows that (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z))) = h(z) + zh (z), hence the subordination (3.2) is equivalent to h(z)+zh (z) q(z)+zq (z). Combining this last relation together with Lemma 2.2 for special case = µ = 1, we obtain our result. 1+Az Taking q(z) = 1+Bz in Theorem 3.1, where -1 B < A 1, the condition (3.1) reduces to (3.4) Re 1 1 - Bz > max 0; - Re 1 + Bz , z E. || < |B|, is convex in It is easy to verify that the function () = (1-) (1+) , E, since () = () for all || < |B| , it follows that (E) is a convex domain symmetric with respect to the real axis, hence (3.5) inf Re 1 - Bz : zE 1 + Bz = 1 - |B| > 0. 1 + |B| |B|-1 |B|+1 , 1 Then, the inequality (3.4) is equivalent to Re following result. hence we have the 1-|B| 1+|B| . 1 Corollary 3.2. Let C , -1 B < A 1 with max{0; - Re } If f A, (1 + ) (3.6) then 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 (A - B)z 1 + Az + , 1 + Bz (1 + Bz)2 1 + Az , 1 + Bz 1+Az 1+Bz is the best dominant of (3.6). 1 Example 3.3. Let C with Re 0. If f A, (3.7) (1+) then 2z (f (z)-f (-z)) 2z 1+z 2z 2 (f (z)+f (-z)) + , 2 ((f (z)-f (-z))) 1-z (1 - z)2 1+z , 1-z 1+z 1-z is the best dominant of (3.7). Theorem 3.4. Let q be univalent in E, with q(0) = 1 q(z) = 0 for all z E. Let , µ C v, C, with v + = 0. Let f A suppose that f q satisfy the following conditions: (3.8) (3.9) If 1 + µ 1 - (3.10) vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) (z) zq 1+ , q(z) Re 1 + zq (z) zq (z) - q (z) q(z) > 0, z E. 2(v + )z = 0, z E, vz (f (z) + f (-z)) + (f (z) - f (-z)) then 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f q(z), q is the best dominant of (3.10). The power is the principal one. Proof. We begin by setting (3.11) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f = h(z), z E, where h(z) is analytic in E with h(0) = 1. Differentiating Equation (3.11) logarithmically with respect to z, we have µ 1- zh (z) vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) . = vz (f (z) + f (-z)) + (f (z) - f (-z)) h(z) To prove our result we use Lemma 2.1. Consider in this Lemma (w) = 1 (w) = w , then is analytic in C (w) = 0 is analytic in C . Also, if we let Q(z) = zq (z)(q(z)) = zq (z) , g(z) = (q(z)) + Q(z) = q(z) 1 + zq (z) , then, since Q(0) = 0 Q (0) = 0, the assumption (3.9) would q(z) yield that Q is a starlike function in E. From (3.9), we have Re zg (z) zq (z) zq (z) = Re 1 + - Q(z) q (z) q(z) > 0, z E, by using Lemma 2.1 we deduce that the subordination (3.10) implies that h(z) q(z), the function q is the best dominant of (3.10). 1+Az In particular, v = 0, = = 1 q(z) = 1+Bz in the above Theorem 3.4, it is easy to see that the assumption (3.9) holds whenever -1 A < B 1, which leads to the next result: Corollary 3.5. Let -1 A < B 1 µ C . Let f A 2z suppose that (f (z)-f (-z)) = 0, z E. If (3.12) then 1+µ 1- z (f (z) + f (-z)) (A - B)z , 1+ (f (z) - f (-z)) (1 + Az)(1 + Bz) 1+Az 1+Bz µ 1 + Az , 1 + Bz is the best dominant of (3.12). The power is the principal one. 1 1 Putting v = 0, = 1, = ab , a, b C , µ = a, q(z) = (1-z)2ab in Theorem 3.4, then combining this together with Lemma 2.5, we have the next result. Corollary 3.6. Let a, b C such that |2ab - 1| 1 or |2ab + 1| 1. 2z Let f A let (f (z)-f (-z)) = 0, for all z E. If 1+ then (3.13) 1 (1-z)2ab 1 b 1- z (f (z) + f (-z)) (f (z) - f (-z)) 1+z , 1-z 1 , (1 - z)2ab is the best dominant of (3.13). The power is the principal one. µ(A-B) Putting v = 0, = = 1 q(z) = (1 + Bz) B , -1 B < A 1, B = 0 in Theorem 3.4, using Lemma 2.5, we have the next result. | µ(A-B) | B-1 Corollary 3.7. Let -1 B < A 1 with B = 0, suppose that 2z 1 or | µ(A-B) | 1. Let f A such that (f (z)-f (-z)) = 0, for all B+1 z E, let µ C . If 1+µ 1- z (f (z) + f (-z)) (f (z) - f (-z)) (3.14) then 1 + [B + µ(A - B)]z , 1 + Bz µ(A-B) B µ(A-B) (1 + Bz) (1 + Bz) B is the best dominant of (3.14). Here the power is the principal one. i By taking v = 0, = 1, = abe , a, b C , || < , µ = a cos 2 1 q(z) = -i in Theorem 3.4, we obtain the following result. 2ab cos e (1-z) Corollary 3.8. Let a, bC ||< , suppose that |2ab cos e-i - 2 2z 1| 1 or |2ab cos e-i + 1| 1. Let f A such that (f (z)-f (-z)) = 0, for all z E. If (3.15) 1+ ei b cos 1- z (f (z) + f (-z)) (f (z) - f (-z)) 1+z , 1-z then (1 - z)2ab cos e-i 1 is the best dominant of (3.15). The power is the principal (1-z)2ab cos e-i one. Theorem 3.9. Let q be univalent in E with q(0) = 1, let µ, C , let µ, C let , , v, C with v + = 0. Let f A suppose that f q satisfy the next two conditions: (3.16) (3.17) If (3.18) (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f 2(v + )z = 0, z E, + + (f (z) - f (-z)) zq (z) Re 1 + > max 0; -Re , z E. q (z) vz (f (z) f (-z)) + µ 1 - (3.19) then vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) + , (z) q(z) + zq (z) + , 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f q(z), q is the best dominant of (3.19). All the powers are the principal ones. Proof. We begin by setting (3.20) 2(v + )z vz (f (z) + f (-z)) + (f (z) - f (-z)) = h(z). Then h(z) is analytic in E with h(0) = 1. Logarithmic differentiating of (3.20) yields µ 1- vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) = zh (z) , h(z) 9 hence µh(z) 1 - vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) = zh (z). Let us consider the functions: (w) = w + , (w) = , w C, Q(z) = zq (z)(q(z) = zq (z), z E, g(z) = (q(z)+Q(z) = q(z)+zq (z)+, z E. From the assumption (3.17) we see that Q is starlike in E , that Re zg (z) = Re Q(z) zq (z) +1+ q (z) > 0, z E, thus, by applying Lemma 2.1 this completes the proof. (1+Az) Taking q(z) = (1+Bz) in Theorem 3.7, where -1 B < A 1 according to (3.5), the condition (3.17) becomes max 0; -Re 1 - |B| . 1 + |B| Hence, for the particular case v = 1 = , = 0, we have the following result: Corollary 3.10. Let -1 B < A 1 let C with max {0; -Re} Let f A suppose that 2 (f (z)+f (-z)) 1 - |B| . 1 + |B| = 0, z E let µ C . If + (3.21) then µ 2 z(f (z) + f (-z)) +µ 1- (f (z) + f (-z)) f (z) - f (-z) z(A - B) 1 + Az ++ , 1 + Bz (1 + Bz)2 2 (f (z) + f (-z)) 1 + Az , 1 + Bz 1+Az 1+Bz is the best dominant of (3.21). All the powers are the principal ones. 1+z 1-z Taking = = 1, v = 0 q(z) = next result. in Theorem 3.7, we obtain the = 0 for all z E, Corollary 3.11. Let f A such that let µ C . If 2z f (z)-f (-z) (3.22) µ z(f (z) + f (-z)) 2z +µ 1- f (z) - f (-z) f (z) - f (-z) 2z 1+z ++ , 1-z (1 - z)2 1+z 1-z 2z then [ f (z)-f (-z) ]µ 1+z , 1-z powers are the principal ones. is the best dominant of (3.22). All the 4. Superordination swich theorems Theorem 4.1. Let q be convex in E with q(0) = 1, let C with 2z Re > 0. Let f A suppose that f (z)-f (-z) H [q(0), 1] Q. If the function (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 is univalent in the unit disc E, (4.1) q(z) + zq (z) (1 + ) then q(z) - 2z 2 (f (z) + f (-z)) , ((f (z) - f (-z)))2 q is the best subordinant of (4.1). Proof. Set = h(z), z E. Then h(z) is analytic in E with h(0) = 1.Taking logarithmic differentiation with respect z, we have (4.2) 1-z f (z) + f (-z) f (z) - f (-z) = zh (z) . h(z) A simple computations show that h(z) + zh (z) = (1 + ) - 2z 2 (f (z) + f (-z)) , ((f (z) - f (-z)))2 now, by using Lemma 2.4, this completes the proof. 1+Az Taking q(z) = 1+Bz in Theorem 4.1, where -1 B < A 1, we obtain the next result. Corollary 4.2. Let q be convex in E with q(0) = 1, let C with 2z Re > 0. Let f A suppose that f (z)-f (-z) H [q(0), 1]Q. If the function (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 is univalent in the unit disc E, 1 + Az (A - B)z + (1 + ) 1 + Bz (1 + Bz)2 2z 2 (f (z) + f (-z)) , - ((f (z) - f (-z)))2 1 + Az 1 + Bz 1+Az 1+Bz is (4.3) then the best subordinant of (4.3), where -1 B < A 1. Using the same techniques as in Theorem 3.8, then applying Lemma 2.3, we have the following theorem. Theorem 4.3. Let q be convex in E with q(0) = 1, let µ, C , let , , v, C with v + = 0 Re > 0. Let f A suppose that f satisfies the following conditions: vz(f (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f f (-z)) 2(v + )z = 0, z E, + (f (z) - f (-z)) H[q(0), 1] Q. If the function given by equation (3.18) is univalent in E, (4.4) then q(z) q(z) + zq (z) + (z), 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f q is the best subordinate of (4.4) (all powers are the principal ones). Note that by combining Theorem 3.1 with Theorem 4.1 Theorem 3.10 with Theorem 4.3, we have, respectively, the following two swich theorems: Theorem 4.4. Let q1 q2 be two convex functions in E with q1 (0) = 2z q2 (0) = 1, let C with Re > 0. Let f A suppose that f (z)-f (-z) H[q(0), 1] Q. If the function (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 is univalent in the unit disc E, q1 (z) + zq1 (z) (1 + ) 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 (4.5) then q2 (z) + zq2 (z), q1 (z) q2 (z), q1 q2 are, respectively, the best subordinate the best dominant of (4.5). Theorem 4.5. Let q1 q2 be two convex functions in E with q1 (0) = q2 (0) = 1, let µ, C , let , , v, C with v + = 0 Re > 0. Let f A satisfy the following conditions: vz(f (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f f (-z)) 2(v + )z = 0, z E, + (f (z) - f (-z)) H[q(0), 1] Q. If the function given by (3.18) is univalent in E, (4.6) then q1 (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f µ q1 (z) + zq1 (z) + (z) q2 (z) + zq2 (z) + , q2 (z), q1 q2 are, respectively, the best subordinate the best dominant of (4.6) (all the powers are the principal ones). http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter

On Differential Subordination, Superordination and Sandwich Theorems of Symmetric Analytic Functions

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Abstract

In this paper, we study the interesting properties of differential subordination superordination for the classes of symmetric functions analytic in the unit disc. We derive swich results on the basis of this theory. Mathematics Subject Classification 2010: 30C45, 30C50. Key words: analytic functions, convex functions, symmetric functions, differential subordination superordination. 1. Introduction Let H(E) denote the class of analytic functions in the open unit disc E = {z : z C |z| < 1} let H[a, 1] denote the subclass of the functions f H(E) of the formf (z) = a + a1 z + a2 z 2 + ..., a C. Also, let A be the subclass of functions f H(E) of the form (1.1) f (z) = z + k=2 ak z k . If f g are analytic in E, we say that f is subordinate to g, written f g or f (z) g(z), if there exists a Schwarz function with w(0) = 0 |w(z)| < 1 in E such that f (z) = g(w(z)). Suppose that h k are two analytic functions in E let (r, s, t; z) : C3 ×EC. If h (h(z), zh (z), z 2 h (z); z) are univalent functions in E if h satisfies the second order superordination (1.2) k(z) (h(z), zh (z), z 2 h (z); z), then k is said to be a solution of the differential superordination (1.2). A function q H(E) is called a subordinant to (1.2), if q(z) h(z) for all the functions h satisfying (1.2). A univalent subordinant q that satisfies q(z) q(z) for all of the subordinants q of (1.2), is said to be the best subordinant. Recently, Miller Mocanu [7] obtained sufficient conditions on the functions k, q for which the following implications hold k(z) (h(z), zh (z), z 2 h (z); z) = q(z) h(z). Using these results, the authors in [1] considered certain classes of first-order differential superordinations, see also [4], as well as superordination-preserving integral operators [3]. Aouf et al. [1, 2], obtained sufficient conditions for certain normalized (z) analytic functions f to satisfyq1 (z) zf(z) q2 (z), where q1 q2 are f given univalent normalized functions in E. Very recently, Shanmugam et al. [12, 13] obtained the such called swich results for certain classes of analytic functions. For interested readers we refer to the work done by the authors [1, 8, 9, 11]. In [14], Sakaguchi defined the class of starlike functions with respect to symmetrical points as follows: Let f A. Then f is said to be starlike with respect to symmetrical zf (z) points in E if, only if,Re f (z)-f (-z) > 0, z E. Obviously, it forms a subclass of close-to-convex functions hence univalent. Moreover, this class includes the class of convex functions odd starlike functions with respect to the origin, see [14]. In this paper, we study several applications of the theory of differential subordination superordination for normalized univalent in the unit disc. On the basis of this theory we also investigate some important swhich results of which will be useful in obtaining applications of geometric function theory. 2. Preliminary results Definition 2.1 ([7]). Let Q be the set of all functions f that are analytic injective on E\U (f ), where U (f ) = { E : limz f (z) = } , are such that f () = 0 for E\U (f ). To establish our main results we need the following Lemmas. Lemma 2.2 (Miller Mocanu [6]). Let q be univalent in the unit disc E, let be analytic in a domain D containing q(E), with (w) = 0 when w q(E). Set Q(z) = zq (z)(q(z)), h(z) = (q(z) + Q(z) suppose that: (i) Q is a starlike function in E, (ii) Re zh (z) > 0, z E. Q(z) If p is analytic in E with p(0) = q(0), p(E) D (2.1) (p(z)) + zp (z)(p(z)) (q(z)) + zq (z)(q(z)), then p(z) q(z), q is the best dominant of (2.1). Lemma 2.3 (Shanmugam et al. [13]). Let µ, C with = 0, let q be a convex function in E with Re 1 + zq (z) q (z) > max 0; -Re µ , z E. If p is analytic in E (2.2) µp(z) + zp (z) µq(z) + zq (z), then p(z) q(z), q is the best dominant of (2.2). Lemma 2.4 (Bulboaca [5]). Let q be a univalent function in the unit disc E, let be analytic in a domain D containing q(E). Suppose that: (q(z)) (i) Re (q(z) > 0, for z E, (ii) h(z) = zq (z)(q(z)) is starlike in E. If p H[q(0), 1] Q with p(E) D, (p(z) + zp (z))(p(z)) is univalent in E, (2.3) (q(z)) + zq (z)(q(z)) (p(z)) + zp (z)(p(z)), then q(z) p(z), q is the best subordinant of (2.3). Note that this result generalize a similar one obtained in [4]. Lemma 2.5 (Miller Mocanu [7]). Let q be convex in E let C, with Re > 0. If p H[q(0), 1] Q p(z) + zp (z) is univalent in E, then (2.4) q(z) + zq (z) p(z) + zp (z), 1 (1-z)2ab implies q(z) p(z), q is the best subordinant of (2.4). Lemma 2.6 (Royster [10]). The function q(z) = in E if only if |2ab - 1| 1 or |2ab + 1| 1. is univalent 3. Main results Theorem 3.1. Let q be univalent in E, with q(0) = 1, suppose that (3.1) Re 1 + zq (z) q(z) > max 0; - Re 1 , z E, where C = C\{0}. If f A satisfies the subordination (3.2) (1 + ) - 2z 2 (f (z) + f (-z)) q(z) + zq (z), ((f (z) - f (-z)))2 2z then ( (f (z)-f (-z)) ) q(z), q is the best dominant of Eq. (3.2). Proof. Set (3.3) = h(z), where h(z) is analytic in E with h(0) = 1. A simple computation shows that (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z))) = h(z) + zh (z), hence the subordination (3.2) is equivalent to h(z)+zh (z) q(z)+zq (z). Combining this last relation together with Lemma 2.2 for special case = µ = 1, we obtain our result. 1+Az Taking q(z) = 1+Bz in Theorem 3.1, where -1 B < A 1, the condition (3.1) reduces to (3.4) Re 1 1 - Bz > max 0; - Re 1 + Bz , z E. || < |B|, is convex in It is easy to verify that the function () = (1-) (1+) , E, since () = () for all || < |B| , it follows that (E) is a convex domain symmetric with respect to the real axis, hence (3.5) inf Re 1 - Bz : zE 1 + Bz = 1 - |B| > 0. 1 + |B| |B|-1 |B|+1 , 1 Then, the inequality (3.4) is equivalent to Re following result. hence we have the 1-|B| 1+|B| . 1 Corollary 3.2. Let C , -1 B < A 1 with max{0; - Re } If f A, (1 + ) (3.6) then 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 (A - B)z 1 + Az + , 1 + Bz (1 + Bz)2 1 + Az , 1 + Bz 1+Az 1+Bz is the best dominant of (3.6). 1 Example 3.3. Let C with Re 0. If f A, (3.7) (1+) then 2z (f (z)-f (-z)) 2z 1+z 2z 2 (f (z)+f (-z)) + , 2 ((f (z)-f (-z))) 1-z (1 - z)2 1+z , 1-z 1+z 1-z is the best dominant of (3.7). Theorem 3.4. Let q be univalent in E, with q(0) = 1 q(z) = 0 for all z E. Let , µ C v, C, with v + = 0. Let f A suppose that f q satisfy the following conditions: (3.8) (3.9) If 1 + µ 1 - (3.10) vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) (z) zq 1+ , q(z) Re 1 + zq (z) zq (z) - q (z) q(z) > 0, z E. 2(v + )z = 0, z E, vz (f (z) + f (-z)) + (f (z) - f (-z)) then 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f q(z), q is the best dominant of (3.10). The power is the principal one. Proof. We begin by setting (3.11) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f = h(z), z E, where h(z) is analytic in E with h(0) = 1. Differentiating Equation (3.11) logarithmically with respect to z, we have µ 1- zh (z) vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) . = vz (f (z) + f (-z)) + (f (z) - f (-z)) h(z) To prove our result we use Lemma 2.1. Consider in this Lemma (w) = 1 (w) = w , then is analytic in C (w) = 0 is analytic in C . Also, if we let Q(z) = zq (z)(q(z)) = zq (z) , g(z) = (q(z)) + Q(z) = q(z) 1 + zq (z) , then, since Q(0) = 0 Q (0) = 0, the assumption (3.9) would q(z) yield that Q is a starlike function in E. From (3.9), we have Re zg (z) zq (z) zq (z) = Re 1 + - Q(z) q (z) q(z) > 0, z E, by using Lemma 2.1 we deduce that the subordination (3.10) implies that h(z) q(z), the function q is the best dominant of (3.10). 1+Az In particular, v = 0, = = 1 q(z) = 1+Bz in the above Theorem 3.4, it is easy to see that the assumption (3.9) holds whenever -1 A < B 1, which leads to the next result: Corollary 3.5. Let -1 A < B 1 µ C . Let f A 2z suppose that (f (z)-f (-z)) = 0, z E. If (3.12) then 1+µ 1- z (f (z) + f (-z)) (A - B)z , 1+ (f (z) - f (-z)) (1 + Az)(1 + Bz) 1+Az 1+Bz µ 1 + Az , 1 + Bz is the best dominant of (3.12). The power is the principal one. 1 1 Putting v = 0, = 1, = ab , a, b C , µ = a, q(z) = (1-z)2ab in Theorem 3.4, then combining this together with Lemma 2.5, we have the next result. Corollary 3.6. Let a, b C such that |2ab - 1| 1 or |2ab + 1| 1. 2z Let f A let (f (z)-f (-z)) = 0, for all z E. If 1+ then (3.13) 1 (1-z)2ab 1 b 1- z (f (z) + f (-z)) (f (z) - f (-z)) 1+z , 1-z 1 , (1 - z)2ab is the best dominant of (3.13). The power is the principal one. µ(A-B) Putting v = 0, = = 1 q(z) = (1 + Bz) B , -1 B < A 1, B = 0 in Theorem 3.4, using Lemma 2.5, we have the next result. | µ(A-B) | B-1 Corollary 3.7. Let -1 B < A 1 with B = 0, suppose that 2z 1 or | µ(A-B) | 1. Let f A such that (f (z)-f (-z)) = 0, for all B+1 z E, let µ C . If 1+µ 1- z (f (z) + f (-z)) (f (z) - f (-z)) (3.14) then 1 + [B + µ(A - B)]z , 1 + Bz µ(A-B) B µ(A-B) (1 + Bz) (1 + Bz) B is the best dominant of (3.14). Here the power is the principal one. i By taking v = 0, = 1, = abe , a, b C , || < , µ = a cos 2 1 q(z) = -i in Theorem 3.4, we obtain the following result. 2ab cos e (1-z) Corollary 3.8. Let a, bC ||< , suppose that |2ab cos e-i - 2 2z 1| 1 or |2ab cos e-i + 1| 1. Let f A such that (f (z)-f (-z)) = 0, for all z E. If (3.15) 1+ ei b cos 1- z (f (z) + f (-z)) (f (z) - f (-z)) 1+z , 1-z then (1 - z)2ab cos e-i 1 is the best dominant of (3.15). The power is the principal (1-z)2ab cos e-i one. Theorem 3.9. Let q be univalent in E with q(0) = 1, let µ, C , let µ, C let , , v, C with v + = 0. Let f A suppose that f q satisfy the next two conditions: (3.16) (3.17) If (3.18) (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f 2(v + )z = 0, z E, + + (f (z) - f (-z)) zq (z) Re 1 + > max 0; -Re , z E. q (z) vz (f (z) f (-z)) + µ 1 - (3.19) then vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) + , (z) q(z) + zq (z) + , 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz (f q(z), q is the best dominant of (3.19). All the powers are the principal ones. Proof. We begin by setting (3.20) 2(v + )z vz (f (z) + f (-z)) + (f (z) - f (-z)) = h(z). Then h(z) is analytic in E with h(0) = 1. Logarithmic differentiating of (3.20) yields µ 1- vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) = zh (z) , h(z) 9 hence µh(z) 1 - vz(z (f (z) + f (-z)) + z (f (z) + f (-z)) vz (f (z) + f (-z)) + (f (z) - f (-z)) = zh (z). Let us consider the functions: (w) = w + , (w) = , w C, Q(z) = zq (z)(q(z) = zq (z), z E, g(z) = (q(z)+Q(z) = q(z)+zq (z)+, z E. From the assumption (3.17) we see that Q is starlike in E , that Re zg (z) = Re Q(z) zq (z) +1+ q (z) > 0, z E, thus, by applying Lemma 2.1 this completes the proof. (1+Az) Taking q(z) = (1+Bz) in Theorem 3.7, where -1 B < A 1 according to (3.5), the condition (3.17) becomes max 0; -Re 1 - |B| . 1 + |B| Hence, for the particular case v = 1 = , = 0, we have the following result: Corollary 3.10. Let -1 B < A 1 let C with max {0; -Re} Let f A suppose that 2 (f (z)+f (-z)) 1 - |B| . 1 + |B| = 0, z E let µ C . If + (3.21) then µ 2 z(f (z) + f (-z)) +µ 1- (f (z) + f (-z)) f (z) - f (-z) z(A - B) 1 + Az ++ , 1 + Bz (1 + Bz)2 2 (f (z) + f (-z)) 1 + Az , 1 + Bz 1+Az 1+Bz is the best dominant of (3.21). All the powers are the principal ones. 1+z 1-z Taking = = 1, v = 0 q(z) = next result. in Theorem 3.7, we obtain the = 0 for all z E, Corollary 3.11. Let f A such that let µ C . If 2z f (z)-f (-z) (3.22) µ z(f (z) + f (-z)) 2z +µ 1- f (z) - f (-z) f (z) - f (-z) 2z 1+z ++ , 1-z (1 - z)2 1+z 1-z 2z then [ f (z)-f (-z) ]µ 1+z , 1-z powers are the principal ones. is the best dominant of (3.22). All the 4. Superordination swich theorems Theorem 4.1. Let q be convex in E with q(0) = 1, let C with 2z Re > 0. Let f A suppose that f (z)-f (-z) H [q(0), 1] Q. If the function (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 is univalent in the unit disc E, (4.1) q(z) + zq (z) (1 + ) then q(z) - 2z 2 (f (z) + f (-z)) , ((f (z) - f (-z)))2 q is the best subordinant of (4.1). Proof. Set = h(z), z E. Then h(z) is analytic in E with h(0) = 1.Taking logarithmic differentiation with respect z, we have (4.2) 1-z f (z) + f (-z) f (z) - f (-z) = zh (z) . h(z) A simple computations show that h(z) + zh (z) = (1 + ) - 2z 2 (f (z) + f (-z)) , ((f (z) - f (-z)))2 now, by using Lemma 2.4, this completes the proof. 1+Az Taking q(z) = 1+Bz in Theorem 4.1, where -1 B < A 1, we obtain the next result. Corollary 4.2. Let q be convex in E with q(0) = 1, let C with 2z Re > 0. Let f A suppose that f (z)-f (-z) H [q(0), 1]Q. If the function (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 is univalent in the unit disc E, 1 + Az (A - B)z + (1 + ) 1 + Bz (1 + Bz)2 2z 2 (f (z) + f (-z)) , - ((f (z) - f (-z)))2 1 + Az 1 + Bz 1+Az 1+Bz is (4.3) then the best subordinant of (4.3), where -1 B < A 1. Using the same techniques as in Theorem 3.8, then applying Lemma 2.3, we have the following theorem. Theorem 4.3. Let q be convex in E with q(0) = 1, let µ, C , let , , v, C with v + = 0 Re > 0. Let f A suppose that f satisfies the following conditions: vz(f (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f f (-z)) 2(v + )z = 0, z E, + (f (z) - f (-z)) H[q(0), 1] Q. If the function given by equation (3.18) is univalent in E, (4.4) then q(z) q(z) + zq (z) + (z), 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f q is the best subordinate of (4.4) (all powers are the principal ones). Note that by combining Theorem 3.1 with Theorem 4.1 Theorem 3.10 with Theorem 4.3, we have, respectively, the following two swich theorems: Theorem 4.4. Let q1 q2 be two convex functions in E with q1 (0) = 2z q2 (0) = 1, let C with Re > 0. Let f A suppose that f (z)-f (-z) H[q(0), 1] Q. If the function (1 + ) - 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 is univalent in the unit disc E, q1 (z) + zq1 (z) (1 + ) 2z 2 (f (z) + f (-z)) ((f (z) - f (-z)))2 (4.5) then q2 (z) + zq2 (z), q1 (z) q2 (z), q1 q2 are, respectively, the best subordinate the best dominant of (4.5). Theorem 4.5. Let q1 q2 be two convex functions in E with q1 (0) = q2 (0) = 1, let µ, C , let , , v, C with v + = 0 Re > 0. Let f A satisfy the following conditions: vz(f (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f f (-z)) 2(v + )z = 0, z E, + (f (z) - f (-z)) H[q(0), 1] Q. If the function given by (3.18) is univalent in E, (4.6) then q1 (z) 2(v + )z (z) + f (-z)) + (f (z) - f (-z)) vz(f µ q1 (z) + zq1 (z) + (z) q2 (z) + zq2 (z) + , q2 (z), q1 q2 are, respectively, the best subordinate the best dominant of (4.6) (all the powers are the principal ones).

Journal

Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter

Published: Nov 24, 2014

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