Differential Subordinations and Harmonic Means

Transcription

1 Bull. Malays. Math. Sci. Soc. (2015) 38: DOI /s Differential Subordinations and Harmonic Means Stanisława Kanas Andreea-Elena Tudor eceived: 9 January 2014 / evised: 3 April 2014 / Published online: 22 January 2015 Malaysian Mathematical Sciences Society and Universiti Sains Malaysia 2015 Abstract The aim of a study of the presented paper is the differential subordination involving harmonic means of the expressions p(), p() + p (), and p() + p () p() when p is an analytic function in the unit disk, such that p(0) 1, p() 1. Several applications in the geometric functions theory are given. Keywords Univalent functions Subordination Differential subordination Harmonic means Mathematics Subject Classification Primary 31A05 Secondary 30C45 1 Introduction The harmonic mean, known also as the subcontrary mean, is one of several kinds of average, and is the special case power mean. It is used for the situations when the average of rates is desired and has several applications in geometry, trigonometry, probabilistics and statistics, algebra, physics, finance, computer science, etc. The harmonic mean is one of the Pythagorean means, along with the arithmetic and the geometric mean, and is no greater than either of them. The harmonic mean H of Communicated by V. avichandran. S. Kanas (B) University of esow, ul. S. Pigonia 1, esow, Poland skanas@ur.edu.pl A.-E. Tudor Babeş-Bolyai University, 1 Kogălniceanu Street, Cluj-Napoca, omania tudor_andreea_elena@yahoo.com

2 1244 S. Kanas, A.-E. Tudor the positive real numbers x 1, x 2,...,x n is defined to be the reciprocal of the arithmetic mean of the reciprocals of x 1, x 2,...,x n H ( 1 n n x k1 k ) 1 1. (1.1) For the special case of just two numbers x 1 and x 2, the harmonic mean can be written H 2x 1x 2 x 1 + x 2. In this special case, the harmonic mean is related to the arithmetic mean A x 1 + x 2, and the geometric mean G x 1 x 2, 2 by H G2 A,orG AH meaning the two numbers geometric mean equals the geometric mean of their arithmetic and harmonic means. Several concepts of the classes involving arithmetic and geometric means were appeared in the literature (see e.g. 1 9,11 14,16,17], for the extensive studies, we refer to the Miller and Mocanu monograph 10]). The purpose of this paper is to study the harmonic mean, as a supplementary to the well-known arithmetics and geometric Pythagorean means. In addition, a new mean brings along a wide range of new possibilities for exploiting harmonic ideas in connection of several quantities or functionals in the geometric function theory. In order to prove our main results, we introduce some fundamental notions and notations. Let A be the class of all analytic functions f in the open unit disk D { C : < 1}, oftheform: + a k k. (1.2) If f and g are two functions analytic in D, we say that f is subordinate to g, written as f g or g(), if there exists a Schwar function ω (i.e., analytic in D, with ω(0) 0 and ω() < 1, for all D) such that g(ω()), D. Furthermore, if g is univalent in D, then we have the following equivalence: k2 g() f (0) g(0) and f (D) g(d). Definition , p. 21] Denote by Q the set of functions q that are analytic and injective on D \ E(q), where E(q) { } ζ D : lim q(), ζ

3 Differential Subordinations and Harmonic Means 1245 such that q (ζ ) 0forζ D \ E(q). If q Q then q(d) is a simply connected domain. In order to prove our main results, we will need the following lemma: Lemma , p. 24] Let q Q, with q(0) a, and let p() a + a n n + be analytic in D with p() a and n 1. If p is not subordinate to q, then there exist points 0 r 0 e iθ 0 D and ζ 0 D \ E(q) and an m n 1 for which p(d r0 ) q(d), (1) p q(ζ 0 ), (2) 0 p mζ 0 q (ζ 0 ), (3) 0 p ζ0 q (ζ 0 ) p + 1 m q (ζ 0 ) ] Harmonic Mean Let p() 1 + a 1 + be analytic in D with p() 1. Then p() + p (), p() + has the same normaliation and play an important role in the theory of differential p () p() subordination. Let f A and set p(). Then p() + p () f (), and if p() f (), then p()+ p () p() 1+ f () f (). Extremal properties of such expressions, as well as several relations, were frequently considered in the theory of univalent functions. In this section, we study the differential subordination involving harmonic means of such expressions, and present some applications in the geometric functions theory. Theorem 2.1 Let p() 1 + a 1 + be analytic in D with p() 1. Then Proof Let { 2p() p() + p () ] } 2p() + p > 0 p() >0. (2.1) () q() q 1 +, (2.2) with q(d) {w :w>0}, q(0) 1, E(q) {1} and q Q. Then, we can rewrite the condition (2.1) as { 2p() p() + p () ] } 2p() + p > 0 p() q(). () Suppose that p() q().then, from Lemma 1.1, there exist a point 0 D and a point ζ 0 D \ {1} such that p q(ζ 0 ) and p() >0 for all D 0.This implies that p 0, therefore we can choose p of the form p : ix,

4 1246 S. Kanas, A.-E. Tudor where x is a real number. Due to symmetry, it is sufficient to consider only the case where x > 0. We have ζ 0 q 1 (p) p( 0) 1 p + 1, then 0 p mζ 0 q (ζ 0 ) m(x 2 + 1)/2 : y, where y < 0. Thus, we obtain: { 2p(0 ) p + 0 p ] } ] 2ix(ix + y) 2p + 0 p 2ix + y 2x2 y 4x 2 + y 2 < 0. This contradicts the hypothesis of the theorem, therefore p q and the proof of Theorem 2.1 is complete. emark 1 We only note that the expression of the left hand side of (2.1) isofthe harmonic form of two elements x 1 p() and x 2 p() + p ()( D). Setting p() in the previous theorem, we obtain the following corollary: Corollary 2.1 Let + a be analytic in D. Then 2 f () () + f > 0 f > 0. () Theorem 2.2 Let p() 1 + a 1 + be analytic in D with p() 1. Then 2p() + 2p ] () 1 + p 2 () + p()p > 0 p() >0. () Proof Following the same steps as in the proof of Theorem 2.1, setting p ix, x > 0 and 0 p y, y < 0, we obtain: 2p p ] 1 + p pp ] 2ix + 2y 1 x 2 + ixy 2y (1 x 2 ) 2 + x 2 y 2 < 0, which completes the proof. Setting p() we obtain Corollary 2.2 Let + a be analytic in D. Then 2 f () + f > 0 > 0. ()

5 Differential Subordinations and Harmonic Means 1247 Theorem 2.3 Let p() 1 + a 1 + be analytic in D with p() 1. Then 2 p() + p () p() 2 + p () p 2 () ] > 0 p() >0. Proof We only need to show that p q, where q is given by (2.2). As in the proof of Theorem 2.1, there exist a point 0 D such that p ix, x > 0 and 0 p y, y < 0. Therefore, we have 2 p + 0 p p p p 2 ] ( 2 ix + y ix 2 y x ) 0 < 0. and the proof of Theorem 2.3 is completed. Setting p() f () we obtain Corollary 2.3 Let + a be analytic in D. Then 2 f () ] 1 + f () f () 1 + f () + f () f () > 0 f () > 0. Theorem 2.4 Let p() 1 + a 1 + be analytic in D with p() 1. Then ] 2 p() + p () p() 1 + p 2 () + p > 0 p() >0. () Proof As in the proof of Theorem 2.1, there exist a point 0 in D such that p ix, x > 0 and 0 p y, y < 0. We have: 2 p + 0 p ] p 1 + p p ( 2 ix + y ix ) 1 x 2 + y 0, which completes the proof. Setting p() f () we obtain

6 1248 S. Kanas, A.-E. Tudor Corollary 2.4 Let + a be analytic in D. Then 2 f 1 + f () () f () 1 + f () + f () f () ] > 0 f () > 0. Theorem 2.5 Let p() 1 + a 1 + a be analytic in D with p() 1, and let 0 < M < 1 3. Then 2p() p() + p () ] 2p() + p 1 < M p() 1 < M. (2.3) () Proof Let q() 1 + M, (2.4) with q(d) {w : w 1 < M}, q(0) 1, E(q) and q Q. Then, the condition (2.3) can be rewritten as 2p() p() + p () ] 2p() + p 1 < M p() q(). () Suppose that p() q().then, from Lemma 1.1, there exist 0 D, ζ 0 D and m 1 such that p q(ζ 0 ) and p 1 < M for all D 0. This implies that p 1 q(ζ 0 ) 1 M, therefore we can choose p of the form p : 1 + Me iθ, where θ is a real number. We have then ζ 0 q 1 (p) p 1] /M, 0 p mζ 0 q (ζ 0 ) mme iθ, m 1. We can write: 2p p+ 0 p ] 2p+ 0 p 1 2p 2 + 2p 0 p 2p 0 p 2p + 0 p 2Me iθ + 2M 2 e 2iθ + mme iθ + 2mM 2 e 2iθ 2 + (2 + m)me iθ Me iθ 1 + m(1 + Meiθ ) 2 + (2 + m)me iθ M 1 + m(1 + Meiθ ) 2 + (2 + m)me iθ.

7 Differential Subordinations and Harmonic Means 1249 In order to obtain the contradiction it suffices to show that the last expression is greater or equal to M that is equivalent to the fact 1 + m(1 + Meiθ ) 2 + (2 + m)me iθ 1 or 2 + m + 2(m + 1)Me iθ (2 + m)me iθ 2. The last inequality holds by virtue of the inequality M 2 (3m+4)+4(2+m)M cos θ + m + 4 M 2 (3m + 4) 4(2 + m)m + m or ( (3m + 4)(M 1) M 4 + m ) m Since M does not exceed 1 3 the above expression is positive for every m 1. This contradicts the hypothesis of the theorem, therefore p q and the proof of Theorem 2.5 is complete. Let p(). Then the previous theorem reduce to the following corollary: Corollary 2.5 Let + a be analytic in D, and let 0 < M < 1. Then 2 f () 2 + f () 1 < M 1 < M. For the case, when p() f (), the Theorem 2.5 gives Corollary 2.6 Let + a be analytic in D, and let 0 < M < 1. Then 2( f () + f ()) f () < M f () 1 < M. f () Also, letting p() f () in Theorem 2.5, we conclude:

8 1250 S. Kanas, A.-E. Tudor Corollary 2.7 Let + a be analytic in D, and let 0 < M < 1. Then 2 f () ( ) 2 + f () f () f () 3 + f () f () f () 1 < M f () 1 < M. Theorem 2.6 Let p() 1 + a 1 + be analytic in D with p() 1 and let γ (0, 1]. Then 2p() p() + p () ] 2p() + p () <γπ 2 p() <γπ 2. (2.5) Proof Let q() ( ) 1 + γ 1 + q 1 +, (2.6) 1 with q(d) { w : w <γ π 2 }, q(0) 1, E(q) {1}, and q Q. Then, the condition (2.5) can be rewritten as 2p() p() + p () ] 2p() + p () <γπ 2 p() q(). Suppose that p() q().then, from Lemma 1.1, there exist 0 D, ζ 0 D \ {1} and m 1 such that p q(ζ 0 ) and 0 p mζ 0 q (ζ 0 ). This implies that p q(ζ 0 ) : (ix) γ x γ e γ π 2 i, (2.7) where x is a real number. Due to symmetry, it is sufficient to consider only the case where x > 0. We have and therefore we obtain ζ 0 q 1 (p) p( γ 0) 1 1, p γ ( x 0 p mζ 0 q (ζ 0 ) mγ x γ 2 ) + 1 e π 2 (γ +1)i. (2.8) 2x

9 Differential Subordinations and Harmonic Means 1251 Taking into consideration (2.7) and (2.8),we have 2p( 0) p(0 ) + 0 p ] 2p p ] 2p + 0 p p p p p p( p + 0 ) p p γ π 1 + ui ui γ π 2 + arctan u 2 + u 2 γ π 2, where ( x 2 ) + 1 u mγ > 0. 2x This contradicts the hypothesis of the theorem, therefore p q and the proof of Theorem 2.1 is complete. Setting p() we obtain the following corollary: Corollary 2.8 Let + a be analytic in D. Then 2 f () + f () <γπ 2 <γπ 2. Theorem 2.7 Let p() 1 + a 1 + be analytic in D with p() 1 and let γ (0, 1]. Then ] 2 p() + p () p() 2 + p () p 2 () <γ π 2 p() <γπ 2. Proof We only need to show that p q, where q is given by (2.6). As in the proof of Theorem 2.6, there exist a point 0 D such that the equalities (2.7) and (2.8) holds. We have

10 1252 S. Kanas, A.-E. Tudor 2 p + 0 p ] p p p p p + 2 p p p 2 2x 2γ + u 2 + 3ux γ sin γ π (xi)γ 2 + 4x 2γ cos 2 γ π 2 + (u + 2xγ sin γ π 2 ) ux γ cos γ π ] 2 +i 4x 2γ cos 2 γ π 2 + (u + 2xγ sin γ π 2 ), ( x 2 ) + 1 where u mγ > 0. 2x We notice that for γ (0, 1] the trigonometric functions sin and cos are located in the first quadrant of trigonometric circle and, therefore, have positive values. It follows that 2 p + 0 p ] p p γ π 2 + arctan u cos γ π 2 2x γ + u 2 x γ + 3u sin γ π γ π p 2 2 2, and this is contradiction with the hypothesis of the theorem. Thus, the proof is complete. Setting p() f () we obtain Corollary 2.9 Let + a be analytic in D. Then 2 f () ] 1 + f () f () 1 + f () + f () f () <γ π 2 f () <γπ 2. Acknowledgments This work was possible with the financial support of the Sectoral Operation Programme for Human esources Development , co-financed by the European Social Fund, under the project number POSDU/107/1.5/S/76841 with the title Modern Doctoral Studies: Internationaliation and Interdisciplinarity. This work was partially supported by the Centre for Innovation and Transfer of Natural Sciences and Engineering Knowledge, Faculty of Mathematics and Natural Sciences, University of esow. eferences 1. Ali.M., Nagpal S., avichandran, V.: Second-order differential subordination for analytic functions with fixed initial coefficient. Bull. Malays. Math. Sci. Soc. (2) 34(3), (2011)

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