Differential Subordinations and Harmonic Means

Bulletin of the Malaysian Mathematical Sciences Society, Mar 2015

The aim of a study of the presented paper is the differential subordination involving harmonic means of the expressions \(p(z)\), \(p(z) + zp'(z)\), and \(p(z) + \frac{zp'(z)}{p(z)}\) when \(p\) is an analytic function in the unit disk, such that \(p(0)=1, p(z)\not \equiv 1\). Several applications in the geometric functions theory are given.

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Differential Subordinations and Harmonic Means

Bull. Malays. Math. Sci. Soc. Differential Subordinations and Harmonic Means Stanisława Kanas 0 1 Andreea-Elena Tudor 0 1 Mathematics Subject Classification 0 1 0 A.-E. Tudor “Babes ̧-Bolyai” University , 1 Koga ̆lniceanu Street, 400084 Cluj-Napoca , Romania 1 Primary 31A05 The aim of a study of the presented paper is the differential subordination involving harmonic means of the expressions p(z), p(z) + zp (z), and p(z) + zpp((zz)) when p is an analytic function in the unit disk, such that p(0) = 1, p(z) ≡ 1. Several applications in the geometric functions theory are given. Univalent functions; Subordination; Differential subordination; Harmonic means 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. Ravichandran. the positive real numbers x1, x2, . . . , xn is defined to be the reciprocal of the arithmetic mean of the reciprocals of x1, x2, . . . , xn H = 1 n 1 n k=1 xk −1 . . For the special case of just two numbers x1 and x2, the harmonic mean can be written In this special case, the harmonic mean is related to the arithmetic mean A = , and the geometric mean G = √x1x2, by H = GA2 , or G = √ A H 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 = {z ∈ C : |z| < 1}, of the form: (1.1) (1.2) f (z) = z + ∞ k=2 ak zk . If f and g are two functions analytic in D, we say that f is subordinate to g, written as f ≺ g or f (z) ≺ g(z), if there exists a Schwarz function ω (i.e., analytic in D, with ω(0) = 0 and |ω(z)| < 1, for all D) such that f (z) = g(ω(z)), z ∈ D. Furthermore, if g is univalent in D, then we have the following equivalence: f (z) ≺ g(z) ⇔ f (0) = g(0) and f (D) ⊂ g(D). Definition 1.1 [10, 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(z) = ∞ z→ζ such that q (ζ ) = 0 for ζ ∈ ∂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 1.1 [10, p. 24] Let q ∈ Q, with q(0) = a, and let p(z) = a + an zn + · · · be analytic in D with p(z) ≡ a and n ≥ 1. If p is not subordinate to q, then there exist points z0 = r0eiθ0 ∈ D and ζ0 ∈ ∂D \ E (q) and an m ≥ n ≥ 1 for which p(Dr0 ) ⊂ q(D), (1) p(z0) = q(ζ0), (2) z0 p (z0) = mζ0q (ζ0), z0 p (z0) (3) p (z0) + 1 ≥ m 2 Harmonic Mean ζ0q (ζ0) q (ζ0) + 1 . Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1. Then p(z) + zp (z), p(z) + zpp((zz)) has the same normalization and play an important role in the theory of differential subordination. Let f ∈ A and set p(z) = f (zz) . Then p(z) + zp (z) = f (z), and if p(z) = z ff ((zz)) , then p(z)+ zpp((zz)) = 1+ zff ((zz)) . 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(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1. Then Proof Let 2 p(z) p(z) + zp (z) 2 p(z) + zp (z) > 0 ⇒ Suppose that p(z) ≺ q(z).Then, from Lemma 1.1, there exist a point z0 ∈ D and a point ζ0 ∈ ∂D \ {1} such that p(z0) = q(ζ0) and p(z) > 0 for all z ∈ D|z0|. This implies that p(z0) = 0, therefore we can choose p(z0) of the form p(z0) := i x , (2.1) (2.2) where x is a real number. Due to symmetry, it is sufficient to consider only the case where x > 0. We have This contradicts the hypothesis of the theorem, therefore p ≺ q and the proof of Theorem 2.1 is complete. Remark 1 We only note that the expression of the left hand side of (2.1) is of the harmonic form of two elements x1 = p(z) and x2 = p(z) + zp (z) (z ∈ D). Corollary 2.1 Let f (z) = z + a2z2 + · · · be analytic in D. Then Theorem 2.2 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1. Then > 0. 2 p(z) + 2zp (z) 1 + p2(z) + zp(z) p (z) p(z) > 0. Proof Following the same steps as in the proof of Theorem 2.1, setting p(z0) = i x , x > 0 and z0 p (z0) = y, y < 0, we obtain: 2 p(z0) + 2z0 p (z0) 1 + p2(z0) + z0 p(z0) p (z0) = 2i x + 2y 1 − x 2 + i x y 2y = (1 − x 2)2 + x 2 y2 < 0, which completes the proof. Corollary 2.2 Let f (z) = z + a2z2 + · · · be analytic in D. Then ⎢ ⎣ ⎡ 2 z f (z) ⎤ f (z) z ⎥ > 0 ⇒ f (z) + f (z) ⎦ Theorem 2.3 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1. Then ⎪⎪⎩ ⎧⎪ 2 p(z) + ⎪⎨ zp (z) ⎫ p(z) ⎪⎬⎪ zp (z) 2 + p2(z) ⎭⎪⎪ > 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 z0 ∈ D such that p(z0) = i x , x > 0 and z0 p (z0) = y, y < 0. Therefore, we have ⎧⎪ 2 p(z0) + ⎪⎨ ⎪⎩⎪ and the proof of Theorem 2.3 is completed. Corollary 2.3 Let f (z) = z + a2z2 + · · · be analytic in D. Then Theorem 2.4 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1. Then ⎧⎪ 2 ⎨⎪ ⎪⎩⎪ 1 + z f (z) f (z) 1 + z f (z) f (z) + ⎪⎧⎪⎨ 2 p(z) + zp (z) ⎫ p(z) ⎪⎬⎪ 1 + p2(z) + zp (z) ⎭⎪⎪ z f (z) f (z) > 0. > 0 ⇒ Proof As in the proof of Theorem 2.1, there exist a point z0 in D such that p(z0) = i x , x > 0 and z0 p (z0) = y, y < 0. We have: ⎪⎧⎪⎨ 2 p(z0) + z0 p (z0) ⎫ p(z0) ⎪⎬⎪ ⎪⎪⎩ 1 + p2(z0) + z0 p (z0) ⎪⎪⎭ = y 2 i x + i x 1 − x 2 + y = 0, which completes the proof. Corollary 2.4 Let f (z) = z + a2z2 + · · · be analytic in D. Then > 0. Theorem 2.5 Let p(z) = 1 + a1z + a2z2 + · · · be analytic in D with p(z) ≡ 1, and let 0 < M < 13 . Then with = q(D) = {w : |w − 1| < M } , q(0) = 1, E (q) = ∅ and q ∈ Q. Then, the condition (2.3) can be rewritten as Suppose that p(z) ≺ q(z).Then, from Lemma 1.1, there exist z0 ∈ D, ζ0 ∈ ∂D and m ≥ 1 such that p(z0) = q(ζ0) and | p(z0) − 1| < M for all z ∈ D|z0|. This implies that | p(z0) − 1| = |q(ζ0) − 1| = M , therefore we can choose p(z0) of the form p(z0) := 1 + M eiθ , where θ is a real number. We have ζ0 = q−1( p(z0)) = [ p(z0) − 1] /M, z0 p (z0) = mζ0q (ζ0) = m M eiθ , m ≥ 1. −1 = = 2 p2(z0) + 2 p(z0)z0 p (z0) − 2 p(z0) − z0 p (z0) 2 p(z0) + z0 p (z0) 2M eiθ + 2M 2e2iθ + m M eiθ + 2m M 2e2iθ 2 + (2 + m)M eiθ m(1 + M eiθ ) = |M eiθ | · 1 + 2 + (2 + m)M eiθ m(1 + M eiθ ) = M 1 + 2 + (2 + m)M eiθ . then We can write: 2 p(z0) p(z0)+z0 p (z0) 2 p(z0)+z0 p (z0) 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 or or 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 + 4 ≥ 0 (3m + 4)(M − 1) 4 + m M − 4 + 3m Since M does not exceed 13 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(z) = . Then the previous theorem reduce to the following corollary: Corollary 2.5 Let f (z) = z + a2z2 + · · · be analytic in D, and let 0 < M < 1. Then 2 + f (z) 2 f (z) z f (z) − 1 < M ⇒ For the case, when p(z) = f (z), the Theorem 2.5 gives Corollary 2.6 Let f (z) = z + a2z2 + · · · be analytic in D, and let 0 < M < 1. Then 2( f (z) + z f (z)) z f (z) 2 + f (z) Also, letting p(z) = z ff ((zz)) in Theorem 2.5, we conclude: Corollary 2.7 Let f (z) = z + a2z2 + · · · be analytic in D, and let 0 < M < 1. Then 2 Theorem 2.6 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1 and let γ ∈ (0, 1]. Then arg and therefore we obtain (2.5) (2.6) (2.7) (2.8) with = q(D) = w : |arg w| < γ π2 , q(0) = 1, E (q) = {1}, and q ∈ Q. Then, the condition (2.5) can be rewritten as arg Suppose that p(z) ≺ q(z).Then, from Lemma 1.1, there exist z0 ∈ D, ζ0 ∈ ∂D \ {1} and m ≥ 1 such that p(z0) = q(ζ0) and z0 p (z0) = mζ0q (ζ0). This implies that where x is a real number. Due to symmetry, it is sufficient to consider only the case where x > 0. We have p(z0) = q(ζ0) := (i x )γ = x γ eγ π2 i , ζ0 = q−1( p(z0)) = 1 p(z0) γ − 1 1 p(z0) γ + 1 , z0 p (z0) = mζ0q (ζ0) = mγ x γ Taking into consideration (2.7) and (2.8),we have arg , 1 + 2 + z0 p (z0) p(z0) z0 p (z0) p(z0) where u = mγ This contradicts the hypothesis of the theorem, therefore p ≺ q and the proof of Theorem 2.1 is complete. arg Theorem 2.7 Let p(z) = 1 + a1z + · · · be analytic in D with p(z) ≡ 1 and let γ ∈ (0, 1]. Then arg 2 p(z) + zp (z) p(z) zp (z) 2 + p2(z) π π < γ 2 ⇒ |arg p(z)| < γ 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 z0 ∈ D such that the equalities (2.7) and (2.8) holds. We have arg 2 p(z0) + , 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. 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Stanisława Kanas, Andreea-Elena Tudor. Differential Subordinations and Harmonic Means, Bulletin of the Malaysian Mathematical Sciences Society, 2015, 1243-1253, DOI: 10.1007/s40840-014-0078-9