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A coefficient inequality for the class of analytic functions in the unit disc

The aim of this paper is to give a coefficient inequality for the class of analytic functions in the unit disc D = {z | |z| < 1}. 2000 Mathematics Subject Classification: 30C45. 1. Introduction. Let Ω be the family of functions ω(z) regular in the disc D and satisfying the conditions ω(0) = 0 and |ω(z)| < 1 for z ∈ D. Next, for arbitrary fixed numbers A and B, −1 < A ≤ 1, −1 ≤ B < A, denote by P (A, B) the family of functions p(z) = 1 + p1 z + p2 z2 + · · · regular in D such that p(z) is in P (A, B) if and only if p(z) = 1 + Aω(z) 1 + Bω(z) (1.2) (1.1) for some function ω(z) ∈ Ω and every z ∈ D. The class P (A, B) was introduced by Janowski [3]. Moreover, let S ∗ (A, B, b) (b ≠0, complex) denote the family of functions f (z) = z + a2 z2 + · · · + an zn + · · · regular in D and such that f (z) is in S ∗ (A, B, b) if and only if 1+ 1 f (z) − 1 = p(z) z b f (z) (1.3) (1.4) for some p(z) in P (A, B) and all z in D. For the aim of this paper we need Jack’s lemma [2]. “Let ω(z) be a regular in the unit disc with ω(0) = 0, then if |ω(z)| attains its maximum value on the circle |z| = r at a point z1 , we can write z1 ω (z1 ) = kω(z1 ), where k is real and k ≥ 1.” 2. Coefficient inequality. The main purpose of this paper is to give sharp upper bound of the modulus of the coefficient an . Therefore, we need the following lemma. Lemma 2.1. The necessary and sufficient condition for g(z) = z +a2 z2 +· · · belongs to S ∗ (A, B, b) is  z · 1 + Bω(z) ⇒ g(z) ∈ S ∗ (A, B, b) ⇐ g(z) = z · ebAω(z) , where ω(z) ∈ Ω. Proof. The proof of this lemma is in four steps. Step 1. Let B ≠0 and g(z) = z · 1 + Bω(z) b(A−B)/B b(A−B)/B B ≠0, B = 0, (2.1) (2.2) If we take the logarithmic derivative from equality (2.2), we obtain g (z) z · ω (z) 1 z· − 1 = (A − B) . b g(z) 1 + Bω(z) If we use Jack’s lemma [2] in equality (2.3), we get (A − B)ω(z) 1 g (z) −1 = . z b g(z) 1 + Bω(z) After the simple calculations from (2.4), we see that 1+ g (z) 1 + Aω(z) 1 z· −1 = . b g(z) 1 + Bω(z) (2.3) (2.4) (2.5) Equality (2.5) shows that g(z) ∈ S ∗ (A, B, b). Step 2. Let B = 0 and g(z) = z · ebAω(z) . Similarly, we obtain 1+ g (z) 1 + Aω(z) 1 z −1 = = 1 + Aω(z). b g(z) 1 + Bω(z) (2.6) (2.7) This shows that g(z) ∈ S ∗ (A, B, b). Step 3. Let g(z) ∈ S ∗ (A, B, b) and B ≠0, then we have 1+ 1 g (z) 1 + Aω(z) z −1 = . b g(z) 1 + Bω(z) (2.8) Equality (2.8) can be written in the from b(A − B) ω(z)/z 1 g (z) = + . g(z) 1 + Bω(z) z If we use Jack’s lemma (2.9), we obtain b(A − B)ω (z) 1 g (z) = + . g(z) 1 + Bω(z) z Integrating both sides of equality (2.10), we get g(z) = z · 1 + Bω(z) b(A−B)/B (2.9) (2.10) (2.11) Step 4. Let g(z) ∈ S ∗ (A, B, b) and B = 0. Similarly, we obtain g(z) = z · ebAω(z) which ends the proof. We note that we choose the branch of (1 + Bw(z))b(A−B)/B such that 1 + Bw(0) b(A−B)/B (2.12) =1 at z = 0. (2.13) Theorem 2.2. If f (z) = z + a2 z2 + · · · + an zn + · · · belongs to S ∗ (A, B, b), then b(A − B) + kB k+1 |bA| k+1 if B = 0. if B = 0, (2.14) These bounds are sharp because the extremal function is    z , (1 − Bδz)−b(A−B)/B  bAz ze , |δ| = 1, if B = 0, if B = 0. f∗ (z) = (2.15) Proof. Let B ≠0. If we use the definition of the class S ∗ (A, B, b), then we write 1+ f (z) 1 z − 1 = p(z). b f (z) (2.16) Equality (2.16) can be written by using the Taylor expansion of f (z) and p(z) in the form z + 2a2 z2 + 3a3 z3 + · · · + nan zn + · · · = z + a2 z2 + · · · + an zn + · · · 1 + bp1 z + bp2 z2 + · · · + bpn zn + · · · . (2.17) Evaluating the coefficient of zn in both sides of (2.17), we get nan = an + bp1 a + bp2 a + · · · + bp . on the other hand, pn ≤ (A − B). (2.19) (2.18) Inequality (2.19) was proved by Aouf [1]. If we consider the relations (2.18) and (2.19) together, then we obtain (n − 1) |b||A − B| 1 + a2 + a3 + · · · + a which can be written in the form 1 (n − 1) (2.20) |b||A − B| ak , a1 = 1. (2.21) To prove (2.14), we will use the induction principle. Now, we consider inequalities (2.21) and b(A − B) + kB . k+1 (2.22) The right-hand sides of these inequalities are the same because (i) for n = 2, |b||A − B| (n − 1) ak , a1 = 1 ⇒ a2 ≤ |b||A − B|, (2.23) ⇒ a2 ≤ |b||A − B|; b(A − B) + kB = b(A − B) k+1 (ii) for n = 3, |b||A − B| (n − 1) a3 ≤ ⇒ a3 a3 ≤ ak = 1 |b||A − B| 1 + a2 2 1 1 2 ≤ |b| |A − B|2 + |b||A − B|, 2 2 b(A − B) + B b(A − B) + kB = |b||A − B| k+1 2 (2.24) ⇒ a3 ≤ ⇒ a3 1 1 |b||A − B| |b||A − B| + |B| ≤ |b||A − B| |b||A − B| + 1 2 2 1 1 2 ≤ |b| |A − B|2 + |b||A + B|. 2 2 Suppose that this result is true for n = p, then we have |b||A − B| (n − 1) ak , a1 = 1 ⇒ ap |b||A − B| 1 + a2 + a3 + · · · + ap−1 ≤ (p − 1) (2.25) , b(A − B) + kB k+1 p−2 ⇒ ap ≤ b(A − B) + kB k+1 (2.26) ⇒ ap ≤ 1 |b||A − B| |b||A − B| + 1 |b||A − B| + 2 (p − 1)! · |b||A − B| + 3 · · · |b||A − B| + (p − 2) from (2.25), (2.26), and induction hypothesis, we have |b||A − B| 1 + a2 + a3 + · · · + ap−1 (p − 1) 1 |b||A − B| |b||A − B| + 1 = (p − 1)! · |b||A − B| + 2 · · · |b||A − B| + (p − 2) . If we write x = |b||A − B| > 0, equality (2.27) can be written in the form. x 1 + a2 + a3 + · · · + ap−1 (p − 1) 1 = x(x + 1)(x + 2) · · · x + (p − 2) . (p − 1)! (2.27) (2.28) After the simple calculation from equality (2.28), we get 1 1 x + (p − 1) 1 + a2 + a3 + · · · + ap−1 p (p − 1) 1 (x + 1)(x + 2)(x + 3) · · · x + (p − 2) x + (p − 1) = p! 1 x ⇒ 1 + a2 + a3 + · · · + ap−1 p p −1 1 + 1 + a2 + a3 + · · · + ap−1 p 1 = (x + 1)(x + 2)(x + 3) · · · x + (p − 2) x + (p − 1) p! 1 1 ⇒ ap + 1 + a2 + a3 + · · · + ap−1 p p 1 (x + 1)(x + 2)(x + 3) · · · x + (p − 2) x + (p − 1) = p! x 1 + a2 + a3 + · · · + ap−1 + ap ⇒ p 1 x(x + 1)(x + 2)(x + 3) · · · x + (p − 2) x + (p − 1) . = p! Equality (2.29) shows that the result is valid for n = p + 1. Therefore, we have (2.14). Corollary 2.3. The first inequality of (2.14) can be rewritten in the form (2.29) b(A − B) + kb k+1 = B(A − B) 1 b(A − B) + B 2 1 1 · b(A − B) + 2B · · · b(A − B) + (n − 2)B 3 (n − 1) 1 b(A − B) · b(A − B) + B = (n − 1)! · b(A − B) + 2B · · · b(A − B) + (n − 2)B ≤ 1 b(A − B) b(A − B) + |B| (n − 1)! · b(A − B) + 2|B| · · · b(A − B) + (n − 2)|B| . (2.30) If A = 1, B = −1, and b = 1, then n! 1 2 · (2 + 1) · (2 + 2) · · · n = = n. (n − 1)! (n − 1)! (2.31) This is the coefficient inequality for the starlike function which is well known. Corollary 2.4. If A = 1, B = −1, an < 1 (n − 1)! |2b + k|. (2.32) This inequality was obtained by Aouf [1]. Therefore, by giving the special value to A, B, and b, we obtain the coefficient inequality for the classes S ∗ (1, −1, β), S ∗ (1, −1, e−iλ Cos λ), S ∗ (1, −1, (1− β)e−iλ Cos λ), S ∗ (1, 0, b), S ∗ (β, 0, b), S ∗ (β, −β, b), S ∗ (1, (−1 + 1/M), b), and S ∗ (1 − 2β, −1, b), where 0 ≤ β < 1, |λ| < π /2, and M > 1. References [1] [2] [3] M. K. Aouf, On a class of p-valent starlike functions of order α, Int. J. Math. Math. Sci. 10 (1987), no. 4, 733–744. I. S. Jack, Functions starlike and convex of order α, J. London Math. Soc. (2) 3 (1971), 469–474. W. Janowski, Some extremal problems for certain families of analytic functions. I, Ann. Polon. Math. 28 (1973), 297–326. Ya¸ar Polato˘lu: Department of Mathematics and Computer Science, Faculty of Scis g ence and Letters, Istanbul Kültür University, Sirinevler 34510, Istanbul, Turkey E-mail address: y.polatoglu@iku.edu.tr Metin Bolcal: Department of Mathematics and Computer Science, Faculty of Science and Letters, Istanbul Kültür University, Sirinevler 34510, Istanbul, Turkey E-mail address: m.bolcal@iku.edu.tr http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png

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