The fifth and sixth coefficients of α-close-to-convex functions, Kragujevac Journal of Mathematics 32 (2009) 5 - 12 by Kunle Oladeji Babalola | Papers by Kunle Oladeji

5 Kragujevac J. Math. 32 (2009) 5–12. THE FIFTH AND SIXTH COEFFICIENTS OF α-CLOSE-TO-CONVEX FUNCTIONS Kunle Oladeji Babalola Department of Mathematics, University of Ilorin, P. M. B. 1515, Ilorin, Nigeria (e-mail: khayrah.babalola@gmail.com) (Received May 02, 2007) Abstract. We obtain sharp bounds on the fifth and sixth coefficients of α-close-to-convex functions introduced in [2]. 1. INTRODUCTION Let A be the class of functions of the form: f (z) = z + a2 z 2 + ... which are analytic in the unit disk U = {z ∈ C: |z| < 1}. In [2], Chichra considered functions f ∈ A for which f (z)f (z)/z = 0 for z ∈ U , and if for some nonnegative real number α, there exists a starlike function: φ(z) = z + b2 z 2 + ... such that Re (1 − α) (zf (z)) zf (z) +α φ(z) φ (z) > 0. 6 The above geometric condition implies that the term in the curly brackets belongs to the class P of analytic functions: p(z) = 1 + p1 z + ... which have positive real part in U . He called functions of this type α-close-to-convex and denoted the class by Cα . Interestingly, the family of functions includes two well known ones. These are the classes of close-to-convex and convex functions, which respectively correspond to C0 and C∞ (see [2]). Among others, he proved the following inequalities: Theorem 1 [2]. Let f (z) ∈ Cα . Then |a2 | ≤ |a3 | ≤ |a4 | ≤ The inequalities are sharp. He remarked that his method might not be easily employed to obtain bounds on higher coefficients in this family of functions. Apart from his efforts, the present author is not aware of any further development on the higher coefficients of this important family of functions. And as it is well known, the coefficient problem in univalent functions theory is ever demanding attention. In this note we visit the old problem of Chichra and obtain the best possible upper bounds for the fifth and sixth coefficients of functions of the class Cα . Our results are the following inequalities: Theorem 2. Let f (z) ∈ Cα . Then |a5 | ≤ |a6 | ≤ 25 + 238α + 755α2 + 902α3 + 120α4 ; 5(1 + α)(1 + 2α)(1 + 3α)(1 + 4α) 2+α ; 1+α 9 + 23α + 6α2 ; 3(1 + α)(1 + 2α) 4 + 22α + 34α2 + 6α3 . 4(1 + α)(1 + 2α)(1 + 3α) 6 + 83α + 418α2 + 951α3 + 955α4 + 120α5 . (1 + α)(1 + 2α)(1 + 3α)(1 + 4α)(1 + 5α) 7 The inequalities are sharp. Equalities in both cases are attained for the function          z 0 0 z f (z) = 1 −1/α t (1 − t)2(1/α−1) α 1+t dt, (1 − t)3 t 0 ξ 1/α−1 (1 + ξ)2 dξ dt, if α = 0; (1 − ξ)2(1/α+1) if α = 0. In our proof, which is presented in the next section, we shall depend on the well known inequalities (Caratheodory lemma and coefficient functionals for starlike functions): |pn | ≤ 2, n ≥ 1; |bn | ≤ n, n ≥ 2; |b3 − λb2 | ≤ 3 − 4λ, λ ≤ 2 3 4 (1) (λ real) and the following lemma which concerns bounds on certain other functionals in the coefficient space of the family of starlike functions. Lemma 1 [1]. For every starlike function φ(z) and real numbers µ, ρ, σ, τ and ω, we have the sharp inequalities: 5 |b4 − µb2 b3 | ≤ 4 − 6µ; µ ≤ , 9 |b4 − µb2 b3 − ρb3 | ≤ 4 − 6µ − 8ρ; 5 − 9µ − 12ρ ≥ 0, 2 2 |b5 − σb2 b3 | ≤ 5 − 12σ; σ ≤ , 2 9 |b5 − τ b2 b4 − ωb2 | ≤ 5 − 8τ − 9ω; 2 − 5τ − 9ω ≥ 0. 3 The proof of the above lemma, which was presented in [1], made use of the equality p2 = 1 2 p 2 1 1 + ε(2 − 2 |p1 |2 ), |ε| ≤ 1, which is a consequence of the well known Caratheodory-Toeplitz inequality |p2 − 1 p2 | ≤ 2 − 1 |p1 |2 . The extremal function is 2 1 2 the Koebe function (up to rotation): k(z) = z/(1 − z)2 . Now the proof of the main result. 8 2. PROOF OF THE MAIN RESULT Let f ∈ Cα . Then there exists an analytic function p ∈ P and a starlike function φ(z) such that (1 − α)zf (z)φ (z) + αφ(z)(zf (z)) = p(z)φ(z)φ (z). The right hand side of the above equation gives p(z)φ(z)φ (z) = z + c2 z 2 + ... where cn = k=0 n−1 qk+1 bn−k ; n ≥ 2, and qk = k−1 (k − j)pj bk−j ; (p0 = b1 = q1 = 1). j=0 (2) Similarly the left hand side yields (1 − α)zf (z)φ (z) + αφ(z)(zf (z)) = z + d2 z 2 + ... where dn = k=0 n−1 (n − k)[(n − 2k − 1)α + k + 1]bk+1 an−k ; n ≥ 2. Comparing the coefficients cn and dn we obtain the recurrence relation: n−1 n−1 n[1 + (n − 1)α]an = k=0 bn−k qk+1 − k=1 (n − k)[(n − 2k − 1)α + k + 1]bk+1 an−k , with q1 = 1, a1 = b1 = 1 and n ≥ 2. Thus we have 2(1 + α)a2 = (α − 1)b2 + q2 , 3(1 + 2α)a3 = 2(α − 1)b3 + b2 q2 + q3 − 4a2 b2 , 4(1 + 3α)a4 = 3(α − 1)b4 + b3 q2 + b2 q3 + q4 + 2(3 − α)a2 b3 − 3(2 + α)a3 b2 (3) (4) (5) 5(1+4α)a5 = 4(α−1)b5 +b4 q2 +b3 q3 +b2 q4 +q5 +4(α−2)a2 b4 −9a3 b3 −8(1+α)a4 b2 , (6) 9 6(1 + 5α)a6 = 5(α − 1)b6 + b5 q2 + b4 q3 + b3 q4 + b2 q5 + q6 + 2(3α − 5)a2 b5 + 3(α − 4)a3 b4 − 4(3 + α)a4 b3 − 5(2 + 3α)a5 b2 , (7) and so on, and where from (2), q2 = 2b2 + p1 , q3 = 3b3 + 2b2 p1 + p2 , q4 = 4b4 + 3b3 p1 + 2b2 p2 + p3 , q5 = 5b5 + 4b4 p1 + 3b3 p2 + 2b2 p3 + p4 and q6 = 6b6 + 5b5 p1 + 4b4 p2 + 3b3 p3 + 2b2 p4 + p5 . Using these in (3) - (7) we get 2(1 + α)a2 = (1 + α)b2 + p1 , 3(1 + α)(1 + 2α)a3 = (1 + α)(1 + 2α)b3 + (1 + 3α)b2 p1 + (1 + α)p2 , 4(1 + α)(1 + 2α)(1 + 3α)a4 = (1 + α)(1 + 2α)(1 + 3α)b4 + (1 + 2α)(1 + 5α)b3 p1 + (1 + α)(1 + 5α)b2 p2 + (1 + α)(1 + 2α)p3 + α(α − 1)b2 p1 , 2 (10) (8) (9) 5A0 a5 = A0 b5 + A1 b4 p1 + A2 b3 p2 + A3 b2 p3 + A4 p4 + A5 b2 b3 p1 + A6 b2 p2 + A7 b3 p1 , (11) 2 2 where A0 = (1 + α)(1 + 2α)(1 + 3α)(1 + 4α), A1 = (1 + 2α)(1 + 3α)(1 + 7α), A2 = (1+α)(1+3α)(1+8α), A3 = (1+α)(1+2α)(1+7α), A4 = (1+α)(1+2α)(1+3α), A5 = 2α(α − 1)(2 + 5α), A6 = 2α(α2 − 1) and A7 = −A6 , and 6B0 a6 = B0 b6 + B1 b5 p1 + B2 b4 p2 + B3 b3 p3 + B4 b2 p4 + B5 p5 + B6 b4 b2 p1 + B7 b2 p1 + B8 b2 b3 p1 + B9 b4 p1 + B10 b3 b2 p2 + B11 b3 p2 + B12 b2 p3 , 3 2 2 2 2 (12) with B0 = (1+α)(1+2α)(1+3α)(1+4α)(1+5α), B1 = (1+2α)(1+3α)(1+4α)(1+9α), B2 = (1 + α)(1 + 3α)(1 + 4α)(1 + 11α), B3 = (1 + α)(1 + 2α)(1 + 4α)(1 + 11α), B4 = (1 + α)(1 + 2α)(1 + 3α)(1 + 9α), B5 = (1 + α)(1 + 2α)(1 + 3α)(1 + 4α), B6 = 6α(α − 1)(1 + 3α)2 , B7 = 4α(α − 1)(1 + 2α)(1 + 4α), B8 = α(1 − α)(11 + 45α + 34α2 ), B9 = 2α(α2 − 1)(2 + 3α), B10 = −10α − 53α2 − 73α3 − 44α4 , B11 = −B9 and B12 = 3α(α2 − 1)(1 + 2α). From (8) - (10) we can obtain the inequalities of Chichra (Theorem 1) using the inequalities (1). 10 Now for n = 5, if α ∈ [1, ∞), we define λ1 = −A6 /A2 and λ2 = −A7 /A5 and then write (11) as 5A0 a5 = A0 b5 + A1 b4 p1 + A2 p2 {b3 − λ1 b2 } + A3 b2 p3 + A4 p4 + A5 b2 p1 {b3 − λ2 b2 }, 2 2 and for α ∈ [0, 1], we define µ = −A5 /A1 and rewrite (11) as 5A0 a5 = A0 b5 + A1 p1 {b4 − µb2 b3 } + A2 p2 {b3 − λ1 b2 } + A3 b2 p3 + A4 p4 + A7 b3 p1 . 2 2 It is easily verified that for all α ≥ 0, the real numbers λ1 , λ2 ≤ equations above. So also µ ≤ 5 . 9 3 4 in the two Thus using the inequalities (1) and the first of Lemma 1, the two equations above both yield the first inequality of our theorem, that is, the upper bound for a5 . Next we proceed to compute the bound for a6 . We define λ1 = −B12 /B3 , λ2 = −B9 /B8 , µ = −B10 /B2 , ρ = −B11 /B2 , σ = −B8 /B1 , τ = −B6 /B1 , and ω = −B7 /B1 , and if α ∈ [1, ∞) we write (12) as 6B0 a6 = B0 b6 + B1 p1 {b5 − σb2 b3 } + B2 p2 {b4 − µb2 b3 − ρb3 } 2 2 + B3 b3 p3 + B4 b2 p4 + B5 p5 + B6 b4 b2 p1 + B7 b2 p1 + B9 b4 p1 + B12 b2 p3 , 3 2 2 and for α ∈ [0, 1], we rewrite (12) as 6B0 a6 = B0 b6 + B1 p1 {b5 − τ b2 b4 − ωb2 } + B2 p2 {b4 − µb2 b3 } 3 + B3 p3 {b3 − λ1 b2 } + B4 b2 p4 + B5 p5 + B8 b2 p1 {b3 − λ2 b2 } + B11 b3 p2 . 2 2 2 2 It is again easy to verify that for all α ≥ 0, the real numbers λ1 , λ2 , µ, ρ, σ, τ , and ω defined in the preceding two equations all satisfy (as appropriate) the conditions of Lemma 1 and the inequalities (1). Thus the bound for a6 follows by appropriately applying the inequalities (1) and Lemma 1. The extremal function is obtained by choosing φ(z) = z/(1 − z)2 and p(z) = (1 + z)/(1 − z) in the integral representation formulae: f (z) =    0     z 0 z 1 −1 t [φ(t)](1/α−1) α t−1 φ(t)p(t)dt t 0 [φ(ξ)](1/α−1) φ (ξ)p(ξ)dξ dt if α = 0, if α = 0. (see [2]). This completes the proof. 11 3. REMARK Here we guide readers to the computation of the coefficients of the given extremal function. After simple calculation, we can write the extremal function in series form as: ∞ f (z) = z + n=2 ηn−1 c + 1 ηn−2 G1 c + 1 η1 Gn−2 c + 1 Gn−1 n z + + ... + + n c+2 n c+n n c+n n 1 − 1, α n where c= ηn = (−1)n n! (2c − j + 1), j=1 G1 = 2 + m1 , Gn = mn−2 + 2mn−1 + mn , n ≥ 2; with mn = 1 n (2c + j + 3), m0 = 1. n! j=1 By careful computation, we find that a5 = c(c − 1)(2c − 1)(2c − 3) c(2c − 1)(2c − 2)(c + 1)(2c + 6) − 30 15(c + 2) 2 c(2c − 1)(c + 1)(2c + 13c + 19) 2c(c + 1)(4c3 + 42c2 + 134c + 132) + − 5(c + 3) 15(c + 4) 4 3 2 (c + 1)(4c + 60c + 311c + 669c + 510) . + 30(c + 5) A − 2B + 6C − 4D + E 30(c + 2)(c + 3)(c + 4)(c + 5) This gives a5 = where A = 4c8 + 44c7 + 127c6 − 85c5 − 629c4 + 41c3 + 858c2 − 360c, B = 8c8 + 116c7 + 596c6 + 1160c5 + 32c4 − 1996c3 − 636c2 − 720c, C = 4c8 + 72c7 + 509c6 + 1769c5 + 3029c4 + 1979c3 − 482c2 − 760c, D = 4c8 + 86c7 + 760c6 + 3572c5 + 9628c4 + 14846c3 + 12072c2 + 3960c, 12 and E = 4c8 + 100c7 + 1051c6 + 6079c5 + 21181c4 + 45505c3 + 58764c2 + 41556c + 12240 so that a5 = 25c4 + 338c3 + 1619c2 + 3226c + 2040 5(c + 2)(c + 3)(c + 4)(c + 5) and finally, setting c = 1/α − 1, we have a5 = 25 + 238α + 755α2 + 902α3 + 120α4 . 5(1 + α)(1 + 2α)(1 + 3α)(1 + 4α) As above, a6 can be computed mutatis mutandis. Bounds on the higher coefficients of functions in the class Cα can also be found using the recurrence relation we have obtained, after the determination of bounds on relevant emerging functionals in the family of starlike functions, which are associated with the desired higher coefficients. References [1] K. O. Babalola, On some functionals associated with certain coefficient problems, J. Nig. Asso. Math. Physics, 11 (November 2007), 67–70. [2] P. N. Chichra, New subclasses of the class of close-to-convex functions, Proc. Amer. Math. Soc. 62 (1) (1977), 37–43. [3] P. L. Duren, Univalent functions, Springer Verlag. New York Inc. 1983.