Results & Lemmas (7)
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PROPOSITION 1. · coeff
PROPOSITION 1. Let f e S and g e X)- M g z) = l/f(l/z) or i£ g(z) + c - l/fc(l/z) ^ 0 for some complex number c, then (a) cnk(f) = -ynk(g)…
PROPOSITION 1. Let f e S and g e X)- M g{z) = l/f(l/z) or i£ g(z) + c - l/fc(l/z) ^ 0 for some complex number c, then (a) cnk(f) = -ynk(g) = 7^(51 + c) = cnk{fc) (n, k > 1) and (b) an{f) = <rn(g) = vn(g + c) - sn(fc) (n ^ 0) PROOF: Since the coefficients ynk(g) = Cnk(f) for all n,fc ^ 1, whenever g(z) — l//(l/z). Part (b) follows from (10) and (16). D
THEOREM 1. · coeff
THEOREM 1. Let an and crn denote the Schwarzian coefficients of f 6 5 and g 6 53, respectively. Then we have the following coefficient…
THEOREM 1. Let an and crn denote the Schwarzian coefficients of f 6 5 and g 6 53, respectively. Then we have the following coefficient bounds: (a) max \<ro(g)\ = max|so(/)| = 6 6 ^ /es https://doi.org/10.1017/S0004972700012053 Published online by Cambridge University Press
THEOREM 2
THEOREM 2. For every n ^ 1, there exists Jn 6 5 and gn G £ suci thai (19) k2n(?»)| - | s2n(/n)| = 2(2n + l)(2n.. 26 8 wiere /3n = — + 3 n(n…
THEOREM 2 . For every n ^ 1, there exists Jn 6 5 and gn G £ suci thai (19) k2n(?»)| - | s2n(/n)| = 2(2n + l)(2n . . 26 8 wiere /3n = — + 3 n(n + 2) ' Furth erzn ore, (20) max|<r2(^)| = max|«2(/)| = 30(l+2c" f t) =30.00071804... . PROOF: Let f(z) = z + a2z2 + a3z3 + ... e S, and let m ^ 2. Then, the mth
PROPOSITION 2. · coeff
PROPOSITION 2. Let ™n = max |an(/)|/n2 Tien, (21) 2.000688... = 2 + Ae~2ils ^ En" mn < 3TT/4 = 2.356194... fl—>oo PROOF: We consider first…
PROPOSITION 2. Let ™n = max{|an(/)|/n2} Tien, (21) 2.000688... = 2 + Ae~2ils ^ En" mn < 3TT/4 = 2.356194... fl—>oo PROOF: We consider first the inequality on the left. For the function /„, defined within the proof of Theorem 2, we conclude from (19) that m 2 n which justifies the conclusion, since /?„ —> —26/3 as n —> oo. To obtain an upper estimate on |an|, we use a well-known [10, p.119] estimate for the Grunsky coefficients. Since |cjtm| ^ 1/y/km for all k,m ^ 1, we have n+l (22)
THEOREM 3 · coeff
THEOREM 3. Let f e S and let an(n = 0,1,2,...) denote the Schwarzian coef- ficients of f. Then, (24) N n = 0 i 2 n n = 0 for every integer…
THEOREM 3 . Let f e S and let an(n = 0,1,2,...) denote the Schwarzian coef- ficients of f. Then, (24) N n = 0 i 2 n n = 0 for every integer N ^ 0 and every complex z. In particular, for z — 1, we get N (25) 71=0 {N 2){N + 3). PROOF: Using (10), we rearrange coefficients to obtain N
COROLLARY 3 · coeff
COROLLARY 3. 1. Let f e S and let «„ (n = 0,1,2,...) denote the Schwarzian coefficients of f. Then, (26) N N N n=0 n=0 V. n=0 n=0 for every…
COROLLARY 3 . 1 . Let f e S and let «„ (n = 0,1,2,...) denote the Schwarzian coefficients of f. Then, (26) N N N n=0 n=0 V. n=0 n=0 for every integer N ^ 0 and every complex z. In particular, if z = 1, tien N •Sr n = 0
COROLLARY 3 · coeff
COROLLARY 3. 2. Let f e S and let an (n = 0,1,2,...) denote tie Sciwarzian coefficients of f. For every N ^ 0 and k ^ 1, (28) N n = 0 n + k…
COROLLARY 3 . 2 . Let f e S and let an (n = 0,1,2,...) denote tie Sciwarzian coefficients of f. For every N ^ 0 and k ^ 1, (28) N n = 0 n + k N n = 0 2n 8B + iMI*|) where Rw(\z\) is a finite sum depending on \z\ and N alone, and as N —> oo. If \z\ < 1, tien, for every positive integer k ^ 1, (29) n = 0 n + k n = 0 2n PROOF: Multiply (24) by zk~l, and integrate the result. Divide by N + 1, rear- range terms and use (26) to obtain (28). Let N -> oo to obtain (29).
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