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Results & Lemmas (7)

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THEOREM 1. · coeff THEOREM 1. Let f eC so that (1) holds. Then for n > 1 we have (2) < ^- PROOF. For / e C let [fk] be a sequence in C which converges…
THEOREM 1. Let f eC so that (1) holds. Then for n > 1 we have (2) \Yn\ < ^- PROOF. For / e C let [fk] be a sequence in C which converges uniformly on compact subsets to / e C. Let also for a fixed n, J(g) = \yn\ where g(z) = log(/(z)/z) = 2]Tn°t1 ynz" and let \og{fk(z)/z) = 2 £ ~ , y^zn. Then, using the coefficient formula we deduce for z = re'e, 0 < r < 1, that \z\=r ^ 0 , since / * - > • / uniformly on \z\ = r as k ->• oo [11, p. 40]. Thus we see that \Ynk)\2 ""*• ly«l2 s o t n a t J(8) is con
Lemma 5.6 Lemma 5.6] we have https://doi.org/10.1017/S1446788700037344 Published online by Cambridge University Press
Lemma 5.6] we have https://doi.org/10.1017/S1446788700037344 Published online by Cambridge University Press
COROLLARY 1. COROLLARY 1. For f eCwe have i r2" 2 ^ Jo f(z) n=l =-f 2TT JO where z = rew, 0<r <,and K(z) = z(l - z)~2.
COROLLARY 1. For f eCwe have i r2" 2 ^ Jo f(z) n=l =-f 2TT JO where z = rew, 0<r <\,and K(z) = z(l - z)~2.
THEOREM 2. THEOREM 2. For f e EHCwe have zK'iz) K(z) dd (4) or equivalently log -, log /(z) *(z) where K z) is as defined above. PROOF. This theorem…
THEOREM 2. For f e EHCwe have zK'iz) K(z) dd (4) or equivalently log -, log /(z) *(z) where K{z) is as defined above. PROOF. This theorem follows from the fact that Re ((1 - bz)/{\ - z)) > 1/2 is equivalent to (1 — bz)/{\ — z) < 1/(1 — z), [11, p. 53], and this is equivalent to log((l - bz)/(l - z)) -< log(l/(l - z)) [17, p. 23]. Thus we have
COROLLARY 2. COROLLARY 2. For f e EHC we see from [11, Theorem 3.3]; [6, Theorem 6.1] and (4) that n2 Jo log /(z) log K(z) where z = rew, 0 < r < 1, >…
COROLLARY 2. For f e EHC we see from [11, Theorem 3.3]; [6, Theorem 6.1] and (4) that n2 Jo log /(z) log K(z) where z = rew, 0 < r < 1, > 0. This extends [10, Theorem I] for f € £//C.
COROLLARY 3. COROLLARY 3. For f e C and yn as defined in Theorem 1 we see from [6, p. 212 (Exercise 7)], (4) and (2)…
COROLLARY 3. For f e C and yn as defined in Theorem 1 we see from [6, p. 212 (Exercise 7)], (4) and (2) https://doi.org/10.1017/S1446788700037344 Published online by Cambridge University Press
THEOREM 3. THEOREM 3. Let f eC and f(z) = z + a2z2 + a3z3 H. Then for n > 2 we have K | - k, - l l | HI- PROOF. For some choice of £ on the boundary,…
THEOREM 3. Let f eC and f(z) = z + a2z2 + a3z3 H . Then for n > 2 we have K | - k , - l l | HI- PROOF. For some choice of £ on the boundary, with |f | = 1, we have Applying the Lebedev-Milininequality \pn\2 < exp {£Li k\ak\2 ~ E*=i l/k] f o r the expansion TZo^k = exP {ELi «*z*}> A> = 1 [6, p. 143]; [5, p. 897], we deduce by using the triangle inequality that (5) \an\ - \an~i\ <exp k=\ We now write f = eu and choose t such that kt + argCy*) = 0. We see that eki'yk — \Yk\- Using this and (2) in (
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