Results & Lemmas (13)
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THEOREM 1.
THEOREM 1. Let fe @(α) (0 ^ a < 1). Then (2.1) f(z) = z exp (2(1 - a) j * log (1 - β"*" 1 where y(t) increases and j(t + 2π) — y(t) = 1.…
THEOREM 1. Let fe @(α) (0 ^ a < 1). Then (2.1) f(z) = z exp (2(1 - a) j * log (1 - β"*" 1 where y(t) increases and j(t + 2π) — y(t) = 1. Also axgf{re iι)—*πt + 2ττ(l — ά)y(t) as r —> 1, and (2.2) z a + ( l a ) Γ ^ f f(z) Jo 1 — e %tz~
COROLLARY 1.
COROLLARY 1. For given fe &(a), there is a sequence of functions fn e @(α) of the form fΛz) = z Π (1 - ZvZ-T 1-*^ (l«vl^l)
COROLLARY 1. For given fe &(a), there is a sequence of functions fn e @(α) of the form fΛz) = z Π (1 - ZvZ-T 1-*^ (l«vl^l)
THEOREM 2.
THEOREM 2. Let f(z) = z + ao + aλz- χ + e @(α) and | = r > 1. Then (3.1) (3.2) z ^ (l + l αol r- 1 + r~ 2Y " ^ (1 + V 1 — a /
THEOREM 2. Let f(z) = z + ao + aλz- χ + e @(α) and \z| = r > 1. Then (3.1) (3.2) z ^ (l + l αol r- 1 + r~ 2Y " ^ (1 + V 1 — a /
THEOREM 3.
THEOREM 3. Let 0 ^ a < 1. Then as r —> 1 2«i—>β-i max max '(z) 1 - r- 1) log 1/(1 - r- 1) The proof will show that the function f(z) for…
THEOREM 3. Let 0 ^ a < 1. Then as r —> 1 2«i—>β-i max max \f'(z)\ 1 - r- 1) log 1/(1 - r- 1) The proof will show that the function f(z) for which \f'(zo)\ be- comes maximal for a given z0 has the form (4.1) f(z) = z(l - ^- 1) 2 ( 1-
THEOREM 4.
THEOREM 4. If fe @(α) (0 S a < 1) then (5.1) ' z) ^a /(*) + (1 - a) cύl l~cύ) 1 / 1 X 1"* L. i > (1 — _ ) iv ~ V 12 iv * Equality can be…
THEOREM 4. If fe @(α) (0 S a < 1) then (5.1) \f'{z)\^a /(*) + (1 - a) cύl{l~cύ) 1 \ / 1 X 1"* L. i > (1 — _ ) \z iv ~ V 12 iv * Equality can be attained in all inequalities.
THEOREM 5.
THEOREM 5. Let fe @(α). (i) If 0 ^ a < ) > 0 s^c/^ that more is ί/iere isα/s: =
THEOREM 5. Let fe @(α). (i) If 0 ^ a < ) > 0 s^c/^ that more is ί/iere isα/s: =
LEMMA 1.
LEMMA 1. // fe @(α) (0 ^ a < 1) and n = 0,1, (w + I) 21 an 2 £ 4(1 - a) 2 - 4(1 - α) Σ(v + α) | αv | 2.
LEMMA 1. // fe @(α) (0 ^ a < 1) and n = 0,1, (w + I) 21 an\ 2 £ 4(1 - a) 2 - 4(1 - α) Σ(v + α) | αv | 2 .
COROLLARY 2.
COROLLARY 2. Let fe @(α:), 0 < α < 1, and let A be the area of the compact complement of the image region f(z): z > 1. Then π ^ A> πa, and…
COROLLARY 2. Let fe @(α:), 0 < α < 1, and let A be the area of the compact complement of the image region {f(z): \ z \ > 1}. Then π ^ A> πa , and these inequalities are best possible.
Lemma 1
Lemma 1 implies Σ (v + OL) I αv | 2 ^ 1 - a, hence A = πf 1 — Σ ^ I αv I 2) ^ TΓ^O: + ^ Σ | αv | 2) ^ 7rα:. V=l / V=0 /
Lemma 1 implies Σ (v + OL) I αv | 2 ^ 1 - a , hence A = πf 1 — Σ ^ I αv I 2) ^ TΓ^O: + ^ Σ | αv | 2) ^ 7rα:. \ V=l / \ V=0 /
THEOREM 6.
THEOREM 6. Let f(z) = z + ΣΓ=o α ^ " u be in @(α) (0 ^ α < 1).
THEOREM 6. Let f(z) = z + ΣΓ=o α ^ " u be in @(α) (0 ^ α < 1).
THEOREM 7.
THEOREM 7. If fe&(a) then (6.3) lim sup n an | < 2(1 - a). For every ε > 0 there is a function f e @(α) such that lim sup n an | > 2(1 — a)…
THEOREM 7. If fe&(a) then (6.3) lim sup n \ an | < 2(1 - a) . For every ε > 0 there is a function f e @(α) such that lim sup n \ an | > 2(1 — a) — ε
THEOREM 8.
THEOREM 8. Let g(ζ) = Σ =ibnζ n be analytic in ζ <l and satisfy (6.4) Reζg'(ξ)lg(ζ) ^ a. and i g(ζ) | < 1. Then (6.5) Σ (n - a) b J 2 ^ 1 -…
THEOREM 8. Let g(ζ) = Σ =ibnζ n be analytic in\ζ\<l and satisfy (6.4) Reζg'(ξ)lg(ζ) ^ a . and i g(ζ) | < 1. Then (6.5) Σ (n - a) \ b J 2 ^ 1 - α . For w = 2, 3, • (6.6) μj
LEMMA 2.
LEMMA 2. Given δ > 0 and η > 0 there exists a function HO = 1 + Σ cnξ* n=l with cn ^ 0 £/&α£ ΐs analytic and has positive real part in ζ <…
LEMMA 2. Given δ > 0 and η > 0 there exists a function HO = 1 + Σ cnξ* n=l with cn ^ 0 £/&α£ ΐs analytic and has positive real part in \ ζ\ < 1 such that (7.1) limsupc^ > 2 - δ , (7.2) Σ — < 7 1 W This lemma was first proved by F. Riesz [8] but only with limsup cn
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