# Download Geometry of Banach spaces: Selected topics by J. Diestel PDF

By J. Diestel

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A product theorem is proved, namely, if α > 0, β > 0, f 0 Fα and g 0 Fβ then f 0 Fα+β. Also f 0 Fα if and only if f ′ 0 Fα+1. The families Fα are strictly increasing in α. Formulas are obtained which give mappings between Fα and Fβ. An analytic condition implying membership in Fα for α > 1 is given. This yields a condition sufficient to imply membership in Fα when α > 0. The latter condition provides a connection between the families Fα and the Besov spaces Bα. Two applications are given. The first shows that if the moduli of the zeros of an infinite Blaschke product f are restricted in a precise way, then f belongs to Fα for suitable α.

Finally, assume 1 – r < θ. Then | 1 – z |2 = (1 – r)2 + 4r sin2 (θ/2) < 2 θ2. This implies (c). 18 Let the function S be defined by ⎡ 1 + z⎤ S(z) = exp ⎢− ⎥ ⎢⎣ 1 − z ⎥⎦ © 2006 by Taylor & Francis Group, LLC Basic Properties of Fα 39 for |z| < 1. Then S 0 Fα for α > ½ and S ⌠ Fα for α < ½. Proof: Let z = reiθ where 0 < r < 1 and –π < θ < π. 38) ⎧⎪ 1 − r 2 ⎫⎪ − exp ⎨ ⎬ dθ . 17) we obtain I( r ) ≤ ∫ ⎧⎪ 1 − r 2 ⎫⎪ exp ⎨− 2 ⎬ dθ ⎪⎩ B (1 − r ) 2 ⎪⎭ (1 − r ) 1− r 1 2 0 and hence I( r ) ≤ ⎫⎪ 1 1 ⎪⎧ exp ⎨− 2 ⎬.

28) is not satisfied when 0 < α < 1. 29) Basic Properties of Fα 35 for |z| < 1 where |c| = 1, m is a nonnegative integer, 0 < |zn| < 1 for n = 1, 2, …, and ∞ ∑ (1 − | z n |) < ∞. 30), and | f (z) | < 1 for |z| < 1. Hence f 0 F1. If the zeros of f have a certain further restriction then f 0 Fα for suitable α with 0 < α < 1. The argument uses the following lemma. 15 Let 0 < α < 1 and for 0 < x < 1 let F (x) = ∫ 1 1 0 1− α (1 − r ) (1 − rx ) dr. 32) (1 − x )1−α for 0 < x < 1. 1) we obtain ∫ 1 0 (1 − r ) α −1 r n dr = n!