# Abstract Analytic Function Theory And Hardy Algebras by K. Barbey, H. König By K. Barbey, H. König

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Extra info for Abstract Analytic Function Theory And Hardy Algebras

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A n d i) s h o w s iii) W e a p p l y the H a h n - B a n a c h that Un6A. Vf6ReC(X). duct vector space V:=ReC(X)×ReC(X) linear of f u n c t i o n s then we can a c h i e v e the s u b l i n e a r shows are ~ 0 S are p r e c i s e l y 0: @(f,g) ~ ~ such that each a sequence T([Un+O])=T(X) for some ~6~(A) i) D e f i n e 0s(f) there exists such that g([Un+1])=~(X) The p r o o f of the a b s t r a c t lemma. as well. I It f o l l o w s v(f,g)6V to o6S and T6T. to 0 and to the c o n v e x t h a t there set e x i s t o6S and T6T such t h a t I := I n f { 0 s ( f ) + @ T ( g ) : O ~ f , g 6 R e C ( X ) = Inf{~(f)+T(g):O~f,g6ReC(X) We a s s e r t g(B)=O t h a t I=O.

Riesz algebra in f a c t situation two is b o u n d of the theorem simple fundamental theorem theorem the m a i n theory Szeg6-Kolmogorov-Krein Szeg~-Kolmogorov-Krein lemma to the o t h e r F. to the H a r d y and algebra Theorem lemmata which theorem to b e c o m e chapter will has be are with intro- core. fix ~ 6 E ( A ) . be d e d u c e d which will its and starts of from The the funda- unrelated to the situation. GEOMETRIC MEAN LEMMA: Assume that m6Prob(X,~) a n d O~f6L1 (m) a n d 37 I 0: 8(t) Then i) 8 is m o n o t o n e when f is c o n s t a n t , called from 8(O+)~I(f) 1 + tflog Thus for O

L. ~(Z) ii) B c ca(X,Z) M(~)cB i) W e m B - mB(X)m 6 B^. m ( ~ a) = ~F(s)P(a,s)dl(s) S = ~a(F) the remark. 8 fv(u-~(u))dm m B = O. 5. consideration. called that reducing M(~) v is We iff a reducing ~6Z(A). i). 5 point evaluations have = ~a(F) the measure if ~6Z(A(D)) some For representing = f(a) representing iii) In the a6D <0a: <0a(F) has o f DUS. and prove , m~ that . Now - ~(u)~vdm in particular and hence so once In particular either Vv6A. from the m~(X)mB(X) m6M(~) in B were some ~£M(~) m6B (u-~(u))m = O Then for each M(~)~cB It Vu6A, follows forces assumption = 0 so t h a t all in B ^ and other then have since this that Thus m B - mB(X)m B = either a6M(~) ~(m+~) ~6Z(A) o r M ( ~ ) V c B ^.