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By Vjačeslav V. Sazonov (auth.)

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Then S IZ-ZoL hence E s ~ BR(Zo). Es : k ½ s < R, The see = E s can be represented in the form {z:zd ck,zj = Zoj + se iej, Oj ~ [-~,~),J = 1,k}. d8 k (2~) -k f (here d' is a nonnegative integral vector). Cb " Together (41) and (42) imply (42) 55 {Dbf(Zo)I = Ibis- bIJ(2~)-k b! d@k I { ] f ( z ) l} . (43) S and o b s e r v e t h a t Eclzo [ c {z:iZ-Zo[ = k½ Oi~ol} l. c {z=lzl ~ [Z-Zol + IZol ~ (l+. ~ c)lzo[} c S r Thus sup {If(z)l} Z~Eclzo I ± sup h([zl) z~ Eclzo I h((l+k ½ c)Izol) and the lemma follows.

Denote t = vT -1. Since R(A) = - R(A c) and and take 30 6 = sup IR(A) I we can choose A such that R(A ) > 6-c, which is either a convex set A~ C or the complement to a convex s e t . 1. Suppose f i r s t that O# A 2 t . E 6 T Z RT(A ~) By Lemma 8, Then + = ( = I 1 + 12. ~ l×l

45) Since V = l,g(t) may be written in the form m-3 -- ~ n-r/2 g(t) r=O Xr+2 m-3 ---r=O ~ n-r/2 ir and we have for IDbg(t) b->O, [Ibll~m -~ I (it)/(r+2)! )td (since IXd] _ O. When g(t) + [t12/21 ~ Choosing c(k,m) shall have b =0 (46) we have in a similar way m-3 ~ (c(k,m))rc(r)It[ 2 r=l smal] enough the right side of (47) to be less than (47) It12/8 we 57 Ig(t)l (48) ~ 5 It12/8 .

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