By I. M. Yaglom
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Example text
4: W e can a s s u m e DP(Fto) ~ DP(~) and Qt6Lq(Ft~) > O. nf{SIPt-ulPFtd~ ii) SuQtFtd~ iii) IPt Ip = PtQ t = to p r o v e that - = Dp(o) - that F~I. 4 we o b t a i n then I~Ft~F and h e n c e functions Pt6LP(Fto) with i) Let us d e f i n e < : u6A} = O, I = ~(u) (DP(Ft~)) p Vu6A, IQt lq =: F t w i t h F t F t o 6 M ( ~ ) . the f u n c t i o n @:e(t) = - l o g DP(Ft~) for O~t~1. O u r a i m is 39 (*) der sup 8 (t) ~ -f (log F) FtFtdo VO~t~1. 2 furnishes a point O~T~I such that m := FTFTo 6 M(~) fulfills the assertion.
3 for t=O to y i e l d IIf~<1 and c o m b i n e the e l e m e n t a r y l
Ii) I Ira(l) I 1=1 21 l+tlm(1)ll L e t T = {m(1) : i = 1 , 2 , . . } 6 POS(X,~) 0` << m for s o m e m6K. 0`,6 ~ O. N o w ~ i). c K be a Decompose Since 6 is s i n g u l a r 0 < 6 < % it f o l l o w s that into e = 6 is s i n g u l a r to e a c h m(1) ii). to m it f o l l o w s and hence 6 = O. T h e r e f o r e after o,+8 singular that to ~. ). For exist m6K} the m6K rest such of t h e p r o o f w e ~ IIell a n d t a k e m ( 1 ) 6 K m£K with m(1)<