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PnZ n fnlp holds for all We include following 1 ( p < ~. a proof which elementary is quite calculus simple. lemma It is based on the left as an exercice to the reader. LEM>L~. ,a N be positive (~N : 1 an)P ~ p ~N : l(~nj : l numbers aj)p-1 an and i ~ p < ~. 3=l of H 6 1 d e r ' s E[ fjlFj] )p-I p11(Zn E[fnlFn] )p-i inequality, it f o l l o w s fnll1 (Zn fn)lll ( pllEn E [ f n l F n] Up-1 p E[Zn f n IIp . So lien E[fnlFn]ll p ( pIIZn fnllp, as r e q u i r e d . 3. O P E R A T O R S FIXING Also section, in this long a l t h o u g h here is [ ~6] Our i n t e r e s t an LP-copy, subspace THEOREM mostly (See.

The reader may consult [31]. 6 k where f = E fn is the diadic d e c o m p o s i t i o n Applying again of f. ~k] 12)I/2Hq Jfnkl2)a/2Uq II

Xn) on X and take E T} for a l l x E X. ~ that (Tx)e = (T~)x 9 Then o[T] Proof : It = sup xEX is e a s i l y If x E X is f i x e d fore (o[T x] x 6 T e+l. + 1) verified and by induction e < o[T x] , t h e n Distinguishing the (Tx)a cases = (T~)x o [ T x] ~ ~ and is not there- a limit ordinal, o[ T x] o[T x] is a limit o[T] >. o[T x] Let conversely ordinal, + i, we So o[T] see ~ sup xEX that (o[T x] ~ : sup (O[Tx] + i). xEX o[ T x] : (T )x = %" But this o[ T x] (T x) x E T and hence + 1). For all means x C X, we h a v e that that no c o m p l e x e s in o[ T x] +I T~ C T starts A tree with x.