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By Charles J. Mozzochi

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45) ~* = (~jvUW ), o j+l,v then no = n[to~] 2v+l Remark. 3 = 4 " 2 ~ ' n j [to~]Ito J ~I some v -> O where implies to* = j that (tojvUtoj+l,v ) nj[to~] = and that n ) .... 2v+l for n. = O if and J only if nj[~*]j = O. 46) Remark. )n 3 j n. < ~ J--i=l < 2 implies (I + bi)n < 2N < 2N. 2~ -- j nj[w~] = 0 ; n. ]3 -< so that 2N ' nj = O. 2) = n and -I m* = [-4~,4~]. 47) that Lemma. (n[~_*l],~_*l) By ( 6 . 4 7 ) the partition by ( 6 . 3 3 ) since x } ~ G~L . 44)we = IS*n o ( x ; ) ~ F ; ~ o ) [ have + O (Lk bk_lY) Suppose n o ~ O.

2~ 26 Let Let Let Gk(~jv ) = { ( n , ~ j v ) ] ~%'(x;Wjv) lc n (wjv ; / [ o _F p k ( ' ; ~ ~y-I 33[ i b k yp/2} i2Vnx = z c °-p (n,mjv)¢Gk(mjv) n(~jv ' X F Pk(X;~jv) = P k ( X ; ~ , v _ l ) • )) ¢ k(''mZv_l + Rk(X;ejv ) • "Ixl < • _ Then Pk(X;~J v) = R k ( X ; ~ r o ) + R k ( X ; ~ s l ) + " ' + R k ( X ; ~ z , v = l ) + R k(x;o~jv) Continuing in this way we define for each k > 1 and for each dyadic interval m a polynomial Rk(X;w), a polynomial Pk(X;~) and a set Gk(~). 1) Remark. an (~) t h e c o e f f i c i e n t To simplify the notation we will often denote by of the ¢ term in Rk(x;~) c o r r e s p o n d i n g t o t h e element (n,~) E Gk(~); times write a ¢ write a ¢ i~x ikx for simplicity It is to be noted that in the sequel whenever we as a term in Pk(X;~) we always assume that the terms in this polynomial have not been "collected"; term of Rk(X; ~') for some ~ ' D have that if a e ]aJ ~ we may a t ~.

I) see product L 1 (a,b). h c inequality Since that For an explanation (x) + i (Hv) [15] page 132. be the pointwise by Holder's (x). In many of our applications of g(t) = c g c L~(a,b), f = gh E 1 L (a,b). in~t and we have of course of the notation used in the following theorem review the first section of chapter 5. 3) Theorem. Let ~ denote a subinterval of w* which X contains x in its middle half. v. CY T = {x ~ ~* We define the maximal J~ U dtl ; x~ ~*. (H'f) (x) > y} fE L (~*) oo x ] Hilbert ~*: x Let dt ; y > O.

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