A Remarkable Rearrangement of the Haar System

We introduce a non standard but to our opinion natural order on the initial segments of the Haar system and investigate the isomorphic classi cation of the linear span in Lp of block bases with respect to this order Introduction In DS it was proved that every unconditional basic sequence fxig i in Lp p which is not equivalent to the natural basis of p has the property that for some K and every positive integer n there are n vectors of the form yi P j i ajxj i n where the sets i are pairwise disjoint and the sequence fyig is equivalent with constant K to the unit vector basis in n See also JMST for a generalization of this fact for more general lattices It was left open in these two papers and speci cally asked in Problem A of DS whether fyig can be chosen to be a block basis i e whether the sets i can be chosen to be successive that is maxfj j kg minfj j lg for all k l The initial motivation of this paper was to solve this question in the negative As is well known the Haar basis fhn ig n n i in Lp p in its common order has a block basis i e the Rademacher functions equivalent to the unit vector basis of It follows from the main results of this paper Theorems and that the initial segments of the Haar system can be rearranged so that they will not have the nite version of this property anymore We denote the new order by Note that the space P N hn i N n n i p is isomorphic to p here xi i A denotes the closed linear span of fxigi A Let fbig i be the natural relative to the order basis of P N hn i N n n i p i e the basis composed of the successive conjunction of the bases fhn ig N n n i each in its order Clearly fbig i is not equivalent to the natural basis of p as long as p and it easily follows from the previous paragraph that For each p p and for each K K there is an n such that no block basis of fbig i of length n is K equivalent to the unit vector basis of n Supported by the Austrian Academy of Sciences APART Program Supported in part by BSF Both authors participated in the Workshop in Linear Analysis and Probability Texas A M University Thus in the case p the sequence fbig i gives the required counterexample to the problem above In Section below we introduce the new order on the initial segments of the Haar system and determine some of its properties In Section we prove the main theorem which states that for p every block basis of such an initial segments of the Haar system is well equivalent to a diagonal of the unit vector basis of p and a subsequence of a permutation of the Haar system In Section we deal with the case p and prove a somewhat weaker property of such block bases in Theorem but which is also enough to deduce above The order An interval of the form I i n i n n i n will be called a dyadic interval The left half of the interval I i n i n is the interval i n i n and the right half of I is i n i n The Haar function hI hn i associated with I is the function hI t if t is in the left half of I if t is in the right half of I The common order on fhIg is the lexicographic order on f n i g Fixing N we would like to introduce a di erent order on the set T TN of dyadic intervals of length larger than or equal to N De nition I J if either I and J are disjoint and I is to the left of J or I is contained in J In terms of the natural dyadic tree structure of T the following picture describes the order for N Proposition below describes the order intervals with respect to the order We rst need two de nitions De nition Given two dyadic intervals I J The cone C C I J of dyadic intervals between I and J is the unique collection of dyadic intervals C fC Ckg satisfying a k log jJj jIj b C I Ck J c jCsj jCs j Cs Cs for s f k g The right ll up of the cone C is the collection of dyadic intervals R R I J Sk s Us where Us if Cs is the right half of Cs and Us fJ jJ j N J Cs n Csg If Cs is the left half of Cs Proposition Let J J TN satisfy J J then there exist a unique collection L fL Lmg of pairwise disjoint dyadic intervals satisfying jLij jLi j if i m jLm j jLmj if m a Li and Li intersect in exactly one point the left endpoint of Li b Li lies right of Li J L J Lm such that fI TN J I J g C J L R J L m i Mi where Mi fI TN I Lig Proof Assume rst that J J Then the assertion is that a dyadic interval I satis es J I J if and only if I C J J R J J The if part is clear Assume now J I J Clearly I can not be disjoint from J otherwise it would have to be to the left of J and thus to the left of J contradicting J I So I J If J I then by the de nition of a cone I C J J if not then I lies to the right of J Then there exists Ci C J J so that I Ci and I Ci We then claim that Ci n Ci is the right half of Ci and then I R J J Indeed if Ci n Ci is the left half of Ci then I lies to the left of Ci and thus to the left of J a contradiction This concludes the proof of this special case and we turn to the general case Let D be the minimal dyadic interval which contains both J and J If D J put m and L D If D strictly contains J then take the left half of D and call it L Observe that J D n L R where R is the right half of D Let I be the dyadic interval of length N which has the same left endpoint as R Let D R be the minimal dyadic interval which contains both I and J If D J put m L D If D strictly contains J then take the left half of D and call it L Observe that J D n L R where R is the right half of D Continuing in that manner let I be the dyadic interval of length N which has the same left endpoint as R Let D R be the minimal dyadic interval which contains both I and J the process stops after nitely many steps and we get a collection L fL Lmg of pairwise disjoint dyadic intervals satisfying through We are left with the task of proving Clearly the right hand side in is contained in the left hand side To prove the other containment note that if J I J then none of the Li i m is strictly contained in I Suppose not then for some i Di I and from J Di I we get J I A contradiction It follows that there exists i so that I Li Suppose not then using that Sm i Li is an interval not necessarily dyadic I is either to the right of Lm or to the left of L Hence I J or J I A contradiction If i then I Mi and we are done so we may suppose J I L This is exactly the special situation described in the beginning of this proof and we conclude that in this case I C J L R J L Next we would like to describe the interaction between two successive order intervals Let I I J J and put B fI TN I I I g B fI TN J I J g Let L fL L m g be the maximal with respect to inclusion elements of B as given in Proposition and let L fL L m g be the maximal elements of B Let Bi I Bi Clearly B and B may intersect Lemma a B may have intersection only with L i e