By J. Bourgain, P. G. Casazza, J. Lindenstrauss

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E. Step II. Pick now an even integer j such t h a t 7 0 ' / 2 ) 1 / 4 > y/2KxK3 and then choose JJ, > 0 so that Assume now that \Vi\i-\ are mutually disjoint blocks on lunin^i s o t h a t H^ill ^ a, for every 1 < i ^ j and some a > 0. 2 JFJ such that (i)6k2ty(j/Z) and || £ > i | l ^ (6k)3/4a/Kv (ii) £ « * * ; / 2 . e. n i: fill >*j"*. where g = 2 i=l < 2. A simple iteration argument shows that if \vi\t=\ are mutually disjoint blocks on fanJ^Li so that ||V{|| ^a, all 1 <£ i -s jf^f /i = 1,2, • • , and some a > 0, then for ii £^11 >o>*/«.

Proof. 2, is actually normalized and has unconditional constant equal to one. We start by proving the existence of a constant C < *» so that every subsequence of | x n } ~ = 1 contains in t u r n another subsequence which is C 2 -equivalent to a permutation of \xn\"=l. Indeed, if such a C does not exist then, for each k, we can construct a subsequence [x*J~ =1 of {x n J~ =1 such that: (i) [ x * + 1 j ~ = 1 is a subsequence of {x*J~=1> for all k, (ii) for any integer k, p e r m u t a t i o n s of the integers and any subsequence fty j/Li of \z*\Z=v the bases te/y{~=1 and bir(n)Jn=i a r e n o t fc3-equivalent.

Since \x\tx% \ is (k+D)2- , xg,x££,x%£, • • • j is clearly a subsequence of {£nltT=i and, for fc large enough, A;3 > (k+D)2, we get a contradiction. Let now (eJiLi be a spreading model of fx n }~ = 1 . More precisely, if \ i\r=\ denote the unit vectors in an abstract vector space and U is a free ultrafilter on the integers we define e m HIE V i l l i = i=1 lim m IIEVnJI. 1 and any integer m . The sequence \ei\t=\ forms clearly a 1-subsymmetric and 1-unconditional basis of its closed linear span.