By Nigel J. Kalton
The final challenge addressed during this paintings is to signify the attainable Banach lattice buildings separable Banach house could have. the elemental questions of area of expertise of lattice constitution for functionality areas were studied earlier than, yet the following the technique makes use of random degree representations for operators in a new method to receive extra robust conclusions. a standard result's the subsequent: If $X$ is a rearrangement-invariant house on $[0,1]$ no longer equivalent to $L_2$, and if $Y$ is an order-continuous Banach lattice which has a complemented subspace isomorphic as a Banach house to $X$, then $Y$ has a complemented sublattice that's isomorphic to $X$ (with considered one of attainable lattice structures). New examples also are given of areas with a special lattice constitution.
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Extra resources for Lattice Structures on Banach Spaces
Now let En = Vri(t)=iGn,i- Then, for 1 < j < ]V, XF^\Tj(fjXEn) - 2Tjfj\ < Du n,j- LATTICE STRUCTURES ON BANACH SPACES It remains to observe that as u fi j ^ u j pointwise and unj < Tjfj we can apply Egoroff's theorem to find a sufficiently large n and a Borel set F C F'n with j(F) so that Dunj > (l — ND~2)j(K) < DUJ + eTjfj on F. The result then follows by taking E = En. 3. Let (K,/J,) be a Polish probability space. Suppose X is a good continuous function space on (A, A) and let Y be an order-continuous (K,fi).
2 for suitably large D (in this case Uj = 0). , e n ), so that j(K \ F) < £ 2 " ^ < § T ( * 0 - Thus 7 ( P ) > | 7 ( K ) > 0. ,e n ) - ^TXA\ Z <( £ k=o 2krjk)2-nTXA < 2-(n+1)TXA so that (*) XFT(XG(£I €B)) < ^2-"XFTXA. Consider the measure defined on A by TTF(H) = JF PQXH dfi. Then irp is nonzero and 0 < 7rp < MA, so that there is a Borel set E of positive measure in A so that for some c > 0, we have TTF(EO) > c\(Eo) for every Borel set Eo C E. Now we define a continuous map a : A —> A so that a(s) = (ek)<^=1 whenever 5 G n ^ = 1 G ( e i , .
PROOF: Suppose / G £ i ,+ with ||/||i = 1. Then, by a well-known factorization theorem due to Lozanovskii (,) there exist