= Solution
Choose a normalized <basic sequence> in the infinite-dimensional <Banach space>, and apply the <Rosenthal l1 theorem>. If a subsequence is equivalent to the unit vector basis of $\ell^1$, any <spreading model> generated from a further subsequence retains its two-sided $\ell^1$ estimates and is therefore unconditional.
Otherwise there is a <weakly Cauchy sequence> as a subsequence $(z_n)$, still basic. Its consecutive differences $d_n=z_{2n}-z_{2n-1}$ are weakly null and are a <block sequence>. The coordinate-functional bound for its normalized parent basis gives $1/(2K)\leq\|d_n\|\leq2$. Hence $d_n/\|d_n\|$ is a normalized weakly null <basic sequence>. Part (a) gives a <spreading model>, and part (b) makes that model one-suppression-unconditional. Thus \b[every infinite-dimensional Banach space has an unconditional spreading model], even though it need not contain an unconditional basic sequence.
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