Results on Markov type, spreading norms, reversible chains of charts, and superreflexivity

The course proves the statements below; each link leads to the statement with its assumptions and its proof. The lessons are self-checked by the writing AI; independent mathematical review is not complete.

• Markov type bounds sums of odd edge labels — Lemma 3.1. Let V have Markov type p with constant K. For a finite stationary reversible marked chain with edges E_1, E_2, ... and a labeling g of the edges with g(e*) = -g(e): E |g(E_1) + ... + g(E_n)|^p <= K^p n E |g(E_1)|^p. The labels need not be differences of a function of the states.
• A finite form of Goldstine's theorem — Lemma 1.2. For w in the bidual of Y with norm at most 1, functionals f_1, ..., f_i and eps > 0 there is y in Y with norm at most 1 and |f_h(y) - w(f_h)| <= eps for all h.
• Sequences that control partial sums in a nonreflexive space — Lemma 2.1. Let Y be a nonreflexive real Banach space and 0 < d < 1. For every M there are z_1, ..., z_M of norm at most 1 with d max_j |a_1 + ... + a_j| - (1/M) sum |a_h| <= |sum a_h z_(i_h)| <= sum |a_h| for all i_1 < ... < i_m and all real a_h.
• The ordered norm — Theorem 5.3. Assume (JE) (James and Enflo; not proved in the programme). If a real Banach space X is not superreflexive, there is a norm N on finitely supported sequences that is spreading invariant, for which merging two adjacent entries does not increase the norm, and such that every (R^m, N) embeds linearly into X with distortion arbitrarily close to 1.
• A law on charts that treats all tuples alike — Lemma 1.1. For a finite ordered set and eps > 0 there are B and a distribution on the strictly increasing maps C into {1..B} such that the laws of C(A) and C(A') are within eps in total variation for any two increasing tuples A, A' of the same length.
• Maximal coupling — Lemma 2.1. Two distributions on a finite set have a coupling whose diagonal has mass 1 - TV.
• Reversible transport of charts — Proposition 3.1. For partial order isomorphisms T_s of a finite ordered set, indexed by a set with a fixed-point-free involution and T_(s^-1) = T_s^-1, and eps > 0: there is a finite stationary reversible marked chain on charts whose letters are independent and uniform and whose every step (C, s, C') satisfies C'(a) = C(T_s(a)) on the domain of T_s with probability at least 1 - eps.
• The reduced length of a random word — Lemma 1.1. For n independent uniform letters from {u, v, u^-1, v^-1} the reduced length l satisfies E l >= n/2 and P(l >= n/4) >= 1/4.
• Four homeomorphisms of the line that play ping-pong — Lemma 2.1. For four disjoint closed intervals in (0,1), indexed by u, v, u^-1, v^-1, with periodic arcs A_s: there are increasing homeomorphisms h_s of the line with h_s(y+1) = h_s(y)+1, h_(s^-1) = h_s^-1, h_s(R minus A_(s^-1)) = A_s and |h_s(y) - y| < 2.
• Two levels — Lemma 3.3. For a word w with reduced form of length l > 0 with n_+ positive and n_- negative letters, the sum c_w of the functions chi_s along the backward orbit satisfies c_w(j) = -n_- and c_w(h_w(0) + j) = n_+ for all integers j.
• A reversible walk with linear displacement — Theorem 5.1. Let N be a spreading invariant norm on finitely supported sequences with contractive merging. For every n there are a finite stationary reversible marked chain and an odd labeling g with N(g(e)) = 1 for all edges such that N(g(E_1) + ... + g(E_n)) >= n/24 with probability at least 3/20. No hypothesis.
• Nontrivial Markov type forces superreflexivity — Theorem 5.2. Assume (JE). Every real Banach space with Markov type p for some p > 1 has an equivalent uniformly convex norm.

Reading order

Five lessons. The first gives the statement and the reduction; lessons 2 to 4 are independent of each other and each proves one tool; the fifth combines the tools of lessons 3 and 4. The theorem is proved from one named hypothesis, the theorem of James and Enflo (a space without an equivalent uniformly convex norm has a nonreflexive space finitely representable in it), which is not proved in the programme. The walk with linear displacement of the fifth lesson needs no hypothesis.