OAI.BoundedTreePotentials.TreeCalculus.main_counterexample

The theorem states that, for every choice of a Boolean flag r (whether the root node, the empty list, is included as a coordinate) and every quadratic kind k among sibling, antichain and global, the following hold for the Banach space X = TestCompletion(treeTestFamily r k). Here nodes are finite lists of natural numbers, the potential of a coefficient function f at a node s is the sum of f over all prefixes of s, and a coefficient function f is a tree test if f vanishes at the root when r is false, every potential satisfies |potential f s| ≤ 1, a quadratic budget holds, and a support condition holds. The budget says that the sum of f(s)² over every admissible finite set B of nodes is at most 1, where admissible means: all of B are children of one common parent (sibling), pairwise prefix-incomparable (antichain), or any finite set (global). The support condition is vacuous for sibling, finite support for antichain, and square-summability for global. The tree test family is the set of such functions on the coordinates, and each finitely supported vector x is given the norm sup over tests f of |Σ x_i f_i|; X is the completion of this normed space. First, X is complete, separable, and not finite-dimensional over ℝ. Second, for every 0 < t < 1, the averaged modulus of X, defined as the infimum over unit vectors x of the supremum over closed finite-codimension subspaces F of the infimum over y in F with ‖y‖ ≥ 1 of (‖x+ty‖+‖x−ty‖)/2 − 1, is at least √(1+t²/4) − 1. Third, for every real normed space Y that is linearly homeomorphic to X, Y does not satisfy IsAUCReal, meaning it is false that the one-sided modulus (defined like the averaged one but with ‖y‖ = 1 and the quantity ‖x+ty‖ − 1) is positive for every t > 0.