Normed versus topological groups: Dichotomy and duality

Normed groups 131Hence,‖x‖ − 2ε ≤ ‖ξ x ‖ ε ≤ ‖x‖ + 2ε,and so ‖x n ‖ → ∞ iff d T (ξ x(n) (t), ξ e (t)) → ∞.Proof. We follow a similar argument to that for the permutation metric. By rightinvariance,sod X (t(x −1 z), t(y −1 z)) ≤ d X (t(x −1 z), x −1 z) + d X (x −1 z, y −1 z) + d X (y −1 z, t(y −1 z))≤ 2‖t‖ + d X (x −1 , y −1 ),d T (ξ x (t), ξ y (t)) = sup z d X (t(x −1 z), t(y −1 z)) ≤ 2‖t‖ + d X (x, e X ).Now, again by right-invariance,Butsoas required.d X (x −1 , y −1 ) ≤ d(x −1 , t(x −1 )) + d(t(x −1 ), t(y −1 )) + d(t(y −1 ), y −1 ).d(t(x −1 ), t(y −1 )) ≤ sup z d X (t(x −1 z), t(y −1 z)),d X (x −1 , y −1 ) ≤ 2‖t‖ + sup z d X (t(x −1 z), t(y −1 z)) = 2‖t‖ + d T (ξ x (t), ξ y (t)),We thus obtain the following result.Theorem 12.7 (Topological Quasi-Duality Theorem). For X a normed group, the seconddual Ξ is a normed group isometric to X which, for any ε ≥ 0, is ε-quasi-isometric to Xin relation to ˜d ε T (ξ x, ξ y ) and the ‖ · ‖ ε norm. Here T = H u (X).Proof. We metrize Ξ by setting d Ξ (ξ x , ξ y ) = d X (x, y). This makes Ξ an isometric copyof X and an ε-quasi-isometric copy in relation to the conjugate metric ˜d ε T (ξ x, ξ y ) whichis given, for any ε ≥ 0, byIn particular for ε = 0 we haved T (ξ −1x˜d ε T (ξ x , ξ y ) := sup ‖t‖≤ε d T (ξ −1x(t), ξy−1 (t)).(e), ξy−1 (e)) = sup z d X (xz, yz) = d(x, y).Assuming d X is right-invariant, d Ξ is right-invariant, sinced Ξ (ξ x ξ z , ξ y ξ z ) = d Ξ (ξ xz , ξ yz ) = d X (xz, yz) = d X (x, y).