Normed versus topological groups: Dichotomy and duality

Normed groups 53By (C-adm), we may apply Lemma 3.33 to deduce for w = y n ym−1 and m, n large that‖x n wx −1m ‖ is small. Hence so also is d R (x n y n , x m y m ). That is, the product of Cauchysequences is Cauchy.Before considering the converse, observe that if w n = y n yn+1 −1 is given with y n Cauchy,then w n is null and with m = n + 1 we have as n → ∞ that‖x n w n x −1n ‖ = ‖x n y n ym −1 x −1m x m x −1n‖ ≤ ‖(x n y n )(x m y m ) −1 ‖ + ‖x m x −1n ‖ → 0,provided x n y n is a Cauchy sequence. We refine this observation below.Returning to the converse: assume that in X the product of Cauchy sequences is Cauchy.Let x n and y n be Cauchy.Let w n be an arbitrary null sequence. By Lemma 3.34 it is enough to show that downa subsequence γ x(n(k)) (w n(k) ) → e X . Since we seek an appropriate subsequence, we mayassume (by passing to a subsequence) that without loss of generality ‖w n ‖ ≤ 2 −n . Wemay now solve the equation w n = z n−1 zn−1 for n = 1, 2, ... with z n null, by taking z 0 = eand inductivelyIndeed z n is null, sinceNow, as n → ∞ we havez n = w −1nz n−1 = wn−1 wn−1 −1 ...w−1 1 .‖z n ‖ ≤ 2 −(1+2+...+n) → 0.‖x n+1 w n+1 x −1n+1 ‖ = ‖x n+1x −1n (x n z n zn−1 x−1 n−1 )x n−1x −1n ‖≤ d(x n z n , x n−1 z n−1 ) + d(x n+1 , x n ) + d(x n , x n−1 ) → 0,since x n and x n z n are Cauchy. By Lemma 3.34 {γ x(n) } is uniformly continuous at e, andso by Lemma 3.33, (C-adm) holds.Remark. The proof in fact shows that it is enough to consider products with x n Cauchyand y n null; since the general case givesand, for y n y −1md R (x n y n , x m y m ) = x n y n y −1m x −1n(x n x −1m )small, this is small by an appeal to (C-adm).Lemma 3.36. (W-adm) is satisfied iff γ x is continuous for all x.Proof. We work in the right norm topology. Asume (W-adm) holds. As the constantsequence x n ≡ x is convergent, it is immediate that γ x is continuous. For the converse,suppose x n → R x and put z n = x n x −1 (which is null); then x n = z n x → R x andx n w n x −1n= z n (xw n x −1 )z −1n → e,by the triangle inequality (since ‖z n (xw n x −1 )z −1n ‖ ≤ ‖xw n x −1 ‖+2‖z n ‖) and so (W-adm)holds.