Normed versus topological groups: Dichotomy and duality

48 N. H. Bingham and A. J. OstaszewskiThus ξ = 0, and hence x n → e. So our assumption of (iii) yieldsa final contradiction.‖g n x n gn−1 ‖(1 + ε) ≤ lim n→∞ = 1,‖x n ‖We note the following variant on Theorem 3.30.Theorem 3.31. Let G be a Lipschitz-normed topological group. The following are equivalent:(i) the mapping g → M(g) is continuous,(ii) the mapping g → M(g) is continuous at e,(iii) the norm is nearly abelian, i.e. for arbitrary g n → e and z n → elim n ‖g n z n g −1n ‖/‖z n ‖ = 1.Proof. Clearly (i)=⇒ (ii). To prove (ii)=⇒ (iii), suppose the mapping is continuous ate, then by the continuity of the maximization operation (cf. [Bor, Ch.12] ) g → M g iscontinuous at e, and Theorem 3.30 applies.To prove (iii)=⇒ (ii), assume the condition; it now suffices by Theorem 3.30 to provelower semicontinuity (lsc) at g = e. So suppose that, for some open U, U ∩ M(e) ≠ ∅.Thus U ∩ (1, ∞) ≠ ∅. Choose m ′ < m ′′ with 1 < m such that (m ′ , m ′′ ) ⊂ U ∩ M(e). IfM is not lsc at e, then there is g n → e such(m ′ , m ′′ ) ∩ M(g n ) = ∅.Take, e.g., m := 1 2 (m′ + m ′′ ). As m ′ < m < m ′′ , there is x n ≠ e such thatm‖x n ‖ < ‖g n x n gn−1 ‖.As before, if ‖x n ‖ is unbounded we may assume ‖x n ‖ → ∞, and so obtain the contradiction‖g n ‖ + ‖x n ‖ + ‖gn−1 ‖1 < m ≤ lim n→∞ = 1.‖x n ‖Now assume ‖x n ‖ → ξ ≥ 0. If ξ > 0 we have the contradiction‖g n ‖ + ‖x n ‖ + ‖gn−1 ‖m ≤ lim n→∞ = 0 + ξ + 0 = 1.‖x n ‖ξThus ξ = 0. So we obtain x n → 0, and now deduce thatagain a contradiction.‖g n x n gn−1 ‖1 < m ≤ lim n→∞ = 1,‖x n ‖