Normed versus topological groups: Dichotomy and duality

Normed groups 47Theorem 3.30. 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. (ne) holds.In particular, if in addition G is compact and condition (ne) holds, then {M g : g ∈ G}is bounded, and so again Theorem 3.24 applies, confirming that ‖x‖ ∞ is an equivalentabelian (hence bi-invariant) norm.Proof. Clearly (i)=⇒ (ii). To prove (ii)=⇒ (i), given continuity at e, we prove continuityat h as follows. Write g = hk; then h = gk −1 and g → h iff k → e iff k −1 → e. Now byLemma 3.23,M h = M gk −1 ≤ M g M k −1,so since M k −1 → M e = 1, we haveSince M k → M e = 1 andwe also haveM h ≤ lim g→h M g .M g = M hk ≤ M h M k ,lim g→h M g ≤ M h .Next we show that (ii)=⇒(iii). By Lemma 3.23, we haveBy assumption, M gn1/M g−1n→ M e = 1 and M g−1n≤ ‖g nz n g −1n ‖/‖z n ‖ ≤ M gn .→ M e = 1, solim n ‖g −1n z n g n ‖/‖z n ‖ = 1.Finally we show that (iii)=⇒(ii). Suppose that the mapping is not continuous at e. AsM e = 1 and M g ≥ 1, for some ε > 0 there is g n → e such that M gn > 1 + ε. Hence thereare x n ≠ e with(1 + ε)‖x n ‖ ≤ ‖g n x n gn−1 ‖.Suppose that ‖x n ‖ is unbounded. We may suppose that ‖x n ‖ → ∞. Hence(1 + ε) ≤ ‖g nx n gn−1 ‖‖x n ‖and so as ‖g n ‖ → 0 and ‖x n ‖ → ∞ we have(1 + ε) ≤ lim n→∞( ‖gn ‖ + ‖x n ‖ + ‖g n ‖‖x n ‖≤ ‖g n‖ + ‖x n ‖ + ‖gn−1‖x n ‖‖,) (= lim n→∞ 1 + 2 )‖x n ‖ · ‖g n‖ = 1,again a contradiction. We may thus now suppose that ‖x n ‖ is bounded and so withoutloss of generality convergent, to ξ ≥ 0 say. If ξ > 0, we again deduce the contraditionthat‖g n ‖ + ‖x n ‖ + ‖gn−1 ‖(1 + ε) ≤ lim n→∞ = 0 + ξ + 0 = 1.‖x n ‖ξ