Normed versus topological groups: Dichotomy and duality

Normed groups 37We now give an explicit construction of a equivalent bi-invariant metric on G whenone exists (compare [HR, Section 8.6]), namely‖x‖ ∞ := sup{‖x‖ g : g ∈ G}.We recall from Section 2 that the group-norm satisfies the norm admissibility condition(on X) if, for z n → e and g n arbitrary,‖g n z n g −1n ‖ G → 0. (n-adm)Evidently in view of the sequence {g n }, this is a sharper version of (adm).Theorem 3.17. For G with group-norm ‖.‖ G , suppose that ‖.‖ ∞ is finite on G. Then‖x‖ ∞ is an equivalent norm iff the ‖.‖ G meets the norm admissibility condition (n-adm).In particular, for |x| := min{‖x‖, 1} the corresponding norm |x| ∞ := sup{|x| g : g ∈ G}is an equivalent abelian norm iff the admissibility condition (n-adm) holds.Proof. First assume (n-adm) holds. As ‖x‖ = ‖x‖ e ≤ ‖x‖ ∞ we need to show that ifz n → e, then ‖z n ‖ ∞ → 0. Suppose otherwise; then for some ε > 0, without loss ofgenerality ‖z n ‖ ∞ ≥ ε, and so there is for each n an element g n such that‖g n z n gn−1 ‖ ≥ ε/2.But this contradicts the admissibility condition (n-adm).As to the abelian property of the norm, we have‖yzy −1 ‖ ∞ = sup{‖gyzy −1 g −1 ‖ : g ∈ G} = sup{‖gyz(gy) −1 ‖ : g ∈ G} = ‖z‖ ∞ ,and so taking z = xy we have ‖yx‖ = ‖xy‖.For the converse, assume ‖x‖ ∞ is an equivalent norm. For g n arbitrary, suppose that‖z n ‖ → 0 and ε > 0. For some N and all n ≥ N we thus have ‖z n ‖ ∞ < ε. Hence forn ≥ N,‖g n z n g −1n ‖ ≤ ‖z n ‖ ∞ < ε,verifying the condition (n-adm).Theorem 3.18. Let G be a normed topological group which is compact under its norm‖.‖ G . Then‖x‖ ∞ := sup{‖x‖ g : g ∈ G}is an abelian (hence bi-invariant) norm topologically equivalent to ‖x‖.Proof. We write ‖.‖ for ‖.‖ G . Suppose, for some x, that {‖x‖ g : g ∈ G} is unbounded.We may select g n with‖g n xg −1n ‖ → ∞.Passing to a convergent subsequence we obtain a contradiction. Thus ‖x‖ ∞ is finite andhence a norm. We verify the admissibility condition. Suppose to the contrary that forsome z n → e, arbitrary g n , and some ε > 0 we have‖g n z n g −1n ‖ > ε.