Normed versus topological groups: Dichotomy and duality

Normed groups 45Proof. Z γ = G (cf. Th. 2.18).The condition M g ≡ 1 is not necessary for the existence of an equivalent bi-invariantnorm, as we see below. The next result is similar to Th. 3.17 (where the Lipschitz propertyis absent).Theorem 3.26. Let G be a Lipschitz-normed topological group. If {M gbounded, then ‖x‖ ∞ is an equivalent abelian (hence bi-invariant) norm.: g ∈ G} isProof. Let M be a bound for the set {M g : g ∈ G}. Thus we have‖x‖ ∞ ≤ M‖x‖,and so ‖x‖ ∞ is again a norm. As we have‖x‖ = ‖x‖ e ≤ ‖x‖ ∞ ≤ M‖x‖,we see that ‖z n ‖ → 0 iff ‖z n ‖ ∞ → 0.Theorem 3.27. Let G be a compact, Lipschitz-normed, topological group. Then {M g :g ∈ G} is bounded, hence ‖x‖ ∞ is an equivalent abelian (hence bi-invariant) norm.Proof. The mapping |γ . | := g → log M g is subadditive. For G a compact metric group,|γ . | is Baire, since, by continuity of conjugacy,{g : a < M g < b} = proj 1 {(g, x) ∈ G 2 : ‖gxg −1 ‖ > a‖x‖} ∩ {g : ‖gxg −1 ‖ < b‖x‖},and so is analytic, hence by Nikodym’s Theorem (see [Jay-Rog, p. 42]) has the Baireproperty. As G is Baire, the subadditive mapping |γ . | is locally bounded (the proof ofProp. 1 in [BOst-GenSub] is applicable here; cf. Th. 4.4), and so by the compactness ofG, is bounded; hence Theorem 3.20 applies.Definition. Let G be a Lipschitz-normed topological group. PutM(g) : = {m : ‖x‖ g ≤ m‖x‖ for all x ∈ G}, and thenM g : = inf{m : m ∈ M(g)},µ(g) : = {m > 0 : m‖x‖ ≤ ‖x‖ g for all x ∈ G}, and thenm g : = sup{m : m ∈ µ(g)}.Proposition 3.28. Let G be a Lipschitz-normed topological group. Thenm −1g = M g −1.Proof. For 0 < m < m g we have for all x that‖x‖ ≤ 1 m ‖gxg−1 ‖.