Normed versus topological groups: Dichotomy and duality

Normed groups 135Definition. Let {ψ u : u ∈ I} for I an open interval be a family of homeomorphisms(cf. [Mon2]). Let u 0 ∈ I. Say that ψ u converges to the identity as u → u 0 iflim u→u0 ‖ψ u ‖ = 0.This property is preserved under topological conjugacy; more precisely we have thefollowing result, whose proof is routine and hence omitted.Lemma 13.2. Let σ ∈ H unif (X) be a homeomorphism which is uniformly continuouswith respect to d X , and write u 0 = σz 0 .If {ψ z : z ∈ B ε (z 0 )} converges to the identity as z → z 0 , then as u → u 0 so does theconjugate {ψ u = σψ z σ −1 : u ∈ B ε (u 0 ), u = σz}.Lemma 13.3. Suppose that the homeomorphisms {ϕ n } are uniformly divergent, {ψ n } areconvergent and σ is bounded, i.e. is in H(X). Then {ϕ n σ} is uniformly divergent andlikewise {σϕ n }. In particular {ϕ n ψ n } is uniformly divergent, and likewise {ϕ n σψ n }, forany bounded homeomorphism σ ∈ H(X).Proof. Consider s := ‖σ‖ = sup d(σ(x), x) > 0. For any M, from some n onwards wehaved ∗ (ϕ n , id) = inf x∈X d(ϕ n (x), x) > M,i.e.d(ϕ n (x), x) > M,for all x. For such n, we have d ∗ (ϕ n σ, id) > M − s, i.e. for all t we haved(ϕ n (σ(t)), t)) > M − s.Indeed, otherwise at some t this last inequality is reversed, and thend(ϕ n (σ(t)), σ(t)) ≤ d(ϕ n (σ(t)), t) + d(σ(t), t)≤ M − s + s = M.But this contradicts our assumption on ϕ n with x = σ(t). Hence d ∗ (ϕ n σ, id) > M − s forall large enough n.The other cases follow by the same argument, with the interpretation that now s > 0 isarbitrary; then we have for all large enough n that d(ψ n (x), x) < s, for all x.Remark. Lemma 13.3 says that the filter of sets (countably) generated from the sets{ϕ|ϕ : X → X is a homeomorphism and ‖ϕ‖ ≥ n}is closed under composition with elements of H(X).We now return to the notion of divergence.