Normed versus topological groups: Dichotomy and duality

52 N. H. Bingham and A. J. OstaszewskiProof. For the direct implication, suppose otherwise. Then for some Cauchy {x n } andsome ε > 0 and each δ = 1/k (k = 1, 2, ...) there is n = n(k) > k and w k with ‖w k ‖ < 1/ksuch that‖x n(k) w k x −1n(k) ‖ > ε.But w k is null and x n(k) is Cauchy, so from (C-adm) it follows that x n(k) w k x −1n(k) → e, acontradiction.The converse is immediate.Definition. For an arbitrary sequence x n , puttingγ x(n) (w) := x n wx −1n ,we say that {γ x(n) } is uniformly continuous at e if the uniformity condition of Lemma3.33 holds. Thus that Lemma may be interpreted as asserting that {γ x(n) } is uniformlycontinuous at e for all Cauchy {x n } iff (C-adm) holds.Our next result strengthens Lemma 3.5 in showing that for inner automorphisms aweak form of continuity implies continuity.Lemma 3.34 (Weak Continuity Criterion). For any fixed sequence x n , if for all nullsequences w n we have γ x(n(k)) (w n(k) ) → e X down some subsequence w n(k) , then {γ x(n) }is uniformly continuous at e.In particular, for a fixed x and all null sequences w n , if γ x (w n(k) ) → e X down somesubsequence w n(k) , then γ x is continuous.Proof. We are to show that for every ε > 0 there is δ > 0 and N such that for all n > Nx n B(δ)x −1n⊂ B(ε).Suppose not. Then there is ε > 0 such that for each k = 1, 2, .. and each δ = 1/k thereis n = n(k) > k and w k with ‖w k ‖ < 1/k and ‖x n(k) w k x −1n(k) ‖ > ε. So w k → 0. Byassumption, down some subsequence k(h) we have ‖x n(k(h)) w k(h) x −1n(k(h))‖ → 0. But thiscontradicts ‖x n(k(h)) w k(h) x −1n(k(h)) ‖ > ε.The last assertion is immediate from taking x n ≡ x, as the uniform continuity conditionat e reduces to continuity at e.Theorem 3.35. In a normed group, the condition (C-adm) holds iff the product ofCauchy sequences is Cauchy.Proof. We work in the right norm topology and refer to d R -Cauchy sequences.First we assume (C-adm). Let x n and y n be Cauchy. For m, n large we are to show thatd R (x n y n , x m y m ) = ‖x n y n ym −1 x −1m ‖ is small. We note that‖x n wx −1m ‖ = ‖x n wx −1n x n x −1m ‖ ≤ ‖x n wx −1n ‖ + ‖x n x −1m ‖≤ ‖x n wx −1n ‖ + d R (x n , x m ).