Normed versus topological groups: Dichotomy and duality

Normed groups 29Theorem 3.4 (Equivalence Theorem). A normed group is a topological group under theright (resp. left) norm topology iff each conjugacyγ g (x) := gxg −1is right-to-right (resp. left-to-left) continuous at x = e (and so everywhere), i.e. forz n → R e and any ggz n g −1 → R e.(adm)Equivalently, it is a topological group iff left/right-shifts are continuous for the right/leftnorm topology, or iff the two norm topologies are themselves equivalent.In particular, if also the group structure is abelian, then the normed group is a topologicalgroup.Proof. Only one direction needs proving. We work with the d R topology, the right topology.By Theorem 3.3 we need only show that under it multiplication is jointly rightcontinuous.First we note that multiplication is right-continuous iffd R (xy, ab) = ‖xyb −1 a −1 ‖, as (x, y) → R (a, b).Here, we may write Y = yb −1 so that Y → R e iff y → R b, and we obtain the equivalentconditiond R (xY b, ab) = d R (xY, a) = ‖xY a −1 ‖, as (x, Y ) → R (a, e).By Theorem 3.3, as inversion is right-to-right continuous, Lemma 3.2 justifies re-writingthe second convergence condition with X = a −1 x and X → R e, yielding the equivalentconditiond R (aXY b, ab) = d R (aXY, a) = ‖aXY a −1 ‖, as (X, Y ) → R (e, e).But, by Lemma 3.1, this is equivalent to continuity of conjugacy.The final is related to a result of Żelazko [Zel] (cf. [Com, §11.6]). We will later applythe Equivalence Theorem several times in conjunction with the following result (see alsoLemma 3.34 for a strengthening).Lemma 3.5 (Weak continuity criterion). For fixed x, if for all null sequences w n , we haveγ x (w n(k) ) → e X down some subsequence 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 > NxB(δ)x −1 ⊂ B(ε).Suppose not. Then there is ε > 0 such that for each k = 1, 2, .. and each δ = 1/kthere is n = n(k) > k and w k with ‖w k ‖ < 1/k and ‖xw k x −1 ‖ > ε. So w k → 0. Byassumption, down some subsequence n(k) we have ‖xw n(k) x −1 ‖ → 0, but this contradicts‖xw n(k) x −1 ‖ > ε.