Normed versus topological groups: Dichotomy and duality

30 N. H. Bingham and A. J. OstaszewskiCorollary 3.6. For X a topological group under its norm, the left-shifts λ a (x) := axare bounded and uniformly continuous in norm.Proof. We have ‖λ a ‖ = ‖a‖ assup x d R (x, ax) = d R (e, a) = ‖a‖.We also haved R (ax, ay) = d R (axy −1 a −1 , e) = ‖γ a (xy −1 )‖.Hence, for any ε > 0, there is δ > 0 such that, for ‖z‖ < δ‖γ a (z)‖ ≤ ε.Thus provided d R (x, y) = ‖xy −1 ‖ < γ, we have d R (ax, ay) < ε.Remarks. 1 (Klee property). If the group has an abelian norm (in particular if thegroup is abelian), then the norm has the Klee property (see [Klee] for the original metricformulation, or Th. 2.18), and then it is a topological group under the norm-topology.Indeed the Klee property is that‖xyb −1 a −1 ‖ ≤ ‖xa −1 ‖ + ‖yb −1 ‖,and so if x → R a and y → R b, then xy → R ab. This may also be deduced from theobservation that γ g is continuous, since here‖gxg −1 ‖ = ‖gxeg −1 ‖ ≤ ‖gg −1 ‖ + ‖xe‖ = ‖x‖.Compare [vM2] Section 3.3, especially Example 3.3.6 of a topological group of real matriceswhich fails to have an abelian norm (see also [HJ, p.354] p.354).2. Theorem 3.4 may be restated in the language of commutators, introduced at the endof Section 2 (see Th. 2.20). These are of interest in Theorems 6.3, 10.7 and 10.9.Corollary 3.7. If the L-commutator is right continuous as a map from (X, d R ) to(X, d R ), then (X, d R ) is a topological group. The same conclusion holds for left continuityand for the R-commutator.Proof. Fix g. We will show that γ g is continuous at e; so let z n → e.First we work with the L-commutator and assume it to be, say right continuous, at e(which is equivalent to being left continuous at e, by Lemma 2.21). From the identityγ g (z n ) := gz n g −1 (z −1n z n ) = [g, z n ] L z n ,the assumed right continuity implies that w n := [g, z n ] L → e; but then w n z n → e, bythe triangle inequality. Thus γ g is continuous. By Theorem 3.4 (X, d R ) is a topologicalgroup.Next we work with the R-commutator and again assume that to be right continuous ate. Noting that [g, z n ] L = [g −1 , z −1n] L and z −1n→ e we may now interpet the previous