Normed versus topological groups: Dichotomy and duality

20 N. H. Bingham and A. J. OstaszewskiThus d X (z n y n , y n ) → 0, as required.(iii) If d X is right-invariant, then d X (z n y n , y n ) = d X (z n , e) → 0 and the admissibilitycondition holds on H. Of course ‖λ x ‖ H = sup z d X (xz, z) = d X (x, e) = ‖x‖ X and soH = X.(iv) If d X is right-invariant, then ¯d H (x, y) := sup z d X (xz, yz) = d X (x, y).(v) If X is compact, then H = H X as z → d X (xz, z) is continuous. If z n → e and y n arearbitrary, suppose that the admissibility condition fails. Then for some ε > 0 we havewithout loss of generalityd X (z n y n , y n ) ≥ ε.Passing down a subsequence y m → y and assuming that X is a topological group weobtain0 = d X (ey, y) ≥ ε,a contradiction.As a corollary we obtain the following known result ([HR, 8.18]; cf. Theorem 3.3.4 in[vM2] p. 101, for a different proof).Proposition 2.15. In a first-countable topological group X the (topological admissibility)condition yn−1 z n y n → e on X as z n → e is equivalent to the existence of an abelian norm(equivalently, a bi-invariant metric).Proof. We shall see below in the Birkhoff-Kakutani Theorem (Th.2. 19) that the topologyof X may be induced by a left-invariant metric, d X L say; we may assume without lossof generality that it is bounded (take d = max{d X L , 1}, which is also left-invariant, cf.Example A6 towards the start of this Section). Then H X = X, and the assumed topologicaladmissibility condition yn−1 z n y n → e on X implies (H-adm), the metric admissibilitycondition on H for d X L . The metric dX L thus induces the norm ‖x‖ H, which is abelian, andin turn, by Proposition 2.3, defines an equivalent bi-invariant metric on X. Conversely,if the norm ‖.‖ X is abelian, then the topological admissibility condition follows from theobservation that‖yn−1 z n y n ‖ = ‖y n yn −1 z n ‖ = ‖z n ‖ → 0.Application. Let S, T be normed groups. For α : S → T an arbitrary function wedefine the possibly infinite number‖α‖ := sup{‖α(s)‖ T /‖s‖ S : s ∈ S} = inf{M : ‖α(s)‖ ≤ M‖s‖ (∀s ∈ S)}.α is called bounded if ‖α‖ is finite. The bounded functions form a group G under thepointwise multiplication (αβ)(t) = α(t)β(t). Clearly ‖α‖ = 0 implies that α(t) = e, forall t. Symmetry is clear. Also‖α(t)β(t)‖ ≤ ‖α(t)‖ + ‖β(t)‖ ≤ [‖α‖ + ‖β‖]‖t‖,