Normed versus topological groups: Dichotomy and duality

Normed groups 109Thus G is locally convex at e if for any ε > 0 there is δ > 0 such that for all g with‖g‖ < ε the equationx 2 = ghas its solutions with ‖x‖ < δ.Remark. Putting u = xt the local convexity equation reduces to u 2 = gt 2 , assertingthe local existence of square roots (local 2-divisibility). If G is abelian the condition at treduces to the condition at e.Theorem 10.9 (Local lower boundedness). Let G be a locally convex group with a 2-Lipschitz norm, i.e. g → g 2 is a bi-Lipschitz isomorphism such that, for some κ > 0,κ‖g‖ ≤ ‖g 2 ‖ ≤ 2‖g‖.For f a 1 2-convex function, if f is locally bounded below at some point, then f is locallybounded below at all points.Proof. Note that by Th. 3.39 the normed group is topological.Case (i) The Abelian case. We change the roles of t and x 0 in the preceeding abeliantheorem, treating tas a reference point, albeit now for lower boundedness, and x 0 assome arbitrary other fixed point. Suppose that f is bounded below by L on B δ (t). Letyx 0 ∈ B κδ (x 0 ), so that 0 < ‖y‖ < κδ. Choose s such that s 2 = y. Then,κ‖s‖ ≤ ‖y‖ < κδ,so ‖s‖ < δ. Thus u = st ∈ B δ (t). Now the identity u 2 = s 2 t 2 = yx 0 z implies thatL ≤ f(u) ≤ 1 2 f(yx 0) + 1 2 f(z t),2L − f(z t ) ≤ f(yx 0 ),i.e. that 2L − f(z t ) is a lower bound for f on B κδ (x 0 ).Case (ii) The general case. Suppose as before that f is bounded below by L on B δ (t).Since the map α(σ) := σtσt −1 σ −2 is continuous (cf. again Th. 3. 7 on commutators) andα(e) = e, we may choose η such that ‖α(σ)‖ < κδ/2, for ‖σ‖ < η. Now choose ε > 0 suchthat, for each y with ‖y‖ < ε, the solution u = σt tou 2 = yt 2has ‖σ‖ < η. Let r = min{κδ/2, ε}.Let yx 0 ∈ B r (x 0 ); then 0 < ‖y‖ < κδ/2 and ‖y‖ < ε. As before put z = z t := x −10 t2 sothat t 2 = x 0 z. Consider u = σt such that u 2 = yx 0 z; thus we haveHence ‖σ‖ < η (as ‖y‖ < ε). Now we writeWe compute thatu 2 = σtσt = yx 0 z = yx 0 x −10 t2 = yt 2 .u 2 = σtσt = [σtσt −1 σ −2 ]σ 2 t 2 = α(σ)σ 2 t 2 = yt 2 .y = α(σ)σ 2