Normed versus topological groups: Dichotomy and duality

Normed groups 43Theorem 3.22 (Lipschitz property in H u ). Let X be a normed group with a rightinvariantmetric d X having a vanishingly small global word-net. Then‖h‖ g ≤ 2δ(g) ‖h‖, for g, h ∈ H u(X),and so H u (X) has the Lipschitz property.Proof. We have for d(z, z ′ ) < δ(g) thatd(g(z), g(z ′ )) < 1.For given x put y = h(x)x −1 . In the definition of the word-net take ε < 1. Now supposethat w(y) = w 1 ...w n(y) with ‖z i ‖ = 1 2 δ(1 + ε i) and |ε i | < ε, where n(y) = n(y, δ) satisfiesPut y 0 = e,1 − ε ≤ n(y)δ(g)‖y‖y i+1 = w i y i≤ 1 + ε.for 0 < i < n(y), and y n(x)+1 = y; the latter is within δ of y. Nowd(y i , y i+1 ) = d(e, w i ) = ‖w i ‖ < δ.Finally put z i = y i x, so that z 0 = x and z n(y)+1 = h(x). Aswe haveHenceThusd(z i , z i+1 ) = d(y i x, y i+1 x) = d(y i , y i+1 ) < δ,d(g(z i ), g(z i+1 )) ≤ 1.d(g(x), g(h(x))) ≤ n(y) + 1 < 2‖y‖/δ(g)= 2 d(h(x), x).δ(g)‖h‖ g = sup x d(g(x), g(h(x))) ≤ 2δ(g) sup x d(h(x), x) = 2δ(g) ‖h‖.Lemma 3.23 (Bi-Lipschitz property). In a Lipschitz-normed group M e = 1 and M g ≥ 1,for each g; moreover M gh ≤ M g M h and for any g and all x in G,1‖x‖ ≤ ‖x‖ g ≤ M g ‖x‖.M g −1Thus in particular ‖x‖ g is an equivalent norm.Proof. Evidently M e = 1. For g ≠ e, as γ g (g) = g, we see that‖g‖ = ‖g‖ g ≤ M g ‖g‖,