Normed versus topological groups: Dichotomy and duality

128 N. H. Bingham and A. J. OstaszewskiBy contrast we have‖f‖ ∞ = sup z sup g d X g (f(z), z).However, for f(z) = λ x (z) := xz, putting s = g ◦ ρ z brings the the two formulas intoalignment, as‖λ x ‖ ∞ = sup z sup g d X (g(xz), g(z)) = sup z sup g d X (g(ρ z (x)), g(ρ z (e))).This motivates the following result.Proposition 12.2. The subgroup H X := {x ∈ X : ‖x‖ ∞ < ∞} equipped with the norm‖x‖ ∞ embeds isometrically under ξ into H u (H u (X)) asΞ H := {ξ x : x ∈ H X }.Proof. Writing y = x −1 z or z = xy, we haveHenced H (ξ x (s), s) = sup z∈X d X (s(x −1 z), s(z)) = sup y∈X d X (s(y), s(xy))= sup y∈X d X s (ρ y e, ρ y x) = sup y d X s-y(e, x).‖ξ x ‖ H = sup s∈H(X) d H (ξ x (s), s) = ‖λ x ‖ ∞ = sup s∈H(X) sup y∈X d X s (y, xy) = ‖x‖ ∞ .Thus for x ∈ H X the map ξ x is bounded over H u (X) and hence is in H u (H u (X)).The next result adapts ideas of Section 3 on the Lipschitz property in H u (Th. 3.22)to the context of ξ x and refers to the inverse modulus of continuity δ(s) which we recall:δ(g) = δ 1 (g) := sup{δ > 0 : d X (g(z), g(z ′ )) ≤ 1, for all d X (z, z ′ ) ≤ δ}.Proposition 12.3 (Further Lipschitz properties of H u ). Let X be a normed group witha vanishingly small global word-net. Then for x, z ∈ X and s ∈ H u (X) the s-z-shiftednorm (recalled below) satisfiesHence‖x‖ s-z := d X s-z(x, e) = d X (s(z), s(xz)) ≤ 2‖x‖/δ(s).‖ξ e ‖ H(Hu (X)) = sup s∈Hu(X) sup z∈X ‖e‖ s-z = 0,and so ξ e ∈ H(H u (X)). Furthermore, if {δ(s) : s ∈ H u (X)} is bounded away from 0,then for x ∈ X‖ξ x ‖ H(Hu (X)) = sup s∈Hu(X) d H(X) (ξ x (s), s) = sup s∈Hu(X) sup z∈X d X (s(x −1 z), s(z))≤ 2‖x‖/ inf{δ(s) : s ∈ H u (X)},and so ξ x ∈ H(H u (X)).In particular this is so if in addition X is compact.Proof. Writing y = x −1 z or z = xy, we haved H (ξ x (s), s) = sup z∈X d X (s(x −1 z), s(z)) = sup y∈X d X (s(y), s(xy)).