Normed versus topological groups: Dichotomy and duality

98 N. H. Bingham and A. J. OstaszewskiNow we argue as in [BGT] page 62-3, though with a normed group as the domain. ChooseX 1 and κ > max{M, 1} such thath(ux)h(x) −1 < κ (u ∈ K, ‖x‖ ≥ X 1 ).Fix v. Now there is some word w(v) = w 1 ...w m(v) using generators in the compact set Z δwith ‖w i ‖ = δ(1 + ε i ) < 2δ, as |ε i | < 1 (so ‖w i ‖ < 2δ < ε), whereandand sod(v, w(v)) < δ1 − ε ≤ m(v)δ‖x‖ ≤ 1 + ε,m + 1 < 2 ‖v‖ + 1 < A‖v‖ + 1, where A = 2/δ.δPut w m+1 = w −1 v, v 0 = e, and for k = 1, ...m + 1,v k = w 1 ...w k ,so that v m+1 = v. Now (v k+1 x)(v k x) −1 = w k+1 ∈ K. So for ‖x‖ ≥ X 1 we have(for large enough ‖v‖), whereIndeed, for ‖v‖ > log κ, we haveh(vx)h(x) −1 = ∏ m+1[h(v kx)h(v k−1 x)] −1k=1≤ κ m+1 ≤ ‖v‖ KK = (A log κ + 1).(m + 1) log κ < (A‖v‖ + 1) log κ < ‖v‖(A log κ + (log κ)‖v‖ −1 ) < log ‖v‖(A log κ + 1).For x 1 with ‖x 1 ‖ ≥ M and with t such that ‖tx −11 ‖ > L, take u = tx−1 1 ; then since‖u‖ > L we havei.e.h(ux 1 )h(x 1 ) −1 = h(t)h(x 1 ) −1 ≤ ‖u‖ K = ‖tx −11 ‖K ,h(t) ≤ ‖tx −11 ‖K h(x 1 ),so that h(t) is bounded away from ∞ on compact t-sets sufficiently far from the identity.Remarks. 1. The one-sided result in Th. 6.11 can be refined to a two-sided one (asin [BGT, Cor. 2.0.5]): taking s = t −1 , g(x) = h(x) −1 for h : X → R + , and using thesubstitution y = tx, yieldsg ∗ (s) = sup ‖y‖→∞ g(sy)g(y) −1 = inf ‖x‖→∞ h(tx)h(x) −1 = h ∗ (s).2. A variant of Th. 7.11 holds with ‖h(ux)h(x) −1 ‖ Y replacing h(ux)h(x) −1 .3. Generalizations of Th. 7.11 along the lines of [BGT] Theorem 2.0.1 may be given for h ∗finite on a ‘large set’ (rather than globally bounded), by use of the Semigroup Theorem(Th. 9.5).