Normed versus topological groups: Dichotomy and duality

68 N. H. Bingham and A. J. OstaszewskiLemma 4.2 (cf. [Kucz, L. 2 p. 403]). For a normed group G, if p : G → R is subadditive,thenm p (x) ≤ M p (x) and M p (x) − m p (x) ≤ M p (e).Proof. Only the second assertion needs proof. For a > m p (x) and b < M p (x), there existu, v ∈ B r (x) witha > p(u) ≥ m p (x), and b < p(v) ≤ M p (x).SoNowb − a < p(v) − p(u) ≤ p(vu −1 u) − p(u) ≤ p(vu −1 ) + p(u) − p(u) = p(vu −1 ).so vu −1 ∈ B 2r (e), and hence‖vu −1 ‖ ≤ ‖v‖ + ‖u‖ < 2r,p(vu −1 ) ≤ sup{p(z) : z ∈ B 2r (e)}.Hence, with r fixed, taking a, b to their respective limits,M p (x) − m p (x) ≤ sup{p(z) : z ∈ B 2r (e)}.Taking limits as r → 0+, we obtain the second inequality.Lemma 4.3. For a normed group G and any subadditive function f : G → R, if f islocally bounded above at a point, then it is locally bounded at every point.Proof. We repeat the proof in [Kucz, Th. 2, p.404], thus verifying that it continues tohold in a normed group.Suppose that p is locally bounded above at t 0 by K. We first show that f is locallybounded above at e. Suppose otherwise that for some t n → e we have p(t n ) → ∞. Nowt n t 0 → et 0 = t 0 , and sop(t n ) = p(t n t 0 t −10 ) ≤ p(t nt 0 ) + p(t −10 ) ≤ K + p(t−1 0 ),a contradiction. Hence p is locally bounded above at e, i.e. M p (e) < ∞. But 0 ≤ M p (x) −m p (x) ≤ M p (e), hence both M p (x) and m p (x) are finite for every x. That is, p is locallybounded above and below at each x.The next result requires that both f(x) and f(x −1 ) be Baire functions; this happensfor instance when (i) f is even, i.e. f(x) = f(x −1 ), with f(x) := ‖gxg −1 ‖ an example ofsome interest here (cf. Th. 3.27 and in connection with the oscillation function of Section3.3), and (ii) both f(x) and x → x −1 are Baire, so that the normed group is a topologicalgroup (Th. 3.41).Proposition 4.4 ([Kucz, Th. 3, p. 404]). For a topologically complete normed group Gand a Baire function f : G → R with x → f(x −1 ) Baire, if f is subadditive, then f islocally bounded.