Normed versus topological groups: Dichotomy and duality

Normed groups 107Proof. Say f is bounded above in B := B δ (x 0 ) by M. Consider u ∈˜d(x 0 , u) = ‖u −1 x 0 ‖ < δ. Put t = u −1 x 2 0; then tx −10 = u −1 x 0 , and soThen t ∈ B, and since x 2 0 = ut we haveord(t, x 0 ) = ‖tx −10 ‖ = ‖u−1 x 0 ‖ = ˜d(u, x 0 ) < δ.2f(x 0 ) ≤ f(u) + f(t) ≤ f(u) + M,f(u) ≥ 2f(x 0 ) − M.Thus 2f(x 0 ) − M is a lower bound for f on the open set ˜B δ (x 0 ).˜B δ (x 0 ). ThusAs a corollary a suitably rephrased Bernstein-Doetsch Theorem ([Kucz], [BOst-Aeq])is thus true.Theorem 10.6 (Bernstein-Doetsch Theorem). In a normed group, for fa 1 2-convex function,if f is locally bounded above at x 0 , then f is continuous at at x 0 .Proof. We repeat the ‘Second proof’ of [Kucz, p. 145]. Choose y n → x 0 with f(y n ) →m f (x 0 ) and z n → x 0 with f(z n ) → M f (x 0 ). Let u n :=ynx 2 −1n . Thus yn 2 = u n x n and so2f(y n ) ≤ f(u n ) + f(z n ),i.e. f(u n ) ≥ 2f(y n ) − f(z n ). Hence in the limit we obtainM f (x 0 ) ≥ lim inf f(u n ) ≥ 2M f (x 0 ) − m f (x 0 ).One thus has that M f (x 0 ) ≤ m f (x 0 ). But m f (x 0 ) ≤ f(x 0 ) ≤ M f (x 0 ), and both hullvalues are finite (by the result above). Thus m f (x 0 ) = f(x 0 ) = M f (x 0 ), from whichcontinuity follows.We now consider the transferability of upper and lower local boundedness. Our proofswork directly with definitions (so are not modelled after those in Kuczma [Kucz]). Wedo not however consider domains other than the whole metric group. For clarity of proofstructure we give separate proofs for the two cases, first when Gis abelian and later forgeneral G.Theorem 10.7 (Local upper boundedness). In a normed topological group G, for fa12 -convex function defined on G, if f is locally bounded above at some point x 0, then f islocally bounded above at all points.Proof. Case (i) The Abelian case. Say f is bounded above in B := B δ (x 0 ) by M. Givena fixed point t, put z = z t := x −10 t2 , so that t 2 = x 0 z. Consider any u ∈ B δ/2 (t).Write u = st with ‖s‖ < δ/2. Now put y = s 2 ; then ‖y‖ = ‖s 2 ‖ ≤ 2‖s‖ < δ. Henceyx 0 ∈ B δ (x 0 ). Nowu 2 = (st) 2 = s 2 t 2 = yx 0 z,