Normed versus topological groups: Dichotomy and duality

Normed groups 39Theorem 3.19 (Abelian normability of H(X) – cf. [BePe, Ch. IV Th 1.1]). For X anormed group, assume that ‖f‖ ∞ is finite for each f in H(X)– for instance if d X isbounded, and in particular if X is compact.Then:(i) H(X) under the abelian norm ‖f‖ ∞ is a topological group.(ii) The norm ‖f‖ ∞ is equivalent to ‖f‖ H iff the admissibility condition (n-adm) holds,which here reads: for ‖f n ‖ H → 0 and any g n in H(X),‖g n f n g −1n ‖ H → 0.Equivalently, for ‖z n ‖ H → 0 (i.e. z n converging to the identity), any g n in H(X), andany y n ∈ X,‖g n (z n (y n ))g n (y n ) −1 ‖ X → 0.(iii) In particular, if X is compact, H(X) = H u (X) is under ‖f‖ H a topological group.Proof. (i) and the first part of (ii) follow from Th. 3.17 (cf. Remarks 1 on the Kleeproperty, after Cor. 3.6); as to (iii), this follows from Th. 3.14 and 3.9. Turning to thesecond part of (ii), suppose first that‖g n z n g −1n ‖ H → 0,and let y n be given. For any ε > 0 there is N such that, for n ≥ N,ε > ‖g n z n g −1nTaking x here as x n = g n (y n ), we obtain‖ H = sup x d(g n z n gn−1 (x), x).ε > d(g n (z n (y n )), g n (y n )) = d(g n z n (y n )g n (y n ) −1 , e X ), for n ≥ N.Hence ‖g n (z n (y n ))g n (y n ) −1 ‖ X → 0, as asserted.For the converse direction, suppose next that‖g n z n g −1n ‖ H ↛ 0.Then without loss of generality there is ε > 0 such that for all nHence, for each n, there exists x n such that‖g n z n gn−1 ‖ H = sup x d(g n z n gn −1 (x), x) > ε.d(g n z n g −1n (x n ), x n ) > ε.Equivalently, setting y n = gn−1 (x n ) we obtaind(g n (z n (y n ))g n (y n ) −1 , e X ) = d(g n (z n (y n )), g n (y n )) > ε.Thus for this sequence y n we have‖g n (z n (y n ))g n (y n ) −1 ‖ X ↛ 0.