28 N. H. Bingham <strong>and</strong> A. J. OstaszewskiLemma 3.2. If inversion is right-to-right continuous, thenx → R a iff a −1 x → R e.Proof. For x → R a, we have d R (e, a −1 x) = d R (x −1 , a −1 ) → 0, assuming continuity.Conversely, for a −1 x → R e we have d R (a −1 x, e) → 0, i.e. d R (x −1 , a −1 ) → 0. So sinceinversion is assumed to be right-continuous <strong>and</strong> (x −1 ) −1 = x, etc, we have d R (x, a) → 0.We now exp<strong>and</strong> this.Theorem 3.3. The following are equivalent:(i) inversion is right-to-right continuous,(ii) left-open sets are right-open,(iii) for each g the conjugacy γ g is right-to-right continuous at e, i.e. for every ε > 0there is δ > 0 such thatgB(δ)g −1 ⊂ B(ε),(iv) left-shifts are right-continuous.Proof. We show that (i)⇐⇒(ii)⇐⇒(iii)⇐⇒(iv).Assume (i). For any a <strong>and</strong> any ε > 0, by continuity of inversion at a, there is δ > 0 suchthat, for x with d R (x, a) < δ, we have d R (x −1 , a −1 ) < ε, i.e. d L (x, a) < ε. ThusB(δ)a = B R (a, δ) ⊂ B L (a, ε) = aB(ε),(incl)i.e. left-open sets are right-open, giving (ii). For the converse, we just reverse the lastargument. Let ε > 0. As a ∈ B L (a, ε) <strong>and</strong> B L (a, ε) is left open, it is right open <strong>and</strong> sothere is δ > 0 such thatB R (a, δ) ⊂ B L (a, ε).Thus for x with d R (x, a) < δ, we have d L (x, a) < ε, i.e. d R (x −1 , a −1 ) < ε, i.e. inversionis right-to-right continuous, giving (i).To show that (ii)⇐⇒(iii) note that the inclusion (incl) is equivalent toi.e. toa −1 B(δ)a ⊂ B(ε),γa−1 [B(δ)] ⊂ B(ε),that is, to the assertion that γ a (x) is continuous at x = e (<strong>and</strong> so continuous, by Lemma3.1). The property (iv) is equivalent to (iii) since the right-shift is right-continuous <strong>and</strong>γ a (x)a = λ a (x) is equivalent to γ a (x) = λ a (x)a −1 .We saw in the Birkhoff-Kakutani Theorem (Th. 2.19) that metrizable <strong>topological</strong><strong>groups</strong> are normable (equivalently, have a right-invariant metric); we now formulate aconverse, showing when the right-invariant metric derived from a group-norm equips itsgroup with a <strong>topological</strong> group structure. As this is a characterization of metric <strong>topological</strong><strong>groups</strong>, we will henceforth refer to them synonymously as normed <strong>topological</strong> <strong>groups</strong>.
<strong>Normed</strong> <strong>groups</strong> 29Theorem 3.4 (Equivalence Theorem). A normed group is a <strong>topological</strong> group under theright (resp. left) norm topology iff each conjugacyγ g (x) := gxg −1is right-to-right (resp. left-to-left) continuous at x = e (<strong>and</strong> so everywhere), i.e. forz n → R e <strong>and</strong> any ggz n g −1 → R e.(adm)Equivalently, it is a <strong>topological</strong> group iff left/right-shifts are continuous for the right/leftnorm topology, or iff the two norm topologies are themselves equivalent.In particular, if also the group structure is abelian, then the normed group is a <strong>topological</strong>group.Proof. Only one direction needs proving. We work with the d R topology, the right topology.By Theorem 3.3 we need only show that under it multiplication is jointly rightcontinuous.First we note that multiplication is right-continuous iffd R (xy, ab) = ‖xyb −1 a −1 ‖, as (x, y) → R (a, b).Here, we may write Y = yb −1 so that Y → R e iff y → R b, <strong>and</strong> we obtain the equivalentconditiond R (xY b, ab) = d R (xY, a) = ‖xY a −1 ‖, as (x, Y ) → R (a, e).By Theorem 3.3, as inversion is right-to-right continuous, Lemma 3.2 justifies re-writingthe second convergence condition with X = a −1 x <strong>and</strong> X → R e, yielding the equivalentconditiond R (aXY b, ab) = d R (aXY, a) = ‖aXY a −1 ‖, as (X, Y ) → R (e, e).But, by Lemma 3.1, this is equivalent to continuity of conjugacy.The final is related to a result of Żelazko [Zel] (cf. [Com, §11.6]). We will later applythe Equivalence Theorem several times in conjunction with the following result (see alsoLemma 3.34 for a strengthening).Lemma 3.5 (Weak continuity criterion). For fixed x, if for all null sequences w n , we haveγ x (w n(k) ) → e X down some subsequence w n(k) , then γ x is continuous.Proof. We are to show that for every ε > 0 there is δ > 0 <strong>and</strong> N such that for all n > NxB(δ)x −1 ⊂ B(ε).Suppose not. Then there is ε > 0 such that for each k = 1, 2, .. <strong>and</strong> each δ = 1/kthere is n = n(k) > k <strong>and</strong> w k with ‖w k ‖ < 1/k <strong>and</strong> ‖xw k x −1 ‖ > ε. So w k → 0. Byassumption, down some subsequence n(k) we have ‖xw n(k) x −1 ‖ → 0, but this contradicts‖xw n(k) x −1 ‖ > ε.
- Page 1 and 2: N. H. BINGHAM and A. J. OSTASZEWSKI
- Page 3 and 4: Normed groups 3ContentsContents . .
- Page 5 and 6: 1. IntroductionGroup-norms, which b
- Page 7 and 8: Normed groups 3Topological complete
- Page 9 and 10: Normed groups 5abelian group has se
- Page 11 and 12: Normed groups 74 (Topological permu
- Page 13 and 14: Normed groups 9The following result
- Page 15 and 16: Normed groups 11Corollary 2.4. For
- Page 17 and 18: Normed groups 13More generally, for
- Page 19 and 20: Normed groups 15definitions, our pr
- Page 21 and 22: Normed groups 17so that fg is in th
- Page 23 and 24: Normed groups 19(iii) The ¯d H -to
- Page 25 and 26: Normed groups 21so‖αβ‖ ≤
- Page 27 and 28: Normed groups 23Remark. Note that,
- Page 29 and 30: Normed groups 25shows that [z n , y
- Page 31: Normed groups 27Denoting this commo
- Page 35 and 36: Normed groups 31argument as again p
- Page 37 and 38: Normed groups 33(ii) For α ∈ H u
- Page 39 and 40: Normed groups 35Definition. A group
- Page 41 and 42: Normed groups 37We now give an expl
- Page 43 and 44: Normed groups 39Theorem 3.19 (Abeli
- Page 45 and 46: Normed groups 412. Further recall t
- Page 47 and 48: Normed groups 43Theorem 3.22 (Lipsc
- Page 49 and 50: Normed groups 45Proof. Z γ = G (cf
- Page 51 and 52: Normed groups 47Theorem 3.30. Let G
- Page 53 and 54: Normed groups 49Remark. On the matt
- Page 55 and 56: Normed groups 51As for the conclusi
- Page 57 and 58: Normed groups 53By (C-adm), we may
- Page 59 and 60: Normed groups 55equipped with an in
- Page 61 and 62: Normed groups 57Proof. To apply Th.
- Page 63 and 64: Normed groups 59Definition. A point
- Page 65 and 66: Normed groups 61Proposition 3.46 (M
- Page 67 and 68: Normed groups 63Thus ω δ (s) ≤
- Page 69 and 70: Normed groups 65Remark. In the penu
- Page 71 and 72: Normed groups 67The result confirms
- Page 73 and 74: Normed groups 69Proof. By the Baire
- Page 75 and 76: Normed groups 715. Generic Dichotom
- Page 77 and 78: Normed groups 73Returning to the cr
- Page 79 and 80: Normed groups 75Examples. Here are
- Page 81 and 82: Normed groups 77cf. [Eng, 4.3.23].)
- Page 83 and 84:
Normed groups 79Remarks. 1. See [Fo
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Normed groups 81Theorem 6.1 (Catego
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Normed groups 83is continuous at th
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Normed groups 85compact. Evidently,
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Normed groups 87j ∈ ω} which enu
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Normed groups 89The result below ge
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Normed groups 91left-shift, not in
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Normed groups 93As a corollary of t
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Normed groups 953. For X a normed g
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Normed groups 97Proof. Note that‖
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Normed groups 99Taking h(x) := ‖
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Normed groups 1019. The Semigroup T
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Normed groups 103Theorem 9.5 (Semig
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Normed groups 105By the Category Em
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Normed groups 107Proof. Say f is bo
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Normed groups 109Thus G is locally
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Normed groups 111Theorem 10.10 (Bar
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Normed groups 113K-analyticity was
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Normed groups 115Theorem 11.6 (Disc
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Normed groups 117restricted to X\M
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Normed groups 119groups need not be
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Normed groups 121Proof. In the meas
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Normed groups 123Hence, as t i n
- Page 129 and 130:
Normed groups 125The corresponding
- Page 131 and 132:
Normed groups 127(t, x) ✛✻Φ T
- Page 133 and 134:
Normed groups 129Fix s. Since s is
- Page 135 and 136:
Normed groups 131Hence,‖x‖ −
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Normed groups 133converging to the
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Normed groups 135Definition. Let {
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Normed groups 137However, whilst th
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Normed groups 139embeddable, 14enab
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Normed groups 141Bibliography[AL]J.
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Normed groups 143Series 378, 2010.[
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Normed groups 145abelian groups, Ma
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Normed groups 147[Kak] S. Kakutani,
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Normed groups 149fields. I. Basic p
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Normed groups 151[So]R. M. Solovay,