Basic Analysis – Gently Done Topological Vector Spaces

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8: Banach Spaces 97 � � Tx� ≤ �T��x�. Therefore � � T� ≤ �T�. But since � T is an extension of T we have that � � T� = sup{� � Tx� : x ∈ X, �x� ≤ 1} ≤ sup{� � Tx� : x ∈ D(T), �x� ≤ 1} = sup{�Tx� : x ∈ X, �x� ≤ 1} = �T�. The equality � � T� = �T� follows. The uniqueness is immediate; if S is also a bounded linear extension of T to the whole of X, then S − � T is a bounded (equivalently, continuous) map on X which vanishes on the dense subset D(T). Thus S − � T must vanish on the whole of X, i.e., S = � T. Remark 8.17 This process of extending a densely-defined bounded linear operator to one on the whole of X is often referred to as ‘extension by continuity’. If T is densely-defined, as above, but is not bounded on D(T), there is no ‘obvious’ way of extending T to the whole of X. Indeed, such a goal may not even be desirable, as we will see later—for example, the Hellinger-Toeplitz theorem. We can use the concept of Hamel basis to give an example of a space which is a Banach space with respect to two inequivalent norms. It is not difficult to give examples of linear spaces with inequivalent norms. For example, C[0,1] equipped with the �·� ∞ and � ·� 1 norms is such an example. It is a little harder to find examples where the space is complete with respect to each of the two inequivalent norms. Example 8.18 Let X = ℓ1 and Y = ℓ2 and, for k ∈ N, let ek be the element ek = (δkm ) m∈N in ℓ1 and let fk denote the corresponding element in ℓ2 . For any t with 0 < t < 1, let bt = (t,t2 ,t3 ,...). Then {ek : k ∈ N}∪{b t : 0 < t < 1} form an independent set in ℓ1 , and {fk : k ∈ N}∪{b t : 0 < t < 1} form an independent set in ℓ2 . These sets can be extended to Hamel bases B1 and B2 of ℓ1 and ℓ2 , respectively. Both B1 and B2 contain a subset of cardinality 2ℵ0 . On the other hand, X ⊆ CN and Y ⊆ CN so we deduce that B1 and B2 both have cardinality equal to 2ℵ0 . In particular, there is an isomorphism ϕ from B1 onto B2 which sends ek into fk , for each k ∈ N. By linearity, ϕ defines a linear isomorphism from ℓ1 onto ℓ2 . For n ∈ N, let an = �n k=1 1 nek ∈ ℓ1 and let bn = �n k=1 1 nfk ∈ ℓ2 . Then �an�1 = 1, for all n ∈ N, whereas �bn�2 = 1/ √ n. We define a new norm |· | on ℓ1 by setting |x| = �ϕ(x)� 2
98 Basic Analysis for x ∈ X = ℓ 1 . This is a norm because ϕ is linear and injective. To see that X is complete with respect to this norm, suppose that (x n ) is a Cauchy sequence with respect to | · |. Then (ϕ(x n )) is a Cauchy sequence in ℓ 2 . Since ℓ 2 is complete, there is some y ∈ ℓ 2 such that �ϕ(x n )−y� 2 → 0. Now, ϕ is surjective and so we may write y as y = ϕ(x) for some x ∈ ℓ 1 . We have �ϕ(x n )−y� 2 = �ϕ(x n )−ϕ(x)� 2 = |x n −x| and it follows that |x n −x| → 0 as n → ∞. In other words, ℓ 1 is complete with respect to the norm |· |. We claim that the norms �·� 1 and |· | are not equivalent norms on X = ℓ 1 . Indeed, we have that ϕ(a n ) = b n and so |a n | = �b n � 2 = 1/ √ n → 0 as n → ∞. However, �a n � 1 = 1 for all n. Department of Mathematics King’s College, London
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Basic Analysis - Gently Done Topolo
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Contents 1 Topological Spaces . . .
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2 Basic Analysis Definition 1.3 A t
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4 Basic Analysis Proof The statemen
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6 Basic Analysis Proposition 1.20 L
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8 Basic Analysis Theorem 1.27 Suppo
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10 Basic Analysis Proposition 1.36
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12 Basic Analysis Thepreviouspropos
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14 Basic Analysis ample, the proble
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16 Basic Analysis Proposition 2.8 L
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18 Basic Analysis Theorem 2.13 Let
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20 Basic Analysis A point z is a cl
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22 Basic Analysis family {U x : x
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24 Basic Analysis and so � A∩B
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26 Basic Analysis Remark 3.2 Let G
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28 Basic Analysis Proposition 3.7 S
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30 Basic Analysis Proof (version 2)
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4. Separation The Hausdorff propert
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34 Basic Analysis Next we show that
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36 Basic Analysis Q x contains no r
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38 Basic Analysis that g 0 (x) =
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5. Vector Spaces We shall collect t
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42 Basic Analysis Zorn’s lemma ca
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44 Basic Analysis Finally, suppose
Page 49 and 50: 46 Basic Analysis Proposition 5.14
Page 51 and 52: 48 Basic Analysis Nextweconsiderthe
Page 53 and 54: 50 Basic Analysis and so, multiplyi
Page 55 and 56: 52 Basic Analysis so that Λ ↾ M
Page 57 and 58: 54 Basic Analysis subsets of X. We
Page 59 and 60: 56 Basic Analysis Remark 6.7 The co
Page 63 and 64: 60 Basic Analysis Corollary 6.20 Su
Page 65 and 66: 62 Basic Analysis For each 1 ≤ i
Page 67 and 68: 64 Basic Analysis Corollary 6.29 Le
Page 69 and 70: 7. Locally Convex Topological Vecto
Page 71 and 72: 68 Basic Analysis To show that each
Page 73 and 74: 70 Basic Analysis vector topology o
Page 75 and 76: 72 Basic Analysis at 0 is to say th
Page 77 and 78: 74 Basic Analysis For the last part
Page 79 and 80: 76 Basic Analysis are equivalent; s
Page 81 and 82: 78 Basic Analysis C = {f ∈ X : p
Page 83 and 84: 80 Basic Analysis Hence, for x ∈
Page 85 and 86: 82 Basic Analysis Thus, for t ∈ K
Page 87 and 88: 8. Banach Spaces In this chapter, w
Page 89 and 90: 86 Basic Analysis 5. The Banach spa
Page 91 and 92: 88 Basic Analysis We shall apply th
Page 93 and 94: 90 Basic Analysis Proposition 8.7 F
Page 97 and 98: 94 Basic Analysis and we deduce tha
Page 99: 96 Basic Analysis Theorem 8.16 Supp
Page 103 and 104: 100 Basic Analysis and so �ϕ x
Page 105 and 106: 102 Basic Analysis The next result
Page 107 and 108: 104 Basic Analysis Now we consider
Page 109 and 110: 106 Basic Analysis Theorem 9.13 For
Page 111 and 112: 108 Basic Analysis Definition 9.19
Page 113 and 114: 110 Basic Analysis According to the
Page 115 and 116: 10. Fréchet Spaces Inthischapter w
Page 117 and 118: 114 Basic Analysis and this is sepa
Page 119 and 120: 116 Basic Analysis For any m,n > N,
Page 121 and 122: 118 Basic Analysis Theorem 10.18 (B
Page 123 and 124: 120 Basic Analysis because X = �
Page 125 and 126: 122 Basic Analysis Corollary 10.23
Page 127 and 128: 124 Basic Analysis Remark 10.27 It
Page 129: 126 Basic Analysis Define P : X →
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Basic Analysis – Gently Done Topological Vector Spaces

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