Basic Analysis – Gently Done Topological Vector Spaces

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6: Topological Vector Spaces 63 Proposition 6.26 For any non-empty sets A and B in a topological vector space X, (i) A+B ⊆ A+B, and (ii) A = � U∈N0 (A+U), where N0 is any neighbourhood base at 0. Proof Let x ∈ A and y ∈ B. Then there are nets (a ν ) and (b ν ), indexed by V 0 , such that a ν → x, and b ν → y. By continuity of addition, it follows that a ν +b ν → x+y. Hence x+y ∈ A+B, which proves (i). To prove (ii), suppose that x ∈ A and let N 0 be any neighbourhood base at 0. For any U ∈ N 0 , −U is a neighbourhood of 0 and so x − U is a neighbourhood of x. Since x ∈ A it follows that (x − U) ∩ A �= ∅, that is, x ∈ A + U, and so A ⊆ � � (A + U). On the other hand, suppose that x ∈ (A + U) and U∈N0 U∈N0 that V is any neighbourhood of x. Then −x+V is a neighbourhood of 0 and so also is −(−x+V) = x−V. Therefore there is some U ∈ N0 such that U ⊆ x−V. Now, x ∈ A + U, by hypothesis, and so x ∈ A + U ⊆ A + x − V. Hence there is a ∈ A and v ∈ V such that x = a + x − v, i.e., a = v. It follows that every neighbourhood of x contains some element of A and therefore x ∈ A. Corollary 6.27 The closed balanced neighbourhoods of 0 form a local neighbourhood base at 0. Proof Let U be any neighbourhood of 0. By Proposition 6.8, there is a neighbourhood V of 0 such that V + V ⊆ U. Also, by Proposition 6.18, there is a balanced neighbourhood of 0, W, say, with W ⊆ V, so that W + W ⊆ U. By Proposition 6.26, W ⊆ W +W ⊆ U, and W is a neighbourhood of 0. We claim that W is balanced. To see this, let x ∈ W and let t ∈ K with |t| ≤ 1. There is a net (w ν ) in W such that w ν → x. Then tw ν → tx, since scalar multiplication is continuous. But tw ν ∈ W, since W is balanced, and so tx ∈ W. It follows that W is balanced, as claimed. Proposition 6.28 If M is a linear subspace of a topological vector space X, then so is its closure M. In particular, any maximal proper subspace is either dense or closed. Proof Let M be a linear subspace of the topological vector space X. We must show that if x,y ∈ M and t ∈ K, then tx+y ∈ M. There are nets (x ν ) and (y ν ) in M, indexed by the base of all neighbourhoods of 0, such that x ν → x and y ν → y. It follows that tx ν → tx and tx ν +y ν → tx+y and we conclude that tx+y ∈ M, as required. If M is a maximal proper subspace, the inclusion M ⊆ M implies that either M = M, in which case M is closed, or M = X, in which case M is dense in X.
64 Basic Analysis Corollary 6.29 Let ℓ : X → K be a linear functional on a topological vector space X. Then either ℓ is continuous or kerℓ is a dense proper subspace of X. Proof If ℓ is zero, it is continuous and its kernel is the whole of X. Otherwise, kerℓ is a maximal proper linear subspace of X which is either closed or dense, by Proposition 6.28. However, ℓ is continuous if and only if its kernel is closed, so if ℓ is not continuous its kernel is a proper dense subspace. It is convenient to introduce here the notion of boundedness in a topological vector space. Definition 6.30 A subset B of a topological vector space is said to be bounded if for each neighbourhood U of 0 there is s > 0 such that B ⊆ tU for all t > s. Example 6.31 Let B be a subset of a normed space X, and, for r > 0, let B r denote the ball B r = {x : �x� < r} in X. Any neighbourhood U of 0 contains some ball B r ′ and so B r ⊆ tB r ′ ⊆ tU for all t > r/r ′ , that is, each B r is bounded according to the definition above. On the other hand, if B is any bounded set, then there is s > 0 such that B ⊆ tB 1 for all t > s, that is, �x� ≤ s for all x ∈ B. Therefore the notion of boundedness given by the above definition coincides with the usual notion of boundedness in a normed space. Proposition 6.32 If (X,T) is a topological vector space and if z ∈ X and A and B are bounded subsets of X, then the sets {z}, A∪B and A+B are bounded. Proof Let U be any neighbourhood of 0. Then U is absorbing and so z ∈ tU for all sufficiently large t > 0, that is, {z} is bounded. By Proposition 6.8, there is a neighbourhood V of 0 such that V + V ⊆ U. Given that A and B are bounded, there is s 1 > 0 such that A ⊆ tV for t > s 1 and there is s 2 > 0 such that B ⊆ tV for t > s 2 . Therefore and for all t > max{s 1 ,s 2 }. A∪B ⊆ tV A+B ⊆ tV +tV ⊆ tU Remark 6.33 It follows, by induction, that any finite set in a topological vector space is bounded. Also, taking A = {z}, we see that any translate of a bounded set is bounded. 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
Page 15 and 16: 12 Basic Analysis Thepreviouspropos
Page 17 and 18: 14 Basic Analysis ample, the proble
Page 19 and 20: 16 Basic Analysis Proposition 2.8 L
Page 21 and 22: 18 Basic Analysis Theorem 2.13 Let
Page 23 and 24: 20 Basic Analysis A point z is a cl
Page 25 and 26: 22 Basic Analysis family {U x : x
Page 27 and 28: 24 Basic Analysis and so � A∩B
Page 29 and 30: 26 Basic Analysis Remark 3.2 Let G
Page 31 and 32: 28 Basic Analysis Proposition 3.7 S
Page 33 and 34: 30 Basic Analysis Proof (version 2)
Page 35 and 36: 4. Separation The Hausdorff propert
Page 37 and 38: 34 Basic Analysis Next we show that
Page 39 and 40: 36 Basic Analysis Q x contains no r
Page 41 and 42: 38 Basic Analysis that g 0 (x) =
Page 43 and 44: 5. Vector Spaces We shall collect t
Page 45 and 46: 42 Basic Analysis Zorn’s lemma ca
Page 47 and 48: 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: 62 Basic Analysis For each 1 ≤ i
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 and 100: 96 Basic Analysis Theorem 8.16 Supp
Page 101 and 102: 98 Basic Analysis for x ∈ X = ℓ
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
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114 Basic Analysis and this is sepa
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116 Basic Analysis For any m,n > N,
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118 Basic Analysis Theorem 10.18 (B
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120 Basic Analysis because X = �
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122 Basic Analysis Corollary 10.23
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124 Basic Analysis Remark 10.27 It
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126 Basic Analysis Define P : X →
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Basic Analysis – Gently Done Topological Vector Spaces

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