Basic Analysis – Gently Done Topological Vector Spaces

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10: Fréchet Spaces 125 linear subspace of X. Furthermore, for any given v ∈ V, there is x ∈ X such that Px = v. Hence Pv = P 2 x = Px = v and we see that (1l−P)v = 0. Hence V = ker(1l−P) and it follows that V is also a linear subspace of X. Now, any x ∈ X can be written as x = Px+(1l−P)x with Px ∈ V = ranP and (1l − P)x ∈ W = kerP. We have seen above that for any v ∈ V, we have v = Pv. If also v ∈ W = kerP then Pv = 0, so that v = Pv = 0. It follows that V ∩W = {0} and so X = V ⊕W. Now suppose that X is a topological vector space and that P : X → X is a continuous linear operator such that P 2 = P. Then both V = ranP = ker(1l−P) and W = kerP are closed subspaces of X and X = V ⊕W. Conversely, suppose that X is a Fréchet space and X = V ⊕ W, where V and W are closed linear subspaces of X. Define P : X → V as above so that P 2 = P and V = ranP = ker(1l−P) and W = kerP. We wish to show that P is continuous. To see this we will show that P is closed and then appeal to the closed-graph theorem. Suppose, then, that x n → x and Px n → y. Now, Px n ∈ V for each n and V is closed, by hypothesis. It follows that y ∈ V and so Py = y. Furthermore, (1l − P)x n = x n − Px n → x − y and (1l − P)x n ∈ W for each n and W is closed, by hypothesis. Hence x−y ∈ W and so P(x−y) = 0, that is, Px = Py. Hence we have Px = Py = y and we conclude that P is closed. Thus P is a closed linear operator from the Fréchet space X onto the Fréchet space V. By the closed-graph theorem, it follows that P is continuous. We have therefore proved the following theorem. Theorem 10.29 Suppose that V is a closed subspace of a Fréchet space X. Then there is a closed subspace W such that X = V ⊕W if and only if there exists a continuous idempotent P with ranP = V. Definition 10.30 We say that a closed subspace V in a topological vector space is complemented if there is a closed subspace W such that X = V ⊕W. Theorem 10.31 Suppose that V is a finite-dimensional subspace of a topological vector space X. Then V is closed and complemented. Proof Let v 1 ,...,v m be linearly independent elements of X which span V. Define ℓ i : V → K by linear extension of the rule ℓ i (v j ) = δ ij for 1 ≤ i,j ≤ m. By Corollary 6.24, each ℓ i is a continuous linear functional on V. By the Hahn- Banach theorem, we may extend each of these to continuous linear functionals on X, which wewillalsodenoteby ℓ i . Then, ifv ∈ V isgivenby v = t 1 v 1 +···+t m v m , we have ℓ i (v) = t i and so v = ℓ 1 (v)v 1 +···+ℓ m (v)v m .
126 Basic Analysis Define P : X → X by Px = ℓ 1 (x)v 1 +···+ℓ m (x)v m . It is clear that P is a continuous linear operator on X with range equal to V. Also we see that P 2 = P (since Pv i = v i for 1 ≤ i ≤ m). Hence V = ker(1l−P) is closed, since (1l−P) is continuous, and W = kerP is a closed complementary subspace for V. We note, in passing, that W = � m i=1 kerℓ i . 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
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48 Basic Analysis Nextweconsiderthe
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50 Basic Analysis and so, multiplyi
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52 Basic Analysis so that Λ ↾ M
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54 Basic Analysis subsets of X. We
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56 Basic Analysis Remark 6.7 The co
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60 Basic Analysis Corollary 6.20 Su
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62 Basic Analysis For each 1 ≤ i
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64 Basic Analysis Corollary 6.29 Le
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7. Locally Convex Topological Vecto
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68 Basic Analysis To show that each
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70 Basic Analysis vector topology o
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72 Basic Analysis at 0 is to say th
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Page 83 and 84: 80 Basic Analysis Hence, for x ∈
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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
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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
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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: 124 Basic Analysis Remark 10.27 It
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Basic Analysis – Gently Done Topological Vector Spaces

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