Basic Analysis – Gently Done Topological Vector Spaces

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10: Fréchet Spaces 119 Example 10.20 Let X be the normed space of those sequences (xk ) of complex numbers with only a finite number of non-zero terms equipped with the usual component-wiseadditionandscalarmultiplicationandnorm�(x k )� = supk |xk |(= maxk |xk |). For each n ∈ N, let Tn be the linear operator Tn : X → X with action given by Tn (xk ) = (akxk ) where ak = min{k,n}. Clearly, each Tn is continuous (�Tn (xk )� ≤ n�(xk )�). Moreover, (Tn (xk )) converges, as n → ∞, to T(xk ), where T : X → X is the linear map T(xk ) = (kxk ). However, the map T is not continuous—for example, if (x (n) ) is the sequence of elements of X with x (n) the complex sequence whose only non-zero term is the nth , which is equal to 1 n , then x (n) → 0 in X whereas �Tx (n) � = 1 for all n ∈ N. Theorem 10.21 (Open Mapping theorem) Suppose that X and Y are Fréchet spaces and that T : X → Y is a continuous linear mapping from X onto Y. Then T is an open mapping. Proof We must show that T(G) is open in Y whenever G is open in X. If G is empty, so is T(G), so suppose that G is a non-empty open set in X. Let b ∈ T(G) with b = T(a) for some a ∈ G. We need to show that b is an interior point of T(G). By translating G by −a and T(G) by T(−a) = −b, this amounts to showing that for any neighbourhood V of 0 in X, 0 is an interior point of T(V). Indeed, since G − a is an open neighbourhood of 0, it would follow that 0 is an interior point of T(G−a), that is, there is an open set W with 0 ∈ W ⊆ T(G−a). Hence T(a)+W ⊆ T(G), which is to say that b = T(a) is an interior point of T(G). Let d be a translation invariant metric compatible with the topology of X, and let V be any neighbourhood of 0. Then there is some r > 0 such that {x ∈ X : d(x,0) < r} ⊆ V. Set V n = {x ∈ X : d(x,0) < 2 −n r}, n = 0,1,2,.... The first step is to show that there is an open neighbourhood W of 0 such that W ⊆ T(V 1 ). It is easy to see that V 2 −V 2 ⊆ V 1 , and so and therefore T(V 2 )−T(V 2 ) ⊆ T(V 1 ) T(V 2 )−T(V 2 ) ⊆ T(V 2 )−T(V 2 ) ⊆ T(V 1 ) using −A = (−A) and A+B ⊆ A+B, for any subsets A,B in a topological vector space. We claim that T(V 2 ) has non-empty interior. To see this, we note that Y = T(X) = ∞� kT(V2 ) k=1
120 Basic Analysis because X = �∞ k=1kV2 , since V2 is a neighbourhood of 0 and so is absorbing. By Baire’s category theorem, k0T(V2 ) has non-empty interior for some k0 ∈ N. But y ↦→ k0y is a homeomorphism of Y onto Y so that k0T(V2 ) = k0T(V2 ) and we deduce that T(V2 ) has non-empty interior, as claimed. In particular, this means that there is an open set U contained in T(V2 ). Putting W = U −U, we see that W is an open neighbourhood of 0 with W ⊆ T(V 2 )−T(V 2 ) ⊆ T(V 1 ) which completes the first part of the argument. Next we shall show that T(V 1 ) ⊆ T(V). To see this, we construct a sequence (y n ) n∈N , with y n ∈ T(V n ), such that y n+1 − y n ∈ T(V n ). The y n are defined recursively. First fix y 1 to be any point of T(V 1 ). Suppose that n ≥ 1 and that y n has been chosen in T(V n ). Arguing as above, but now with V n+1 instead of V 1 , we see that T(V n+1 ) is a neighbourhood of 0. Hence y n −T(V n+1 ) is a neighbourhood of y n , and, since y n belongs to the closure of T(V n ), we have � yn −T(V n+1 ) � ∩T(V n ) �= ∅. Hence there is x n ∈ V n such that T(x n ) ∈ y n −T(V n+1 ). Set y n+1 = y n −T(x n ). Then y n+1 ∈ T(V n+1 ) which completes the construction of the sequence (y n ). Let s n = x 1 +x 2 +···+x n , for n ∈ N. Then for n > m, d(s n ,s m ) = d(s n −s m ,0) = d(x n +···+x m+1 ,0) ≤ d(xn +···+x m+1 ,xm+1 )+d(x m+1 ,0) < d(xn +···+x m+2 ,0)+ 1 2m+1 since xm+1 ∈ Vm+1 , < 1 1 +···+ 2n 2m+1. Hence (s n ) is a Cauchy sequence in X and therefore converges (because X is complete) to some x ∈ X, with d(x,0) < r (since d(x,0) ≤ d(x,x 1 )+d(x 1 ,0) < d(x,x 1 ) + r/2 and d(s n ,x 1 ) = d(x n + ··· + x 2 ,0) ≤ r/2 for all n > 1, so that d(x,x 1 ) = limd(s n ,x 1 ) ≤ r/2). It follows that x ∈ V 0 ⊆ V. Furthermore, T(s n ) = n� T(xj ) = j=1 n� j=1 (y j −y j+1 ) = y 1 −y n+1 . Now, y n ∈ T(V n ) and so there is some v n ∈ V n such that d(y n ,T(v n )) < 1 n and hence d(y n ,0) ≤ d(y n ,T(v n ))+d(T(v n ),0) < 1 n +d(T(v n ),0). 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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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 95 and 96: 92 Basic Analysis Proposition 8.10
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
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: 118 Basic Analysis Theorem 10.18 (B
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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