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By G. Temple

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14 (Separability and classical spaces) The comments above show that the space ∞ of bounded sequences is not a separable space. On the other hand, the space p of p-summable sequences is separable when p ∈ [1, ∞), because the set {ek : k ∈ N} is a countable dense subset of p , where ek is the sequence with 1 in the k th coordinate and zero in every other coordinate. The space C[0, 1] of continuous functions on [0, 1] is separable, because the set t k : k ∈ N ∪ {0} is dense in C[0, 1], by the Weierstrass Approximation Theorem.

17(c) provides a bounded linear functional on C[0, 1] because 1 1 f (t) g(t) dt ≤ sup |f (t)| · t∈[0,1] 0 |g(t)| dt = f ∞ g 1, 0 for f ∈ C(K) and g ∈ L1 (0, 1). This, too, can be realized as integration against a measure. Let ν(A) = g(t) dt, A ∈ B, A where B denotes the collection of Borel measurable subsets of [0, 1]. Then 1 f (t) g(t) dt = 0 f dν. 20, below), but they do give a hint to the true nature of (C[0, 1])∗ . Before we identify this space, let us recall several definitions from measure theory.

With |aj | ≤ 1 for all j ≥ 2 will determine a norming functional for ξ . (iii) In 2 , the norming functional is always unique. Let ξ = (ξ1 , ξ2 , . . ) be an ∞ 2 1/2 element of 2 . Then ξ = and the norming element is ξ/ ξ . ) (iv) Consider the space ∞ and let ξ = (1, 1, 1, . . ). Any linear functional φ on ∞ will be a norming element for ξ provided both φ(ξ ) = 1 and φ = 1.

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