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Unformatted text preview: SOLUTIONS to PROBLEM SHEET 6. Problem 6.1. (1) Let X be a normed space and let C be a convex subset of X . Use the HB Separation Theorem to show that the (norm) closure C is the same as the closure C w , taken in the weak topology. Taking X = ℓ 1 give an example to show that for a convex subset D of X * we need not have D = D w * . (2) Let us say that a vector y is a convex combination of vectors z 1 ,...,z m if there exist nonnegative real numbers α i with y = ∑ m i =1 α i z i and ∑ n i =1 α i = 1. For a subset A of a real vector space we write co A for the set of all y that are expressible as a convex combination of (finitely many) elements of A . [c.f. the definition of the linear span of A ]. Let ( x n ) be a sequence in a normed space X which converges weakly to x ∈ X . Show that we can find an increasing sequence of natural numbers n k and vectors y k ∈ co { x j : n k 1 &lt; j ≤ n k } such that bardbl y k − x bardbl → 0. Solution. (1) Since the weak topology is coarser than the norm topology we have C ⊆ C w . If x ∈ X \ C then by the HBST there exist f ∈ X * and r ∈ R such that f ( y ) ≤ r for all y ∈ C while f ( x ) &gt; r . The set U = { z ∈ X : f ( z ) &gt; r } is a weak neighbourhood of x disjoint from C , so x / ∈ C w . Now consider X = ℓ 1 and identify the dual space as usual with ℓ ∞ . Take D to be the unit ball of c , which is normclosed because c is a Banach space. If y ∈ B ℓ ∞ then, for all x ∈ ℓ 1 , we have ( P n ( y ) ,x ) = P n j =1 x j y j → P ∞ j =1 x j y j = ( y,x ) . Thus y is the w* limit of the sequence P n ( y ) in D , showing that D is not weak* closed. [Equally we could just quote Goldstine.] (2) The set co A is convex, so, by (1) we have x ∈ A w = ⇒ x ∈ C w = ⇒ x ∈ C . Assuming now that x n converges weakly to x we see that x ∈ { x j : n &lt; j ∈ N } for all n . We proceed with a recursive construction. First choose....
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 Spring '10
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 Topology

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