# SolnsSix (1) - SOLUTIONS to PROBLEM SHEET 6 Problem 6.1(1...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 non-negative 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 &amp;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 ) &amp;gt; r . The set U = { z ∈ X : f ( z ) &amp;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 norm-closed 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 &amp;lt; j ∈ N } for all n . We proceed with a recursive construction. First choose....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

SolnsSix (1) - SOLUTIONS to PROBLEM SHEET 6 Problem 6.1(1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online