Convex hulls and closed convex hulls in topological vector spaces
[15/0%]Recall Corollary 3.11: if is a real LCS and , then is the intersection of the closed half-spaces containing .
State and prove a version of this corollary for a complex LCS.
If is a LCS and are compact convex subsets of , then and this convex hull is compact.
If is a Banach space and is a compact subset of , prove that is compact. (This will be proved in Theorem VI.4.8 below.)
In the proof of (13.4) (where is a weakly compact subset of a Banach space , , and is defined by , the element of such that for all in ), if the probability measures on , show that .
Let be any set in a linear space , the intersection of all the convex sets in containing , and the collection of all vectors of the form , where and the coefficients are nonnegative and have the property that . Prove that . This set is called the convex hull of and denoted by .
Prove that in every LCS the convex hull of a bounded set is bounded. Give an example showing that this property may fail to hold in an arbitrary LTS.
The convex hull of a set in a vector space is the set of all convex combinations of members of , that is, the set of all sums
in which is arbitrary. Prove that the convex hull of is convex and that it is the intersection of all convex sets that contain .
Let be a topological vector space. All sets mentioned below are understood to be the subsets of . Prove the following statements.
The convex hull of every open set is open.
If is locally convex then the convex hull of every bounded set is bounded. (This is false without local convexity; see Section 1.47).
If and are bounded, so is .
If and are compact, so is .
If is compact and is closed, then is closed.
The sum of two closed sets may fail to be closed.
In part (f): Theorem 1.13(b) states that if and , then ; part (f) shows that this inclusion may therefore be strict.
Define functionals on (see Exercise 5) by
Define by if . Let consist of , Prove that is compact. Compute . Show that is bounded for each but is not. Convexity can therefore not be omitted from the hypotheses of Theorem 2.9.
Choose so that . Take . Show that lies in the closed convex hull of (by definition, this is the closure of the convex hull) and that is not bounded.
Show that the convex hull of is not closed.
Let be a sequence of pairwise orthogonal unit vectors in a Hilbert space. Let consist of the vectors 0 and . Show that
is compact;
is bounded;
is not closed.
Find all extreme points of .
If , show that contains a compact set whose convex hull is unbounded. This happens in spite of the fact that separates points on ; see Exercise 5. Suggestion: Define by
Let consist of If
show that is unbounded in .
Suppose that are convex sets in a vector space . Prove that every can be represented in the form
with and for all .
Let be the Banach space of all continuous complex functions on the closed unit interval , with the supremum norm. Let , the space of all complex Borel measures on . Give the weak*-topology induced by .
For each , let be the "evaluation functional" defined by , and define by .
Show that is a continuous map from into and that is a compact set in .
Show that .
Find all .
Let be the subspace of consisting of all finite linear combinations
with complex coefficients . Note that and that is the closed convex hull of within . Prove that is an extreme point of , even though is not in .
Suppose is a vector space, is affine, and . Prove that . (This was tacitly used in the proof of Theorem 5.11).
Prove the statement about convex hulls made in the last sentence of the proof of Theorem 8.4.