Existence of Efficient Points in Vector Optimization and Generalized Bishop-Phelps Theorem^{1}

K. F. NG AND X. Y. ZHENG

1 The authors gratefully acknowledge financial support by an Earmarked Grant from the Research Grant Council of Hong Kong. Research of the second author was also supported by the National Natural Science Foundation of PRC and ABSF of Yunnan Province, PRC.

Abstract

Abstract. In a set without linear structure equipped with a preorder, we give a general existence result for efficient points. In a topological vector space equipped with a partial order induced by a closed convex cone with a bounded base, we prove another kind of existence result for efficient points. This result does not depend on the Zorn lemma. As applications,we study the solution problem in vector optimization and generalize the Bishop-Phelps theorem to topological vector spaces setting by showing that the B-support points of any sequentially complete closed subset A of a topological vector space E is dense in ∂A, where B is any bounded convex subset of E.

Key Words. Preorder, efficient points, vector optimization, support points.

1 Introduction

Let E be a topological vector space and C ⊂ E a convex cone. C specifies a preorder in E as usual: for x, y ∈ E,

x ⩽_{C} y if and only if y − x ∈ C.

Clearly, ⩽_{C} satisfies (i) x ⩽_{C} x for each x ∈ E, and (ii) if x, y, z ∈ E, x ⩽_{C} y and y ⩽_{C} z then x ⩽_{C} z. If C is pointed (i.e. C ⋂ −C = {0}), ⩽_{C} also satisfies (iii) if x ⩽_{C} y and y ⩽_{C} x then x = y, that is, ⩽_{C} is a partial order. A convex set Θ is called a base of C if 0 ∉ cl(Θ) and C = cone(Θ); and Θ is called a bounded base of C if Θ is a base of C and is a bounded set. Clearly, C is pointed if C has a base. Let A be a subset of E. A point a ∈ A is called an efficient point of A if a ⩽_{C} x whenever x ⩽_{C} a with x ∈ A, that is,

A ⋂(a − C) ⊂ a + C  (1)

When C is pointed, (1) is equivalent to

A ⋂(a − C) = {a}.

Let Min(A, ⩽_{C}) (or Min(A, C)) denote the set of all efficient points of A.

Let X be a set and F : X → 2^{E} a set-valued mapping. Consider the following vector optimization problem:

(VOP) MinF (x), x ∈ X,

which means to find x_{0} ∈ X with the property that there exists y_{0} ∈ F (x_{0}) such that y_{0} ∈ Min(F (X), ⩽_{C}); in this case, x_{0} is called a solution of (VOP). It is clear that (VOP) has a solution if and only if Min(F (X), ⩽_{C}) ≠ ∅.

An important problem in vector optimization is to find conditions which guarantee Min(A, C) ≠ ∅. The following theorem (Ref. 1) is a well-known and useful result in this regard.

Theorem 1.1. If A is compact then Min(A, C) ≠ ∅.

As compactness is a very restrictive assumption, several authors have tried to weaken it (Ref. 2-10). Substituting cone-compactness for compactness in Theorem 1.1, Hartley (Ref. 6) obtained an existence result. Wagner (Ref. 10) and Corley (Ref. 4) weakened further cone-compactness to conesemicompactness. A is called C-complete (Ref. 8) if A has no open cover of the form {x_{α} − cl(C)^{c}|α ∈ Λ}, where {x_{α}}_{α∈Λ} is a decreasing net in A and cl(C)^{c} = E \cl(C) denotes the complement of the closure cl(C) in E; a general existence theorem generalizing all above-mentioned results was given by Luc in Ref. 8. In the case when the order cone C is closed, it is clear that the C-completeness of A implies the following property (Z): A is said to have (Z) if every totally ordered subset of A is bounded below by an element in A; thus, in this case, the above-mentioned existence results are all immediate from the Zorn lemma. In this paper, we prove that if E is an abstract set equipped with a preorder ⪯ and A ⊂ X has a maximal totally subset which is minorized by an order-s-semicompact subset of A (for definitions see Section 2) then Min(A, ⪯) ≠ ∅.

Another notable existence result is the Bishop and Phelps lemma, which was first given in Banach spaces and is of important applications to geometry of Banach spaces. Using the Zorn lemma, Jameson (Ref. 11) established the following existence result (Corollary 3.8.10, Ref. 11), which generalizes the Bishop-Phelps lemma to a topological vector space.

Theorem 1.2. If A is complete and C has a bounded base, then Min(A, C) ≠ ∅.

To the best of our knowledge, in the case when the space E is a topological vector space, all known existence results are obtained by using the Zorn lemma. In the paper, we give a generalization of Theorem 1.2, and our proof is free from the Zorn lemma. As an application we show in section 4 that, for any bounded convex subset B of a topological vector space E, the B-support points of any sequentially complete closed convex set A in E are dense in its topological boundary ∂A.

2 Existence Result without Linear Structure

Stadler (Ref. 12) and Hazen and Morin (Ref. 7) investigated existence of efficient points for the orders not necessarily induced by convex cones. In this section, we will consider existence without linear structure and always assume that E is a set equipped with a preorder ⪯, which is not assumed to be a partial order. For A, B ⊂ E, we say that A is minorized by B if for each a ∈ A there exists b ∈ B such that b ⪯ a. For each x ∈ E, let L_{x} = {y ∈ E|y ⪯ x} and S_{x} = {y ∈ E|x ⪯ y}. For a subset A of E, let S_{A} = ⋃_{x∈A} S_{x}. If E is

a topological vector space and ⪯ is induced by a convex cone C in E, then L_{x} = x − C and S_{A} = A + C.

Definition 2.1. A is said to be order-semicompact if every cover of A of the form {L_{x}^{c}| x ∈ D} with D ⊂ A has a finite subcover, and A is said to be order-s-semicompact if every cover of A of the form {L_{x}^{c}| x ∈ D} with D ⊂ E has a finite subcover.

Clearly, A is order-semicompact if A is order-s-semicompact; and in the case when E is a topological vector space and ⪯ is induced by a closed convex cone C ⊂ E, order-semicompactness is identical with C-semicompactness in Corley (Ref. 4).

Lemma 2.1. For each x ∈ E, S_{L_{x}^{c}} = L_{x}^{c}.

Proof. Since A ⊂ S_{A} in general, we need only show that L_{x}^{c} ⊃ S_{L_{x}^{c}}. Suppose on the contrary that there exists y ∈ S_{L_{x}^{c}} such that y ∉ L_{x}^{c}. Then y ⪯ x, and there exists z ∈ L_{x}^{c} such that, z ⪯ y. Hence z ⪯ x, that is, z ∈ L_{x}. This contradicts z ∈ L_{x}^{c}.

Proposition 2.1. Let A_{1} and A_{2} be subsets of E.

(i) A_{1} ⋃ A_{2} is order-s-semicompact if both A_{1} and A_{2} are order-s-semicompact.

(ii) If A_{1} ⊂ A_{2} ⊂ S_{A_{1}}, then A_{1} is order-s-semicompact if and only if A_{2} is order-s-semicompact.

(iii) If S_{A} is order-semicompact then A is order-s-semicompact.

Consequently, for sets of form S_{A}, the order-s-semicompactness is equivalent to the order-semicompactness.

Proof. (i) is immediate from Definition 2.1.

(ii). Suppose that A_{1} is order-s-semicompact. Let D be a subset of E such

that A_{2} ⊂ ⋃_{y∈D} L_{y}^{c}. Then A_{1} ⊂ ⋃_{y∈D} L_{y}^{c}. Hence there exist y_{1}, ⋯, y_{n} ∈ D such

that A_{1} ⊂ ⋃_{i=1}^{n} L_{y_{i}}^{c}, and so

S_{A_{1}} ⊂ ⋃_{i=1}^{n} S_{L_{y_{i}}^{c}}.

By Lemma 2.1 and A_{2} ⊂ S_{A_{1}}, A_{2} ⊂ ⋃_{i=1}^{n} L_{y_{i}}^{c}. Hence A_{2} is order-s-semicompact.

Conversely, suppose that A_{2} is order-s-semicompact. Let D ⊂ E satisfy

A_{1} ⊂ ⋃_{y∈D} L_{y}^{c}. Then S_{A_{1}} ⊂ ⋃_{y∈D} S_{L_{y}^{c}}. By Lemma 2.1 and A_{2} ⊂ S_{A_{1}}, A_{2} ⊂ ⋃_{y∈D} L_{y}^{c}.

Hence there exist y_{1} ⋯, y_{n} ∈ D such that A_{2} ⊂ ⋃_{i=1}^{n} L_{y_{i}}^{c}. It follows from

A_{1} ⊂ A_{2} that A_{1} ⊂ ⋃_{i=1}^{n} L_{y_{i}}^{c}.

(iii). Let D ⊂ E satisfy A ⊂ ⋃_{y∈D} L_{y}^{c}. By Lemma 2.1, S_{A} ⊂ ⋃_{y∈D} L_{y}^{c}. If D ⊂

S_{A}, it follows from the order-compactness of S_{A} that there exist y_{1}, ⋯, y_{n} ∈ D such that

A ⊂ S_{A} ⊂ ⋃_{i=1}^{n} L_{y_{i}}^{c}.

If D is not a subset of S_{A}, then there exists y_{0} ∈ D such that y_{0} ∉ S_{A}. Hence, for each x ∈ A, x ⪯̸ y_{0}, that is, x ∈ L_{y_{0}}^{c}, and so A ⊂ L_{y_{0}}^{c}. This completes the proof.

Corollary 2.1. Let A, A_{1} be subsets of E such that A ⊂ A_{1} ⊂ S_{A}. Then A_{1} is order-s-semicompact in each of the following cases:

(i). E is a topological space, A is compact, and L_{x} is closed for each x ∈ E.

(ii). E is a locally convex space, A is weakly compact and ⪯ is induced by a closed convex cone in E.

Proof. In view of (ii) of Proposition 2.1, we need only show that A is orders-semicompact. Thus, (i) is clear. For (ii), let C denote the closed convex cone in the locally convex space E inducing the preorder ⪯. Then C is weakly closed (see Ref. 13) and hence L_{y}^{c} = y − C^{c} is weakly open for each y ∈ E. This implies that the given weakly compact set A is order-s-semicompact.

Theorem 2.1. If A has a maximal totally ordered subset minorized by an order-s-semicompact subset of S_{A}, then Min(A, ⪯) ≠ ∅.

Proof. Let D be a maximal totally ordered subset of A and H be an order-s-semicompact subset of S_{A} such that D ⊂ S_{H}. By the maximality of D, it suffices to show that D is bounded below by an element in A. Suppose on the contrary that for each x ∈ A there exists y ∈ D such that x ⪯̸ y, that

is, x ∈ L_{y}^{c}. Then A ⊂ ⋃_{y∈D} L_{y}^{c}, and so

H ⊂ S_{A} ⊂ ⋃_{y∈D} S_{L_{y}^{c}}.

By Lemma 2.1 and the order-s-semicompactness of H, there exist y_{1}, ⋯, y_{n} ∈

D such that H ⊂ ⋃_{i=1}^{n} L_{y_{i}}^{c}. Since D is totally ordered, without loss of generality

we can assume that y_{1} ⪯ y_{i} and so L_{y_{1}} ⊂ L_{y_{i}} for each i. Hence H ⊂ L_{y_{1}}^{c}. By Lemma 2.1, one has S_{H} ⊂ L_{y_{1}}^{c}. Since y_{1} ∈ D ⊂ S_{H}, it follows that y_{1} ∈ L_{y_{1}}^{c}, which is absurd.

Let X be a topological space and F : X → 2^{E} a set-valued mapping. For each e ∈ E, the level set of F at e is defined by

L(F, e) = {x ∈ X|there exists y ∈ F (x) such that y ⪯ e};

F is said to be inf-compact if there exists e_{0} ∈ F (X) such that L(F, e) is compact for each e ∈ E with e ⪯ e_{0}.

Corollary 2.2. Let X be a toplogical space and F : X → 2^{E} an inf-compact mapping with order-s-semicompact values. Then (VOP) has a solution.

Proof. Since F is inf-compact, there exists e_{0} ∈ F (X) such that for each e ∈ E with e ⪯ e_{0}, L(F, e) is compact. By Zorn’s Lemma, F (X) has a maximal totally ordered subset D containing e_{0}. By definition of level sets of F and since D is totally ordered, it is clear that

⋂_{e∈D} L(F, e) = ⋂_{e∈D,e⪯e_{0}} L(F, e)  (2)

Since D ⊂ F (X), L(F, e) is nonempty for each e ∈ D . Since D is totally ordered, for any finite elements e_{1}, ⋯, e_{n} ∈ D with e_{i} ⪯ e_{0} (1 ⩽ i ⩽ n), we can assume that e_{1} ⪯ e_{i} (1 ⩽ i ⩽ n), and so

⋂_{i=1}^{n} L(F, e_{i}) = L(F, e_{1}) ≠ ∅.

Hence {L(F, e)|e ∈ D, e ⪯ e_{0}} is a family with the finite intersection property. By compactness, ⋂_{e∈D,e⪯e_{0}} L(F, e) ≠ ∅. By (2), pick x_{0} ∈ ⋂_{e∈D} L(F, e).

This implies that D is minorized by F (x_{0}). Since F (x_{0}) is assumed order-s-semicompact, one may apply Theorem 2.1 to conclude that Min(F (X), ⪯) ≠ ∅. Therefore, (VOP) has a solution.

3 Existence Theorem Independent of the Zorn Lemma

Throughout this section, we assume that E is a topological vector space, C ⊂ E is a closed convex cone and that A is a subset of E.

Definition 3.1. A is said to be C-sequentially complete if every decreasing Cauchy sequence in A converges to some point in A.

Clearly, sequential completeness implies C-sequential completeness.

Proposition 3.1. Let C be a closed convex cone with a bounded sequentially complete base. Then every sequentially closed subset of E is C-sequentially complete.

Proof. Let A be a sequentially closed subset of E and {x_{n}} a decreasing Cauchy sequence in A. We need only to show that {x_{n}} converges in E. Note that {x_{n}} is a bounded sequence in A ⋂(x_{1} − C). Let Θ be a sequentially complete bounded base of C. Pick a sequence {t_{n}} in R^{+} and a sequence {θ_{n}} in Θ such that x_{n} = x_{1} − t_{n}θ_{n} for each n ∈ N, where N denotes the set of all natural numbers. Since 0 ∉ cl(Θ), there exists a balanced neighborhood V of 0 such that V ⋂ −Θ = ∅. By the boundedness of {x_{n}}, there exists δ > 0 such that x_{n} − x_{1} = −t_{n}θ_{n} ∈ δV for each n ∈ N. This and V ⋂ −Θ = ∅ imply that t_{n} ⩽ δ for each n ∈ N. Without loss of generality we can assume that t_{n} → t_{0} ∈ [0, δ]. If t_{0} = 0, then t_{n}θ_{n} → 0 and so x_{n} → x_{1}. If t_{0} > 0, θ_{n} = \frac{x_{1}−x_{n}}{t_{n}} is Cauchy in Θ. Hence θ_{n} → θ_{0} for some θ_{0} ∈ Θ, and so x_{n} → x_{1} − t_{0}θ_{0}.

Lemma 3.1. Let C be a closed convex cone with a bounded base Θ. If there exists x ∈ A such that A ⋂(x − C) is bounded, then for any ε > 0 there exists a ∈ A ⋂(x − C) such that

A ⋂(a − C) ⊂ a − [0, ε]Θ,

where [0, ε]Θ := {tθ|0 ⩽ t ⩽ ε, θ ∈ Θ}.

Proof. Without loss of generality we can assume that x = 0. Take a balanced neighborhood V of 0 such that V ⋂ Θ = ∅. Since both (−A) ⋂ C and Θ are bounded, there exist r > 0 and δ > 0 such that

(−A) ⋂ C + [0, δ]Θ ⊂ rV.

Note that V is disjoint from [1, +∞)Θ, δ belongs to the set T defined by

T = {t ⩾ 0|(−A) ⋂ C + [0, t]Θ and [r, +∞)Θ are disjoint}.

Since 0 ∈ (−A) ⋂ C, T is bounded above by r. Denoting l := sup{t|t ∈ T }, we have δ ⩽ l ⩽ r. Write l_{ε} := l(1 + \frac{ε}{r}). Since l_{ε} > l,

(−A) ⋂ C + [0, l_{ε}]Θ intersects [r, +∞)Θ;

hence there exist a ∈ A ⋂ −C, t_{1} ∈ [r, +∞) and θ_{1} ∈ Θ such that

a + t_{1}θ_{1} ∈ [0, l_{ε}]Θ  (3)

To complete the proof, we will show that

A ⋂(a − C) ⊂ a − [0, ε]Θ.

To do this, let y ∈ A ⋂(a−C) such that y = a−t_{2}θ_{2} for some t_{2} ⩾ 0 and θ_{2} ∈ Θ. It remains to show that t_{2} ⩽ ε. Since a ∈ −C, −y = −a + t_{2}θ_{2} ∈ [t_{2}, +∞)Θ and −y ∈ (−A) ⋂ C; since

(−A) ⋂ C ⊂ (−A) ⋂ C + [0, δ]Θ

and the latter set is disjoint from [r, +∞)Θ, it follows that −y ∉ [r, +∞)Θ and consequently that t_{2} < r. Hence

r = t_{2} + (1 − \frac{t_{2}}{r})r ⩽ t_{2} + (1 − \frac{t_{2}}{r})t_{1}

and so

t_{2}θ_{2} + (1 − \frac{t_{2}}{r})t_{1}θ_{1} ∈ (t_{2} + (1 − \frac{t_{2}}{r})t_{1})Θ ⊂ [r, +∞)Θ.

Writing z for −\frac{t_{2}}{r} a + t_{2}θ_{2} + (1 − \frac{t_{2}}{r})t_{1}θ_{1}, and noting −a ∈ C, we then have that

z ∈ [r, +∞)Θ.

Note that, by (3),

z + y = z + a − t_{2}θ_{2} = (1 − \frac{t_{2}}{r})(t_{1}θ_{1} + a) ∈ (1 − \frac{t_{2}}{r})[0, l_{ε}]Θ;

since −y ∈ (−A) ⋂ C it follows that z belongs to (−A) ⋂ C + [0, t^{∗}]Θ, where t^{∗} := (1 − \frac{t_{2}}{r})l_{ε}. Since z ∈ [r, +∞)Θ, t^{∗} ∉ T. By the definition of l, one must have l ⩽ t^{∗}, so 1 ⩽ (1 − \frac{t_{2}}{r})(1 + \frac{ε}{r}) and t_{2} ⩽ ε.

Theorem 3.1. Let C be a closed convex cone with a bounded base. If there exists x ∈ A such that A ⋂(x − C) is C-sequentially complete and bounded, then there exists a ∈ A with a ⩽_{C} x such that a ∈ Min(A, C).

Proof. Let Θ be a bounded base of C. By Lemma 3.1, one can construct a sequence {x_{n}} in A such that

(i) x_{1} ⩽_{C} x and x_{n+1} ⩽_{C} x_{n} for each n ∈ N;

(ii) A ⋂(x_{n} − C) ⊂ x_{n} − [0, \frac{1}{n}]Θ for each n ∈ N.

Note that for any a_{1}, a_{2} ∈ A with a_{2} ⩽_{C} a_{1},

A ⋂(a_{2} − C) ⊂ A ⋂(a_{1} − C)  (4)

It follows from (i) and (ii) that for any n, k ∈ N, A ⋂(x_{n+k}−C) ⊂ x_{n}−[0, \frac{1}{n}]Θ; in particular x_{n+k} − x_{n} ∈ −[0, \frac{1}{n}]Θ. This and the boundedness of Θ imply that {x_{n}} is a decreasing Cauchy sequence in A ⋂(x − C). Hence there exists a ∈ A ⋂(x − C) such that x_{n} → a. By (i), (4) and the closedness of C, a ⩽_{C} x_{n} ⩽_{C} x for each n ∈ N. For each y ∈ A ⋂(a − C), by (ii) and (4), y ∈ A ⋂(x_{n} − C) ⊂ x_{n} − [0, \frac{1}{n}]Θ. It follows from the boundedness of Θ that x_{n} → y, and so a = y. This shows that A ⋂(a − C) = {a}, that is, a ∈ Min(A, C).

Corollary 3.1. Let C be a closed convex cone with a bounded base. Assume that there exists x ∈ A such that A ⋂(x − C) is bounded and sequentially closed. Then Min(A, C) ≠ ∅ if any one of the following conditions (i) and (ii) holds:

(i) A is sequentially complete,

(ii) Θ is sequentially complete.

The above corollary is immediate from Theorem 2.1 and Proposition 2.1.

Corollary 3.2. Let C be a closed convex cone with a sequentially complete bounded base and X be a topological space; let F : X → 2^{E} be an infcompact and upper semicontinuous mapping with nonempty bounded closed values. Then (VOP) has a solution.

Proof. Let e_{0} ∈ F (X) such that for each e ∈ E with e ⩽_{C} e_{0}, L(F, e) is a compact subset of X. By Corollary 3.1, it suffices to show that F (X) ⋂(e_{0}−C) is bounded and closed. It is clear that

F (X) ⋂(e_{0} − C) ⊂ F (L(F, e_{0}))  (5)

Since F is a upper semicontinuous mapping with closed values, it is easy from the compactness of L(F, e_{0}) and (5) to verify that F (X) ⋂(e_{0} −C) is closed. It remains to prove that F (X) ⋂(e_{0} − C) is bounded. By (5), we need only show that F (L(F, e_{0})) is bounded. To do this, for any neighborhood U of 0, pick a balanced open neighborhood V of 0 such that V + V ⊂ U. Since F is upper semicontinuous, for each x ∈ L(F, e_{0}) there exists an open neighborhood W_{x} of x in X such that F (W_{x}) ⊂ F (x) + V. Note that L(F, e_{0}) is compact and that {W_{x}|x ∈ L(F, e_{0})} is an open cover of L(F, e_{0}). There exist x_{1}, ⋯, x_{n} ∈

L(F, e_{0}) such that L(F, e_{0}) ⊂ ⋃_{i=1}^{n} W_{x_{i}} and so

F (L(F, e_{0})) ⊂ ⋃_{i=1}^{n} F (W_{x_{i}}) ⊂ ⋃_{i=1}^{n} F (x_{i}) + V  (6)

Since F (x_{i}) is bounded, there exists r > 0 such that tF (x_{i}) ⊂ V for each 1 ⩽ i ⩽ n and 0 < t ⩽ r. It follows from (6) that for each 0 < t ⩽ r,

tF (L(F, e_{0})) ⊂ V + tV ⊂ (1 + r)(V + V) ⊂ (1 + r)U,

and so for each 0 < t ⩽ \frac{r}{1+r}, tF (L(F, e_{0})) ⊂ U. This shows that F (L(F, e_{0})) is bounded.

Remark 3.1. Jameson presented an interesting result (Theorem 3.8.7, Ref.

11) that every decreasing net in E having a bounded residual subset is Cauchy if the order cone C has a bounded base. Using this result, he (Ref. 11) proved that every complete bounded subset of E has (Z) and so Theorem 1.2 follows from Zorn’s Lemma. However if the completeness assumption of A is replaced by the sequential completeness assumption then his method does not work here as one can only assert from the sequential completeness of A that every countable totally ordered subset of A has a lower bound. Thus, Theorem 3.1 and Corollary 3.1 cannot be proved by Jameson’s method.

Remark 3.2. Let Λ be a noncountable index set and E_{λ} = R for each λ ∈ Λ, and let E be the product space of {E_{λ}} equipped with the product topology. Take A = {x = (r_{λ})_{λ∈Λ} ∈ E| |r_{λ}| ⩽ 1 for all λ ∈ Λ and there exists a countable subset Λ_{0} of Λ such that r_{λ} = 0 for each λ ∈ Λ \ Λ_{0}}. Then A is bounded and sequentially complete. Let Θ = {x = (r_{λ})_{λ∈Λ} ∈ E|1 ⩽ r_{λ} ⩽ 2 for each λ ∈ Λ} and C = cone(Θ). Then C is a closed convex cone with a bounded base Θ. By Theorem 3.1, Min(A, C) ≠ ∅. But A is not complete. [Take y ∈ E such that each component of y is 1. Then y ∈ ∂A \ A. This implies that A is not complete]. Thus, one cannot induce Min(A, C) ≠ ∅ from Theorem 1.2. Therefore, Theorem 3.1 is a proper generalization of Theorem 1.2.

Remark 3.3. In 1962, Bishop and Phelps (Ref. 14) proved by Zorn’s Lemma Theorem 1.2 for the case when E is a normed space and C = {x ∈ E|∥x∥ ⩽ f(x)} with f ∈ E^{∗}; later Phelps (Ref. 15) gave another proof of Theorem 1.2 in which Zorn’s Lemma is not used explicitly. But, Phelps’ later proof relied on the fact that there exists f ∈ E^{∗} \ {0} such that C ⊂ {x ∈ X|∥x∥ ⩽ f(x)}. Existence of such a f in E^{∗} depends on Hahn-Banach Theorem, and the proof of Hahn-Banach Theorem depends on Zorn’s Lemma. Our proof is different from both Jameson’s and Phelps’ and is completely free from Zorn’s Lemma.

4 Application

The Bishop-Phelps theorem is an important result in geometry of Banach spaces, which asserts that if A is a closed convex subset of a Banach space then support points of A are dense in the boundary ∂A of A. In this section, we will apply our results in the preceding section to generalize the Bishop-Phelps theorem to sets in general topological vector spaces. In what follows, we assume that E is a topological vector space. For a subset A of E, we denote by ∂_{a}A the algebraic boundary of A.

Definition 4.1. Let A be a subset of E and B be a bounded convex subset of E. We say that a ∈ A is a B-support point of A if there exist a linear functional f on E and a subspace E_{0} containing B ⋃{a} such that f is bounded on B and is not constantly zero on E_{0}, and such that

(f + δ_{E_{0}})(a) = inf{(f + δ_{E_{0}})(x)|x ∈ A}.

Here δ_{E_{0}} is the indicate function of E_{0}: δ_{E_{0}}(x) = 0 if x ∈ E_{0} and δ_{E_{0}}(x) = +∞ if x ∈ E \ E_{0}.

Theorem 4.1. Let A, B be convex subsets of a topological vector space E such that B is bounded and that A is closed and sequentially complete. Then B-support points of A are dense in ∂A.

If E is a normed space and B is the unit ball of E, then the B-support points of A are just the usual support points of A. Hence Theorem 4.1 is a generalization of the Bishop-Phelps theorem. To prove Theorem 4.1, we need the following lemmas, which are of some independent interest.

Lemma 4.1. Let A ⊂ E be closed. Then cl(∂_{a}A) = ∂A.

Proof. Clearly, ∂_{a}A ⊂ ∂A and so cl(∂_{a}A) ⊂ ∂A. It remains to show that for any x ∈ ∂A and any balanced neighborhood V of 0 in E, (x + V) ⋂ ∂_{a}A ≠ ∅. Since x + V is not a subset of A, there exists h ∈ V such that x + h ∉ A. Let

t_{0} = sup{0 ⩽ t < 1|x + th ∈ A}.

By the closedness of A and x + h ∉ A, one has t_{0} < 1 and x + t_{0}h ∈ A. It is clear that

(x + t_{0}h + (0, 1 − t_{0}]h) ⋂ A = (x + (t_{0}, 1]h) ⋂ A = ∅.

It follows that x + t_{0}h ∈ ∂_{a}A. By the balancedness of V, x + t_{0}h ∈ x + V. Therefore x + t_{0}h ∈ (x + V) ⋂ ∂_{a}A.

Remark 4.1. Let E be an infinite dimensional normed space, let A be a bounded and weakly closed set in E. Then A has no interior points with respect to the weak topology. It follows from Lemma 4.1 that A equals to the weak-closure of ∂_{a}A, and hence also equals to that of ∂A because the normboundary ∂A certainly contains ∂_{a}A. In particular, the sphere of every closed ball in E is dense in the ball with respect to the weak topology.

Lemma 4.2. Let A, B be convex subsets of E such that B is bounded and that A is closed and sequentially complete. Let z ∈ ∂A and let U be a neighborhood of 0 in E. Then there exist a bounded balanced closed convex subset Θ of E, r > 0 and x_{0}, a ∈ spanΘ satisfying the properties:

(i) B ⊂ Θ.

(ii) x_{0} ∉ rΘ.

(iii) a ∈ (z + U) ⋂ A.

(iv) a ∈ Min(A, cone(x_{0} − rΘ)).

Proof. By Lemma 4.1, without loss of generality we can assume that z ∈ ∂_{a}A. Thus, by the convexity of A, there exist δ > 0 and h ∈ E such that

(z + (0, δ]h) ⋂ A = ∅.

Let

Θ = cl({t_{1}b_{1} − t_{2}b_{2} + t_{3}z + t_{4}h|b_{1}, b_{2} ∈ B, ∑_{i=1}^{4} |t_{i}| ⩽ 1}).

Thus z, h ∈ Θ. By the convexity of B, it is clear that Θ is a balanced bounded closed convex set. In particular, there exists ε > 0 such that [0, ε]Θ ⊂ U. Let l = min\{δ, \frac{ε}{2}\} and y = z + lh. Then y ∉ A. Since E \ A is open, there exists 0 < r < \frac{ε}{2} such that

y + rΘ ⊂ E \ A  (7)

It follows from z ∈ A that

x_{0} := z − y = −lh ∉ rΘ.

This implies that C := cone(x_{0} − rΘ) is a convex cone with a bounded base x_{0} − rΘ. Hence C is closed (Ref. 16). For each w ∈ (z − C) ⋂ A, there exist θ ∈ Θ and t ⩾ 0 such that w = z − t(x_{0} − rθ). We claim that t ⩽ 1. Indeed, if this is not the case, by the convexity of A and z ∈ A, one has

w_{1} := z − (x_{0} − rθ) = y + rθ ∈ A.

It follows that w_{1} ∈ (y + rΘ) ⋂ A, contradicting (7). Hence t ⩽ 1 and so

w − z = −t(x_{0} − rθ) = t(y − z + rθ) = t(lh + rθ)
∈ t(l + r)Θ ⊂ [0, ε]Θ.

Therefore, w ∈ z + [0, ε]Θ. This shows that

(z − C) ⋂ A ⊂ z + [0, ε]Θ  (8)

Since C is closed and A is sequentially complete, (z − C) ⋂ A is seen a sequentially complete bounded set. By Corollary 3.1, there exists a ∈ Min((z − C) ⋂ A, C), and so a ∈ Min(A, C). Noting that (z − C) ⊂ spanΘ, one has a ∈ spanΘ. By (8) and [0, ε]Θ ⊂ U, one has a ∈ z + U.

Proof of Theorem 4.1 Let z ∈ ∂A and let U be a neighborhood of 0 in E. By Lemma 4.1, take the corresponding Θ, r > 0, x_{0} and a. It suffices to show that a is a B-support point of A. Write E_{0} for spanΘ and let A_{0} = A ⋂ E_{0}. Then

(a − cone(x_{0} − rΘ)) ⋂ A_{0} = {a}  (9)

Note that a − x_{0} is an algebraic interior point of a − cone(x_{0} − rΘ) in E_{0}. By (9) and the Separation Theorem (Ref. 11), there exists a non-zero linear functional on E_{0} such that

sup{f_{0}(x)|x ∈ a − cone(x_{0} − rΘ)} ⩽ inf{f_{0}(x)|x ∈ A_{0}}.

It follows that

f_{0}(a) = inf{f_{0}(x)|x ∈ A_{0}}  (10)

and for each x ∈ cone(x_{0} −rΘ), f_{0}(x) ⩾ 0. Hence sup\{f_{0}(θ)|θ ∈ Θ\} ⩽ \frac{1}{r}f_{0}(x_{0}). Since Θ is balanced, this implies that f_{0} is bounded on Θ, and so on B as B ⊂ Θ. Pick a linear functional f on E such that f(x) = f_{0}(x) for each x ∈ E_{0}. By (10),

(f + δ_{E_{0}})(a) = inf{(f + δ_{E_{0}})(x)|x ∈ A}.

Therefore, a is a B-support point of A.

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