Abstract
Suppose is a cone contained in real vector space . When does contain a hyperplane that intersects each of the 0-rays in exactly once? We build on results found in Aliprantis, Tourky, and Klee Jr.’s work to give a partial answer to this question. We also present an example of a salient, closed Banach space cone for which there does not exist a hyperplane that intersects each 0-ray in exactly once.
1. Introduction
Let be a vector space of finite or infinite dimension over the reals. A 0-ray is an open ray whose source is the origin. We consider the origin of , , to be a 0-ray. For us, a cone is any union of 0-rays (precise definitions for 0-rays and cones are given in Section 2). Many results involving cones, especially convex cones, require the existence of a hyperplane which intersects each -ray of exactly once. For example, see Garrett Birkhoff’s original proof of his Projective Contraction Mapping Theorem [1], which is discussed in detail in [2]. There seems to be a relatively small amount of literature on the existence of such hyperplanes. Perhaps the most accessible source is Aliprantis and Tourky’s, “Cones and Duality” [3]. On the other hand, there is a large body of literature on closely related topics: on the separation of convex bodies by hyperplanes (the various separation versions of the Hann-Banach Theorem) and on the support of cones and convex sets by hyperplanes: Aliprantis and Border [4] or Klee Jr. [5].
Sections 2 and 3 consist of definitions and lemmas which lead to our main results, which are found in Section 4. In Theorem 21 we show that there exists a linear functional such that on if and only if there exists a hyperplane which intersects each -ray exactly once. Our theorem is a slight generalization of a similar result for cone bases (cone bases are defined in Definition 16), given by Aliprantis and Tourky: Theorem 1.47, page 40 of [3], which we present as Theorem 20. Aliprantis and Tourky’s results, which involve cone bases, require convexity, whereas ours, which replace cone bases with hyperplanes, do not. We explore the relationship of hyperplanes to cone bases, Lemmas 14 and 15, as well as Corollary 22. Theorem 24 gives an alternative proof of Aliprantis and Tourky’s previously cited Theorem 1.47. These results, combined with results of Klee Jr. [5, 6], on linear functionals, give a partial answer to when one should expect such a cone-intersecting hyperplane to exist. We end our short paper in Section 5 with an interesting counterexample involving Banach space cones. Before we prove our results, we include the following illustrative example.
Example 1. Let . Let be the cone of nonnegative vectors: Let be the linear functional defined as follows: So, on . Let It is easy to see that the hyperplane intersects each 0-ray of exactly once, at the point .
The following is clear. If and we let , then .
Notes. has a nice geometric interpretation; it is standard simplex; in probability theory is the space of all possible probability distributions for processes with possible outcomes.
2. Cones, 0-Rays, and Hyperplanes
Let be a real linear space, that is, any real vector space of finite or infinite dimension. Geometrically speaking, a cone is a subset of which can be represented as a union of rays emanating from a single source point. If that source point is considered to be part of the cone, we say the cone is pointed. In this paper we will always assume that the source of the cone is the null vector (the origin ) of . This assumption leads to the concise algebraic definition.
Definition 2. A subset of a vector space is called a cone if it is closed under positive scaling, that is, if whenever .
Definition 3. The ray emanating from the origin and passing through the point is denoted by . As a point set, Such a ray will be called a 0-ray to emphasize its source.
The reason for the strict inequality in Definition 3 is to make 0-rays into equivalence classes.
Proposition 4. Let be a vector space of any dimension.(1)The 0-rays partition into equivalence classes.(2)If is an element of cone , then . Hence, any cone is partitioned by its 0-rays.(3)Suppose . Then, if and only if for some .(4)0-ray consists of single point .
Proof.
Proof of (1). Let . Since it follows that If , then . But then for some . This allows us to write . But then Proof of (2) and (3). These two results are a trivial consequence of part and the definitions of cones and 0-rays.
Proof of (4). This result is a trivial consequence of the definition of a 0-ray.
Definition 5. The cone opposite to is denoted by , algebraically:
Definition 6. The cone is salient (or bounded) if .
Definition 7. One will say that the cone is closed if it is a closed subset in ’s topology.
Definition 8. If , then is the smallest vector space in containing . Alternatively, is the set of all finite linear combinations of vectors from .
The following proposition indicates why salient cones are sometimes called bounded.
Proposition 9. If is a salient and closed cone contained in normed vector space , then contains no lines.
Proof. Let and be any two distinct points on a line contained in . Then . Since is closed under positive scaling the following two sequences are contained in . They also are contained in , which, being a finite dimensional, is complete. Since is complete and is closed, these two sequences converge, respectively, to the following two points in : This contradicts being salient.
In Proposition 9 the requirement “ is closed” is necessary as the following example shows.
Example 10. Consider the open upper half plane: . is closed under positive scaling so is a cone. Since it follows that and so . Thus is a salient cone contained in , a Banach space. However, cone contains every line for each . is not a counterexample to Proposition 9 since is not topologically closed.
The topologically closed upper half plane, , is a Banach space cone which contains every line for each . is not a counterexample to Proposition 9 since is not salient: and for each .
Definition 11. If and is a vector subspace of , then one calls a hyperplane.
The following standard result, regarding the smallest hyperplane generated by a subset , of a vector space , is useful.
Lemma 12. Let be any subset of vector space . Let be any fixed element of and let Then is the smallest hyperplane containing .
Proof. is a vector subspace of . So is a hyperplane. Let . Then, . So . Now, let be any hyperplane that contains . Since , we can write , for some vector subspace . But then, . This implies . If , then So . Hence, .
The following standard result about convex sets is stated without proof. See Klee [7] or Lay [8] for details.
Proposition 13. Suppose is a convex set contained in . If and are positive real numbers, then
3. Intersection Lemmas
Lemma 14. Suppose that is a cone contained in , a vector space over the reals of finite or infinite dimension, and that there exists linear functional from to the reals such that for each . Let . For each , let be the central projection of onto . Then (1) is salient.Suppose ; then (2) if and only if ;(3);(4);(5) intersects each 0-ray in once and only once; in particular, ;(6), so that is a hyperplane in .
Proof. (1) If , then . But then and , which is impossible. So , implying is salient.
(2) Let . Then . If , then , which implies . By Proposition 4, part , the 0-rays are equivalence classes. Hence, .
On the other hand, if , then there exists such that . But then (3) (4) , so . The 0-rays are equivalence classes, by Proposition 4, part , so .
(5) by part of this lemma. becauseSo . The following shows . Let . By part of this lemma, . So, for some . is in . So, . By (14), , so and .
(6) We can use the following elementary, standard result: if is any linear functional (not identically zero) from to , then is a hyperplane . Moreover, can always be written in form , where is any element of . To prove this, simply use the linearity of and note that being nonidentically zero implies . To prove , note that and .
Lemma 15. Suppose that is a cone contained in , an arbitrary vector space of finite or infinite dimension; that contains at least one nonzero vector; and that there exists hyperplane which intersects every 0-ray in exactly once. (1)Then there exists linear functional mapping to the reals such that on .One can sharpen this result. For each , let be the unique intersection of 0-ray and hyperplane . In other words, . (2)Linear functional , mentioned in part (1), can be chosen so that ; we have . With this choice of , we have on and on .
Proof. The representative Figure 1 serves to illustrate some of the arguments detailed in this proof. In Figure 1 the vector space is represented as ; the cone as ; the hyperplane as the solution to ; the nonzero element of as ; and the vector subspace as the solution to . In our proof the basis for subspace will be denoted by ; in the figure it is represented by . In our proof the basis for will be denoted by ; in Figure 1 it is represented by . The linear functional , mentioned above (in part of this lemma), is given by .
Let , such a exists by this lemma’s main assumption.
Claim 1. is a vector subspace of . Proof is as follows: since is hyperplane there exists vector and vector subspace such that . But then, for some . Hence .
Claim 2. . Proof is as follows: if then for some , but then . That means . This contradicts with the fact that each ray intersects exactly once. So claim 2 is proven.
Let be a basis for . Since , which is a vector subspace of , the set forms a basis for . Since it follows that .
If , let . If is a proper vector subspace of we can extend the basis to a basis for . See Theorem 4.72 in [9] regarding extensions of bases.
For each we define as follows. We can write uniquely as a finite linear combination of basis elements from : where . Define .
If then, by this lemma’s assumption, the set contains a single element, which we will call . So . Since we must have with . By Proposition 4, part , , so there exists such that . So . So and . Hence, if , then and .
The following definition comes from Aliprantis and Tourky [3].
Definition 16. is a base for the cone if is a convex subset of and if for each there exists a unique and a unique such that .
Proposition 17. If is a base for cone and , then intersects 0-ray in exactly one point.
Proof. By the definition of a cone base, for a unique and a unique . But then . If , then since is a 0-ray, for some . But then . Uniqueness forces .
Lemma 18. Suppose is a base for cone . Let be the smallest hyperplane containing . Then intersects each 0-ray in exactly once. Moreover, .
Proof. Suppose is a 0-ray in . Since is a cone base for a unique and a unique . So we can write in terms of : . This means every element in can be written in the form for some . We will show that if , then , which will prove the lemma.
By Lemma 12, we can write in the form , where . So, if , we can writewith , , and , and . A little algebra transforms (16) intoprovided . We can rewrite (17) in the following form:where all coefficients are positive by letting and by relabeling if necessary, that is, by switching with when .
We can simplify (18) by letting and rearranging terms:where is convex. So Proposition 13 implies . Since , if , the convexity of impliesWe expand (22) using (20):We find a such that the coefficient of in the third line of (23) is zero. That is, we find the intersection of line segment , which is contained in by convexity, with the 0-ray . See Figure 2.
Here is the algebra: So, with this choice of , (23) becomes But , which implies . So , which means we can write as multiples of two distinct elements from , namely, and . That is, and . This contradicts being a cone base. Hence, our assumption that must have been wrong.
Lemma 19. Suppose that is a convex cone and is a hyperplane which intersects each 0-ray in exactly once. Then is cone base for .
Proof. is convex by assumption. Hyperplanes are always convex. Since the intersection of convex sets is convex, is convex. If then set contains a single, unique point, say . Since , there exists such that . Since , is unique. Let . Then with being unique with respect to and being unique with respect to .
4. Intersection Theorems
Theorem 20, below, can be found in Aliprantis and Tourky [3]. It relies on convexity.
Theorem 20. Suppose that convex cone is a subset of , an arbitrary vector space of finite or infinite dimension, and that contains at least one nonzero vector. Then the following are equivalent. (1)There exists base for cone .(2)There exists a linear functional on which is strictly positive on .
Proof.
Sketch. The linear functional is defined as follows. Suppose . Since is a cone base for there exists a unique and a unique such that . Define . Aliprantis and Tourky use the convexity of to show that functional is linear. For details and the rest of the proof, see Aliprantis and Tourky [3], Theorem 1.47, page 40.
Theorem 21, below, is a slight generalization of Theorem 20. Our version of the theorem does not require convexity and it replaces the cone base by an intersecting hyperplane.
Theorem 21. Suppose that cone is a subset of , an arbitrary vector space of finite or infinite dimension, and that contains at least one nonzero vector. Then the following are equivalent. (1)There exists a linear functional on which is strictly positive on .(2)There exists hyperplane such that intersects each 0-ray in exactly once.
Proof. (1) is equivalent to (2) is proven in Lemmas 14 and 15.
Corollary 22. If cone has a cone base, then both of the following are true. (1)There exists a linear functional on which is strictly positive on .(2)There exists hyperplane such that intersects each 0-ray in exactly once.
Proof. Lemma 18 combined with Theorem 21 proves Corollary 22.
The converse of Corollary 22 is not true as the following example shows.
Example 23. Let and . Let be the nonconvex cone . Hyperplane intersects each 0-ray in exactly once. However no cone base exists for .
If the cone is convex, then the converse of Corollary 22 is true. See Theorem 24.
Theorem 24. Suppose that convex cone is a subset of , an arbitrary vector space of finite or infinite dimension, and that contains at least one nonzero vector. Then the following are equivalent. (1)There exists base for cone .(2)There exists a linear functional on which is strictly positive on .(3)There exists hyperplane such that intersects each 0-ray in exactly once.Suppose (1), (2), or (3) holds. Let be the smallest hyperplane containing base . Then (i);(ii) will intersect each 0-ray in exactly once.
Proof. (1) is equivalent to is proven in Lemmas 18 and 19. For an alternative proof that is equivalent to see Aliprantis and Tourky [3], Theorem 1.47, page 40. is equivalent to as proven in Lemmas 14 and 15. (i) and (ii) are proven in Lemma 18.
Corollary 25. If is a convex, salient, closed cone in finite dimensional normed linear space , then there exists hyperplane such that intersects each 0-ray in exactly once. is compact.
Proof. According to Corollary of Aliprantis and Tourky [3], Klee Jr. [6] proved that every convex, salient, closed cone in a finite dimensional normed linear space has a compact base, . But then Theorem 24 implies there exists a hyperplane, , which intersects each 0-ray in exactly once and .
Remark 26. Klee Jr.’s paper, “Separation Properties of Convex Cones” [5], much referenced in the literature, shows that a closed, salient, convex cone in a separable normed linear space will have associated to it a linear functional which is strictly positive on . However, the following example shows that given an arbitrary salient cone in a Banach space , we cannot always find a strictly positive linear functional on . By Theorem 24, this means we cannot always find hyperplane which intersects every 0-ray in exactly once.
5. Banach Space Counterexample
We cannot always find hyperplane which intersects every 0-ray of a closed cone exactly once, even if the underlying vector space is Banach, as the following example, based upon Problem 6, page 42 of [3], shows. By Theorem 21, the existence of such a hyperplane is equivalent to the existence of a nonzero linear functional on which is strictly positive on .
Example 27. Let the set of all bounded functions from an uncountable set, to . equipped with the sup norm (if , then ), is an Banach Space. Let all the bounded nonnegative functions from to . For each let Then , for each . Note that , and if , then . If , are subsets of , we have the following identity:If , then we can write as disjoint union . Equation (27) impliesSuppose is any linear functional on ; that ; and that is any finite subset of . Then (28) impliesLet us suppose that linear functional on all of . If is strictly contained in , then (29) impliesLet . If the cardinality of is infinite, then must be infinite. We can see this as follows. Let be a finite set. By (30) and the definition of , , where is the size of . By (31), . So, if is infinite, then must be infinite. However, . So must be real, which means must be finite. But then is at most countable. This implies . Suppose . Then , which contradicts on .
Competing Interests
The author declares that there are no competing interests.