Abstract

Let be complete, simply connected n-dimensional Riemannian manifolds without conjugate points. Assume that is starshaped where . For every point , define : y lies on some geodesic segment in S from x to a point of kerS. There is a finite collection of all maximal sets whose union is S. Further, ker in .

1. Introduction

We begin with some definitions from [1–3]. A subset in a Riemannian manifold is convex if, for each pair point , there is a unique minimal geodesic segment from to and this segment is in . When dealing with a subset , the word “a unique minimal geodesic segment” should be replaced by “the geodesic.” A subset in a Riemannian manifold is starshaped if there is a point such that, for all , there is a unique minimal geodesic segment from to and this segment is in . The subset of consisting of all points like is called the kernel of   . In , a subset is starshaped if there is a point such that, for all , the geodesic segment joining and is contained in . The subset is convex if and only if . Throughout the paper, and will denote the interior and the boundary of the subset , respectively. Let denote the distance between the two points . For the pair of points , will denote the geodesic segment joining , . For and , . The distance to is locally maximal at if there is some neighborhood of such that   . Finally, all manifolds, maps, fields, and so forth are discussions to make sense. All curves are parametrized by arc-length.

Readers may refer to [4] for discussions concerning visibility via geodesic segment and starshaped sets. The behavior geodesic in has been discussed by many geometrers as Eberlein [2] and Goto [5]. For more properties of convexity and starshapedness in , see [6–9].

2. Main Results

Let be 2-dimensional manifolds without conjugate points, and assume that is starshaped. For every , we define . Clearly .

Lemma 1. Let be starshaped in 2-dimensional manifolds without conjugate point. Then (a)the set is convex,(b)if is any geodesic segment in joining points of , then ,(c)for every point , there is a unique point of closest to ; that is, there is a unique point of such that .

Proof. We have the following.(a)Since is a starshaped subset of , by [10], is convex.(b)Since is convex, clearly any geodesic segment in joining points of is in .(c)Let be an arbitrary point in . If , then the result is trivial. . Let there be two points and in closest to , that is, , to show that . Suppose on the contrary that . Since and are in , then . Let , such that and . However, is closer to than and , contradicting the assumption that and are closest to ; then .

Proposition 2. Let be starshaped in 2-dimensional manifolds without conjugate points. For every point , one defines . For any pair of points , if , then .

Proof. Let ; then where . Since , then where . This implies that ; then . We conclude that and the proof is complete.

Proposition 3. Let be a starshaped set. Let . If , then there is some point such that and hence . Moreover, is not locally maximal at .

Proof. Let and . Let be a geodesic segment from to where . Then is a geodesic segment in from to to . Hence . Then, by Proposition 2, we have . Moreover, , because . Then , and . Finally, it is clear that, for all points , . Then the distance to is not locally maximal at .

Proposition 4. Let . If is a maximal A set, then is locally maximal at .

Proof. Assume on the contrary that every neighborhood of contains some point for which . By Proposition 3, let . Then where . Then , so . By Proposition 2, . Since is maximal, then which means that where , which contradicts

Proposition 5. Let be a starshaped set. Let such that is locally maximal at . Then (a)the point is a vertex of ,(b)for every point near , , is not locally maximal at ,(c) for any point .

Proof. We have the following.(a)Since is locally maximal at , then and every neighborhood of contains some point such that . By Proposition 3, . Then must be a vertex of .(b)It is direct from (a).(c), because for any . Then .

Theorem 6. Let be a starshaped set. For every point , there is a finite subcollection of all maximal . Then, the following conditions hold:(a),(b)for every point , in if and only if is a maximal set.

Proof. Assume that .(a)For , belongs to at least one set . Either is in or where is in . This implies that ; then .The reverse inclusion is immediate, so the sets are equal.(b). Let be in ; then is a maximal set. To prove that is in , assume on the contrary that is not in . Since , by (a), there is in such that , but is a maximal set; this implies that . Then, is not subset of any set , so A(z) is in .

Lemma 7. Let be a starshaped set. Each point . Then (a)the set is convex,(b)if is any geodesic segment in joining points of , then .

Proof. We have the following.(a)Let ; then , because is convex. Moreover . Let , such that , respectively. We will prove that . Assume on the contrary that is not in ; then there is and . Let be an arbitrary point of such that , but which means that is not in , which is a contradiction.(b)It is clear that any geodesic segment in S joining points of is in , because is convex.

Theorem 8. Let be a starshaped set. Then .

Proof. Clearly . Now, we will prove that . Let ; then there is a point such that does not see via . Since , by Theorem 6, there is in such that ; by Lemma 7, is convex; then ; this implies that ; then and the proof is complete.

3. Conclusion

All results of the present work are valid in Euclidean space as manifolds without conjugate points [5], but the generalization of these results to , is more difficult and is left as open problem.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors are very grateful to the anonymous referees for their valuable suggestions and comments, which helped the authors to improve this paper.