Abstract
We prove a new result of existence of equilibria for an u.s.c. set-valued mapping on a compact set of which is epi-Lipschitz and satisfies a weak tangential condition. Equivalently this provides existence of fixed points of the set-valued mapping . The main point of our result lies in the fact that we do not impose the usual tangential condition in terms of the Clarke tangent cone. Illustrative examples are stated showing the importance of our results and that the existence of such equilibria does not need necessarily such usual tangential condition.
1. Introduction
Let be a set-valued mapping defined from into . A familiar result on existence of equilibria on convex compact sets is formulated as follows (see [1]).
Theorem 1. Let be an u.s.c. set-valued mapping defined on a convex compact set and suppose that is nonempty convex compact, . If then has equilibria on ; that is, such that .
Here is the tangent cone in the sense of convex analysis defined as , where denotes the set of all nonnegative real numbers and denotes the closure in . Assume now that is not necessarily convex and assume that in the tangential condition (1) the tangent cone is replaced by the Clarke tangent cone ; that is, In order to get the same conclusion of Theorem 1 in the nonconvex case, we need one more assumption on , which is the epi-Lipschitzness of . The following theorem is an extension of Theorem 1 to the nonconvex case (see [1]).
Theorem 2. Let be homeomorphic to a convex compact set in and let be an u.s.c. set-valued mapping with nonempty closed convex values. Assume that is epi-Lipschitz and (2) holds. Then has equilibria on .
It is very important to point out that for epi-Lipschitz sets the tangential condition (2) cannot be weakened to where is (generally greater) the contingent cone defined below in Section 2 (see Example in [1]).
Let and . Clearly is an epi-Lipschitz compact convex set in and is u.s.c. on with for , , and . However, the point satisfies . This shows that the tangential condition (2) in Theorem 2 is not necessary to get equilibria on . Our main purpose in this work is to replace in (2) by a new tangent set which is always larger than and to prove the existence of equilibria under the new tangential condition.
The main result of the paper is read as follows.
Theorem 3. Let be homeomorphic to a convex compact set in which is epi-Lipschitz and let be an u.s.c. set-valued mapping with nonempty closed convex values. Let be a convex compact set in with . Assume that there exists some such that for any If is tangentially regular, then there exists such that
We point out that the price that we pay for replacing (2) by (4) is the tangential regularity of the set (i.e., ) and it is an open question to extend the result of Theorem 3 to the general case without the tangential regularity assumption. The first corollary of Theorem 3 is the following result.
Corollary 4. Let be homeomorphic to a convex compact set in which is epi-Lipschitz and tangentially regular and let be an u.s.c. set-valued mapping on with nonempty closed convex values. Assume that (2) holds. Then has equilibria on .
Proof. Take in Theorem 3. All the assumptions are fulfilled and hence has equilibria on .
The paper is organised as follows. The next section is devoted to some preliminary concepts and results needed in the development of our approach. In Section 3 we prove our main result stated in Theorem 3 and we present an illustrative example showing the importance of our main result in Theorem 3. An application to fixed point results is presented at the end of paper.
2. Preliminaries
Throughout this section, we assume that is a Hausdorff topological vector space. We will denote by the topological dual of and by the pairing between the spaces and .
Let be a nonempty closed subset of and let be a point in . Let us recall [2, 3] the following classical tangent cones of Bouligand and of Clarke.(i)The Bouligand tangent cone (also called contingent cone) to at is the set of all such that, for every neighborhood of in and for every , there exists such that (ii)The Clarke tangent cone to at is the set of all such that for every neighborhood of in there exist a neighborhood of in and a real number such that (iii)The Rockefeller hypertangent cone to at is the set of all for which there exist a neighborhood of in , a neighborhood of in , and a real number such that Clearly . Recall that (see [3, 4]) is a closed convex cone and is a closed cone (that may be nonconvex) while is an open convex cone.
For a closed convex set and , both Clarke tangent cone and Bouligand tangent cone coincide and they are equal to the convex tangent cone . The class of nonempty closed sets satisfying the equality is called the class of tangentially regular sets.
In [5], we introduced the concept of -epi-Lipschitz sets as follows.
Definition 5. Let be a closed subset of and , and let be a bounded set in . One will say that is -epi-Lipschitz at in a direction if and only if the minimal time function associated with and is directionally Lipschitz at in the direction in the sense of Rockefeller [6], that is, if and only if there exists such that , whereand means and . Recall that the minimal time function is defined as follows:A geometric characterization of -epi-Lipschitz sets has been established in [5] saying that a set is -epi-Lipschitz at in a direction if and only if there exist , , , and such that In our analysis in this work we need the constant to be uniform with respect to the directions; that is, is only dependent on and in this case we say that is -epi-Lipschitz at in the direction . Since does not depend on the directions, we define the set of all directions satisfying ; that is,
Following the same lines of the proof of Proposition in [5] we can prove the following result.
Proposition 6. Let be a closed subset of and , , and let be a bounded set in . (i)If there exist , , , and such that then is -epi-Lipschitz at in the direction .(ii)Conversely, if there exists for which is -epi-Lipschitz at in the direction , then there exist , , and such that
Remark 7. Clearly, if is -epi-Lipschitz at , then is -epi-Lipschitz at for any . Consequently, any epi-Lipschitz set in the sense of Rockefeller [4] is -epi-Lipschitz for any and for any with and the constant is uniform for any . We note that the notion of -epi-Lipschitz sets recovers some well known concepts in variational analysis. (i)Obviously a closed set in is epi-Lipschitz in the sense of Rockefeller [4] if and only if it is -epi-Lipschitz in the sense of Definition 5 and if and only if it is -epi-Lipschitz and if and only if it is -epi-Lipschitz for any .(ii)In normed spaces, any compactly epi-Lipschitz set at with respect to a convex compact set in the sense of Borwein and Strójwas [7] is -epi-Lipschitz in the direction with , for any . We recall that is compactly epi-Lipschitz at with respect to a convex compact set in the sense of [7] provided that there exists such that (iii)Assume that is a normed space and and are closed sets in . If is -directionally Lipschitz in the sense of [8], then is -epi-Lipschitz with , for any .
Our main tools in the present work are two tangent sets associated with -epi-Lipschitz sets. The first tangent set is defined above and will be called the -hypertangent set and it characterizes the class of -epi-Lipschitz sets by its nonemptiness. The second tangent set will be called the -Clarke tangent set and is defined as the set of all satisfying that for any neighborhood of there exist and such that Observe that for both sets and coincide, respectively, with the hypertangent cone and the Clarke tangent cone .
Obviously, we always have the following inclusions: and . Consequently, any epi-Lipschitz set is -epi-Lipschitz. In our analysis we need to prove many properties for the tangent sets and . We notice that is an open set in and is a closed set in and both are not necessarily convex.
Proposition 8. Let be a Hausdorff topological vector space, let be a nonempty closed subset of , and let be a convex bounded set in with . Let . (1) is -epi-Lipschitz at if and only if .(2).(3).(4)If is epi-Lipschitz at , then .(5)If is epi-Lipschitz at , then .
Proof. (1) It follows directly from the definition of .
(2) Let and . By definition of there exist , , and such thatChoose a symmetric neighborhood of 0 in such that . By definition of there exist and such thatChoose now and such thatPut and . Fix any , any , and any . Then by (18) there exist and such that Put . Then the previous inclusion and (19) ensure that and hence we can use (17) to get Therefore, Thus, for any , any , and any we have This ensures by definition of the -hypertangent cone that .
(3) It is a direct consequence of Part (2) and the inclusions .
(4) First observe that since is epi-Lipschitz at we have that is nonempty. Hence, there exists some . Now, since and as is open, it is enough to show the inclusion . Consider . Then there exists a positive number such that . Since is a cone, we have and hence by Part (2) we obtain and so the proof of Part (4) is finished.
(5) Taking into account the inclusion and the fact that is closed, it is sufficient to show the inclusion .
Fix again some element . For any we have by Part (2) This ensures that because . So the proof is complete.
We require the notion of lower semicontinuity for set-valued mappings. Let be a set-valued mapping on , with values that are subsets of another Hausdorff topological vector space . We say that is lower semicontinuous (l.s.c.) at provided that, for every , there exists such that We say that is upper semicontinuous (u.s.c.) at provided that, for every , there exists such that In [6], the author proved the l.s.c. of the set-valued mapping whenever is epi-Lipschitz. In the following proposition we extend Theorem in [6] to the set-valued mapping with . This l.s.c. property is very important in the proof of our main result in Theorem 3.
Proposition 9. Let be a Hausdorff locally convex topological vector space, let be a nonempty closed subset of , and let , be a convex bounded in with . If is epi-Lipschitz at , then the set-valued mapping is lower semicontinuous at .
Proof. Let and . Assume that is epi-Lipschitz at . Then there exists . Let be any neighborhood of zero in . Since is locally convex we can choose a balanced neighborhood such that . Set . Clearly is a symmetric convex neighborhood of zero in . Fix any where is the Minkowski function associated with . Since is a cone, we have and hence by Part (2) in Proposition 8 we obtain Thus, there exists , , and such thatLet such that . Fix any . Let any . Then and so by (28) we get Therefore, for any we have . Hence, . Observe that which implies that and hence . Thus, . Consequently, for any and any there exists such that This proves that the set-valued mapping is lower semicontinuous at and the proof is complete.
Remark 10. Following the proof of the previous proposition, we can prove the lower semicontinuity of the set-valued mapping at whenever is epi-Lipschitz set at .
Lemma 11. For any and any one has
Proof. Let . Without loss of generality we assume that . Then, for any , there exists and such that . By definition of the hypertangent there exist , , and such that By adding to both sides we obtain This ensures by definition of that Therefore, and hence the proof is complete.
The next lemma establishes an analogue result for the -Clarke tangent set.
Lemma 12. Assume that is a convex bounded set with and that is epi-Lipschitz at . Then
Proof. Let . Since is epi-Lipschitz at , we have by Part (5) in Proposition 8 the equalities and . Therefore, the conclusion follows directly from the previous lemma.
It is a natural question to ask whether the inclusion in the previous lemma becomes an equality. The next lemma establishes a positive answer whenever the set is assumed to be tangentially regular. Its proof needs the following proposition which is also needed in the proof of Theorem 19. It has been proved in Proposition in [5] for -epi-Lipschitz sets. The proof stated below is a direct adaptation of the proof in [5] that we give for the sake of completeness of the paper. To do that, we need the following characterization of in terms of nets (see, e.g., [9]). A vector if and only if there exist a net of positive real numbers converging to zero and a net in converging to such that
Proposition 13. Let be a Hausdorff topological vector space, let be a nonempty closed subset of , let be a convex compact set in with , and let . Assume that for some one has ; then
Proof. By definition of , there exist , , and such that Choose such that and let be a net converging to in . Applying (38) we get, for any , the existence of such that Since is compact, we may extract a subnet of converging to some point . Put , , and . Then (39) ensures Since and , we deduce from the characterization of the Bouligand cone that . Thus, This completes the proof.
Lemma 14. Let . Assume that is convex compact with and that is epi-Lipschitz at . If, in addition, is assumed to be tangentially regular at , then one has
Proof. Let . Since is epi-Lipschitz at , then is -epi-Lipschitz at and hence . Fix any . By Proposition 13, we have . Consequently, the tangential regularity of at implies that Since is taken to be arbitrary in , then Taking the closure of both sides of the previous inclusion and taking into account the fact that , we obtain Therefore, the proof of this lemma is finished since the reverse inclusion is always true by Lemma 12.
A direct and very important result on the convexity of the set can be deduced from the previous lemma.
Corollary 15. Let , let be convex compact with , and let be epi-Lipschitz at . If is tangentially regular at , then and are both convex.
Using Lemma 14 we can easily construct many examples of closed sets and set-valued mappings for which the tangential condition (2) is not satisfied and the new tangential condition (4) is satisfied.
Example 16. Let , , (see Figure 1), and . Let , , and . Simple computations yield the following: Hence, , for And , for and for any . Although the set is closed convex, the tangential condition (2) is not satisfied while the tangential condition (4) is satisfied.
Consider now another example with that is a nonconvex tangentially regular set.
Example 17. Let and (see Figure 2). Clearly is a closed nonconvex set in . Also the set is epi-Lipschitz and tangentially regular. Let . We can check the following: Let and . Then using Lemma 14 we obtain Also, we can verify that for and for some points in we have For the rest of points in the Clarke tangent cone is strictly included in . Define now the set-valued mapping as follows: Clearly is an u.s.c. set-valued mapping with closed convex values. For this couple and we have the following facts:
(i)(ii) Thus, for the nonconvex epi-Lipschitz tangentially regular set , the tangential condition (2) is not satisfied and the tangential condition (4) is satisfied.
3. Existence Results
Throughout this section . Let us recall the following important facts needed in our next proofs.
Fact 1 (see [10]). For any finite covering of a compact set , with each being open and bounded, there exists a Lipschitz partition of unity ; that is, each is a Lipschitz function and the following hold: (i), , .(ii), .(iii), .
Fact 2 (see [11]). Let be a compact subset of and suppose that is an u.s.c. set-valued mapping on with images that are nonempty closed convex sets in . Let be given. Then there exists a sequence of set-valued mappings with closed convex values such that the following hold: (a), , ;(b) is l.s.c.
Fact 3 (see [1]). Let be homeomorphic to a convex compact set and let be an upper semicontinuous set-valued mapping on with nonempty closed convex values in . Assume that is epi-Lipschitz and suppose that (2) holds. Then has a zero on .
In order to prove the first main result we need to prove the following lemma.
Lemma 18. Let be a compact set. Assume that is an epi-Lipschitz compact set, , is l.s.c. on , and for any there exists such that Then there exists (not depending on nor on ) and a Lipschitz function such that
Proof. Since is epi-Lipschitz, we have that by Proposition 9 the set-valued mapping is l.s.c. at any . Let be too small so that . Thus, for any and any , the l.s.c. of both and imply the existence of such that Hence, the family of open ballsforms an open covering of . By compactness of there exists a finite subcover of and let be a Lipschitz partition of unity subordinate to this subcover (by Fact 1). Define now the Lipschitz function Then Therefore, the convexity of both -Clarke tangent sets (by Corollary 15) and images of the set-valued mappings imply that the selection satisfies Set . Then the fact that and for any and the fact that is balanced ensure the inclusion By Proposition 8 we have , and consequently we obtain which completes the proof.
The following result can be seen as an approximate equilibria result. It will be used to prove the main result of the paper.
Theorem 19. Let be homeomorphic to a convex compact set in which is epi-Lipschitz and let be an u.s.c. set-valued mapping with nonempty closed convex values. Let be a convex compact set in with . Assume that for any there exist some such that If is tangentially regular, then there exist and such that
Proof.
Case 1. First assume that is l.s.c. on and Then for any there exists, by the previous lemma, a Lipschitz function and a constant such that Let us introduce the ordinary differential equation In view of (65) we have for any that is, for any the set is -epi-Lipschitz at in the direction . Use now Proposition 13 to deduce that Since is tangentially regular, we get Define the set-valued mapping . Clearly this set-valued mapping satisfies the hypothesis of Fact 3 and so there exists such that . Clearly, Case 2. Assume now that is u.s.c. Let as and let be a sequence of l.s.c. approximations of as in Fact 2. Let as and let . Clearly is l.s.c. and , . Indeed, by Part (b) in Fact 2 we have and so obviously we obtain And hence Applying Case for all we obtain a constant and an element such that Consequently, from the monotonicity of (by Part (a) in Fact 1), we have Since is assumed to be compact, we can extract a subsequence of (still denoted by ) converging to a limit . Let be such that . Thenand therefore by Part (a) in Fact 1 we get Upon letting we obtain and hence the proof is complete.
Observe that in the previous theorem cannot be controlled since it depends on the pointwise constants in (4). However, if we assume that the tangential condition (4) is satisfied with a uniform (i.e., does not depend on and it is the same for any ), then we get the following first corollary in which the constant is the same satisfying (4).
Corollary 20. Let be homeomorphic to a convex compact set in and let be an u.s.c. set-valued mapping with nonempty closed convex values. Let be a convex compact set in with . Assume that is epi-Lipschitz and for some one has If is tangentially regular, then there exists such that
Using this corollary we prove our main result establishing an existence result of exact equilibria of on .
Theorem 21. Let be homeomorphic to a convex compact set in and let be an u.s.c. set-valued mapping with nonempty closed convex values. Let be a convex compact set in with . Assume that is epi-Lipschitz and for some one has If is tangentially regular, then there exists such that
Proof. We proceed by approximation. Fix any . First observe the following fact due to the balanced property of : Then (80) ensures Applying now Theorem 19 we obtain for any the existence of some point with . Using the fact that is compact we can extract a subsequence of converging to some limit and since obviously our assumptions on ensure that the limit satisfies , then the proof is complete.
We apply this existence result to the following two examples for which we cannot apply the results in [1] because the tangential condition (2) is not satisfied.
Example 22. Let , , , and . Since for any (by Example 16) and since all the assumptions on and in Theorem 21 are fulfilled, then there exists some with . In this example the equilibrium is and it is unique.
Example 23. Let , , and The set is the intersection of the set given in Example 17 with the first quarter in (see Figure 3). Using the computations presented in Example 17 we can check that, for any , Also, the set is homeomorphic to a convex compact set. Indeed, can be mapped continuously and with continuous inverse to the segment by projection. Thus, all the assumptions of Theorem 21 are satisfied and so there exists such that . In this example we do not have the uniqueness of the equilibria and there is an infinity of them. Indeed, for any with we have .
Let us now apply our main result in Theorem 21 for the existence of fixed points for set-valued mappings on nonconvex sets. It extends Theorems and in [1] from the case of and satisfying the tangential condition (80) with to the general case with any .
Theorem 24. Let be homeomorphic to a convex compact set in and let be an u.s.c. set-valued mapping with nonempty closed convex values. Let be a convex compact set in with . Assume that is epi-Lipschitz and tangentially regular and for some one has Then has a fixed point in ; that is, there exists such that .
Proof. It follows directly from Theorem 21 by taking .
Conflict of Interests
The author, Professor Messaoud Bounkhel, declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The author would like to extend his sincere appreciations to the Deanship of Scientific Research at King Saud University for funding this Research Group no. RGP-024.