On the translations of quasimonotone maps and monotonicity
© Farajzadeh et al.; licensee Springer. 2012
Received: 4 April 2012
Accepted: 15 August 2012
Published: 31 August 2012
We show that given a convex subset K of a topological vector space X and a multivalued map , if there exists a nonempty subset S of with the surjective property on K and is quasimonotone for each , then T is monotone. Our result is a new version of the result obtained by N. Hadjisavvas (Appl. Math. Lett. 19:913-915, 2006).
1 Introduction and some definitions
It is clear that a monotone map is pseudomonotone, while a pseudomonotone map is quasimonotone. The converse is not true. If T is pseudomonotone (quasimonotone) and , then is not pseudomonotone (quasimonotone) in general. In the case of a single-valued linear map T defined on the whole space , it is known that if is quasimonotone, then T is monotone . Many authors (see, e.g., [4, 5]) extended this result for a nonlinear Gateaux differentiable map defined on a convex subset K (of a Hilbert space) with a nonempty interior.
Recently, Hadjisavvas  extended the above result to the multivalued maps defined on a convex subset of a real topological vector space with no assumption of differentiability or even continuity on the map T whose domain need not have a nonempty interior. In this paper, we first introduce the surjective property of a subset of on a segment of K. By using this concept, we can extend the corresponding result obtained in . Before stating the main result, we recall some definitions.
Hence if S has the surjective property on x, y, then the image of S under the linear functional is all of the real numbers, and that is why we used the phrase surjective property.
Definition 2 Let be a nonempty set and . We say that S has the surjective property on K if for every there exists such that S has the surjective property on x and y.
Also the straight line , where with , is said to be perpendicular to K if v is perpendicular to K.
Remark 1 If is a nonempty convex set and with v is not perpendicular to K, then the straight line has the surjective property on K. Indeed, let be an arbitrary member of K. Because v is not perpendicular to K, there exists such that . For each , we put and so . Hence . This means that S has the surjective property. Therefore, v being not perpendicular to K implies the surjective property while the simple example , and shows that the converse does not hold in general. In this example, one can see that S has the surjective property and is perpendicular to K (note ). The notion v is not perpendicular to K, which plays a crucial rule in proving the main results in ; while in this note, the surjective property has an essential rule in the main result. Hence one can consider this paper as an improvement of  (slightly, of course).
We need the following lemma in the sequel.
Lemma 1 Let X be a real topological vector space, K a nonempty convex subset of X anda multivalued map. Suppose, has the surjective property on x, y andis quasimonotone on the line segmentfor all. Then T is monotone on.
which is a contradiction. This completes the proof. □
Now we are ready to present the main result.
Theorem 1 Let X be a real topological vector space, K a nonempty convex subset of X anda multivalued map. Assumeis connected and has the surjective property on K. Ifis quasimonotone for all, then T is monotone on K.
This means T is monotone and the proof is now complete. □
Remark 1 shows that Theorem 1 is a new version of Theorem 1 in , although our proof is, in fact, completely similar to it.
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