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# On the translations of quasimonotone maps and monotonicity

- AP Farajzadeh
^{1}, - A Karamian
^{1}and - S Plubtieng
^{2}Email author

**2012**:192

https://doi.org/10.1186/1029-242X-2012-192

© Farajzadeh et al.; licensee Springer. 2012

**Received:**4 April 2012**Accepted:**15 August 2012**Published:**31 August 2012

## Abstract

We show that given a convex subset *K* of a topological vector space *X* and a multivalued map $T:K\rightrightarrows {X}^{\ast}$, if there exists a nonempty subset *S* of ${X}^{\ast}$ with the surjective property on *K* and $T+w$ is quasimonotone for each $w\in S$, then *T* is monotone. Our result is a new version of the result obtained by N. Hadjisavvas (Appl. Math. Lett. 19:913-915, 2006).

## Keywords

- monotone map
- pseudomonotone map
- quasimonotone map
- surjective property

## 1 Introduction and some definitions

*X*and ${X}^{\ast}$ denote a real topological vector space and the dual space of

*X*, respectively. Suppose $K\subseteq X$ is a nonempty subset of

*X*and $T:K\rightrightarrows {X}^{\ast}$ is a multivalued map from

*K*to ${X}^{\ast}$. Recall that

*T*is said to be monotone if for all ${x}^{\ast}\in T(x)$, ${y}^{\ast}\in T(y)$ one has

*T*is said to be pseudomonotone and quasimonotone, in the sense of Karamardian (see [1, 2]), respectively, if for any ${x}^{\ast}\in T(x)$, ${y}^{\ast}\in T(y)$ the following implications hold:

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 $w\in {X}^{\ast}\mathrm{\setminus}\{0\}$, then $T+w$ is not pseudomonotone (quasimonotone) in general. In the case of a single-valued linear map *T* defined on the whole space ${\mathbb{R}}^{n}$, it is known that if $T+w$ is quasimonotone, then *T* is monotone [2]. 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 [3] 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 ${X}^{\ast}$ on a segment of *K*. By using this concept, we can extend the corresponding result obtained in [3]. Before stating the main result, we recall some definitions.

**Definition 1**Let

*x*,

*y*be two elements of

*K*. We say that $S\subseteq {X}^{\ast}$ has the surjective property on

*x*and

*y*whenever the following equality holds:

Hence if *S* has the surjective property on *x*, *y*, then the image of *S* under the linear functional $\stackrel{\u02c6}{x-y}$ is all of the real numbers, and that is why we used the phrase surjective property.

**Definition 2** Let $K\subseteq X$ be a nonempty set and $S\subseteq {X}^{\ast}$. We say that *S* has the surjective property on *K* if for every $x\in K$ there exists $y\in K$ such that *S* has the surjective property on *x* and *y*.

**Definition 3**[3]

*K*be a convex subset of

*X*. An element

*v*of ${X}^{\ast}$ is called perpendicular to

*K*if

*v*is constant on

*K*,

*i.e.*,

Also the straight line $S=\{u+tv:t\in \mathbb{R}\}$, where $u,v\in {X}^{\ast}$ with $v\ne 0$, is said to be perpendicular to *K* if *v* is perpendicular to *K*.

**Remark 1** If $K\subseteq X$ is a nonempty convex set and $u,v\in {X}^{\ast}$ with *v* is not perpendicular to *K*, then the straight line $S=\{u+tv:t\in \mathbb{R}\}$ has the surjective property on *K*. Indeed, let $x\in K$ be an arbitrary member of *K*. Because *v* is not perpendicular to *K*, there exists $y\in K$ such that $c=\u3008v,x-y\u3009\ne 0$. For each $a\in \mathbb{R}$, we put $t=\frac{a-\u3008u,x-y\u3009}{c}$ and so $a=\u3008u+tv,x-y\u3009$. Hence $\u3008S,x-y\u3009=\mathbb{R}$. This means that *S* has the surjective property. Therefore, *v* being not perpendicular to *K* implies the surjective property while the simple example $X={\mathbb{R}}^{2}$, $S=\{(t,t)=(0,0)+(1,1)t:t\in \mathbb{R}\}$ and $K=\{(x,-x):x\in \mathbb{R}\}$ shows that the converse does not hold in general. In this example, one can see that *S* has the surjective property and $v=(1,1)$ is perpendicular to *K* (note $\u3008v=(1,1),(x,-x)\u3009=\u3008v=(1,1),(y,-y)\u3009=0$). The notion *v* is not perpendicular to *K*, which plays a crucial rule in proving the main results in [3]; 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 [3] (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* *and*$T:K\rightrightarrows {X}^{\ast}$*a multivalued map*. *Suppose*$x,y\in K$, $S\subseteq {X}^{\ast}$*has the surjective property on* *x*, *y* *and*$T+w$*is quasimonotone on the line segment*$[x,y]=\{tx+(1-t)y:t\in [0,1]\}$*for all*$w\in S$. *Then* *T* *is monotone on*$[x,y]$.

*Proof*We can define an order on $[x,y]$ as follows:

*T*is not monotone on $[x,y]$. So there exist $a,b\in [x,y]$ and ${a}^{\ast}\in T(a)$, ${b}^{\ast}\in T(b)$ with $a\prec b$ and $\u3008{a}^{\ast}-{b}^{\ast},a-b\u3009<0$. Hence we have

*S*is surjective on

*x*,

*y*, there exists $w\in S$ such that

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* *and*$T:X\rightrightarrows {X}^{\ast}$*a multivalued map*. *Assume*$S\subseteq {X}^{\ast}$*is connected and has the surjective property on* *K*. *If*$T+w$*is quasimonotone for all*$w\in S$, *then* *T* *is monotone on* *K*.

*Proof*Let $x,y\in K$, ${x}^{\ast}\in T(x)$ and ${y}^{\ast}\in T(y)$ be arbitrary elements. If

*S*is surjective on

*x*,

*y*then, by Lemma 1,

*T*is monotone on $[x,y]$ and the proof is complete. Assume

*S*does not have the surjective property on

*x*,

*y*. So $S(x-y)\ne \mathbb{R}$. Since

*S*has the surjective property on

*K*, then there exists $z\in K$ such that

*S*is surjective on

*x*,

*z*; and since

*S*is connected, then $\u3008S,y-z\u3009$ is a connected subset of the real numbers unbounded from above and below, and so it is equal to the real numbers. This means that

*S*has the surjective property on

*y*,

*z*and also on $\frac{x+y}{2}$,

*z*. Therefore, it follows from Lemma 1 that

*T*is monotone on $[y,z]$ and $[\frac{x+y}{2},z]$. Similarly,

*T*is monotone on the segments $[x,{z}_{s}]$ and $[y,{z}_{s}]$, for all $s\in \phantom{\rule{0.2em}{0ex}}]0,1[$, where ${z}_{s}=sz+(1-s)\frac{x+y}{2}$. Therefore, for any ${z}_{s}^{\ast}\in T({z}_{s})$ and ${z}^{\ast}\in T(z)$, we have

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 [3], although our proof is, in fact, completely similar to it.

## Declarations

## Authors’ Affiliations

## References

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## Copyright

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