On the translations of quasimonotone maps and monotonicity

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).

1 Introduction and some definitions

Throughout the paper, 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

$$〈x^{\ast }-y^{\ast },x-y〉\ge 0.$$

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:

$$〈y^{\ast },x-y〉\ge 0\Rightarrow 〈x^{\ast },x-y〉\ge 0$$

and

$$〈y^{\ast },x-y〉>0\Rightarrow 〈x^{\ast },x-y〉\ge 0.$$

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:

$$〈S,x-y〉=\{〈x^{\ast },x-y〉:x^{\ast }\in S\}=\mathbb{R}.$$

Remark that we can consider $x-y$ as a linear functional (denoted by $\widehat{x-y}$) on $X^{\ast }$ which is defined by

$$〈\widehat{x-y},f〉=〈f,x-y〉.$$

Hence if S has the surjective property on x, y, then the image of S under the linear functional $\widehat{x-y}$ 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]

Let 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.,

$$〈v,x〉=〈v,y〉,\mathrm{\forall }x,y\in K.$$

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 $c=〈v,x-y〉\ne 0$. For each $a\in \mathbb{R}$, we put $t=\frac{a-〈u,x-y〉}{c}$ and so $a=〈u+tv,x-y〉$. Hence $〈S,x-y〉=\mathbb{R}$. 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 $〈v=(1,1),(x,-x)〉=〈v=(1,1),(y,-y)〉=0$). 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:

$$a⪯b\iff t_1\le t_2,\text{where }a=x+t_1(y-x),b=x+t_2(y-x).$$

On the contrary, assume 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 $〈a^{\ast }-b^{\ast },a-b〉<0$. Hence we have

$$〈a^{\ast },y-x〉>〈b^{\ast },y-x〉.$$

Since S is surjective on x, y, there exists $w\in S$ such that

$$〈a^{\ast },y-x〉>〈w,x-y〉>〈b^{\ast },y-x〉.$$

Therefore,

$$〈a^{\ast }+w,y-x〉>0,〈b^{\ast }+w,y-x〉<0,$$

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 $〈S,y-z〉$ 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 $s\in ]0,1[$, 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

(1)

(2)

(3)

From (1) and (2), we deduce that

$$2s〈z_s^{\ast },z-\frac{x+y}{2}〉\ge 〈x^{\ast },z_s-x〉+〈y^{\ast },z_s-y〉.$$

(4)

Now from $z-z_s=(1-s)(z-\frac{x+y}{2})$ and (3), we obtain

$$〈z^{\ast },z-\frac{x+y}{2}〉\ge 〈z_s^{\ast },z-\frac{x+y}{2}〉.$$

(5)

Combining (4) and (5), we have

$$2s〈z^{\ast },z-\frac{x+y}{2}〉\ge 〈x^{\ast },z_s-x〉+〈y^{\ast },z_s-y〉.$$

So if in the previous inequality we tend $s\to 0$, then $z_s\to \frac{x+y}{2}$, and hence we deduce

$$0\ge 〈x^{\ast },y-x〉+〈y^{\ast },y-x〉.$$

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.