and $x_2^*+\epsilon<0.$ By the lower semicontinuity of $F$ there exists an open neighborhood $U_x$ of $x$, such that for every $y\in U_x\cap I$ one has $F(y)\cap V_1\neq\emptyset$ and $F(y)\cap V_2\neq\emptyset.$ Let $y\in U_x\cap J.$ Then, according to the hypothesis of the lemma, there exists an open neighborhood $U_y$ of $y$ such that $F|_{U_y\cap J}$ is quasimonotone.

Let $U=U_x\cap U_y\cap J$ and let $u,v\in U,\, u\neq v.$ Then, there exist $u_1^*,u_2^*\in F(u),\, v_1^*,v_2^*\in F(v)$ such that $u_1^*>0,v_1^*>0,\mbox{ and }u_2^*<0,v_2^*<0.$

Hence, $$\<u_1^*,v-u\>>0,\mbox{ or } \<u_2^*,v-u\>>0,$$ on the other hand $$\<v_1^*,v-u\><0,\mbox{ or } \<v_2^*,v-u\><0,$$ which contradicts the fact that $F$ is quasimonotone in $U.$

Consequently, for any $x\in I$ one has $F(x)\ge 0,$ or $F(x)\le 0.$

Further, we need the following result which proof is straightforward.

\[p32\] Let $F:I\subseteq{\mathbb{R}}\rightrightarrows {\mathbb{R}}$ a set-valued function.

-   If $F(x)\ge 0$ for every $x\in I$, (respectively $F(x)\le0$ for every $x\in I$) , then $F$ is quasimonotone on $I$.

-   If $F(x)=\{0\}$ for some $x\in I$, and $F(z)\le 0$ for every $z<x,\,z\in I,$ respectively $F(y)\ge 0$ for every $x< y,\,y\in I,$ then $F$ is quasimonotone.

Now we are able to state and prove the following result concerning on the strict quasimonotonicity of a set-valued function.

\[t31\] Let $I\subseteq 