is strictly quasimonotone, which in particular shows that for some $u\in \operatorname*{int}(I)\cap U_x$ there exists $u^*\in
F(u),\, u^*>0.$ On the other hand, it can be deduced in the same manner, that there exists an open neighborhood $U_y$ of $y$, such that $F|_{U_y\cap I}
$ is strictly quasimonotone, which in particular shows that for some $v\in \operatorname*{int}(I)\cap U_y$ there exists $v^*\in F(v),\, v^*<0.$ Obviously we can assume $u<v.$ But then, $\<u^*,v-u\>>0$ and $\<v^*,v-u\><0$ which is in contradiction with the fact that $F$ is strictly quasimonotone on $\operatorname*{int}(I).$

The case when $I$ is closed from right can be treated similarly and we omit it.

\[r31\] Note that for the case when $F$ is locally monotone, locally pseudomonotone, respectively locally strictly pseudomonotone, one may use Proposition \[p31\] $(a),\,(c)$ and $(d)$ respectively.

The next example shows, that locally quasimonotone set-valued operators are not globally quasimonotone in general. Even more, our operator is also lower semicontinuous, hence searching for conditions that ensure global quasimonotonicity of a lower semicontinuous set-valued operator, based on its local quasimonotonicity, is meaningless.

\[ex1\] Consider the operator $$F:\mathbb{R} \rightrightarrows\mathbb{R} ,\, F(x):=\left\{
\begin{array}{lll}
[0,-x-1],\mbox{ if } x<-1, &  &  \\
\, 0,\mbox{ if } x\in[-1,1], &  &  \\
\, [-x+1,0],\mbox{ if } 