For example if I had the 2 dimensional rectangle R formed by the vertices $z_1,\ldots,z_4$ and we have a system of linear differential equations $\dot x=f(x)$ with $x \in \mathbb{R}^{n}$.
If we want that our system evolves inside this rectangle, then the system should not across the facets F1,F2,F3,F4 of this rectangle, which means the product of normal vector of Fi of unit length and pointing out of R with the dynamics of the system $\dot x$ has the same sign for all points on a facet Fi for example $n^Tf(x) \le 0$. For example if we want to verify this condition for every point $x$ on the facet defined by the vertices $z_1$ und $z_2$, it is sufficient to verify this condition : $n^Tf(x) \le 0$ just at the vertices $z_1$ and $z_2$, that is $n^Tf(z_1) \le 0$, $n^Tf(z_2) \le 0$, and not for every $x$ on the facet since we have a linear function $f$.
My question is: in case that the function $f(x)$ is a nonlinear function, why it is not sufficient to verify the above condition just at the vertices to say that is true for every $x$ on the facet?
I will be very grateful for your help.
Let us consider a polytope $P$ with extremal vertices $\{z_1,\ldots,z_N\}$. In such a case, we have that $P=\mathbf{co}\{z_1,\ldots,z_N\}$. In other words, we have that
$$P=\left\{\sum_{i}^N\lambda_iz_i:\lambda\ge0,||\lambda||_1=1\right\}.$$
Let us consider the facet given by the line that connects, for instance, $z_1$ and $z_2$, and consider a vector $v$. If both $v^TAz_1\le0$ and $v^TAz_2\le0$, then we have that $v^TAz\le0$ for all $z\in\mathbf{co}\{z_1,z_2\}$.
In the case of a more general function $f$, we do not necessarily have that $v^Tf(z_1)\le0$ and $v^Tf(z_2)\le0$ imply that $v^Tf(z)\le0$ for all $z\in\mathbf{co}\{z_1,z_2\}$. This will be the case, however, whenever the function $v^Tf$ is convex since we will have that
$$v^Tf(\lambda z_1+(1-\lambda)z_2)\le \lambda v^Tf(z_1)+(1-\lambda)v^Tf(z_2).$$
Therefore, if both $v^Tf(z_1)\le0$ and $v^Tf(z_2)\le0$, then we will have that $v^Tf(z)\le0$ for all $z\in\mathbf{co}\{z_1,z_2\}$.