Consider a convex function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, such that for all $x \in \mathbb{R}^n$ we have
$$ a_1^\top x + b_1 \leq f(x) \leq a_2^\top x + b_2 $$
for some given $a_1, a_2 \in \mathbb{R}^n$, $b_1, b_2 \in \mathbb{R}$.
Assume that $f(x) \leq 0$ for all $x \in \text{conv}( \{x^1, ..., x^N \} )$.
Say if there exists $( c,d ) \in \mathbb{R}^{n} \times \mathbb{R}$ such that the halfspace $$H(c,d) := \{ x \in \mathbb{R}^n \mid c^\top x \leq d \}$$ is such that
$$ f(x) \leq 0 \ \ \forall x \in H(c,d) \\ x^i \in H(c,d) \ \ \forall i=1,...,N $$
Comment. The statement is clearly true if $f$ is affine. I am wondering about this more general case.
I read the beginning of your question and the comment. :-) It seems that the function $f$ should be affine, by the following. If $a_1\not=a_2$, then there is a vector $x=\lambda(a_1-a_2)\in\mathbb R^n$ such that $a_1^\top x+b_1>a_2^\top x+b_2$. Moreover, if $ a^\top x + b_1 \leq f(x) \leq a^\top x + b_2$ for all $x \in \mathbb{R}^n$ and $f$ is a convex, then $f$ should be affine, because $f$ is affine on each one-dimensional subspace of $\mathbb R^n$. Indeed, let $g$ be a convex function on $\mathbb R$ such that for all $x\in\mathbb R$ we have $ax + b_1 \leq g(x) \leq ax + b_2$ for some given $a,b_1,b_2\in\mathbb R$. Then a function $h:\mathbb R\to\mathbb R$ such that $h(x)=g(x)-ax$ for every $x\in\mathbb R$ is a convex bounded function on $\mathbb R$, and, hence, constant.