Let $P \subset \Bbb R^n$ a polytope and $f : P \to \Bbb R$ a convex piecewise-linear function. This means there is polytopes $P_1, \dots, P_m$ such that $P = \cup_i P_i$ and $f_{|P_i}$ is linear.
Since $f$ is determined by its value on the vertices of each $P_i$ we get a vector $v_f \in \Bbb R^{|V|}$, where $V$ is the set of all vertices of the $P_i$'s.
We define $C \subset \Bbb R^{|V|}$ to be the set of vectors $v$ such that there is a function $g : P \to \Bbb R$ which is piecewise linear, convex; its domain of linearity exactly coincide with $f$, and such that $v_g = v$.
For example, if $P = [0,1]$ and $f$ is piecewise-linear on $[0,a_1], \dots, [a_n,1]$ then $C$ is exactly the cone $C = \{ (b_0, b_1, \dots, b_n, b_{n+1}) : \frac{b_i - b_{i-1}}{a_i - a_{i-1}} < \frac{b_{i+1} - b_i}{a_{i+1} - a_i }\}$.
This is a cone, i.e if $v,w \in C$ then for any positive real $a,b >0$ one has $av + bw \in C$.
Does such $C$ always contains an open ball $B$ ?
I did realize that for trivial reasons it's not possible. The condition that $f_{|P_i}$ is affine already forces $C$ to be contained in some hyperplane, so $C$ can't be fully-dimensional.