Let $\mathcal{A}$ be a set of convex subsets of $\mathbb{R}^n$ and for each $A\in\mathcal{A}$ let there be an affine linear map $f_A\colon \mathbb{R}^n\rightarrow\mathbb{R}$. Moreover, assume that for any $A,B\in \mathcal{A}$ and $v\in A\cap B$, it is $f_A(v)=f_B(v)$ (*).
I am interested in finding a necessary and sufficient condition on $\mathcal{A}$ such that for all choices of functions $(f_A)_{A\in\mathcal{A}}$ satisfying (*), there is an affine map $g$ such that $$g(v) = f_A(v) \text{ for all }A\in\mathcal{A}\text{ and }v\in A.\quad (1)$$
Motivation: this question arises in decision theory. One can think of the points in $A\in\mathcal{A}$ as utilities individuals could get under a certain condition, and the functions $f_A$ as aggregating these utilities into a social welfare function. There are results that for each of the sets $A\in\mathcal{A}$, the social welfare function is an affine linear function $f_A$ on the individual utilities. Moreover, I know that for two different sets, the functions need to coincide on their intersection. I now want to derive a necessary and sufficient condition for the social welfare function to be representable by one affine linear function of individual utilities for all sets in $\mathcal{A}$. I also know that all $A\in\mathcal{A}$ must be convex, although I haven't used that fact yet.
My own work: It is easy to find a counter-example that shows that some condition is needed. E.g., take two sets $A,B$ which are disjoint and both have full dimensionality in $\mathbb{R}^n$, two differing functions $f_A\neq f_B$ and $\mathcal{A}=\{A,B\}$.
As a means to finding such a condition I tried introducing sets $\Phi(A,B)\subseteq\mathbb{R}^n$ which include all points on which $f_A$ and $f_B$ must coincide. For instance, it is $f_A=f_B$ on $\mathrm{aff}(A\cap B)$ (where aff denotes the affine span) and hence $\mathrm{aff}(A\cap B)\subseteq\Phi(A,B)$. Also, $\Phi(A,B)\cap\Phi(B,C)\subseteq\Phi(A,C)$ for $A,B,C\in\mathcal{A}$, and $\Phi(A,B)$ is always an affine span.
A sufficient condition would be if for any strict subset $\mathcal{G}\subsetneq\mathcal{A}$, there is a condition $A\in\mathcal{A}\setminus\mathcal{G}$ with $$\mathrm{aff}(A)\cap\mathrm{aff}\left(\bigcup_{G\in\mathcal{G}}G\right)\subseteq \mathrm{aff}\left(\bigcup_{G\in\mathcal{G}}\Phi(A,G)\right).$$
Then we can inductively choose some function $g$ which fulfills (1) on $\mathcal{A}$ (or via some other method if $\mathcal{A}$ is uncountable). To illustrate the idea, assume there is such a function $g$ which works for a subset $\mathcal{G}\subsetneq \mathcal{A}$. Then the condition provides a set $A$ such that $g$ and $f_A$ are equal on $\mathrm{aff}(A)\cap\mathrm{aff}(\bigcup_{G\in\mathcal{G}}G)$. Hence, we can choose a new function $g'$ which equals both $f_A$ and $g$ on this set and extend this function as equal to $f_A$ on $\mathrm{aff}(A)$ and equal to $g$ on $\mathrm{aff}(\bigcup_{G\in\mathcal{G}}G)$. Then $g'$ fulfills (1) for $\mathcal{G}\cup\{A\}$.
Unfortunately, I have not been able to find a sufficient condition which is also necessary or to prove that the above condition is necessary.
