There is one step in a proof which I don't manage to show, although it seems to be very easy.
Let $A, B$ be real vector spaces, let $S \subset A$ be a convex set and let $\text{aff}(S)$ be its affine hull, \begin{align} \text{aff}(S) := \left\{ \sum\limits_{i=1}^n \alpha_i v_i \ \middle\vert \ n \in \{0, 1, 2, \ldots \}, v_i \in S, \alpha_i \in \mathbb{R}, \sum\limits_{i=1}^n \alpha_i = 1 \right\} \,. \end{align} Say that a map $\phi: S \rightarrow B$ is convex-linear if $\phi(\lambda x + (1-\lambda) y) = \lambda \phi(x) + (1-\lambda) \phi(y)$ for all $x, y \in A, \lambda \in [0,1]$, and say that a map $\Phi: \text{aff}(S) \rightarrow B$ is affine if $\phi(\lambda x + (1-\lambda) y) = \lambda \phi(x) + (1-\lambda) \phi(y)$ for all $x, y \in A$, $\lambda \in \mathbb{R}$.
I want to prove the following
Claim: For a convex-linear function $\phi: S \rightarrow B$, there is a unique affine map $\Phi: \text{aff}(S) \rightarrow B$ such that its restriction to $S$ coincides with $\phi$.
Proof: Uniqueness. I can show that: Assume there are two such maps $\Phi, \tilde \Phi$. Let $x \in \text{aff}(S)$, i.e. $x$ is an affine sum $x = \sum_i \alpha_i x_i$ of elements $x_i \in S$. Then $\Phi(x) = \Phi(\sum_i \alpha_i x_i) = \sum_i \alpha_i \Phi(x_i) = \sum_i \alpha_i \tilde \Phi(x_i) = \tilde \Phi(\sum_i \alpha_i x_i) = \tilde \Phi(x)$.
Existence. For $x \in \text{aff}(S)$, i.e. $x$ is an affine sum $x = \sum_i \alpha_i x_i$ of elements $x_i \in S$, define $\Phi(x) := \sum_i \alpha_i \phi(x_i)$.
Question: How can I show that $\Phi(x)$ is well-defined, i.e. that $\Phi(x) := \sum_i \alpha_i \phi(x_i)$ does not depend on the choice of the affine combination $\sum_i \alpha_i x_i$? In other words, if $x = \sum_i \alpha_i x_i = \sum_j \beta_j y_j$ (both being affine sums), why is then $\sum_i \alpha_i \phi(x_i) = \sum_j \beta_j \phi(y_j)$?