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 $\sum_i \alpha_i \phi(x_i) = \sum_j \beta_j \phi(y_j)$?

  • 1
    $\begingroup$ Move negative $\alpha$'s to the right and negative $\beta$'s to the left. Normalize the sums of coefficients to $1$. Once you have two convex combinations of points of $S$ representing the same point, the corresponding combinations of $\varphi$'s are nothing but the value at that point (which, by convexity, lies in $S$). $\endgroup$
    – fedja
    Nov 2, 2012 at 3:49
  • $\begingroup$ Oh, I thought that one cannot normalize the coefficients of both sides to one simultaneously, but one can! Thanks. Make your comment an answer, and I will accept it. $\endgroup$ Nov 2, 2012 at 6:04
  • $\begingroup$ How can you simultaneously normalize both sides? $\endgroup$
    – copper.hat
    Nov 2, 2012 at 6:35
  • 1
    $\begingroup$ If the alphas and betas both sum up to one (which they do since they are affine coefficients), then by moving the negative alphas to the right and the negative betas to the left, one can see that the positive alphas plus the (absolute of) the negative betas sum up to the same value as the positive betas plus the (absolute of) the negative alphas. $\endgroup$ Nov 2, 2012 at 7:19
  • $\begingroup$ Of course. I missed that completely. $\endgroup$
    – copper.hat
    Nov 2, 2012 at 7:27

1 Answer 1


Each point $a$ in $\text{Aff}(S)$ has a representation $$a=\sum_{i=1}^n\alpha_i v_i,\qquad \sumα_i=1,\qquad v_i\in S$$ where neither the $v_i$ nor the $α_i$ need to be unique. We define $$\Phi(a)=∑_{i=1}^n α_i \phi(v_i)$$ If $ a = ∑_{i=1}^n α_i v_i = ∑_{i=1}^m \beta_i w_i$, we can move all negative $β$ to the left side and all negative $α$ to the right side of the equation. The sum of the coefficients on the left will then equal the sum on the right, and we can divide each coefficient by that sum. We then obtain two convex combinations of points in $S$. If we apply $ϕ=Φ$ which respects convex combinations, and perform all these steps backwards, we end up with the equality of both images. Hence $Φ$ is well-defined.

To show that $Φ$ is affine, consider points $a = ∑_{i=1}^n α_i v_i$ and $b=∑_{i=1}^n \beta_i v_i$ which are affine combinations of vectors in $S$. We can assume that both have the same vectors by filling the coefficients up with zeros. Then $$\begin{eqnarray} Φ(\lambda a+(1-λ)b) &=& Φ\left(∑_{i=1}^n (λα_i+(1-λ)β_i) v_i\right) \\ &=& ∑_{i=1}^n (λα_i+(1-λ)β_i) ϕ(v_i)\\ &=& λ∑_{i=1}^n α_i ϕ(v_i) + (1-λ)∑_{i=1}^n β_i ϕ(v_i)\\ &=& λΦ\left(∑_{i=1}^n α_i v_i\right) + (1-λ)Φ\left(∑_{i=1}^n β_i v_i\right) \end{eqnarray}$$ Your proof of the uniqueness is fine.


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