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Let's say that a map $f: V \rightarrow W$ between finite-dimensional real vector spaces is convex-linear if $f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y)$ for all $\lambda \in [0,1]$.

Let's say that a map $f: V \rightarrow W$ between finite-dimensional real vector spaces is affine if $f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y)$ for all $\lambda \in \mathbb{R}$.

From the definition, it seems that the requirement of being convex-linear is weaker than the requirement of being affine. However, I can't think of an example of a map which is convex-linear but not affine, but I also can't prove that convex-linearity implies affinity.

Can someone show me an example of a convex-linear map which is not affine? Or tell me how to prove that every convex-linear map is affine? Or give me an appropriate reference?

EDIT: With the intuition of Qiaochu Yuan's comment in mind, I've come up with the following proof:

Claim: Every convex-linear map is affine.

Proof: Let $f$ be convex-linear. For $\lambda \in [0, 1]$, We have that $f(\lambda x + (1-\lambda) y) = \lambda f(x) + (1-\lambda) f(y)$. For $\lambda \notin [0,1]$, we can assume without loss of generality that $\lambda < 0$ (in the other case where $\lambda > 1$, we can interchange the role of $x$ and $y$). We can write \begin{align} f(y) = f\left( \underbrace{\frac{1}{1-\lambda}}_{\in [0,1]}(\lambda x + (1-\lambda) y) + \left( 1 - \frac{1}{1-\lambda} \right) x \right). \label{bla} \end{align} By the convex-linearity of $f$, this reduces to \begin{align} &f(y) = \frac{1}{1-\lambda} f(\lambda x + (1-\lambda) y) + \left( 1-\frac{1}{1-\lambda} \right) f(x) \end{align} which in turn can be reduced to \begin{align} f(\lambda x + (1-\lambda x)) = \lambda f(x) + (1-\lambda) f(y). \end{align}

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The first condition says that $f$ preserves line segments and the second condition says that $f$ preserves lines. Can you see geometrically why these conditions should be equivalent? (Imagine taking longer and longer line segments.) – Qiaochu Yuan Dec 9 '11 at 3:49

If $c>1$, and $f$ is a convex linear, then $z:=cx +(1-c)y$ Then $$ x=\frac{1}{c}z + \frac{-1+c}{c}y $$

Hence $$f (\frac{1}{c}z + \frac{-1+c}{c}y )=\frac{1}{c} f(z) + \frac{-1+c}{c} f(y) $$ so that $$ f(z)=cf(x) + (1-c) f(y) $$

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