Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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}

share|cite|improve this question
4  
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 $$ f(z)=cf(x) + (1-c) f(y) $$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.