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An affine transformation on $\mathbb{R}$ preserves mid-points. That is, if $f$ is affine then for any $x, z, q$, if $x$ is the mid-point of $[z, q]$, then $f(x)$ is the mid-point of $[f(z), f(q)]$.

Question: Are there any non-affine transformations on $\mathbb{R}$ with this same property? Or is it only affine transformations that preserve mid-points.

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The answer to the title question is "no" and the answer to the body question is "yes." $\mathbb{R}$ is a $\mathbb{Q}$-vector space, and any $\mathbb{Q}$-linear function $\mathbb{R} \to \mathbb{R}$ (that is, a solution to the Cauchy functional equation) preserves midpoints. Most such functions are not $\mathbb{R}$-affine.

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    Here's the link: http://math.stackexchange.com/questions/248344/affine-transformation2012-12-03
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The answer is no. If $f$ preserves midpoints, it must be midpoint convex. However, there are examples of midpoint-convex functions that are not convex (and hence not affine). A classical one is this. Take a Hamel basis $\cal B$ of $\mathbb{R}$ (as a vector space over $\mathbb{Q}$). Let $u$ be a nonzero elements in $\cal B$. For any real number $x$, define $f(x)=q_u$ whenever $x=\sum_{r\in{\cal B}}q_r r$ is the unique representation of $x$ as linear combination of "vectors" in $\cal B$. Then $f$ is a linear function on the vector space $\mathbb{R}$ over $\mathbb{Q}$ (hence midpoint preserving). However, it is not an affine function on the field $\mathbb{R}$: we have $f(0)=0$ and $f(u)=1$, but for every $v\in{\cal B}\setminus\{0,u\}$, we also have $f(v)=0$.

That said, practically, most convex functions are midpoint convex.