# [Math] Proof of the Mazur-Ulam Theorem

functional-analysisproof-verification

The Mazur-Ulam Theorem Theorem $$2.1$$ states that any surjective isometry between any two real normed spaces $$f:X \rightarrow Y$$ is affine.

In the proof of the theorem, the author mentioned that it suffices to show that for any $$x, y \in X$$,
$$f\left(\dfrac{x+y}{2}\right) = \dfrac{f(x)+f(y)}{2}$$

Why is it the case? How to conclude $$f$$ is affine from equation above?

Recall that $$f:X \rightarrow Y$$ is an affine function if for all $$x,y \in X$$ and $$0 \leq t \leq 1$$,
$$f[(1-t)x+ty] = (1-t)f(x) + t f(y)$$

The midpoint-affine property $$f\left(\dfrac{x+y}{2}\right) = \dfrac{f(x)+f(y)}{2} \tag{1}$$ implies being affine under the assumption that $f$ is continuous (which it is, being an isometry). As stated, $(1)$ amounts to the case $t=1/2$ of $$f[(1-t)x+ty] = (1-t)f(x) + t f(y)\tag{2}$$ But applying $(1)$ again, the second time to $x$ and $(x+y)/2$, yields $(2)$ for $t=1/4$. Similarly, applying $(1)$ to $y$ and $(x+y)/2$ yields $(2)$ for $t=3/4$.
Continuing this process, we obtain $(2)$ for all dyadic rationals in $(0,1)$: numbers of the form $k/2^m$, $0<k<2^m$. These are dense in $[0,1]$ and since both sides of $(2)$ are continuous with respect to $t$, equality $(2)$ holds for all $t\in [0,1]$.