# Mazur–Ulam Theorem - Why is Isometry of Normed Vector Spaces over R only Affine and not Linear?

The Mazur–Ulam theorem shows that a surjective isometry between normed spaces is affine. My problem is not with the proof (which I haven't yet understood), simply how an affine isometric function between normed spaces can be anything other than linear ? My reasoning (which is presumably wrong in view of the literature) is.....

Let $f: A \to B$ be affine, then by definition, $f(a) = g(a) + f(0)$ where $g$ is linear. Since $f$ is isometric, $||f(0)|| = ||0|| = 0$, so by the property of the norm, $||f(0)||= 0 \implies$f(0) = 0 $and so$f(a) = g(a)$a linear function. Ah ... I think I found my error (thanks for comment). If$f$is an isometry (not necessarily linear), it doesn't follow that$||f(a)||= ||a ||$, only that$||f(a_1) - f(a_2)||=||a_1 - a_2 ||$and so not necessarily$||f(0)||= 0 $. ( I see while I was typing this the comment expanded to say the same). • consider translations. your reasoning is false because$||f(a)-f(b)||=||a-b||$does not imply$||f(a)||=||a||\$(i.e., an isometry need not be norm-preserving). – HyJu Jan 6 '17 at 9:46
• @HyJu. many thanks. – Tom Collinge Jan 6 '17 at 9:57