To show that $f$ is an orthogonal linear map, if $f(0) = 0$ and $f$ preserves the distance

Question

Let $V$ be a finite-dimensional inner product space, and $f:V\to V$ be a map such that $f(0) = 0$ and $\forall x,y\in V$, $||f(x) - f(y)|| = ||x-y||$. Show that $f$ is an orthogonal linear map.

I've been able to show only a part of what's required, and my work is as follows:

Since $||f(x) - f(y)|| = ||x-y||$ holds for all $x,y$, put $y=0$. This gives $||f(x)|| = ||x||$ for all $x\in V$. $f:V\to V$ is a norm-preserving map, and so it is orthogonal if it is linear (proof of this is given at the end).

Now, how do I show that $f$ is linear? Is there a nice way of doing this? I'd appreciate any hints.

Proof of statement above:
I know that a map $T:V\to W$ is orthogonal if $\forall x,y\in V$, we have $\langle Tx,Ty\rangle = \langle x,y\rangle$.
$f$ is norm-preserving, so for arbitrary $x,y\in V$, we have $||f(x+y)|| = ||x+y||$ and hence $||f(x+y)||^2 = ||f(x)+f(y)||^2 = ||x+y||^2$. Expanding, we get $||f(x)||^2 + ||f(y)||^2 + 2\langle f(x),f(y)\rangle = ||x||^2 + ||y||^2 + 2\langle x,y\rangle$ and using $||f(x)||=||x||$ and $||f(y)|| = ||y||$ we get $\langle f(x),f(y)\rangle = \langle x,y\rangle$ as desired.

Answer 1

First, let's show that $$ f(x+y)=f(x)+f(y). $$ We have \begin{align*} || (f(x+y)-f(x))-f(y)||^2&= || f(x+y)-f(x) ||^2 + ||f(y)||^2 -2<f(x+y)-f(x),f(y)>\\ &= 2||y||^2 -2<f(x+y),f(y)>+2<x,y>\\ &= ||y||^2 +||x||^2-||x+y||^2+2<x,y>\\ &=0 \end{align*} since $$ ||f(x+y)-f(y) ||^2=||x+y||^2 + || y||^2 -2<f(x+y),f(y)>=|| x||^2. $$ Using, the same method we have $$ f(\lambda x)= \lambda f(x). $$ Indeed, we have $$ || f(\lambda x)- \lambda f(x) || ^2= 2\lambda^2||x ||^2-\lambda<f(\lambda x),f(x)> $$ but $$ ||f(\lambda x)-f(x) ||^2=|| (\lambda -1)x||^2=(\lambda -1)^2||x||^2 = (\lambda^2+1)||x||^2 - 2<f(\lambda x),f(x)> $$ so $$ 2\lambda||x||^2 = - 2<f(\lambda x),f(x)> $$ thus $$ 2\lambda^2||x||^2 = - 2\lambda<f(\lambda x),f(x)>. $$ So $$ || f(\lambda x)- \lambda f(x) || ^2= 2\lambda^2||x ||^2-2\lambda^2||x||^2=0. $$

Answer 2

Hint

All isometry can be written as $$f(v)=\ell(v)+b,$$ where $\ell:V\to V$ is a linear map.

Answer 3

For any $x$ and $y$ in $V$, $(x+y)/2$ is the unique $z\in V$ such that $\|z-x\| = \|z-y\|= \|x-y\|/2$. Since $f$ is an isometry, $\|f(z) - f(x)\| = \|f(z) - f(y)\| = \|f(x)-f(y)\|/2$, therefore $$f((x+y)/2) = (f(x) + f(y))/2$$ The case $y=0$ gives us $f(x/2) = f(x)/2$ so we have $$f(x+y) = f(x) + f(y)$$ Next prove that $f(n x) = n f(x)$ for any integer $n$, then for rational numbers, then (by continuity) for reals.