Q:

Map of Pre-Hilbert Space is affine if it is an isometry

Let $(H, <\cdot ,\cdot>)   $ be a real pre-Hilbert space and let $T: H \to H$ be an isometry so that $||T(x)-T(y)|| = ||x - y||$ for all $x, y \in H$. Prove $T$ is affine.  
How do I prove this?

A:

Replace $T$ by $S(x) = T(x) - T(0)$.  $S$ is an isometry with
 $S(0) = 0$, and  the task is to prove that $S$ is linear.  
Now $S$ preserves both distances and norms.  Writing out
$$
||S(a)-S(b)||^2 = d(S(a), S(b))^2 = d(a, b)^2  = ||a-b||^2,
$$
and expanding gives $$\langle S(a), S(b)\rangle  = \langle a, b \rangle.$$
  Using this, compute 
$$
\langle S(a + \lambda b) - S(a) - \lambda S(b), 
 S(a + \lambda b) - S(a) - \lambda S(b)\rangle = 0.
$$
Hence $S(a + \lambda b) - S(a) - \lambda S(b) = 0$, and $S$ is linear.
Addendum:  Maybe the last step is more palatable in the following form:
For all elements $S(z)$ in the range of $S$, one has
$$
\langle S(a + \lambda b) - S(a) - \lambda S(b), S(z) \rangle = 0.
$$
Hence for all $y$ in the linear span of the range (which we have to consider as we don't yet know the range is a linear space), 
$$
\langle S(a + \lambda b) - S(a) - \lambda S(b), y \rangle = 0.
$$
Hence, in particular,
$$
\langle S(a + \lambda b) - S(a) - \lambda S(b), 
 S(a + \lambda b) - S(a) - \lambda S(b)\rangle = 0.
$$
Comment:  Every complex inner product spaces is in particular a real inner product space under $(a, b) \mapsto \operatorname{Re}\langle a, b\rangle $, and the real inner product gives, of course, the same norm.  So an isometry is necessarily real affine.

