I know that the following theorem holds true:
Let $S$ be a subset of $\mathbb R^n$, and let $f:S\to \mathbb R^n$ a map such that $d(p,q)=d(f(p),f(q))$ for every $p,q \in S$ (here $d$ is the usual euclidean distance). Suppose that $0 \in S$ and $f(0)=0$. Then $f$ can be extended to an orthogonal transformation of $\mathbb R^n$.
I have two questions:
Can someone give me a reference of a proof of this theorem?
I believe that the previous theorem holds true also if we change $\mathbb R^n$ with $\mathbb C^n$ and "orthogonal" with "unitary", with nearly the same proof. Does it holds true also in the $ \ell^2$ space? With the same proof?
Infinite-dimensional case
False. Consider the backward shift operator on $\ell^2$: $$S^*(x) = (x_2,x_3,\dots)$$ which is an isometry of the subspace $\{x:x_1=0\}$ onto $\ell^2$. It can't be extended further since it's already surjective.
Complex case
False. For example, there is no unitary transformation $T:\mathbb{C}\to \mathbb{C}$ such that $T(1)=1$ and $T(i)=-i$. The issue is that as long as Euclidean metric is concerned, $\mathbb{C}^n$ is just $\mathbb{R}^{2n}$. But the unitary group $U(n)$ is much smaller than $O(2n)$, due to the $\mathbb{C}$-linearity requirement.
Real case
True. A reference is Theorem 11.4 in Embeddings and Extensions in Analysis by J.H. Wells and L.R. Williams. Terminological note: they say that a pair of spaces $(X,Y)$ has the isometric extension property if every isometry from a subset of $X$ into $Y$ can be extended to an isometry of $X$ into $Y$.