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I remember that I read somewhere that the following theorem is true:

Let $A\subseteq H$ be a subset of a real Hilbert space $H$ and let $f : A \to A$ be a distance-preserving bijection, i.e. a bijective map with $\|f(x)-f(y)\|=\|x-y\|$ for all $x,y\in A$. Then it is always possible to find a distance-preserving bijection $\phi: H \to H$ such that $\phi|_A = f$.

But I am not able to find it anywhere online (probably, because I remembered it incorrectly and it is false?). Is this theorem true and if so, can someone tell me a reference where to find it?

(I know that this $\phi$ will then be the composition of a translation and a linear map, but that is not important to me at the moment.)

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    $\begingroup$ This can probably be attributed to Schoenberg in the 1930s. I'll look for a reference. $\endgroup$
    – YCor
    Commented Jul 1 at 14:38
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    $\begingroup$ Note, however, that starting with a distance-preserving bijection $A_1 \to A_2$, we may not be able to extend to $H \mapsto H$. So Hilbert space differs from (finite-dimensional) Euclidean space, which does have this property. $\endgroup$
    – Gerald Edgar
    Commented Jul 1 at 17:05
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    $\begingroup$ @Gerald: wow, is that really true in $\mathbb{R}^n$? Do you have a proof or reference? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 2 at 2:55
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    $\begingroup$ @QiaochuYuan Pick any maximal subset of up to $n+1$ points in general position. Take any isometry of $\mathbb{R}^n$ that works for these points. This works for all the other points because their positions are determined by their distances to our chosen points. $\endgroup$
    – alphacapture
    Commented Jul 2 at 8:50
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    $\begingroup$ @Gerald: somehow I don't think Euclid proves or makes use of this fact for arbitrary subsets of $\mathbb{R}^n$! $\endgroup$
    – Qiaochu Yuan
    Commented Jul 2 at 16:04

1 Answer 1

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This is a consequence of the GNS construction for negative definite kernel, see, e.g., Theorem C.2.3 in the Bekka–Harpe–Valette book.

In the present case, let $H'$ be the closed affine span of $A$: the uniqueness ensures that $f$ extends to a bijective self-isometry $f'$ of $H'$. Then it is easy to extends to all of $H$ (up to translate, we can suppose $0\in H'$, then define the orthogonal $V$ (so $H=H'\oplus V$, orthogonal sum) and use the map $f'\oplus\mathrm{id}_V$).

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