IMC 2008 · Day 2 · P12

It is clear that, if $B$ is an orthonormal system in a Hilbert space $\mathcal{H}$, then $\{ (d/\sqrt{2}) e : e \in B \}$ is a set of points in $\mathcal{H}$, any two of which are at distance $d$ apart. We need to show that every set $S$ of equidistant points is a translate of such a set.

We begin by noting that if $x_1, x_2, x_3, x_4 \in S$ are four distinct points, then $$\begin{align*} \langle x_2 - x_1, x_2 - x_1 \rangle &= d^2, \\ \langle x_2 - x_1, x_3 - x_1 \rangle &= \tfrac{1}{2} \bigl( \|x_2 - x_1\|^2 + \|x_3 - x_1\|^2 - \|x_2 - x_3\|^2 \bigr) = \tfrac{d^2}{2}, \\ \langle x_2 - x_1, x_4 - x_3 \rangle &= \langle x_2 - x_1, x_4 - x_1 \rangle - \langle x_2 - x_1, x_3 - x_1 \rangle = \tfrac{d^2}{2} - \tfrac{d^2}{2} = 0. \end{align*}$$ This shows that scalar products among vectors which are finite linear combinations of the form $$\lambda_1 x_1 + \lambda_2 x_2 + \dots + \lambda_n x_n,$$ where $x_1, x_2, \dots, x_n$ are distinct points in $S$ and $\lambda_1, \lambda_2, \dots, \lambda_n$ are integers with $\lambda_1 + \lambda_2 + \dots + \lambda_n = 0$, are universal across all such sets $S$ in all Hilbert spaces $\mathcal{H}$; in particular, we may conveniently evaluate them using examples of our choosing, such as the canonical example above in $\mathbb{R}^n$. In fact this property trivially follows also when coefficients $\lambda_i$ are rational, and hence by continuity any real numbers with sum 0.

If $S = \{x_1, x_2, \dots, x_n\}$ is a finite set, we form $x = \frac{1}{n}(x_1 + x_2 + \dots + x_n)$, pick a non-zero vector $z \in [\operatorname{span}(x_1 - x, x_2 - x, \dots, x_n - x)]^{\perp}$ and seek $y$ in the form $y = x + \lambda z$ for a suitable $\lambda \in \mathbb{R}$. We find that $$\langle x_1 - y, x_2 - y \rangle = \langle x_1 - x - \lambda z, x_2 - x - \lambda z \rangle = \langle x_1 - x, x_2 - x \rangle + \lambda^2 \|z\|^2.$$ $\langle x_1 - x, x_2 - x \rangle$ may be computed by our remark above as $$\langle x_1 - x, x_2 - x \rangle = \frac{d^2}{2} \left\langle \left( \tfrac{n-1}{n}, -\tfrac{1}{n}, -\tfrac{1}{n}, \dots, -\tfrac{1}{n} \right)^{\!\top}, \left( -\tfrac{1}{n}, \tfrac{n-1}{n}, -\tfrac{1}{n}, \dots, -\tfrac{1}{n} \right)^{\!\top} \right\rangle_{\mathbb{R}^n} = \frac{d^2}{2} \left( -\frac{2(n-1)}{n^2} + \frac{n-2}{n^2} \right) = -\frac{d^2}{2n}.$$ So the choice $\lambda = \dfrac{d}{\sqrt{2n}\, \|z\|}$ will make all vectors $\frac{\sqrt{2}}{d}(x_i - y)$ orthogonal to each other; it is easily checked as above that they will also be of length one.

Let now $S$ be an infinite set. Pick an infinite sequence $T = \{x_1, x_2, \dots, x_n, \dots\}$ of distinct points in $S$. We claim that the sequence $$y_n = \frac{1}{n} (x_1 + x_2 + \dots + x_n)$$ is a Cauchy sequence in $\mathcal{H}$. (This is the crucial observation.) Indeed, for $m > n$, the norm $\|y_m - y_n\|$ may be computed by the above remark as $$\|y_m - y_n\|^2 = \frac{d^2}{2} \left\| \left( \tfrac{1}{m} - \tfrac{1}{n}, \dots, \tfrac{1}{m} - \tfrac{1}{n}, \tfrac{1}{m}, \dots, \tfrac{1}{m} \right)^{\!\top} \right\|^2_{\mathbb{R}^m} = \frac{d^2}{2} \left( \frac{n(m-n)^2}{m^2 n^2} + \frac{m-n}{m^2} \right) = \frac{d^2}{2} \cdot \frac{m-n}{mn} = \frac{d^2}{2} \left( \frac{1}{n} - \frac{1}{m} \right) \to 0, \quad m, n \to \infty.$$ By completeness of $\mathcal{H}$, it follows that there exists a limit $$y = \lim_{n \to \infty} y_n \in \mathcal{H}.$$ We claim that $y$ satisfies all conditions of the problem. For $m > n > p$, with $n, p$ fixed, we compute $$\|x_n - y_m\|^2 = \frac{d^2}{2} \left\| \left( -\tfrac{1}{m}, \dots, -\tfrac{1}{m}, 1 - \tfrac{1}{m}, -\tfrac{1}{m}, \dots, -\tfrac{1}{m} \right)^{\!\top} \right\|^2_{\mathbb{R}^m} = \frac{d^2}{2} \left( \frac{m-1}{m^2} + \frac{(m-1)^2}{m^2} \right) = \frac{d^2}{2} \cdot \frac{m-1}{m} \to \frac{d^2}{2}, \quad m \to \infty,$$ showing that $\|x_n - y\| = d/\sqrt{2}$, as well as $$\langle x_n - y_m, x_p - y_m \rangle = \frac{d^2}{2} \left( \frac{m-2}{m^2} - \frac{2}{m} \left( 1 - \frac{1}{m} \right) \right) = -\frac{d^2}{2m} \to 0, \quad m \to \infty,$$ showing that $\langle x_n - y, x_p - y \rangle = 0$, so that $$\left\{ \frac{\sqrt{2}}{d} (x_n - y) : n \in \mathbb{N} \right\}$$ is indeed an orthonormal system of vectors.

This completes the proof in the case when $T = S$, which we can always take if $S$ is countable. If it is not, let $x', x''$ be any two distinct points in $S \setminus T$. Then applying the above procedure to the set $T' = \{x', x'', x_1, x_2, \dots, x_n, \dots\}$ it follows that the limit $$y = \lim_{n \to \infty} \frac{x' + x'' + x_1 + x_2 + \dots + x_n}{n + 2} = \lim_{n \to \infty} \frac{x_1 + x_2 + \dots + x_n}{n} = y$$ satisfies that $$\left\{ \frac{\sqrt{2}}{d} (x' - y), \frac{\sqrt{2}}{d} (x'' - y) \right\} \cup \left\{ \frac{\sqrt{2}}{d} (x_n - y) : n \in \mathbb{N} \right\}$$ is still an orthonormal system.

This it true for any distinct $x', x'' \in S \setminus T$; it follows that the entire system $$\left\{ \frac{\sqrt{2}}{d} (x - y) : x \in S \right\}$$ is an orthonormal system of vectors in $\mathcal{H}$, as required.