Bourbaki

Exercise 14, § 2

General Topology, VI, p. 25

Let x1,…,xnx_1, \ldots, x_n be a finite number of points in a Euclidean space Rm\mathbf{R}^m. Put

s=1n∑k=1nxk,r2=1n∑k=1n∥xk∥2(r⩾0) s = \frac{1}{n} \sum_{k=1}^n x_k, \qquad r^2 = \frac{1}{n} \sum_{k=1}^n \|x_k\|^2 \quad (r \geqslant 0)

and one has ∥s∥⩽r\|s\| \leqslant r.

a) Now prove that for every index k, one has

∥xk−s∥⩽((n−1)(r2−∥s∥2))1/2 \|x_k - s\| \leqslant ((n-1)(r^2 - \|s\|^2))^{1/2}

(observe that one has ∑k=1n∥xk−s∥2=n(r2−∥s∥2)\sum_{k=1}^n \|x_k - s\|^2 = n(r^2 - \|s\|^2) and order the ∥xk−s∥\|x_k - s\| in increasing order).

b) Now prove that one has ∑j<i∥xi−xj∥2=n∑k=1n∥xk−s∥2\sum_{j < i} \|x_i - x_j\|^2 = n \sum_{k=1}^n \|x_k - s\|^2. Deduce that one has

sup⁡i<j∥xi−xj∥⩽(2n(r2−∥s∥2))1/2. \sup_{i < j} \|x_i - x_j\| \leqslant (2n(r^2 - \|s\|^2))^{1/2}.

(Number the xkx_k so that the greatest value M of the ∥xi−xj∥\|x_i - x_j\| is ∥xn−1−xn∥\|x_{n-1} - x_n\|. Put tk=(2xk−xn−1−xn)/Mt_k = (2x_k - x_{n-1} - x_n)/M, and calculate the sum

∑k=1n−2∥tk−tn−1∥2+∑k=1n−1∥tk−tn∥2 \sum_{k=1}^{n-2} \|t_k - t_{n-1}\|^2 + \sum_{k=1}^{n-1} \|t_k - t_n\|^2

by observing that tn−1+tn=0t_{n-1} + t_n = 0.)

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.