Bourbaki

Exercise 28, § 1

Integration, VII, p. 86

Bourbaki marks this one as harder than the rest.

With hypotheses as in Exer. 27 a), assume in addition that there exists a μ\mu-integrable G-paving F.

a) For every μ\mu-integrable numerical function f≥0f \geq 0 defined on X, set f~(x)=∑s∈Gf(sx)\widetilde{f}(x) = \sum_{s \in G} f(sx), so that ∫Ff~(x)dμ(x)=∫Xf(x)dμ(x)\int_F \widetilde{f}(x) d\mu(x) = \int_X f(x) d\mu(x) (§ 2, No. 10, Prop. 15). Let (fi)1≤i≤n(f_i)_{1 \leq i \leq n} be a family of numerical functions ≥0\geq 0, μ\mu-integrable on X; for i≠ji \neq j set

mij=∫Ff~i(x)f~j(x)dμ(x), m_{ij} = \int_F \widetilde{f}_i(x) \widetilde{f}_j(x) d\mu(x),
and let ci=sup⁡x∈Xf~i(x)c_i = \sup_{x \in X} \widetilde{f}_i(x). Show that there exists at least one pair of distinct indices (i,j)(i, j) such that
mij≥1n(n−1)μ(F)((∑i=1n∫Xfi(x)dμ(x))2−μ(F)∑i=1nci∫Xfi(x)dx). m_{ij} \geq \frac{1}{n(n-1)\mu(F)} \left( \left( \sum_{i=1}^n \int_X f_i(x) d\mu(x) \right)^2 - \mu(F) \sum_{i=1}^n c_i \int_X f_i(x) dx \right).

(5) Cf. TVS II, §5, No. 2, Prop. 3; a closed convex set with nonempty interior was called a convex body (corps convexe) in the first edition of Ch. II of Esp. vect. top. (§ 3, No. 2, Def. 4).

(Bound the sum ∑i≠jmij\sum_{i \neq j} m_{ij} from below, using the Cauchy–Schwarz inequality.) From this, deduce that if (Ai)1≤i≤n(A_i)_{1 \leq i \leq n} is a finite family of μ\mu-integrable G-fillings, and if

∑i=1nμ(Ai)>μ(F), \sum_{i=1}^n \mu(A_i) > \mu(F),
then there exist a pair (i,j)(i, j) of distinct indices and an s∈Gs \in G such that μ(Ais∩Aj)>0\mu(A_i s \cap A_j) > 0.

b) If G0G_0 is a subgroup of GG of finite index (G:G0)=h(G : G_0) = h, and if (s1,…,sh)(s_1, \ldots, s_h) is a system of representatives of the right cosets of G0G_0 in GG, then F0=⋃1≤i≤hsiFF_0 = \bigcup_{1 \leq i \leq h} s_i F is a G0G_0-paving.

c) In X=RnX = \mathbf{R}^n, let AA be a symmetric closed convex set with nonempty interior; let ui:(xj)↦∑j=1ncijxju_i : (x_j) \mapsto \sum_{j=1}^n c_{ij} x_j be mm linear forms on XX, with integer coefficients cijc_{ij} (m<nm < n). Show that for every integer p≥1p \geq 1 and every number r≥0r \geq 0 such that μ(A)rn≥2npm\mu(A) r^n \geq 2^n p^m, there exists a point x∈rA∩Znx \in rA \cap \mathbf{Z}^n distinct from 0 and such that ui(x)≡0(modp)u_i(x) \equiv 0 \pmod{p} for 1≤i≤m1 \leq i \leq m (apply Minkowski’s theorem (Exer. 27 c)) to the subgroup G0G_0 of Zn\mathbf{Z}^n formed by the z∈Znz \in \mathbf{Z}^n such that ui(z)≡0(modp)u_i(z) \equiv 0 \pmod{p} for 1≤i≤m1 \leq i \leq m, and make use of b)). Special case that n=2,m=1n = 2, m = 1 and AA is defined by ∣x1∣≤1,∣x2∣≤1|x_1| \leq 1, |x_2| \leq 1 (Thue’s theorem).

Solution

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Transcribed from the printed volume. Not translated.