Bourbaki

Exercise 4, § 1

Integration, III, p. 56

Bourbaki marks this one as harder than the rest.

a) Let BnB_n be the closed unit ball in the Euclidean space Rn\mathbf{R}^n, and let AnA_n be the set of functions f∈C(Bn;R)f \in \mathscr{C}(B_n; \mathbf{R}) of the form

f(x)=λ(∥x∥2−1)+(x∣a) f(x) = \lambda(\|x\|^2 - 1) + (x|a)
with ∣λ∣≤1/n|\lambda| \leq 1/n and ∥a∥≤1/n\|a\| \leq 1/n. The set AnA_n is convex and compact in the Banach space C(Bn;R)\mathscr{C}(B_n; \mathbf{R}), and possesses the following property: for any functions f1,…,fnf_1, \ldots, f_n belonging to AnA_n, there exists a point x∈Bnx \in B_n such that f1(x)=…=fn(x)=0f_1(x) = \ldots = f_n(x) = 0; however, there exist n+1n+1 functions g1,…,gn+1g_1, \ldots, g_{n+1} belonging to AnA_n that are not simultaneously zero at any point of BnB_n (one may consider the mapping x↦x/(1−∥x∥2)x \mapsto x/(1-\|x\|^2) of B∘n\overset{\circ}{B}_n into Rn\mathbf{R}^n). b) Let Bn′B'_n be the set BnB_n equipped with the discrete topology, and let Dn=βBn′D_n = \beta B'_n. Let Y (resp. Y′Y') be the locally compact topological sum space of the spaces DnD_n (resp. Bn′B'_n). Show that βY=βY′\beta Y = \beta Y'. c) For every integer nn, let Kn⊂C(Dn;R)K_n \subset \mathscr{C}(D_n; \mathbf{R}) be the set of all real-valued functions whose restriction to Bn′=BnB'_n = B_n belongs to AnA_n. Let K be the subset of C(βY;R)\mathscr{C}(\beta Y; \mathbf{R}) formed by the functions whose restriction to DnD_n belongs to KnK_n for every n∈Nn \in \mathbf{N}. Show that K is a compact convex subset of the Banach space C(βY;R)\mathscr{C}(\beta Y; \mathbf{R}). For every finite family (fi)1≤i≤n(f_i)_{1 \leq i \leq n} of functions in K, let W(f1,…,fn)W(f_1, \ldots, f_n) be the set of y∈Y′y \in Y' such that fi(y)=0f_i(y) = 0 for 1≤i≤n1 \leq i \leq n. Show that the W(f1,…,fn)W(f_1, \ldots, f_n) form a base of a filter W\mathfrak{W} on the discrete space Y′Y'. From this, conclude that there exists a point ω∈βY−Y\omega \in \beta Y - Y such that the trace on Y′Y' of the filter of neighborhoods of ω\omega in βY\beta Y is a filter finer than W\mathfrak{W}. Setting X=βY−{ω}X = \beta Y - \{ \omega \}, deduce from this that K⊂K(X;R)K \subset \mathscr{K}(X; \mathbf{R}) and, with the help of Exercise 2, conclude that in K(X;R)\mathscr{K}(X; \mathbf{R}) the set K is convex and compact, but not strictly compact.

Solution

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Transcribed from the printed volume. Not translated.