Bourbaki

Exercise 9, § 8

Tag 02HN, permanent. Cite it as [Bourbaki, Tag 02HN].

Spectral Theories, I, p. 194

Bourbaki marks this one as harder than the rest.

Let X be a compact topological space and B a unital Banach subalgebra of C(X)\mathscr{C}(X) (for the induced norm). Every element xx of X defines a character χx\chi_x of B. If x,x′x,x' are in X, we write x∼x′x\sim x' when ∥χx−χx′∥≠2\|\chi_x-\chi_{x'}\| \not= 2. a) Let xx and x′x' be in X. If there exists a sequence (fn)(f_n) of elements of B such that ∥fn∥⩽1\|f_n\|\leqslant 1, that ∣fn(x)∣|f_n(x)| tends to 1 and that lim infn→∞∣fn(x)−fn(x′)∣>0_{n\rightarrow \infty}|f_n(x)-f_n(x')|>0, then one does not have x∼x′x\sim x'. (One may suppose that fn(x)f_n(x) tends to 1. Consider gn=(fn−λn)(1−λnfn)−1g_n= (f_n-\lambda_n)(1-\lambda_nf_n)^{-1}, where (λn)(\lambda_n) is a sequence of numbers of [0,1[[0,1[ tending to 1. One has ∥gn∥⩽1\|g_n\|\leqslant 1. If the sequence (λn)(\lambda_n) is suitably chosen, gn(x)g_n(x) tends to 1 and gn(x′)g_n(x') tends to −1-1.)

b) Let R be the set of real parts of the elements of B. If x∼x′x\sim x', there exists c∈]0,1[c\in ]0,1[ such that f(x)⩾cf(x′)f(x)\geqslant cf(x') and f(x′)⩾cf(x)f(x')\geqslant cf(x) for every f⩾0f\geqslant 0 in R. (It suffices to obtain the first condition; if it is not true, there exists a sequence (fn)(f_n) in R with fn⩾0,fn(x′)=1,fn(x)f_n\geqslant 0,f_n(x') = 1,f_n(x) tending to 0; let gn∈Bg_n\in B with R(gn)=fn\mathscr{R}(g_n) =f_n. Then e−gn∈B,∥e−gn∥⩽1,∣e−gn(x′)∣=1/e,∣e−gn(x)∣e^{-g_n}\in B,\|e^{-g_n}\|\leqslant 1,|e^{-g_n(x')}|= 1/e,|e^{-g_n(x)}| tends to 1, which contradicts a).)

c) There exist positive measures α\alpha and β\beta on X such that f(x)−cf(x′)=α(f)f(x)-cf(x') =\alpha (f) and f(x′)−cf(x)=β(f)f(x')-cf(x) =\beta (f) for f∈Rf\in R. Setting μx=(1−c2)−1(cβ+α)\mu_x= (1-c^2)^{-1}(c\beta +\alpha ), μx′=(1−c2)−1(cα+β)\mu_{x'}= (1-c^2)^{-1}(c\alpha +\beta ), one has μx(f)=f(x)\mu_x(f) =f(x) and μx′(f)=f(x′)\mu_{x'}(f) =f(x') for every f∈Bf\in B, and cμx⩽μx′,cμx′⩽μxc\mu_x\leqslant \mu_{x'},c\mu_{x'}\leqslant \mu_x.

d) Suppose again that x∼x′x\sim x'. Prove that if μ\mu is a positive measure on X such that μ(f)=f(x)\mu(f) =f(x) for every f∈Bf\in B, there exists a positive measure μ′\mu' on X such that μ′(f)=f(x′)\mu'(f) =f(x') for every f∈Bf\in B and cμ⩽μ′c\mu\leqslant \mu' for a c∈]0,1[c\in ]0,1[. (With the notation of c), set μ′=μx′−cμx+cμ\mu'=\mu_{x'}-c\mu_x+c\mu.)

e) The relation x∼x′x\sim x' is an equivalence relation in X. (Use d).)

Solution

No solution has been written yet.

Cited by

Nothing.

Cites

Nothing.

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.