Bourbaki

Exercise 18, § 3

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

Spectral Theories, II, p. 317

Let G be a topological commutative discrete ordered group. We write the law of the group of G additively, and we denote its unit by 0. We denote by G+G^+ the set of elements x⩾0x\geqslant 0 of G.

a) The space L+1L^1_+ of functions f∈L1(G)f\in L^1(G) vanishing outside G+G^+ is a closed subalgebra of L1(G)L^1(G).

b) Suppose that G is not Archimedean (TG, V, p. 16, exercise 1 of §3). The algebra L+1L^1_+ is not maximal in L1(G)L^1(G).

Suppose now that G is Archimedean. Let A be a closed subalgebra of L1(G)L^1(G) containing L+1L^1_+. For every x∈Gx\in G, we denote by φx∈L1(G)\varphi_x\in L^1(G) the characteristic function of x{x}. It is invertible in L1(G)L^1(G) and its inverse is φ−x\varphi_{-x}.

c) There exists y∈G+y\in G_+ such that φy∈A\varphi_y\in A and φ−y=φy−1∉A\varphi_{-y}=\varphi^{-1}_y\notin A and there exists a character χ≠0\chi \not= 0 of A such that χ(y)=0\chi (y) = 0.

d) For every x>0x >0 in G, one has χ(x)=0\chi (x) = 0. (There exists an integer k⩾1k\geqslant 1 such that −y+kx>0-y+kx >0.)

e) There exists a measure μ∈M1(G)\mu\in \mathscr{M}^1(G) of norm 1 such that χ\chi is the restriction of μ\mu to the algebra A viewed as a subspace of C(G)\mathscr{C}(G). (For f∈Af\in A, one has ∣χ(f)∣⩽ϱ(f)|\chi (f)|\leqslant \varrho (f) and moreover ϱ(f)=∣f∣∞.\varrho (f) =|f|_{\infty}.)

f) Let μ^:G→C\widehat{\mu}: G\rightarrow \mathbf{C} be the Fourier cotransform of μ\mu. One has μ^(x)=0\widehat{\mu}(x) = 0 if x>0x >0.

g) The measure μ\mu is the counting measure on G.

h) The subalgebra L+1L^1_+ is maximal in L1(G)L^1(G).

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.