Tag 02Q7, permanent. Cite it as [Bourbaki, Tag 02Q7].
Spectral Theories, III, p. 112
a) Let N⩾1 be an integer. We denote by GN (resp. HN) the set {−1,1}N (resp. the set {−1,2}N). We denote by μN (resp. νN) the positive measure of total mass 1 on GN (resp. on HN) such that, for every (ti)1⩽i⩽N in GN (resp. in HN), one has μN((ti)1⩽i⩽N)=21N, resp.
1 j 2 N−j
νN((ti)1⩽i⩽N)=
3 3 where j is the number of indices i such that ti=2.
There exists a real number c>0 such that, for every family (ai)1⩽i⩽N of complex numbers, one has
(One may suppose that ai∈R for every i and that ∑∣ai∣2=1; for the first inequality, prove and use the upper bound
∫GN exp(λ∣∑i=1Naiti∣)dμN⩽2 exp(λ2)
for every real λ>0; argue similarly for the second upper bound by using the inequality 31(e2x+2e−x)⩽e2x2 for every x∈R.)
b) For k∈N, let Ak=Z/(3⋅2k)Z. There exists a real number c′>0, independent of k, and a function fk:Ak→{−1,2} such that
χ∈Ak∑fk(χ)=0
and, for every x∈Ak, one has
∣χ∈Ak∑χ(x)fk(χ)∣⩽c′(k+1)1/22k/2
(With the aid of a), establish the existence of a function fk satisfying the second condition, then modify fk to obtain the first.)
We denote by Bk (resp. Ck) the set of characters χ of Ak such that fk(χ)=2 (resp. fk(χ)=−1). One has Card(Bk) =2k and Card(Ck) =2k+1.
For every k⩾1, we fix a bijection ιk:Bk→Ck−1.
We denote by A the sum set of the Ak for k∈N; we endow A with the discrete topology, and we denote by E=Cb(A).
We fix a family ε=(εk)k⩾N of functions εk:Bk→{−1,1}. We then define functions ek,χ∈E, for k∈N and χ∈Bk, by
ιk(χ)(x) if k⩾1 and x∈Ak−1,
ek,χ(x)=εk(χ)χ(x) if x∈Ak
0 otherwise.
We denote by Eε the closed subspace of E generated by the functions (ek,χ)k∈N,χ∈Bk. It is a complex Banach space of countable type. The rest of the exercise is intended to prove that, for a suitable choice of ε, the space Eε does not have the approximation property; it does not admit a Banach basis.
c) For k∈N and χ∈Bk, there exists a continuous linear form λk,χ∈Eε′ such that, for every f∈Eε,
λk,χ(f)= Card(A1 k)∑εk(χ)χ(x−1)f(x).
x∈Ak
For k⩾1 and f∈Eε, one has
λk,χ(f)= Card(A2 k)∑ιk(χ)(x−1)f(x).
x∈Ak
d) For every k∈N, let βk be the continuous linear form on L(Eε) defined by
One has ∣φk,x(y)∣⩽c′(k+1)1/22−k/2 for (x,y)∈Ak2.
f) There exists a real number c′′⩾c′ such that, for every k⩾1, there exists a function εk:Bk→{−1,1} satisfying ∣φk,x(y)∣⩽c′′3(k+1)1/22−k/2 for (x,y)∈Ak×Ak−1. (Apply the first inequality of a).) We henceforth suppose that ε=(εk) satisfies these properties.
g) For every k∈N and x∈Ak, one has ∥φk,x∥⩽c′′(k+1)1/22−k/2.
h) Let K be the subset of Eε whose elements are e0,χ0, where χ0 is the unique element of B0, and the functions (k+1)2φk,x for k∈N and x∈Ak. The set K is relatively compact in Eε, and one has
∣βk+1(u)−βk(u)∣⩽ 1 2 sup∥u(x)∥
(k+1)x∈K
for every k∈N and every u∈L(Eε).
i) The sequence (βk)k∈N converges in the dual of the space L(Eε) endowed with the topology of simple convergence. Let β denote its limit. For every u∈L(Eε), one has ∣β(u)∣⩽3 supx∈K∥u(x)∥.
j) One has β(1Eε)=1 and β(u)=0 if u is of finite rank.
k) The space Eε does not have the approximation property. l) Let p∈]2,+∞[. Endow A with the counting measure and denote by Eεp the closed subspace of Lp(A) generated by the (ek,χ)k∈N,χ∈Bk. The space Eεp does not have the approximation property. (Verify that the cardinal of the support of φk,x is ⩽21⋅2k and deduce that ∥φk,x∥p=O((k+1)1/22−k(p−2)/2p) for k∈N and x∈Ak.)