Bourbaki

Exercise 25, § 1

Tag 02Q7, permanent. Cite it as [Bourbaki, Tag 02Q7].

Spectral Theories, III, p. 112

a) Let N⩾1N\geqslant 1 be an integer. We denote by GNG_N (resp. HNH_N) the set {−1,1}N\{-1,1\}^N (resp. the set {−1,2}N\{-1,2\}^N). We denote by μN\mu_N (resp. νN\nu_N) the positive measure of total mass 1 on GNG_N (resp. on HNH_N) such that, for every (ti)1⩽i⩽N(t_i)_{1\leqslant i\leqslant N} in GNG_N (resp. in HNH_N), one has μN((ti)1⩽i⩽N)=21N\mu_N((t_i)_{1\leqslant i\leqslant N}) =_2^{1_N}, resp.

1 j^j 2 N−j^{N-j}

νN((ti)1⩽i⩽N)= \nu_N((t_i)_{1\leqslant i\leqslant N}) =

3 3 where jj is the number of indices ii such that ti=2t_i= 2.

There exists a real number c>0c >0 such that, for every family (ai)1⩽i⩽N(a_i)_{1\leqslant i\leqslant N} of complex numbers, one has

μN({(ti)1⩽i⩽N∈GN∣∣∑Naiti∣>c\mu_N(\{(t_i)_{1\leqslant i\leqslant N}\in G_N||\sum^Na_it_i|> c(log NN)1/2(∑N∣ai∣2)1/2})<cN3^{1/2}(\sum^N|a_i|^2)^{1/2}\})<\frac{c}{N^3},

i=1i=1i=1i=1

and

νN({(ti)1⩽i⩽N∈HN∣∣∑Naiti∣>c\nu_N(\{(t_i)_{1\leqslant i\leqslant N}\in H_N||\sum^Na_it_i|> c(log NN)1/2(∑N∣ai∣2)1/2})<cN3^{1/2}(\sum^N|a_i|^2)^{1/2}\})<\frac{c}{N^3}.

i=1i=1i=1i=1

(One may suppose that ai∈Ra_i\in \mathbf{R} for every ii and that ∑∣ai∣2=1\sum|a_i|^2= 1; for the first inequality, prove and use the upper bound

∫GN\int_{G_N} exp(λ∣∑i=1Naiti∣)dμN⩽2(\lambda |\sum_{i=1}^Na_it_i|)d\mu_N\leqslant 2 exp(λ2\lambda^2)

for every real λ>0\lambda >0; argue similarly for the second upper bound by using the inequality 13(e2x+2e−x)⩽e2x2\frac{1}{3}(e^{2x}+ 2e^{-x})\leqslant e^{2x^2} for every x∈R.)x\in \mathbf{R}.)

b) For k∈Nk\in \mathbf{N}, let Ak=Z/(3⋅2k)ZA_k=\mathbf{Z}/(3\cdot 2^k)\mathbf{Z}. There exists a real number c′>0c'>0, independent of kk, and a function fk:A^k→{−1,2}f_k:\widehat{A}_k\rightarrow \{-1,2\} such that

∑χ∈A^kfk(χ)=0 \sum_{\chi\in\widehat{A}_k}f_k(\chi ) = 0

and, for every x∈Akx\in A_k, one has

∣∑χ∈A^kχ(x)fk(χ)∣⩽c′(k+1)1/22k/2 |\sum_{\chi\in\widehat{A}_k}\chi (x)f_k(\chi )|\leqslant c'(k+ 1)^{1/2}2^{k/2}

(With the aid of a), establish the existence of a function fkf_k satisfying the second condition, then modify fkf_k to obtain the first.)

We denote by BkB_k (resp. CkC_k) the set of characters χ\chi of AkA_k such that fk(χ)=2f_k(\chi ) = 2 (resp. fk(χ)=−1f_k(\chi ) =-1). One has Card(Bk_k) =2k= 2^k and Card(Ck_k) =2k+1= 2^{k+1}.

For every k⩾1k\geqslant 1, we fix a bijection ιk:Bk→Ck−1\iota_k: B_k\rightarrow C_{k-1}.

We denote by A the sum set of the AkA_k for k∈Nk\in \mathbf{N}; we endow A with the discrete topology, and we denote by E=Cb(A)E =\mathscr{C}_b(A).

We fix a family ε=(εk)k⩾N\varepsilon = (\varepsilon_k)_{k\geqslant\mathbf{N}} of functions εk:Bk→{−1,1}\varepsilon_k: B_k\rightarrow \{-1,1\}. We then define functions ek,χ∈Ee_{k,\chi}\in E, for k∈Nk\in \mathbf{N} and χ∈Bk\chi \in B_k, by

ιk(χ)(x)\iota_k(\chi )(x) if k⩾1k\geqslant 1 and x∈Ak−1x\in A_{k-1},

ek,χ(x)=εk(χ)χ(x)e_{k,\chi}(x) =\varepsilon_k(\chi )\chi (x) if x∈Akx\in A_k

0 otherwise.

We denote by EεE_{\varepsilon} the closed subspace of E generated by the functions (ek,χ)k∈N,χ∈Bk(e_{k,\chi})_{k\in\mathbf{N},\chi\in B_k}. 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\varepsilon, the space EεE_{\varepsilon} does not have the approximation property; it does not admit a Banach basis.

c) For k∈Nk\in \mathbf{N} and χ∈Bk\chi \in B_k, there exists a continuous linear form λk,χ∈Eε′\lambda_{k,\chi}\in E'_{\varepsilon} such that, for every f∈Eεf\in E_{\varepsilon},

λk,χ(f)=\lambda_{k,\chi}(f) = Card(A1 k_k)∑εk(χ)χ(x−1)f(x)\sum\varepsilon_k(\chi )\chi (x^{-1})f(x).

x∈Akx\in A_k

For k⩾1k\geqslant 1 and f∈Eεf\in E_{\varepsilon}, one has

λk,χ(f)=\lambda_{k,\chi}(f) = Card(A2 k_k)∑ιk(χ)(x−1)f(x)\sum\iota_k(\chi )(x^{-1})f(x).

x∈Akx\in A_k

d) For every k∈Nk\in \mathbf{N}, let βk\beta_k be the continuous linear form on L(Eε)\mathscr{L}(E_{\varepsilon}) defined by

βk(u)=12k∑λk,χ(u(ek,χ)) \beta_k(u) =\frac{1}{2^k}\sum\lambda_{k,\chi}(u(e_{k,\chi}))
χ∈Bk\chi \in B_k

One has βk(1Eε)=1\beta_k(1_{E_{\varepsilon}}) = 1.

e) Let u∈L(Eε)u\in \mathscr{L}(E_{\varepsilon}) and k∈Nk\in \mathbf{N}. One has

βk+1(u)−βk(u)=\beta_{k+1}(u)-\beta_k(u) = Card(A1 k_k)∑u(φk,x)(x)\sum u(\varphi_{k,x})(x)

x∈Akx\in A_k

where, for x∈Akx\in A_k, we put

φk,x=12k+1∑ιk+1(χ)(x−1)ek+1,χ−12k∑εk(χ)χ(x−1)ek,χ \varphi_{k,x}=\frac{1}{2^{k+1}}\sum\iota_{k+1}(\chi )(x^{-1})e_{k+1,\chi}-\frac{1}{2^k}\sum\varepsilon_k(\chi )\chi (x^{-1})e_{k,\chi}
χ∈Bk+1χ∈Bk\chi \in B_{k+1}\chi \in B_k

One has ∣φk,x(y)∣⩽c′(k+1)1/22−k/2|\varphi_{k,x}(y)|\leqslant c'(k+ 1)^{1/2}2^{-k/2} for (x,y)∈Ak2(x, y)\in A^2_k.

f) There exists a real number c′′⩾c′c''\geqslant c' such that, for every k⩾1k\geqslant 1, there exists a function εk:Bk→{−1,1}\varepsilon_k: B_k\rightarrow \{-1,1\} satisfying ∣φk,x(y)∣⩽c′′3(k+1)1/22−k/2|\varphi_{k,x}(y)|\leqslant c''3(k+ 1)^{1/2}2^{-k/2} for (x,y)∈Ak×Ak−1(x, y)\in A_k\times A_{k-1}. (Apply the first inequality of a).) We henceforth suppose that ε=(εk)\varepsilon = (\varepsilon_k) satisfies these properties.

g) For every k∈Nk\in \mathbf{N} and x∈Akx\in A_k, one has ∥φk,x∥⩽c′′(k+1)1/22−k/2\|\varphi_{k,x}\|\leqslant c''(k+ 1)^{1/2}2^{-k/2}.

h) Let K be the subset of EεE_{\varepsilon} whose elements are e0,χ0e_{0,\chi_0}, where χ0\chi_0 is the unique element of B0B_0, and the functions (k+1)2φk,x(k+ 1)^2\varphi_{k,x} for k∈Nk\in \mathbf{N} and x∈Akx\in A_k. The set K is relatively compact in EεE_{\varepsilon}, and one has

∣βk+1(u)−βk(u)∣⩽|\beta_{k+1}(u)-\beta_k(u)|\leqslant 1 2_2 sup∥u(x)∥\|u(x)\|

(k+1)x∈K (k+ 1)_{x\in K}

for every k∈Nk\in \mathbf{N} and every u∈L(Eε)u\in \mathscr{L}(E_{\varepsilon}).

i) The sequence (βk)k∈N(\beta_k)_{k\in\mathbf{N}} converges in the dual of the space L(Eε)\mathscr{L}(E_{\varepsilon}) endowed with the topology of simple convergence. Let β\beta denote its limit. For every u∈L(Eε)u\in \mathscr{L}(E_{\varepsilon}), one has ∣β(u)∣⩽3|\beta (u)|\leqslant 3 supx∈K∥u(x)∥_{x\in K}\|u(x)\|.

j) One has β(1Eε)=1\beta (1_{E_{\varepsilon}}) = 1 and β(u)=0\beta (u) = 0 if uu is of finite rank.

k) The space EεE_{\varepsilon} does not have the approximation property. l) Let p∈]2,+∞[p\in ]2,+\infty [. Endow A with the counting measure and denote by EεpE^p_{\varepsilon} the closed subspace of Lp(A)L^p(A) generated by the (ek,χ)k∈N,χ∈Bk(e_{k,\chi})_{k\in\mathbf{N},\chi\in B_k}. The space EεpE^p_{\varepsilon} does not have the approximation property. (Verify that the cardinal of the support of φk,x\varphi_{k,x} is ⩽21⋅2k\leqslant 21\cdot 2^k and deduce that ∥φk,x∥p=O((k+1)1/22−k(p−2)/2p)\|\varphi_{k,x}\|_p= O((k+ 1)^{1/2}2^{-k(p-2)/2p}) for k∈Nk\in \mathbf{N} and x∈Ak.x\in A_k.)

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Machine translation of the English, by gpt-5.4. Not checked by a person.