Maximal eigenvalue and norm of the product of Toeplitz matrices. Study of a particular case.

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Maximal eigenvalue and norm of the product of Toeplitz matrices. Study of a particular case.

Philippe Rambour

To cite this version:

Philippe Rambour. Maximal eigenvalue and norm of the product of Toeplitz matrices. Study of a particular case.. 2012. �hal-00746484�

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Maximal Eigenvalue and norm of the product of Toeplitz matrices. Study of a particular case.

Philippe Rambour^∗

Abstract

Maximal eigenvalue and norm of the product of Toeplitz matrices. Study of a particular case

In this paper we describe the asymptotic behaviour of the spectral norm of the product of two finite Toeplitz matrices as the matrix dimension goes to infinity. These Toeplitz matrices are generated by positive functions with Fisher-Hartwig singularities of negative order. Since we have positive operators it is known that the spectral norm is also the largest eigenvalue of this product.

1 Introduction

Iff ∈L¹(T) the Toeplitz matrix with symbolf denoted byT_N(f) is the (N+ 1)×(N+ 1) matrix such that

(T_N(f))_i+1,j+1 = ˆf(j−i) ∀i, j 0≤i, j≤N

(see, for instance, [6],[7]). We say that a function h is regular if h ∈ L^∞(T) and h > 0.

Otherwise the function h is said singular. Ifb is a regular function continuous in e^iθ^r we call Fisher-Hartwig symbols the functions

f(e^iθ) =b(e^iθ) YR

r=1

|eîθ−eîθ^r|^2α^rϕ_β_r_,θ_r(eîθ) where

• the complex numbers α_r and β_r are subject to the constraints −¹₂ < α_r < ¹₂ and

−¹₂ < β_r < ¹₂,

• the functions ϕ_β_r_,θ_r are defined as ϕ_β_r_,θ_r(e^iθ) =e^iβ^r^(π+θ−θ^r⁾.

The problem of the extreme eigenvalues of a Toeplitz matrix is well known (see [12] and [1]).

Ifλ_k,N1≤k≤N + 1 are the eigenvalues ofT_N(f) with λ_1,N ≤λ_2,N· · · ≤λ_N_+1,N we have

N→=+∞lim λ_1,N =m_f and lim

N→+∞λ_N+1,N =M_f

with m_f = essinff and M_f = essupf. In [5] and [8] B¨ottcher and Grudsky on one hand and B¨ottcher and Virtanen in the other hand give an asymptotic estimation of the maximal eigenvalue in the case of one Toeplitz matrix when the symbol has one or several zeros of

∗Universit´e de Paris Sud, Bˆatiment 425; F-91405 Orsay Cedex; tel : 01 69 15 57 28 ; fax 01 69 15 60 19 e-mail : philippe.rambour@math.u-psud.fr

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negative order. In [13] we have obtained the asymptotic of the minimal eigenvalue of one Toeplitz matrix when the symbol has one zero of orderα withα > ¹₂.

But estimatig the eigenvalues of the product of two Toeplitz matrices is more delicate.

Effectively it is clear that a product of Toeplitz matrices is generally not a Toeplitz matrix.

In the first part of this paper we consider the productT_N(f₁)T_N(f₂) of two Toeplitz matrices where f₁(eîθ) = |1−eîθ|^−2α¹c₁(eîθ), and f₂(eîθ) = |1−eîθ|^−2α²c₂(eîθ) with 0 < α₁, α₂ < ¹₂ and c₁, c₂ are two regular continuous functions on the torus. For these symbols we obtain the norm of the matrixT_N(f₁)T_N(f₂). Owing to an important result of Widom (see Lemma 3 and also [18], [17], [16], [9]), which connects the norm of an operator and the norm of a matrix.

A proof of this result can be found in [8]. Since T_N(f₁)TN(f₂) is a positive matrix the norm is also the maximal eigenvalue of this matrix. Hence our main result (see Theorem 3) can be also stated as

Theorem 1 Let f₁(eîθ) = |1 −eîθ|^−2α¹c₁(eîθ) and f₂(eîθ) = |1−eîθ|^−2α²c₂(eîθ) with 0 <

α₁, α₂ < ¹₂ and c₁, c₂ ∈L^∞(T) continuous and nonzero in 1. Then ifΛα₁,α₂,N is the maximal eigenvalue of T_N(f₁)T_N(f₂) we have

Λα₁,α₂,N =N^2α¹^+2α²Cα₁Cα₂c₁(1)c₂(1)kKα₁,α₂k+o(N^2α¹^+2α²).

with

∀α∈]0,1

2[C_α= Γ(1−2α) sin(πα) π

and K_α₁_,α₂ the integral operator on L²[0,1] with kernel (x, y)→R₁

0 |x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt.

Then we obtain bounds on kKα1,α2kwhich provides bounds on Λα1,α2,N (see the theorem 4).

In a second part we apply this result to obtain the maximal eigenvalue Λ_α,β,N of the more general symbols

f˜₁(e^iθ) =|1−e^iθ|^−2α Yp

j=1

|eîθ^j−eîθ|^−2α^jc₁(eîθ) and f˜₂(eîθ) =|1−eîθ|^−2β Yq

j=1

|eîθ^j−eîθ|^−2α^jc₂(eîθ) (1) with 0 < α, β < ¹₂, α > max

1≤j≤p(αj), β > max

1≤j≤q(βj) and where c₁, c₂ are two regular functions satisfying precise hypotheses. We obtain

Λ_α,β,N ∼CN^2α+2βkK_α,βk (see Theorem 5 for the expression ofC).

Remark 1 To get Theorem 5 we give in Lemma 2 an asymptotic of the Fourier coefficients of the symbols f˜₁ and f˜₂ of (1). We may observe that this lemma provides a statement that slightly differs from Theorem 4.2. in [8].

This statement will be Theorem 2 Put σ=QR

j=1|χ−χ_j|^−2α^jcwhere ∀j, χ_j ∈T and i) 0< α₁< ¹₂

ii) α₁> max

2≤j≤R(αj) .

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If c is a regular positive function with c∈A(T, r) for 1> r >0 (see the point 2.2) we have ΛN ∼HkKα₁kN^2α¹

where ΛN is the maximal eigenvalue of TN(σ), H =Cα₁c(χ₁)QR

j=2|1−χj|−2αj and Kα₁ is the integral operator on L²(0,1) with kernel (x, y)→ |x−y|^−2α¹⁻¹.

An important application of the knowledge of the maximal eigenvalue of the product of two Toeplitz matricesT_N(f₁) andT_N(f₂) is the application of the G¨artner-Ellis Theorem to obtain a large deviation principle and([11]). Here we consider the case of long memory (see also [15]).

For the application of the G¨artner-Ellis Theorem in the case wheref₁andf₂ belong toL^∞(T) [2], [3],[4] are good references.

Remark 2 For the case where f, g ∈ L^∞(T) is it not true in general that the maximal eigenvalue of T_N(f)T_N(g) goes to essup(f g). Likewise it is not always true that the minimal eigenvalue of T_N(f)TN(g) goes to essinf(f g). If we denote these maximal and minimal eigenvalues by Λ_max,N and Λ_min,N Bercu, Bony and Bruneau give in [4] an example of two functions f, g∈C⁰(T), g≥0 such that lim

N→+∞Λ_max,N exists but is greater than sup_θ∈T(f g)(θ) and another example where lim

N→+∞Λ_min,N is defined but is smaller than inf_θ∈T(f g)(θ). Howe- ver iff, g∈L^∞(T) since essup(f)essup(g)−T_N(f)T_N(g) is a nonnegative operator it is quite easy to obtain, from the results of [2], that

essup(f)essup(g) = essup(f g)⇒ lim

N→+∞Λ_max,N = essup(f g).

2 Main result

In the rest of this paper we denote by χ the functionθ→e^iθ. 2.1 Single Fisher-Hartwig singularities.

Theorem 3 Let f₁ =|1−χ|^−2α¹c₁ and f₂ =|1−χ|^−2α²c₂ with 0 < α₁, α₂ < ¹₂ and c₁, c₂ ∈ L^∞(T) that are continuous and nonzero in 1. We have

kTN(f₁)TN(f₂)k=N^2α¹^+2α²Cα₁Cα₂c₁(1)c₂(1)kKα₁,α₂k+o(N^2α¹^+2α²).

withC_α₁, C_α₂ and K_α₁_,α₂ as in Theorem 1.

Now we give a lemma which is useful to prove Theorem 4.

Lemma 1 There exits a constant H_α₁_α₂ such that for all (x, y)∈[0,1]², x6=y

|y−x|^2α¹^+2α²⁻¹ ≤ Z ₁

0

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt≤H_α₁_α₂|x−y|^2α¹^+2α²⁻¹,

with

H_α₁_α₂ =B(2α₁,2α₂) + Z +∞

0

(v^2α¹⁻¹(1 +v)^2α²⁻¹+v^2α²⁻¹(1 +v)^2α¹⁻¹)dv that is also

Hα₁α₂ =B(2α₁,2α₂) +B(2α₂,3−2α₁−2α₂) +B(2α₁,3−2α₁−2α₂).

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Then we have, as corollary of Theorem 3

Theorem 4 With the hypotheses of Theorem 3, if γ_α₁_,α₂ is such that kTN(f1)TN(f2)k ∼ N^2α¹^+2α²c₁(1)c₂(1)γα₁α₂ we have the bounds

ψ(α₁+α₂)C_α₁C_α₂ Cα₁+α2

≤γ_α₁_,α₂ ≤H_α₁_α₂C_α₁C_α₂ Cα₁+α2

1 α₁+α₂, withψ(α) = _2α¹

2

4α+1+ 2^Γ_Γ(4α+2)²^(2α+1)¹₂ .

If we consider now the two symbolsf_1,χ₀ =|χ₀−χ|^−2α¹c₁ and f_2,χ₀ =|χ₀−χ|^−2α²c₂ with χ₀ ∈T it is known (see [14]) that

T_N(|χ₀−χ|^−2αc) = ∆₀(χ₀)T_N |1−χ|^−2αc_χ₀

∆⁻¹₀ (χ₀)

wherec_χ₀(χ) =c(χ0χ) and where ∆0(χ0) is the diagonal matrix defined by (∆0(χ0))_i,j = 0 if i6=jand (∆₀(χ₀))_i,i=χⁱ₀. Hence we have the following corollary of Theorems 3 and 4 Corollary 1 With the previous notations and hypotheses we have

kTN(f_1,χ₀)TN(f_2,χ₀)k ∼N^−2α¹^−2α²Cα₁Cα₂c₁(χ₀)c₂(χ₀)kKα₁,α₂k 2.2 Several Fisher-Hartwig singularities

Let r >0, we denote by A(T, r) the set{g∈L¹(T) |P

u∈Z|u|^r|ˆg(u)|<∞}. We first state the following lemma

Lemma 2 Put σ= YR

j=1

|χ−χj|^−2α^jc where ∀j, χj ∈T and α₁ > max

2≤j≤R(αj). If c is a regular positive function with c ∈A(T, r) (1 ≥r > 0 if ¹₂ > α₁ >0 and r ≥ 2 if 0 > α₁ >−¹₂) we have

b

σ(M) =C_α₁c(χ₁) YR

j=2

|1−χ_j|^−2α^jM^2α¹⁻¹+o(M^2α¹⁻¹) uniformly in M.

This lemma and the proof of Theorem 3 allow us to obtain Theorem 5 Let f˜₁ = |1−χ|^−2α

Yp

j=1

|χ_j −χ|^−2α^jc₁ and f˜₂ = |1−χ|^−2β Yq

j=1

|χ˜_j−χ|^−2α^jc₂ with 0 < α, β < ¹₂, α > max

1≤j≤p(αj), β > max

1≤j≤q(βj), χ_j 6= 1, χ˜_j 6= 1 and c₁, c₂ two regular functions with c₁∈A(T, r₁), c₂∈A(T, r₂) for 1≥r₁, r₂>0. Then

kTN( ˜f₁)TN( ˜f₂)k ∼CN^2α+2βkKα,βk with

C=c₁(1)c2(1)CαC_β Yp

j=1

|1−χ_j|^+2α^j Yq

j=1

|1−χ˜_j|^+2β^j.

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With the same hypotheses on α and β we can now considerTN( ˜f_1,χ₀)TN( ˜f_2,χ₀) with ˜f_1,χ₀ =

|χ₀−χ|^−2α Yp

j=1

|χ_j−χ|^−2α^jc₁ and ˜f_2,χ₀ =|χ₀−χ|^−2β Yq

j=1

|˜χ_j−χ|^−2α^jc₂, with∀j∈ {1,· · · , p}

χ_j 6=χ₀ and ∀h∈ {1,· · · , q} χ˜_h 6=χ₀ . We obtain the corollary Corollary 2 With the previous notations and hypotheses we have

kTN( ˜f_1,χ₀)TN( ˜f_2,χ₀)k ∼N^2α+2βC_χ₀kKα1,α2k with

C_χ₀ =C_αC_βc₁(χ₀)c₂(χ₀) Yp

j=1

|χ₀−χ_j|^−2α^j Yq

j=1

|χ₀−χ_j|^−2β^j.

3 Demonstration of Theorem 3

Let us recall the following Widom’s result ( see, for instance, [9]).

Lemma 3 Let A_N = (a_i,j)^N_i,j=0⁻¹ be an N×N matrix with complex entries. We denote by G_N the integral operator on L²[0,1] with kernel

g_N(x, y) =a[N x],[N y], (x, y)∈(0,1)².

Then the spectral norm ofA_N and the operator norm ofG_N are related by the equalitykANk= NkG_Nk.

Denote byKN andKα₁,α₂ the integral operators onL²(0,1) with the kernels, defined forx6=y by

k_N(x, y) =N^−2α¹^−2α²⁺¹ X

0≤u≤N,u6=[N x],u6=[N y]

[N x]−u^2α¹⁻¹[N y]−u^2α²⁻¹ and

k_α₁_,α₂(x, y) = Z 1

0

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt.

To prove Theorem 3 we first assume that the following lemma is true.

Lemma 4 The operator K_N converges to K_α₁_,α₂ in the operator norm on L²(0,1).

Assume Lemma 4 is true. Supposec₁ =c₂ = 1. Then putT_1,N, T_2,N, D_1,N, D_2,N the (N+ 1)× (N + 1) matrices defined by if k6=l

(T1,N)_(k+1,l+1)=C_α₁|k−l|^2α¹⁻¹ (T2,N)_(k+1,l+1) =C_α₂|k−l|^2α²⁻¹, and (T_1,N)_(k+1,k+1)= 0,(T_2,N)_(k+1,k+1)= 0. On the other hand

(D_1,N)_(k+1,l+1)= (TN(f₁))_(k+1,l+1)−(T_1,N)_(k+1,l+1), (D_2,N)_(k+1,l+1)= (TN(f₂))_(k+1,l+1)−(T_2,N)_(k+1,l+1).

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We can remark that D_1,N and D_2,N are Toeplitz matrices such that (D_1,N)_(k+1,l+1) =o(|k− l|^2α¹⁻¹) and (D_2,N)_(k+1,l+1)=o(|k−l|^2α²⁻¹) (see[10]) and this implies (see [8])

kD_1,Nk=o(N^2α¹) and kD_2,Nk=o(N^2α²).

Then we have the upper bound

kTN(f₁)TN(f₂)−T_1,NT_2,Nk ≤ kD_1,NT_2,Nk+kD_2,NT_1,Nk+kD_1,ND_2,Nk and (see [8])

kD_1,NT_2,Nk ≤ kD_1,NkkT_2,Nk=o(N^2α¹)O(N^2α²) =o(N^2α¹^+2α²) kD2,NT_1,Nk ≤ kD2,NkkT1,Nk=o(N^2α²)O(N^2α¹) =o(N^2α¹^+2α²).

kD_1,ND_2,Nk ≤ kD_1,NkkD_2,Nk=o(N^2α¹)o(N^2α²) =o(N^2α¹^+2α²).

Hence

kT_N(f₁)T_N(f₂)k=kT_1,NT_2,Nk+o(N^2α¹^+2α²).

Lemma 3 implies

T_1,NT_2,N N

=kN^2α¹^+2α²⁻¹K_Nk and with Lemma 4 we obtain lim

N→+∞kKNk=kKα₁α₂k that ends the proof in the case where the regular function equals 1. Now assume that c₁, c₂ are any continuous positive functions in L^∞(T). Let ˜c₁ and ˜c₂ defined by ∀j ∈ {1,2} ˜cj(θ) = cj(θ) if θ 6= 1 and ˜cj(1) = 0. If f˜₁ =|1−χ|^−2α¹˜c₁ and ˜f₂ =|1−χ|^−2α²˜c₂ we have (see [8])

kT_Nf˜₁k=o(N^2α¹) kT˜_Nf˜₂k=o(N^2α²).

HencekT_Nf˜₁T_Nf˜₂k=o(N^−2α¹^−2α²). Sincef₁=|1−χ|^−2α¹(˜c₁+c₁(1)) andf₂ =|1−χ|^−2α²(˜c₂+ c₂(1)) we have

kT_N(f₁)T_N(f₂)−T_N c₁(1)|1−χ|^−2α¹

T_N c₂(1)|1−χ|^−2α²

k=o(N^2α¹^+2α²) and we finally get, via the beginning of the proof

kT_N(f₁)T_N(f₂)k=N^2α¹^+2α²C_α₁C_α₂c₁(1)c₂(1)kK_α₁_α₂k+o(N^2α¹^+2α²) which is the expected formula. We are therefore left with proving Lemma 4.

Proof of the lemma 4:

Fix µ, 0< µ <1 sufficiently close to 1 such that µ >max(1−2α₁,1−2α₂,¹₂). Put k¹_N(x, y) =

k_N(x, y) if |x−y|> N^µ−1,

0 otherwise

k²_N(x, y) =

k_N(x, y) if |x−y|< N^µ−1,

0 otherwise

k_α¹₁_,α₂_,N(x, y) =

kα₁,α₂(x, y) if |x−y|> N^µ−1,

0 otherwise.

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k_α²₁_,α₂_,N(x, y) =

k_α₁_,α₂(x, y) if |x−y|< N^µ−1,

0 otherwise.

If we denote byK_N¹, K_α¹₁_,α₂_,N, K_α²₁_,α₂_,N the integral operator onL²(0,1) with the kernels h¹_N, h¹_N,k¹_Nk¹_α

1,α₂,N, k_α²

1,α₂,N respectively. We have

kKα1,α2−K_Nk ≤ kK_α¹₁_,α₂_,N−K_N¹k+kK_α²₁_,α₂_,Nk+kK_N²k.

Hence we have to show that

N→+∞lim kK_α¹₁_,α₂_,N−K_N¹k= 0, lim

N→+∞kK_α²₁_,α₂_,Nk= 0, lim

N→+∞kK_N²k= 0.

First we prove the following lemma.

Lemma 5 When N goes to the infinity kK_α¹₁_,α₂_,N−K_N¹k →0

Proof : To prove that kK_α¹₁_,α₂_,N −K_N¹k → 0 it suffices to show that |k¹_N(x, y) −k¹_α₁_,α₂| converges uniformly to zero for |x−y| > N^µ−1. First we may assume that x < y and we consider the case [N x]> N^µ and [N y]< N−N^µ.Next we study the cases [N x]≤N^µ and [N y]≥N−N^µ.

For |x−y|> N^µ−1 we have to consider the difference S_N(x, y) = 1

N

XN

u=0,u6=[N x],u6=[N y]

[N x]

N − u N ^2α¹⁻¹

[N y]

N − u N ^2α²⁻¹

− Z ₁

0

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt.

LetS_i,N(x, y), 1≤i≤7 be the following differences S_1,N(x, y) = 1

N

[N x]−NX^µ¹

u=0

[N x]

N − u N ^2α¹⁻¹

[N y]

N − u N ^2α²⁻¹

− Z ^{[N x]}

N −N^µ¹⁻¹ 0

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

S_2,N(x, y) = 1 N

[N x]−1X

[N x]−N^µ¹+1

[N x]

N − u N

^2α¹⁻¹[N y]

N − u N ^2α²⁻¹

−

Z ^{[N x]−1}

N

[N x]

N −N^µ¹⁻¹+_N¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

S_3,N(x, y) = 1 N

[N x]+NX^µ²

[N x]+1

[N x]

N − u N ^2α¹⁻¹

[N y]

N − u N ^2α²⁻¹

− Z ^{[N x]}

N +N^µ²⁻¹

[N x]+1 N

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

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S_4,N(x, y) = 1 N

[N y]−NX^µ³

[N x]+1+N^µ²

[N x]

N − u N ^2α¹⁻¹

[N y]

N − u N ^2α²⁻¹

− Z ^{[N y]}

N −N^µ³⁻¹

[N x]+1 N +N^µ²⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

S_5,N(x, y) = 1 N

[N y]−1X

[N y]−N^µ³+1

[N x]

N − u N

^2α¹⁻¹[N y]

N − u N ^2α²⁻¹

−

Z ^{[N y]−1}

N

[N y]+1 N −N^µ³⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

S_6,N(x, y) = 1 N

[N y]+NX^µ⁴

[N y]+1

[N x]

N − u N ^2α¹⁻¹

[N y]

N − u N ^2α²⁻¹

−

Z ^{[N y]}_N +N^µ⁴⁻¹

[N y]+1 N

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

S_7,N(x, y) = 1 N

XN

[N y]+N₄^µ+1

[N x]

N − u N

^2α¹⁻¹[N y]

N − u N ^2α²⁻¹

− Z ₁

[N y]+1 N +N^µ⁴⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt

with 0< µ₁ < µ, 0< µ₂< µ, 0< µ₃< µ, 0< µ₄ < µ.We can remark that S_1,N(x, y)∼

Z ^{[N x]}

N −N^µ¹⁻¹ 0

[N x]

N −t

2α1−1 [N y]

N −t 2α2−1

−(x−t)^2α¹⁻¹(y−t)^2α²⁻¹dt.

We may study the two differences S_1,N^′ (x, y) =

Z ^{[N x]}

N −N^µ¹⁻¹ 0

[N x]

N −t 2α1−1

−(x−t)^2α¹⁻¹

! [N y]

N −t 2α2−1

dt

and

S_1,N^′′ (x, y) =

Z ^{[N x]}_N −N^µ¹⁻¹ 0

(x−t)^2α¹⁻¹ [N y]

N −t 2α2−1

−(y−t)^2α²⁻¹

! dt.

Since |^{[N x]−x}_x−t | ≤N^−µ¹ we have_{[N x]}

N −t2α1−1

−(x−t)^2α¹⁻¹ =O(N^−µ¹) and

|S_1,N^′ (x, y)| ≤O(N^−µ¹) Z ^{[N x]}

N −N^µ¹⁻¹ 0

[N y]

N −t 2α2−1

dt =O(N^−µ¹) =o(1).

(10)

The same method provides

S^′′_1,N(x, y)|=O(N^−µ) =o(1).

As previously we have now S_2,N(x, y)∼

Z ^{[N x]−1}

N

[N x]

N −N^µ¹⁻¹+_N¹

[N x]

N −t

2α1−1 [N y]

N −t 2α2−1

−(x−t)^2α¹⁻¹(y−t)^2α²⁻¹ dt.

Obviously we have to consider the differences S^′_2,N(x, y) =

Z ^{[N x]−1}

N

[N x]

N −N^µ¹⁻¹+_N¹

[N x]

N −t 2α1−1

−(x−t)^2α¹⁻¹

! [N y]

N −t 2α2−1

dt

and

S_2,N^′′ (x, y) =

Z ^{[N x]−1}

N

[N x]

N −N^µ¹⁻¹+_N¹

(x−t)^2α¹⁻¹ [N y]

N −t 2α2−1

−(y−t)^2α²⁻¹

! dt.

With the main value theorem we can write S_2,N^′ (x, y) =−(−2α₁+ 1)

[N x]

N −x

Z ^{[N x]−1}

N

[N x]

N −N^µ¹⁻¹+_N¹

c^2α_x,N¹⁻²(t) [N y]

N −t 2α2−1

dt

withc_x,N(t)> N⁻¹ and Z ^{[N x]−1}

N

[N x]

N −N^µ¹⁻¹+_N¹

[N y]

N −t 2α2−1

dt=O(N^µ¹^−µ).

SoS_2,N^′ (x, y) =O(N^µ¹^−µ)−2α¹⁺¹).We can remark that−2α₁+ 1−µ <0 ⇐⇒ −2α₁+ 1< µ.

Hence if −2α₁ + 1 < µ and µ₁ sufficiently little we have S_2,N^′ (x, y) = o(1). Likewise we have S_2,N^′′ (x, y) = O N−1+(µ−1)(2α2−2)

. Hence µ > ^−2α_−2α²⁺¹

2+2 ⇒ S_2,N^′′ (x, y) = o(1), and since

−2α₂+ 1> ^−2α_−2α²⁺¹

2+2 we have S_2,N^′′ (x, y) =o(1).

We prove exactly as previously

µ >−2α₁+ 1 and µ >−2α₂+ 1⇒S_3,N =o(1) µ₂ >0 and µ₃ >0⇒S_4,N =o(1)

Swappingx and y we obtain

– µ >−2α₂+ 1 andµ > ^−2α_−2α¹⁺¹

1+2 thenS_5,N(x, y) =o(1).

– µ >−2α₂+ 1 andµ > ^−2α_−2α¹⁺¹

1+2 thenS_6,N(x, y) =o(1).

– µ >0 and µ₄>0 then S_7,N(x, y) =o(1).

To complete the proof we have still to bound the integrals Z ^{[N x]}

N −N^µ¹⁻¹+_N¹

[N x]

N −N^µ¹⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

Z ^{[N x]+1}

N

[N x]−1 N

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt

(11)

Z ^{[N x]}

N +N^µ²⁻¹+_N¹

[N x]

N +N^µ²⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

Z ^{[N y]}

N −N^µ³⁻¹+_N¹

[N y]

N −N^µ³⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt Z ^{[N y]+1}

N

[N y]−1 N

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

Z ^{[N y]}

N +N^µ⁴⁻¹+_N¹

[N y]

N +N^µ⁴⁻¹

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt

which are obviously in o(1)with the hypotheses onµ.

Assume now 1−N^µ−1 > y > N^µ−1 > x > 0. For this case we have to consider the decomposition S_N(x, y) =

X6

i=1

S_i,N(x, y) with

S_1,N(x, y) = 1 N

[N x]−1X

u=0

[N x]

N − u N

^2α¹⁻¹[N y]

N − u N

^2α²⁻¹−

Z ^{[N x]−1}

N

0

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt,

and S_i,N defined asS_i+1,N in the previous case. We still consider the two differences S_1,N^′ (x, y) =

Z ^{[N x]−1}

N

0

[N x]

N −t 2α1−1

−(x−t)^2α¹⁻¹

! [N y]

N −t 2α2−1

dt

and

S^′′_1,N(x, y) =

Z ^{[N x]−1}

N

0

(x−t)^2α¹⁻¹ [N y]

N −t 2α2−1

−(y−t)^2α²⁻¹

! dt.

We have

S^′_1,N(x, y)≤N^(µ−1)(2α²⁻¹⁾O

([N x]

N )^2α¹ −x^2α¹

=N^(µ−1)(2α²⁻¹⁾O(N^−2α¹).

We can remark that (µ−1)(2α2−1)−2α1 <0 ⇐⇒ µ > _2α^2α¹

2−1+ 1. Since 1−2α1 > _2α^2α¹

2−1+ 1 the hypotheses on µ give S_1,N^′ (x, y) = o(1). Moreover _{[N y]}

N −t2α2−1

−(y−t)^2α²⁻¹

= O(N^−µ) and S_1,N^′′ (x, y) = O(N^−µ) = o(1). The differences S_i,N for 2 ≤ i ≤ 6 are as in the first case.

The caseN −N^µ<[N y]< N can be tackled identically.

2 Proof of kK_α²₁_,α₂_,Nk →0

From the lemma 1 we have, for g∈L²(T) et y∈[0,1]

K_α²₁_,α₂_,N(g)(x) =

Z ₁

0

kα₁,α₂(x, y)g(y)dy =

Z _x+N^µ+1

x−N^µ⁻¹

fα₁,α₂(x, y)g(y)dy

≤

Z x+N^µ+1

x−N^µ−1

H_α₁_,α₂|x−y|^2α¹^+2α²⁻¹g(y)dy =K_α^′2₁_,α₂_,N(g)(y) whereK_α^′2

1,α2,N is the integral operator on L²(0,1) with kernel k_α^′₁_,α₂_,N(x, y) =H_α₁_,α₂|x−y|^2α¹^2α²⁻¹

(12)

if |x−y|< N^µ−1 and k_N^α¹^,α²^,′(x, y) = 0 otherwise.

Ifkgk₂ = 1we have Z 1

0

|K_α²₁_,α₂_,N(g)(x)|²dx= = Z 1

0

Z 1

0

k_α₁_,α₂_,N(x, y)g(y)dy²dx

≤ Z 1

0

Z ¹

0

k_α₁_,α₂_,N(x, y)|g(y)|dy2

dx

≤ Z 1

0

Z ¹

0

k^′_α₁_,α₂_,N(x, y)|g(y)|dy2

dx≤ kK_α^′2₁_,α₂_,Nk².

Hence kK_α²₁_,α₂_,Nk ≤ kK_α^′2₁_,α₂_,Nk=O N^(µ−1)(2α¹^2α²⁾

=o(1) (see [8]).

Proof of kK_N²k →0

As in [8] we define the integral operator ˜K_N² on L²(0,1) with the kernel ˜k²_N defined by k_N² in the staircase-like bordered strip |[N x]−[N y]| < N^µ and be zero otherwise. On the squares where ˜k_N² (x, y)−k_N²(x, y)6= 0 we have|[N x]−[N y]| ∼N^µand, as for the proof of the lemma 4, for (x, y) in this squares

k˜²_N(x, y)−k_N² (x, y)∼ Z 1

0

|x−t|^2α¹⁻¹|y−t|^2α²⁻¹dt and always with the lemma 1

|˜k_N² (x, y)−k²_N(x, y)| ≤H_α₁_,α₂|x−y|^2α¹^+2α²⁻¹ =O(N^(µ−1)(2α¹^+2α²⁻¹⁾).

As the difference ˜h²_N(x, y)−h²_N(x, y) is supported in about 4(N −N^µ) = O(N) squares of side length _N¹ we have the squared Hilbert-Schmidt norm

kK˜_N² −K_N²k=O

N 1

N²N^(µ−1)(4α¹^+4α²⁻²⁾

.

If 2α₁+ 2α₂−1>0 we have (µ−1)(4α₁+ 4α₂−2)−1<0 and

kK˜_N² −K_N²k →0. (2) Otherwise since µ > ¹₂ > ^−4α_−4α¹^−4α²⁺¹

1−4α2+2 we have also (2).

We are therefore with proving kK˜_N²k →0. LetB_N be the matrix such (B_N)_k+1,l+1=C_α₁C_α₂ X

u=0,u6=k,u6=l

|k−u|^2α¹⁻¹|l−u|^2α²⁻¹ if |k−l| ≤N^µ and (B_N)_k+1,l+1= 0 otherwise. We have to prove the following technical lemma

Lemma 6 ∃M_α₁_,α₂ >0 such for k6=l

B_k+1,l+1≤M_α₁_,α₂|k−l|^2α¹^+2α²⁻¹ Proof :

Assume l > k and write XN

u=0,u6=k,u6=l

|k−u|^2α¹⁻¹|l−u|^2α²⁻¹ =

k−1X

u=0,

|k−u|^2α¹⁻¹|l−u|^2α²⁻¹+

+ Xl−1

k+1

|k−u|^2α¹⁻¹|l−u|^2α²⁻¹+ XN

l+1

|k−u|^2α¹⁻¹|l−u|^2α²⁻¹.