Asmptotic of the eigenvalues of Toeplitz matrices with even symbol
aa r X i v : . [ m a t h . C A ] J a n Asymptotic of the eigenvalues of Toeplitz matrices with evensymbol.
Philippe Rambour ∗ January 28, 2021
AbstractAsymptotic of the eigenvalues of Toeplitz matrices with even symbol.
In this article we first obtain an asymptotic expression of the N + 1 eigenvalues of theToeplitz matrix T N ( f ) where f is a strictly increasing or strictly decreasing, differentiableand even function on [ − π, π ]. In the particular case where the symbol only verifies theseassumptions locally we specify the eigenvalues that belong to a certain interval. In thecase where the symbol of the Toeplitz matrix is of the form h α = (1 − cos θ ) α c with α > and c a regular function we give the general expression of the eigenvalues and a secondorder of this asymptotic for a subset of the small eigenvalues as well as an approximationof the eigenvectors. Mathematical Subject Classification (2000)
Primary 35S05 ; Secondary 47G30.
Keywords
Toeplitz matrices, eigenvalues, Fractional Laplacian operators.
For h ∈ L ( T ) we denote by T N ( h ) the Toeplitz matrix of order N and with symbol h . Itis the ( N + 1) × ( N + 1) matrix such that for N ≥ k, l ≥ T N ( h )) k +1 ,l +1 = ˆ h ( k − l )where ˆ h ( u ) is the Fourier coefficient of order u of h ([9],[6]). The search for eigenvalues (orthe minimum or maximum eigenvalue) of Toeplitz matrices is an old and important problem([27, 26, 25, 9, 23, 21, 14, 16, 15] and more recently, for instance, [3, 5, 4, 2, 8, 7, 18] . For f ∈ L ( T ) we denote by λ N ≤ λ N ≤ · · · ≤ λ N +1 N the eigenvalues of the ( N + 1) × ( N + 1)matrix T N ( f ). Lastly we denote by A + ( T ) (resp. A − ( T )) be the set of the even differentiablesfunctions on [ − π, π ], such that f ′ ( θ ) > f ′ ( θ ) <
0) for all θ in [0 , π ]. These functionsare simple loop functions on [0 , π ]. In 2007 we have obtained for f belonging to A + ( T ) or A − ( T ) an asymptotic expression of the eigenvalue λ ( k ) N of T N ( f ) for k ∈ { , , · · · , N } suchthat kN → x for x ∈ ]0 ,
1[ (see [18]). ∗ Universit´e de Paris Saclay, Bˆatiment 307; F-91405 Orsay Cedex; tel : 01 69 15 57 28 ; fax 01 69 15 60 19e-mail : [email protected] f = g g , g ( θ ) = χ (1 − r ¯ χ χ ) ( P N +1 ) − , g ( θ ) = (1 − r ¯ χ ¯ χ ) ( P N +1 ) − where 0 < r < P N +1 a polynomial with degree N + 1 without zeros on ¯ D = { z/ | z | ≤ } and where χ is thefunction θ e iθ and χ = e iθ , θ ∈ R . In the appendix to this article we give this formulaand briefly recall how we use it to calculate the inverse of a Toeplitz matrix ( N + 1) × ( N + 1)of symbol χ (1 − ¯ χ χ ) (1 − ¯ χ ¯ χ ) | P N +1 | .Here we take up this demonstration in a more general context and with greater precision andthis allows us to obtain a precise distribution of the N + 1 eigenvalues of the matrices T N ( f )for functions f in A + ( T ).Finally we need to recall the following definition. D´efinition 1
For s ≥ we denote by A ( T , ν ) the weighted Wiener algebra of all functions ψ T C whose Fourier coefficients verify P j ∈ Z | ˆ ψ ( j ) | ( | j | ) s < ∞ . Now we can state the following result
Theorem 1
Let f ∈ A + ( T ) a regular function in A ( T , s ) s > such that f ∈ C ( I ) for I aneigbourhood of zero and π with f ′′ (0) > and f ′′ ( π ) > . Then for a sufficiently large N wehave for k ∈ { , , · · · , N + 1 } λ ( k ) N = f (cid:18) kπN + 2 + ρ N ( k ) N + 2 (cid:19) , where the quantity | ρ N ( k ) | is bounded by a real M no depending from k, N. Our result can also be compared with that of Trench ([24]) where it is proved that for thisclass of symbols the eigenvalues are all distincts.We can notice that the hypotheses and results of this theorem are relatively close to thoseof [3]. But the demonstration methods are totally different in the both cases. We use hereany formulas for inverting Toeplitz matrices that provide us an equation that generates theeigenvalues, while J.M. Bogoya, A. B¨ottcher, S.M. Grudsky and E.A Maximenko use the firstSzeg¨o limit theorem and criteria for weak convergence of probability measures.Also the two papers study two different sets of symbols. In the present paper the symbolsconsidered are functions in A ( T , s ), s > while those considered in [3] are in a subset of A ( T , θ
7→ | − e iθ | α c where c is a regular function, thatis not possible with the tools developed in [3].Then in cases where the symbols do not verify the hypotheses of the theorem 1 we can stillspecify the expression of eigenvalues belonging to certain intervals which is not done in [3].This is the purpose of the following statement. To state this theorem we must consider aregular function f ∈ L ( T ) which is a even and derivable on [ − π, π ] and which belongs to A ( T , s ), s > . So if f is defined as above we assume that f ′ ( t ) > t in an interval[ δ , δ ] ⊂ ]0 ,
2[ to an interval [ a, b ]. We assume also min t ∈ [0 , f ( t ) < f ( t ) < a ⇐⇒ < t < δ and b < f ( t ) < max t ∈ [0 , f ( t ) ⇐⇒ δ < t <
2. Now we can give the following theorem
Theorem 2
Let N be a sufficiently large integer and λ an eigenvalue of T N ( f ) in ] f ( δ ) , f ( δ )[ .Then there is an integer k with kπN +2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ such that λ = f (cid:18) kπ + ρ N ( k ) N + 2 (cid:19) . eciprocally for all k such that kπN +2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ we have an ˜ λ N ( k ) aneigenvalue of f such that λ kN = f (cid:18) kπ + ρ N ( k ) N + 2 (cid:19) . with ρ N ( k ) as in the theorem (1). Remark 1
Of course we have the statement corresponding to the theorems 1 and 2 for thecase where f ′ ( t ) < for t ∈ [0 , or for all t in an interval [ δ , δ ] ⊂ ]0 , Remark 2
Of course we cannot state that ˜ λ ( k ) N is not necessarily the k-th largest eigenvalueof f . This is nevertheless true for the two cases δ = 0 , < δ < or < δ < , δ = 2 . Finally we can point out an interesting consequence of the theorem 1. We consider a real α > h α = θ
7→ | − e iθ | α c ( θ ) where c is a function in C ( T ). Then Corollary 1
For α > and with the notation of the introduction for all k in [1 , N + 1] thek-th largest eigenvalue of T N ( h α ) is λ ( k ) N,α = h α (cid:18) kπ + ρ N ( k ) N + 2 (cid:19) , with ρ N ( k ) as in the theorem (1). For α ∈ ]0 ,
1[ we have only the following result
Corollary 2
Let δ ∈ ]0 , and N a sufficiently large integer. Then for all integer k such that kπN +2 ∈ ] arccos(1 − δ ) , π [ we have λ ( k ) N = h α (cid:18) kπ + ρ N ( k ) N + 2 (cid:19) , where | ρ N,α ( k ) | is bounded by a real M α no depending from k, N . The proof of this corollary is founded on the same ideas as the corollary 1.In the second part of this paper we give a more precise approximation of some eigenvaluesand eigenvectors of the Toeplitz matrices with symbol | − χ | α c with α ∈ ]0 , \{ } and c aregular function. In [20] and the appendix of the present paper we have established the deepconnection that exists between the fractional Laplace operator on ]0 ,
1[ denoted by ( − ∆) α \ [0 , and the ( N + 1) × ( N + 1) Toeplitz matrix of symbol θ
7→ | − e iθ | α for a sufficiently large N and for α ∈ ]0 , \{ } (we have obtained in [19] that in the case α = the Toeplitz matrix ofsymbol θ
7→ | − e iθ | is associated with a derivation and not with a Laplacian). It is thereforenatural to investigate how the eigenvalues and the eigenvectors of these two operators can belinked. In the theorem 4 we see that the eigenvalues of the fractional Laplace operator in ]0 , T N ( h α ). In thetheorem 5 we specify the deep links that exist between the eigenvectors of T N ( h α ) and thoseof ( − ∆) α \ [0 , .The factional Laplace operator of order 2 α on ]0 ,
1[ is defined on f ∈ C ∞ c ([0 , (cid:16) ( − ∆) α \ ]0 , ( f ) (cid:17) ( x ) = C α P.V. Z + ∞−∞ f ( x ) − f ( y ) | x − y | α dy, for x ∈ ]0 , C α = α Γ( α ) √ π | Γ( α ) | . We have seen in [20] that for all f ∈ C ∞ c (]0 , x ∈ ]0 , heorem 3 For α ∈ ]0 , \{ } we have lim N → + ∞ N α N X l =0 (cid:16) T N ( | − χ | ) α (cid:17) k +1 ,l +1 f ( lN ) = (cid:16) ( − ∆) α \ [0 , ( f ) (cid:17) ( x ) , for lim N → + ∞ kN = x, uniformly in x ∈ [ δ , δ ] with [[ δ , δ ] ⊂ ]0 , . The proof of the theorems 4 and 5 is based on the idea that the eigenvalue and theeigenvectors of order k of T N ( | − χ | ) α are closed to those of ( − ∆) α \ ]0 , .In the theorem (4) we have to consider a ( N + 1) × ( N + 1) Toeplitz matrix with symbol h α = | − χ | α c where c is a even regular function in A ( T , s ), and α ∈ ]0 , \{ } .To obtain a general result for the functions we need the following definition and the lemma1 which is stated in the appendix. D´efinition 2
For a real s > we denote by A ( T , s ) the set { h ∈ ( T ) | X u ∈ Z | u s ˆ h ( u ) | < ∞} . Lemma 1
For c an even regular function such that c ∈ A ( T , s ) with s > we have theasymptotic c h α ( u ) = C α | u | − α − c (0) (1 + o (1))According to this lemma 1 we can , with the same demonstration as for the theorem 3 we canprove Lemma 2
For a regular function c in A ( T , s ) , s > and f ∈ C ∞ ( T ) , we have, for α ∈ ]0 , \{ } , and for lim N → + ∞ kN = x, < x < . lim N → + ∞ N α N X l =0 ( T N ( | − χ | ) α c ) k +1 ,l +1 f ( lN ) = c (0) (cid:16) ( − ∆) α ]0 , ( f ) (cid:17) ( x ) , (1) uniformly in x ∈ [ δ , δ ] with [[ δ , δ ] ⊂ ]0 , . With this link we can now state the theorem 4 which is a direct consequence of Theorem 1 of[12] and of the lemma 2. In the next statement we denote by C a constant no depending from N and k , and by L α = ( C (2 α ) − / ) α . The constant C will be specified in the remark 3. Theorem 4
We assume α ∈ ]0 , \{ } and h α = | − χ | α c with c a regular even function in A ( T , s ) , s > . Then we have for a fixed integer k ≤ L α and for N sufficiently large there isat least one eigenvalue ˜ µ kN of T N ( h α such that (cid:12)(cid:12)(cid:12) ˜ µ kN − (cid:18) kπN − (1 − α ) π N (cid:19) α c (0) (cid:12)(cid:12)(cid:12) ≤ α − C − α √ αk N − α . Remark 3
By following carefully the proof of Theorem 1 of [12] and also the proof of thetheorem 4 we can obtain that C ≤ . We can also remark that the error term tends to zeroas α approaches 1. k , 1 ≤ k ≤ N + 1 we denote by Y kN the eigenvector of T N ( ϕ α ) correspondingto the eigenvalue λ kN such that ( Y N,k ) ≤ k ≤ N +1 form a complete orthonormal set in R N +1 .The following theorem is a consequence of the proof of theorem 4. For the next statementwe may consider the function ϕ ⋆ , ,k defined by ϕ ⋆ , ,k ( x ) = ± cos (cid:0)(cid:0) kπ − ( − α ) π (cid:1) (1 − x ) (cid:1) forodd k and ϕ ⋆ , ,k ( x ) = ± sin (cid:0)(cid:0) kπ − ( − α ) π (cid:1) (1 − x ) (cid:1) for even k . We also need to define theconstant L ′ α = max( L α , ( C ′ α − ) α ). where C ′ is a constant no depending from k and N which is specified in the proof of the theorem 5. Theorem 5
For k , L ′ α ≤ k ≤ N + 1 and with the same hypothesis as in the theorem 4 let Z N,k be the vector such that ( Z N,k ) m +1 = √ N ϕ ⋆ , ,k ( mN ) c (0) for ≤ m ≤ N . Then we have1. (cid:13)(cid:13)(cid:13) Z N,k k Z N,k k − Y N,k (cid:13)(cid:13)(cid:13) ≤ O (cid:18) − α √ αk α (cid:19) for 0 < α < (cid:13)(cid:13)(cid:13) Z N,k k Z N,k k − Y N,k (cid:13)(cid:13)(cid:13) ≤ O (cid:18) − α √ k (cid:19) for 14 ≤ α < , α = 12 First of all we must state the following lemma, which will be used in the demonstration. Inthis lemma we consider a regular function f = g ¯ g , g ∈ H + ( T ). This lemma will be prove inthe appendix of the article. We also need to recall the following definition and property D´efinition 3
We call the degree predictor polynomial M of a regular function h the trigono-metric polynomial K M defined by K M = M X k =0 (cid:0) T − M (cid:1) k +1 , q(cid:0) T − M (cid:1) , χ k . Property 1
For all the integer j , such that − M ≤ j ≤ M we have \ | K M | j = b h ( j ) . For an integer k we denote by β k the Fourier coefficient d g − ( k ) and by β k,N the coefficient of χ k in the predictor polynomial P N of f . Lemma 3
For f ∈ A ( T , s ) with s > , and k an integer in [0 , N ] we have β k,N = ¯ β β k + O ( N + 1) − s +3 / . The reader can consult [13] for the predictor polynomials. Using the assumptions we can write f ( θ ) = f (1 − cos θ ) where f is a differentiable function strictly increasing on [0 , λ ∈ f ([0 , f (1 − cos θ ) − λ = (cid:0) (1 − cos θ ) − f − ( λ ) (cid:1) H λ ′ ( θ )5here H λ is a regular function on the torus. Put λ ′ = f − ( λ ). We can write (cid:0) (1 − cos θ ) − λ ′ (cid:1) = 12 (cid:0) | − χ | − λ ′ (cid:1) . (2)If χ λ ′ = (1 − λ ′ ) + i p − ( λ ′ − we can write the equation (2) as (cid:0) (1 − cos θ ) − λ ′ (cid:1) = − χ λ ′ (1 − ¯ χ λ ′ χ )(1 − ¯ χ λ ′ ¯ χ ) . (3) Remark 4
If we denote by P N +1 ,λ ′ the predictor polynomial of H λ ′ the main property of thesepolynomials (see the property 1) allows to write that the ( N + 1) × ( N + 1) Toeplitz matricesof the respective symbol ((1 − cos θ ) − λ ′ ) H λ ′ and − χ λ ′ (1 − ¯ χ λ ′ χ )(1 − ¯ χ λ ′ ¯ χ ) | P N +1 ,λ ′ | are thesame. In all the following of the proof we assume that N is a fixed integer.If T ,N,λ = (cid:0) ( T N ( f ) − λI N ) − (cid:1) , we have T ,N,λ = det ( T N − ( f ) − λI N − )det ( T N ( f ) − λI N )and, since the eigenvalues of ( T N − ( f )) are not in Spec ( T N ( f )) (see [10, 1]), we have λ ∈ S pec ( T N ( f )) ⇐⇒ T ,N,λ = 0 . Using the remark (4) and the inversion formula of Toeplitz matrices obtained in [18] (thereader can consult the appendix to have an explicit expression of this formula) we obtain theentry (cid:0) T − N ( f ) (cid:1) (1 , . Hence with the results of the appendix we have1 T ,N,λ = 1 − ¯ χ N +2) λ ′ N ( χ λ ′ ) (cid:16) − ¯ χ N +2) λ ′ τ N ( χ λ ′ ) (cid:17) B ,N,λ − B ,N,λ . (4)With τ N ( χ λ ′ ) = ¯ P N +1 ,λ ′ ( χ λ ′ ) P N +1 ,λ ′ ( χ λ ′ )¯ P N +1 ,λ ′ ( χ λ ′ ) P N +1 ,λ ′ ( χ λ ′ ) , and B ,N,λ = (cid:12)(cid:12)(cid:12) P N +1 ,λ ′ (0) P N +1 ,λ ′ ( χ λ ′ )¯ P N +1 ,λ ′ ( χ λ ′ ) (cid:12)(cid:12)(cid:12) (1 − ¯ χ λ ′ ) − . For χ λ ′ = − , B ,N,λ is defined and the definition of the predictor polynomialprovides that P N +1 ,λ ′ (0) = det( T N ( H λ ′ ))det( T N +1 ( H λ ′ )) . Since det ( T N ( H λ ′ )) = 0 for all N ∈ N ∗ (see [9])and P N +1 ,λ ( e iθ ) = 0 for all θ ∈ R (see [13]) we have B ,N,λ = 0 for χ λ ′ = − ,
1. The constant B ,N,λ will be is defined for χ λ ′ = − ,
1. In any case this constant does not matter in ourdemonstration. Indeed if (cid:16) − ¯ χ N +2) λ ′ τ N ( χ λ ′ ) (cid:17) B ,N,λ − B ,N,λ = 0 this equality means thatdet ( T N − ( f ) − λI N − ) = 0 and λ is an eigenvalue of T N − ( f ) so it cannot be an eigenvalue6f T N ( f ).Since χ λ ′ ∈ {− , } ⇐⇒ λ ′ ∈ { , } ⇐⇒ λ ∈ { min T f, max T f } . and with the additional remark that neither min f nor max f cannot be eigenvalues of f wecan conclude that λ ∈ S pec ( T N ( f )) ⇐⇒ χ N +2) λ ′ = τ N ( χ λ ′ ) , χ λ ′ = − , . (5)Since the function λ ′ τ N ( λ ′ ) is continuous from [0 ,
2] to { z || z | = 1 } we have a non negativefunction ˜ ρ N defined and continuous on [0 ,
2] such that τ N ( λ ′ ) = e i ˜ ρ N ( λ ′ ) Then the equation 5can be write λ ∈ S pec ( T N ( f )) ⇐⇒ χ N +2) λ ′ = e i ˜ ρ N ( λ ′ ) , χ λ ′ = − , . (6)Now according to the construction of χ λ ′ we have necessary λ ∈ S pec ( T N ( f )) ⇐⇒ (cid:18) λ ′ = 1 − cos (cid:18) ˜ ρ N ( λ ′ ) + 2 kπ N + 4 (cid:19) for k ∈ [0 , N + 3] (cid:19) , χ λ ′ = − , . (7)And since Im χ λ ′ ≥ (cid:18) λ ′ = 1 − cos (cid:18) ρ N ( λ ′ ) + kπN + 2 (cid:19) for k ∈ [1 , N + 1] (cid:19) ⇒ λ ∈ S pec ( T N ( f )) (8)with ρ N ( λ ′ ) = ˜ ρ N ( λ ′ )2 . On the other hand we can remark that ( λ ′ , θ ) H λ ′ ( θ ) is positiveand continuous on [0 , × [ − π, π ] we have τ > τ > λ ′ ∈ [0 , H λ ′ is the restriction to { z || z | = 1 } of a continuous and positive function on { z | τ ≤ | z | ≤ τ } . Using [17], Corollaire 1, we can claim that for a sufficiently large k wehave ( T N ( H λ ′ )) − k, = O (cid:18)(cid:16) ρ (cid:17) k (cid:19) . This result and the lemma 3 provide the existence of a nonnegative function ˜ ρ defined on [0 ,
2] such that ˜ ρ N ( λ ′ ) → N →∞ ˜ ρ ( λ ′ ) uniformly in N . Hence wehave two real M , M with M ≥ M ≥ M ≤ ρ N ( λ ′ ) ≤ M for all λ ′ ∈ [0 ,
2] andall N ∈ N .Now for a sufficiently large N let δ N be a positive real such that, for 1 ≤ k ≤ N + 11 − cos( ρ N ( λ ′ ) + kπN + 2 ) ≥ − cos( π + M ( N + 2) ) > δ N . (9)and 1 − cos( ρ N ( λ ′ ) + kπN + 2 ) ≤ − cos( M + ( N + 1) πN + 2 ) < − δ N (10)We can say that or a fixed k ∈ { , · · · , N + 1 } the function Ψ k : λ ′ − cos (cid:16) ρ N ( λ ′ )+ kπN +2 (cid:17) − λ ′ changes sign on [ δ N , − δ N ] then, by continuity of the function ρ on this interval, the equation 8has at less one solution on this interval and this solution is a λ ′ j = f − ( λ Nj ). Also if 1 ≤ k , k ≤ N , k = k and λ ′ k = 1 − cos (cid:16) ρ N ( λ ′ )+ k πN +2 (cid:17) , λ ′ k = 1 − cos (cid:16) ρ N ( λ ′ )+ k πN +2 (cid:17) we have necessary λ ′ k = λ ′ k with λ N = f − ( λ ′ k ) and λ N = f − ( λ ′ k ) two different eigenvalues of T N ( f ). Hencewe have obtained N + 1 eigenvalues of the matrice. Lastly for k ∈ { , , · · · , N + 1 } we put ρ N ( k ) = ρ N ( λ ′ ) where λ ′ is the solution of the equation (8).7 Demonstration of the Theorem 2
Let λ in [ f ( δ ) , f ( δ )], and λ ′ = f − ( λ ). As previously we can also define the function H λ ′ by the relation f (1 − cos θ ) − λ = (cid:0) (1 − cos θ ) − λ ′ (cid:1) H λ ′ ( θ ) . We can remark that the function H λ ′ is an even and regular on [ − π, + π ].We can always define χ λ ′ with the relations (2) and 3 and with the same definition of T ,N,λ as n the proof of the theorem 1 we can write for λ ′ ∈ [ δ , δ ] λ = f ( λ ′ ) ∈ S pec ( T N ( f )) ⇐⇒ T ,N,λ = 0 . Of course the inversion formula gives always the relation (4) with the constant τ N ( χ λ ′ ) , B ,N,λ , B ,N,λ defined as previously. Lastly we can write f ( λ ′ ) = λ ∈ S pec ( T N ( f )) \ ] f ( δ ) , f ( δ )[ ⇐⇒ χ N +2) λ ′ = τ N ( χ λ ′ ) , λ ′ ∈ ] δ , δ [ . Using the same notations as in the proof of the theorem (1) we can define the function ˜ ρ N , ρ N and ρ on [ δ , δ ] and we can write λ ∈ S pec ( T N ( f )) \ ] f ( δ ) , f ( δ )[ ⇐⇒ (cid:18) λ ′ = 1 − cos (cid:18) ˜ ρ N ( λ ′ , N ) + 2 kπ N + 4 (cid:19) for k ∈ [ k ,N , k ,N ] (cid:19) , (11)with k ,N = min { k | kπN + 2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ } and k ,N = max { k | kπN + 2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ } . Indeed if the two positive reals M , M ,are as in the proof of the theorem (1), if N is sufficientlylarge and kπN +2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ we have also, since M no depending from N , kπ + M N + 2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ , and kπ + ρ N ( λ ′ ) N + 2 ∈ ] arccos(1 − δ ) , arccos(1 − δ )[ . Reciprocally we can adapt the previous reasoning to the present situation. Indeed we can findtwo reals δ N and δ ′ N such that such that δ < δ N < δ ′ N < δ and for all integer k ∈ [ k , k ]1 − cos (cid:18) kπ + ρ N ( λ ′ ) N + 2 (cid:19) ≥ − cos (cid:18) k π − π N + 2 (cid:19) > δ N > δ and 1 − cos (cid:18) kπ + ρ N ( λ ′ ) N + 2 (cid:19) ≤ − cos (cid:18) k π + π N + 2 (cid:19) < δ ′ N < δ Then for an integer k in [ k , k ] the function λ ′ − cos (cid:16) kπ + ρ N ( λ ) N +2 (cid:17) − λ ′ is continuous on[ δ , δ ] and changes at least once of sign in this interval. Then the end of the demonstrationis quite identical to the previous one, and the quantity ρ N ( k ) is defined as in the proof of thetheorem (1). 8 Demonstration of the corollary 1
In all the following of the proof we assume that N is a fixed integer. For all r ∈ ]0 , θ ∈ [ − π, + π ] and λ ′ ∈ [0 ,
2] we define the functions h α,r : θ (1 − r cos θ ) α , h α : θ (1 − cos θ ) α ,and h ,α,r λ ′ (1 − r + rλ ′ ) α , h ,α λ ′ ( λ ′ ) α .We can remark that h ,α,r (1 − cos θ ) = h α,r ( θ ). For all λ = f ( λ ′ ), for r ∈ ]0 ,
1[ and λ ′ ∈ [0 , H λ ′ ,α,r ( θ ) = h α,r ( θ ) − h ,α,r ( λ ′ )(1 − cos θ ) − λ ′ , and for λ ′ ∈ ]0 , H λ ′ ,α ( θ ) = h α ( θ ) − ( λ ′ ) α (1 − cos θ ) − λ ′ . As previously we obtain τ N,α,r ( λ ′ ) = (cid:18) P N +1 ,λ ′ ,α,r ( χ λ ′ ) P N +1 ,λ ′ ,α,r ( χ λ ′ ) (cid:19) where P N +1 ,λ ′ ,α,r is the predictor polynomial of the function H λ ′ ,α,r and we have a family ofcontinuous function ρ N,α,r such that τ N,α,r ( λ ′ ) = e iρ N,α,r ( λ ′ ) . Since we can show the lemma (see the proof in the appendix)
Lemma 4
The Fourier coefficients of H λ ′ ,α,r converge to those of H λ ′ ,α uniformly in λ ′ ∈ [0 , and r ∈ ]0 , r → (cid:16)(cid:0) T − N (cid:0) H λ ′ ,α,r (cid:1)(cid:1) ,k +1 (cid:17) = (cid:0) T − N (cid:0) H λ ′ ,α (cid:1)(cid:1) ,k +1 ∀ k ∈ [0 , N ] uniformly in λ ′ ∈ [0 , ρ N,α such that ρ N,α,r ( λ ′ ) → ρ N,α ( λ ′ ) uniformlyin λ ′ ∈ [0 , ρ N,α,r is continuous on [0 ,
2] we can conclude that for all k ∈ [1 , N + 1] the solution λ ′ ( k ) N,α,r of λ ′ = 1 − cos (cid:18) ρ N,α,r ( λ ′ ) + 2 kπ N + 4 (cid:19) have a limit λ ′ ( k ) N,α when r →
1. If λ ( k ) N,α = h ,α, ( λ ′ ( k ) N,α ) λ ( k ) N is an eigenvalue of T N ( h α ). Wehave necessary λ ( k ) N,α = λ ( k +1) N,α . Indeed since (see [10, 1]) λ ( k ) N,r < λ ( k ) N − ,r < λ ( k +1) N,r (where λ ( k ) N,r denote the k-th eigenvalue of T N ( h α,r ) we have λ ( k ) N,α = λ ( k +1) N,α = ⇒ λ N,αk = λ N − ,αk that is a contradiction with λ ( k ) N,α < λ ( k ) N − ,α < λ ( k +1) N,α . Hence we can conclude that the eigen-values of T N ( h α,r ) are λ ( k ) N,α = h α (cid:18) kπ + ρ N,α ( k ) N + 2 (cid:19) , with ρ N,α ( k ) = ρ N,α ( λ ′ ( k ) N,α ). On the other hand we know that for a fixed r the quantities ρ N,r ( k ) in the expression of the eigenvalues of T N ( h α,r ) are bounded independently of k and N . Hence we have a real M α no depending from k and N such that ρ N,α ( k ) ≤ M α for all k, N that ends our proof. 9 Demonstration of the theorem 4
Let c a regular function which satisfies the assumption of the theorem and h α = | − χ | α c with α ∈ ]0 , \{ } . In the following we use the formula 2 by assuming that c (0) = 1. On theother hand it is known that there exists an infinite sequence of eigenvalues 0 < λ < λ ≤ λ ≤ · · · for ( − ∆) α and the corresponding eigenvectors ϕ n form a complete orthonormal setin L (]0 , k ≥ L α be a fixed positive integer and we denoted by µ k the real kπ − ( − α ) π . Nowlet the function ˜ ϕ k defined for all x ∈ ] − ,
1[ by˜ ϕ k ( x ) = q ( − x ) F µ k (1 + x ) + ( − k q ( x ) F µ k (1 − x ) , with q ( t ) = t ∈ ( −∞ , − ) , ( t + ) for t ∈ ( − , , − ( t − ) for t ∈ (0 , ) , x ∈ ( , ∞ ) . and F λ ( t ) = sin( λt + 1 − α − G ( t )where G is a completely monotone function ( see ([12]) for a complete expression of G ). It isknown that ˜ ϕ k is in L (] − , − ∆) α \ ] − , ( ˜ ϕ k ) is defined on ] − ,
1[ (see [12] lemma 1)2. k ( − ∆) α \ ]0 , ˜ ϕ k − µ αk ϕ k k ≤ C (2 − α ) √ αk (13)(see [12] lemma 1)3. for k ≥ L α we have k ˜ ϕ k k ≥ , (see [12] lemma 2).where C is the constant of the statement of the theorem. In the following of the proof we put R k = C (2 − α ) √ αk , and ˜ µ k = 2 α µ αk . If we put ˜ ϕ k ( x ) = ˜ ϕ k ( − x ) we have1. ( − ∆) α ]0 , ( ˜ ϕ k ) is defined on ]0 , k ( − ∆) α \ ]0 , ˜ ϕ k − ˜ µ k ˜ ϕ k k ≤ α − R k (14)3. for k ≥ L α we have k ˜ ϕ k k ≥ . (15)By construction of the fractional Laplacian we have a sequence (Φ k,n ) n ∈ N of functions in C ∞ c (]0 , n → + ∞ k Φ k,n − ˜ ϕ k k = 0 , (16)10. lim n → + ∞ k ( − ∆) α Φ k,n − ( − ∆) α ˜ ϕ k k = 0 . (17)Hence for ǫ > ?? ) we have a function Φ k ∈ C ∞ c (]0 , k ( − ∆) α Φ k − ˜ µ k Φ k k ≤ α − R k + ǫ. (18)Since Φ k ∈ C ([0 , x ∈ ]0 , (cid:16) ( − ∆) α [0 , Φ k ( x ) − ˜ µ k Φ k ( x ) (cid:17) = lim N → + ∞ N α N X l =0 ( T N ( f )) [ Nx ]+1 ,l +1 )Φ k ( lN ) − ˜ µ k Φ k ( [ N x ] N ) ! , uniformly in x for x ∈ [ δ , δ ] ⊂ ]0 , k ∈ C ∞ c (]0 , < δ < inf (Supp(Φ k )) and sup (Supp(Φ k )) < δ < Z (cid:16) ( − ∆) α [0 , Φ k ( x ) − ˜ µ k ˜Φ k ( x ) (cid:17) dx == Z − δ δ (cid:16) ( − ∆) α [0 , Φ k ( x ) − ˜ µ k ˜Φ k ( x ) (cid:17) dx == lim N → + ∞ Z − δ δ N α N X l =0 ( T N ( f )) [ Nx ]+1 ,l +1 )Φ k ( lN ) − ˜ µ k Φ k (cid:18) [ N x ] N (cid:19)! dx = lim N → + ∞ N N − [ Nδ ] X m =[ Nδ ] N α N X l =0 ( T N ( f )) m +1 ,l +1 )Φ k ( lN ) − ˜ µ k Φ k (cid:16) mN ) (cid:17)! . And we have lim N → + ∞ √ N k N α ( T N ( f )) ( X N,k ) − ˜ µ k X N,k k ≤ α − R k + ǫ with ( X N,k ) m +1 = Φ k (cid:0) mN (cid:1) , for m = 0 , , · · · , N . Finally we can conclude : k ( T N ( f )) ( X N,k ) − ˜ µ k N α X N,k k ≤ α − R k + ǫN α − / (19)Now denote by ˜ µ kN the eigenvalue of T N ( h α ) nearest of ˜ µ k N α . We have k ( T N ( h α )) ( X N,k ) − ˜ µ k N α X N,k k ≥ | ˜ µ kN − ˜ µ k N α |k X N,k k . Since k X N,k k = (cid:16)P Nl =0 (Φ k ) (cid:0) lN (cid:1)(cid:17) and, using (15) lim N →∞ N N X l =0 Φ k (cid:18) lN (cid:19)! = Z Φ k ( t ) dt ≥ N such that for all N ≥ N we have N N X j =0 (cid:12)(cid:12)(cid:12) Φ k (cid:18) jN (cid:19)(cid:12)(cid:12)(cid:12) > ǫ k X N,k k = (cid:16)P Nl =0 (Φ k ) ( lN ) (cid:17) .Hence for ǫ sufficiently close of zero, we have k X N,k k ≥ √ N . That provides with (19)that for N sufficiently large | ˜ µ kN − ˜ µ k N α | ≤ α − N − α ( R k + ǫ ) (20)and finally, since (20) is true for all ǫ > | ˜ µ kN − ˜ µ k N α | ≤ α − N − α ( R k ) . (21)for N sufficiently large, that is the expected result. Let ϕ α be the function | − χ | α c with c a regular function as in the hypotheses of the theoremand always we assume α ∈ ]0 , \{ } . IAs in the previous proof we assume c (0) = 1. In the restof this proof k a fixed integer. We define the vector ˜ X kN by (cid:16) ˜ X N,k (cid:17) m +1 = Φ( mN ) √ N , ≤ l ≤ N .We have ˜ X kN = P Nj =1 a j Y N,j where ( Y N,j ) ≤ j
0. Thenwe may write k ˜ X N,k − k ˜ X N,k k Y N,k k ≤ k ˜ X N,k − a k Y N,k k + (cid:12)(cid:12)(cid:12) a k − k ˜ X N,k k (cid:12)(cid:12)(cid:12) and • (cid:12)(cid:12)(cid:12) a k − k ˜ X N,k k (cid:12)(cid:12)(cid:12) ≤ k ˜ X N,k k − a k = X j = k a j . • k ˜ X N,k − a k Y N,k k ≤ X j = k a j . Hence k ˜ X N,k − k ˜ X N,k k Y N,k k ≤ O ( 1 − α √ αk α ) (24)or else k ˜ X N,k k ˜ X N,k k − Y N,k k ≤ O ( 1 − α √ αk α ) . Now let ϕ ⋆ , ,k be the function ϕϕ ⋆ , ,k as in the statement of the theorem. Using the Proposition1 of [12] we can write1. for 0 < α < k ˜ ϕ k − ϕ ⋆ , ,k k ≤ C (2 − α ) α k α
2. for ≤ α < α = k ˜ ϕ k − ϕ ⋆ , k k ≤ C (2 − α ) √ k for two constants C and C two constants no depending from k . First we assume 0 < α < .We have seen in the previous demonstration that for a real ǫ > k ∈ C ∞ c (]0 , k ˜ ϕ k − Φ k k ≤ ǫ that implies, k ϕ ⋆ , ,k − Φ k k ≤ ǫ + C (2 − α ) α k α . Then for an integer N sufficiently large N +1 X m =1 (cid:16) Z N,k ( m ) − ˜ X N,k ( m ) (cid:17) ! ≤ ǫ + C (2 − α ) α k α with Z N,k ( m + 1) = ˜ ϕ , ,k ( mN ) √ N for 0 ≤ m ≤ N + 1. And with (24) we have k Z N,k − k ˜ X N,k k Y N,k k ≤ ǫ + O ( 1 − α √ αk α ) . Since (cid:12)(cid:12)(cid:12) k Z N,k k − k ˜ X N,k k (cid:12)(cid:12)(cid:12) ∼ (cid:12)(cid:12)(cid:12) k ϕ ⋆ , ,k k − k ˜ ϕ , ,k k (cid:12)(cid:12)(cid:12) N → + ∞ and (cid:12)(cid:12)(cid:12) k ϕ ⋆ , ,k k − k ˜ ϕ , ,k k (cid:12)(cid:12)(cid:12) ≤ k ϕ ⋆ , ,k − ˜ ϕ , ,k k ≤ C (2 − α ) α k α we have finally, for N sufficiently large (cid:13)(cid:13)(cid:13) Z N,k k Z N,k k − Y N,k (cid:13)(cid:13)(cid:13) ≤ O (cid:18) − α √ αk α (cid:19) that is the expected bound. The same methods provides also (cid:13)(cid:13)(cid:13) Z N,k k Z N,k k − Y N,k (cid:13)(cid:13)(cid:13) ≤ O (cid:16) − α √ k (cid:17) for α > , α = . We denote by π + and π − the operator defined by π + X n ∈ Z a n χ n ! = X n ∈ N a n χ n and π − X n ∈ Z a n χ n ! = X n ∈ N ∗ a − n χ − n .In [18] we have state the following inversion formula Theorem 6
Let P N +1 a trigonometric polynomial with degree N + 1 and without zeros onthe united disc ¯ D . Let ω = r ¯ χ , < r < , | χ | = 1 and also f r = g , g , with g = χ (1 − ωχ )( P N +1 ) − and g = (1 − ω ¯ χ ) P N +1 − . Then for all polynomial P in P N we have T N ( f r ) − ( P ) = 1 g π + (cid:18) Pg (cid:19) − g π + Φ N + ∞ X s =0 (cid:0) H ∗ Φ N H Φ N (cid:1) s π + (cid:18) ˜Φ N π + (cid:18) Pg (cid:19)(cid:19)! . with Φ N = g g χ N +1 , ˜Φ N = g g χ − ( N +1) ,H Φ N (Ψ) = π − (Φ N Ψ) H ∗ Φ N (Ψ) = π + ( ˜Φ N Ψ) for Ψ ∈ H + = { X j ≥ a n χ n | a n ∈ C for all n ∈ N } . For the proof of the theorem 4 we have to know T N ( f r ) − , that is also h T N ( f r ) − (1) | i . Write h T N ( f ) − (1) | i = x − y . The previous theorem provides x = h π + (cid:18) g (cid:19) | g i = χ (cid:12)(cid:12)(cid:12) P N +1 (0) (cid:12)(cid:12)(cid:12) . To obtain y we need the terms π + (cid:16) ˜Φ N π + (cid:16) g (cid:17)(cid:17) and π + (cid:16) ¯Φ N π + (cid:16) g (cid:17)(cid:17) . We have π + (cid:18) ˜Φ N π + (cid:18) g (cid:19)(cid:19) = P N +1 (0) π + (cid:18) g g χ − N − (cid:19) = C − ωχ C = P N +1 (0) ¯ χ P N +1 ( ω )¯ P N +1 ( ω ) ! ω N +1 (1 − ω ) . Likewise we can write π + (cid:18) ¯Φ N π + (cid:18) g (cid:19)(cid:19) = C ′ − ¯ ωχ , with C ′ = P N +1 (0) ¯ χ P N +1 ( ω )¯ P N +1 (¯ ω ) ! ¯ ω N +1 (1 − ¯ ω ) . Hence y = C C ′ D ( I − H Φ ∗ N H Φ N ) 11 − ωχ (cid:12)(cid:12) − ¯ ωχ E . We have now to use the lemma
Lemma 5 − ωχ is an eigenvector of H ⋆ Φ N H Φ N for the eigenvalue τ N,r ( ω ) ω N +2) with τ N,r ( ω ) = P N +1 ( ω ) P N +1 ( ω ) P N +1 ( ω ) P N +1 ( ω ) , with | τ N,r ( ω ) | = 1 for < r < . It is the lemma 1 of [18]. We obtain y = C C ′ − ω N +2 τ N,r ( ω ) 11 − ω . If now we define the function f by f = ˜ g ˜ g with ˜ g = χ (1 − ¯ χ χ ) P N +1 and ˜ g = (1 − ¯ χ ¯ χ ) P N +1 , then for a fixed N lim r → ( T N f r ) − , = ( T N f ) − , . Indeed( T N f r ) − ( T N f ) = ( T N f r ) − ( T N f r ) + ( T N f r ) − ( T N ( f − f r )) . And lim r → ( T N ( f − f r )) = 0 that implies lim r → ( T N f r ) − ( T N f ) = I N . Hence we can con-clude that ( T N ( f )) − , = (1 − χ N +20 τ N ) B ,N − B ,N − χ N +2)0 τ N , with B ,N = C C ′ (1 − ¯ χ ), B ,N = χ (cid:12)(cid:12)(cid:12) P N +1 (0) (cid:12)(cid:12)(cid:12) , and τ N = P N +1 (¯ χ ) P N +1 (¯ χ ) P N +1 ( χ ) P N +1 ( χ ) With the notation of the previous point the inversion formula provides, for f = g ¯ g , g ∈ H + ( T ). (cid:0) T − N (cid:1) l +1 ,k +1 = h π + (cid:18) χ l ¯ g (cid:19) | χ k ¯ g i − h + ∞ X s =0 (cid:0) H ∗ Φ N H Φ N (cid:1) s π + ¯Φ N π + (cid:18) χ l ¯ g (cid:19) | π + ¯Φ N π + (cid:18) χ k ¯ g (cid:19) i . with Φ N = g ¯ g χ N +1 . For l = 0 this formula gives us (cid:0) T − N (cid:1) ,k +1 = h π + (cid:18) g (cid:19) | χ k ¯ g i − h + ∞ X s =0 (cid:0) H ∗ Φ N H Φ N (cid:1) s π + ¯Φ N π + (cid:18) g (cid:19) | π + ¯Φ N π + (cid:18) χ k ¯ g (cid:19) i .
15n the next of the proof we use the following notations :1 g = X u ≥ β u χ u g ¯ g = X u ∈ Z γ u χ u . Since the hypothesis f ∈ A ( T , s ) implies that g and g ¯ g ∈ A ( T , s ) (see [22] or [11])we have twopositive constants K and K ′ such that | β u | ≤ Ku s ∀ u ∈ N ⋆ and | γ u | ≤ K ′ u s ∀ u ∈ Z ⋆ . With these notations we obtain h π + (cid:18) g (cid:19) | χ k ¯ g i = ¯ β β k ,π + ¯Φ N π + (cid:18) g (cid:19) = π + (cid:0) ¯Φ N ¯ β (cid:1) = ¯ β X v ≥ N +1 ¯ γ − v χ v − N − ,π + ¯Φ N π + (cid:18) χ k ¯ g (cid:19) = k X w =0 ¯ β w X v ≥ N +1 − k + w ¯ γ v χ v − N − k − w . On the other hand for ψ = X w ≥ α w χ w a function in H + we have H Φ N ( ψ ) = X w ≥ α w X v>N +1+ w γ − v χ v + w + N +1 ! that provides k H Φ N ( ψ ) k ≤ X w ≥ | α w X v>N +1+ w | γ − v | ! ≤ k ψ k X w ≥ X v>N +1+ w | γ − v | ! / ≤ K k ψ k ( N + 1) − s +3 / that means k H Φ N k ≤ K ( N + 1) − s +3 / . Clearly we have also k H ⋆ Φ N k ≤ K ′ ( N + 1) − s +3 / andwe can write k + ∞ X s =0 (cid:0) H ∗ Φ N H Φ N (cid:1) s π + ¯Φ N π + (cid:18) g (cid:19) k ≤ K ′ (cid:0) − K ′ ( N + 1) − s +3 / (cid:1) ( N + 1) − s +1 . Since for all fixed integer k in [0,N] we have k π + ¯Φ N π + (cid:18) χ k ¯ g (cid:19) k ≤ k X w =0 | β w | X v ≥ N +1 − k + w | γ v | = O (1)we can write (cid:0) T − N (cid:1) ,k +1 = ¯ β β k + O (cid:16) ( N + 1) − s +3 / (cid:17) that is the expected result. 16 .3 Demonstration of the lemma 4 For 0 ≤ k ≤ N we have to consider the quantity I k ( r, λ ′ ) = (cid:12)(cid:12)(cid:12)Z π (cid:18) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α ( θ ) + h ,α ( λ ′ )(1 − cos θ ) − λ ′ (cid:19) e − ikθ dθ (cid:12)(cid:12)(cid:12) where r ∈ ]0 , λ ′ ∈ [0 , I k ( r, λ ′ ) ≤ Z π (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α ( θ ) + h ,α ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ. In the following we assume λ ′ = 1 − cos θ ′ and π > θ > θ ′ ∈ [0 , θ ], θ ′ ∈ [ θ , π ] and θ ′ ∈ [ π , π ] . First we assume θ ′ ∈ [0 , θ ], and we use the decomposition I k ( r, λ ′ ) ≤ I ( r, , λ ′ ) + I ( r, , λ ′ ) + I ( r, , λ ′ ) with I ( r, , λ ′ ) = Z θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α ( θ ) + h ,α ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθI ( r, , λ ′ ) = Z π θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α,r ( θ ) + h ,α,r ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and I ( r, , λ ′ ) = Z π π (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α ( θ ) + h ,α ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ. We have clearly I ( r, , λ ′ ) ≤ Z θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ + Z θ h α ( θ ) − h ,α ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and I ( r, , λ ′ ) ≤ O (cid:0) ( θ (1 − r ) α − (cid:1) + O (cid:0) θ (1 − cos 2 θ ) α − (cid:1) . Then I ( r, , λ ′ ) ≤ Z π θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h α ( θ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ + Z π θ (cid:12)(cid:12)(cid:12) h ,α,r ( θ ′ ) − h ,α ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and I ( r, , λ ′ ) ≤ O (cid:18) (1 − r )(1 − cos θ ) α − θ (cid:19) . And lastly I ( r, , λ ′ ) ≤ Z π π (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h α ( θ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ + Z π π (cid:12)(cid:12)(cid:12) h ,α,r ( θ ′ ) − h ,α ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and I ( r, , λ ′ ) = O (1 − r ) . For θ ′ ∈ [ θ , pi ] we use the decomposition I k ( r, λ ′ ) = J ( r, , λ ′ ) + J ( r, , λ ′ ) + J ( r, , λ ′ ) with J ( r, , λ ′ ) = Z θ ′ − θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α ( θ ) + h ,α ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ ( r, , λ ′ ) = Z θ ′ + θ θ ′ − θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α,r ( θ ) + h ,α,r ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and J ( r, , λ ′ ) = Z πθ ′ + θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( λ ′ ) − h α ( θ ) + h ,α ( λ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ. As pr´eviously we write J ( r, , λ ′ ) ≤ Z θ ′ − θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h α ( θ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ + Z θ ′ − θ h ,α,r ( θ ′ ) − h ,α ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and J ( r, , λ ′ ) ≤ O (cid:18) (1 − r ) 1 θ (cid:19) . Then J ( r, , λ ′ ) ≤ Z θ ′ + θ θ ′ − θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h ,α,r ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ + Z θ ′ + θ θ ′ − θ (cid:12)(cid:12)(cid:12) h α ( θ ) − h ,α ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and J ( r, , λ ′ ) ≤ O ( θ ) . And lastly J ( r, , λ ′ ) ≤ Z πθ ′ + θ (cid:12)(cid:12)(cid:12) h α,r ( θ ) − h α ( θ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ + Z πθ ′ + θ (cid:12)(cid:12)(cid:12) h ,α,r ( θ ′ ) − h ,α ( θ ′ )(1 − cos θ ) − λ ′ (cid:12)(cid:12)(cid:12) dθ and J ( r, , λ ′ ) = O (cid:18) (1 − r ) θ (cid:19) . The case θ ′ ∈ [ π , π ] is quite identical.To conclude we see that for r such that 1 − r = θ we have | I k ( r, λ ′ ) | ≤ θ for all k ∈ Z and λ ′ ∈ [0 , Proof : If δ u = \ | − χ | α ( u ) we can write, for all j ∈ Z , c h α ( j ) = X u
0, provides X u
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