Edge Universality for Orthogonal Ensembles of Random Matrices

arXiv:0812.3228v1 [math-ph] 17 Dec 2008

Edge Universality for Orthogonal Ensembles of Random Matrices M. Shcherbina Institute for Low Temperature Physics, Kharkov, Ukraine. E-mail: [email protected]

Abstract We prove edge universality of local eigenvalue statistics for orthogonal invariant matrix models with real analytic potentials and one interval limiting spectrum. Our starting point is the result of [21] on the representation of the reproducing matrix kernels of orthogonal ensembles in terms of scalar reproducing kernel of corresponding unitary ensemble.

1

Introduction and main results

We study ensembles of n × n real symmetric (or Hermitian) matrices M with the probability distribution nβ −1 (1.1) Pn (M )dM = Zn,β exp{− TrV (M )}dM, 2 where Zn,β is a normalization constant, V : R → R+ is a H¨older function satisfying the condition |V (λ)| ≥ 2(1 + ǫ) log(1 + |λ|). (1.2) A positive parameter β here assumes the values β = 1 (in the case of real symmetric matrices) or β = 2 (in the Hermitian case), and dM means the Lebesgue measure on the algebraically independent entries of M . The joint eigenvalue distribution corresponding to (1.1) has the form (see [14]) pn,β (λ1 , ..., λn ) =

Q−1 n,β

n Y

e−nβV (λi )/2

i=1

Y

|λi − λj |β ,

(1.3)

1≤j 0.

(1.15)

C3. DOS ρ(λ) is strictly positive in the internal points λ ∈ (−2, 2) and ρ(λ) ∼ |λ ∓ 2|1/2 , as λ ∼ ±2. C4. The function u(λ) = 2

Z

log |µ − λ|ρ(µ)dµ − V (λ)

achieves its maximum if and only if λ ∈ σ. Note (see [2]) that under conditions C1-C4 the limiting density of states (DOS) ρ has the form p 1 (1.16) P (λ) 4 − λ2 1|λ| 0,

λ ∈ [−2, 2].

(1.18)

We will use also that under conditions C1-C4 the entries of semi infinite Jacoby matrix J (n) , generated by the recursion relations for orthogonal polynomials (1.8) (n)

(n)

(n)

(n)

(n)

(n)

(n)

Jl+1 ψl+1 (λ) + ql ψl (λ) + Jl ψl−1 (λ) = λψl (λ), (n)

satisfy the relations (see [1, 2]): ql (n) J n+k − 1 −

(n)

J0

= 0,

l = 0, ....

(1.19)

= 0 and

k |k|2 + n2/3 ≤ C , 2nP (2) n2

|k| ≤ 2n1/2 ,

(1.20)

where P is defined by (1.17). Here and everywhere below we denote by C, C0 , C1 , c, ... positive n-independent constants (different in different formulas). The main result of the paper is Theorem 1 Consider an orthogonal ensemble of random matrices (1.3) with β = 1, and V satisfying conditions C1-C4. Set γ = P 2/3 (2), where P is defined by (1.17). Then for even n we have: (n)

(n)

(i) if pl1 is the lth marginal of (1.3), then nl/3 pl1 (2+x1 /γn2/3 , . . . , 2+xl /γn2/3 ) converges uniformly in xj ≥ s > −∞, j = 1, ..., l to the limits coinciding with that for GOE and given in terms of b Ai (x, y) = lim n−2/3 γ −1 K b n (2 + x/γn2/3 , 2 + y/γn2/3 ), Q n→∞

b Ai (x, y) = Q



SAi (x, y) DAi (x, y) IAi (x, y) − ǫ(x − y) SAi (y, x)

4



(1.21) (1.22)

with

  Z ∞ 1 Ai(z)dz , SAi (x, y) = QAi (x, y) + Ai(x) 1 − 2 y 1 DAi (x, y) = −∂y QAi (x, y) − Ai(x)Ai(y), 2 Z

(1.23)

∞

IAi (x, y) = −

QAi (z, y)dy  Z y Z ∞ Z ∞ 1 Ai(z)dz , Ai(z)dz Ai(z)dz + + 2 y x x x

and QAi (x, y) of (1.9); (ii) if En,1 is the gap
probability (1.14) of (1.3), corresponding to the semi-infinite interval 2/3 2 + s/γn , ∞ , then   (edge) bAi (s)), (s) = det1/2 (I − Q (1.24) lim En,1 2 + s/γn2/3 , ∞ := E1 n→∞

b Ai (s) is the integral operator, defined in L2 (s, ∞; w)⊕L2 (s, ∞; w−1 ) by the 2×2 where Q matrix kernel (1.22), (1.23) with w(x) = x2 + 1.

Remark 1 According to the results of [2] and [16], if we restrict the integration in (1.3) by (n,L) |λi | ≤ L = 2 + d1 /2, consider the polynomials {pk }∞ k=0 orthogonal on the interval [−L, L] (n,L) (n,L) −nV −nV /2 with the weight e and set ψk =e pk , then for k ≤ n(1 + ε) with some ε > 0 (n,L)

sup|λ|≤L |ψk

(n)

(λ) − ψk (λ)| ≤ e−nC ,

(n)

|ψk (±L)| ≤ e−nC

(1.25)

with some absolute C. Therefore from the very beginning we can take all integrals in (1.3), (1.8), (1.13) and (1.12) over the interval [−L, L]. Note also that since V is an analytic function in Ω[d1 , d2 ] (see (1.15)), for any m ∈ N there exists a polynomial Vm of the (2m)th degree such that |Vm (z)| ≤ C0 ,

|V (z) − Vm (z)| ≤ e−Cm ,

z ∈ Ω[d1 /2, d2 /2].

(1.26)

Take m = [log2 n]

(1.27)

(n,L,m)

}∞ and consider the system of polynomials {pk k=0 orthogonal in the interval [−L, L] with (0,n) (n,L,m) (n,L,m) −nV (λ) m e−nVm /2 and construct Mm by (1.12) = pk respect to the weight e . Set ψk (n,L,m) with ψk . Then for any k ≤ n + 2n1/2 and uniformly in λ ∈ [−L, L] (n,L)

(n,L,m)

(n,L)

2

(n,L,m)

|ψk (λ) − ψk (λ)| ≤ e−C log n , |εψk (λ) − εψk (λ)| ≤ e−C log 2 2 (0,n) (0,n) ||Mm − M(0,n) || ≤ e−C log n , ||(Mm )−1 − (M(0,n) )−1 || ≤ e−C log n .

2

n

(1.28)

The proof of the first bound here is identical to the proof of (1.25) (see [16]). The second bound follows from the first one because the operator ε : L2 [−L, L] → C[−L, L] is bounded by L. The third bound in (1.28) follows from the first, and the last bound follows from the third (0,n) one and from the fact that ||(Mm )−1 || is uniformly bounded (see [21]). Hence |Sn,m (λ, µ) − Sn (λ, µ)| ≤ Cn4 e−C log

2

n

≤ e−C

′

log2 n ,

(1.29)

and below we will study Sn,m (λ, µ) instead of Sn (λ, µ). To simplify notations we omit the indexes m, L, but keep the dependence on m in the estimates. For more detail of the replacement see [21]. 5

Our starting point is the representation of the matrix (M(0,n) )−1 valid under conditions C1-C4 (see Corollary 1 in [21]). To formulate this result we introduce a few Toeplitz matrices. Consider the infinite matrix P = {Pj,k }∞ j,k=−∞ in l2 [−∞, ∞] with entries Pj,k

1 = 2π

Z

π

P (2 cos y)ei(j−k)y dy,

(1.30)

−π

and R = P −1 , (0,n)

R

=

n−1 {Rj,k }j,k=0

Rj,k = Rj−k

Z

1 = 2π

π −π

ei(j−k)x dx . P (2 cos x)

(1.31)

It is important that (see [21]), Proposition 1) −c|j−k| |((R(0,n) ))−1 , j,k | ≤ e

|Rj,k | ≤ e−c|j−k|,

(1.32)

where c > 0 is some n-independent constant. Remark also that if we denote by J ∗ an infinite Jacobi matrix with constant coefficients ∗ = δj+1,k + δj−1,k , Jj,k

∗ ∞ }j,k=−∞, J ∗ = {Jj,k

(1.33)

then the spectral theorem yields that P = P (J ∗ ), R = P −1 (J ∗ ). Two more matrices which we use below have the form n−1 , D (0,n) = {Dj,k }j,k=0

Dj,k = δj+1,k − δj−1,k .

(1.34)

and V (0,∞) = {Vj,l }∞ j,l=0 , where Vj,l = sign(l −

(n) (n) j)(ψj , V ′ ψl )2

2 = n

(

(n)

(n)

(ψj , (ψl )′ )2 ,

(n) (n) (ψj , (ψl )′ )2

j > l,

2 + O(e−C log n ),

j ≤ l.

(1.35)

2

Here O(e−C log n ) appears because of the integration by parts and bounds (1.25), (1.28). According to Corollary 1 from [21], under conditions C1-C4 (0,n)

(M(0,n) )−1 j,k = Qj,k

1 + aj bk + O(n−1/2 log6 n), 2

(1.36)

where (0,n)

Qj,k

1 = 2

(

(0,∞)

Vj,k , (R(0,n) )−1 D (0,n) )j,k ,

for for

0 ≤ j ≤ n − 2m, 0 ≤ k < n, n − 2m < j < n, 0 ≤ k < n.

(1.37)

and aj = ((R(0,n) )−1 en−1 )j ,

bk = ((R(0,n) )−1 r ∗ )k ,

∗ rn−i = Ri

(1.38)

with Ri defined by (1.31). Note that since (R)−1 j,k = Pj,k = 0 for |j − k| > 2m − 2, the standard linear algebra yields (0,n) −1 that (R ) possesses the same property, i.e., (0,n)

(R(0,n) )−1 j,k = 0, for |j − k| > 2m − 2 ⇒ Qj,k

6

= 0, for |j − k| > 2m − 2.

(1.39)

2

Proof of Theorem 1.

b n (2+x/γn2/3 , λ0 + Remark 2 (1.) It is easy to see that the integral operator with the kernel K y/γn2/3 ) defined in (1.10) is not a trace class operator in L2 (s, ∞) ⊕ L2 (s, ∞) (recall that A : H1 → H2 is a trace class operator if ||A||1 := Tr (A∗ A)1/2 < ∞). To take care of this problem we follow the approach of [26] and use weighted L2 spaces. If we take any w such that b n is a Hilbert-Schmidt w−1 ∈ L1 and grows at infinity not faster than exponentially, then K 2 2 −1 b n are finite rank, hence trace class. on L (s, ∞; w) ⊕ L (s, ∞; w ). The diagonal entries of K Now the definition of determinant extends to Hilbert-Schmidt operator matrices Tb with trace class diagonal entries by setting b det(I − Tb) = det2 (I − Tb)e−Tr T ,

where Tr denotes the sum of the traces of the diagonal entries of Tb and the det2 is the regularized 2-determinant, defined for the Hilbert-Schmidt operator T with eigenvalues tk as Y det2 (I − Tb) = (1 − tk )etk

(see [13], Section IV.2). It follows from the identity for 2-determinant

b b det2 (I − Tb1 )(I − Tb2 )eTr T1 T2 = det2 (I − Tb1 ) det2 (I − Tb2 )

that for this extended definition we still have the relation

det(I − Tb1 )(I − Tb2 ) = det(I − Tb1 ) det(I − Tb2 )

(2.) Consider a rank one kernel u(x)v(y), where u ∈ L2 (s, ∞; w2 ) and v ∈ L2 (s, ∞; w1−1 ) and u ⊗ v : L2 (s, ∞; w1 ) → L2 (s, ∞; w2 ) Z ∞ h(y)v(y)dy. (2.1) (u ⊗ v h)(x) = u(x) s

Then we have ||u ⊗ v||1 ≤ ||u||L2 (w2 ) ||v||L2 (w−1 )

(2.2)

1

Introduce the scaled kernels: Sn (x, y) Dn (x, y) In (x, y) Kn (x, y)

= = = =

(n2/3 γ)−1 Sn (2 + x/γn2/3 , 2 + y/γn2/3 ), (n2/3 γ)−2 Dn (2 + x/γn2/3 , 2 + y/γn2/3 ), In (2 + x/γn2/3 , 2 + y/γn2/3 ), (n2/3 γ)−1 Kn (2 + x/γn2/3 , 2 + y/γn2/3 ),

(2.3)

where Sn is defined by (1.11) and Kn is defined in (1.6). We prove first assertion (ii). Observe that the determinant in (1.14) is the same if we b n by replace the interval (2 + s/n2/3 , 2 + ε) by (s, εn2/3 ) and the kernel K   Sn (x, y) Dn (x, y) b Kn (x, y) = . (2.4) In (x, y) − ǫ(x − y) Sn (y, x)

Lemma 1 If we denote

ϕn (x) = n−1/6 γ −1/2 ψn(n) (2 + x/γn2/3 ),

(n)

ψn (x) = n−1/6 γ −1/2 ψn−1 (2 + x/γn2/3 ), 7

(2.5)

then 1 Sn (x, y) = Kn (x, y) + ψn (x)ǫϕn (y) + rn (x, y) 2 1 ∂ ∂ rn (x, y) Dn (x, y) = − Kn (x, y) − ψn (x)ϕn (y) − ∂y 2 ∂y 1 In (x, y) = IKn (x, y) + ǫψn (x)ǫϕn (y) + (ǫrn )(x, y) 2 where ||rn (x, y)||1 , ||

(2.6)

∂ rn (x, y)||1 , ||(ǫrn )(x, y)||1 ≤ Cn−1/3 log6 n. ∂y

(2.7)

Proof. Since (1.36), (1.37) and (1.35) imply for j ≤ n − 2m n

n−1 X (n) (n) −c log2 n ), (M(0,n) )−1 j,k ǫψk (µ) = −ψj (µ) + O(e k=0

we have −n

n−2m X n−1 X

(n) (n) ψj (λ)(M(0,n) )−1 j,k ǫψk (µ)

=

j=0 k=0

n−2m X

(n)

(n)

ψj (λ)ψj (µ) + O(e−c log

2

n

).

(2.8)

j=0

For n − 1 ≤ j > n − 2m we need to use the result of [21] (see Eq. (68)), according to which for any |p − n| ≤ 4 log2 n we have −

∞  X 1  (n) (n) (n) ǫψp+1 (µ) − ǫψp−1 (µ) = n−1 Rp,l ψl (µ) + n−1 ep (µ) 2

=

l=0 n−1 X (0,n) (n) Rp,l ψl (µ) n−1 l=0

+n

−1

∞ X

(n)

Rp,l ψl (µ) + n−1 ep (µ), (2.9)

l=n

where the remainder terms ep (µ) satisfy the bounds

||ep ||L2 [−L,L] ≤ Cn−1/2 log4 n. Therefore the scaled functions eep (x) = n−1/3 ep (2 + x/γn2/3 ) admit the bounds ||e ep ||L2 (w−1 ) ≤ Cn−1/2 log4 n.

Hence, using the definition of D (0,n) (1.34) and (2.9), we obtain −n

n−1 X

n−1 X

j=n−2m+1 k=0

+

  (n) ψj (λ) (R(0,n) )−1 D (0,n) n (n) ǫψ (µ) 2 n

n−1 X

(n)

j,k

ǫψk (µ) =

n−1 X

(n)

(n)

ψj (λ)ψj (µ)

j=n−2m+1

(n)

(1) (2) ψj (λ)(R(0,n) )−1 j,n−1 + rn (λ, µ) + rn (λ, µ), (2.10)

j=n−2m+1

(1)

where rn (λ, µ) collects the terms, which appear because of the second sum in the r.h.s. of (2) (2.9), and rn (λ, µ) collects the remainder terms ep of (2.9): rn(1) (λ, µ)

:=

n−1 X

n−1 ∞ XX

(n)

j=n−2m+1 p=0 l=n

rn(2) (λ, µ)

:=

n−1 X

n−1 X

(n)

(R(0,n) )−1 j,p ψj (λ)ep (µ).

j=n−2m+1 p=0

8

(n)

(R(0,n) )−1 j,p Rp,l ψj (λ)ψl (µ),

(α)

Definition 1 We will say that some remainder kernel rn (λ, µ) (α = 1, 2, . . . ) satisfies the bound B with exponents κ1 , κ2 and κ3 , if ||n−2/3 rn(α) (2 + x/γn2/3 , 2 + y/γn2/3 )||1 ≤ Cn−κ1 mκ2 logκ3 n.

(B)

According to the results of [7], we have (n)

n−1/6 γ −1/2 ψj (2 + x/γn2/3 ) = Ai(x + (n − j)/c∗ n1/3 )(1 + O(n−1/3 )), (n) |n−1/6 γ −1/2 ψj (2

(2.11)

+ x/γn2/3 )| ≤ Ce−x ,

where c∗ is some constant not important for us. These asymptotic implies, in particular, that (n)

||n−1/6 γ −1/2 ψj (2 + x/γn2/3 )||L2 (w) ≤ C, (n)

||n−1/6 γ −1/2 ψj (2 + x/γn2/3 )||L2 (w−1 ) ≤ C,

(2.12)

(1)

Using the asymptotic, (1.32) and (2.2), we obtain that rn (λ, µ) satisfies (B) with κ1 = 1/3, κ2 = 1, κ3 = 0. (2) Similarly, using (2.9), (2.11), (1.32) and (2.2), we get that rn (λ, µ) satisfies (B) with κ1 = 2/3, κ2 = 1, κ3 = 4. Moreover, if we denote n−1 X

rn(3) (λ, µ) = nǫψn(n) (µ)

(n)

(n)

(ψj (λ) − ψn−1 (λ))(R(0,n) )−1 j,n−1

(2.13)

j=n−2m+1

then (2.8) and (2.10) give us −n

n−1 n−1 X X

(n)

(n)

ψj (λ)(M(0,n) )−1 j,k ǫψk (µ) = Kn (λ, µ)

j=0 k=0

+

n (n) (n) ǫψ (µ)ψn−1 (λ)((R(0,n) )−1 en−1 , u) + rn(1) (λ, µ) + rn(2) (λ, µ) + rn(3) (λ, µ), (2.14) 2 n

where u is a vector, whose components are given by ui = 1,

i ∈ [n − 2m, n − 1],

ui = 0,

i 6∈ [n − 2m, n − 1],

(2.15)

(3)

and the remainder term rn in view of (2.11), (1.32) and (2.2) satisfies the bound (B) with κ1 = 1/3, κ2 = 1, κ3 = 0. Now we consider the term (see (1.36)) A(λ, µ) =

n 2

n−1 X

(n)

ak ψk (λ)

n−1 X

(n)

bj ǫψj (µ).

j=n−2m

k=n−2m

Using (2.9) and (2.11), similarly to the above it is easy to obtain that A(λ, µ) =

n (n) (a, u)(b, u) ψn−1 (λ) ǫψn(n) (µ) + rn(4) (λ, µ), 2 (4)

(2.16)

where u is defined in (2.15), and the remainder rn satisfies the bound (B) with κ1 = 1/3, κ2 = 0, κ3 = 0. 9

Let us find (a, u)(b, u). Making transposition in (1.36) and taking into account that (M(0,n) )−1 and D (0,n) are skew symmetric matrices, we get 1 1 (0,n) (0,n) −1 (R ) )j,k + ak bj + O(n−1/2 log n). −(M(0,n) )−1 j,k = − 2 (D 2 Taking the sum of the equation with (1.36) and applying the result to u we get (a, u)(b, u) =

1 ([D (0,n) , (R(0,n) )−1 ]u, u) = −((R(0,n) )−1 u, D (0,n) u) + O(n−1/2 m2 log n). 2

But it is easy to see that D (0,n) u = −en−1 + en−2m + en−2m−1 Hence, (a, u)(b, u) = ((R(0,n) )−1 en−1 , u) − ((R(0,n) )−1 (en−2m + en−2m−1 ), u) + O(n−1/2 m2 log n). Moreover, since P = R−1 has only 2m − 2 nonzero diagonals, the standard linear algebra argument yields that for j ≤ n − 2m (R(0,n) )−1 ej = Pej . Then, using (1.30) and (2.15), we obtain (P(en−2m + en−2m−1 ), u) = P (2) Finally (a, u)(b, u) = ((R(0,n) )−1 en−1 , u) − P (2) + O(n−1/2 m2 log n).

(2.17)

(α)

Then, combining (2.14) with (2.17) and bounds (B) for rn (λ, µ) with α = 1, 2, 3, 4, we get the first line of (2.6). The second line of (2.6) can be proved similarly, if we use that (2.11) can be differentiated. To prove the last line of (2.6) we used that (2.11) implies that for |k − n| = o(n) |ǫψk (µ)| ≤ Cn−1/2 . (2.18) Hence ||ǫψk (2 + x/γn2/3 )||L2 (w−1 ) ≤ Cn−1/2 .

(2.19)

Using these bounds we get the estimates for ǫr (1) (λ, µ) and ǫr (2) (λ, µ). The bound for ǫr (3) (λ, µ) and ǫr (4) (λ, µ) follow from (2.9) (2.2), (2.12) and (1.32).  Let us transform the kernel Kn . We use the representation n−1 ∞

n X X ′ (n) V (J )j,k Kn (λ, µ) = 2 k=0 j=n

Z

0

∞

  (n) (n) (n) (n) dν ψk (λ + ν)ψj (µ + ν) + ψk (µ + ν)ψj (λ + ν) .

(2.20) ∂ ∂ The representation can be obtained by taking + from both sides of (2.20) and using ∂λ ∂µ ∂ (n) (n) ψ (λ) with respect to the basis {ψj }∞ of (1.35) to expand j=1 . ∂λ k Using the same trick as above, on the basis of (2.11) and (2.2) it is easy to show that   Z n−1 ∞ n X X ′ (n)  ∞  (n) (n) Kn (λ, µ) = V (J )j,k dν ψn (λ + ν)ψn−1 (µ + ν) 2 0 k=0 j=n  (n) +ψn(n) (µ + ν)ψn−1 (λ + ν) + rn(5) (λ, µ). (2.21) 10

(5)

where rn (λ, µ) satisfies (B) with κ1 = 1/3, κ2 = κ3 = 0. Moreover, using (1.20), we have Vns :=

n−1 ∞ XX

V ′ (J (n) )j,k =

n−1 ∞ XX

V ′ (J ∗ )j,k + O(n−1 ) =

k=0 j=n

k=0 j=n

where Vk′

1 = 2π

Z

π

∞ X

kVk′ + O(n−1 ),

k=1

V ′ (2 cos x)eikx dx.

−π

On the other hand, it is evident that if we consider V(x) =

n−1 X

Vk′ sin kx,

k=0

then Vns

d = V(x) + O(n−1 ). dx x=0

But it was proved in [21] (see Lemma 1) that V(x) = sin xP (2 cos x). Thus we get Vns = P (2) + O(n−1 ), and we obtain from (2.21) that the kernel Kn from (2.3) can be represented in the form Z 1 ∞ dz (ψn (x + z)φn (y + z) + ψn (y + z)φn (x + z)) + rn(6) (x, y), (2.22) Kn (x, y) = 2 0 (6)

where ||rn ||1 ≤ Cn−1/3 . Hence, the kernel Sn is represented in the form 1 Sn (x, y) = 2

Z

∞

dz (ψn (x + z)φn (y + z) + ψn (y + z)φn (x + z)) 0

+

n ψn (x)ǫϕn (y) + rn (x, y), 2

||rn ||1 ≤ Cmn−1/3 . (2.23)

Thus we can prove that Sn converges in the trace norm to SAi , repeating almost literally argument of [26], but using (2.11) instead of classical asymptotic for Hermite polynomials. Indeed, relations (2.11) yield lim ||ϕn (· + z) − Ai(· + z)||L2 (w) = lim ||ψn (· + z) − Ai(· + z)||L2 (w) = 0.

n→∞

n→∞

(2.24)

Let us prove that Kn : L2 (w) → L2 (w) of (2.3) converges in the ||...||1 norm to QAi : L2 (w) → L2 (w). where QAi is defined in (1.9). Using (2.11), (1.9), and (2.2) we have ||Kn − QAi ||1 ≤ (2.25) Z ∞
1 ||ϕn (· + z) − Ai(· + z)||L2 (w) + ||ψn (· + z) − Ai(· + z)||L2 (w) 2 0
× ||ϕn (· + z)||L2 (w) + ||ψn (· + z)||L2 (w) + ||Ai(· + z)||L2 (w) dz Z ∞
≤C ||ϕ(· + z) − Ai(· + z)||L2 (w) + ||ψ(· + z) − Ai(· + z)||L2 (w) e−z dz 0

for some n-independent C > 0. Here we have again used (2.11), implying ||ϕn (· + z)||L2 (w) ≤ Ce−z ,

||ψn (· + z)||L2 (w) ≤ Ce−z . 11

(2.26)

Now we can use the dominated convergence theorem to make the limit n → ∞ in (2.25). To pass to the limit n → ∞ in ψn (x)ǫϕn (y), remark that uniformly in y > s Z

ǫϕn (y) = cϕn − cϕn

1 = 2

Z

n2/3 γ(L−2)

ϕn (z)dz,

y

n2/3 γ(L−2)

n1/2 ϕn (z)dz = 2 −n2/3 γ(L+2)

Z

L

−L

ψn(n) (λ)dλ,

(2.27)

where L was defined in Remark 1. But according to the results of [7], n1/2 n1/2

Z

2−n−1/4 −2+n−1/4

Z

ψn(n) (λ)dλ → 0, ! Z −1/4 −2−n

L

+

2+n−1/4

−L

(n)

Thus, (2.11) and the evenness of ψn lim cϕn = lim n1/2

n→∞

n→∞

ψn(n) (λ)dλ → 0,

n → ∞.

yield

Z

2+n−1/4

2−n−1/4

ψn(n) (λ)dλ =

Z

∞

Ai(x)dx = 1. −∞

Moreover, (2.11) allow us to pass to the limit n → ∞ in the second term of the representation (2.27) of ǫϕn . Thus we have uniformly in y ≥ s > −∞ Z ∞ Ai(z)dz. lim ǫϕn (y) = 1 − n→∞

y

Now (2.2) implies     Z ∞ 1 1 lim Kn (x, y) + ψn (x)ǫϕn (y) = QAi (x, y) + Ai(x) 1 − Ai(z)dz , n→∞ 2 2 y where the limit is understood in the || . . . ||1 -norm. To prove that −∂y Kn : L2 (w−1 ) → L2 (w) converges in the ||...||1 norm to −∂y QAi : 2 L (w−1 ) → L2 (w), we repeat the argument used in (2.25), taking into account (2.2) with w1 = w−1 and w2 = w. Besides, we have the relations ||ϕn (· + z)||L2 (w−1 ) ≤ C ′ e−z ,

||ψn (· + z)||L2 (w−1 ) ≤ C ′ e−z ,

lim ||ϕn (· + z) − Ai(· + z)||L2 (w−1 ) = lim ||ψn (· + z) − Ai(· + z)||L2 (w−1 ) = 0,

n→∞

n→∞

and lim ||ϕ′n (· + z) − Ai′ (· + z)||L2 (w−1 ) = lim ||ψn′ (· + z) − Ai′ (· + z)||L2 (w−1 ) = 0.

n→∞

n→∞

We obtain then 1 lim (−∂y Kn (x, y) − ψn (x)ϕn (y)) = −∂y QAi (x, y) − Ai(x)Ai(y), n→∞ 2 where the limit is understood in the ||...||1 norm.

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(2.28)

We are left to prove that In : L2 (w) → L2 (w−1 ) converges in the ||...||1 norm to the operator from L2 (w) to L2 (w−1 ), defined by the kernel Z ǫ(x − x′ )QAi (x′ , y)dx′ . To this end denote Φn (x) =

Z

∞

′

′

ϕn (x )dx ,

Ψn (x) =

Z

∞

ψn (x′ )dx′ .

x

x

Then ǫϕn (x) = ǫψn (x) = ǫKn (x, y) = =

Z 1 ∞ ϕn (x′ )dx′ − Φn (x) = cϕn − Φn (x), 2 Z−∞ 1 ∞ ψn (x′ )dx′ − Ψn (x) = −Ψn (x) 2 Z−∞ 1 ∞ dz(ǫψn (x + z)ϕn (y + z) + ǫϕn (x + z)ψn (y + z)) 2 0Z ∞ cϕ 1 − dz(Ψn (x + z)ϕn (y + z) + Φn (x + z)ψn (y + z)) + n Ψn (y). 2 0 2 (n)

Here the second relation follows from the fact that ψn−1 is an odd function, and the third one follows from the first, and the second, combined with (2.27). Hence, repeating again the argument used in (2.25) and taking into account that (2.11) implies Z ∞ Z ∞ ′ ′ Ai(x′ )dx′ ||L2 (w−1 ) = 0, Ai(x )dx ||L2 (w−1 ) = lim ||Φn (x) − lim ||Ψn (x) − n→∞

n→∞

x

x

and ||Ψn (· + z)||L2 (w−1 ) ≤ Ce−z ,

||Φn (· + z)||L2 (w−1 ) ≤ Ce−z ,

we obtain ∞

QAi (x′ , y)dx′ x  Z x Z ∞ Z ∞ 1 ′ ′ ′ ′ ′ ′ Ai(x )dx Ai(x )dx + Ai(y )dy , + 2 x y y

lim In (x, y) = −

n→∞

Z

where the limit is understood in the || . . . ||1 norm. Thus we have proved assertion (ii). b n (x, y) = Q b Ai (x, y) uniformly in x, y ∈ Note that we have also proved that limn→∞ K (s, ∞). Hence assertion (i) is also proved. 

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