ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
arXiv:1810.09223v1 [math.PR] 22 Oct 2018
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU Abstract. Our first result states that the orthogonal and symplectic Bessel processes are rigid in the sense of Ghosh and Peres. Our argument in the Bessel case proceeds by an estimate of the variance of additive statistics in the spirit of Ghosh and Peres. Second, a sufficient condition for number rigidity of stationary Pfaffian processes, relying on the Kolmogorov criterion for interpolation of stationary processes and applicable, in particular, to pfaffian sine-processes, is given in terms of the asymptotics of the spectral measure for additive statistics.
1. Introduction 1.1. Point processes and number rigidity. Let Conf(R) be the set of non-negative integer-valued Radon measures on the real line R. Elements of Conf(R) are called (locally finite) configurations on R. The space Conf(R) is a Polish space equipped with the vague topology generated by the maps: Z f dξ for compactly supported continuous functions f : R → C. ξ 7→ R
By definition, a point process on R is a Borel probability measure on Conf(R). A configuration ξ ∈ Conf(R) is called simple if ξ({x}) ∈ {0, 1} for all x ∈ R. A point process P is called simple, if P-almost every configuration is simple. Given an element ξ ∈ Conf(R) and any Borel subset S ⊂ R, we denote ξ|S the restriction of the measure ξ on S. Definition 1.1 (Ghosh [7], Ghosh-Peres[8]). A point process P on R is called number rigid if for any bounded Borel subset B ⊂ R, there exists a Borel function FB : Conf(R) → Z such that ξ(B) = FB (ξ|R\B ), for P-almost every ξ ∈ Conf(R). 1.2. Pfaffian point processes. Recall that for a simple point process P on R, the k(k) point correlation function ρP of P with respect to the Lebesgue measure, if it exists, is the (k) non-negative function ρP : Rk → R such that for any continuous compactly supported function ϕ : Rk → C, we have Z Z ∗ X (k) ϕ(x1 , . . . , xk )P(dX ) = ϕ(x1 , . . . , xk )ρP (x1 , . . . , xk )dx1 · · · dxk , Conf(R)
where
∗ P
x1 ,...,xk ∈X
Rk
denotes the sum over all ordered k-tuples of distinct points (x1 , . . . , xk ) ∈ X k .
2010 Mathematics Subject Classification. Primary 60G55; Secondary 60G10. Key words and phrases. Pfaffian point process, stationary point process, number rigidity. 1
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
A simple point process P on R is said to be a Pfaffian point process if there exists a matrix kernel K : R × R → C2×2 such that for all positive integers k, the k-point correlation functions of P exist and have the form (k)
ρP (x1 , · · · , xk ) = Pf[K(xi , xj )J]1≤i,j≤k . Here Pf(A) is the Pfaffian of an antisymmetric matrix and the matrix kernel K must satisfy the condition 0 1 t (1.1) (K(x, y)J) = −K(y, x)J, where J = −1 0 to ensure that the 2k × 2k matrix [K(xi , xj )J]1≤i,j≤k is antisymmetric. In this situation, we say that the point process P is the Pfaffian point process induced by the matrix kernel K and is denoted PK . We can write the matrix kernel K as K11 (x, y) K12 (x, y) K(x, y) = (1.2) , K21 (x, y) K22 (x, y) where the entries Kij : R × R → C are scalar functions and then the condition (1.1) says that (1.3)
K22 (x, y) = K11 (y, x), K12 (x, y) = −K12 (y, x), K21 (x, y) = −K21 (y, x).
We recall the general structure of the Pfaffian kernels for β = 4 (symplectic) ensembles and β = 1 (orthogonal) ensembles and their scaling limits: R 1 K4 (x, y) − xy K4 (x, t)dt (1.4) , K4 (x, y) = ∂ K4 (x, y) K4 (y, x) 2 ∂x Ry K1 (x, y) − x K1 (x, t)dt − 1/2sgn(x − y) K1 (x, y) = ∂ (1.5) , K (x, y) K1 (y, x) ∂x 1 for some particular K1 (x, y), K4 (x, y). In the integrable case a kernel has the following form A(x)B(y) − B(x)A(y) Kβ (x, y) = + C(x)D(y). x−y
Section 2 is devoted to the Pfaffian Bessel processes. Recall that a classical polynomial β-ensemble is defined by the probability density function const(β, wβ )
N Y i=1
wβ (xi )
Y
1≤i 0. Regarding the formula for KsBessel (x, y), see Proposition 2.15. The orthogonal Bessel kernel KBessel (x, y) is given by the formula 1,s 1/2 Z x Js+1 (y 1/2 ) ∞ Bessel K1,s (x, y) = K2,s+1 (x, y) + Js+1 (t)dt, y 4y 1/2 x1/2 Ry K1,s (x, y) − x K1,s (x, t)dt − 1/2sgn(x − y) Bessel K1,s (x, y) = ∂ . K (x, y) K1,s (y, x) ∂x 1,s Theorem 1.2. (i) The symplectic Bessel process is number rigid. (ii) The orthogonal Bessel process is number rigid.
The number rigidity of the Pfaffian Bessel processes is proved in subsections 2.3, 2.4. We use the following sufficient condition for the number rigidity of a point process due to Ghosh and Peres. Proposition 1.3 (Ghosh [7], Ghosh and Peres [8]). Let M be a complete metric separable space. Let P be a Borel probability measure on Conf(M). Assume that for any ε > 0 and any bounded subset B ⊂ M there exists a bounded measurable P function f of bounded support such that f ≡ 1 on B and VarP Sf < ε, where Sf (X) = x∈X f (x), X ∈ Conf(M). Then the measure P is number rigid. We give an explicit formula for the variance of an additive functional of a Pfaffian point process in (2.11), subsection 2.1. Preliminary integral estimates are discussed in subsection 2.2. A difference with the determinantal case can be seen as follows. Consider a point process on R with the first correlation function ρ(1) (x), second corrleation function ρ(2) (x, y) and truncated second correlation function ρ(2,T ) (x, y) = ρ(2) (x, y)−ρ(1) (x)ρ(1) (y). For a determinantal process governed by an orthogonal projection, we have Z (1.6) ρ(2,T ) (x, y)dy = −ρ(1) (x), R
In a discussion of perfect screening in [6], 14.1, p.660, Forrester writes that property (1.6) should “remain valid in the thermodynamic limit”. We show that (1.6) is not valid for the Pfaffian Bessel processes, see (2.33), (2.39). Nonetheless, a weaker integral property holds (see Proposition 2.16 and Proposition 2.19) and suffices for our purposes. We next give a general sufficient condition for the number rigidity of a stationary point process convenient for working with stationary Pfaffian processes. Let P be a stationary (1) (2) point process on R admitting the first and the second correlation functions ρP and ρP . The first correlation function is a constant, and we set (1)
ρ = ρP (x).
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Denote also
(2)
F (x) = ρP (x, 0) − ρ2 .
Proposition 1.4. Assume that there exists C > 0 such that the Fourier transform Fb of F satisfies (1.7) 0 ≤ Fb(λ) + ρ ≤ C|λ| for all λ ∈ R. Then the point process P is number rigid.
For example, Pfaffian sine processes arise as bulk scaling limits of the Pfaffian Gaussian ensembles, see [6], 7.6.1. and 7.8.1. Let sin(πx) Ksine,2 (x, y) = S(x − y), S(x) := πx be the standard sine-kernel. Then, the orthogonal sine process or Sine1 -process, is the Pfaffian point process on R with a matrix correlation kernel S(x − y) IS(x − y) − ε(x − y) Ksine,1 (x, y) = , S ′ (x − y) S(x − y)
where
Z
x
1 S(t)dt and ε(x) := sgn(x), 2 0 and the symplectic sine process, the Sine4 -process, is the Pfaffian point process on R with a matrix correlation kernel 1 S(x − y) IS(x − y) Ksine,4 (x, y) = . 2 S ′ (x − y) S(x − y) (1.8)
IS(x) :=
In these cases, the quantity Fb(λ) − ρ = Fb(λ) − Fb(0) is computed in Forrester [6]. For the orthogonal sine process, Forrester [6, formula (7.136)] gives ( 2|λ| − |λ| log(1 + 2|λ|) if |λ| ≤ 1 . Fb(λ) − Fb(0) = if |λ| ≥ 1 2 − |λ| log 2|λ|+1 2|λ|−1 For the symplectic sine process, Forrester [6, formula (7.95)] gives |λ| − |λ| log 1 − 2|λ| if |λ| ≤ 1 2 4 Fb(λ) − Fb(0) = . 1 if |λ| ≥ 1 2
In both the orthogonal and symplectic case, we have 0 ≤ Fb(λ) − Fb(0) ≤ C|λ|,
and Proposition 1.4 yields
Proposition 1.5. The orthogonal sine process is number rigid. Proposition 1.6. The symplectic sine process is number rigid. Remark. For the general Sineβ processes, rigidity is due to Chhaibi and Najnudel [5], and Propositions 1.5, 1.6 of course follow from their result. Their argument is quite different. It would be interesting to obtain a spectral asymptotics at zero for general Sineβ processes. Remark. The soft edge scaling limit yields Pfaffian Airy kernels, and it would be interesting to prove the rigidity of the corresponding point processes.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
5
2. Pfaffian Bessel point processes 2.1. Variance of an additive functional. P For a general point process we have the following formula for the variance of Sf (X) = x∈X f (x), X ∈ Conf(M). Var(Sf ) = EPK (|Sf |2 ) − |EPK (Sf )|2 = X X 2 = EPK |f (x)| + EPK x∈X
x,y∈X,x6=y
X 2 f (x)f (y) − EPK f (x) = x∈X
2 Z (1) (2) 2 (1) = |f (x)| ρPK (x)dx + f (x)f (y)ρPK (x, y)dxdy − f (x)ρPK (x)dx = R R R2 Z Z (1) (2,T ) = |f (x)|2ρPK (x)dx + f (x)f (y)ρPK (x, y)dxdy = 2 R R Z Z Z 1 (1) (2,T ) (2,T ) 2 ρPK (x, y)dy dx − |f (x) − f (y)|2ρPK (x, y)dxdy, = |f (x)| ρPK (x) + 2 2 2 R R R Z
(2,T )
Z
(2)
(1)
(1)
where ρPK (x, y) = ρPK (x, y) − ρPK (x)ρPK (y) is the truncated second correlation function of PK . If we additionally have (1.6), then we finally obtain the formula Z 1 (2,T ) |f (x) − f (y)|2ρPK (x, y)dxdy. (2.9) Var(Sf ) = − 2 R2 For a Pfaffian point process induced by a matrix kernel K, by definition, we have (1)
ρPK (x) = K1,1 (x, x),
(2,T )
ρPK (x, y) = − det K(x, y).
Thus using (2.9) we obtain that if the condition (1.6) holds, then Z 1 (2.10) Var(Sf ) = |f (x) − f (y)|2 det K(x, y)dxdy 2 R2
and in general case, we have Z Z 2 (2.11) Var(Sf ) = |f (x)| K1,1 (x, x) − det K(x, y)dy dx+ R R Z 1 |f (x) − f (y)|2 det K(x, y)dxdy. + 2 R2
We recall once again the formulas for the kernels for β = 4 (symplectic ensembles) and β = 1 (orthogonal ensembles): R 1 K4 (x, y) − xy K4 (x, t)dt K4 (x, y) = , ∂ K4 (x, y) K4 (y, x) 2 ∂x Ry K1 (x, y) − x K1 (x, t)dt − 12 sgn(x − y) . K1 (x, y) = ∂ K (x, y) K1 (y, x) ∂x 1
Remark 2.1. We will consider Pfaffian sine and Bessel processes, they arise as limits of the Pfaffian polynomial ensembles. Kernels of these polynomial ensembles are skewsymmetric by construction, therefore the limit kernels are also skew-symmetric (it is not obvious from the definition of the Pfaffian Bessel kernel). We note also that det K(x, y) is a symmetric function if K(x, y)J is skew-symmetric.
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Proposition 2.2. If K(x, y) is a projection, for any x ∈ R we have lim K(y, x) = 0,
∂ K(x, y) ∂x
lim K(y, x)
y→±∞
x→±∞
Z
is skew-symmetric function, and
y
K(x, t)dt = 0,
x
then the condition (1.6) holds for the kernels K4 (x, y) and K1 (x, y). Proof. For K4 (x, y) we have
Z y 1 ∂ det K4 (x, y) = K(x, y)K(y, x) + K(x, y) K(x, t)dt , 4 ∂x x
and we use the equality Z R
∂ K(x, y) ∂x
∂ = − ∂y K(y, x) to write
Z y Z ∂ K(y, x) K(x, t)dt dy = − K(x, t)dt dy = x R ∂y x ∞ Z Z Z y K(x, y)K(y, x)dy, − K(y, x) K(x, t)dt + K(y, x)K(x, y)dy =
∂ K(x, y) ∂x
Z
y
x
and finally
Z
−∞
1 det K4 (x, y)dy = 2 R
R
R
Z
1 K(x, y)K(y, x)dy = K(x, x). 2 R
For K1 (x, y) we have Z y ∂ K(x, y) K(x, t)dt + det K1 (x, y) = K(x, y)K(y, x) + ∂x x ∂ 1 ∂ 1 + sgn(x − y) K(x, y) = 4 det K4 (x, y) − sgn(x − y) K(y, x). 2 ∂x 2 ∂y Integrating the last term we obtain Z Z 1 ∂ 1 x ∂ − sgn(x − y) K(y, x)dy = − K(y, x)dy+ 2 R ∂y 2 −∞ ∂y Z 1 ∞ ∂ K(y, x)dy = −K(x, x), + 2 x ∂y thus
Z
R
det K1 (x, y)dy = 2K(x, x) − K(x, x) = K(x, x).
For a Pfaffian process with a kernel of the form K1 (x, y) or K4 (x, y), let us define the defect of the process as the difference Z Def K (x) := K(x, y)K(y, x)dy − K(x, x). R
We have the following corollary from the proof of the previous proposition; it will be useful for the analysis of the Bessel processes later.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
7
Corollary 2.3. For K = K4 we have y=∞ i Z y Z 1h (2,T ) (1) . K(x, t)dt − ρPK (x, y)dy − ρPK (x) = 2 Def K (x) − K(y, x) 4 x R y=−∞ For K = K1 , if for any x ∈ R we have
lim K(y, x) = 0,
y→±∞
then −
Z
R
(2,T ) ρPK (x, y)dy
−
(1) ρPK (x)
= 2 Def K (x) − K(y, x)
Z
x
y
y=∞ . K(x, t)dt y=−∞
2.2. Preliminary propositions. We plan to estimate the variance (2.11) term by term. From the formulas in the previous section we see that it leads to the estimates of the integrals of the type Z (2.12) |f (x) − f (y)|2Π(x, y)dxdy, R2+
where for most of the summands we will have Π(x, y) = A(x, y)/(x − y)2, or Π(x, y) = B(x, y)/(x − y) or Π(x, y) = C(x)D(y) for some reasonable functions A(x, y), B(x, y), C(x), D(y). The corresponding integrals usually will be not absolutely convergent, and we will use the oscillation of Π(x, y) to obtain a required estimate. In this subsection sufficient conditions on a kernel Π(x, y) in these three cases are stated in Propositions 2.5, 2.9 and 2.12 respectively. Remark 2.4. We have noted that det K(x, y) is symmetric for a skew-symmetric kernel, therefore in our case it is sufficient to estimate all the integrals only for y ≥ x. For the general case one can split the domain of integration into two parts and change the variables. Moreover, in the present paper we consider the case of R only for the sine process; the proof of the rigidity in this case is much simpler than for Bessel process, and we can also use the stationarity to give an alternative proof as presented in section 3. Thus we decided to present the detailed proof for the case of the processes on R+ . Analogous propositions are true when we consider R instead of R+ , with almost the same arguments. We will use the family of functions that were used in [2] for the determinantal point processes. Take R > 0, T > R and set 1, x ≤ R; log(x − R + 1) if R ≤ x ≤ T ; ϕ(R,T ) (x) = 1 − log(T − R + 1) 0, T ≤ x. We split the domain of integration in the following way: D = {(x, y)|y ≥ x ≥ 0} ⊂ R+ × R+ ,
D>R = {(x, y)|y ≥ x ≥ R},
D = D>R ⊔ D y ≥ x}, D x}.
The difference ϕ(R,T ) (x) − ϕ(R,T ) (y) is zero on {(x, y)|R > y ≥ x}, therefore it is sufficient to consider the two other domains D>R and D 0, a constant const(R) > 0 such that for any y ≥ x ≥ R we have (x/y)2α + (y/x)2α |Π(x, y)| ≤ const(R) · . (x − y)2 Then we have Z D>R
|ϕ(R,T ) (x) − ϕ(R,T ) (y)|2|Π(x, y)|dxdy −−−→ 0. T →∞
(ii) Assume that there exists ε > 0, and, for any R > 0, a constant const(R) > 0 such that for any y ≥ R we have Z R const(R) |Π(x, y)|dx ≤ . y 1+ε −R Then we have Z D 0 such that for any c ∈ [a, b) we have Z c ≤ M. f (x)dx a
Then
Z b ≤ M · |g(a)|. f (x)g(x)dx
(2.13)
a
We will also use the following simple analog of Riemann-Lebesgue lemma: Lemma 2.8. Let f (x) be a differentiable function on [a, b], then for every α ∈ R we have Z b Z b const α α−1 ′ f (x) sin(mx + n)x dx ≤ max |f (x)| + |f (x)|dx . a≤x≤b m a
a
¯ λ) = Proposition 2.9. Let Π(x, y) be a function on D. We set λ = y/x and Π(x, Π(x, λx).
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
9
(i) Assume that there exist constants 1 ≥ ε1 > ε4 ≥ 0, 1 > ε2 > ε5 ≥ 0, ε3 > ε6 ≥ 0, λ1 > 1, λ2 < 1, a positive function ψ, ψ(T ) = o(log2 (T )) when T → ∞, and, for any R > 0, a constant const(R) > 0 such that the following holds: (2.14)
Z b λε 4 λε 5 λε 6 ¯ max Π(x, λ)dx ≤ const(R)ψ(max(a, b)) , + + a,b>R a |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3 (2.15) |Π(x, y)| ≤ const(R) for y > x > R. (2.16)
Then we have Z
D>R
|ϕ(R,T ) (x) − ϕ(R,T ) (y)|2
Π(x, y) dxdy −−−→ 0. T →∞ y−x
(ii) If ε2 = 1 and ψ(T ) = o(log(T )) when T → ∞, then (2.16) also holds. (iii) Assume that there exist constants ε7 > −1, ε8 > 0, and, for any R > 0, a constant const(R) > 0 such that the following holds: |Π(x, y)| ≤ const(R)xε7 y −ε8 for x < R, y > R.
(2.17)
(2.18)
Then we have Z
DR into three parts: D>R = {x, y ∈ D : R ≤ x ≤ y < T } ⊔ {x, y ∈ D : R ≤ x < T ≤ y} ⊔
⊔ {x, y ∈ D : T ≤ x ≤ y}.
Note that the integral is zero on {x, y ∈ D : T ≤ x ≤ y}. The First Case: R ≤ x ≤ y < T . T T Z Z 2 const Π(x, y) ≤ (2.19) dydx log(x − R + 1) − log(y − R + 1) x−y log2 (T − R + 1) R x T T /x Z Z ¯ λ) const(R) Π(x, 2 + dλdx log (λ) (log T )2 λ−1 R 1 T T /x Z Z ¯ Π(x, λ) const(R) −1 −1 + dλdx log(λ) log(1 − (R − 1)x ) − log(1 − (R − 1)(λx) ) + (log T )2 λ−1 R 1 T T Z Z 2 const(R) Π(x, y) −1 −1 . + log(1 − (R − 1)x ) − log(1 − (R − 1)(y) ) dydx (log T )2 y−x R
x
where we have used a simple estimate log(T − R + 1) ≥ const log(T ) for T sufficiently large.
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
For the first term we change the order of integration to obtain (2.20)
T T /x T /R T /λ Z Z Z Z 2 ¯ λ) const(R) log (λ) const(R) Π(x, 2 = ¯ λ)dx dλ ≤ Π(x, dλdx log (λ) (log T )2 λ−1 (log T )2 λ−1 R
1
1
const(R)ψ(T ) (log T )2
T /R Z 1
R
log2 (λ) λε 4 λε 5 λε 6 dλ −−−→ 0. + + T →∞ λ − 1 |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3
We have | log(1 − (R − 1)y −1) − const(R)y −1| ≤ const(R)y −2 , therefore for the second term in (2.19) we obtain T T /x Z Z ¯ Π(x, λ) const(R) −1 −1 ≤ log(λ) log(1 − (R − 1)x ) − log(1 − (R − 1)(λx) ) dλdx (log T )2 λ−1 R
1
const(R) (log T )2
T /R Z 1
const(R) (log T )2
T /R Z 1
const(R)ψ(T ) (log T )2
1
ZT /λ
log(λ) λ(λ − 1)
T /R Z
ZT /λ log(λ) −1 ¯ log(1 − (R − 1)x )Π(x, λ)dx dλ+ λ−1
R
R
T /λ T /R Z Z ¯ λ) log(λ) const(R) Π(x, dx dλ + dxdλ ≤ 2 2 x (log T ) x λ2 (λ − 1) 1
R
log(λ) λε 4 const(R) λε 5 λε 6 dλ + + + −−−→ 0, λ − 1 |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3 (log T )2 T →∞
where we have used (2.13) and (2.14) to estimate the first two terms, and (2.15) to bound |Π(x, y)| in the last term. We have −1 1 − (R − 1)x const(R)(y − x) , for y > x > R, ≤ log 1 − (R − 1)y −1 xy
therefore for the last term in (2.19) we obtain
T T Z Z 2 Π(x, y) const(R) −1 −1 log(1 − (R − 1)x ) − log(1 − (R − 1)(y) ) dydx ≤ 2 (log T ) y−x R x T T Z Z const(R) (y − x)|Π(x, y)| −−−→ 0, ≤ dydx T →∞ (log T )2 x2 y 2 R
where we have used (2.15) once again. The Second Case: R ≤ x < T ≤ y.
x
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
11
We need to estimate the integral ∞ T Z Z 2 log(x − R + 1) Π(x, y) 1− const(R) dxdy ≤ log(T − R + 1) y−x T R Z T /R ZT const(R) dλ 2 x−R+1 ¯ Π(x, λ)dx ≤ + log T −R+1 λ − 1 log2 (T ) 1 T /λ Z∞ ZT const(R) dλ x − R + 1 ¯ λ)dx log2 + Π(x, . T −R+1 λ − 1 log2 (T ) T /R R
We use (2.13) to bound both summands from above: Z T /R ZT const(R) x − R + 1 dλ 2 ¯ λ)dx log Π(x, ≤ 2 T −R+1 λ − 1 log (T ) 1 T /λ T /R Z ε ε ε λ4 λ5 λ6 dλ const(R)ψ(T ) 2 T /λ − R + 1 log · + + ≤ T −R+1 |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3 λ − 1 log2 (T ) 1 T /R Z ε5 ε6 ε4 dλ λ λ λ const(R)ψ(T ) 2 log λ · + + −−−→ 0. ≤ |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3 λ − 1 T →∞ log2 (T ) 1
and
Z∞ ZT dλ const(R) x − R + 1 2 ¯ Π(x, λ)dx log ≤ 2 T −R+1 λ − 1 log (T ) T /R R Z∞ ε5 ε6 ε4 const(R)ψ(T ) dλ λ λ λ 2 + + log (T − R + 1) · −−−→ 0. |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3 λ − 1 T →∞ log2 (T ) T /R
If ε2 = 1 and ψ(T ) = o(log(T )) then we split the interval [1, T /R] into two parts: P1 = [1, λ1 − T −1 ] ∪ [λ1 + T −1 , T /R] and P2 = [λ1 − T −1 , λ1 + T −1 ]. For the first part we have Z logj (λ) λε 4 λε 5 λε 6 (2.21) dλ ≤ const log(T ), j ∈ Z+ , + + |λ − 1|ε1 |λ − λ1 |ε2 |λ − λ2 |ε3 P1 λ − 1 and for the second part we combine the estimate (2.15) with an obvious estimate (2.22)
ZT Z R
P2
F (x, λ)dλdx ≤ const(R),
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
where F (x, λ) is bounded on P2 , |F (x, λ)| ≤ const(R). The cases (i) and (ii) are fully proved. Now we prove (iii). We split the domain D R. We use this fact many times below. In some of the applications below the condition (2.17) doesn’t hold. In this case we will apply the following obvious corollary from the proof of the previous theorem.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
13
Corollary 2.11. Let Π(x, y) be a function on D. If we have T R Z Z 1 Π(x, y) 2 −−−→ 0, log (y − R + 1) dxdy T →∞ (log T )2 y−x R 0 ∞ R Z Z Π(x, y) dxdy −−−→ 0, y−x T →∞ T
0
then the convergence (2.18) holds.
We use similar but simpler arguments to show the convergence to zero of the required integrals in case when the variables are split: Proposition 2.12. Let Π(x, y) = Π1 (x) · Π2 (y) be a function on D. ˜ ψ(T ˜ ) = o(log(T )) when T → ∞, (i) Assume that there exists a positive function ψ, and, for any R > 0, a constant const(R) > 0 such that for m ∈ {0, 1, 2}, we have Z b max logm (y − R + 1)Π2 (y)dy ≤ const(R)ψ˜m (T ), R 0. Then we have Z |ϕ(R,T ) (x) − ϕ(R,T ) (y)|2Π(x, y)dxdy −−−→ 0. D 1 such that R b • maxa,b>R a fi (x)dx ≤ const(R), for i ∈ {1, 2}; • • • •
gi (x) log2 (x − R + 1) are decreasing to zero for x sufficiently large, i ∈ {1, 2}; |hi (x)| ≤ const(R)x−ε2 for x > R, i ∈ {1, 2}; |Π1 (x)| ≤ const(R), x ≥ R, |Π2 (y)| ≤ const(R), y ≥ R; |Π1 (x)| ≤ const(R)xε1 for x < R.
Then we have
Z
D
|ϕ(R,T ) (x) − ϕ(R,T ) (y)|2Π1 (x)Π2 (y)dxdy −−−→ 0. T →∞
Proof. Let g2 (y) logm (t − R + 1) be decreasing for y > R1 > R, m ∈ {0, 1, 2}. We have Z b max logm (y − R + 1)Π2 (y)dy ≤ R −1 Thus Z ∞ Z Bessel Bessel 2 Ks (x, y)Ks (y, x)dy = 0
∞
0
KsBessel (x, y)4
y 1/2 x
KBessel 2,2s−1 (4x, 4y)dy−
1/2 Z y1/2 x J2s−1 (2x ) Bessel J2s−1 (2t)dtdy+ K2,2s−1 (4x, 4y) 4 − 2x1/2 y 0 0 Z 1/2 Z 1/2 Z ∞ J2s−1 (2x1/2 ) x J2s−1 (2y 1/2 ) y + J2s−1 (2t)dt J2s−1 (2t)dtdy = x1/2 2y 1/2 0 0 0 Z 1/2 3 J2s−1 (2x1/2 ) x Bessel J2s−1 (2t)dt, Ks (x, x) − 8 x1/2 0 1/2
Z
∞
and 3 J2s−1 (2x1/2 ) 2 Def KBessel (x) = − 4,s 8 x1/2 Now since KsBessel (0, x) = 0 and
Z
x1/2
J2s−1 (2t)dt. 0
lim KsBessel (y, x) = −
y→∞
J2s−1 (2x1/2 ) , 4x1/2
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
19
we have KsBessel (y, x)
∞ Bessel Ks (y, t)dt = x y=0
Z
y
J2s−1 (2x1/2 ) 4x1/2
Z
y
y 1/2
KBessel 2,2s−1 (4y, 4t)− t x y=∞ Z 1/2 Z y 1/2 J2s−1 (2x1/2 ) y J2s−1 (2t1/2 ) y lim = KBessel J2s−1 (2p)dp dt 2,2s−1 (4y, 4t)dt− 1/2 1/2 y→∞ 2t 2x t x 0 Z J2s−1 (2x1/2 ) ∞ − J2s−1 (2p)dp. 8x1/2 x1/2 −
2
We write Z y 1/2 Z ∞ 1/2 y y Bessel K2,2s−1 (4y, 4t)dt = KBessel 2,2s−1 (4y, 4t)dt− t t x 0 Z x 1/2 Z ∞ 1/2 y y Bessel K2,2s−1 (4y, 4t)dt − KBessel 2,2s−1 (4y, 4t)dt, t t 0 y where we have Z lim y→∞
x
0
y 1/2 t
KBessel 2,2s−1 (4y, 4t)dt
const(x)y 3/4 = 0, y→∞ y−x
≤ lim
and we use Lemma 2.8 to obtain Z ∞ 1/2 y Bessel lim K2,2s−1 (4y, 4t)dt = y→∞ y t Z ∞ √ √ −1/4 y − sπ − π/4)v cos( yv − sπ + π/4) cos( − const lim y→∞ 1 v 1/2 (v − 1) √ √ cos( y − sπ + π/4)v 1/4 cos( yv − sπ − π/4) − dv = 0. v 1/2 (v − 1)
We also have Z ∞ 1/2 √ Z ∞ Z 1 √ y √ y dt Bessel J2s−1 (2 uy)J2s−1 (2 ut)du √ = K2,2s−1 (4y, 4t)dt = t 4 0 t 0 0 Z y1/2 Z ∞ √ √ Z 1 √ Z 1 √ J2s−1 (2 uy) y y √ 1 √ J2s−1 (2p)dp, J2s−1 (2 uy) du = J2s−1 (2 up)dpdu = 2 0 2 0 2 u 2 0 0 therefore lim
y→∞
Z
y
y 1/2 t
x
KBessel 2,2s−1 (4y, 4t)dt
1 = lim 2 y→∞
Z
0
y 1/2
1 J2s−1 (2p)dp = , 4
and −KsBessel (y, x)
∞ Bessel Ks (y, t)dt = x y=0
Z
y
J2s−1 (2x1/2 ) J2s−1 (2x1/2 ) − 8x1/2 8x1/2
Z
∞
J2s−1 (2p)dp. x1/2
20
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Write Z ∞ 0
det KBessel (x, y)dy − KsBessel (x, x) = 4,s 2 Def KBessel (x) − 4,s
KsBessel (y, x)
∞ Bessel Ks (y, t)dt = x y=0
Z
y
Z x1/2 Z ∞ J2s−1 (2x1/2 ) J2s−1 (2t)dt + 1 − J2s−1 (2p)dp = = −3 8x1/2 0 x1/2 Z 1/2 J2s−1 (2x1/2 ) J2s−1 (2x1/2 ) x = J2s−1 (2t)dt. − 16x1/2 4x1/2 0 We directly see that (2.33)
Z
J2s−1 (2x1/2 ) J2s−1 (2x1/2 ) − 16x1/2 4x1/2
0
x1/2
J2s−1 (2t)dt 6= 0
. Nonetheless, we have and therefore the relation (1.6) does not hold for the kernel KBessel 4,s Z
0
∞
Z 1/2 J2s−1 (2x1/2 ) J2s−1 (2x1/2 ) x − J2s−1 (2t)dt dx = 16x1/2 4x1/2 0 Z x1/2 2 ∞ Z ∞ 1 1 1 J2s−1 (2x1/2 ) 1 dx − − = 0. J2s−1 (2t)dt = = 1/2 16 0 x 4 0 16 16 0
Remark 2.17. The assumptions of Proposition 2.2 do hold for the Pfaffian Laguerre kernel S˜4 (x, y), for the finite N (see the definition of the kernel in (2.29)), and therefore the condition (1.6) holds also. In this case, first of all, 1/2 x (L) ˜ 2S4 (x, y) = K2N (x, y) + fN (x)gN (y) y 1/2 (L) (L) has a reproducing property: Π(x, y) = xy K2N (x, y) is a projection because K2N (x, y) R∞ is a Christoffel–Darboux kernel, fN (x) = x t(s−1)/2 e−t/2 Ls2N −1 (t)dt lies in the image of (2N )! y (s−1)/2 e−y/2 Ls2N (y) is orthogonal to the image of Π(x, y). Π(x, y) and gN (y) = 2Γ(s+2N ) And we also have S˜4 (0, x) = 0, lim S˜4 (y, x) = 0, y→∞
y 1/2
(L)
because the same is true for the kernel x K2N (y, x) and because the integral (2.30) is zero. Neither property holds when we consider the limiting kernel 2KsBessel (x, y). First, there is no reproducing property: scaling limits for Ls2N −1 (y) and Ls2N (y) are the same, therefore the scaling limit of gN (y) is not orthogonal to the image of the limiting projection. And, second, as we see from (2.32), the integral (2.30) is not zero after the scaling limit, and limy→∞ KsBessel (y, x) 6= 0.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
Lemma 2.18.
21
(i) The convergence to zero of the integrals (2.16), (2.18) holds for ε1 /2 √ y √ √ Π(x, y) = Jt (2 y)Jt+ε1 (2 y)Jv (2 x), x
where ε1 ∈ {0, 1} and t > −1, v > −1. (ii) The convergence to zero of the integrals (2.16), (2.18) holds for ε1 /2−1/2 √ √ y √ Π(x, y) = Jt (2 y)Jt+ε1 (2 x)Jv (2 x), x where ε1 ∈ {0, 1} and t > −1, v > −1. ¯ λ) = Π(x, λx) as a sum Proof. (i). We set λ = y/x, we write Π(x, (2.34) λ
ε1 /2−1/2
√ √ √ Jt (2 λx)Jt+ε1 (2 x)Jv (2 x) =
λε1 /2−1/2 √ − π x +
+
√ √ √ √ √ λε1 /2 Jt (2 λ x)Jt+ε1 (2 λ x)Jv (2 x)− √ √ √ √ √ cos(2 λ x − tπ/2 − π/4) cos(2 λ x − (t + ε1 )π/2 − π/4)Jv (2 x) +
√ √ √ √ λε1 /2−1/2 √ cos(2 λ x − tπ/2 − π/4) cos(2 λ x − (t + ε1 )π/2 − π/4)× π x ! √ √ 1 + × Jv (2 x) − √ 1/4 cos(2 x − vπ/2 − π/4) πx
√ √ √ √ λε1 /2−1/2 √ cos(2 λ x − tπ/2 − π/4) cos(2 λ x − (t + ε1 )π/2 − π/4)× π x √ 1 × √ 1/4 cos(2 x − vπ/2 − π/4), πx
and we check the convergence to zero of the integrals (2.16) and (2.18) term by term. We fix arbitrary R > 0 and for the first term in (2.34) we obtain √ √ √ √ √ (2.35) λε1 /2 Jt (2 λ x)Jt+ε1 (2 λ x)Jv (2 x)−
√ √ √ √ √ λε1 /2−1/2 √ − cos(2 λ x − tπ/2 − π/4) cos(2 λ x − (t + ε1 )π/2 − π/4)Jv (2 x) ≤ π x √ const(R)Jv (2 x) √ ≤ , λx
for y = λx ≥ R, where we have used the estimate (2.24) and the fact that λ ≥ 1. Now we can combine this estimate with (2.23) to see that the assumptions of Proposition 2.9 are satisfied for the difference (2.35).
22
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
As a next step we need to estimate the intergal Z b λε1 /2−1/2 √ √ √ √ √ cos(2 λ x − tπ/2 − π/4) cos(2 λ x − (t + ε1 )π/2 − π/4)× a π x √ √ 1 × Jv (2 x) − √ 1/4 cos(2 x − vπ/2 − π/4) dx ≤ πx Z √ √ 1 λε1 /2−1/2 b 1 √ cos(ε1 π/2) Jv (2 x) − √ 1/4 cos(2 x − vπ/2 − π/4) dx + 2π a x πx Z √ √ λε1 /2−1/2 b 1 √ cos(4 λ x − (2t + ε1 + 1)π/2)× + 2π a x √ √ 1 const(R) √ , × Jv (2 x) − √ 1/4 cos(2 x − vπ/2 − π/4) dx ≤ πx λ
where we have used estimates (2.24) and (2.27) and also (2.13) for non-zero first term, and Lemma 2.8 for the second term. Next, for a, b > R for the main part we can write Z b constλε1 /2−1/2 √ √ √ √ cos(2 λ x − tπ/2 − π/4) cos(2 λ x − (t + ε1 )π/2 − π/4)× a x3/4 Z b constλε1 /2−1/2 √ const × cos(2 x − vπ/2 − π/4) dx ≤ √ + × a x3/4 λ √ √ √ × cos(4 λ x − (2t + ε1 + 1)π/2) cos(2 x − vπ/2 − π/4)dx ≤ 1 1 1 ≤ const(R) √ + √ , + √ λ 2 λ+1 2 λ−1 where we have used (2.13) once again. Therefore we can apply Proposition 2.9(i) in this case. If ε1 = 0 then Proposition 2.9(iii) obviously holds, therefore we can now put ε1 = 1. We have R T √ Z Z √ cos(4 y − tπ)Jv (2 x) const(R) 2 √ dydx ≤ log (y − R + 1) 2 (log T ) x(y − x) 0 R ! √ ZR ZR′ 2 ′ log2 (R′ − R + 1) R const(R) log (y − R + 1) const(R) (v−1)/2 ≤ x dx dy+ , ≤ (log T )2 y−R R′ − R (log T )2 0
√
R
2
where the function y log (y − R + 1)/(y − R) is decreasing for y ≥ R′ . And R ∞ √ ZR Z Z √ √ const(R) T y − tπ)J (2 x) cos(4 v √ x(v−1)/2 dx −−−→ 0, dydx ≤ T →∞ T − R x(y − x) 0
T
0
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
23
thus we can use Corollary 2.11. (ii). We fix again arbitrary R > 0 and write √ √ ε /2−1/2 √ √ λ 1 J (2 λ x)Jt+ε1 (2 x)Jt (2 x)− t
√ √ √ √ λε1 /2−3/4 − √ 1/4 cos(2 λ x − tπ/2 − π/4)Jt+ε1 (2 x)Jv (2 x) ≤ πx √ √ const(R)Jt+ε1 (2 x)Jv (2 x) , ≤ λ1/4 x3/4 for y = λx ≥ R, where we have used the estimate (2.24) and the fact that λ ≥ 1. Now we can combine this estimate with (2.23) to see that the assumptions of Proposition 2.9 are satisfied for the considered difference. As a next step we need to estimate the intergal Z b λε1 /2−3/4 √ √ √ √ √ √ cos(2 λ x − tπ/2 − π/4) x1/4 Jt+ε1 (2 x)Jv (2 x)− a π x √ √ 1 const(R) , − 1/4 cos(2 x − (t + ε1 )π/2 − π/4) cos(2 x − vπ/2 − π/4) dx ≤ πx λ3/4 where we have used estimates (2.24) and (2.27) and then Lemma 2.8. Next, for a, b > R for the main part we can write Z b constλε1 /2−3/4 √ √ √ cos(2 λ x − tπ/2 − π/4) cos(2 x − (t + ε1 )π/2 − π/4)× 3/4 a x Z b √ √ √ constλε1 /2−3/4 × cos(2 x − vπ/2 − π/4) dx = cos(2 λ x − tπ/2 − π/4)× a x3/4 √ × cos(4 x − (t + v + ε1 + 1)π/2) + cos((t − v + ε1 )π/2) dx ≤ 1 1 const(R) 1 √ +√ +√ , ≤ λ1/4 λ λ+2 λ−2
where we have used (2.13) for the last estimate. Therefore we can apply Proposition 2.9(i) and (ii) in this case. Finally for x < R, y > R we have ε /2−3/4 √ √ λ 1 √ √ const(R)x(t+v+ε1 )/2 √ , πx1/4 cos(2 λ x − tπ/2 − π/4)Jt+ε1 (2 x)Jv (2 x) ≤ y 1/4 therefore the assumptions of Proposition 2.9(iii) hold for this term and Lemma is fully proved. Proof of the Theorem 1.2(i). Our plan is to expand the determinant, Z x ∂ Bessel Bessel Bessel Bessel det K4,s (x, y) = Ks (x, y)Ks (y, x) − Ks (x, y) KsBessel (x, t)dt ∂x y
24
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
and then to show that all the summands in the formula (2.11) for the variance, with f (x) = ϕ(R,T ) (x), tend to zero term by term.
Z
The first part,
R+
|ϕ
(R,T )
(x)|
2
KsBessel (x, x)
−
Z
det KBessel (x, y)dy 4,s R
dx.
We set
Π(x) =
KsBessel (x, x)
−
Z
∞
0
(x, y)dy = det KBessel 4,s
J2s−1 (2x1/2 ) = 4x1/2
Z
0
x1/2
J2s−1 (2x1/2 ) = J2s−1 (2t)dt − 16x1/2 Z J2s−1 (2x1/2 ) J2s−1 (2x1/2 ) ∞ = − J2s−1 (2t)dt, 16x1/2 4x1/2 x1/2
we use Proposition 2.16 to write
Z
∞ 0
|ϕ
(R,T )
2
(x)| Π(x)dx =
Z
0
∞
|ϕ
(R,T )
(x)| − 1 Π(x)dx 2
and then we use estimates (2.24) and (2.28) and Corollary 2.14 to see that
Z ∞ (R,T ) 2 |ϕ (x)| − 1 Π(x)dx −−−→ 0. T →∞ 0
The second part,
Z
D
|ϕ(R,T ) (x) − ϕ(R,T ) (y)|2KsBessel (x, y) · KsBessel (y, x)dxdy.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
25
We have KsBessel (x, y) · KsBessel (y, x) = 1/2 ! Z ∞ 1/2 x 1 J (2y ) 2s−1 2 − + KBessel J2s−1 (2t)dt × 2,2s−1 (4x, 4y) + y 2y 1/2 2 x1/2 ! Z ∞ y 1/2 1/2 1 J (2x ) 2s−1 × 2 − + KBessel J2s−1 (2t)dt = 2,2s−1 (4y, 4x) + x 2x1/2 2 y 1/2 1/2 2 x J2s−1(2x1/2 ) Bessel 4 K2,2s−1 (4x, 4y) + 2 × KBessel (4x, 4y) 2,2s−1 y 2x1/2 ! Z Z ∞ 2 J2s (2x1/2 )J2s−1 (2y 1/2 ) ∞ J2s−1 (2t)dt + × J2s−1 (2t)dt + 4(x − y) x1/2 y 1/2 Z y 1/2 x−1/2 J2s (2y 1/2 )J2s−1 (2x1/2 )J2s−1 (2y 1/2) ∞ + J2s−1 (2t)dt− 4(x − y) x1/2 ! 1/2 y 1/2 1/2 1/2 J (2x ) J (2y ) x 2s−1 2s−1 − + KBessel + KBessel 2,2s−1 (4x, 4y) 2,2s−1 (4y, 4x) y 2x1/2 x 2y 1/2 Z 1/2 Z 1/2 J2s−1 (2x1/2 ) x J2s−1 (2y 1/2 ) y + J2s−1 (2t)dt J2s−1 (2t)dt =: 2x1/2 2y 1/2 0 0 S1 (x, y) S2 (x, y) S3 (x, y) S4 (x, y) + + + + S5 (x, y). =: (x − y)2 x−y x−y x−y 2 Bessel The integral for the first term, S1 (x, y)/(x − y) = 4 K2,2s−1 (4x, 4y) , was estimated 2
in [2] by Proposition 2.5. We use (2.24) and (2.28) to obtain
S2 (x, y) = x1/2 J2s (2x1/2 )J2s−1 (2y 1/2 ) − y 1/2 J2s (2y 1/2 )J2s−1 (2x1/2 ) × Z Z 2 J2s (2x1/2 )J2s−1 (2y 1/2 ) ∞ 1 J2s−1 (2x1/2 ) ∞ J (2t)dt J (2t)dt + ≤ const 1/2 , 2s−1 2s−1 1/2 4y 4 λ x x1/2 y 1/2
where we have set, as usual, λ = y/x. Thus 2.14 is satisfied for x, y > R and any R > 0. Also we easily obtain S2 (x, y) ≤ const(R, s)y −1/2 for x < R, y > R,
and we can apply Proposition 2.9.
26
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
We also have 1 S3 (x, y) − √ cos(2y 1/2 − sπ − π/4) cos(2y 1/2 − (2s − 1)π/2 − π/4)J2s−1 (2x1/2 )× 2π x R Z ∞ J2s−1 (2x1/2 ) ∞1/2 J2s−1 (2t)dt × J2s−1 (2t)dt ≤ const , √x xy 1/2 x
thus we can combine this estimate with (2.23), (2.24), (2.28) to use Proposition 2.9 in this case. For the main term of S3 (x, y) we have Z b 1 √ cos(2y 1/2 − sπ − π/4) cos(2y 1/2 − (2s − 1)π/2 − π/4)J2s−1 (2x1/2 )× a 2π x Z Z ∞ b 1 √ cos(4λ1/2 x1/2 − 2sπ)J2s−1(2x1/2 )× × J2s−1 (2t)dtdx = 4π x 1/2 x a Z ∞ const log(b) √ × J2s−1 (2t)dtdx ≤ λ x1/2 by Lemma 2.8, and we can use Proposition 2.9(i). We also have R T R∞ Z Z 1/2 1/2 cos(4y − 2sπ)J (2x ) J (2t)dt const(R) 2s−1 1/2 2s−1 2 x ≤ log (y − R + 1) dydx (log T )2 x1/2 (y − x) 0
const(R) ≤ (log T )2
R
ZR
xs−1 dx
0
ZR′ R
√
log2 (y − R + 1) dy + y−R
√
R′ log2 (R′ − R + 1) R′ − R
!
≤
const(R) , (log T )2
2
where the function y log (y − R + 1)/(y − R) is decreasing for y ≥ R′ . And R ∞ R∞ Z Z 1/2 1/2 cos(4y − 2sπ)J (2x ) J (2t)dt 2s−1 1/2 2s−1 x ≤ dydx x1/2 (y − x) 0
T
√ ZR const(R) T xs−1 dx −−−→ 0, T →∞ T −R 0
thus we can use Corollary 2.11. For the term S4 (x, y) we use Lemma 2.18 for all the four summands. In the last term S5 (x, y) the variables are split, and we have Z 1/2 Z 1/2 J2s−1 (2x1/2 ) x J2s−1 (2y 1/2 ) y S5 (x, y) = J2s−1 (2t)dt × J2s−1 (2t)dt = 2x1/2 2y 1/2 0 0 Z ∞ Z ∞ J2s−1 (2x1/2 ) 1 J2s−1 (2y 1/2 ) 1 = − − J2s−1 (2t)dt × J2s−1 (2t)dt . 2x1/2 2 2y 1/2 2 x1/2 y 1/2
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
27
We use the estimates (2.23), (2.24) to obtain Z J2s−1 (2x1/2 ) x1/2 ≤ const(R)xs−1 for x < R. J (2t)dt 2s−1 1/2 2x 0 Moreover, from the estimates (2.24), (2.28) we see that
(2.36) Z ∞ J 1 2s−1 (2x1/2 ) 1 − J2s−1 (2t)dt − √ 3/4 cos(2x1/2 − (2s − 1)π/2 − π/4)× 1/2 2x 2 2 πx x1/2 1 1 const(R) × − √ 1/4 sin(2x1/2 − (2s − 1)π/2 − π/4) ≤ , 2 2 πx x5/4
for x > R, therefore the conditions of Corollary 2.13 are satisfied and we have proved the required convergence to zero of the intergal (2.12) for S5 (x, y). Thus the required convergence is proved for the whole first term KsBessel (x, y) · KsBessel (y, x). The third part, Z
|ϕ
D>R
(R,T )
(x) − ϕ
(R,T )
∂ Bessel (y)| K (x, y) ∂x s 2
Z
x
y
KsBessel (x, t)dtdxdy.
We will use the following notation: ∂ Bessel K (x, y) = ∂x s xy −1/2 J2s (2x1/2 )J2s−1 (2y 1/2 ) − x1/2 J2s (2y 1/2 )J2s−1 (2x1/2 ) + − 2(x − y)2
′ y −1/2 J2s (2x1/2 )J2s−1 (y 1/2 ) + x1/2 y −1/2 J2s (2x1/2 )J2s−1 (2y 1/2 ) − 2(x − y) ! ′ 1/2x−1/2 J2s (2y 1/2 )J2s−1 (2x1/2 ) + J2s (2y 1/2 )J2s−1 (2x1/2 ) − + 2(x − y)
And Z y x
2
Z
0
J2s−1 (2y 1/2 ) J2s−1 (2x1/2 ) D1 (x, y) D2 (x, y) + =: + + D3 (x, y). 1/2 1/2 2y 2x (x − y)2 (x − y)
KsBessel (x, t)dt ∞
y−x 0
KsBessel (x, t + x)dt =
1/2 x + x))dt − 2 KBessel 2,2s−1 (4x, 4(t + x))dt+ t+x y−x Z 1/2 Z y1/2 Z ∞ 1 y + J2s−1 (2p)dp =: J2s−1 (2p)dp J2s−1 (2p)dp − 2 x1/2 x1/2 x1/2 I˜1 (x) + I2 (x, y) + I3 (x, y) + I4 (x, y).
x t+x
1/2
=
Z
KBessel 2,2s−1 (4x, 4(t
Z
∞
28
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
We will first separate the main part of I˜1 (x), because the corresponding integrals are not absolutely convergent. We have 1/2 Z ∞ x x1/2 J2s (2x1/2 )J2s−1 (2(t + x)1/2 ) ˜ I1 (x) = 2 − t+x 4t 0 ! (t + x)1/2 J2s (2(t + x)1/2 )J2s−1 (2x1/2 ) − dt, 4t and we use (2.24) to estimate it as follows: Z ∞ 3/4 x cos(2x1/2 − sπ − π/4) cos(2(t + x)1/2 − (2s − 1)π/2 − π/4) 1 ˜ − I1 (x) − 2π 0 t(t + x)3/4 ! x1/4 cos(2(t + x)1/2 − sπ − π/4) cos(2x1/2 − (2s − 1)π/2 − π/4) − dt ≤ t(t + x)1/4 Z 1 Z ∞ const const dt −1/4 √ dt + constx ≤ √ . 1/4 x t(t + x) x 0 1
We also have Z ∞ x1/4 cos(2x1/2 − sπ − π/4) cos(2(t + x)1/2 − (2s − 1)π/2 − π/4) × 0 t(t + x)1/4 ! Z 1 dt x1/2 1/4 − 1 dt ≤ x + × 1/2 3/4 1/2 (t + x) (x + (t + x)1/2 ) 0 (t + x) Z ∞ cos(2(t + x)1/2 − (2s − 1)π/2 − π/4) const(R) √ + x1/4 dt ≤ , for x > R, 1 (t + x)3/4 (x1/2 + (t + x)1/2 ) x
where we have used (2.13) once again to estimate the second term. Now we estimate the main term: Z ∞ 1 x1/4 cos(2x1/2 − sπ − π/4) cos(2(t + x)1/2 − (2s − 1)π/2 − π/4)− 1/4 2π 0 t(t + x) 1/2 1/2 − cos(2(t + x) − sπ − π/4) cos(2x − (2s − 1)π/2 − π/4) dt = Z Z −1 ∞ (u + 1)1/2 x1/4 −1 ∞ 1/2 1/2 sin(2(t + x) − 2x )dt = sin(2x1/2 u)du = = 2π 0 t(t + x)1/4 π 0 (u + 1)2 − 1 Z Z −1 ∞ sin(2x1/2 u) (u + 1)1/2 −1 ∞ sin(u) 1 1 1 du = du + O √ =− +O √ , π 0 u u+2 2π 0 u 4 x x where we have put u = (t/x + 1)1/2 − 1 and then used Lemma 2.8 at the last step. After all, we can write Z y 1 1 ˜ KsBessel (x, t)dt = − + I1 (x) + I2 (x, y) + I3 (x, y) + I4 (x, y), I1 (x) = − + I1 (x), 4 4 x √ where |I1 (x)| ≤ const(R)/ x for x ≥ R.
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Now we write D1 (x, y) D2 (x, y) + + D3 (x, y) × = (x − y)2 (x − y) x 1 × − + I1 (x) + I2 (x, y) + I3 (x, y) + I4 (x, y) = 4 ! 1 D2 (x, y) D1 (x, y) × − + I1 (x) + I3 (x, y) + I4 (x, y) + × I2 (x, y) + = (x − y)2 4 (x − y) ! D2 (x, y) D1 (x, y) + × I1 (x) + I3 (x, y) + D3 (x, y) × I2 (x, y) + × I2 (x, y)+ (x − y) (x − y)2 D2 (x, y) 1 1 + − + I4 (x, y) + D3 (x, y) × − + I1 (x) + I3 (x, y) + I4 (x, y) , (x − y) 4 4
∂ Bessel K (x, y) ∂x s
Z
y
KsBessel (x, t)dt
and we estimate the summands one by one. We use (2.24), (2.27), (2.28) and (2.13) to obtain the following simple estimates D1 (x, y) ≤
const(R) const(R) const(R) , D2 (x, y) ≤ √ 1/4 , D3 (x, y) ≤ 3/2 3/4 , 1/4 λ x λ xλ const(R) const(R) √ I3 (x, y) ≤ . , I4 (x, y) ≤ x1/4 x
We additionally use (2.13) to obtain the following estimate: Z ∞ x 1/2 x1/2 J (x1/2 )J 1/2 ) 2s 2s−1 ((t + x) (2.37) |I2 (x, y)| = 2 − y−x t + x 2t ! (t + x)1/2 J2s ((t + x)1/2 )J2s−1 (x1/2 ) − dt ≤ 2t Z ∞ 1/2 J ((t + x) ) 1 2s−1 1/2 dt + ≤ const · x|J2s (x )| 1/4 1/4 y−x t(t + x) (t + x) Z ∞ (t + x)1/4 J ((t + x)1/2 ) 2s + const · x1/2 |J2s−1 (x1/2 )| dt ≤ 1/4 y−x t (t + x)
√ 1/4 x3/4 y −1/4 + x1/4 y 1/4 xλ ≤ const(R) ≤ const(R) . y−x y−x
From the estimates above we obtain const(R)λ−1/4 D1 (x, y) 1 D (x, y) 2 ≤ × − + I (x) + I (x, y) + I (x, y) + × I (x, y) × 1 3 4 2 (x − y)2 4 (x − y) (x − y)2 √ 1/4 1 const(R) const(R) const(R) const(R) xλ √ × − + + √ 1/4 × + ≤ , 1/4 2 x xλ 4 x (x − y) (x − y)2 and we can apply Proposition 2.5(i).
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Then we have D2 (x, y) ≤ √ const(R) × × I (x) + I (x, y) + D (x, y) × I (x, y) 1 3 3 2 (x − y) xλ1/4 (y − x) √ 1/4 xλ const(R) const(R) const(R) const(R) √ √ × ≤ , + 3/2 3/4 × + x λ y−x xλ1/4 (y − x) x x and we can apply Proposition 2.9(i). For D1 (x, y)/(x − y)2 × I2 (x, y) we obtain an estimate √ 1/4 √ const(R)λ−1/4 D1 (x, y) xλ x const(R) × ≤ . (x − y)2 × I2 (x, y) ≤ 2 3 (y − x) y−x (y − x)
The corresponding integral diverges in the neighborhood of the point λ = y/x = 1, so we split the domain into two parts: y − x ≤ x1/2 and y − x > x1/2 . For the first part we have Z
0
y−x
x t+x
1/2
KBessel 2,2s−1 (4x, 4(t
+ x))dt ≤ x1/2 max KBessel 2,2s−1 (4x, 4(ξ + x)) = 0≤ξ≤x1/2
J2s−1 (2(ξ + x)1/2 ) − J2s−1 (2x1/2 ) − 4ξ 0≤ξ≤x1/2 1/2 J2s (2(ξ + x)1/2 ) − x1/2 J2s (2x1/2 ) 1/2 (ξ + x) ≤ − J2s−1 (2x ) 4ξ ′ 1 1/2 1 1/2 1/2 ≤x |J2s (2x )| max J2s−1 (2(ξ + x) ) + |J2s−1(2x1/2 )|× 4 4 0≤ξ≤x1/2 ! 1 −1/2 ′ (2(ξ + x)1/2 ) ≤ const(R), x max J2s (2(ξ + x)1/2 ) + max J2s × 2 0≤ξ≤x1/2 0≤ξ≤x1/2 = x1/2 max
x1/2 J2s (2x1/2 )
and we can apply Proposition 2.5(i) for this part. For the second part we have D1 (x, y) const(R) (x − y)2 × I2 (x, y) ≤ (y − x)2 , and we apply Proposition 2.5(i) once again. For D2 (x, y)/(x − y)(−1/4 + I4 (x, y)) we have 2D2 (x, y) =
!
y −1/2 J2s (2x1/2 )J2s−1 (y 1/2 ) − 1/2x−1/2 J2s (2y 1/2)J2s−1 (2x1/2 ) + +
!
′ ′ x1/2 y −1/2 J2s (2x1/2 )J2s−1 (2y 1/2) + J2s (2y 1/2)J2s−1 (2x1/2 ) ,
and for the first term there is an easy estimate const −1/2 J2s (2x1/2 )J2s−1 (y 1/2 ) − 1/2x−1/2 J2s (2y 1/2)J2s−1 (2x1/2 ) ≤ . y xλ1/4
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
31
For the second term we have Z b −1/2 ′ 1/2 1/2 1/2 1/2 1/2 ′ 1/2 λ J2s (2x )J2s−1 (2λ x ) + J2s (2λ x )J2s−1 (2x ) × a Z 1/2 1/2 1 1 1 λ x J2s−1 (2p)dp + 1/4 1/2 λ−1/2 sin(2x1/2 − sπ − π/4)× × − − 4 2 x1/2 πλ x
× cos(2λ1/2 x1/2 − (2s − 1)π/2 − π/4) + cos(2λ1/2 x1/2 − sπ − π/4)× 1 1 1/2 × sin(2x − (2s − 1)π/2 − π/4) − − √ 1/4 sin(2x1/2 − (2s − 1)π/2 − π/4)+ 4 4 πx 1 const(R) log(b) 1/2 1/2 sin(2λ x − (2s − 1)π/2 − π/4) dx , + √ ≤ 4 π(λx)1/4 λ1/4
and
λ−1/2 sin(2x1/2 − sπ − π/4) cos(2λ1/2 x1/2 − (2s − 1)π/2 − π/4)+ 1/4 1/2 πλ x 1/2 1/2 1/2 + cos(2λ x − sπ − π/4) sin(2x − (2s − 1)π/2 − π/4) = 1 −λ−1/2 sin(2x1/2 − sπ − π/4) sin(2λ1/2 x1/2 − sπ − π/4)+ πλ1/4 x1/2 1/2 1/2 1/2 + cos(2λ x − sπ − π/4) cos(2x − sπ − π/4) = 1 −1/2 1/2 1/2 (1 + λ ) sin 2x (λ + 1) − 2sπ + 2πλ1/4 x1/2 1
+ (1 − λ
−1/2
) cos 2(λ
1/2
1/2
− 1)x
,
thus we have Z b const(R) log(b) ≤ D (x, λx) −1/4 + I (x, λx) dx + 2 4 λ1/4 a const(R) 1 1 1 1 1 1 √ +√ + √ + √ +√ +√ for R < x < y, λ1/4 λ+1 λ−1 2 λ+1 2 λ−1 λ+2 λ−2
and we can use Proposition 2.9(i) and (ii). Regarding the product D3 (x, y) × −1/4 + I1 (x) + I3 (x, y) + I4 (x, y) , we can split the
variables for all the summands. For D3 (x, y)I1(x) we have
J2s−1 (2y 1/2 ) J2s−1 (2x1/2 ) I1 (x), D3 (x, y)I1 (x) = 2y 1/2 2x1/2 where
J2s−1 (2x1/2 ) const(R) ≤ I (x) 1 2x1/2 x5/4
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
and (2.38)
J2s−1 (2y 1/2 ) cos(2y 1/2 − (2s − 1)π/2 − π/4) const ≤ √ − , 2y 1/2 2 πy 3/4 y 5/4
and we can apply Corollary 2.13. For D3 (x, y)I3 (x, y) we write
Z Z ∞ J2s−1 (2y 1/2 ) J2s−1 (2x1/2 ) ∞ J2s−1 (2p)dp J2s−1 (2p)dp− D3 (x, y)I3(x, y) = 2y 1/2 2x1/2 x1/2 x1/2 Z Z ∞ J2s−1 (2y 1/2 ) J2s−1 (2x1/2 ) ∞ − J2s−1 (2p)dp J2s−1 (2p)dp 2y 1/2 2x1/2 y 1/2 x1/2 R∞ where we can use the estimate (2.36) for J2s−1 (2x1/2 )/2x1/2 × x1/2 J2s−1 (2p)dp. We also have Z ∞ 2 J 1/2 (2x ) const(R) 2s−1 , J (2p)dp ≤ 2s−1 2x1/2 x5/4 x1/2 and we apply Corollary 2.13 once again. For D3 (x, y)I4 (x, y) we have
Z 1 J2s−1 (2y 1/2 ) J2s−1 (2x1/2 ) ∞ D3 (x, y)I3(x, y) = − J2s−1 (2p)dp+ 2 2y 1/2 2x1/2 x1/2 Z 1 J2s−1 (2y 1/2 ) J2s−1 (2x1/2 ) ∞ J2s−1 (2p)dp. + 2 2y 1/2 2x1/2 y 1/2 We use the estimates (2.38) and Z J2s−1 (2y 1/2 ) ∞ J2s−1 (2p)dp− 2y 1/2 y 1/2
cos(2y 1/2 − (2s − 1)π/2 − π/4) sin(2y 1/2 − (2s − 1)π/2 − π/4) − = 2πy Z 1/2 J2s−1 (2y 1/2 ) ∞ const cos(4y − 2sπ) ≤ = J (2p)dp − , 2s−1 2y 1/2 4πy y 3/2 y 1/2
and we apply Corollary 2.13 one more time. Finally for D3 (x, y) we use the estimate (2.38) and then Corollary 2.13 for the fourth time. We have checked that the integral (2.12) tends to zero for Z y ∂ Bessel Π(x, y) = K (x, y) KsBessel (x, t)dt, R < x < y. ∂x s x The fourth part, Z
D R we have
|D1 (x, y)| ≤ const(R)y −1/4 ,
|D2 (x, y)| ≤ const(R)xs−1 y −1/4 ,
|D3 (x, y)| ≤ const(R)xs−1 y −3/4 . We also have Z 1/2 ∞ x ˜ I1 (x) = 2 0 t+x
x1/2 J2s (2x1/2 )J2s−1 (2(t + x)1/2 ) − 4t Z ! 1/2 x x (t + x)1/2 J2s (2(t + x)1/2 )J2s−1 (2x1/2 ) Bessel dt ≤ 2 K2,2s−1 (4x, 4(t + x))dt + − 0 4t t+x Z 1/2 ∞ x Bessel + 2 K2,2s−1 (4x, 4(t + x))dt , x t+x where Z 1/2 x x 0 t+x
x1/2 J2s (2x1/2 )J2s−1 (2(t + x)1/2 ) − 2t
! (t + x)1/2 J2s (2(t + x)1/2 )J2s−1 (2x1/2 ) − dt ≤ 2t 1/2 Z x 1/2 1/2 x J (2(t + x) ) − J (2x ) 1/2 2s−1 2s−1 dt + x J2s (2x1/2 ) t + x 2t 0 1/2 Z x x (t + x)1/2 − x1/2 1/2 1/2 dt + J2s (2x )J2s−1 (2x ) t+x 2t 0 Z x J2s (2(t + x)1/2 ) − J2s (2x1/2 ) 1/2 1/2 dt ≤ x J2s−1 (2x ) 2t 0
1 ′ ≤ x|J2s (2x1/2 )| max |J2s−1 (2(x + ξ)1/2 )| + const · x1/2 |J2s (2x1/2 )J2s−1 (2x1/2 )|+ ξ∈[0,x] 2 1 ′ (2(x + ξ)1/2 )| ≤ const(R), + x|J2s−1 (2x1/2 )| max |J2s ξ∈[0,x] 2
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
and Z 1/2 ∞ x x t+x
x1/2 J2s (2x1/2 )J2s−1 (2(t + x)1/2 ) − 2t
! (t + x)1/2 J2s (2(t + x)1/2 )J2s−1 (2x1/2 ) − dt ≤ 2t Z ∞ 1 J2s−1 (2(t + x)1/2 ) 1/2 x J2s (2x ) dt + 2t (t + x)1/2 x Z ∞ 1/2 J (2(t + x) ) (t + x) 2s dt ≤ + x1/2 J2s−1 (2x1/2 ) 2t t+x x
const|J2s (2x1/2 )| + const · x1/2 |J2s−1(2x1/2 )| ≤ const(R)xs ≤ const(R), where we have used (2.13) to estimate both terms. Finally we have ˜ I1 (x) ≤ const(R).
We use the estimate (2.37) for x < R, y > R to obtain Z ∞ 1/2 1 J (2(t + x) ) 2s−1 |I2 (x, y)| ≤ const · x|J2s (2x1/2 )| dt + 1/4 1/4 y−x t(t + x) (t + x) Z ∞ (t + x)1/4 J (2(t + x)1/2 ) 2s dt ≤ + const · x1/2 |J2s−1 (2x1/2 )| 1/4 y−x t (t + x)
xs+1 y −1/4 + xs y 1/4 y 1/4 ≤ const(R) ≤ const(R) . y−x y−x
For I3 (x, y) + I4 (x, y) = Z ∞ Z = J2s−1 (2p)dp y 1/2
we have
and
x1/2
J2s−1 (2p)dp −
0
Z Z
∞
J2s−1 (2p)dp x1/2 ∞
Z
∞
J2s−1 (2p)dp x1/2
Z
x1/2
J2s−1 (2p)dp
0
x1/2
0
Z
Z
J2s−1 (2p)dp ≤ const(R)
x1/2
const(R) . y 1/4 0 y 1/2 2 ˜ Thus we can use Proposition 2.5(ii) for D1 (x, y)/(x − y) × I1 (x) + I3 (x, y) + I4 (x, y) J2s−1 (2p)dp
J2s−1 (2p)dp ≤
and for D2 (x, y)/(x − y) × I2 (x, y). For D1 (x, y)/(x − y)2 × I2 (x, y) we obtain an estimate const(R)xs y −1/4 xs y 1/4 D1 (x, y) const(R) ≤ × I (x, y) × ≤ . 2 (x − y)2 2 (y − x) y−x (y − x)3
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We split the domain into two parts: x < R < y < 2R and x < R < 2R < y. For the first part we have Z y−x x 1/2 Bessel max K2,2s−1 (4x, 4(t + x))dt ≤ const(R), x R, and where we have set, as usual, λ = y/x. We see that the conditions of Proposition 2.92.14 are satisfied. Also we easily obtain S2 (x, y) ≤ const(R, s)y −1/2 for x < R, y > R,
and we can apply Proposition 2.9.
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We also have 1 S3 (x, y) − √ cos(y 1/2 − (s + 2)π/2 − π/4) cos(y 1/2 − (s + 1)π/2 − π/4)Js+1(x1/2 )× 4π x R Z ∞ Js+1 (x1/2 ) ∞1/2 Js+1 (t)dt × Js+1 (t)dt ≤ const , √x xy 1/2 x
thus we can combine this estimate with (2.23), (2.24), (2.28) to use Proposition 2.9 in this case. For the main term of S3 (x, y) we have Z b 1 √ cos(y 1/2 − (s + 2)π/2 − π/4) cos(y 1/2 − (s + 1)π/2 − π/4)Js+1(x1/2 )× a 4π x Z Z ∞ b 1 √ cos(2λ1/2 x1/2 − (s + 2)π)Js+1(x1/2 )× × Js+1 (t)dtdx = a 8π x x1/2 Z ∞ const log(b) √ × Js+1 (t)dtdx ≤ λ x1/2 by Lemma 2.8, and we can use Proposition 2.9(i). We also have R T R∞ Z Z 1/2 1/2 cos(2y − (s + 2)π)Js+1(x ) x1/2 Js+1 (t)dt const(R) 2 ≤ log (y − R + 1) dydx (log T )2 x1/2 (y − x) 0
const(R) ≤ (log T )2
R
ZR
xs/2 dx
0
ZR′ R
√
log2 (y − R + 1) dy + y−R
√
R′ log2 (R′ − R + 1) R′ − R
!
≤
const(R) , (log T )2
2
where the function y log (y − R + 1)/(y − R) is decreasing for y ≥ R′ . And R ∞ R∞ Z Z 1/2 1/2 cos(2y − (s + 2)π)J (x ) s+1 1/2 Js+1 (t)dt x ≤ dydx x1/2 (y − x) 0
T
√ ZR const(R) T xs/2 dx −−−→ 0, T →∞ T −R 0
thus we can use Corollary 2.11. For the last term S4 (x, y) we have Z Z Js+1 (x1/2 ) ∞ Js+1 (y 1/2 ) ∞ S4 (x, y) = Js+1 (t)dt × Js+1 (t)dt, 4x1/2 4y 1/2 x1/2 y 1/2 where the variables are split. We use the estimates (2.23), (2.24) to obtain Z Js+1 (x1/2 ) ∞ ≤ const(R)xs/2 for x < R. J (t)dt s+1 4x1/2 1/2 x
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ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Moreover, from the estimates (2.24), (2.28) we see that Z Js+1 (x1/2 ) ∞ Js+1 (t)dt− (2.40) 4x1/2 x1/2 const(R) 1 1/2 1/2 − sin(x − (s + 1)π/2 − π/4) cos(x − (s + 1)π/2 − π/4) ≤ , 2πx x3/2
for x > R thus the conditions of Corollary 2.13 are satisfied and we have proved the required convergence to zero of the intergal (2.12) for S4 (x, y), and therefore for all the first term K1,s (x, y) · K1,s (y, x).
The third part, ! Z Z y ∂ sgn(x − y) |ϕ(R,T ) (x) − ϕ(R,T ) (y)|2 K1,s (x, y) dxdy. K1,s (x, t)dt + ∂x 2 D>R x We will use the following notation: ∂ K1,s (x, y) = ∂x −
xy −1/2 Js+2 (x1/2 )Js+1 (y 1/2 ) − x1/2 Js+2 (y 1/2 )Js+1 (x1/2 ) + 2(x − y)2
′ y −1/2 Js+2 (x1/2 )J2s−1 (y 1/2 ) + x1/2 y −1/2 Js+2 (x1/2 )Js+1 (y 1/2 ) − 2(x − y) ! ′ 1/2x−1/2 Js+2 (y 1/2 )Js+1(x1/2 ) + Js+2 (y 1/2 )Js+1 (x1/2 ) + − 2(x − y)
+
Js+1 (y 1/2 ) Js+1 (x1/2 ) D1 (x, y) D2 (x, y) =: + + D3 (x, y). 4y 1/2 2x1/2 (x − y)2 (x − y)
And Z
x
y
Z y−x 1 1 K1,s (x, t)dt + sgn(x − y) = K1,s (x, t + x)dt − = 2 2 0 1/2 1/2 Z ∞ Z ∞ x x KBessel KBessel 2,s+1 (x, t + x)dt − 2,s+1 (x, t + x)dt+ t+x t+x 0 y−x Z ∞ Z 1/2 1 1 1 y Js+1 (p)dp Js+1(p)dp − =: I˜1 (x) + I2 (x, y) + I3 (x, y) − , + 2 x1/2 2 2 x1/2
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41
where we have used that sgn(x − y) = −1 for y > x. We will first separate the main part of I˜1 (x), because the corresponding integrals are not absolutely convergent. We have I˜1 (x) =
Z
∞ 0
x t+x
1/2
x1/2 Js+2 (x1/2 )Js+1 ((t + x)1/2 ) − 2t
! (t + x)1/2 Js+2 ((t + x)1/2 )Js+1(x1/2 ) dt, − 2t
and we use (2.24) to estimate it as follows: Z 2 ∞ x3/4 cos(x1/2 − (s + 2)π/2 − π/4) cos((t + x)1/2 − (s + 1)π/2 − π/4) ˜ − I1 (x) − π 0 2t(t + x)3/4 ! x1/4 cos((t + x)1/2 − (s + 2)π/2 − π/4) cos(x1/2 − (s + 1)π/2 − π/4) − dt ≤ 1/4 2t(t + x) Z ∞ Z 1 dt const const −1/4 dt + constx ≤ . x1/2 t(t + x)1/4 x1/2 1 0
We also have Z ! ∞ x1/4 cos(x1/2 − (s + 2)π/2 − π/4) cos((t + x)1/2 − (s + 1)π/2 − π/4) x1/2 −1 dt ≤ 1/4 1/2 0 2t(t + x) (t + x) Z 1 Z ∞ cos((t + x)1/2 − (s + 1)π/2 − π/4) dt 1/4 1/4 x +x dt ≤ 3/4 (x1/2 + (t + x)1/2 ) 3/4 (x1/2 + (t + x)1/2 ) 2(t + x) 2(t + x) 0 1 ≤
const(R) √ , x
for x > R,
where we have used (2.13) once again to estimate the second term. We proceed now with the estimate of the main term Z 2 ∞ x1/4 cos(x1/2 − (s + 2)π/2 − π/4) cos((t + x)1/2 − (s + 1)π/2 − π/4)− π 0 2t(t + x)1/4 1/2 1/2 − cos((t + x) − (s + 2)π/2 − π/4) cos(x − (s + 1)π/2 − π/4) dt = Z Z −2 ∞ −2 ∞ (u + 1)1/2 x1/4 1/2 1/2 = sin((t + x) − x )dt = sin(x1/2 u)du = π 0 2t(t + x)1/4 π 0 (u + 1)2 − 1 Z Z 1 −1 ∞ sin(u) 1 1 −2 ∞ sin(x1/2 u) (u + 1)1/2 =− +O √ , du = du + O √ π 0 u u+2 π 0 u x 2 x
where we have put u = (t/x + 1)1/2 − 1 and then used Lemma 2.8 at the last step. After all, we can write Z y 1 1 ˜ K1,s (x, t)dt − = −1 + I1 (x) + I2 (x, y) + I3 (x, y), I1 (x) = − + I1 (x), 2 2 x √ where |I1 (x)| ≤ const(R)/ x for x ≥ R.
42
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Now we write
!
D1 (x, y) D2 (x, y) 1 ∂ = K1,s (x, y) + + D3 (x, y) × K1,s (x, t)dt − ∂x 2 (x − y)2 (x − y) x D1 (x, y) × −1 + I1 (x) + I2 (x, y) + I3 (x, y) = × −1 + I1 (x) + I3 (x, y) + (x − y)2 ! ! D2 (x, y) D2 (x, y) × I2 (x, y) + × I1 (x) + I3 (x, y) + D3 (x, y) × I2 (x, y) + (x − y) (x − y) D1 (x, y) D2 (x, y) + × I2 (x, y) − + D3 (x, y) × −1 + I1 (x) + I3 (x, y) , (x − y)2 (x − y) Z
y
and we estimate the summands one by one. We use (2.24), (2.27), (2.28) and (2.13) to obtain the following simple estimates D1 (x, y) ≤
const(R) const(R) const(R) const(R) √ , D2 (x, y) ≤ √ 1/4 , D3 (x, y) ≤ 3/2 3/4 , I3 (x, y) ≤ . 1/4 λ xλ x λ x
We additionally use (2.13) to obtain the following estimate: Z ∞ x 1/2 x1/2 J (x1/2 )J ((t + x)1/2 ) s+2 s+1 (2.41) |I2 (x, y)| = − y−x t + x 2t ! (t + x)1/2 Js+2 ((t + x)1/2 )Js+1 (x1/2 ) dt ≤ − 2t Z ∞ 1/2 1 J ((t + x) ) s+1 ≤ const · x|Js+2 (x1/2 )| dt + y−x t(t + x)1/4 (t + x)1/4 Z ∞ (t + x)1/4 J ((t + x)1/2 ) s+2 + const · x1/2 |Js+1(x1/2 )| dt ≤ y−x t (t + x)1/4
√ 1/4 xλ x3/4 y −1/4 + x1/4 y 1/4 ≤ const(R) . ≤ const(R) y−x y−x From the estimates above we obtain D1 (x, y) const(R)λ−1/4 D (x, y) 2 × (x − y)2 × −1 + I1 (x) + I3 (x, y) + (x − y) × I2 (x, y) ≤ (x − y)2 √ 1/4 const(R) const(R) xλ const(R) const(R) √ √ × 1+ ≤ , + √ 1/4 × + 2 (x − y) (x − y)2 x x xλ
and we can apply Proposition 2.5(i). Then we have D2 (x, y) const(R) (x − y) × I1 (x) + I3 (x, y) + D3 (x, y) × I2 (x, y) ≤ √xλ1/4 (y − x) × √ 1/4 xλ const(R) const(R) const(R) const(R) √ √ + 3/2 3/4 × + ≤ , × x λ y−x xλ1/4 (y − x) x x
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
43
and we can apply Proposition 2.9(i). For D1 (x, y)/(x − y)2 × I2 (x, y) we obtain an estimate √ 1/4 √ D1 (x, y) const(R)λ−1/4 xλ const(R) x × ≤ . (x − y)2 × I2 (x, y) ≤ (y − x)2 y−x (y − x)3
The corresponding integral diverges in the neighborhood of the point λ = y/x = 1, so we split the domain into two parts: y − x ≤ x1/2 and y − x > x1/2 . For the first part we have Z y−x 1/2 x Bessel K2,s+1 (x, t + x)dt ≤ x1/2 max KBessel 2,s+1 (x, ξ + x) = t+x 0≤ξ≤x1/2 0 J2s−1 ((ξ + x)1/2 ) − J2s−1 (x1/2 ) 1/2 =x max x1/2 J2s (x1/2 ) − ξ 0≤ξ≤x1/2 1/2 J2s ((ξ + x)1/2 ) − x1/2 J2s (x1/2 ) 1/2 (ξ + x) ≤ − J2s−1 (x ) ξ ′ 1 1/2 1 1/2 1/2 ≤x |J2s (x )| max J2s−1 ((ξ + x) ) + |J2s−1 (x1/2 )|× 2 2 0≤ξ≤x1/2 ! ′ ((ξ + x)1/2 ) ≤ const(R), × x−1/2 max J2s ((ξ + x)1/2 ) + max J2s 0≤ξ≤x1/2
0≤ξ≤x1/2
and we can apply Proposition 2.5(i) for this part. For the second part we have D1 (x, y) const(R) (x − y)2 × I2 (x, y) ≤ (y − x)2 , and we apply Proposition 2.5(i) once again. For D2 (x, y)/(x − y) we have 2D2 (x, y) =
y
−1/2
1/2
Js+2 (x
)Js+1 (y
+
1/2
−1/2
) − 1/2x
Js+2 (y
1/2
1/2
)Js+1 (x
!
) + !
′ ′ x1/2 y −1/2 Js+2 (x1/2 )Js+1(y 1/2 ) + Js+2(y 1/2 )Js+1 (x1/2 ) ,
and for the first term there is an easy estimate const −1/2 1/2 1/2 −1/2 1/2 1/2 Js+2 (x )Js+1(y ) − 1/2x Js+2 (y )Js+1 (x ) ≤ . y xλ1/4
For the second term we have Z b ′ ′ λ−1/2 Js+2 (x1/2 )Js+1 (λ1/2 x1/2 ) + Js+2 (λ1/2 x1/2 )Js+1 (x1/2 )+ a 2 + 1/4 1/2 λ−1/2 sin(x1/2 − (s + 2)π/2 − π/4) cos(λ1/2 x1/2 − (s + 1)π/2 − π/4)+ πλ x const(R) log(b) , + cos(λ1/2 x1/2 − (s + 2)π/2 − π/4) sin(x1/2 − (s + 1)π/2 − π/4) dx ≤ λ1/4
44
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
and
λ−1/2 sin(x1/2 − (s + 2)π/2 − π/4) cos(λ1/2 x1/2 − (s + 1)π/2 − π/4)+ πλ1/4 x1/2 1/2 1/2 1/2 + cos(λ x − (s + 2)π/2 − π/4) sin(x − (s + 1)π/2 − π/4) = 2 −λ−1/2 sin(x1/2 − (s + 2)π/2 − π/4) sin(λ1/2 x1/2 − (s + 2)π/2 − π/4)+ πλ1/4 x1/2 1/2 1/2 1/2 + cos(λ x − (s + 2)π/2 − π/4) cos(x − (s + 2)π/2 − π/4) = 2 (1 + λ−1/2 ) sin x1/2 (λ1/2 + 1) − (s + 2)π + 1/4 1/2 πλ x −1/2 1/2 1/2 , + (1 − λ ) cos (λ − 1)x 2
thus we have
Z b const(R) log(b) D2 (x, λx)dx ≤ for R < x < y, λ1/4 a and we can use Proposition 2.9(i). Regarding the product D3 (x, y) × −1 + I1 (x) + I3 (x, y) , we can split the variables for all the summands. For D3 (x, y)I1 (x) we have D3 (x, y)I1 (x) = where
and (2.42)
Js+1 (y 1/2 ) Js+1 (x1/2 ) I1 (x), 4y 1/2 2x1/2
const(R) Js+1 (x1/2 ) ≤ I (x) 1 2x1/2 x5/4
Js+1(y 1/2 ) cos(y 1/2 − (s + 1)π/2 − π/4) const ≤ √ , 4y 1/2 − y 5/4 2 2πy 3/4
and we can apply Corollary 2.13. For D3 (x, y)I3 (x, y) we write
Z Z ∞ Js+1 (y 1/2 ) Js+1 (x1/2 ) ∞ D3 (x, y)I3(x) = Js+1 (p)dp Js+1 (p)dp− 4y 1/2 4x1/2 x1/2 x1/2 Z Z ∞ Js+1 (y 1/2 ) Js+1(x1/2 ) ∞ − Js+1(p)dp Js+1 (p)dp 4y 1/2 4x1/2 y 1/2 x1/2 R∞ where we can use the estimate (2.40) for Js+1 (x1/2 )/4x1/2 × x1/2 Js+1 (p)dp. We also have 2 J (x1/2 ) Z ∞ const(R) s+1 , Js+1 (p)dp ≤ 1/2 4x x5/4 x1/2
and we apply Corollary 2.13 once again.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
45
Finally for D3 (x, y) we use the estimate (2.42) and then Corollary 2.13 for the third time. We have checked that the integral (2.12) tends to zero for
∂ K1,s (x, y) Π(x, y) = ∂x
Z
x
y
! 1 , K1,s (x, t)dt − 2
R < x < y.
The fourth part, ! Z Z y sgn(x − y) (R,T ) (R,T ) 2 ∂ dxdy. |ϕ (x) − ϕ (y)| K1,s (x, y) K1,s (x, t)dt + ∂x 2 D R we have
|D1 (x, y)| ≤ const(R)y −1/4 ,
|D2 (x, y)| ≤ const(R)y −1/4 ,
|D3 (x, y)| ≤ const(R)y −3/4 .
We also have
Z ∞ x 1/2 ˜ I1 (x) = 0 t+x
x1/2 Js+2 (x1/2 )Js+1 ((t + x)1/2 ) − 2t ! Z 1/2 x (t + x)1/2 Js+2((t + x)1/2 )Js+1 (x1/2 ) x − dt ≤ KBessel (x, t + x)dt + 2,s+1 0 2t t+x Z 1/2 ∞ x KBessel (x, t + x)dt + , 2,s+1 x t+x
46
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
where Z 1/2 x x 0 t+x
≤
x1/2 Js+2 (x1/2 )Js+1 ((t + x)1/2 ) − 2t
! (t + x)1/2 Js+2 ((t + x)1/2 )Js+1 (x1/2 ) − dt ≤ 2t 1/2 Z x x1/2 1/2 1/2 Js+1((t + x) ) − Js+1 (x ) x 1/2 Js+2 (x ) dt + 2 t+x t 0 Z x 1/2 1 (t + x)1/2 − x1/2 x 1/2 1/2 dt + Js+2 (x )Js+1 (x ) 2 t+x t 0 Z x x1/2 Js+2 ((t + x)1/2 ) − Js+2 (x1/2 ) 1/2 Js+1 (x ) dt ≤ 2 t 0
|J ′ ((x + ξ)1/2 )| x3/2 |Js+2(x1/2 )| max s+1 + const · x1/2 Js+2 (x1/2 )Js+1 (x1/2 )+ ξ∈[0,x] 2 2(x + ξ)1/2 |J ′ ((x + ξ)1/2 )| x3/2 + |Js+1(x1/2 )| max s+2 ≤ const(R), ξ∈[0,x] 2 2(x + ξ)1/2
and Z 1/2 ∞ x x t+x
x1/2 Js+2 (x1/2 )Js+1 ((t + x)1/2 ) − 2t
! (t + x)1/2 Js+2 ((t + x)1/2 )Js+1 (x1/2 ) − dt ≤ 2t Z ∞ x 1 Js+1 ((t + x)1/2 ) 1/2 dt + Js+2 (x ) 2 t (t + x)1/2 x Z ∞ 1/2 x1/2 (t + x) J ((t + x) ) s+2 1/2 + J (x ) dt s+1 ≤ 2 t t + x x const|Js+2 (x1/2 )| + const · x1/2 |Js+1(x1/2 )| ≤
≤ const(R)xs/2+1 ≤ const(R),
where we have used (2.13) to estimate both terms. Finally we have ˜ I1 (x) ≤ const(R).
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
47
We use the estimate (2.41) for x < R, y > R to obtain Z ∞ 1/2 1 J ((t + x) ) s+1 |I2 (x, y)| ≤ const · x|Js+2 (x1/2 )| dt + y−x t(t + x)1/4 (t + x)1/4 Z ∞ (t + x)1/4 J ((t + x)1/2 ) s+2 + const · x1/2 |Js+1(x1/2 )| dt ≤ 1/4 y−x t (t + x) ≤ const(R)
For 1 I3 (x, y) = 2 we have
and
Z
Z
∞
∞
xs/2+2 y −1/4 + xs/2+1 y 1/4 const(R)y 1/4 ≤ . y−x y−x
1 Js+1 (p)dp Js+1 (p)dp − 2 x1/2 x1/2 Z
Z
∞
Js+1 (p)dp x1/2
Z
Z
∞
Js+1(p)dp y 1/2
Z
∞
Js+1 (p)dp
x1/2
∞
x1/2
∞
Z
Js+1 (p)dp ≤ const(R)
∞
const(R) . y 1/4 y 1/2 x1/2 2 Thus we can use Proposition 2.5(ii) for D1 (x, y)/(x − y) × I˜1 (x) + I3 (x, y) and for Js+1 (p)dp
Js+1 (p)dp ≤
D2 (x, y)/(x − y) × I2 (x, y). For D1 (x, y)/(x − y)2 × I2 (x, y) we have obtained an estimate const(R)y −1/4 D1 (x, y) y 1/4 const(R) ≤ × I (x, y) × ≤ . 2 (x − y)2 2 (y − x) y−x (y − x)3
We split the domain into two parts: x < R < y < 2R and x < R < 2R < y. For the first part we have Z y−x x 1/2 KBessel (x, t + x)dt max ≤ const(R), 2,s+1 x 0 and any n ∈ Z, set In(λ) := [nλ − λ/2, nλ + λ/2).
The sequence (4.50)
(Xn(λ) )n∈Z := (#In(λ) )n∈Z
defines a stationary stochastic process on the probability space (Conf(R), P). Assume (λ) that P admits up to second order correlation measures. Then in particular, Xn are square integrable for any λ > 0. The number rigidity of P directly follows once there exists a sequence of positive real numbers (λk )k∈N such that λk → +∞ and for any k ∈ N, we have n oL2 (λk ) (λk ) (λk ) (λk ) (4.51) . X0 − E(X0 ) ∈ span Xn − E(Xn ) : n ∈ Z \ {0}
52
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Proposition 4.1 ([3, Theorem 3.1]). Let Z = (Zn )n∈Z be a stationary stochastic process. If X sup N |Cov(Z0 , Zn )| < ∞, (4.52) N ≥1
|n|≥N
and
X
(4.53)
Cov(Z0 , Zn ) = 0.
n∈Z
Then
n oL2 . Z0 − E(Z0 ) ∈ span Zn − E(Zn ) : n ∈ Z \ {0} P Remark 4.2. The absolute summability n∈Z |Cov(Z0 , Zn )| < ∞ is clear from (4.52).
Consider now a Pfaffian point process PK induced by a matrix kernel K such that (1.3) is satisfied. We have 0 K11 (x, x) (1) (4.54) ρPK (x) = Pf = K11 (x, x), −K11 (x, x) 0 and also
(4.55)
0 K11 (x, x) −K12 (x, y) K11 (x, y) −K11 (x, x) 0 −K22 (x, y) K21 (x, y) (2) ρPK (x, y) = Pf −K12 (y, x) K11 (y, x) 0 K11 (y, y) −K22 (y, x) K21 (y, x) −K11 (y, y) 0
= K11 (x, x)K11 (y, y) − K11 (x, y)K22 (x, y) + K12 (x, y)K21 (x, y) (1)
(1)
= ρPK (x)ρPK (y) − det K(x, y).
In what follows, we denote the truncated second correlation function (4.56)
(2,T )
(2)
(1)
(1)
ρPK (x, y) := ρPK (x, y) − ρPK (x)ρPK (y) = − det K(x, y). (λ)
Lemma 4.3. Fix λ > 0. Then for (Xn )n∈Z defined by (4.50) we have Z Z (1) (2,T ) ρ (x)dx + ρPK (x, y)dxdy, n = 0; (λ) (λ) I (λ) PK I0 ×I0 0 (λ) Cov(X0 , Xn(λ) ) = . Z (2,T ) ρPK (x, y)dxdy, n= 6 0. (λ)
(λ)
I0 ×In
(λ)
(λ)
Proof. Fix λ > 0. For brevity, in this proof, we denote Xn , In by Xn , In respectively. Let X denote a random configuration on R with distribution PK . If n 6= 0, then X X X II0 (x)IIn (y) IIn (y) = E II0 (x) E(X0 Xn ) = E = =
Z
y∈X
x∈X
(2)
ZI0 ×In I0 ×In
ρPK (x, y)dxdy = (2,T )
Z
I0 ×In
x,y∈X ,x6=y
h i (2,T ) (1) (1) ρPK (x, y) + ρPK (x)ρPK (y) dxdy
ρPK (x, y)dxdy + E(X0 )E(Xn ).
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
It follows that Cov(X0 , Xn ) = E(X0 Xn ) − E(X0 )E(Xn ) =
Z
(2,T )
I0 ×In
ρPK (x, y)dxdy.
If n = 0, then
X X E(X02 ) = E II0 (x)II0 (y) = E II0 (x) + E x,y∈X
53
x∈X
X
x,y∈X ,x6=y
II0 (x)II0 (y) .
By similar computation as above, we obtain Z Z (2,T ) (1) 2 ρPK (x, y)dxdy + E(X0 )E(X0 ) ρPK (x)dx + E(X0 ) = I0 ×I0
I0
and thus Cov(X0 , X0 ) =
Z
I0
(1) ρPK (x)dx
+
Z
(2,T )
I0 ×I0
ρPK (x, y)dxdy.
4.1. The orthogonal sine process. Proof of Proposition 1.5. It suffices to show that for any λ > 0, the hypothesis of Propo(λ) sition 4.1 is satisfied for the stochastic process (Xn )n∈Z defined in (4.50). For brevity, in (λ) (λ) this proof, we will omit the superscript λ in the notation Xn , In . Claim A1: We have
X sup N |Cov(X0 , Xn )| < ∞.
(4.57)
N ∈N
|n|≥N
(2,T )
We will use the estimate of ρsine,1 (x, y) in Forrester [6, formula (7.135)]: 1 (2,T ) ρsine,1 (x, 0) = O 2 as |x| → ∞. x In other words, there exists C > 0, such that (2,T )
|ρsine,1 (x, 0)| ≤ (2,T )
C . x2
(2,T )
By Lemma 4.3 and note that ρsine,1 (x, y) = ρsine,1 (x − y, 0), we have Z Z X X Z (2,T ) dx |ρsine,1 (x, y)|dxdy = |Cov(X0 , Xn )| ≤ |n|≥N
|n|≥N
≤ ≤
Z
sup Z Z I0 x∈I0
I0
I0
I0 ×In
Z
|y|≥N λ−λ/2
− y, 0)|dy dx |y|≥N λ−λ/2 Z (2,T ) |ρsine,1 (z, 0)|dz dx ≤ λ
|z|≥N λ−λ
We thus obtain the inequality (4.57).
(2,T )
|ρsine,1 (x−y, 0)|dy
(2,T ) |ρsine,1 (x
2C C dz = . 2 N −1 |z|≥N λ−λ z
54
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Claim B1: We have
X
(4.58)
Cov(X0 , Xn ) = 0.
n∈Z
Indeed, by (4.57), we already know that the above series converges absolutely. And by Lemma 4.3, we have Z Z X (1) (2,T ) ρsine,1 (x)dx + ρsine,1 (x, y)dxdy. Cov(X0 , Xn ) = I0 ×R
I0
n∈Z
Then by using the equality (3.47), we obtain the desired equality (4.58). The proof of Proposition 1.5 is complete.
4.2. The symplectic sine process. In what follows, we will use the following estimate (2,T ) of ρsine,4 (x, y) in Forrester [6, formula (7.94)]: 1 cos(πx) (2,T ) as |x| → ∞. +O 2 ρsine,4 (x, 0) = 8x x That is, there exists C > 0, such that cos(πx) C (2,T ) ρ (x, 0) − (4.59) sine,4 ≤ 2. 8x x Note that Z Z (2,T ) (2,T ) |ρsine,4 (x, y)|dy = |ρsine,4 (t, 0)|dt = ∞. R
R
Proof of Proposition 1.6. It suffices to show that for any positive even integer 2k ∈ 2N, (2k) the hypothesis of Proposition 4.1 is satisfied for the stochastic process (Xn )n∈Z defined in (4.50). Let us now fix the positive even integer 2k ∈ 2N. For brevity, in this proof, we (2k) (2k) will omit the superscript k in the notation Xn , In . Claim A2: We have (4.60)
X sup N |Cov(X0 , Xn )| < ∞.
N ∈N
|n|≥N
(2,T )
(2,T )
By (4.59) and Lemma 4.3 and note that ρsine,4 (x, y) = ρsine,4 (x − y, 0), we have X Z X (2,T ) ρsine,4 (x, y)dxdy |Cov(X0 , Xn )| = |n|≥N
|n|≥N
I0 ×In
X Z X Z cos(π(x − y)) 1 ≤ dxdy + C dxdy. 2 8(x − y) I0 ×In (x − y) |n|≥N | I0 ×In |n|≥N {z } denoted σn
For the term σn , we have Z Z Z Z cos(π(x − y)) d sin(π(y − x)) dx σn = dx dy = = 8(x − y) 8π(x − y) In I0 In I0 Z Z sin(π(y − x)) sin(π(y − x)) 2nk+k − dy . dx = 2 8π(x − y) y=2nk−k In 8π(x − y) I0
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
55
Note that since the function y 7→ sin(π(y − x)) is 2π-periodic, we have Z sin(π(y − x)) 2nk+k | sin(π(2nk + k − x)| Z 1 1 dy ≤ dy. = 2 2 8π(x − y) y=2nk−k 8π In (y − x) In 8π(y − x)
Therefore, we have Z Z Z 1 | sin(π(y − x))| 1 |σn | ≤ dxdy + dxdy ≤ dxdy. 2 2 2 8π(x − y) I0 ×In 8π(x − y) I0 ×In I0 ×In 4π(x − y) Consequently, X
Z 1 X 1 |Cov(X0 , Xn )| ≤ (C + ) dxdy 4π (x − y)2 I ×I n 0 |n|≥N |n|≥N Z Z 1 1 = (C + dx ) dy 2 4π I0 |y|≥N k−k/2 (x − y) Z Z 1 C + 4π 1 1 ) dz dx = . ≤ (C + 2 4π I0 N −1 |z|≥N k−k z
Claim B2: We have
X
(4.61)
Cov(X0 , Xn ) = 0.
n∈Z
Indeed, by (4.60), we already know that the above series converges absolutely. And by Lemma 4.3, we have Z XZ X (2,T ) (1) ρsine,4 (x, y)dxdy ρsine,4 (x)dx + Cov(X0 , Xn ) = I0
n∈Z
=
Z
I0
n∈Z
(1) ρsine,4 (x)dx
+
Z
I0
I0 ×In
dx
Z
R
(2,T )
ρsine,4 (x, y)dy.
Then by using the equality (3.48), we obtain the desired equality (4.61). The proof of Proposition 1.6 is complete.
4.2.1. Comments. For the symplectic sine process, let us now consider the stationary (1) stochastic process (Xn )n∈Z . We show that X (1) (4.62) |Cov(X0 , Xn(1) )| = ∞. n∈Z
If follows that the assumptions of Proposition 4.1 doesn’t hold in this case, see Remark 4.2. Indeed, by Lemma 4.3 and the inequality (4.59), for n 6= 0, we have Z (1) (2,T ) (1) ρsine,4 (x, y)dxdy ≥ (4.63) |Cov(X0 , Xn )| = (1)
(1)
I0 ×In
Z Z 1 cos(π(x − y)) dxdy − C dxdy. ≥ 2 (1) (1) (1) (1) 8(x − y) I0 ×In (x − y) I0 ×In {z } | (1)
denoted σn
56
ALEXANDER I. BUFETOV, PAVEL P. NIKITIN, AND YANQI QIU
Note that by integration by parts, we obtain Z Z sin(π(y − x)) sin(π(y − x)) n+1/2 (1) σn = dx − dy . (1) (1) 8π(x − y) y=n−1/2 8π(x − y)2 I0 In
Using the identity h i h i sin π(n + 1/2 − x) = − sin π(n − 1/2 − x) = (−1)n cos(πx), we obtain that
i (−1)n cos(πx) h sin(π(y − x)) n+1/2 1 1 = . + 8π(x − y) y=n−1/2 8π x − n − 1/2 x − n + 1/2
For any positive integer n > 0, we have Z
(1)
I0
Z 1/2 cos(πx) h i sin(π(y − x)) n+1/2 1 1 dx ≥ dx = + 8π(x − y) y=n−1/2 8π n + 1/2 − x n − 1/2 − x −1/2 Z 1/3 Z 1/3 1 1 1 1 cos(πx) · dx ≥ · dx ≥ . ≥ 8π n − 1/2 − x 16π n − 1/2 − x 48π(n − 1/2) 0 0
Therefore, |σn(1) |
Z ≥
(1)
I0
Z Z sin(π(y − x)) n+1/2 sin(π(y − x)) dx − dx dy ≥ (1) (1) 8π(x − y) y=n−1/2 8π(x − y)2 I0 In Z Z 1 1 − dy. dx ≥ 2 (1) (1) 48π(n − 1/2) I0 In 8π(x − y)
Combining the above inequality with (4.63), we get (1) |Cov(X0 , Xn(1) )|
1 48π(n − 1/2)
≥
− (C
+
1 ) 8π
Z
(1)
I0
dx
Z
(1)
In
1 dy (x − y)2
By the argument in the proof of Proposition 1.6, we know that Z XZ 1 dx dy < ∞. (1) (1) (x − y)2 I I n 0 n∈Z Therefore, we get the claimed divergence X n∈Z
(1) |Cov(X0 , Xn(1) )|
≥
X n≥1
Z Z 1 1 X 1 −(C+ ) dy = ∞. dx 2 (1) 48π(n − 1/2) 8π n≥1 I0(1) In (x − y)
Acknowledgements. We are deeply grateful to Alexei Klimenko for useful discussions. The research of A. Bufetov on this project has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme under grant agreement No 647133 (ICHAOS). Yanqi Qiu’s research is supported by the National Natural Science Foundation of China, grants NSFC Y7116335K1, NSFC 11801547 and NSFC 11688101. The research of P. Nikitin is supported by the RFBR grant 17-01-00433.
ON NUMBER RIGIDITY FOR PFAFFIAN POINT PROCESSES
57
References [1] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Department of Commerce of the United States of America, Tenth Printing, 1972. [2] Alexander I. Bufetov, Rigidity of determinantal point processes with the Airy, the Bessel and the Gamma kernel, Bull. Math. Sci., 6:1 (2016), 163172. [3] Alexander I. Bufetov, Yoann Dabrowski, and Yanqi Qiu. Linear rigidity of stationary stochastic processes. Ergodic Theory Dynam. Systems, 38 (2018), no.7, 2493–2507. [4] Alexander I. Bufetov and Yanqi Qiu. J-Hermitian determinantal point processes: balanced rigidity and balanced Palm equivalence. Math. Ann., 371 (2018), no. 1-2, 127–188. [5] Reda Chhaibi, Joseph Najnudel, Rigidity of the Sineβ process. arXiv:1804.01216. [6] Peter J. Forrester. Log-gases and random matrices, volume 34 of London Mathematical Society Monographs Series. Princeton University Press, Princeton, NJ, 2010. [7] Subhroshekhar Ghosh. Determinantal processes and completeness of random exponentials: the critical case. Probab. Theory Related Fields, pages 1–23, 2014. [8] Subhroshekhar Ghosh and Yuval Peres. Rigidity and tolerance in point processes: Gaussian zeros and Ginibre eigenvalues. Duke Math. J. 166 (2017), no.10, 1789–1858. [9] C. A. Tracy and H. Widom. Level spacing distributions and the Bessel kernel. Comm. Math. Phys. 161, no. 2 (1994), 289309. Alexander I. BUFETOV: Aix-Marseille Universit´ e, Centrale Marseille, CNRS, Institut de Math´ ematiques de Marseille, UMR7373, 39 Rue F. Joliot Curie 13453, Marseille, France; Steklov Mathematical Institute of RAS, Moscow, Russia E-mail address:
[email protected],
[email protected] Pavel P. Nikitin: St. Petersburg Department of V.A.Steklov Institute of Mathematics of the Russian Academy of Sciences, 27 Fontanka, 191023, St. Petersburg, Russia; St. Petersburg State University, St. Petersburg, Russia E-mail address:
[email protected] Yanqi QIU: Institute of Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China E-mail address:
[email protected] 
