Introduction
The quantum unique ergodicity problem asks whether high-energy Laplace eigenfunctions become equidistributed on the underlying surface. For the modular surface X = Γ \ H X=\Gamma\backslash\mathbb{H} X = Γ\ H H = { x + i y : y > 0 } \mathbb{H}=\{x+iy:y>0\} H = { x + i y : y > 0 } Γ = PSL 2 ( Z ) \Gamma=\operatorname{PSL}_{2}(\mathbb{Z}) Γ = PSL 2 ( Z ) L 2 L^2 L 2 [10 ] and Soundararajan [13 ] . In particular, for every ψ ∈ C c ∞ ( X ) \psi\in C^\infty_c(X) ψ ∈ C c ∞ ( X ) t j → ∞ t_j\to\infty t j → ∞
∫ X ψ ( z ) ∣ u j ( z ) ∣ 2 d μ ( z ) ⟶ 1 vol ( X ) ∫ X ψ ( z ) d μ ( z ) , \int_X \psi(z)|u_j(z)|^2\,d\mu(z)\longrightarrow\frac{1}{\operatorname{vol}(X)}\int_X\psi(z)\,d\mu(z), ∫ X ψ ( z ) ∣ u j ( z ) ∣ 2 d μ ( z ) ⟶ vol ( X ) 1 ∫ X ψ ( z ) d μ ( z ) , where the hyperbolic area measure d μ ( z ) d\mu(z) d μ ( z ) d μ ( z ) = d x d y / y 2 d\mu(z)=dx\,dy/y^2 d μ ( z ) = d x d y / y 2 vol ( X ) = π / 3 \operatorname{vol}(X)=\pi/3 vol ( X ) = π /3
A natural question is whether the restriction mass of these eigenfunctions also becomes equidistributed along a fixed geodesic. Since the usual QUE result concerns integration against smooth functions on the surface, it does not directly give information about the restriction mass along a geodesic. Young formulated the following conjecture for the vertical geodesic [16 Conjecture 1.1] .
Conjecture 1.1. Let u j u_j u j L 2 L^2 L 2 X X X 1 / 4 + t j 2 1/4+t_j^2 1/4 + t j 2 ψ ∈ C c ∞ ( 0 , ∞ ) \psi\in C^\infty_c(0,\infty) ψ ∈ C c ∞ ( 0 , ∞ )
lim t j → ∞ u j e v e n ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 d y y = 6 π ∫ 0 ∞ ψ ( y ) d y y . (1) \lim_{\substack{t_j\to\infty\\u_j\ \mathrm{even}}}\int_0^\infty\psi(y)|u_j(iy)|^2\frac{dy}{y}=\frac{6}{\pi}\int_0^\infty\psi(y)\frac{dy}{y}. \tag*{(1)} t j → ∞ u j even lim ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 y d y = π 6 ∫ 0 ∞ ψ ( y ) y d y . (1) One way to study this conjecture is through quantum variance. In the holomorphic setting, Zenz [17 ] obtained an asymptotic formula for the quantum variance along the vertical geodesic, averaged over Hecke cusp forms of even weights k ≍ K k \asymp K k ≍ K K 3 / 2 log K K^{3/2}\log K K 3/2 log K [14 ] extended this result to short weight intervals of length G = K 9 / 10 G=K^{9/10} G = K 9/10 K 1 / 2 G log K K^{1/2}G\log K K 1/2 G log K
Let { u j } \{u_j\} { u j } X X X 1 / 4 + t j 2 1/4+t_j^2 1/4 + t j 2 [7 ] )
u j ( x + i y ) = y ∑ n ≠ 0 ρ j ( n ) K i t j ( 2 π ∣ n ∣ y ) e ( n x ) , u_j(x+iy)=\sqrt{y}\sum_{n\ne0}\rho_j(n)K_{it_j}(2\pi|n|y)e(nx), u j ( x + i y ) = y n = 0 ∑ ρ j ( n ) K i t j ( 2 π ∣ n ∣ y ) e ( n x ) , where, as usual, e ( x ) = e 2 π i x e(x)=e^{2\pi ix} e ( x ) = e 2 π i x ρ j ( ± n ) = ρ j ( 1 ) λ j ( n ) \rho_j(\pm n)=\rho_j(1)\lambda_j(n) ρ j ( ± n ) = ρ j ( 1 ) λ j ( n ) n ≥ 1 n\ge1 n ≥ 1 λ j ( n ) \lambda_j(n) λ j ( n )
λ j ( m ) λ j ( n ) = ∑ d ∣ ( m , n ) λ j ( m n / d 2 ) . \lambda_j(m)\lambda_j(n)=\sum_{d\mid(m,n)}\lambda_j(mn/d^2). λ j ( m ) λ j ( n ) = d ∣ ( m , n ) ∑ λ j ( mn / d 2 ) . The L 2 L^2 L 2
∣ ρ j ( 1 ) ∣ 2 cosh ( π t j ) = 2 L ( 1 , sym 2 u j ) . \frac{|\rho_j(1)|^2}{\cosh(\pi t_j)}=\frac{2}{L(1,\operatorname{sym}^2u_j)}. cosh ( π t j ) ∣ ρ j ( 1 ) ∣ 2 = L ( 1 , sym 2 u j ) 2 . Our main result is the following.
Theorem 1.2. For T ≥ 2 T\ge2 T ≥ 2 1 ≤ U ≤ T 1\le U\le T 1 ≤ U ≤ T ψ ∈ C c ∞ ( 0 , ∞ ) \psi\in C^\infty_c(0,\infty) ψ ∈ C c ∞ ( 0 , ∞ ) ε > 0 \varepsilon>0 ε > 0
∑ u j e v e n T < t j ≤ T + U L ( 1 , sym 2 u j ) ∣ ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 d y y − 6 π ∫ 0 ∞ ψ ( y ) d y y ∣ 2 ≪ ψ , ε T 1 + ε . (2) \sum_{\substack{u_j\ \mathrm{even}\\ T<t_j\le T+U}}L(1,\operatorname{sym}^2u_j)\left|\int_0^\infty\psi(y)|u_j(iy)|^2\frac{dy}{y}-\frac{6}{\pi}\int_0^\infty\psi(y)\frac{dy}{y}\right|^2\ll_{\psi,\varepsilon}T^{1+\varepsilon}. \tag*{(2)} u j even T < t j ≤ T + U ∑ L ( 1 , sym 2 u j ) ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 y d y − π 6 ∫ 0 ∞ ψ ( y ) y d y 2 ≪ ψ , ε T 1 + ε . (2) Here the implied constant depends only on ψ \psi ψ ε \varepsilon ε 1 ≤ U ≤ T 1\le U\le T 1 ≤ U ≤ T
For U ≥ T θ U\ge T^\theta U ≥ T θ θ > 0 \theta>0 θ > 0 Lemma A.1 ) gives
# { j e v e n : T < t j ≤ T + U } ≍ T U . \#\{j\ \mathrm{even}:T<t_j\le T+U\}\asymp TU. # { j even : T < t j ≤ T + U } ≍ T U . Thus the upper bound in (2) implies the average order of the weighted squares L ( 1 , sym 2 u j ) ∣ μ j ( ψ ) − M ( ψ ) ∣ 2 L(1,\operatorname{sym}^2u_j)|\mu_j(\psi)-M(\psi)|^2 L ( 1 , sym 2 u j ) ∣ μ j ( ψ ) − M ( ψ ) ∣ 2 O ψ , ε ( T ε / U ) O_{\psi,\varepsilon}(T^\varepsilon/U) O ψ , ε ( T ε / U ) k ≍ K k\asymp K k ≍ K [17 ] and the authors [14 ] give an average of order K − 1 / 2 log K K^{-1/2}\log K K − 1/2 log K U = T θ U=T^\theta U = T θ θ > 1 / 2 \theta>1/2 θ > 1/2 (2) gives T − θ + ε T^{-\theta+\varepsilon} T − θ + ε
As a direct consequence of Theorem 1.2 and the even-spectrum Weyl law in Lemma A.1 , we obtain density-one restricted equidistribution in short spectral intervals.
Corollary 1.3. Fix a real-valued ψ ∈ C c ∞ ( 0 , ∞ ) \psi\in C^\infty_c(0,\infty) ψ ∈ C c ∞ ( 0 , ∞ ) 0 < θ ≤ 1 0<\theta\le1 0 < θ ≤ 1 0 < η < θ / 2 0<\eta<\theta/2 0 < η < θ /2 T θ ≤ U ≤ T T^\theta\le U\le T T θ ≤ U ≤ T ε > 0 \varepsilon>0 ε > 0
O ψ , θ , η , ε ( T 1 + 2 η + ε ) (3) O_{\psi,\theta,\eta,\varepsilon}(T^{1+2\eta+\varepsilon}) \tag*{(3)} O ψ , θ , η , ε ( T 1 + 2 η + ε ) (3) even forms with T < t j ≤ T + U T<t_j\le T+U T < t j ≤ T + U
∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 d y y = 6 π ∫ 0 ∞ ψ ( y ) d y y + O ( T − η ) . \int_0^\infty\psi(y)|u_j(iy)|^2\frac{dy}{y}=\frac{6}{\pi}\int_0^\infty\psi(y)\frac{dy}{y}+O(T^{-\eta}). ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 y d y = π 6 ∫ 0 ∞ ψ ( y ) y d y + O ( T − η ) . In particular, taking 0 < ε < θ − 2 η 0<\varepsilon<\theta-2\eta 0 < ε < θ − 2 η
Proof. Fix ε > 0 \varepsilon> 0 ε > 0 E \mathcal{E} E L ( 1 , sym 2 u j ) ≫ ε T − ε / 2 L(1,\operatorname{sym}^{2}u_j) \gg_{\varepsilon} T^{-\varepsilon/2} L ( 1 , sym 2 u j ) ≫ ε T − ε /2 Theorem 1.2 , we obtain
# E ≪ ψ , ε T 1 + 2 η + ε . \#\mathcal{E} \ll_{\psi,\varepsilon} T^{1+2\eta+\varepsilon}. # E ≪ ψ , ε T 1 + 2 η + ε . By Lemma A.1 in Appendix A, the even-spectrum count satisfies
# { j e v e n : T < t j ≤ T + U } = 2 T U + U 2 24 + O ( T log T ) ≍ θ T U \#\{j\ \mathrm{even}: T < t_j \leq T+U\} = \frac{2TU+U^2}{24}+O(T\log T)\asymp_{\theta} TU # { j even : T < t j ≤ T + U } = 24 2 T U + U 2 + O ( T log T ) ≍ θ T U uniformly for T θ ≤ U ≤ T T^{\theta} \leq U \leq T T θ ≤ U ≤ T T → ∞ T \to\infty T → ∞ 0 < ε < θ − 2 η 0 < \varepsilon< \theta- 2\eta 0 < ε < θ − 2 η
O ψ , θ , η , ε ( T 2 η + ε − θ ) = o ( 1 ) . O_{\psi,\theta,\eta,\varepsilon}(T^{2\eta+\varepsilon-\theta})=o(1). O ψ , θ , η , ε ( T 2 η + ε − θ ) = o ( 1 ) . □ \square □
Outline of the proof of Theorem 1.2 . We substitute the Fourier–Whittaker expansion of the Maass forms into the restriction integral and separate the resulting sum into diagonal and off-diagonal parts. The diagonal part has a direct interpretation in terms of an incomplete Eisenstein series. For h ( y ) = y ψ ( y ) h(y)=y\psi(y) h ( y ) = y ψ ( y )
E ( z ∣ h ) = ∑ γ ∈ Γ ∞ \ Γ h ( Im γ z ) . E(z\mid h)=\sum_{\gamma\in\Gamma_{\infty}\backslash\Gamma}h(\operatorname{Im}\gamma z). E ( z ∣ h ) = γ ∈ Γ ∞ \Γ ∑ h ( Im γ z ) . Unfolding shows that the diagonal part is exactly
2 ⟨ ∣ u j ∣ 2 , E ( ⋅ ∣ y ψ ) ⟩ X . 2\langle|u_j|^2,E(\cdot\mid y\psi)\rangle_X. 2 ⟨ ∣ u j ∣ 2 , E ( ⋅ ∣ y ψ ) ⟩ X . Since vol ( X ) = π / 3 \operatorname{vol}(X)=\pi/3 vol ( X ) = π /3
2 vol ( X ) ⟨ E ( ⋅ ∣ y ψ ) , 1 ⟩ X = 6 π ∫ 0 ∞ ψ ( y ) d y y . \frac{2}{\operatorname{vol}(X)}\langle E(\cdot\mid y\psi),1\rangle_X=\frac{6}{\pi}\int_0^\infty\psi(y)\frac{dy}{y}. vol ( X ) 2 ⟨ E ( ⋅ ∣ y ψ ) , 1 ⟩ X = π 6 ∫ 0 ∞ ψ ( y ) y d y . Thus the constant part of the incomplete Eisenstein series gives the predicted mass. The diagonal fluctuations are controlled by the quantum variance estimate of Luo and Sarnak [12 Proposition 3.1] .
The off-diagonal terms involve the normalised Bessel product integral
B m , n ( t ) = 16 cosh ( π t ) ∫ 0 ∞ ψ ( y ) K i t ( 2 π m y ) K i t ( 2 π n y ) d y , m ≠ n . B_{m,n}(t)=16\cosh(\pi t)\int_0^\infty\psi(y)K_{it}(2\pi my)K_{it}(2\pi ny)\,dy,\qquad m\ne n. B m , n ( t ) = 16 cosh ( π t ) ∫ 0 ∞ ψ ( y ) K i t ( 2 π m y ) K i t ( 2 π n y ) d y , m = n . In the holomorphic case, the restriction integral involves exponential functions, which are relatively easy to integrate. For Maass forms, the corresponding integrals involve K K K K K K § 2 and § 3 , we show that the off-diagonal sum can be restricted to
1 ≤ m < n ≪ ψ T , n 2 − m 2 ≪ T 1 + ε , 1\leq m<n\ll_{\psi}T,\qquad n^2-m^2\ll T^{1+\varepsilon}, 1 ≤ m < n ≪ ψ T , n 2 − m 2 ≪ T 1 + ε , with a negligible remainder in the off-diagonal contribution.
Applying the Hecke relation and making the change of variables m = d a m=da m = d a n = d ( a + b ) n=d(a+b) n = d ( a + b )
∑ j e v e n T < t j ≤ T + U 1 L ( 1 , sym 2 u j ) ∣ ∑ d , a , b ≥ 1 λ j ( a ( a + b ) ) B d a , d ( a + b ) ( t j ) ∣ 2 . \sum_{\substack{j\ \mathrm{even}\\ T<t_j\leq T+U}} \frac{1}{L(1,\operatorname{sym}^{2}u_j)}\left|\sum_{d,a,b\geq1}\lambda_j\left(a(a+b)\right)\mathcal{B}_{da,d(a+b)}(t_j)\right|^{2}. j even T < t j ≤ T + U ∑ L ( 1 , sym 2 u j ) 1 d , a , b ≥ 1 ∑ λ j ( a ( a + b ) ) B d a , d ( a + b ) ( t j ) 2 . The coefficients in the inner sum still depend on t j t_j t j I I I t I t_I t I
B m , n ( t ) ⇝ ∑ ∣ k ∣ ≤ K c m , n , I ( k ) e ( k ( t − t I ) ∣ I ∣ ) , t ∈ I . \mathcal{B}_{m,n}(t)\leadsto\sum_{|k|\leq K}c_{m,n,I}(k)e\left(\frac{k(t-t_I)}{|I|}\right),\qquad t\in I. B m , n ( t ) ⇝ ∣ k ∣ ≤ K ∑ c m , n , I ( k ) e ( ∣ I ∣ k ( t − t I ) ) , t ∈ I . For each Fourier frequency, the coefficients on the right are independent of the varying spectral parameter. We may therefore apply Jutila’s spectral large sieve on each short interval. Combining the resulting estimates over the dyadic ranges proves the off-diagonal bound O ψ , ε ( T 1 + ε ) O_{\psi,\varepsilon}(T^{1+\varepsilon}) O ψ , ε ( T 1 + ε )
Together with the diagonal variance estimate, this proves Theorem 1.2 .
Organisation of the paper. Section 2 records the spectral normalisations and the large-sieve input. Section 3 proves the arithmetic truncation, and Section 4 develops the Bessel analysis and the spectral derivative estimates. Section 5 gives the Fourier localisation needed for the local spectral large sieve. Section 6 establishes the dyadic coefficient estimates and completes the proof of the main theorem. Appendix A records the parity-specific Weyl law used in Corollary 1.3.
Preliminaries
Throughout the paper, write
μ j ( ψ ) = ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 d y y , M ( ψ ) = 6 π ∫ 0 ∞ ψ ( y ) d y y , L j = L ( 1 , sym 2 u j ) . \mu_j(\psi)=\int_0^\infty\psi(y)|u_j(iy)|^2\frac{dy}{y},\qquad M(\psi)=\frac{6}{\pi}\int_0^\infty\psi(y)\frac{dy}{y},\qquad L_j=L(1,\operatorname{sym}^{2}u_j). μ j ( ψ ) = ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 y d y , M ( ψ ) = π 6 ∫ 0 ∞ ψ ( y ) y d y , L j = L ( 1 , sym 2 u j ) . where u j u_j u j L 2 L^2 L 2 [6 ] and Li [9 ] ,
t j − ε ≪ ε L j ≪ ε t j ε . t_j^{-\varepsilon}\ll_{\varepsilon} L_j\ll_{\varepsilon} t_j^{\varepsilon}. t j − ε ≪ ε L j ≪ ε t j ε . Recall the Fourier–Whittaker expansion
u j ( z ) = ∑ n ≠ 0 ρ j ( n ) y K i t j ( 2 π ∣ n ∣ y ) e ( n x ) , ρ j ( ± n ) = ρ j ( 1 ) λ j ( n ) ( n ≥ 1 ) . (4) u_j(z)=\sum_{n\ne0}\rho_j(n)\sqrt{y}K_{it_j}(2\pi|n|y)e(nx),\qquad\rho_j(\pm n)=\rho_j(1)\lambda_j(n)\quad(n\geq1). \tag*{(4)} u j ( z ) = n = 0 ∑ ρ j ( n ) y K i t j ( 2 π ∣ n ∣ y ) e ( n x ) , ρ j ( ± n ) = ρ j ( 1 ) λ j ( n ) ( n ≥ 1 ) . (4) where ∥ u j ∥ 2 = 1 \|u_j\|_2=1 ∥ u j ∥ 2 = 1
∣ ρ j ( 1 ) ∣ 2 cosh ( π t j ) = 2 L ( 1 , sym 2 u j ) . \frac{|\rho_j(1)|^2}{\cosh(\pi t_j)}=\frac{2}{L(1,\operatorname{sym}^{2}u_j)}. cosh ( π t j ) ∣ ρ j ( 1 ) ∣ 2 = L ( 1 , sym 2 u j ) 2 . The restriction integral μ j ( ψ ) \mu_j(\psi) μ j ( ψ )
For positive integers m , n m,n m , n
B m , n ( t ) = 16 cosh ( π t ) ∫ 0 ∞ ψ ( y ) K i t ( 2 π m y ) K i t ( 2 π n y ) d y . (5) \mathcal{B}_{m,n}(t)=16\cosh(\pi t)\int_0^\infty\psi(y)K_{it}(2\pi my)K_{it}(2\pi ny)\,dy. \tag*{(5)} B m , n ( t ) = 16 cosh ( π t ) ∫ 0 ∞ ψ ( y ) K i t ( 2 π m y ) K i t ( 2 π n y ) d y . (5) Substituting the Fourier expansion into the restriction integral gives
∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 d y y = 4 ∣ ρ j ( 1 ) ∣ 2 ∑ m , n ≥ 1 λ j ( m ) λ j ( n ) ∫ 0 ∞ ψ ( y ) K i t j ( 2 π m y ) K i t j ( 2 π n y ) d y = 1 2 L j ∑ m , n ≥ 1 λ j ( m ) λ j ( n ) B m , n ( t j ) . (6) \begin{aligned}
\int_{0}^{\infty} \psi(y)|u_j(iy)|^2 \frac{dy}{y}
=4|\rho_j(1)|^2\sum_{m,n\geq1}\lambda_j(m)\lambda_j(n)\int_{0}^{\infty}\psi(y)K_{it_j}(2\pi my)K_{it_j}(2\pi ny)\,dy \\
&=\frac{1}{2L_j}\sum_{m,n\geq1}\lambda_j(m)\lambda_j(n)B_{m,n}(t_j). \tag*{(6)}
\end{aligned} ∫ 0 ∞ ψ ( y ) ∣ u j ( i y ) ∣ 2 y d y = 4∣ ρ j ( 1 ) ∣ 2 m , n ≥ 1 ∑ λ j ( m ) λ j ( n ) ∫ 0 ∞ ψ ( y ) K i t j ( 2 π m y ) K i t j ( 2 π n y ) d y = 2 L j 1 m , n ≥ 1 ∑ λ j ( m ) λ j ( n ) B m , n ( t j ) . (6) Write μ j = D j + S j \mu_j=D_j+S_j μ j = D j + S j D j D_j D j m = n m=n m = n
Let
E ( z ∣ h ) = ∑ γ ∈ Γ ∞ \ Γ h ( ℑ γ z ) , h ( y ) = y ψ ( y ) . (7) E(z\mid h)=\sum_{\gamma\in\Gamma_\infty\backslash\Gamma}h(\Im\gamma z),\qquad h(y)=y\psi(y). \tag*{(7)} E ( z ∣ h ) = γ ∈ Γ ∞ \Γ ∑ h ( ℑ γ z ) , h ( y ) = y ψ ( y ) . (7) Unfolding (7) , using both signs in (4) , gives the exact identity
D j ( ψ ) = 2 ⟨ ∣ u j ∣ 2 , E ( ⋅ ∣ y ψ ) ⟩ X . (8) D_j(\psi)=2\langle|u_j|^2,E(\cdot\mid y\psi)\rangle_X. \tag*{(8)} D j ( ψ ) = 2 ⟨ ∣ u j ∣ 2 , E ( ⋅ ∣ y ψ ) ⟩ X . (8) Moreover,
⟨ E ( ⋅ ∣ y ψ ) , 1 ⟩ X = ∫ 0 ∞ ψ ( y ) d y y . \langle E(\cdot\mid y\psi),1\rangle_X=\int_{0}^{\infty}\psi(y)\frac{dy}{y}. ⟨ E ( ⋅ ∣ y ψ ) , 1 ⟩ X = ∫ 0 ∞ ψ ( y ) y d y . The mean of the incomplete Eisenstein series therefore determines the natural centering constant
M ( ψ ) = 2 vol ( X ) ∫ 0 ∞ ψ ( y ) d y y = 6 π ∫ 0 ∞ ψ ( y ) d y y . (9) M(\psi)=\frac{2}{\operatorname{vol}(X)}\int_{0}^{\infty}\psi(y)\frac{dy}{y}
=\frac{6}{\pi}\int_{0}^{\infty}\psi(y)\frac{dy}{y}. \tag*{(9)} M ( ψ ) = vol ( X ) 2 ∫ 0 ∞ ψ ( y ) y d y = π 6 ∫ 0 ∞ ψ ( y ) y d y . (9) The diagonal variance estimate
The diagonal contribution satisfies the following estimate.
Lemma 2.1. Let ψ ∈ C c ∞ ( 0 , ∞ ) \psi\in C^\infty_c(0,\infty) ψ ∈ C c ∞ ( 0 , ∞ ) T ≥ 2 T\geq2 T ≥ 2 1 ≤ U ≤ T 1\leq U\leq T 1 ≤ U ≤ T
∑ j e v e n T < t j ≤ T + U L j ∣ D j ( ψ ) − M ( ψ ) ∣ 2 ≪ ψ , ε T 1 + ε . (10) \sum_{\substack{j\ \mathrm{even}\\ T<t_j\leq T+U}}L_j|D_j(\psi)-M(\psi)|^2\ll_{\psi,\varepsilon}T^{1+\varepsilon}. \tag*{(10)} j even T < t j ≤ T + U ∑ L j ∣ D j ( ψ ) − M ( ψ ) ∣ 2 ≪ ψ , ε T 1 + ε . (10) Proof. By (8) and (9) ,
D j ( ψ ) − M ( ψ ) = 2 ( ⟨ ∣ u j ∣ 2 , E ( ⋅ ∣ y ψ ) ⟩ X − ⟨ E ( ⋅ ∣ y ψ ) , 1 ⟩ X vol ( X ) ) . D_j(\psi)-M(\psi)
=2\left(\langle|u_j|^2,E(\cdot\mid y\psi)\rangle_X-\frac{\langle E(\cdot\mid y\psi),1\rangle_X}{\operatorname{vol}(X)}\right). D j ( ψ ) − M ( ψ ) = 2 ( ⟨ ∣ u j ∣ 2 , E ( ⋅ ∣ y ψ ) ⟩ X − vol ( X ) ⟨ E ( ⋅ ∣ y ψ ) , 1 ⟩ X ) . Apply [12 Proposition 3.1] to the incomplete Eisenstein series E ( z ∣ h ) E(z\mid h) E ( z ∣ h ) (7) , with h ( y ) = y ψ ( y ) h(y)=y\psi(y) h ( y ) = y ψ ( y ) P h , 0 P_{h,0} P h , 0 E ( z ∣ h ) E(z\mid h) E ( z ∣ h ) κ j \kappa_j κ j t j t_j t j h h h ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) ψ \psi ψ
∑ t j ≤ 2 T ∣ D j ( ψ ) − M ( ψ ) ∣ 2 ≪ ψ , ε T 1 + ε . \sum_{t_j\leq2T}|D_j(\psi)-M(\psi)|^2\ll_{\psi,\varepsilon}T^{1+\varepsilon}. t j ≤ 2 T ∑ ∣ D j ( ψ ) − M ( ψ ) ∣ 2 ≪ ψ , ε T 1 + ε . The asserted bound follows by restricting to the even forms in T < t j ≤ T + U ≤ 2 T T<t_j\leq T+U\leq2T T < t j ≤ T + U ≤ 2 T L j ≪ ε T ε L_j\ll_\varepsilon T^\varepsilon L j ≪ ε T ε ε \varepsilon ε
Using the symmetry in m , n m,n m , n
λ j ( m ) λ j ( n ) = ∑ d ∣ ( m , n ) λ j ( m n d 2 ) (11) \lambda_j(m)\lambda_j(n)=\sum_{d\mid(m,n)}\lambda_j\left(\frac{mn}{d^2}\right) \tag*{(11)} λ j ( m ) λ j ( n ) = d ∣ ( m , n ) ∑ λ j ( d 2 mn ) (11) and making the substitution m = d a m = da m = d a n = d ( a + b ) n = d(a+b) n = d ( a + b )
S j ( ψ ) = 1 L j ∑ d , a , b ≥ 1 λ j ( a ( a + b ) ) B d a , d ( a + b ) ( t j ) , (12) S_j(\psi) = \frac{1}{L_j}\sum_{d,a,b\geq1}\lambda_j\bigl(a(a+b)\bigr)\mathcal{B}_{da,d(a+b)}(t_j), \tag*{(12)} S j ( ψ ) = L j 1 d , a , b ≥ 1 ∑ λ j ( a ( a + b ) ) B d a , d ( a + b ) ( t j ) , (12) Truncation of the off-diagonal sum
We next make a preliminary truncation of the off-diagonal sum. Write supp ψ ⊂ [ y 0 , y 1 ] \operatorname{supp}\psi\subset[y_0,y_1] supp ψ ⊂ [ y 0 , y 1 ] 0 < y 0 < y 1 0<y_0<y_1 0 < y 0 < y 1 E j f a r \mathcal{E}^{\mathrm{far}}_j E j far S j ( ψ ) S_j(\psi) S j ( ψ ) max { m , n } ≫ ψ T \max\{m,n\}\gg_\psi T max { m , n } ≫ ψ T
Lemma 2.2. For every A > 0 A>0 A > 0 T < t j ≤ T + U ≤ 2 T T<t_j\leq T+U\leq2T T < t j ≤ T + U ≤ 2 T
E j f a r ≪ A , ψ T − A . \mathcal{E}^{\mathrm{far}}_j\ll_{A,\psi}T^{-A}. E j far ≪ A , ψ T − A . Consequently, the off-diagonal sum can be restricted to
m , n ≪ ψ T . m,n\ll_\psi T. m , n ≪ ψ T . Proof. The integral representation (see [4 8.432.1] )
K i t ( x ) = ∫ 0 ∞ e − x cosh u cos ( t u ) d u K_{it}(x)=\int_0^\infty e^{-x\cosh u}\cos(tu)\,du K i t ( x ) = ∫ 0 ∞ e − x c o s h u cos ( t u ) d u and the inequality cosh u ≥ 1 + u 2 / 2 \cosh u\geq1+u^2/2 cosh u ≥ 1 + u 2 /2
∣ K i t ( x ) ∣ ≤ K 0 ( x ) ≤ e − x ∫ 0 ∞ e − x u 2 / 2 d u = π 2 x e − x . |K_{it}(x)|\leq K_0(x)\leq e^{-x}\int_0^\infty e^{-xu^2/2}\,du=\sqrt{\frac{\pi}{2x}}e^{-x}. ∣ K i t ( x ) ∣ ≤ K 0 ( x ) ≤ e − x ∫ 0 ∞ e − x u 2 /2 d u = 2 x π e − x . Since y ≥ y 0 y\geq y_0 y ≥ y 0 supp ψ \operatorname{supp}\psi supp ψ t j ≤ 2 T t_j\leq2T t j ≤ 2 T n ≫ ψ T n\gg_\psi T n ≫ ψ T
e π t j / 2 ∣ K i t j ( 2 π n y ) ∣ ≪ ψ n − 1 / 2 exp { π T − 2 π y 0 n } ≪ A , ψ T − A n − A . e^{\pi t_j/2}|K_{it_j}(2\pi n y)|\ll_\psi n^{-1/2}\exp\{\pi T-2\pi y_0n\}\ll_{A,\psi}T^{-A}n^{-A}. e π t j /2 ∣ K i t j ( 2 π n y ) ∣ ≪ ψ n − 1/2 exp { π T − 2 π y 0 n } ≪ A , ψ T − A n − A . The same estimate holds with m m m n n n (6) , followed by absolute summation, proves the assertion. □
Fix a sufficiently large constant C ψ C_\psi C ψ m , n ≤ C ψ T m,n\leq C_\psi T m , n ≤ C ψ T
The spectral large sieve
We record Jutila’s spectral large-sieve inequality [8 Theorem 1.1] in the form needed for the off-diagonal contribution.
Lemma 2.3. Let T ≥ 2 T\geq2 T ≥ 2 1 ≤ H ≤ T 1\leq H\leq T 1 ≤ H ≤ T ( a n ) (a_n) ( a n ) n ≤ N n\leq N n ≤ N
∑ T ≤ t j ≤ T + H 1 L j ∣ ∑ n ≤ N a n λ j ( n ) ∣ 2 ≪ ε ( T H + N ) ( T N ) ε ∑ n ≤ N ∣ a n ∣ 2 . \sum_{T\leq t_j\leq T+H}\frac{1}{L_j}\left|\sum_{n\leq N}a_n\lambda_j(n)\right|^2\ll_\varepsilon(TH+N)(TN)^\varepsilon\sum_{n\leq N}|a_n|^2. T ≤ t j ≤ T + H ∑ L j 1 n ≤ N ∑ a n λ j ( n ) 2 ≪ ε ( T H + N ) ( T N ) ε n ≤ N ∑ ∣ a n ∣ 2 . The sum may be restricted to any subfamily, in particular to the even forms.
Note that the truncation m , n ≪ ψ T m,n\ll_\psi T m , n ≪ ψ T T ≤ t ≤ 2 T T\leq t\leq2T T ≤ t ≤ 2 T
d ≍ D , a ≍ X , b ≍ Y , N : = X ( X + Y ) , d\asymp D,\qquad a\asymp X,\qquad b\asymp Y,\qquad N:=X(X+Y), d ≍ D , a ≍ X , b ≍ Y , N := X ( X + Y ) , for which, uniformly on supp ψ \operatorname{supp}\psi supp ψ
T / 4 ≤ 2 π d a y ≤ 2 π d ( a + b ) y ≤ T / 2. T/4\leq2\pi day\leq2\pi d(a+b)y\leq T/2. T /4 ≤ 2 π d a y ≤ 2 π d ( a + b ) y ≤ T /2. Thus both arguments lie in the oscillatory range at a fixed proportional distance from the turning point. The standard oscillatory asymptotic for the K K K [3 Lemma 3.1] then gives
e π t / 2 ∣ K i t ( 2 π d a y ) ∣ + e π t / 2 ∣ K i t ( 2 π d ( a + b ) y ) ∣ ≪ ψ T − 1 / 2 . e^{\pi t/2}|K_{it}(2\pi day)|+e^{\pi t/2}|K_{it}(2\pi d(a+b)y)|\ll_{\psi} T^{-1/2}. e π t /2 ∣ K i t ( 2 π d a y ) ∣ + e π t /2 ∣ K i t ( 2 π d ( a + b ) y ) ∣ ≪ ψ T − 1/2 . Hence, by (5) ,
B d a , d ( a + b ) ( t ) ≪ ψ T − 1 . \mathcal{B}_{da,d(a+b)}(t)\ll_{\psi} T^{-1}. B d a , d ( a + b ) ( t ) ≪ ψ T − 1 . For each A A A a ∣ A a\mid A a ∣ A b b b a ( a + b ) = A a(a+b)=A a ( a + b ) = A O ( D ) O(D) O ( D ) d d d
∑ A ≍ X ( X + Y ) d ≍ D , a ≍ X , b ≍ Y a ( a + b ) = A ∣ ∑ d ≍ D , a ≍ X , b ≍ Y B d a , d ( a + b ) ( t ) ∣ 2 ≪ ε , ψ D T ε ∑ d ≍ D , a ≍ X , b ≍ Y ∣ B d a , d ( a + b ) ( t ) ∣ 2 ≪ ε , ψ T ε D 2 X Y T 2 . \sum_{\substack{A\asymp X(X+Y)\\d\asymp D,\ a\asymp X,\ b\asymp Y\\a(a+b)=A}}
\left|\sum_{\substack{d\asymp D,\ a\asymp X,\ b\asymp Y}}
\mathcal{B}_{da,d(a+b)}(t)\right|^2
\ll_{\varepsilon,\psi} DT^\varepsilon
\sum_{\substack{d\asymp D,\ a\asymp X,\ b\asymp Y}}
|\mathcal{B}_{da,d(a+b)}(t)|^2
\ll_{\varepsilon,\psi} T^\varepsilon\frac{D^2XY}{T^2}. A ≍ X ( X + Y ) d ≍ D , a ≍ X , b ≍ Y a ( a + b ) = A ∑ d ≍ D , a ≍ X , b ≍ Y ∑ B d a , d ( a + b ) ( t ) 2 ≪ ε , ψ D T ε d ≍ D , a ≍ X , b ≍ Y ∑ ∣ B d a , d ( a + b ) ( t ) ∣ 2 ≪ ε , ψ T ε T 2 D 2 X Y . The spectral large sieve can not be used here directly, but the structure of it would therefore produce the scale
T ε ( T U + X ( X + Y ) ) D 2 X Y T 2 . (13) T^\varepsilon\bigl(TU+X(X+Y)\bigr)\frac{D^2XY}{T^2}. \tag*{(13)} T ε ( T U + X ( X + Y ) ) T 2 D 2 X Y . (13) For D X ≍ D Y ≍ T DX\asymp DY\asymp T D X ≍ D Y ≍ T (13) is of size
T ε ( T U + T 2 / D 2 ) . T^\varepsilon(TU+T^2/D^2). T ε ( T U + T 2 / D 2 ) . In particular, blocks with D ≍ 1 D\asymp1 D ≍ 1 T ε ( T U + T 2 ) T^\varepsilon(TU+T^2) T ε ( T U + T 2 ) T 1 + ε T^{1+\varepsilon} T 1 + ε m , n ≪ ψ T m,n\ll_{\psi} T m , n ≪ ψ T d ∣ b ∣ = ∣ m − n ∣ d|b|=|m-n| d ∣ b ∣ = ∣ m − n ∣
Two auxiliary calculus lemmas
Theorem 2.4 (Faà di Bruno formula, [2 ] ). Let f , g ∈ C ℓ f,g\in C^\ell f , g ∈ C ℓ ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
d ℓ d h ℓ f ( g ( h ) ) = ℓ ! ∑ m 1 , … , m ℓ ≥ 0 m 1 + 2 m 2 + ⋯ + ℓ m ℓ = ℓ f ( m 1 + ⋯ + m ℓ ) ( g ( h ) ) ∏ r = 1 ℓ 1 m r ! ( g ( r ) ( h ) r ! ) m r . \frac{d^\ell}{dh^\ell}f(g(h))=\ell!\sum_{\substack{m_1,\ldots,m_\ell\ge0\\m_1+2m_2+\cdots+\ell m_\ell=\ell}}
f^{(m_1+\cdots+m_\ell)}(g(h))\prod_{r=1}^{\ell}\frac{1}{m_r!}\left(\frac{g^{(r)}(h)}{r!}\right)^{m_r}. d h ℓ d ℓ f ( g ( h )) = ℓ ! m 1 , … , m ℓ ≥ 0 m 1 + 2 m 2 + ⋯ + ℓ m ℓ = ℓ ∑ f ( m 1 + ⋯ + m ℓ ) ( g ( h )) r = 1 ∏ ℓ m r ! 1 ( r ! g ( r ) ( h ) ) m r . Lemma 2.5. Let f : R → [ 0 , ∞ ) f:\mathbb{R}\to[0,\infty) f : R → [ 0 , ∞ ) N ≥ 0 N\ge0 N ≥ 0
f ( x ) ≪ N ⟨ x ⟩ − N . f(x)\ll_N\langle x\rangle^{-N}. f ( x ) ≪ N ⟨ x ⟩ − N . Then there is a nonnegative ρ ∈ S ( R ) \rho\in\mathcal{S}(\mathbb{R}) ρ ∈ S ( R ) f ≤ ρ f\le\rho f ≤ ρ
Proof. Choose χ ∈ C c ∞ ( ( − 1 , 1 ) ) \chi\in C^\infty_{\mathrm{c}}((-1,1)) χ ∈ C c ∞ (( − 1 , 1 )) χ ≥ 0 \chi\ge0 χ ≥ 0 ∫ R χ = 1 \int_{\mathbb{R}}\chi=1 ∫ R χ = 1
f ♯ ( x ) : = sup ∣ y − x ∣ ≤ 1 f ( y ) , ρ = f ♯ ∗ χ . f^\sharp(x):=\sup_{|y-x|\le1}f(y),\qquad\rho=f^\sharp*\chi. f ♯ ( x ) := ∣ y − x ∣ ≤ 1 sup f ( y ) , ρ = f ♯ ∗ χ . Then f ♯ ( x ) ≪ N ⟨ x ⟩ − N f^\sharp(x)\ll_N\langle x\rangle^{-N} f ♯ ( x ) ≪ N ⟨ x ⟩ − N N N N χ ( x − y ) ≠ 0 \chi(x-y)\ne0 χ ( x − y ) = 0 ∣ x − y ∣ < 1 |x-y|<1 ∣ x − y ∣ < 1 f ♯ ( y ) ≥ f ( x ) f^\sharp(y)\ge f(x) f ♯ ( y ) ≥ f ( x )
ρ ( x ) = ∫ R f ♯ ( y ) χ ( x − y ) d y ≥ f ( x ) . \rho(x)=\int_{\mathbb{R}}f^\sharp(y)\chi(x-y)\,dy\ge f(x). ρ ( x ) = ∫ R f ♯ ( y ) χ ( x − y ) d y ≥ f ( x ) . Finally,
ρ ( k ) = f ♯ ∗ χ ( k ) , \rho^{(k)}=f^\sharp\ast\chi^{(k)}, ρ ( k ) = f ♯ ∗ χ ( k ) , and the compact support of χ ( k ) \chi^{(k)} χ ( k ) f ♯ f^\sharp f ♯
∣ ρ ( k ) ( x ) ∣ ≪ k , N ⟨ x ⟩ − N \left|\rho^{(k)}(x)\right|\ll_{k,N}\langle x\rangle^{-N} ρ ( k ) ( x ) ≪ k , N ⟨ x ⟩ − N for every k , N ≥ 0 k,N\ge0 k , N ≥ 0 ρ ∈ S ( R ) \rho\in\mathcal{S}(\mathbb{R}) ρ ∈ S ( R )
Truncation of the shift
We now prove a further restriction on the off-diagonal sum. Up to a negligible error, the off-diagonal terms may be restricted to
∣ m 2 − n 2 ∣ ≪ T 1 + ε . \left|m^2-n^2\right|\ll T^{1+\varepsilon}. m 2 − n 2 ≪ T 1 + ε . The proof uses the fact that the differential operator arising from the modified Bessel equation is formally symmetric with respect to the weighted inner product defined below; see, for example, [15 § 5.4, (5.53)] . The modified Bessel function satisfies
x 2 K i t ′ ′ ( x ) + x K i t ′ ( x ) − ( x 2 − t 2 ) K i t ( x ) = 0 ; (14) x^2K_{it}^{\prime\prime}(x)+xK_{it}^{\prime}(x)-\left(x^2-t^2\right)K_{it}(x)=0; \tag*{(14)} x 2 K i t ′′ ( x ) + x K i t ′ ( x ) − ( x 2 − t 2 ) K i t ( x ) = 0 ; (14) see [4 8.494(1)] . If L x = x d d x L_x=x\frac{d}{dx} L x = x d x d (14) becomes
( L x 2 + t 2 ) K i t ( x ) = x 2 K i t ( x ) . \left(L_x^2+t^2\right)K_{it}(x)=x^2K_{it}(x). ( L x 2 + t 2 ) K i t ( x ) = x 2 K i t ( x ) . After putting x = 2 π n y x=2\pi ny x = 2 π n y x ∂ x = y ∂ y x\partial_x=y\partial_y x ∂ x = y ∂ y
( ( y ∂ y ) 2 + t 2 ) K i t ( 2 π n y ) = ( 2 π n ) 2 y 2 K i t ( 2 π n y ) . \left((y\partial_y)^2+t^2\right)K_{it}(2\pi ny)=(2\pi n)^2y^2K_{it}(2\pi ny). ( ( y ∂ y ) 2 + t 2 ) K i t ( 2 π n y ) = ( 2 π n ) 2 y 2 K i t ( 2 π n y ) . In Sturm–Liouville form, this leads to the differential expression
P t : = 1 y ( d d y y d d y + t 2 y ) = ∂ y 2 + y − 1 ∂ y + t 2 y − 2 , P_t:=\frac{1}{y}\left(\frac{d}{dy}y\frac{d}{dy}+\frac{t^2}{y}\right)=\partial_y^2+y^{-1}\partial_y+t^2y^{-2}, P t := y 1 ( d y d y d y d + y t 2 ) = ∂ y 2 + y − 1 ∂ y + t 2 y − 2 , for which
P t K i t ( 2 π n y ) = ( 2 π n ) 2 K i t ( 2 π n y ) . P_tK_{it}(2\pi ny)=(2\pi n)^2K_{it}(2\pi ny). P t K i t ( 2 π n y ) = ( 2 π n ) 2 K i t ( 2 π n y ) . It is formally symmetric with respect to
⟨ f , g ⟩ = ∫ 0 ∞ f ( y ) g ( y ) ‾ y d y . \langle f,g\rangle=\int_0^\infty f(y)\overline{g(y)}\,y\,dy. ⟨ f , g ⟩ = ∫ 0 ∞ f ( y ) g ( y ) y d y . With respect to this inner product, P t P_t P t f , g f,g f , g ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
⟨ P t f , g ⟩ = − ∫ 0 ∞ y f ′ ( y ) g ′ ( y ) ‾ d y + t 2 ∫ 0 ∞ f ( y ) g ( y ) ‾ d y y = ⟨ f , P t g ⟩ . \langle P_tf,g\rangle=-\int_0^\infty yf'(y)\overline{g'(y)}\,dy+t^2\int_0^\infty f(y)\overline{g(y)}\,\frac{dy}{y}=\langle f,P_tg\rangle. ⟨ P t f , g ⟩ = − ∫ 0 ∞ y f ′ ( y ) g ′ ( y ) d y + t 2 ∫ 0 ∞ f ( y ) g ( y ) y d y = ⟨ f , P t g ⟩ . A uniform bound for the K K K
We use the following uniform estimate throughout this section.
Lemma 3.1. Uniformly for t ≥ 1 t \ge1 t ≥ 1 x > 0 x > 0 x > 0
e π t / 2 ∣ K i t ( x ) ∣ ≪ t − 1 / 3 . e^{\pi t/2}|K_{it}(x)| \ll t^{-1/3}. e π t /2 ∣ K i t ( x ) ∣ ≪ t − 1/3 . Proof. For x > 1 x > 1 x > 1 [11 Proposition 4.3] . For 0 < x ≤ 1 0 < x \le1 0 < x ≤ 1
K i t ( x ) = π 2 i sinh ( π t ) ( I − i t ( x ) − I i t ( x ) ) . (15) K_{it}(x)=\frac{\pi}{2i\sinh(\pi t)}\left(I_{-it}(x)-I_{it}(x)\right). \tag*{(15)} K i t ( x ) = 2 i sinh ( π t ) π ( I − i t ( x ) − I i t ( x ) ) . (15) which is formula [4 8.485] . We also have the power series for I i t ( x ) I_{it}(x) I i t ( x ) [4 8.445] ,
I i t ( x ) = ( x 2 ) i t ∑ k = 0 ∞ ( x 2 / 4 ) k k ! Γ ( k + 1 + i t ) . I_{it}(x)=\left(\frac{x}{2}\right)^{it}\sum_{k=0}^{\infty}\frac{(x^2/4)^k}{k!\,\Gamma(k+1+it)}. I i t ( x ) = ( 2 x ) i t k = 0 ∑ ∞ k ! Γ ( k + 1 + i t ) ( x 2 /4 ) k . Since
∣ Γ ( k + 1 + i t ) ∣ = ∣ Γ ( 1 + i t ) ∣ ∏ j = 1 k ∣ j + i t ∣ ≥ ∣ Γ ( 1 + i t ) ∣ k ! , |\Gamma(k+1+it)|=|\Gamma(1+it)|\prod_{j=1}^{k}|j+it|\ge|\Gamma(1+it)|\,k!, ∣Γ ( k + 1 + i t ) ∣ = ∣Γ ( 1 + i t ) ∣ j = 1 ∏ k ∣ j + i t ∣ ≥ ∣Γ ( 1 + i t ) ∣ k ! , we obtain, uniformly for 0 < x ≤ 1 0 < x \le1 0 < x ≤ 1
∣ I i t ( x ) ∣ + ∣ I − i t ( x ) ∣ ≪ ∣ Γ ( 1 + i t ) ∣ − 1 . |I_{it}(x)|+|I_{-it}(x)|\ll|\Gamma(1+it)|^{-1}. ∣ I i t ( x ) ∣ + ∣ I − i t ( x ) ∣ ≪ ∣Γ ( 1 + i t ) ∣ − 1 . Using ∣ Γ ( 1 + i t ) ∣ 2 = π t / sinh ( π t ) |\Gamma(1+it)|^2=\pi t/\sinh(\pi t) ∣Γ ( 1 + i t ) ∣ 2 = π t / sinh ( π t ) (15) gives
e π t / 2 ∣ K i t ( x ) ∣ ≪ t − 1 / 2 ≪ t − 1 / 3 . e^{\pi t/2}|K_{it}(x)|\ll t^{-1/2}\ll t^{-1/3}. e π t /2 ∣ K i t ( x ) ∣ ≪ t − 1/2 ≪ t − 1/3 . We also need bounds for derivatives with respect to x x x e π t / 2 e^{\pi t/2} e π t /2 x x x (14) also holds for e π t / 2 K i t ( x ) e^{\pi t/2}K_{it}(x) e π t /2 K i t ( x )
Lemma 3.2. Fix x 0 > 0 x_0>0 x 0 > 0 M > 0 M>0 M > 0 ℓ ≥ 0 \ell\ge0 ℓ ≥ 0 t ≥ 1 t\ge1 t ≥ 1 x 0 ≤ x ≤ M t x_0\le x\le Mt x 0 ≤ x ≤ M t
L x ℓ ( e π t / 2 K i t ( x ) ) ≪ ℓ , x 0 , M t ℓ − 1 / 3 . (16) L_x^\ell\left(e^{\pi t/2}K_{it}(x)\right)\ll_{\ell,x_0,M}t^{\ell-1/3}. \tag*{(16)} L x ℓ ( e π t /2 K i t ( x ) ) ≪ ℓ , x 0 , M t ℓ − 1/3 . (16) Proof. We first treat ℓ = 1 \ell=1 ℓ = 1 x = e s x=e^s x = e s u ( s ) = e π t / 2 K i t ( e s ) u(s)=e^{\pi t/2}K_{it}(e^s) u ( s ) = e π t /2 K i t ( e s )
u ′ ( s ) = L x ( e π t / 2 K i t ( x ) ) , u ′ ′ ( s ) = ( e 2 s − t 2 ) u ( s ) . u'(s)=L_x\left(e^{\pi t/2}K_{it}(x)\right),\qquad u''(s)=\left(e^{2s}-t^2\right)u(s). u ′ ( s ) = L x ( e π t /2 K i t ( x ) ) , u ′′ ( s ) = ( e 2 s − t 2 ) u ( s ) . With h = t − 1 h=t^{-1} h = t − 1
u ′ ( s ) = u ( s + h ) − u ( s ) h − 1 h ∫ 0 h ( h − r ) u ′ ′ ( s + r ) d r . u'(s)=\frac{u(s+h)-u(s)}{h}-\frac{1}{h}\int_0^h(h-r)u''(s+r)\,dr. u ′ ( s ) = h u ( s + h ) − u ( s ) − h 1 ∫ 0 h ( h − r ) u ′′ ( s + r ) d r . For 0 ≤ r ≤ h 0\le r\le h 0 ≤ r ≤ h x 0 ≤ e s + r ≤ e M t x_0\le e^{s+r}\le eMt x 0 ≤ e s + r ≤ e M t u ( s + r ) ≪ t − 1 / 3 u(s+r)\ll t^{-1/3} u ( s + r ) ≪ t − 1/3 u ′ ′ ( s + r ) ≪ t 5 / 3 u''(s+r)\ll t^{5/3} u ′′ ( s + r ) ≪ t 5/3
L x ( e π t / 2 K i t ( x ) ) = u ′ ( s ) ≪ t 2 / 3 . L_x\left(e^{\pi t/2}K_{it}(x)\right)=u'(s)\ll t^{2/3}. L x ( e π t /2 K i t ( x ) ) = u ′ ( s ) ≪ t 2/3 . For the higher derivatives, we claim that for every ℓ ≥ 0 \ell\ge0 ℓ ≥ 0 ω ℓ , 0 ( x , t ) \omega_{\ell,0}(x,t) ω ℓ , 0 ( x , t ) ω ℓ , 1 ( x , t ) \omega_{\ell,1}(x,t) ω ℓ , 1 ( x , t )
L x ℓ ( e π t / 2 K i t ( x ) ) = ω ℓ , 0 ( x , t ) e π t / 2 K i t ( x ) + ω ℓ , 1 ( x , t ) L x ( e π t / 2 K i t ( x ) ) , (17) L_x^\ell\left(e^{\pi t/2}K_{it}(x)\right)=\omega_{\ell,0}(x,t)e^{\pi t/2}K_{it}(x)+\omega_{\ell,1}(x,t)L_x\left(e^{\pi t/2}K_{it}(x)\right), \tag*{(17)} L x ℓ ( e π t /2 K i t ( x ) ) = ω ℓ , 0 ( x , t ) e π t /2 K i t ( x ) + ω ℓ , 1 ( x , t ) L x ( e π t /2 K i t ( x ) ) , (17) where, for x ≤ M t x \le Mt x ≤ M t
ω ℓ , 0 ( x , t ) ≪ ℓ , M t ℓ \omega_{\ell,0}(x,t) \ll_{\ell,M} t^\ell ω ℓ , 0 ( x , t ) ≪ ℓ , M t ℓ ω ℓ , 1 ( x , t ) ≪ ℓ , M t ℓ − 1 \omega_{\ell,1}(x,t) \ll_{\ell,M} t^{\ell-1} ω ℓ , 1 ( x , t ) ≪ ℓ , M t ℓ − 1 ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
Indeed, take ω 0 , 0 = 1 \omega_{0,0}=1 ω 0 , 0 = 1 ω 0 , 1 = 0 \omega_{0,1}=0 ω 0 , 1 = 0
ω ℓ + 1 , 0 = L x ω ℓ , 0 + ( x 2 − t 2 ) ω ℓ , 1 , ω ℓ + 1 , 1 = ω ℓ , 0 + L x ω ℓ , 1 . \omega_{\ell+1,0}=L_x\omega_{\ell,0}+(x^2-t^2)\omega_{\ell,1},\qquad\omega_{\ell+1,1}=\omega_{\ell,0}+L_x\omega_{\ell,1}. ω ℓ + 1 , 0 = L x ω ℓ , 0 + ( x 2 − t 2 ) ω ℓ , 1 , ω ℓ + 1 , 1 = ω ℓ , 0 + L x ω ℓ , 1 . The stated polynomial bounds follow by induction. The estimate (16) then follows from (17) , Lemma 3.1, and the bound for L x ( e π t / 2 K i t ) L_x(e^{\pi t/2}K_{it}) L x ( e π t /2 K i t )
We also need ordinary derivatives. For every integer ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
x ℓ d ℓ d x ℓ = L x ( L x − 1 ) ⋯ ( L x − ℓ + 1 ) . x^\ell\frac{d^\ell}{dx^\ell}=L_x(L_x-1)\cdots(L_x-\ell+1). x ℓ d x ℓ d ℓ = L x ( L x − 1 ) ⋯ ( L x − ℓ + 1 ) . Indeed, this follows by induction from
x ℓ + 1 d ℓ + 1 d x ℓ + 1 = ( L x − ℓ ) x ℓ d ℓ d x ℓ . x^{\ell+1}\frac{d^{\ell+1}}{dx^{\ell+1}}=(L_x-\ell)x^\ell\frac{d^\ell}{dx^\ell}. x ℓ + 1 d x ℓ + 1 d ℓ + 1 = ( L x − ℓ ) x ℓ d x ℓ d ℓ . Consequently, for suitable constants c ℓ , k c_{\ell,k} c ℓ , k
d ℓ d x ℓ = x − ℓ ∑ k = 0 ℓ c ℓ , k L x k . \frac{d^\ell}{dx^\ell}=x^{-\ell}\sum_{k=0}^{\ell}c_{\ell,k}L_x^k. d x ℓ d ℓ = x − ℓ k = 0 ∑ ℓ c ℓ , k L x k . Since x ≥ x 0 x \ge x_0 x ≥ x 0
e π t / 2 d ℓ d x ℓ K i t ( x ) ≪ ℓ , x 0 , M t ℓ − 1 / 3 , x 0 ≤ x ≤ M t . e^{\pi t/2}\frac{d^\ell}{dx^\ell}K_{it}(x)\ll_{\ell,x_0,M}t^{\ell-1/3},\qquad x_0\le x\le Mt. e π t /2 d x ℓ d ℓ K i t ( x ) ≪ ℓ , x 0 , M t ℓ − 1/3 , x 0 ≤ x ≤ M t . For a positive integer n n n
F n ( y ) = e π t / 2 K i t ( 2 π n y ) . F_n(y)=e^{\pi t/2}K_{it}(2\pi ny). F n ( y ) = e π t /2 K i t ( 2 π n y ) . If L y = y ∂ y L_y=y\partial_y L y = y ∂ y x = 2 π n y x=2\pi ny x = 2 π n y
L y ℓ F n ( y ) = L x ℓ ( e π t / 2 K i t ( x ) ) ∣ x = 2 π n y . (18) L_y^\ell F_n(y)=\left.L_x^\ell\left(e^{\pi t/2}K_{it}(x)\right)\right|_{x=2\pi ny}. \tag*{(18)} L y ℓ F n ( y ) = L x ℓ ( e π t /2 K i t ( x ) ) x = 2 π n y . (18) Moreover,
y j ∂ y j = L y ( L y − 1 ) ⋯ ( L y − j + 1 ) . y^j\partial_y^j=L_y(L_y-1)\cdots(L_y-j+1). y j ∂ y j = L y ( L y − 1 ) ⋯ ( L y − j + 1 ) . Since supp ψ \operatorname{supp}\psi supp ψ ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) (18) imply that, for every fixed j ≥ 0 j\ge0 j ≥ 0
∂ y j F n ( y ) ≪ ψ , j t j − 1 / 3 , n ≪ ψ t , y ∈ supp ψ . (19) \partial_y^jF_n(y)\ll_{\psi,j}t^{j-1/3},\qquad n\ll_\psi t,\quad y\in\operatorname{supp}\psi. \tag*{(19)} ∂ y j F n ( y ) ≪ ψ , j t j − 1/3 , n ≪ ψ t , y ∈ supp ψ . (19) The Sturm–Liouville truncation
Lemma 3.3. Let
T 0 = W , T R + 1 = [ P t , T R ] = P t T R − T R P t , W ( y ) = ψ ( y ) y . T_0=W,\qquad T_{R+1}=[P_t,T_R]=P_tT_R-T_RP_t,\qquad W(y)=\frac{\psi(y)}{y}. T 0 = W , T R + 1 = [ P t , T R ] = P t T R − T R P t , W ( y ) = y ψ ( y ) . For every integer R ≥ 0 R\ge0 R ≥ 0
T R = ∑ j = 0 R c R , j ( y ; t ) ∂ y j , (20) T_R=\sum_{j=0}^{R}c_{R,j}(y;t)\partial_y^j, \tag*{(20)} T R = j = 0 ∑ R c R , j ( y ; t ) ∂ y j , (20) where every coefficient c R , j c_{R,j} c R , j supp ψ \operatorname{supp}\psi supp ψ ℓ ≥ 0 \ell\ge0 ℓ ≥ 0
∂ y ℓ c R , j ( y ; t ) ≪ ψ , R , ℓ t R − j . (21) \partial_y^\ell c_{R,j}(y;t)\ll_{\psi,R,\ell}t^{R-j}. \tag*{(21)} ∂ y ℓ c R , j ( y ; t ) ≪ ψ , R , ℓ t R − j . (21) Proof. We use induction on R R R (20) , the support condition
supp c R , j ⊆ supp ψ , \operatorname{supp} c_{R,j} \subseteq\operatorname{supp}\psi, supp c R , j ⊆ supp ψ , and the estimates (21) for every ℓ ≥ 0 \ell\ge0 ℓ ≥ 0 R = 0 R=0 R = 0 T 0 = W = ψ / y T_0=W=\psi/y T 0 = W = ψ / y supp ψ \operatorname{supp}\psi supp ψ
Suppose that they hold for T R T_R T R
T R + 1 = ∑ j = 0 R [ P t , c R , j ( y ; t ) ∂ y j ] . T_{R+1}=\sum_{j=0}^{R}[P_t,c_{R,j}(y;t)\partial_y^j]. T R + 1 = j = 0 ∑ R [ P t , c R , j ( y ; t ) ∂ y j ] . Every coefficient occurring in these commutators is a linear combination of derivatives of c R , j c_{R,j} c R , j y y y supp ψ \operatorname{supp}\psi supp ψ P t P_t P t
First, Leibniz’ rule gives
[ ∂ y 2 , c R , j ∂ y j ] = 2 c R , j ′ ∂ y j + 1 + c R , j ′ ′ ∂ y j . [\partial_y^2,c_{R,j}\partial_y^j]=2c_{R,j}'\partial_y^{j+1}+c_{R,j}''\partial_y^j. [ ∂ y 2 , c R , j ∂ y j ] = 2 c R , j ′ ∂ y j + 1 + c R , j ′′ ∂ y j . Thus the first term contributes to c R + 1 , j + 1 c_{R+1,j+1} c R + 1 , j + 1 c R + 1 , j c_{R+1,j} c R + 1 , j ℓ ≥ 0 \ell\ge0 ℓ ≥ 0
∂ y ℓ c R , j ≪ ψ , R , ℓ t R − j = t R + 1 − ( j + 1 ) , \partial_y^\ell c_{R,j}\ll_{\psi,R,\ell}t^{R-j}=t^{R+1-(j+1)}, ∂ y ℓ c R , j ≪ ψ , R , ℓ t R − j = t R + 1 − ( j + 1 ) , while
∂ y ℓ c R , j ′ ′ ≪ ψ , R , ℓ t R − j ≤ t R + 1 − j . \partial_y^\ell c_{R,j}''\ll_{\psi,R,\ell}t^{R-j}\leq t^{R+1-j}. ∂ y ℓ c R , j ′′ ≪ ψ , R , ℓ t R − j ≤ t R + 1 − j . Both contributions therefore satisfy the required coefficient bounds at their respective differential orders.
For the first-order term in P t P_t P t
[ 1 y ∂ y , c R , j ∂ y j ] = c R , j ′ y ∂ y j − c R , j ∑ ν = 1 j ( − 1 ) ν ν ! ( j ν ) y − ν − 1 ∂ y j − ν + 1 . \left[\frac{1}{y}\partial_y,c_{R,j}\partial_y^j\right]=\frac{c_{R,j}'}{y}\partial_y^j-c_{R,j}\sum_{\nu=1}^{j}(-1)^\nu\nu!\binom{j}{\nu}y^{-\nu-1}\partial_y^{j-\nu+1}. [ y 1 ∂ y , c R , j ∂ y j ] = y c R , j ′ ∂ y j − c R , j ν = 1 ∑ j ( − 1 ) ν ν ! ( ν j ) y − ν − 1 ∂ y j − ν + 1 . The first term contributes to c R + 1 , j c_{R+1,j} c R + 1 , j ν \nu ν c R + 1 , k c_{R+1,k} c R + 1 , k
k = j − ν + 1 ≤ j . k=j-\nu+1\leq j. k = j − ν + 1 ≤ j . By Leibniz’ rule and the induction hypothesis, for every ℓ ≥ 0 \ell\geq0 ℓ ≥ 0 ℓ \ell ℓ
≪ ψ , R , ℓ t R − j ≤ t R + 1 − k . \ll_{\psi,R,\ell}t^{R-j}\leq t^{R+1-k}. ≪ ψ , R , ℓ t R − j ≤ t R + 1 − k . Finally,
[ t 2 y 2 , c R , j ∂ y j ] = − c R , j ∑ ν = 1 j ( − 1 ) ν ( ν + 1 ) ! ( j ν ) t 2 y − ν − 2 ∂ y j − ν . \left[\frac{t^2}{y^2},c_{R,j}\partial_y^j\right]=-c_{R,j}\sum_{\nu=1}^{j}(-1)^\nu(\nu+1)!\binom{j}{\nu}t^2y^{-\nu-2}\partial_y^{j-\nu}. [ y 2 t 2 , c R , j ∂ y j ] = − c R , j ν = 1 ∑ j ( − 1 ) ν ( ν + 1 )! ( ν j ) t 2 y − ν − 2 ∂ y j − ν . A term indexed by ν \nu ν c R + 1 , k c_{R+1,k} c R + 1 , k
k = j − ν ≤ j − 1. k=j-\nu\leq j-1. k = j − ν ≤ j − 1. Again by Leibniz’ rule and the induction hypothesis, for every ℓ ≥ 0 \ell\geq0 ℓ ≥ 0 ℓ \ell ℓ
≪ ψ , R , ℓ t R − j + 2 . \ll_{\psi,R,\ell}t^{R-j+2}. ≪ ψ , R , ℓ t R − j + 2 . Since k ≤ j − 1 k\leq j-1 k ≤ j − 1
R − j + 2 ≤ R + 1 − k , R-j+2\leq R+1-k, R − j + 2 ≤ R + 1 − k , so this is O ψ , R , ℓ ( t R + 1 − k ) O_{\psi,R,\ell}(t^{R+1-k}) O ψ , R , ℓ ( t R + 1 − k ) T R + 1 T_{R+1} T R + 1 R + 1 R+1 R + 1 k k k ∂ y k \partial_y^k ∂ y k
∂ y ℓ c R + 1 , k ( y ; t ) ≪ ψ , R , ℓ t R + 1 − k , \partial_y^\ell c_{R+1,k}(y;t) \ll_{\psi,R,\ell} t^{R+1-k}, ∂ y ℓ c R + 1 , k ( y ; t ) ≪ ψ , R , ℓ t R + 1 − k , for every ℓ ≥ 0 \ell\ge0 ℓ ≥ 0
Lemma 3.4. Let ψ ∈ C 0 ∞ ( 0 , ∞ ) \psi\in C^\infty_0(0,\infty) ψ ∈ C 0 ∞ ( 0 , ∞ ) m , n ≪ ψ t m,n \ll_\psi t m , n ≪ ψ t R ≥ 0 R \ge0 R ≥ 0
e π t ∣ ∫ 0 ∞ ψ ( y ) K i t ( 2 π m y ) K i t ( 2 π n y ) d y ∣ ≪ ψ , R t − 2 / 3 ( 1 + ∣ m 2 − n 2 ∣ t ) − R . (22) e^{\pi t}\left|\int_0^\infty\psi(y)K_{it}(2\pi my)K_{it}(2\pi ny)\,\mathrm{d}y\right| \ll_{\psi,R} t^{-2/3}\left(1+\frac{|m^2-n^2|}{t}\right)^{-R}. \tag*{(22)} e π t ∫ 0 ∞ ψ ( y ) K i t ( 2 π m y ) K i t ( 2 π n y ) d y ≪ ψ , R t − 2/3 ( 1 + t ∣ m 2 − n 2 ∣ ) − R . (22) Proof. Put λ n = ( 2 π n ) 2 \lambda_n=(2\pi n)^2 λ n = ( 2 π n ) 2 T R T_R T R
P t F n = λ n F n P_tF_n=\lambda_nF_n P t F n = λ n F n and the coefficients of T r T_r T r supp ψ \operatorname{supp}\psi supp ψ P t P_t P t
⟨ T r + 1 F n , F m ⟩ = ⟨ P t T r F n , F m ⟩ − ⟨ T r P t F n , F m ⟩ = ⟨ T r F n , P t F m ⟩ − λ n ⟨ T r F n , F m ⟩ = ( λ m − λ n ) ⟨ T r F n , F m ⟩ . \begin{aligned}
\langle T_{r+1}F_n,F_m\rangle
&=\langle P_tT_rF_n,F_m\rangle-\langle T_rP_tF_n,F_m\rangle\\
&=\langle T_rF_n,P_tF_m\rangle-\lambda_n\langle T_rF_n,F_m\rangle\\
&=(\lambda_m-\lambda_n)\langle T_rF_n,F_m\rangle.
\end{aligned} ⟨ T r + 1 F n , F m ⟩ = ⟨ P t T r F n , F m ⟩ − ⟨ T r P t F n , F m ⟩ = ⟨ T r F n , P t F m ⟩ − λ n ⟨ T r F n , F m ⟩ = ( λ m − λ n ) ⟨ T r F n , F m ⟩ . Iterating this identity yields
( λ m − λ n ) R ⟨ W F n , F m ⟩ = ⟨ T R F n , F m ⟩ , W ( y ) = ψ ( y ) y . (23) (\lambda_m-\lambda_n)^R\langle WF_n,F_m\rangle=\langle T_RF_n,F_m\rangle,\qquad W(y)=\frac{\psi(y)}{y}. \tag*{(23)} ( λ m − λ n ) R ⟨ W F n , F m ⟩ = ⟨ T R F n , F m ⟩ , W ( y ) = y ψ ( y ) . (23) By Lemma 3.3 and (3.6),
∣ T R F n ′ ( y ) ∣ ≪ ψ , R ∑ j = 0 R t R − j t j − 1 / 3 ≪ ψ , R t R − 1 / 3 ( y ∈ supp ψ ) . |T_RF_n'(y)|\ll_{\psi,R}\sum_{j=0}^{R}t^{R-j}t^{j-1/3}\ll_{\psi,R}t^{R-1/3}\qquad(y\in\operatorname{supp}\psi). ∣ T R F n ′ ( y ) ∣ ≪ ψ , R j = 0 ∑ R t R − j t j − 1/3 ≪ ψ , R t R − 1/3 ( y ∈ supp ψ ) . Together with Lemma 3.1, this gives
∣ ⟨ T R F n , F m ⟩ ∣ ≪ ψ , R t R − 2 / 3 . (24) |\langle T_RF_n,F_m\rangle|\ll_{\psi,R}t^{R-2/3}. \tag*{(24)} ∣ ⟨ T R F n , F m ⟩ ∣ ≪ ψ , R t R − 2/3 . (24) Hence, for m ≠ n m\ne n m = n
∣ ⟨ W F n , F m ⟩ ∣ ≪ ψ , R t − 2 / 3 ( t ∣ m 2 − n 2 ∣ ) R . |\langle WF_n,F_m\rangle|\ll_{\psi,R}t^{-2/3}\left(\frac{t}{|m^2-n^2|}\right)^R. ∣ ⟨ W F n , F m ⟩ ∣ ≪ ψ , R t − 2/3 ( ∣ m 2 − n 2 ∣ t ) R . On the other hand, Lemma 3.1 gives
∣ ⟨ W F n , F m ⟩ ∣ ≪ ψ t − 2 / 3 . |\langle WF_n,F_m\rangle|\ll_\psi t^{-2/3}. ∣ ⟨ W F n , F m ⟩ ∣ ≪ ψ t − 2/3 . Combining these two estimates, and using ∣ λ m − λ n ∣ = ( 2 π ) 2 ∣ m 2 − n 2 ∣ |\lambda_m-\lambda_n|=(2\pi)^2|m^2-n^2| ∣ λ m − λ n ∣ = ( 2 π ) 2 ∣ m 2 − n 2 ∣
∣ ⟨ W F n , F m ⟩ ∣ ≪ ψ , R t − 2 / 3 ( 1 + ∣ m 2 − n 2 ∣ t ) − R . |\langle WF_n,F_m\rangle|\ll_{\psi,R}t^{-2/3}\left(1+\frac{|m^2-n^2|}{t}\right)^{-R}. ∣ ⟨ W F n , F m ⟩ ∣ ≪ ψ , R t − 2/3 ( 1 + t ∣ m 2 − n 2 ∣ ) − R . This proves (3.9).
The truncation error on a spectral interval
Define
E j s h i f t = 1 2 L j ∑ m ≠ n m , n ≤ C ψ T ∣ m 2 − n 2 ∣ > T 1 + ε λ j ( m ) λ j ( n ) B m , n ( t j ) . \mathcal{E}^{\mathrm{shift}}_j=\frac{1}{2L_j}\sum_{\substack{m\ne n\\m,n\le C_\psi T\\|m^2-n^2|>T^{1+\varepsilon}}}\lambda_j(m)\lambda_j(n)B_{m,n}(t_j). E j shift = 2 L j 1 m = n m , n ≤ C ψ T ∣ m 2 − n 2 ∣ > T 1 + ε ∑ λ j ( m ) λ j ( n ) B m , n ( t j ) . Lemma 3.5. Fix ε > 0 \varepsilon>0 ε > 0 E j s h i f t \mathcal{E}^{\mathrm{shift}}_j E j shift A > 0 A>0 A > 0 1 ≤ U ≤ T 1\le U\le T 1 ≤ U ≤ T
∑ j e v e n T < t j ≤ T + U L j ∣ E j s h i f t ∣ 2 ≪ A , ψ , ε T − A . \sum_{\substack{j\ \mathrm{even}\\T<t_j\le T+U}}L_j|\mathcal{E}^{\mathrm{shift}}_j|^2\ll_{A,\psi,\varepsilon}T^{-A}. j even T < t j ≤ T + U ∑ L j ∣ E j shift ∣ 2 ≪ A , ψ , ε T − A . Proof. This is an immediate application of Lemma 3.4. □
By Lemmas 2.2 and 3.5 , we may restrict the off-diagonal sum to
m , n ≤ C ψ T , ∣ m 2 − n 2 ∣ ≤ T 1 + ε . m,n\le C_\psi T,\qquad|m^2-n^2|\le T^{1+\varepsilon}. m , n ≤ C ψ T , ∣ m 2 − n 2 ∣ ≤ T 1 + ε . Applying (11) and writing m = d a m=da m = d a n = d ( a + b ) n=d(a+b) n = d ( a + b )
d 2 b ( 2 a + b ) = ∣ d 2 ( a + b ) 2 − d 2 a 2 ∣ ≪ T 1 + ε . (25) d^2b(2a+b)=|d^2(a+b)^2-d^2a^2|\ll T^{1+\varepsilon}. \tag*{(25)} d 2 b ( 2 a + b ) = ∣ d 2 ( a + b ) 2 − d 2 a 2 ∣ ≪ T 1 + ε . (25) Bounds for the Bessel integral and its t t t
We use the truncation established in § 3 , with the same parameter ε \varepsilon ε
p = d a , q = d ( a + b ) , p=da,\qquad q=d(a+b), p = d a , q = d ( a + b ) , so that
q 2 − p 2 = d 2 b ( 2 a + b ) ≪ T 1 + ε . q^2-p^2=d^2b(2a+b)\ll T^{1+\varepsilon}. q 2 − p 2 = d 2 b ( 2 a + b ) ≪ T 1 + ε . Our goal is the uniform spectral derivative estimate in Lemma 4.7, with Λ \Lambda Λ
∣ ∂ t r B p , q ( t ) ∣ ≪ r , ψ , ε log T T Λ ( q − p p , T ) r . |\partial_t^r B_{p,q}(t)|\ll_{r,\psi,\varepsilon}\frac{\log T}{T}\Lambda\left(\frac{q-p}{p},T\right)^r. ∣ ∂ t r B p , q ( t ) ∣ ≪ r , ψ , ε T log T Λ ( p q − p , T ) r . We begin by establishing the Fourier representation
B p , q ( t ) = 2 ( 1 + e − 2 π t ) ∫ R e i t ξ A p , q ( ξ ) d ξ . (26) B_{p,q}(t)=2(1+e^{-2\pi t})\int_{\mathbb{R}}e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi. \tag*{(26)} B p , q ( t ) = 2 ( 1 + e − 2 π t ) ∫ R e i t ξ A p , q ( ξ ) d ξ . (26) where, for ξ ≠ ± log ( q / p ) \xi\ne\pm\log(q/p) ξ = ± log ( q / p )
A p , q ( ξ ) : = ∫ R ψ ^ ( p sinh u + q sinh ( ξ − u ) ) d u \mathcal{A}_{p,q}(\xi):=\int_{\mathbb{R}}\widehat{\psi}(p\sinh u+q\sinh(\xi-u))\,du A p , q ( ξ ) := ∫ R ψ ( p sinh u + q sinh ( ξ − u )) d u with the convention
ψ ^ ( ζ ) = ∫ R ψ ( y ) e − 2 π i y ζ d y . \widehat{\psi}(\zeta)=\int_{\mathbb{R}}\psi(y)e^{-2\pi iy\zeta}\,dy. ψ ( ζ ) = ∫ R ψ ( y ) e − 2 π i y ζ d y . Then the analysis rests on the kernel A p , q ( ξ ) \mathcal{A}_{p,q}(\xi) A p , q ( ξ )
ξ = ± log ( q / p ) , \xi=\pm\log(q/p), ξ = ± log ( q / p ) , which separate the regimes of the kernel. We study its behaviour near these points as well as in their complement, before turning to the spectral derivatives of B p , q ( t ) B_{p,q}(t) B p , q ( t )
The function A p , q \mathcal{A}_{p,q} A p , q
We begin with a change of variables that describes the structure of A p , q \mathcal{A}_{p,q} A p , q
Lemma 4.1. Let q > p q>p q > p
ϑ = log ( q / p ) , D p , q ( ξ ) : = p 2 + q 2 − 2 p q cosh ξ . \vartheta=\log(q/p),\qquad D_{p,q}(\xi):=p^2+q^2-2pq\cosh\xi. ϑ = log ( q / p ) , D p , q ( ξ ) := p 2 + q 2 − 2 pq cosh ξ . Then
A p , q ( ξ ) = ∫ R ψ ^ ( s ) s 2 + D p , q ( ξ ) d s , ∣ ξ ∣ < ϑ . (27) \mathcal{A}_{p,q}(\xi)=\int_{\mathbb{R}}\frac{\widehat{\psi}(s)}{\sqrt{s^2+D_{p,q}(\xi)}}\,ds,\qquad|\xi|<\vartheta. \tag*{(27)} A p , q ( ξ ) = ∫ R s 2 + D p , q ( ξ ) ψ ( s ) d s , ∣ ξ ∣ < ϑ . (27) A p , q ( ξ ) = 2 ∫ − D p , q ( ξ ) ∞ ψ ^ ( s ) s 2 + D p , q ( ξ ) d s , ξ > ϑ . (28) \mathcal{A}_{p,q}(\xi)=2\int_{\sqrt{-D_{p,q}(\xi)}}^{\infty}\frac{\widehat{\psi}(s)}{\sqrt{s^2+D_{p,q}(\xi)}}\,ds,\qquad\xi>\vartheta. \tag*{(28)} A p , q ( ξ ) = 2 ∫ − D p , q ( ξ ) ∞ s 2 + D p , q ( ξ ) ψ ( s ) d s , ξ > ϑ . (28) A p , q ( ξ ) = 2 ∫ − ∞ − − D p , q ( ξ ) ψ ^ ( s ) s 2 + D p , q ( ξ ) d s , ξ < − ϑ . (29) \mathcal{A}_{p,q}(\xi)=2\int_{-\infty}^{-\sqrt{-D_{p,q}(\xi)}}\frac{\widehat{\psi}(s)}{\sqrt{s^2+D_{p,q}(\xi)}}\,ds,\qquad\xi<-\vartheta. \tag*{(29)} A p , q ( ξ ) = 2 ∫ − ∞ − − D p , q ( ξ ) s 2 + D p , q ( ξ ) ψ ( s ) d s , ξ < − ϑ . (29) The square roots in (4.3)–(4.4) are positive on the domains of integration.
Proof. Put
F ξ ( u ) = p sinh u + q sinh ( ξ − u ) . F_\xi(u)=p\sinh u+q\sinh(\xi-u). F ξ ( u ) = p sinh u + q sinh ( ξ − u ) . We can write
F ξ ( u ) = 1 2 { α ( ξ ) e u + β ( ξ ) e − u } , α ( ξ ) = p − q e − ξ , β ( ξ ) = q e ξ − p . F_\xi(u)=\frac{1}{2}\{\alpha(\xi)e^u+\beta(\xi)e^{-u}\},\qquad\alpha(\xi)=p-qe^{-\xi},\quad\beta(\xi)=qe^\xi-p. F ξ ( u ) = 2 1 { α ( ξ ) e u + β ( ξ ) e − u } , α ( ξ ) = p − q e − ξ , β ( ξ ) = q e ξ − p . Then
∂ u F ξ ( u ) = 1 2 { α ( ξ ) e u − β ( ξ ) e − u } , \partial_uF_\xi(u)=\frac{1}{2}\{\alpha(\xi)e^u-\beta(\xi)e^{-u}\}, ∂ u F ξ ( u ) = 2 1 { α ( ξ ) e u − β ( ξ ) e − u } , and hence
{ ∂ u F ξ ( u ) } 2 − F ξ ( u ) 2 = − α ( ξ ) β ( ξ ) = p 2 + q 2 − 2 p q cosh ξ = D p , q ( ξ ) . \{\partial_uF_\xi(u)\}^2-F_\xi(u)^2=-\alpha(\xi)\beta(\xi)=p^2+q^2-2pq\cosh\xi=D_{p,q}(\xi). { ∂ u F ξ ( u ) } 2 − F ξ ( u ) 2 = − α ( ξ ) β ( ξ ) = p 2 + q 2 − 2 pq cosh ξ = D p , q ( ξ ) . Thus
{ ∂ u F ξ ( u ) } 2 − F ξ ( u ) 2 = D p , q ( ξ ) . (30) \{\partial_uF_\xi(u)\}^2-F_\xi(u)^2=D_{p,q}(\xi). \tag*{(30)} { ∂ u F ξ ( u ) } 2 − F ξ ( u ) 2 = D p , q ( ξ ) . (30) Also,
D p , q ( ξ ) = 4 p q sinh ϑ + ξ 2 sinh ϑ − ξ 2 . D_{p,q}(\xi)=4pq\sinh\frac{\vartheta+\xi}{2}\sinh\frac{\vartheta-\xi}{2}. D p , q ( ξ ) = 4 pq sinh 2 ϑ + ξ sinh 2 ϑ − ξ . Thus D p , q D_{p,q} D p , q ξ = ± ϑ \xi=\pm\vartheta ξ = ± ϑ
D p , q ′ ( ϑ ) = − ( q 2 − p 2 ) , D p , q ′ ( − ϑ ) = q 2 − p 2 . D'_{p,q}(\vartheta)=-(q^2-p^2),\qquad D'_{p,q}(-\vartheta)=q^2-p^2. D p , q ′ ( ϑ ) = − ( q 2 − p 2 ) , D p , q ′ ( − ϑ ) = q 2 − p 2 . The points ξ = ± ϑ \xi=\pm\vartheta ξ = ± ϑ F ξ F_\xi F ξ
Suppose first that ∣ ξ ∣ < ϑ |\xi|<\vartheta ∣ ξ ∣ < ϑ
α ( ξ ) < 0 , β ( ξ ) > 0 , α ( ξ ) β ( ξ ) < 0. \alpha(\xi)<0,\qquad\beta(\xi)>0,\qquad\alpha(\xi)\beta(\xi)<0. α ( ξ ) < 0 , β ( ξ ) > 0 , α ( ξ ) β ( ξ ) < 0. By (4.5),
∣ ∂ u F ξ ( u ) ∣ = F ξ ( u ) 2 + D p , q ( ξ ) > 0. |\partial_uF_\xi(u)|=\sqrt{F_\xi(u)^2+D_{p,q}(\xi)}>0. ∣ ∂ u F ξ ( u ) ∣ = F ξ ( u ) 2 + D p , q ( ξ ) > 0. Moreover,
lim u → − ∞ F ξ ( u ) = + ∞ , lim u → + ∞ F ξ ( u ) = − ∞ . \lim_{u\to-\infty}F_\xi(u)=+\infty,\qquad\lim_{u\to+\infty}F_\xi(u)=-\infty. u → − ∞ lim F ξ ( u ) = + ∞ , u → + ∞ lim F ξ ( u ) = − ∞. Thus F ξ F_{\xi} F ξ R \mathbb{R} R R \mathbb{R} R s = F ξ ( u ) s=F_{\xi}(u) s = F ξ ( u )
∣ d u ∣ = d s ∣ ∂ u F ξ ( u ) ∣ = d s s 2 + D p , q ( ξ ) . |du|=\frac{ds}{|\partial_uF_{\xi}(u)|}=\frac{ds}{\sqrt{s^2+D_{p,q}(\xi)}}. ∣ d u ∣ = ∣ ∂ u F ξ ( u ) ∣ d s = s 2 + D p , q ( ξ ) d s . Substituting this into the integral defining A p , q ( ξ ) \mathcal{A}_{p,q}(\xi) A p , q ( ξ )
If ξ > ϑ \xi>\vartheta ξ > ϑ α ( ξ ) , β ( ξ ) > 0 \alpha(\xi),\beta(\xi)>0 α ( ξ ) , β ( ξ ) > 0 F ξ F_{\xi} F ξ
e 2 u 0 = β ( ξ ) α ( ξ ) . e^{2u_0}=\frac{\beta(\xi)}{\alpha(\xi)}. e 2 u 0 = α ( ξ ) β ( ξ ) . At this point,
F ξ ( u 0 ) = α ( ξ ) β ( ξ ) = − D p , q ( ξ ) . F_{\xi}(u_0)=\sqrt{\alpha(\xi)\beta(\xi)}=\sqrt{-D_{p,q}(\xi)}. F ξ ( u 0 ) = α ( ξ ) β ( ξ ) = − D p , q ( ξ ) . Hence the range is [ − D p , q ( ξ ) , ∞ ) [\sqrt{-D_{p,q}(\xi)},\infty) [ − D p , q ( ξ ) , ∞ ) ( s 2 + D p , q ( ξ ) ) − 1 / 2 (s^2+D_{p,q}(\xi))^{-1/2} ( s 2 + D p , q ( ξ ) ) − 1/2
If ξ < − ϑ \xi<-\vartheta ξ < − ϑ α ( ξ ) , β ( ξ ) < 0 \alpha(\xi),\beta(\xi)<0 α ( ξ ) , β ( ξ ) < 0 − − D p , q ( ξ ) -\sqrt{-D_{p,q}(\xi)} − − D p , q ( ξ ) ( − ∞ , − − D p , q ( ξ ) ] (-\infty,-\sqrt{-D_{p,q}(\xi)}] ( − ∞ , − − D p , q ( ξ ) ]
The representations in Lemma 4.1 reduce the behaviour of A p , q \mathcal{A}_{p,q} A p , q D p , q D_{p,q} D p , q
Lemma 4.2. Let φ ∈ S ( R ) \varphi\in\mathcal{S}(\mathbb{R}) φ ∈ S ( R ) h > 0 h>0 h > 0
Φ 0 ( h ) = ∫ R φ ( s ) s 2 + h d s , \Phi_0(h)=\int_{\mathbb{R}}\frac{\varphi(s)}{\sqrt{s^2+h}}\,ds, Φ 0 ( h ) = ∫ R s 2 + h φ ( s ) d s , Φ + ( h ) = 2 ∫ h ∞ φ ( s ) s 2 − h d s , \Phi_+(h)=2\int_{\sqrt{h}}^{\infty}\frac{\varphi(s)}{\sqrt{s^2-h}}\,ds, Φ + ( h ) = 2 ∫ h ∞ s 2 − h φ ( s ) d s , Φ − ( h ) = 2 ∫ − ∞ − h φ ( s ) s 2 − h d s . \Phi_-(h)=2\int_{-\infty}^{-\sqrt{h}}\frac{\varphi(s)}{\sqrt{s^2-h}}\,ds. Φ − ( h ) = 2 ∫ − ∞ − h s 2 − h φ ( s ) d s . For every ν ∈ { 0 , + , − } \nu\in\{0,+,-\} ν ∈ { 0 , + , − } c ν ( φ ) c_\nu(\varphi) c ν ( φ ) R ν ∈ C ∞ ( ( 0 , ∞ ) ) R_\nu\in C^\infty((0,\infty)) R ν ∈ C ∞ (( 0 , ∞ ))
Φ ν ( h ) = − φ ( 0 ) log h + c ν ( φ ) + R ν ( h ) (31) \Phi_\nu(h)=-\varphi(0)\log h+c_\nu(\varphi)+R_\nu(h) \tag*{(31)} Φ ν ( h ) = − φ ( 0 ) log h + c ν ( φ ) + R ν ( h ) (31) and, for 0 < h ≤ 1 0<h\leq1 0 < h ≤ 1 ℓ ≥ 0 \ell\geq0 ℓ ≥ 0
∣ R ν ( ℓ ) ( h ) ∣ ≪ ℓ , φ h 1 / 2 − ℓ ( 1 + ∣ log h ∣ ) . (32) |R_\nu^{(\ell)}(h)|\ll_{\ell,\varphi}h^{1/2-\ell}(1+|\log h|). \tag*{(32)} ∣ R ν ( ℓ ) ( h ) ∣ ≪ ℓ , φ h 1/2 − ℓ ( 1 + ∣ log h ∣ ) . (32) For h > 1 h>1 h > 1 Φ ± \Phi_\pm Φ ± h h h O N , ℓ , φ ( h − N ) O_{N,\ell,\varphi}(h^{-N}) O N , ℓ , φ ( h − N ) ∫ R s k φ ( s ) d s \int_{\mathbb{R}}s^k\varphi(s)\,ds ∫ R s k φ ( s ) d s Φ 0 \Phi_0 Φ 0
Proof. We first prove (4.6)–(4.7) for 0 < h ≤ 1 / 2 0<h\leq1/2 0 < h ≤ 1/2
We begin with
Φ 0 ( h ) = ∫ R φ ( s ) s 2 + h d s . \Phi_0(h)=\int_{\mathbb{R}}\frac{\varphi(s)}{\sqrt{s^2+h}}\,ds. Φ 0 ( h ) = ∫ R s 2 + h φ ( s ) d s . Choose an even χ ∈ C c ∞ ( ( − 2 , 2 ) ) \chi\in C^\infty_c((-2,2)) χ ∈ C c ∞ (( − 2 , 2 )) χ = 1 \chi=1 χ = 1 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
g ( s ) = φ ( s ) − φ ( 0 ) χ ( s ) . g(s)=\varphi(s)-\varphi(0)\chi(s). g ( s ) = φ ( s ) − φ ( 0 ) χ ( s ) . Then g ( s ) = O φ ( ∣ s ∣ ) g(s) = O_{\varphi}(|s|) g ( s ) = O φ ( ∣ s ∣ )
Φ 0 ( h ) = φ ( 0 ) ∫ R χ ( s ) s 2 + h d s + ∫ R g ( s ) s 2 + h d s . \Phi_0(h) = \varphi(0) \int_{\mathbb{R}} \frac{\chi(s)}{\sqrt{s^2+h}}\,ds + \int_{\mathbb{R}} \frac{g(s)}{\sqrt{s^2+h}}\,ds. Φ 0 ( h ) = φ ( 0 ) ∫ R s 2 + h χ ( s ) d s + ∫ R s 2 + h g ( s ) d s . Since χ = 1 \chi= 1 χ = 1 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
∫ R χ ( s ) s 2 + h d s = 2 ∫ 0 1 d s s 2 + h + ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) s 2 + h d s = − log h + 2 log 2 + ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ∣ s ∣ d s + 2 log 1 + 1 + h 2 + ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( 1 s 2 + h − 1 ∣ s ∣ ) d s , \begin{aligned}
\int_{\mathbb{R}} \frac{\chi(s)}{\sqrt{s^2+h}}\,ds = 2\int_0^1 \frac{ds}{\sqrt{s^2+h}} + \int_{1\leq|s|\leq2} \frac{\chi(s)}{\sqrt{s^2+h}}\,ds \\
&= -\log h + 2\log2 + \int_{1\leq|s|\leq2} \frac{\chi(s)}{|s|}\,ds \\
&+ 2\log\frac{1+\sqrt{1+h}}{2} + \int_{1\leq|s|\leq2} \chi(s)\left(\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right)\,ds,
\end{aligned} ∫ R s 2 + h χ ( s ) d s = 2 ∫ 0 1 s 2 + h d s + ∫ 1 ≤ ∣ s ∣ ≤ 2 s 2 + h χ ( s ) d s = − log h + 2 log 2 + ∫ 1 ≤ ∣ s ∣ ≤ 2 ∣ s ∣ χ ( s ) d s + 2 log 2 1 + 1 + h + ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( s 2 + h 1 − ∣ s ∣ 1 ) d s , where
2 ∫ 0 1 d s s 2 + h = 2 arsinh ( h − 1 / 2 ) = − log h + 2 log ( 1 + 1 + h ) . 2\int_0^1 \frac{ds}{\sqrt{s^2+h}} = 2\operatorname{arsinh}(h^{-1/2}) = -\log h + 2\log(1+\sqrt{1+h}). 2 ∫ 0 1 s 2 + h d s = 2 arsinh ( h − 1/2 ) = − log h + 2 log ( 1 + 1 + h ) . Define
c 0 ( φ ) : = φ ( 0 ) ( 2 log 2 + ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ∣ s ∣ d s ) + ∫ R g ( s ) ∣ s ∣ d s c_0(\varphi) := \varphi(0)\left(2\log2+\int_{1\leq|s|\leq2}\frac{\chi(s)}{|s|}\,ds\right)+\int_{\mathbb{R}}\frac{g(s)}{|s|}\,ds c 0 ( φ ) := φ ( 0 ) ( 2 log 2 + ∫ 1 ≤ ∣ s ∣ ≤ 2 ∣ s ∣ χ ( s ) d s ) + ∫ R ∣ s ∣ g ( s ) d s and
R 0 ( h ) : = 2 φ ( 0 ) log 1 + 1 + h 2 + φ ( 0 ) ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( 1 s 2 + h − 1 ∣ s ∣ ) d s + ∫ R g ( s ) ( 1 s 2 + h − 1 ∣ s ∣ ) d s . \begin{aligned}
R_0(h) := {}& 2\varphi(0)\log\frac{1+\sqrt{1+h}}{2} \\
&+ \varphi(0)\int_{1\leq|s|\leq2}\chi(s)\left(\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right)\,ds \\
&+ \int_{\mathbb{R}}g(s)\left(\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right)\,ds.
\end{aligned} R 0 ( h ) := 2 φ ( 0 ) log 2 1 + 1 + h + φ ( 0 ) ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( s 2 + h 1 − ∣ s ∣ 1 ) d s + ∫ R g ( s ) ( s 2 + h 1 − ∣ s ∣ 1 ) d s . Thus
Φ 0 ( h ) = − φ ( 0 ) log h + c 0 ( φ ) + R 0 ( h ) . \Phi_0(h)=-\varphi(0)\log h+c_0(\varphi)+R_0(h). Φ 0 ( h ) = − φ ( 0 ) log h + c 0 ( φ ) + R 0 ( h ) . For 0 < h ≤ 1 / 2 0<h\leq1/2 0 < h ≤ 1/2
log 1 + 1 + h 2 ≪ h , ∂ h ℓ log 1 + 1 + h 2 ≪ ℓ 1 ( ℓ ≥ 1 ) , \log\frac{1+\sqrt{1+h}}{2}\ll h,\qquad\partial_h^\ell\log\frac{1+\sqrt{1+h}}{2}\ll_\ell1\quad(\ell\geq1), log 2 1 + 1 + h ≪ h , ∂ h ℓ log 2 1 + 1 + h ≪ ℓ 1 ( ℓ ≥ 1 ) , and
∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( 1 s 2 + h − 1 ∣ s ∣ ) d s ≪ χ h , \int_{1\leq|s|\leq2}\chi(s)\left(\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right)\,ds\ll_\chi h, ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( s 2 + h 1 − ∣ s ∣ 1 ) d s ≪ χ h , ∂ h ℓ ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( 1 s 2 + h − 1 ∣ s ∣ ) d s ≪ ℓ , χ 1 ( ℓ ≥ 1 ) . \partial_h^\ell\int_{1\leq|s|\leq2}\chi(s)\left(\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right)\,ds\ll_{\ell,\chi}1\quad(\ell\geq1). ∂ h ℓ ∫ 1 ≤ ∣ s ∣ ≤ 2 χ ( s ) ( s 2 + h 1 − ∣ s ∣ 1 ) d s ≪ ℓ , χ 1 ( ℓ ≥ 1 ) . For the last term, splitting at ∣ s ∣ = h |s|=\sqrt{h} ∣ s ∣ = h
∣ 1 s 2 + h − 1 ∣ s ∣ ∣ ≪ { ∣ s ∣ − 1 , ∣ s ∣ ≤ h , h ∣ s ∣ − 3 , ∣ s ∣ > h , \left|\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right|\ll
\begin{cases}
|s|^{-1}, & |s|\leq\sqrt{h},\\
h|s|^{-3}, & |s|>\sqrt{h},
\end{cases} s 2 + h 1 − ∣ s ∣ 1 ≪ { ∣ s ∣ − 1 , h ∣ s ∣ − 3 , ∣ s ∣ ≤ h , ∣ s ∣ > h , gives
∫ R ∣ g ( s ) ∣ ∣ 1 s 2 + h − 1 ∣ s ∣ ∣ d s ≪ φ h 1 / 2 . \int_{\mathbb{R}} |g(s)|\left|\frac{1}{\sqrt{s^2+h}}-\frac{1}{|s|}\right|\,ds \ll_{\varphi} h^{1/2}. ∫ R ∣ g ( s ) ∣ s 2 + h 1 − ∣ s ∣ 1 d s ≪ φ h 1/2 . For every ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
∂ h ℓ ∫ R g ( s ) s 2 + h d s = c ℓ ∫ R g ( s ) ( s 2 + h ) − ℓ − 1 / 2 d s ≪ ℓ , φ h 1 / 2 − ℓ , \partial_h^\ell\int_{\mathbb{R}} \frac{g(s)}{\sqrt{s^2+h}}\,ds
= c_\ell\int_{\mathbb{R}} g(s)(s^2+h)^{-\ell-1/2}\,ds
\ll_{\ell,\varphi} h^{1/2-\ell}, ∂ h ℓ ∫ R s 2 + h g ( s ) d s = c ℓ ∫ R g ( s ) ( s 2 + h ) − ℓ − 1/2 d s ≪ ℓ , φ h 1/2 − ℓ , where c ℓ c_\ell c ℓ ℓ \ell ℓ
∣ R 0 ( ℓ ) ( h ) ∣ ≪ ℓ , φ h 1 / 2 − ℓ ( 1 + ∣ log h ∣ ) , 0 < h ≤ 1 / 2. \left|R_0^{(\ell)}(h)\right| \ll_{\ell,\varphi} h^{1/2-\ell}(1+|\log h|), \qquad0<h\le1/2. R 0 ( ℓ ) ( h ) ≪ ℓ , φ h 1/2 − ℓ ( 1 + ∣ log h ∣ ) , 0 < h ≤ 1/2. We next consider
Φ + ( h ) = 2 ∫ h ∞ φ ( s ) s 2 − h d s . \Phi_+(h)=2\int_{\sqrt{h}}^\infty\frac{\varphi(s)}{\sqrt{s^2-h}}\,ds. Φ + ( h ) = 2 ∫ h ∞ s 2 − h φ ( s ) d s . On 0 ≤ s ≤ 1 0\le s\le1 0 ≤ s ≤ 1
φ + ( s ) = { φ ( s ) − φ ( 0 ) s , s > 0 , φ ′ ( 0 ) , s = 0 , \varphi_+(s)=
\begin{cases}
\dfrac{\varphi(s)-\varphi(0)}{s}, & s>0,\\
\varphi'(0), & s=0,
\end{cases} φ + ( s ) = ⎩ ⎨ ⎧ s φ ( s ) − φ ( 0 ) , φ ′ ( 0 ) , s > 0 , s = 0 , so that φ + ∈ C ∞ ( [ 0 , 1 ] ) \varphi_+\in C^\infty([0,1]) φ + ∈ C ∞ ([ 0 , 1 ]) φ ( s ) = φ ( 0 ) + s φ + ( s ) \varphi(s)=\varphi(0)+s\varphi_+(s) φ ( s ) = φ ( 0 ) + s φ + ( s ) s = 1 s=1 s = 1 u = s 2 − h u=\sqrt{s^2-h} u = s 2 − h
Φ + ( h ) = 2 φ ( 0 ) arccosh ( h − 1 / 2 ) + 2 ∫ 0 1 − h φ + ( u 2 + h ) d u \Phi_+(h)=2\varphi(0)\operatorname{arccosh}(h^{-1/2})+2\int_0^{\sqrt{1-h}}\varphi_+\left(\sqrt{u^2+h}\right)\,du Φ + ( h ) = 2 φ ( 0 ) arccosh ( h − 1/2 ) + 2 ∫ 0 1 − h φ + ( u 2 + h ) d u + 2 ∫ 1 ∞ φ ( s ) s 2 − h d s . \qquad+2\int_1^\infty\frac{\varphi(s)}{\sqrt{s^2-h}}\,ds. + 2 ∫ 1 ∞ s 2 − h φ ( s ) d s . Since
2 arccosh ( h − 1 / 2 ) = − log h + 2 log 2 + 2 log 1 + 1 − h 2 , 2\operatorname{arccosh}(h^{-1/2})=-\log h+2\log2+2\log\frac{1+\sqrt{1-h}}{2}, 2 arccosh ( h − 1/2 ) = − log h + 2 log 2 + 2 log 2 1 + 1 − h , define
c + ( φ ) : = 2 φ ( 0 ) log 2 + 2 ∫ 0 1 φ + ( u ) d u + 2 ∫ 1 ∞ φ ( s ) s d s c_+(\varphi):=2\varphi(0)\log2+2\int_0^1\varphi_+(u)\,du+2\int_1^\infty\frac{\varphi(s)}{s}\,ds c + ( φ ) := 2 φ ( 0 ) log 2 + 2 ∫ 0 1 φ + ( u ) d u + 2 ∫ 1 ∞ s φ ( s ) d s and
R + ( h ) : = 2 φ ( 0 ) log 1 + 1 − h 2 + 2 { ∫ 0 1 − h φ + ( u 2 + h ) d u − ∫ 0 1 φ + ( u ) d u } R_+(h):=2\varphi(0)\log\frac{1+\sqrt{1-h}}{2}
+2\left\{\int_0^{\sqrt{1-h}}\varphi_+\left(\sqrt{u^2+h}\right)\,du-\int_0^1\varphi_+(u)\,du\right\} R + ( h ) := 2 φ ( 0 ) log 2 1 + 1 − h + 2 { ∫ 0 1 − h φ + ( u 2 + h ) d u − ∫ 0 1 φ + ( u ) d u } + 2 ∫ 1 ∞ φ ( s ) ( 1 s 2 − h − 1 s ) d s . \qquad+2\int_1^\infty\varphi(s)\left(\frac{1}{\sqrt{s^2-h}}-\frac{1}{s}\right)\,ds. + 2 ∫ 1 ∞ φ ( s ) ( s 2 − h 1 − s 1 ) d s . Then
Φ + ( h ) = − φ ( 0 ) log h + c + ( φ ) + R + ( h ) . \Phi_+(h)=-\varphi(0)\log h+c_+(\varphi)+R_+(h). Φ + ( h ) = − φ ( 0 ) log h + c + ( φ ) + R + ( h ) . For 0 < h ≤ 1 / 2 0<h\le1/2 0 < h ≤ 1/2
log 1 + 1 − h 2 ≪ h , ∂ h ℓ log 1 + 1 − h 2 ≪ ℓ 1 ( ℓ ≥ 1 ) , \log\frac{1+\sqrt{1-h}}{2}\ll h,\qquad
\partial_h^\ell\log\frac{1+\sqrt{1-h}}{2}\ll_\ell1\quad(\ell\ge1), log 2 1 + 1 − h ≪ h , ∂ h ℓ log 2 1 + 1 − h ≪ ℓ 1 ( ℓ ≥ 1 ) , and
∫ 1 ∞ ∣ φ ( s ) ∣ ∣ 1 s 2 − h − 1 s ∣ d s ≪ φ h , \int_{1}^{\infty} \left|\varphi(s)\right| \left|\frac{1}{\sqrt{s^{2}-h}}-\frac{1}{s}\right|\,ds \ll_{\varphi} h, ∫ 1 ∞ ∣ φ ( s ) ∣ s 2 − h 1 − s 1 d s ≪ φ h , ∂ h ℓ ∫ 1 ∞ φ ( s ) ( 1 s 2 − h − 1 s ) d s ≪ ℓ , φ 1 ( ℓ ≥ 1 ) . \partial_{h}^{\ell}\int_{1}^{\infty}\varphi(s)\left(\frac{1}{\sqrt{s^{2}-h}}-\frac{1}{s}\right)\,ds \ll_{\ell,\varphi} 1 \qquad(\ell\ge1). ∂ h ℓ ∫ 1 ∞ φ ( s ) ( s 2 − h 1 − s 1 ) d s ≪ ℓ , φ 1 ( ℓ ≥ 1 ) . Moreover,
∫ 0 1 − h ∣ φ + ( u 2 + h ) − φ + ( u ) ∣ d u + ∫ 1 − h 1 ∣ φ + ( u ) ∣ d u ≪ φ h ( 1 + ∣ log h ∣ ) . \int_{0}^{\sqrt{1-h}}\left|\varphi_{+}\left(\sqrt{u^{2}+h}\right)-\varphi_{+}(u)\right|\,du+\int_{\sqrt{1-h}}^{1}\left|\varphi_{+}(u)\right|\,du\ll_{\varphi} h(1+|\log h|). ∫ 0 1 − h φ + ( u 2 + h ) − φ + ( u ) d u + ∫ 1 − h 1 ∣ φ + ( u ) ∣ d u ≪ φ h ( 1 + ∣ log h ∣ ) . By Theorem 2.4, for ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
∂ h ℓ φ + ( u 2 + h ) = ∑ k = 1 ℓ c ℓ , k φ + ( k ) ( u 2 + h ) ( u 2 + h ) k / 2 − ℓ , \partial_{h}^{\ell}\varphi_{+}\left(\sqrt{u^{2}+h}\right)=\sum_{k=1}^{\ell}c_{\ell,k}\varphi_{+}^{(k)}\left(\sqrt{u^{2}+h}\right)(u^{2}+h)^{k/2-\ell}, ∂ h ℓ φ + ( u 2 + h ) = k = 1 ∑ ℓ c ℓ , k φ + ( k ) ( u 2 + h ) ( u 2 + h ) k /2 − ℓ , where the constants c ℓ , k c_{\ell,k} c ℓ , k ℓ \ell ℓ k k k u 2 + h = 1 u^{2}+h=1 u 2 + h = 1
∂ h ℓ ∫ 0 1 − h φ + ( u 2 + h ) d u ≪ ℓ , φ { 1 + ∣ log h ∣ , ℓ = 1 , h 1 − ℓ , ℓ ≥ 2. \partial_{h}^{\ell}\int_{0}^{\sqrt{1-h}}\varphi_{+}\left(\sqrt{u^{2}+h}\right)\,du\ll_{\ell,\varphi}
\begin{cases}
1+|\log h|,&\ell=1,\\
h^{1-\ell},&\ell\ge2.
\end{cases} ∂ h ℓ ∫ 0 1 − h φ + ( u 2 + h ) d u ≪ ℓ , φ { 1 + ∣ log h ∣ , h 1 − ℓ , ℓ = 1 , ℓ ≥ 2. Consequently,
∣ R + ( ℓ ) ( h ) ∣ ≪ ℓ , φ h 1 / 2 − ℓ ( 1 + ∣ log h ∣ ) , 0 < h ≤ 1 / 2. \left|R_{+}^{(\ell)}(h)\right|\ll_{\ell,\varphi}h^{1/2-\ell}(1+|\log h|),\qquad0<h\le1/2. R + ( ℓ ) ( h ) ≪ ℓ , φ h 1/2 − ℓ ( 1 + ∣ log h ∣ ) , 0 < h ≤ 1/2. Replacing φ ( s ) \varphi(s) φ ( s ) φ ( − s ) \varphi(-s) φ ( − s ) R − ( h ) R_{-}(h) R − ( h ) R 0 , R + , R − ∈ C ∞ ( ( 0 , ∞ ) ) R_{0},R_{+},R_{-}\in C^{\infty}((0,\infty)) R 0 , R + , R − ∈ C ∞ (( 0 , ∞ )) [ 1 / 2 , 1 ] [1/2,1] [ 1/2 , 1 ] 0 < h ≤ 1 0<h\le1 0 < h ≤ 1
It remains to consider h ≥ 1 h\ge1 h ≥ 1
Φ ± ( h ) = 2 ∫ 0 ∞ φ ( ± h cosh v ) d v . \Phi_{\pm}(h)=2\int_{0}^{\infty}\varphi(\pm\sqrt{h}\cosh v)\,dv. Φ ± ( h ) = 2 ∫ 0 ∞ φ ( ± h cosh v ) d v . For ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
∂ h ℓ φ ( ± h cosh v ) = ∑ k = 1 ℓ c ℓ , k ± h k / 2 − ℓ ( cosh v ) k φ ( k ) ( ± h cosh v ) . \partial_{h}^{\ell}\varphi(\pm\sqrt{h}\cosh v)=\sum_{k=1}^{\ell}c_{\ell,k}^{\pm}h^{k/2-\ell}(\cosh v)^{k}\varphi^{(k)}(\pm\sqrt{h}\cosh v). ∂ h ℓ φ ( ± h cosh v ) = k = 1 ∑ ℓ c ℓ , k ± h k /2 − ℓ ( cosh v ) k φ ( k ) ( ± h cosh v ) . Together with the case ℓ = 0 \ell=0 ℓ = 0 φ \varphi φ
Φ ± ( ℓ ) ( h ) ≪ N , ℓ , φ h − N \Phi_{\pm}^{(\ell)}(h)\ll_{N,\ell,\varphi}h^{-N} Φ ± ( ℓ ) ( h ) ≪ N , ℓ , φ h − N for every N > 0 N>0 N > 0
Finally,
Φ 0 ( ℓ ) ( h ) = c ℓ ∫ R φ ( s ) ( s 2 + h ) − ℓ − 1 / 2 d s . \Phi_{0}^{(\ell)}(h)=c_{\ell}\int_{\mathbb{R}}\varphi(s)(s^{2}+h)^{-\ell-1/2}\,ds. Φ 0 ( ℓ ) ( h ) = c ℓ ∫ R φ ( s ) ( s 2 + h ) − ℓ − 1/2 d s . Assume that all moments of φ \varphi φ ∣ s ∣ ≤ h / 2 |s|\le\sqrt{h}/2 ∣ s ∣ ≤ h /2 s 2 / h s^{2}/h s 2 / h k ≥ 0 k\ge0 k ≥ 0
∫ ∣ s ∣ ≤ h / 2 s 2 k φ ( s ) d s = − ∫ ∣ s ∣ > h / 2 s 2 k φ ( s ) d s ≪ N , k , φ h − N . \int_{|s|\le\sqrt{h}/2}s^{2k}\varphi(s)\,ds=-\int_{|s|>\sqrt{h}/2}s^{2k}\varphi(s)\,ds\ll_{N,k,\varphi}h^{-N}. ∫ ∣ s ∣ ≤ h /2 s 2 k φ ( s ) d s = − ∫ ∣ s ∣ > h /2 s 2 k φ ( s ) d s ≪ N , k , φ h − N . The expansion remainder and the range ∣ s ∣ > h / 2 |s| > \sqrt{h/2} ∣ s ∣ > h /2 φ \varphi φ
Φ 0 ( ℓ ) ( h ) ≪ N , ℓ , φ h − N \Phi_0^{(\ell)}(h) \ll_{N,\ell,\varphi} h^{-N} Φ 0 ( ℓ ) ( h ) ≪ N , ℓ , φ h − N for every N > 0 N > 0 N > 0
In our application φ = ψ ^ \varphi= \widehat{\psi} φ = ψ supp ψ ⋐ ( 0 , ∞ ) \operatorname{supp}\psi\Subset(0,\infty) supp ψ ⋐ ( 0 , ∞ ) k ≥ 0 k \ge0 k ≥ 0
∫ R s k ψ ^ ( s ) d s = ( 2 π i ) − k ψ ( k ) ( 0 ) = 0. \int_{\mathbb{R}} s^k\widehat{\psi}(s)\,\mathrm{d}s = (2\pi i)^{-k}\psi^{(k)}(0) = 0. ∫ R s k ψ ( s ) d s = ( 2 π i ) − k ψ ( k ) ( 0 ) = 0. We first recall the definition of the total variation used in the next lemma. If I = [ α , β ] I=[\alpha,\beta] I = [ α , β ] F : I → C F:I\to\mathbb{C} F : I → C
Var I ( F ) : = sup α = x 0 < ⋯ < x m = β ∑ k = 1 m ∣ F ( x k ) − F ( x k − 1 ) ∣ , \operatorname{Var}_{I}(F):=\sup_{\alpha=x_0<\cdots<x_m=\beta}\sum_{k=1}^{m}|F(x_k)-F(x_{k-1})|, Var I ( F ) := α = x 0 < ⋯ < x m = β sup k = 1 ∑ m ∣ F ( x k ) − F ( x k − 1 ) ∣ , where the supremum is over all finite partitions of I I I F ∈ B V ( I ) F\in BV(I) F ∈ B V ( I ) F F F
Var I ( F ) = ∫ I ∣ F ′ ( x ) ∣ d x . \operatorname{Var}_{I}(F)=\int_I|F'(x)|\,\mathrm{d}x. Var I ( F ) = ∫ I ∣ F ′ ( x ) ∣ d x . If h h h I I I G G G h ( I ) h(I) h ( I )
Var I ( G ∘ h ) ≤ ∫ h ( I ) ∣ G ′ ( u ) ∣ d u . (33) \operatorname{Var}_{I}(G\circ h)\leq\int_{h(I)}|G'(u)|\,\mathrm{d}u. \tag*{(33)} Var I ( G ∘ h ) ≤ ∫ h ( I ) ∣ G ′ ( u ) ∣ d u . (33) Lemma 4.3. Assume q > p ≥ 1 q>p\geq1 q > p ≥ 1 p , q ≪ T p,q\ll T p , q ≪ T 1 ≤ q 2 − p 2 ≪ T 1 + ε 1\leq q^2-p^2\ll T^{1+\varepsilon} 1 ≤ q 2 − p 2 ≪ T 1 + ε r ≥ 0 r\geq0 r ≥ 0
∫ R ∣ ξ ∣ r ∣ A p , q ( ξ ) ∣ d ξ < ∞ . (34) \int_{\mathbb{R}}|\xi|^r|\mathcal{A}_{p,q}(\xi)|\,\mathrm{d}\xi<\infty. \tag*{(34)} ∫ R ∣ ξ ∣ r ∣ A p , q ( ξ ) ∣ d ξ < ∞. (34) Let 0 < Δ ≤ 1 / 4 0<\Delta\leq1/4 0 < Δ ≤ 1/4 j , N ≥ 0 j,N\geq0 j , N ≥ 0
∫ dist ( ξ , { ± ϑ } ) ≥ Δ ( 1 + ∣ ξ ∣ ) N ∣ ∂ ξ j A p , q ( ξ ) ∣ d ξ ≪ j , N , ψ ( log T ) N Δ − j . (35) \int_{\operatorname{dist}(\xi,\{\pm\vartheta\})\geq\Delta}(1+|\xi|)^N|\partial_\xi^j\mathcal{A}_{p,q}(\xi)|\,\mathrm{d}\xi\ll_{j,N,\psi}(\log T)^N\Delta^{-j}. \tag*{(35)} ∫ dist ( ξ , { ± ϑ }) ≥ Δ ( 1 + ∣ ξ ∣ ) N ∣ ∂ ξ j A p , q ( ξ ) ∣ d ξ ≪ j , N , ψ ( log T ) N Δ − j . (35) For σ ∈ { + , − } \sigma\in\{+,-\} σ ∈ { + , − }
I σ = [ σ ϑ − 2 Δ , σ ϑ + 2 Δ ] . I_\sigma=[\sigma\vartheta-2\Delta,\sigma\vartheta+2\Delta]. I σ = [ σ ϑ − 2Δ , σ ϑ + 2Δ ] . There are explicitly constructed functions H σ ∈ B V ( I σ ) H_\sigma\in BV(I_\sigma) H σ ∈ B V ( I σ ) ξ ∈ I σ ∖ { ± ϑ } \xi\in I_\sigma\setminus\{\pm\vartheta\} ξ ∈ I σ ∖ { ± ϑ }
A p , q ( ξ ) = − ψ ^ ( 0 ) log ∣ D p , q ( ξ ) ∣ + H σ ( ξ ) , (36) \mathcal{A}_{p,q}(\xi)=-\widehat{\psi}(0)\log|\mathcal{D}_{p,q}(\xi)|+H_\sigma(\xi), \tag*{(36)} A p , q ( ξ ) = − ψ ( 0 ) log ∣ D p , q ( ξ ) ∣ + H σ ( ξ ) , (36) and
∥ H σ ∥ L ∞ ( I σ ) + Var I σ ( H σ ) ≪ ψ log T . (37) \|H_\sigma\|_{L^\infty(I_\sigma)}+\operatorname{Var}_{I_\sigma}(H_\sigma)\ll_\psi\log T. \tag*{(37)} ∥ H σ ∥ L ∞ ( I σ ) + Var I σ ( H σ ) ≪ ψ log T . (37) Proof. Put
h ( ξ ) : = ∣ D p , q ( ξ ) ∣ , ν ( ξ ) = { − , ξ < − ϑ , 0 , ∣ ξ ∣ < ϑ , + , ξ > ϑ . h(\xi):=|\mathcal{D}_{p,q}(\xi)|,\qquad
\nu(\xi)=
\begin{cases}
-, & \xi<-\vartheta,\\
0, & |\xi|<\vartheta,\\
+, & \xi>\vartheta.
\end{cases} h ( ξ ) := ∣ D p , q ( ξ ) ∣ , ν ( ξ ) = ⎩ ⎨ ⎧ − , 0 , + , ξ < − ϑ , ∣ ξ ∣ < ϑ , ξ > ϑ . We take φ = ψ ^ \varphi=\widehat{\psi} φ = ψ
A p , q ( ξ ) = Φ ν ( ξ ) ( h ( ξ ) ) , ξ ≠ ± ϑ . (38) \mathcal{A}_{p,q}(\xi)=\Phi_{\nu(\xi)}(h(\xi)),\qquad\xi\neq\pm\vartheta. \tag*{(38)} A p , q ( ξ ) = Φ ν ( ξ ) ( h ( ξ )) , ξ = ± ϑ . (38) Thus the proof follows the change of the single parameter h ( ξ ) h(\xi) h ( ξ ) ξ \xi ξ ± ϑ \pm\vartheta ± ϑ 0 0 0 h ( ξ ) h(\xi) h ( ξ )
Figure 1.
We first prove (4.11) and (4.12). On I + ∪ I − I_{+}\cup I_{-} I + ∪ I − ∣ ξ ∣ ≤ ϑ + 1 / 2 |\xi|\leq\vartheta+1/2 ∣ ξ ∣ ≤ ϑ + 1/2 p , q ≪ T p,q\ll T p , q ≪ T
0 ≤ h ( ξ ) ≤ C T 2 0\leq h(\xi)\leq CT^{2} 0 ≤ h ( ξ ) ≤ C T 2 for a constant C C C 0 < h ≤ C T 2 0<h\leq CT^{2} 0 < h ≤ C T 2
For ν ∈ { 0 , + , − } \nu\in\{0,+,-\} ν ∈ { 0 , + , − }
F ν ( h ) : = Φ ν ( h ) + ψ ^ ( 0 ) log h . (39) F_{\nu}(h):=\Phi_{\nu}(h)+\widehat{\psi}(0)\log h. \tag*{(39)} F ν ( h ) := Φ ν ( h ) + ψ ( 0 ) log h . (39) Lemma 4.2 gives
F ν ( h ) = c ν ( ψ ^ ) + R ν ( h ) , 0 < h ≤ 1 , F_{\nu}(h)=c_{\nu}(\widehat{\psi})+R_{\nu}(h),\qquad0<h\leq1, F ν ( h ) = c ν ( ψ ) + R ν ( h ) , 0 < h ≤ 1 , and therefore
∥ F ν ∥ L ∞ ( 0 , 1 ] + ∫ 0 1 ∣ F ν ′ ( u ) ∣ d u ≪ ψ 1. \lVert F_{\nu}\rVert_{L^{\infty}(0,1]}+\int_{0}^{1}|F_{\nu}'(u)|\,du\ll_{\psi}1. ∥ F ν ∥ L ∞ ( 0 , 1 ] + ∫ 0 1 ∣ F ν ′ ( u ) ∣ d u ≪ ψ 1. For 1 ≤ h ≤ C T 2 1\leq h\leq CT^{2} 1 ≤ h ≤ C T 2
∣ F ν ( h ) ∣ ≪ ψ 1 + log h , ∣ F ν ′ ( h ) ∣ ≪ ψ h − 1 . |F_{\nu}(h)|\ll_{\psi}1+\log h,\qquad|F_{\nu}'(h)|\ll_{\psi}h^{-1}. ∣ F ν ( h ) ∣ ≪ ψ 1 + log h , ∣ F ν ′ ( h ) ∣ ≪ ψ h − 1 . Hence
∥ F ν ∥ L ∞ ( 0 , C T 2 ] + ∫ 0 C T 2 ∣ F ν ′ ( u ) ∣ d u ≪ ψ log T . (40) \lVert F_{\nu}\rVert_{L^{\infty}(0,CT^{2}]}+\int_{0}^{CT^{2}}|F_{\nu}'(u)|\,du\ll_{\psi}\log T. \tag*{(40)} ∥ F ν ∥ L ∞ ( 0 , C T 2 ] + ∫ 0 C T 2 ∣ F ν ′ ( u ) ∣ d u ≪ ψ log T . (40) Extend each F ν F_{\nu} F ν h = 0 h=0 h = 0 F ν ( 0 ) = c ν ( ψ ^ ) F_{\nu}(0)=c_{\nu}(\widehat{\psi}) F ν ( 0 ) = c ν ( ψ ) σ ∈ { + , − } \sigma\in\{+,-\} σ ∈ { + , − }
H σ ( ξ ) : = F ν ( ξ ) ( h ( ξ ) ) , ξ ∈ I σ ∖ { ± ϑ } , H_{\sigma}(\xi):=F_{\nu(\xi)}(h(\xi)),\qquad\xi\in I_{\sigma}\setminus\{\pm\vartheta\}, H σ ( ξ ) := F ν ( ξ ) ( h ( ξ )) , ξ ∈ I σ ∖ { ± ϑ } , and at an interior zero choose its right-hand trace. At either endpoint of I σ I_{\sigma} I σ I σ I_{\sigma} I σ H σ H_{\sigma} H σ
A p , q ( ξ ) = − ψ ^ ( 0 ) log h ( ξ ) + H σ ( ξ ) = − ψ ^ ( 0 ) log ∣ D p , q ( ξ ) ∣ + H σ ( ξ ) , \mathcal{A}_{p,q}(\xi)=-\widehat{\psi}(0)\log h(\xi)+H_{\sigma}(\xi)=-\widehat{\psi}(0)\log|\mathcal{D}_{p,q}(\xi)|+H_{\sigma}(\xi), A p , q ( ξ ) = − ψ ( 0 ) log h ( ξ ) + H σ ( ξ ) = − ψ ( 0 ) log ∣ D p , q ( ξ ) ∣ + H σ ( ξ ) , which proves (4.11). This identity is only asserted away from the zeros, or equivalently almost everywhere on I σ I_{\sigma} I σ
We next follow h ( ξ ) h(\xi) h ( ξ ) I σ I_{\sigma} I σ I σ I_{\sigma} I σ − ϑ , 0 , ϑ -\vartheta,0,\vartheta − ϑ , 0 , ϑ J J J ν ( ξ ) \nu(\xi) ν ( ξ ) h ( ξ ) h(\xi) h ( ξ )
Var J ( H σ ) ≤ ∫ 0 C T 2 ∣ F ν ′ ( u ) ∣ d u ≪ ψ log T . \operatorname{Var}_{J}(H_{\sigma})\leq\int_{0}^{CT^{2}}|F_{\nu}'(u)|\,du\ll_{\psi}\log T. Var J ( H σ ) ≤ ∫ 0 C T 2 ∣ F ν ′ ( u ) ∣ d u ≪ ψ log T . At ξ = ϑ \xi=\vartheta ξ = ϑ − ϑ -\vartheta − ϑ F ν ( h ) → c ν ( ψ ^ ) F_{\nu}(h)\to c_{\nu}(\widehat{\psi}) F ν ( h ) → c ν ( ψ ) h ↓ 0 h\downarrow0 h ↓ 0
∣ c + ( ψ ^ ) − c 0 ( ψ ^ ) ∣ , ∣ c − ( ψ ^ ) − c 0 ( ψ ^ ) ∣ . |c_{+}(\widehat{\psi})-c_{0}(\widehat{\psi})|,\qquad|c_{-}(\widehat{\psi})-c_{0}(\widehat{\psi})|. ∣ c + ( ψ ) − c 0 ( ψ ) ∣ , ∣ c − ( ψ ) − c 0 ( ψ ) ∣. There are only finitely many pieces, so (37) follows. When I + I_{+} I + I − I_{-} I − − ϑ , 0 , ϑ -\vartheta,0,\vartheta − ϑ , 0 , ϑ ν \nu ν h h h
It remains to prove (35) . Throughout this part suppose
dist ( ξ , { ± ϑ } ) ≥ Δ . \operatorname{dist}(\xi,\{\pm\vartheta\})\ge\Delta. dist ( ξ , { ± ϑ }) ≥ Δ. The factorisation of D p , q D_{p,q} D p , q
log h ( ξ ) = log ( 4 p q ) + log ∣ sinh ϑ + ξ 2 ∣ + log ∣ sinh ϑ − ξ 2 ∣ . \log h(\xi)=\log(4pq)+\log\left|\sinh\frac{\vartheta+\xi}{2}\right|+\log\left|\sinh\frac{\vartheta-\xi}{2}\right|. log h ( ξ ) = log ( 4 pq ) + log sinh 2 ϑ + ξ + log sinh 2 ϑ − ξ . and hence, for every fixed ℓ ≥ 1 \ell\ge1 ℓ ≥ 1
∣ ∂ ξ ℓ log h ( ξ ) ∣ ≪ ℓ Δ − ℓ . (41) \left|\partial_{\xi}^{\ell}\log h(\xi)\right|\ll_{\ell}\Delta^{-\ell}. \tag*{(41)} ∂ ξ ℓ log h ( ξ ) ≪ ℓ Δ − ℓ . (41) On each interval determined by − ϑ , 0 , ϑ -\vartheta,0,\vartheta − ϑ , 0 , ϑ h h h h = exp ( log h ) h=\exp(\log h) h = exp ( log h )
∣ h ( ℓ ) ( ξ ) ∣ ≪ ℓ h ( ξ ) Δ − ℓ . (42) \left|h^{(\ell)}(\xi)\right|\ll_{\ell}h(\xi)\Delta^{-\ell}. \tag*{(42)} h ( ℓ ) ( ξ ) ≪ ℓ h ( ξ ) Δ − ℓ . (42) We first treat 0 < h ( ξ ) ≤ 1 0<h(\xi)\le1 0 < h ( ξ ) ≤ 1
A p , q ( ξ ) = − ψ ^ ( 0 ) log h ( ξ ) + c ν ( ξ ) ( ψ ^ ) + R ν ( ξ ) ( h ( ξ ) ) . \mathcal{A}_{p,q}(\xi)=-\widehat{\psi}(0)\log h(\xi)+c_{\nu(\xi)}(\widehat{\psi})+R_{\nu(\xi)}(h(\xi)). A p , q ( ξ ) = − ψ ( 0 ) log h ( ξ ) + c ν ( ξ ) ( ψ ) + R ν ( ξ ) ( h ( ξ )) . For j ≥ 1 j\ge1 j ≥ 1 (41) . For the remainder, Theorem 2.4, (42) , and (32) show that every term in ∂ ξ j R ν ( h ( ξ ) ) \partial_{\xi}^{j}R_{\nu}(h(\xi)) ∂ ξ j R ν ( h ( ξ ))
h ( ξ ) 1 / 2 − k ( 1 + ∣ log h ( ξ ) ∣ ) ∏ ℓ = 1 j ( h ( ξ ) Δ − ℓ ) m ℓ = h ( ξ ) 1 / 2 ( 1 + ∣ log h ( ξ ) ∣ ) Δ − j ≪ Δ − j , h(\xi)^{1/2-k}(1+|\log h(\xi)|)\prod_{\ell=1}^{j}(h(\xi)\Delta^{-\ell})^{m_{\ell}}=h(\xi)^{1/2}(1+|\log h(\xi)|)\Delta^{-j}\ll\Delta^{-j}, h ( ξ ) 1/2 − k ( 1 + ∣ log h ( ξ ) ∣ ) ℓ = 1 ∏ j ( h ( ξ ) Δ − ℓ ) m ℓ = h ( ξ ) 1/2 ( 1 + ∣ log h ( ξ ) ∣ ) Δ − j ≪ Δ − j , where k = ∑ ℓ = 1 j m ℓ k=\sum_{\ell=1}^{j}m_{\ell} k = ∑ ℓ = 1 j m ℓ ∑ ℓ = 1 j ℓ m ℓ = j \sum_{\ell=1}^{j}\ell m_{\ell}=j ∑ ℓ = 1 j ℓ m ℓ = j j ≥ 1 j\ge1 j ≥ 1
∣ ∂ ξ j A p , q ( ξ ) ∣ ≪ j , ψ Δ − j , 0 < h ( ξ ) ≤ 1. (43) \left|\partial_{\xi}^{j}\mathcal{A}_{p,q}(\xi)\right|\ll_{j,\psi}\Delta^{-j},\qquad0<h(\xi)\le1. \tag*{(43)} ∂ ξ j A p , q ( ξ ) ≪ j , ψ Δ − j , 0 < h ( ξ ) ≤ 1. (43) For j = 0 j=0 j = 0 h h h
∣ A p , q ( ξ ) ∣ ≪ ψ 1 + ∣ log h ( ξ ) ∣ , 0 < h ( ξ ) ≤ 1 , (44) |\mathcal{A}_{p,q}(\xi)|\ll_{\psi}1+|\log h(\xi)|,\qquad0<h(\xi)\le1, \tag*{(44)} ∣ A p , q ( ξ ) ∣ ≪ ψ 1 + ∣ log h ( ξ ) ∣ , 0 < h ( ξ ) ≤ 1 , (44) since h 1 / 2 ( 1 + ∣ log h ∣ ) h^{1/2}(1+|\log h|) h 1/2 ( 1 + ∣ log h ∣ ) ( 0 , 1 ] (0,1] ( 0 , 1 ]
We next treat h ( ξ ) ≥ 1 h(\xi)\ge1 h ( ξ ) ≥ 1 (38) directly, the large-h h h (42) , and Theorem 2.4 give, for every fixed A > 0 A>0 A > 0
∣ ∂ ξ j A p , q ( ξ ) ∣ ≪ A , j , ψ Δ − j h ( ξ ) − A . (45) \left|\partial_{\xi}^{j}\mathcal{A}_{p,q}(\xi)\right|\ll_{A,j,\psi}\Delta^{-j}h(\xi)^{-A}. \tag*{(45)} ∂ ξ j A p , q ( ξ ) ≪ A , j , ψ Δ − j h ( ξ ) − A . (45) It remains to integrate these bounds. We use the elementary inequality
h ( ξ ) = 2 p q ∣ cosh ξ − cosh ϑ ∣ ≥ p q ∣ ξ 2 − ϑ 2 ∣ ≥ ∣ ξ 2 − ϑ 2 ∣ . (46) h(\xi)=2pq|\cosh\xi-\cosh\vartheta|\ge pq|\xi^{2}-\vartheta^{2}|\ge|\xi^{2}-\vartheta^{2}|. \tag*{(46)} h ( ξ ) = 2 pq ∣ cosh ξ − cosh ϑ ∣ ≥ pq ∣ ξ 2 − ϑ 2 ∣ ≥ ∣ ξ 2 − ϑ 2 ∣. (46) Indeed, ∣ cosh x − cosh y ∣ ≥ ∣ x 2 − y 2 ∣ / 2 |\cosh x-\cosh y|\ge|x^{2}-y^{2}|/2 ∣ cosh x − cosh y ∣ ≥ ∣ x 2 − y 2 ∣/2 x , y ≥ 0 x,y\ge0 x , y ≥ 0 u > 0 u>0 u > 0
∣ { ξ : h ( ξ ) < u } ∣ ≤ ∣ { ξ : ∣ ξ 2 − ϑ 2 ∣ < u } ∣ ≤ 4 u . (47) |\{\xi:h(\xi)<u\}|\le|\{\xi:|\xi^{2}-\vartheta^{2}|<u\}|\le4\sqrt{u}. \tag*{(47)} ∣ { ξ : h ( ξ ) < u } ∣ ≤ ∣ { ξ : ∣ ξ 2 − ϑ 2 ∣ < u } ∣ ≤ 4 u . (47) To see the last inequality, if ϑ 2 ≤ u \vartheta^{2}\le u ϑ 2 ≤ u 2 2 u 2\sqrt{2u} 2 2 u 2 ( ϑ 2 + u − ϑ 2 − u ) ≤ 2 2 u 2(\sqrt{\vartheta^{2}+u}-\sqrt{\vartheta^{2}-u})\le2\sqrt{2u} 2 ( ϑ 2 + u − ϑ 2 − u ) ≤ 2 2 u h ( ξ ) < u h(\xi)<u h ( ξ ) < u ∣ ξ ∣ ≤ ϑ + u |\xi|\le\vartheta+\sqrt{u} ∣ ξ ∣ ≤ ϑ + u 0 < h ( ξ ) ≤ 1 0 < h(\xi) \le1 0 < h ( ξ ) ≤ 1 ∣ ξ ∣ ≤ ϑ + 1 |\xi| \le\vartheta+ 1 ∣ ξ ∣ ≤ ϑ + 1 j ≥ 1 j \ge1 j ≥ 1
∫ 0 < h ( ξ ) ≤ 1 dist ( ξ , { ± ϑ } ) ≥ Δ ( 1 + ∣ ξ ∣ ) N ∣ ∂ ξ j A p , q ( ξ ) ∣ d ξ ≪ j , N , ψ ( 1 + ϑ ) N Δ − j . \int_{\substack{0<h(\xi)\le1\\ \operatorname{dist}(\xi,\{\pm\vartheta\})\ge\Delta}} (1+|\xi|)^N \left|\partial_\xi^j \mathcal{A}_{p,q}(\xi)\right|\,d\xi\ll_{j,N,\psi} (1+\vartheta)^N \Delta^{-j}. ∫ 0 < h ( ξ ) ≤ 1 dist ( ξ , { ± ϑ }) ≥ Δ ( 1 + ∣ ξ ∣ ) N ∂ ξ j A p , q ( ξ ) d ξ ≪ j , N , ψ ( 1 + ϑ ) N Δ − j . For j = 0 j=0 j = 0 p , q p,q p , q
∫ { 0 < h ( ξ ) ≤ 1 } ∣ log h ( ξ ) ∣ d ξ = ∫ 0 ∞ ∣ { ξ : 0 < h ( ξ ) < e − v } ∣ d v ≤ 4 ∫ 0 ∞ e − v / 2 d v = 8. (48) \int_{\{0<h(\xi)\le1\}} |\log h(\xi)|\,d\xi
= \int_0^\infty\left|\{\xi:0<h(\xi)<e^{-v}\}\right|\,dv
\le4\int_0^\infty e^{-v/2}\,dv=8. \tag*{(48)} ∫ { 0 < h ( ξ ) ≤ 1 } ∣ log h ( ξ ) ∣ d ξ = ∫ 0 ∞ { ξ : 0 < h ( ξ ) < e − v } d v ≤ 4 ∫ 0 ∞ e − v /2 d v = 8. (48) Together with (4.19), this proves the same weighted integral bound for j = 0 j=0 j = 0 Δ 0 = 1 \Delta^0=1 Δ 0 = 1
For h ( ξ ) ≥ 1 h(\xi)\ge1 h ( ξ ) ≥ 1 Ω k = { ξ : 2 k ≤ h ( ξ ) < 2 k + 1 } \Omega_k=\{\xi:2^k\le h(\xi)<2^{k+1}\} Ω k = { ξ : 2 k ≤ h ( ξ ) < 2 k + 1 } k ≥ 0 k\ge0 k ≥ 0 ∣ Ω k ∣ ≤ 4 2 k + 1 ≪ 2 k / 2 |\Omega_k|\le4\sqrt{2^{k+1}}\ll2^{k/2} ∣ Ω k ∣ ≤ 4 2 k + 1 ≪ 2 k /2 ξ ∈ Ω k \xi\in\Omega_k ξ ∈ Ω k
∣ ξ ∣ 2 ≤ ϑ 2 + h ( ξ ) < ϑ 2 + 2 k + 1 , ∣ ξ ∣ ≤ ϑ + 2 ( k + 1 ) / 2 . |\xi|^2\le\vartheta^2+h(\xi)<\vartheta^2+2^{k+1},\qquad|\xi|\le\vartheta+2^{(k+1)/2}. ∣ ξ ∣ 2 ≤ ϑ 2 + h ( ξ ) < ϑ 2 + 2 k + 1 , ∣ ξ ∣ ≤ ϑ + 2 ( k + 1 ) /2 . Consequently,
( 1 + ∣ ξ ∣ ) N ≤ ( 1 + ϑ + 2 ( k + 1 ) / 2 ) N ≪ N ( 1 + ϑ ) N 2 k N / 2 . (1+|\xi|)^N\le(1+\vartheta+2^{(k+1)/2})^N\ll_N (1+\vartheta)^N2^{kN/2}. ( 1 + ∣ ξ ∣ ) N ≤ ( 1 + ϑ + 2 ( k + 1 ) /2 ) N ≪ N ( 1 + ϑ ) N 2 k N /2 . On Ω k \Omega_k Ω k ∣ ∂ ξ j A p , q ( ξ ) ∣ ≪ B , j , ψ Δ − j 2 − k B |\partial_\xi^j\mathcal{A}_{p,q}(\xi)|\ll_{B,j,\psi}\Delta^{-j}2^{-kB} ∣ ∂ ξ j A p , q ( ξ ) ∣ ≪ B , j , ψ Δ − j 2 − k B B > ( N + 1 ) / 2 B>(N+1)/2 B > ( N + 1 ) /2
∫ h ( ξ ) ≥ 1 dist ( ξ , { ± ϑ } ) ≥ Δ ( 1 + ∣ ξ ∣ ) N ∣ ∂ ξ j A p , q ( ξ ) ∣ d ξ ≪ B , j , N , ψ ( 1 + ϑ ) N Δ − j ∑ k ≥ 0 2 k N / 2 2 k / 2 2 − k B ≪ j , N , ψ ( 1 + ϑ ) N Δ − j , \begin{aligned}
\int_{\substack{h(\xi)\ge1\\ \operatorname{dist}(\xi,\{\pm\vartheta\})\ge\Delta}}
(1+|\xi|)^N|\partial_\xi^j\mathcal{A}_{p,q}(\xi)|\,d\xi
&\ll_{B,j,N,\psi}(1+\vartheta)^N\Delta^{-j}
\sum_{k\ge0}2^{kN/2}2^{k/2}2^{-kB}\\
&\ll_{j,N,\psi}(1+\vartheta)^N\Delta^{-j},
\end{aligned} ∫ h ( ξ ) ≥ 1 dist ( ξ , { ± ϑ }) ≥ Δ ( 1 + ∣ ξ ∣ ) N ∣ ∂ ξ j A p , q ( ξ ) ∣ d ξ ≪ B , j , N , ψ ( 1 + ϑ ) N Δ − j k ≥ 0 ∑ 2 k N /2 2 k /2 2 − k B ≪ j , N , ψ ( 1 + ϑ ) N Δ − j , since the resulting geometric series converges. Combining the two ranges and using ϑ = log ( q / p ) ≪ ψ log T \vartheta=\log(q/p)\ll_\psi\log T ϑ = log ( q / p ) ≪ ψ log T N N N
Finally, near ± ϑ \pm\vartheta ± ϑ h ( ξ ) h(\xi) h ( ξ ) Φ ν ( h ( ξ ) ) \Phi_\nu(h(\xi)) Φ ν ( h ( ξ )) − ϑ , 0 , ϑ -\vartheta,0,\vartheta − ϑ , 0 , ϑ ν \nu ν h h h
A Fourier representation of the Bessel integral
By the preceding lemmas, for every fixed integer m ≥ 0 m\ge0 m ≥ 0
∫ R ( 1 + ∣ ξ ∣ ) m ∣ A p , q ( ξ ) ∣ d ξ < ∞ . (49) \int_{\mathbb{R}}(1+|\xi|)^m|\mathcal{A}_{p,q}(\xi)|\,d\xi<\infty. \tag*{(49)} ∫ R ( 1 + ∣ ξ ∣ ) m ∣ A p , q ( ξ ) ∣ d ξ < ∞. (49) Lemma 4.4. For q > p ≥ 1 q>p\ge1 q > p ≥ 1
B p , q ( t ) : = 16 cosh ( π t ) ∫ 0 ∞ ψ ( y ) K i t ( 2 π p y ) K i t ( 2 π q y ) d y . \mathcal{B}_{p,q}(t):=16\cosh(\pi t)\int_0^\infty\psi(y)K_{it}(2\pi py)K_{it}(2\pi qy)\,dy. B p , q ( t ) := 16 cosh ( π t ) ∫ 0 ∞ ψ ( y ) K i t ( 2 π p y ) K i t ( 2 π q y ) d y . Then
B p , q ( t ) = 2 ( 1 + e − 2 π t ) ∫ R e i t ξ A p , q ( ξ ) d ξ . (50) \mathcal{B}_{p,q}(t)=2(1+e^{-2\pi t})\int_{\mathbb{R}}e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi. \tag*{(50)} B p , q ( t ) = 2 ( 1 + e − 2 π t ) ∫ R e i t ξ A p , q ( ξ ) d ξ . (50) For every fixed integer r ≥ 0 r \ge0 r ≥ 0
∂ t r B p , q ( t ) = 2 ( 1 + e − 2 π t ) ∫ R ( i ξ ) r e i t ξ A p , q ( ξ ) d ξ + 2 e − 2 π t ∑ ℓ = 1 r ( r ℓ ) ( − 2 π ) ℓ ∫ R ( i ξ ) r − ℓ e i t ξ A p , q ( ξ ) d ξ , (51) \begin{aligned}
\partial_t^r \mathcal{B}_{p,q}(t)={}&2(1+e^{-2\pi t})\int_{\mathbb{R}}(i\xi)^r e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi\\
&+2e^{-2\pi t}\sum_{\ell=1}^{r}\binom{r}{\ell}(-2\pi)^\ell\int_{\mathbb{R}}(i\xi)^{r-\ell}e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi, \tag*{(51)}
\end{aligned} ∂ t r B p , q ( t ) = 2 ( 1 + e − 2 π t ) ∫ R ( i ξ ) r e i t ξ A p , q ( ξ ) d ξ + 2 e − 2 π t ℓ = 1 ∑ r ( ℓ r ) ( − 2 π ) ℓ ∫ R ( i ξ ) r − ℓ e i t ξ A p , q ( ξ ) d ξ , (51) with an empty second line when r = 0 r=0 r = 0
Proof. For 0 < η < π / 2 0<\eta<\pi/2 0 < η < π /2
K i t ( x ) = 1 2 ∫ R e − x cosh z e i t z d z K_{it}(x)=\frac{1}{2}\int_{\mathbb{R}}e^{-x\cosh z}e^{itz}\,dz K i t ( x ) = 2 1 ∫ R e − x c o s h z e i t z d z to ℑ z = π / 2 − η \Im z=\pi/2-\eta ℑ z = π /2 − η
2 e ( π / 2 − η ) t K i t ( x ) = ∫ R e − x sin η cosh u e i { t u − x cos η sinh u } d u . (52) 2e^{(\pi/2-\eta)t}K_{it}(x)=\int_{\mathbb{R}}e^{-x\sin\eta\cosh u}e^{i\{tu-x\cos\eta\sinh u\}}\,du. \tag*{(52)} 2 e ( π /2 − η ) t K i t ( x ) = ∫ R e − x s i n η c o s h u e i { t u − x c o s η s i n h u } d u . (52) Indeed, for fixed η > 0 \eta>0 η > 0 (52) with x = 2 π p y x=2\pi py x = 2 π p y x = 2 π q y x=2\pi qy x = 2 π q y B p , q ( t ) \mathcal{B}_{p,q}(t) B p , q ( t ) supp ψ ⊂ [ y 0 , y 1 ] ⋐ ( 0 , ∞ ) \operatorname{supp}\psi\subset[y_0,y_1]\Subset(0,\infty) supp ψ ⊂ [ y 0 , y 1 ] ⋐ ( 0 , ∞ ) η > 0 \eta>0 η > 0 ξ = u + v \xi=u+v ξ = u + v
B p , q ( t ) = P η ( t ) ∫ R e i t ξ A p , q ( η ) ( ξ ) d ξ , P η ( t ) : = 4 e − ( π − 2 η ) t cosh ( π t ) , (53) \mathcal{B}_{p,q}(t)=P_\eta(t)\int_{\mathbb{R}}e^{it\xi}\mathcal{A}_{p,q}^{(\eta)}(\xi)\,d\xi,\qquad P_\eta(t):=4e^{-(\pi-2\eta)t}\cosh(\pi t), \tag*{(53)} B p , q ( t ) = P η ( t ) ∫ R e i t ξ A p , q ( η ) ( ξ ) d ξ , P η ( t ) := 4 e − ( π − 2 η ) t cosh ( π t ) , (53) where
A p , q ( η ) ( ξ ) : = ∫ R ∫ 0 ∞ ψ ( y ) e − 2 π y sin η { p cosh u + q cosh ( ξ − u ) } × e − 2 π i y cos η { p sinh u + q sinh ( ξ − u ) } d y d u . \mathcal{A}_{p,q}^{(\eta)}(\xi):=\int_{\mathbb{R}}\int_0^\infty\psi(y)e^{-2\pi y\sin\eta\{p\cosh u+q\cosh(\xi-u)\}}
\times e^{-2\pi iy\cos\eta\{p\sinh u+q\sinh(\xi-u)\}}\,dy\,du. A p , q ( η ) ( ξ ) := ∫ R ∫ 0 ∞ ψ ( y ) e − 2 π y s i n η { p c o s h u + q c o s h ( ξ − u )} × e − 2 π i y c o s η { p s i n h u + q s i n h ( ξ − u )} d y d u . It remains to let η ↓ 0 \eta\downarrow0 η ↓ 0 ϕ ∈ C ∞ ( 0 , ∞ ) \phi\in C^\infty(0,\infty) ϕ ∈ C ∞ ( 0 , ∞ )
Φ ( s ) = ∫ 0 ∞ ϕ ( t ) e i t s d t . \Phi(s)=\int_0^\infty\phi(t)e^{its}\,dt. Φ ( s ) = ∫ 0 ∞ ϕ ( t ) e i t s d t . Then Φ \Phi Φ ϕ ( t ) P η ( t ) − 1 \phi(t)P_\eta(t)^{-1} ϕ ( t ) P η ( t ) − 1 t t t
∫ 0 ∞ ϕ ( t ) P η ( t ) − 1 B p , q ( t ) d t = ∫ R Φ ( ξ ) A p , q ( η ) ( ξ ) d ξ . (54) \int_0^\infty\phi(t)P_\eta(t)^{-1}\mathcal{B}_{p,q}(t)\,dt
=\int_{\mathbb{R}}\Phi(\xi)\mathcal{A}_{p,q}^{(\eta)}(\xi)\,d\xi. \tag*{(54)} ∫ 0 ∞ ϕ ( t ) P η ( t ) − 1 B p , q ( t ) d t = ∫ R Φ ( ξ ) A p , q ( η ) ( ξ ) d ξ . (54) Write
F ξ ( u ) = p sinh u + q sinh ( ξ − u ) , F_\xi(u)=p\sinh u+q\sinh(\xi-u), F ξ ( u ) = p sinh u + q sinh ( ξ − u ) , and put
A η = 2 π sin η { p cosh u + q cosh ( ξ − u ) } , B η = 2 π cos η F ξ ( u ) . A_\eta=2\pi\sin\eta\{p\cosh u+q\cosh(\xi-u)\},\qquad B_\eta=2\pi\cos\eta F_\xi(u). A η = 2 π sin η { p cosh u + q cosh ( ξ − u )} , B η = 2 π cos η F ξ ( u ) . Since supp ψ ⊂ [ y 0 , y 1 ] ⊂ ( 0 , ∞ ) \operatorname{supp}\psi\subset[y_0,y_1]\subset(0,\infty) supp ψ ⊂ [ y 0 , y 1 ] ⊂ ( 0 , ∞ ) M M M y y y M ≥ 0 M\ge0 M ≥ 0
∣ ∫ 0 ∞ ψ ( y ) e − A η y e − i B η y d y ∣ ≪ M , ψ ⟨ B η ⟩ − M . \left|\int_0^\infty\psi(y)e^{-A_\eta y}e^{-iB_\eta y}\,dy\right|
\ll_{M,\psi}\langle B_\eta\rangle^{-M}. ∫ 0 ∞ ψ ( y ) e − A η y e − i B η y d y ≪ M , ψ ⟨ B η ⟩ − M . Indeed, differentiating ψ ( y ) e − A η y \psi(y)e^{-A_\eta y} ψ ( y ) e − A η y A η k e − A η y 0 A_\eta^k e^{-A_\eta y_0} A η k e − A η y 0 A η k e − A η y 0 ≪ k , y 0 1 A_\eta^k e^{-A_\eta y_0} \ll_{k,y_0} 1 A η k e − A η y 0 ≪ k , y 0 1 0 < η ≤ η 0 < π / 2 0<\eta\leq\eta_0<\pi/2 0 < η ≤ η 0 < π /2
∣ ∫ 0 ∞ ψ ( y ) e − 2 π y sin η { p cosh u + q cosh ( ξ − u ) } e − 2 π i y cos η F ξ ( u ) d y ∣ ≪ M , ψ , η 0 ⟨ F ξ ( u ) ⟩ − M ( M ≥ 0 ) , (55) \left|\int_0^\infty\psi(y)e^{-2\pi y\sin\eta\{p\cosh u+q\cosh(\xi-u)\}}e^{-2\pi i y\cos\eta F_\xi(u)}\,dy\right|\ll_{M,\psi,\eta_0}\langle F_\xi(u)\rangle^{-M}\qquad(M\geq0), \tag*{(55)} ∫ 0 ∞ ψ ( y ) e − 2 π y s i n η { p c o s h u + q c o s h ( ξ − u )} e − 2 π i y c o s η F ξ ( u ) d y ≪ M , ψ , η 0 ⟨ F ξ ( u ) ⟩ − M ( M ≥ 0 ) , (55) since cos η ≥ cos η 0 > 0 \cos\eta\geq\cos\eta_0>0 cos η ≥ cos η 0 > 0
The constants in these estimates depend only on M M M ψ \psi ψ η 0 \eta_0 η 0 C M C_M C M (55) is bounded by C M ⟨ F ξ ( u ) ⟩ − M C_M\langle F_\xi(u)\rangle^{-M} C M ⟨ F ξ ( u ) ⟩ − M
g ( s ) : = inf M ≥ 0 C M ⟨ s ⟩ − M . g(s):=\inf_{M\geq0} C_M\langle s\rangle^{-M}. g ( s ) := M ≥ 0 inf C M ⟨ s ⟩ − M . Then g g g g ( s ) ≪ M ⟨ s ⟩ − M g(s)\ll_M\langle s\rangle^{-M} g ( s ) ≪ M ⟨ s ⟩ − M M M M (55) is bounded by g ( F ξ ( u ) ) g(F_\xi(u)) g ( F ξ ( u )) ρ ∈ S ( R ) \rho\in\mathcal{S}(\mathbb{R}) ρ ∈ S ( R ) ψ \psi ψ η 0 \eta_0 η 0 g ≤ ρ g\leq\rho g ≤ ρ
∣ ∫ 0 ∞ ψ ( y ) e − 2 π y sin η { p cosh u + q cosh ( ξ − u ) } e − 2 π i y cos η F ξ ( u ) d y ∣ ≤ ρ ( F ξ ( u ) ) . (56) \left|\int_0^\infty\psi(y)e^{-2\pi y\sin\eta\{p\cosh u+q\cosh(\xi-u)\}}e^{-2\pi i y\cos\eta F_\xi(u)}\,dy\right|\leq\rho(F_\xi(u)). \tag*{(56)} ∫ 0 ∞ ψ ( y ) e − 2 π y s i n η { p c o s h u + q c o s h ( ξ − u )} e − 2 π i y c o s η F ξ ( u ) d y ≤ ρ ( F ξ ( u )) . (56) By Lemmas 4.1 and 4.2, applied with φ = ρ \varphi=\rho φ = ρ
ξ ⟼ ∫ R ρ ( F ξ ( u ) ) d u \xi\longmapsto\int_{\mathbb{R}}\rho(F_\xi(u))\,du ξ ⟼ ∫ R ρ ( F ξ ( u )) d u is locally integrable, with at most logarithmic singularities at ξ = ± ϑ \xi=\pm\vartheta ξ = ± ϑ ∣ ξ ∣ → ∞ |\xi|\to\infty ∣ ξ ∣ → ∞ Φ \Phi Φ
∫ R ∣ Φ ( ξ ) ∣ ( ∫ R ρ ( F ξ ( u ) ) d u ) d ξ < ∞ . \int_{\mathbb{R}}|\Phi(\xi)|\left(\int_{\mathbb{R}}\rho(F_\xi(u))\,du\right)d\xi<\infty. ∫ R ∣Φ ( ξ ) ∣ ( ∫ R ρ ( F ξ ( u )) d u ) d ξ < ∞. Thus (56) gives an η \eta η (54) . Dominated convergence yields
lim η → 0 ∫ R Φ ( ξ ) A p , q ( η ) ( ξ ) d ξ = ∫ R Φ ( ξ ) ( ∫ R ψ ^ ( F ξ ( u ) ) d u ) d ξ = ∫ R Φ ( ξ ) A p , q ( ξ ) d ξ . (57) \lim_{\eta\to0}\int_{\mathbb{R}}\Phi(\xi)\mathcal{A}_{p,q}^{(\eta)}(\xi)\,d\xi
=\int_{\mathbb{R}}\Phi(\xi)\left(\int_{\mathbb{R}}\widehat{\psi}(F_\xi(u))\,du\right)d\xi
=\int_{\mathbb{R}}\Phi(\xi)\mathcal{A}_{p,q}(\xi)\,d\xi. \tag*{(57)} η → 0 lim ∫ R Φ ( ξ ) A p , q ( η ) ( ξ ) d ξ = ∫ R Φ ( ξ ) ( ∫ R ψ ( F ξ ( u )) d u ) d ξ = ∫ R Φ ( ξ ) A p , q ( ξ ) d ξ . (57) Since P η − 1 → P 0 − 1 P_\eta^{-1}\to P_0^{-1} P η − 1 → P 0 − 1 supp φ \operatorname{supp}\varphi supp φ P 0 ( t ) = 2 ( 1 + e − 2 π t ) P_0(t)=2(1+e^{-2\pi t}) P 0 ( t ) = 2 ( 1 + e − 2 π t ) (54) and (57) give
B p , q ( t ) P 0 ( t ) = ∫ R e i t ξ A p , q ( ξ ) d ξ \frac{\mathcal{B}_{p,q}(t)}{P_0(t)}=\int_{\mathbb{R}}e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi P 0 ( t ) B p , q ( t ) = ∫ R e i t ξ A p , q ( ξ ) d ξ in the distributional sense on ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) (49) with m = 0 m=0 m = 0 (50) .
Finally, (49) permits differentiation under the integral:
d m d t m ∫ R e i t ξ A p , q ( ξ ) d ξ = ∫ R ( i ξ ) m e i t ξ A p , q ( ξ ) d ξ . \frac{d^m}{dt^m}\int_{\mathbb{R}}e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi
=\int_{\mathbb{R}}(i\xi)^m e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi. d t m d m ∫ R e i t ξ A p , q ( ξ ) d ξ = ∫ R ( i ξ ) m e i t ξ A p , q ( ξ ) d ξ . Applying Leibniz’ rule to (50) gives (51) . □
Estimates near and away from the zeros of D p , q D_{p,q} D p , q
Lemma 4.5. Let I = [ α , β ] I=[\alpha,\beta] I = [ α , β ] H ∈ B V ( I ) H\in BV(I) H ∈ B V ( I ) m ∈ C 1 ( I ) m\in C^1(I) m ∈ C 1 ( I )
Var I ( m H ) ≤ ∥ m ∥ L ∞ ( I ) Var I ( H ) + ∥ H ∥ L ∞ ( I ) ∫ I ∣ m ′ ( s ) ∣ d s . (58) \operatorname{Var}_{I}(mH)\leq\|m\|_{L^\infty(I)}\operatorname{Var}_{I}(H)+\|H\|_{L^\infty(I)}\int_I |m'(s)|\,ds. \tag*{(58)} Var I ( m H ) ≤ ∥ m ∥ L ∞ ( I ) Var I ( H ) + ∥ H ∥ L ∞ ( I ) ∫ I ∣ m ′ ( s ) ∣ d s . (58) For every t ≠ 0 t\ne0 t = 0
∣ ∫ α β e i t s m ( s ) H ( s ) d s ∣ ≤ 1 ∣ t ∣ ( { 2 ∥ m ∥ ∞ + ∥ m ′ ∥ L 1 ( I ) } ∥ H ∥ ∞ + ∥ m ∥ ∞ Var I ( H ) ) . (59) \left|\int_\alpha^\beta e^{its}m(s)H(s)\,ds\right|
\leq\frac{1}{|t|}\left(\{2\|m\|_\infty+\|m'\|_{L^1(I)}\}\|H\|_\infty+\|m\|_\infty\operatorname{Var}_{I}(H)\right). \tag*{(59)} ∫ α β e i t s m ( s ) H ( s ) d s ≤ ∣ t ∣ 1 ( { 2∥ m ∥ ∞ + ∥ m ′ ∥ L 1 ( I ) } ∥ H ∥ ∞ + ∥ m ∥ ∞ Var I ( H ) ) . (59) Proof. For a partition α = x 0 < ⋯ < x n = β \alpha=x_0<\cdots<x_n=\beta α = x 0 < ⋯ < x n = β
∣ m ( x k ) H ( x k ) − m ( x k − 1 ) H ( x k − 1 ) ∣ ≤ ∣ m ( x k ) ∣ ∣ H ( x k ) − H ( x k − 1 ) ∣ + ∣ H ( x k − 1 ) ∣ ∣ m ( x k ) − m ( x k − 1 ) ∣ . |m(x_k)H(x_k)-m(x_{k-1})H(x_{k-1})|
\leq|m(x_k)||H(x_k)-H(x_{k-1})|+|H(x_{k-1})||m(x_k)-m(x_{k-1})|. ∣ m ( x k ) H ( x k ) − m ( x k − 1 ) H ( x k − 1 ) ∣ ≤ ∣ m ( x k ) ∣∣ H ( x k ) − H ( x k − 1 ) ∣ + ∣ H ( x k − 1 ) ∣∣ m ( x k ) − m ( x k − 1 ) ∣. After summing and taking the supremum, the identity Var I ( m ) = ∫ I ∣ m ′ ∣ \operatorname{Var}_{I}(m)=\int_I|m'| Var I ( m ) = ∫ I ∣ m ′ ∣ (58) .
Let F = m H F=mH F = m H d F dF d F ∣ d F ∣ ( I ) = Var I ( F ) |dF|(I)=\operatorname{Var}_{I}(F) ∣ d F ∣ ( I ) = Var I ( F )
∫ α β e i t s F ( s ) d s = e i t β F ( β − ) − e i t α F ( α + ) i t − 1 i t ∫ ( α , β ] e i t s d F ( s ) . \int_\alpha^\beta e^{its}F(s)\,ds
=\frac{e^{it\beta}F(\beta-)-e^{it\alpha}F(\alpha+)}{it}
-\frac{1}{it}\int_{(\alpha,\beta]}e^{its}\,dF(s). ∫ α β e i t s F ( s ) d s = i t e i tβ F ( β − ) − e i t α F ( α + ) − i t 1 ∫ ( α , β ] e i t s d F ( s ) . Taking absolute values and using ∣ e i t s ∣ = 1 |e^{its}|=1 ∣ e i t s ∣ = 1
∣ ∫ α β e i t s F ( s ) d s ∣ ≤ 2 ∥ F ∥ ∞ + Var I ( F ) ∣ t ∣ . \left|\int_\alpha^\beta e^{its}F(s)\,ds\right|
\leq\frac{2\|F\|_\infty+\operatorname{Var}_{I}(F)}{|t|}. ∫ α β e i t s F ( s ) d s ≤ ∣ t ∣ 2∥ F ∥ ∞ + Var I ( F ) . Together with ∥ F ∥ ∞ ≤ ∥ m ∥ ∞ ∥ H ∥ ∞ \|F\|_\infty\leq\|m\|_\infty\|H\|_\infty ∥ F ∥ ∞ ≤ ∥ m ∥ ∞ ∥ H ∥ ∞ (58) , this proves (59) . □
Lemma 4.6. Fix 0 < ε < 1 0<\varepsilon<1 0 < ε < 1 T ≥ 2 T\geq2 T ≥ 2
1 ≤ p < q ≤ C ψ T , 1 ≤ q 2 − p 2 ≪ T 1 + ε , 1\leq p<q\leq C_\psi T,\qquad1\leq q^2-p^2\ll T^{1+\varepsilon}, 1 ≤ p < q ≤ C ψ T , 1 ≤ q 2 − p 2 ≪ T 1 + ε , and put
ϑ = log ( q / p ) , Δ = T − 1 + ε . \vartheta=\log(q/p),\qquad\Delta=T^{-1+\varepsilon}. ϑ = log ( q / p ) , Δ = T − 1 + ε . For every fixed integer r ≥ 0 r\geq0 r ≥ 0 t ≍ T t\asymp T t ≍ T
∫ R ( i ξ ) r e i t ξ A p , q ( ξ ) d ξ ≪ r , ψ , ε log T T ( ϑ + Δ ) r . (60) \int_{\mathbb{R}}(i\xi)^r e^{it\xi}A_{p,q}(\xi)\,d\xi
\ll_{r,\psi,\varepsilon}\frac{\log T}{T}(\vartheta+\Delta)^r. \tag*{(60)} ∫ R ( i ξ ) r e i t ξ A p , q ( ξ ) d ξ ≪ r , ψ , ε T log T ( ϑ + Δ ) r . (60) Proof. Choose χ ∈ C ∞ ( ( − 2 , 2 ) ) \chi\in C^\infty((-2,2)) χ ∈ C ∞ (( − 2 , 2 )) χ = 1 \chi=1 χ = 1 [ − 1 , 1 ] [-1,1] [ − 1 , 1 ]
χ + ( ξ ) = χ ( ( ξ − ϑ ) / Δ ) , χ − ( ξ ) = χ ( ( ξ + ϑ ) / Δ ) . \chi_+(\xi)=\chi((\xi-\vartheta)/\Delta),\qquad
\chi_-(\xi)=\chi((\xi+\vartheta)/\Delta). χ + ( ξ ) = χ (( ξ − ϑ ) /Δ ) , χ − ( ξ ) = χ (( ξ + ϑ ) /Δ ) . Use the partition
η + = χ + , η − = ( 1 − χ + ) χ − , η 0 = ( 1 − χ + ) ( 1 − χ − ) , η + + η − + η 0 = 1. \eta_+=\chi_+,\qquad
\eta_-=(1-\chi_+)\chi_-,\qquad
\eta_0=(1-\chi_+)(1-\chi_-),\qquad
\eta_++\eta_-+\eta_0=1. η + = χ + , η − = ( 1 − χ + ) χ − , η 0 = ( 1 − χ + ) ( 1 − χ − ) , η + + η − + η 0 = 1. For every ℓ ≥ 0 \ell\geq0 ℓ ≥ 0
∣ ∂ ξ ℓ η ± ( ξ ) ∣ + ∣ ∂ ξ ℓ η 0 ( ξ ) ∣ ≪ ℓ Δ − ℓ . (61) |\partial_\xi^\ell\eta_\pm(\xi)|+|\partial_\xi^\ell\eta_0(\xi)|\ll_\ell\Delta^{-\ell}. \tag*{(61)} ∣ ∂ ξ ℓ η ± ( ξ ) ∣ + ∣ ∂ ξ ℓ η 0 ( ξ ) ∣ ≪ ℓ Δ − ℓ . (61) Since t ≍ T t\asymp T t ≍ T
t Δ ≍ T ε ≥ 1 t\Delta\asymp T^\varepsilon\geq1 t Δ ≍ T ε ≥ 1 for large T T T T T T
We first estimate the two terms supported near the zeros ± ϑ \pm\vartheta ± ϑ (36) : the singular logarithm is treated directly, while the remaining term is handled by the BV estimate (37) . Suppose first that the two Δ \Delta Δ
ϑ ≥ 4 Δ , ξ = ϑ + s , ∣ s ∣ ≤ 2 Δ . \vartheta\ge4\Delta,\qquad\xi=\vartheta+s,\qquad|s|\le2\Delta. ϑ ≥ 4Δ , ξ = ϑ + s , ∣ s ∣ ≤ 2Δ. Using
D p , q ( ξ ) = 4 p q sinh ϑ + ξ 2 sinh ϑ − ξ 2 , q 2 − p 2 = 2 p q sinh ϑ , D_{p,q}(\xi)=4pq\sinh\frac{\vartheta+\xi}{2}\sinh\frac{\vartheta-\xi}{2},\qquad q^2-p^2=2pq\sinh\vartheta, D p , q ( ξ ) = 4 pq sinh 2 ϑ + ξ sinh 2 ϑ − ξ , q 2 − p 2 = 2 pq sinh ϑ , we obtain
∣ D p , q ( ϑ + s ) ∣ = ( q 2 − p 2 ) ∣ s ∣ e b + ( s ) . |D_{p,q}(\vartheta+s)|=(q^2-p^2)|s|e^{b_{+}(s)}. ∣ D p , q ( ϑ + s ) ∣ = ( q 2 − p 2 ) ∣ s ∣ e b + ( s ) . where
b + ( s ) = log ( 2 ∣ sinh ( s / 2 ) ∣ ∣ s ∣ ) + log ( sinh ( ϑ + s / 2 ) sinh ϑ ) , b_{+}(s)=\log\left(\frac{2|\sinh(s/2)|}{|s|}\right)+\log\left(\frac{\sinh(\vartheta+s/2)}{\sinh\vartheta}\right), b + ( s ) = log ( ∣ s ∣ 2∣ sinh ( s /2 ) ∣ ) + log ( sinh ϑ sinh ( ϑ + s /2 ) ) , with b + ( 0 ) = 0 b_{+}(0)=0 b + ( 0 ) = 0
log ∣ D p , q ( ϑ + s ) ∣ = log ( q 2 − p 2 ) + log ∣ s ∣ + b + ( s ) . \log|D_{p,q}(\vartheta+s)|=\log(q^2-p^2)+\log|s|+b_{+}(s). log ∣ D p , q ( ϑ + s ) ∣ = log ( q 2 − p 2 ) + log ∣ s ∣ + b + ( s ) . Write
b + , 1 ( s ) = log ( 2 ∣ sinh ( s / 2 ) ∣ ∣ s ∣ ) , b + , 2 ( s ) = log ( sinh ( ϑ + s / 2 ) sinh ϑ ) . b_{+,1}(s)=\log\left(\frac{2|\sinh(s/2)|}{|s|}\right),\qquad b_{+,2}(s)=\log\left(\frac{\sinh(\vartheta+s/2)}{\sinh\vartheta}\right). b + , 1 ( s ) = log ( ∣ s ∣ 2∣ sinh ( s /2 ) ∣ ) , b + , 2 ( s ) = log ( sinh ϑ sinh ( ϑ + s /2 ) ) . On ∣ s ∣ ≤ 1 / 2 |s|\le1/2 ∣ s ∣ ≤ 1/2
b + , 1 ( s ) = O ( s 2 ) , b + , 1 ′ ( s ) = O ( ∣ s ∣ ) . b_{+,1}(s)=O(s^2),\qquad b'_{+,1}(s)=O(|s|). b + , 1 ( s ) = O ( s 2 ) , b + , 1 ′ ( s ) = O ( ∣ s ∣ ) . Moreover,
b + , 2 ′ ( s ) = 1 2 coth ( ϑ + s / 2 ) . b'_{+,2}(s)=\frac{1}{2}\coth(\vartheta+s/2). b + , 2 ′ ( s ) = 2 1 coth ( ϑ + s /2 ) . Since
ϑ + s / 2 ≥ ϑ − Δ ≥ 3 4 ϑ , \vartheta+s/2\ge\vartheta-\Delta\ge\frac{3}{4}\vartheta, ϑ + s /2 ≥ ϑ − Δ ≥ 4 3 ϑ , if 0 < ϑ ≤ 1 0<\vartheta\le1 0 < ϑ ≤ 1
∣ b + , 2 ′ ( s ) ∣ ≪ ϑ − 1 , ∣ b + , 2 ( s ) ∣ ≪ Δ ϑ ≪ 1. |b'_{+,2}(s)|\ll\vartheta^{-1},\qquad|b_{+,2}(s)|\ll\frac{\Delta}{\vartheta}\ll1. ∣ b + , 2 ′ ( s ) ∣ ≪ ϑ − 1 , ∣ b + , 2 ( s ) ∣ ≪ ϑ Δ ≪ 1. If ϑ > 1 \vartheta>1 ϑ > 1
∥ b + ∥ L ∞ ( ∣ s ∣ ≤ 2 Δ ) + Δ ∥ b + ′ ∥ L ∞ ( ∣ s ∣ ≤ 2 Δ ) ≪ 1. (62) \|b_{+}\|_{L^\infty(|s|\le2\Delta)}+\Delta\|b'_{+}\|_{L^\infty(|s|\le2\Delta)}\ll1. \tag*{(62)} ∥ b + ∥ L ∞ ( ∣ s ∣ ≤ 2Δ ) + Δ∥ b + ′ ∥ L ∞ ( ∣ s ∣ ≤ 2Δ ) ≪ 1. (62) At the zero − ϑ -\vartheta − ϑ s = ξ + ϑ s=\xi+\vartheta s = ξ + ϑ D p , q D_{p,q} D p , q b − ( s ) = b + ( − s ) b_{-}(s)=b_{+}(-s) b − ( s ) = b + ( − s )
We now estimate the contribution of these two neighbourhoods. On the support of either η + \eta_{+} η + η − \eta_{-} η −
∣ ξ ∣ r ≤ C r ( ϑ + Δ ) r . (63) |\xi|^r\le C_r(\vartheta+\Delta)^r. \tag*{(63)} ∣ ξ ∣ r ≤ C r ( ϑ + Δ ) r . (63) If
m r , + ( s ) = ( i ( ϑ + s ) ) r η + ( ϑ + s ) , m_{r,+}(s)=\bigl(i(\vartheta+s)\bigr)^r\eta_{+}(\vartheta+s), m r , + ( s ) = ( i ( ϑ + s ) ) r η + ( ϑ + s ) , then
∥ m r , + ∥ ∞ + Δ ∥ m r , + ′ ∥ ∞ ≪ r ( ϑ + Δ ) r . (64) \|m_{r,+}\|_\infty+\Delta\|m'_{r,+}\|_\infty\ll_r(\vartheta+\Delta)^r. \tag*{(64)} ∥ m r , + ∥ ∞ + Δ∥ m r , + ′ ∥ ∞ ≪ r ( ϑ + Δ ) r . (64) The same estimate holds near the zero − ϑ -\vartheta − ϑ
Let c ∈ C 1 ( [ − 2 Δ , 2 Δ ] ) c\in C^1([-2\Delta,2\Delta]) c ∈ C 1 ([ − 2Δ , 2Δ ]) ∣ s ∣ ≤ t − 1 |s|\le t^{-1} ∣ s ∣ ≤ t − 1
∫ ∣ s ∣ ≤ t − 1 ∣ log ∣ s ∣ ∣ d s = 2 t − 1 ( 1 + log t ) ≪ t − 1 log ( 2 + t ) . \int_{|s|\le t^{-1}}|\log|s||\,ds=2t^{-1}(1+\log t)\ll t^{-1}\log(2+t). ∫ ∣ s ∣ ≤ t − 1 ∣ log ∣ s ∣∣ d s = 2 t − 1 ( 1 + log t ) ≪ t − 1 log ( 2 + t ) . On the two intervals t − 1 ≤ ∣ s ∣ ≤ 2 Δ t^{-1} \leq|s| \leq2\Delta t − 1 ≤ ∣ s ∣ ≤ 2Δ
∫ e i t s log ∣ s ∣ c ( s ) d s = e i t s log ∣ s ∣ c ( s ) i t − 1 i t ∫ e i t s ( c ( s ) s + log ∣ s ∣ c ′ ( s ) ) d s . \int e^{its}\log|s|c(s)\,ds = \frac{e^{its}\log|s|c(s)}{it} - \frac{1}{it}\int e^{its}\left(\frac{c(s)}{s}+\log|s|c'(s)\right)\,ds. ∫ e i t s log ∣ s ∣ c ( s ) d s = i t e i t s log ∣ s ∣ c ( s ) − i t 1 ∫ e i t s ( s c ( s ) + log ∣ s ∣ c ′ ( s ) ) d s . Hence
∣ ∫ ∣ s ∣ ≤ 2 Δ e i t s log ∣ s ∣ c ( s ) d s ∣ ≪ t − 1 log ( 2 + t ) { ∥ c ∥ ∞ + Δ ∥ c ′ ∥ ∞ } . (65) \left|\int_{|s|\leq2\Delta} e^{its}\log|s|c(s)\,ds\right| \ll t^{-1}\log(2+t)\{\|c\|_{\infty}+\Delta\|c'\|_{\infty}\}. \tag*{(65)} ∫ ∣ s ∣ ≤ 2Δ e i t s log ∣ s ∣ c ( s ) d s ≪ t − 1 log ( 2 + t ) { ∥ c ∥ ∞ + Δ∥ c ′ ∥ ∞ } . (65) Apply (36) . At the zero ϑ \vartheta ϑ
H + ( s ) = H ( ϑ + s ) , ∣ s ∣ ≤ 2 Δ . \mathcal{H}_{+}(s)=H(\vartheta+s), \qquad|s|\leq2\Delta. H + ( s ) = H ( ϑ + s ) , ∣ s ∣ ≤ 2Δ. Translation invariance of total variation, (37) , (64) , and Lemma 4.5 give
∣ ∫ ∣ s ∣ ≤ 2 Δ e i t s m r , + ( s ) H + ( s ) d s ∣ ≪ 1 t { ∥ m r , + ∥ ∞ + Δ ∥ m r , + ′ ∥ ∞ } { ∥ H + ∥ ∞ + Var + ( H + ) } \begin{aligned}
\left|\int_{|s|\leq2\Delta}e^{its}m_{r,+}(s)\mathcal{H}_{+}(s)\,ds\right| \\
&\ll\frac{1}{t}\{\|m_{r,+}\|_{\infty}+\Delta\|m'_{r,+}\|_{\infty}\}\{\|H_{+}\|_{\infty}+\operatorname{Var}_{+}(H_{+})\}
\end{aligned} ∫ ∣ s ∣ ≤ 2Δ e i t s m r , + ( s ) H + ( s ) d s ≪ t 1 { ∥ m r , + ∥ ∞ + Δ∥ m r , + ′ ∥ ∞ } { ∥ H + ∥ ∞ + Var + ( H + )} ≪ r , ψ , ε T − 1 ( log T ) ( ϑ + Δ ) r . \ll_{r,\psi,\varepsilon} T^{-1}(\log T)(\vartheta+\Delta)^r. ≪ r , ψ , ε T − 1 ( log T ) ( ϑ + Δ ) r . The term log ∣ s ∣ \log|s| log ∣ s ∣ (65) and (64) .
Since m r , + ( ± 2 Δ ) = 0 m_{r,+}(\pm2\Delta)=0 m r , + ( ± 2Δ ) = 0
∣ ∫ ∣ s ∣ ≤ 2 Δ e i t s m r , + ( s ) { log ( q 2 − p 2 ) + b + ( s ) } d s ∣ ≤ 1 t ∫ ∣ s ∣ ≤ 2 Δ ∣ d d s [ m r , + ( s ) { log ( q 2 − p 2 ) + b + ( s ) } ] ∣ d s ≪ 1 t [ Δ ∥ m r , + ′ ∥ ∞ { log T + ∥ b + ∥ ∞ } + Δ ∥ m r , + ∥ ∞ ∥ b + ′ ∥ ∞ ] \begin{aligned}
\left|\int_{|s|\leq2\Delta}e^{its}m_{r,+}(s)\{\log(q^2-p^2)+b_{+}(s)\}\,ds\right| \\
&\leq\frac{1}{t}\int_{|s|\leq2\Delta}\left|\frac{d}{ds}\left[m_{r,+}(s)\{\log(q^2-p^2)+b_{+}(s)\}\right]\right|\,ds \\
&\ll\frac{1}{t}\left[\Delta\|m'_{r,+}\|_{\infty}\{\log T+\|b_{+}\|_{\infty}\}+\Delta\|m_{r,+}\|_{\infty}\|b'_{+}\|_{\infty}\right]
\end{aligned} ∫ ∣ s ∣ ≤ 2Δ e i t s m r , + ( s ) { log ( q 2 − p 2 ) + b + ( s )} d s ≤ t 1 ∫ ∣ s ∣ ≤ 2Δ d s d [ m r , + ( s ) { log ( q 2 − p 2 ) + b + ( s )} ] d s ≪ t 1 [ Δ∥ m r , + ′ ∥ ∞ { log T + ∥ b + ∥ ∞ } + Δ∥ m r , + ∥ ∞ ∥ b + ′ ∥ ∞ ] ≪ r , ψ , ε T − 1 ( log T ) ( ϑ + Δ ) r . \ll_{r,\psi,\varepsilon} T^{-1}(\log T)(\vartheta+\Delta)^r. ≪ r , ψ , ε T − 1 ( log T ) ( ϑ + Δ ) r . Here
1 ≤ q 2 − p 2 ≪ T 1 + ε , ∣ log ( q 2 − p 2 ) ∣ ≪ ε log T , 1\leq q^2-p^2\ll T^{1+\varepsilon}, \qquad|\log(q^2-p^2)|\ll_{\varepsilon}\log T, 1 ≤ q 2 − p 2 ≪ T 1 + ε , ∣ log ( q 2 − p 2 ) ∣ ≪ ε log T , and (62) were used. The argument at the zero − ϑ -\vartheta − ϑ
It remains to treat the case in which the two Δ \Delta Δ ϑ < 4 Δ \vartheta<4\Delta ϑ < 4Δ b + b_{+} b +
log ∣ D p , q ( ξ ) ∣ = log ( 4 p q ) + log ∣ sinh ξ + ϑ 2 ∣ + log ∣ sinh ξ − ϑ 2 ∣ . \log|\mathcal{D}_{p,q}(\xi)|=\log(4pq)+\log\left|\sinh\frac{\xi+\vartheta}{2}\right|+\log\left|\sinh\frac{\xi-\vartheta}{2}\right|. log ∣ D p , q ( ξ ) ∣ = log ( 4 pq ) + log sinh 2 ξ + ϑ + log sinh 2 ξ − ϑ . Both zeros lie in ∣ ξ ∣ ≤ 4 Δ |\xi|\leq4\Delta ∣ ξ ∣ ≤ 4Δ η + + η − \eta_{+}+\eta_{-} η + + η − ∣ ξ ∣ ≤ 6 Δ |\xi|\leq6\Delta ∣ ξ ∣ ≤ 6Δ
log ∣ sinh ξ − ξ 0 2 ∣ = log ∣ ξ − ξ 0 ∣ + log ∣ sinh ( ( ξ − ξ 0 ) / 2 ) ξ − ξ 0 ∣ , ξ 0 = ± ϑ . \log\left|\sinh\frac{\xi-\xi_{0}}{2}\right|=\log|\xi-\xi_{0}|+\log\left|\frac{\sinh((\xi-\xi_{0})/2)}{\xi-\xi_{0}}\right|,\qquad\xi_{0}=\pm\vartheta. log sinh 2 ξ − ξ 0 = log ∣ ξ − ξ 0 ∣ + log ξ − ξ 0 sinh (( ξ − ξ 0 ) /2 ) , ξ 0 = ± ϑ . The second term and its first derivative are uniformly bounded, while the first is handled by (65) . Together with (63) and the BV estimate, this gives
T − 1 ( log T ) ( ϑ + Δ ) r . T^{-1}(\log T)(\vartheta+\Delta)^r. T − 1 ( log T ) ( ϑ + Δ ) r . Thus
∣ ∫ e i t ξ ( i ξ ) r ( η + + η − ) A p , q ( ξ ) d ξ ∣ ≪ r , ψ , ε T − 1 ( log T ) ( ϑ + Δ ) r . (66) \left|\int e^{it\xi}(i\xi)^r(\eta_+ + \eta_-)\mathcal{A}_{p,q}(\xi)\,d\xi\right| \ll_{r,\psi,\varepsilon} T^{-1}(\log T)(\vartheta+\Delta)^r. \tag*{(66)} ∫ e i t ξ ( i ξ ) r ( η + + η − ) A p , q ( ξ ) d ξ ≪ r , ψ , ε T − 1 ( log T ) ( ϑ + Δ ) r . (66) We now turn to the complementary term containing η 0 \eta_0 η 0 ≫ Δ \gg\Delta ≫ Δ L 1 L^1 L 1 (35) can be combined with repeated integration by parts. Leibniz’ formula gives
∂ ξ J { ( i ξ ) r η 0 A p , q } = ∑ j 1 + j 2 + j 3 = J c j 1 , j 2 , j 3 ∂ ξ j 1 ( i ξ ) r ∂ ξ j 2 η 0 ∂ ξ j 3 A p , q . \partial_\xi^J\{(i\xi)^r\eta_0\mathcal{A}_{p,q}\}
=
\sum_{j_1+j_2+j_3=J}c_{j_1,j_2,j_3}\,\partial_\xi^{j_1}(i\xi)^r\partial_\xi^{j_2}\eta_0\,\partial_\xi^{j_3}\mathcal{A}_{p,q}. ∂ ξ J {( i ξ ) r η 0 A p , q } = j 1 + j 2 + j 3 = J ∑ c j 1 , j 2 , j 3 ∂ ξ j 1 ( i ξ ) r ∂ ξ j 2 η 0 ∂ ξ j 3 A p , q . Using (61) and (35) ,
∫ R ∣ ∂ ξ J { ( i ξ ) r η 0 ( ξ ) A p , q ( ξ ) } ∣ d ξ ≪ J , r , ψ ( log T ) r + 1 Δ − J . \int_{\mathbb{R}}\left|\partial_\xi^J\{(i\xi)^r\eta_0(\xi)\mathcal{A}_{p,q}(\xi)\}\right|\,d\xi
\ll_{J,r,\psi}(\log T)^{r+1}\Delta^{-J}. ∫ R ∂ ξ J {( i ξ ) r η 0 ( ξ ) A p , q ( ξ )} d ξ ≪ J , r , ψ ( log T ) r + 1 Δ − J . Here one uses the weighted L 1 L^1 L 1 (35) with weight at most ( 1 + ∣ ξ ∣ ) r (1+|\xi|)^r ( 1 + ∣ ξ ∣ ) r ∂ ξ j 1 ( i ξ ) r \partial_\xi^{j_1}(i\xi)^r ∂ ξ j 1 ( i ξ ) r ( log T ) r + 1 (\log T)^{r+1} ( log T ) r + 1 η 0 \eta_0 η 0 A p , q \mathcal{A}_{p,q} A p , q Δ − J \Delta^{-J} Δ − J J J J
∣ ∫ e i t ξ ( i ξ ) r η 0 ( ξ ) A p , q ( ξ ) d ξ ∣ ≪ J , r , ψ ( log T ) r + 1 ( t Δ ) − J ≪ J , r , ψ ( log T ) r + 1 T − J ε . \left|\int e^{it\xi}(i\xi)^r\eta_0(\xi)\mathcal{A}_{p,q}(\xi)\,d\xi\right|
\ll_{J,r,\psi}(\log T)^{r+1}(t\Delta)^{-J}
\ll_{J,r,\psi}(\log T)^{r+1}T^{-J\varepsilon}. ∫ e i t ξ ( i ξ ) r η 0 ( ξ ) A p , q ( ξ ) d ξ ≪ J , r , ψ ( log T ) r + 1 ( t Δ ) − J ≪ J , r , ψ ( log T ) r + 1 T − J ε . Since
ϑ + Δ ≥ Δ = T − 1 + ε , \vartheta+\Delta\geq\Delta=T^{-1+\varepsilon}, ϑ + Δ ≥ Δ = T − 1 + ε , we may choose J = J ( r , ε ) J=J(r,\varepsilon) J = J ( r , ε )
( log T ) r + 1 T − J ε ≪ T − 1 ( log T ) ( ϑ + Δ ) r . (\log T)^{r+1}T^{-J\varepsilon}\ll T^{-1}(\log T)(\vartheta+\Delta)^r. ( log T ) r + 1 T − J ε ≪ T − 1 ( log T ) ( ϑ + Δ ) r . Combining this with (66) proves (60) . □
Lemma 4.7. Fix 0 < ε < 1 0<\varepsilon<1 0 < ε < 1 t ≍ T t\asymp T t ≍ T p = d a p=da p = d a q = d ( a + b ) q=d(a+b) q = d ( a + b ) ρ = b / a \rho=b/a ρ = b / a Lemma 4.6 . Then, for every fixed integer r ≥ 0 r\geq0 r ≥ 0
d r d t r B d a , d ( a + b ) ( t ) ≪ r , ψ , ε log T T Λ ( ρ , T ) r , (67) \frac{d^r}{dt^r}B_{da,d(a+b)}(t)\ll_{r,\psi,\varepsilon}\frac{\log T}{T}\Lambda(\rho,T)^r, \tag*{(67)} d t r d r B d a , d ( a + b ) ( t ) ≪ r , ψ , ε T log T Λ ( ρ , T ) r , (67) where
Λ ( ρ , T ) = T − 1 + ε + { ρ , 0 < ρ ≤ 1 , log ( 2 + ρ ) , ρ > 1. (68) \Lambda(\rho,T)=T^{-1+\varepsilon}+
\begin{cases}
\rho, & 0<\rho\leq1,\\
\log(2+\rho), & \rho>1.
\end{cases} \tag*{(68)} Λ ( ρ , T ) = T − 1 + ε + { ρ , log ( 2 + ρ ) , 0 < ρ ≤ 1 , ρ > 1. (68) Proof. Fix the ε > 0 \varepsilon>0 ε > 0 ε 0 > 0 \varepsilon_0>0 ε 0 > 0 ε 0 < ε / 2 \varepsilon_0<\varepsilon/2 ε 0 < ε /2
p = d a ≤ C ψ T , q = d ( a + b ) ≤ C ψ T , p=da\leq C_\psi T,\qquad q=d(a+b)\leq C_\psi T, p = d a ≤ C ψ T , q = d ( a + b ) ≤ C ψ T , and
1 ≤ q 2 − p 2 = d 2 b ( 2 a + b ) ≪ T 1 + ε 0 . 1\leq q^2-p^2=d^2b(2a+b)\ll T^{1+\varepsilon_0}. 1 ≤ q 2 − p 2 = d 2 b ( 2 a + b ) ≪ T 1 + ε 0 . Hence Lemma 4.6 applies. Put
ϑ = log ( q / p ) , Δ = T − 1 + ε 0 , \vartheta=\log(q/p),\qquad\Delta=T^{-1+\varepsilon_0}, ϑ = log ( q / p ) , Δ = T − 1 + ε 0 , and for 0 ≤ m ≤ r 0\leq m\leq r 0 ≤ m ≤ r
I m ( t ; p , q ) = ∫ R ( i ξ ) m e i t ξ A p , q ( ξ ) d ξ . \mathcal{I}_m(t;p,q)=\int_{\mathbb{R}}(i\xi)^m e^{it\xi}\mathcal{A}_{p,q}(\xi)\,d\xi. I m ( t ; p , q ) = ∫ R ( i ξ ) m e i t ξ A p , q ( ξ ) d ξ . e n d s m i d = 1 {ends_mid=1} e n d s m i d = 1 By (60) ,
I m ( t ; p , q ) ≪ m , ψ , ε 0 T − 1 ( log T ) ( ϑ + Δ ) m . \mathcal{I}_m(t;p,q) \ll_{m,\psi,\varepsilon_0} T^{-1}(\log T)(\vartheta+\Delta)^m. I m ( t ; p , q ) ≪ m , ψ , ε 0 T − 1 ( log T ) ( ϑ + Δ ) m . The differentiation formula (51) gives
∣ ∂ t r B p , q ( t ) ∣ ≤ 2 ( 1 + e − 2 π t ) ∣ I r ( t ; p , q ) ∣ + 2 e − 2 π t ∑ ℓ = 1 r ( r ℓ ) ( 2 π ) ℓ ∣ I r − ℓ ( t ; p , q ) ∣ . (69) \left|\partial_t^r B_{p,q}(t)\right| \le2(1+e^{-2\pi t})|\mathcal{I}_r(t;p,q)|+2e^{-2\pi t}\sum_{\ell=1}^{r}\binom{r}{\ell}(2\pi)^\ell|\mathcal{I}_{r-\ell}(t;p,q)|. \tag*{(69)} ∣ ∂ t r B p , q ( t ) ∣ ≤ 2 ( 1 + e − 2 π t ) ∣ I r ( t ; p , q ) ∣ + 2 e − 2 π t ℓ = 1 ∑ r ( ℓ r ) ( 2 π ) ℓ ∣ I r − ℓ ( t ; p , q ) ∣. (69) The first term is
≪ r , ψ , ε 0 T − 1 ( log T ) ( ϑ + Δ ) r . \ll_{r,\psi,\varepsilon_0} T^{-1}(\log T)(\vartheta+\Delta)^r. ≪ r , ψ , ε 0 T − 1 ( log T ) ( ϑ + Δ ) r . Since
0 < ϑ = log ( q / p ) ≤ log ( C ψ T ) ≪ ψ log T , 0<\vartheta=\log(q/p)\leq\log(C_\psi T)\ll_\psi\log T, 0 < ϑ = log ( q / p ) ≤ log ( C ψ T ) ≪ ψ log T , for every fixed B > 0 B>0 B > 0 (69) is
≪ r , ψ , ε 0 e − c T T − 1 ( log T ) r + 1 \ll_{r,\psi,\varepsilon_0}e^{-cT}T^{-1}(\log T)^{r+1} ≪ r , ψ , ε 0 e − c T T − 1 ( log T ) r + 1 ≪ B , r , ψ T − B . (70) \ll_{B,r,\psi}T^{-B}. \tag*{(70)} ≪ B , r , ψ T − B . (70) Thus
∣ ∂ t r B p , q ( t ) ∣ ≪ r , ψ , ε 0 T − 1 ( log T ) ( ϑ + T − 1 + ε 0 ) r + O B , r , ψ ( T − B ) . \left|\partial_t^r B_{p,q}(t)\right|\ll_{r,\psi,\varepsilon_0}T^{-1}(\log T)(\vartheta+T^{-1+\varepsilon_0})^r+O_{B,r,\psi}(T^{-B}). ∣ ∂ t r B p , q ( t ) ∣ ≪ r , ψ , ε 0 T − 1 ( log T ) ( ϑ + T − 1 + ε 0 ) r + O B , r , ψ ( T − B ) . Finally,
q p = 1 + ρ , ϑ = log ( 1 + ρ ) . \frac{q}{p}=1+\rho,\qquad\vartheta=\log(1+\rho). p q = 1 + ρ , ϑ = log ( 1 + ρ ) . If 0 < ρ ≤ 1 0<\rho\leq1 0 < ρ ≤ 1
log ( 1 + ρ ) ≤ ρ , \log(1+\rho)\leq\rho, log ( 1 + ρ ) ≤ ρ , whereas if ρ > 1 \rho>1 ρ > 1
log ( 1 + ρ ) ≤ log ( 2 + ρ ) . \log(1+\rho)\leq\log(2+\rho). log ( 1 + ρ ) ≤ log ( 2 + ρ ) . Therefore
ϑ + T − 1 + ε ≤ Λ ( ρ , T ) . \vartheta+T^{-1+\varepsilon}\leq\Lambda(\rho,T). ϑ + T − 1 + ε ≤ Λ ( ρ , T ) . Since Λ ( ρ , T ) ≥ T − 1 + ε \Lambda(\rho,T)\geq T^{-1+\varepsilon} Λ ( ρ , T ) ≥ T − 1 + ε B B B (70) into T − 1 ( log T ) Λ ( ρ , T ) r T^{-1}(\log T)\Lambda(\rho,T)^r T − 1 ( log T ) Λ ( ρ , T ) r ε 0 < ε \varepsilon_0<\varepsilon ε 0 < ε (67) .
The local spectral large sieve with varying coefficients
To apply the spectral large sieve in Lemma 2.3 to the t t t
Lemma 5.1. Let N \mathcal{N} N J J J L ≍ H L\asymp H L ≍ H t J t_J t J
F : J ⟶ ℓ 2 ( N ) , F ( t ) = ( F ν ( t ) ) ν ∈ N , F:J\longrightarrow\ell^2(\mathcal{N}),\qquad F(t)=(F_\nu(t))_{\nu\in\mathcal{N}}, F : J ⟶ ℓ 2 ( N ) , F ( t ) = ( F ν ( t ) ) ν ∈ N , be smooth and vanish near the endpoints. Suppose that, for every fixed r ≥ 0 r\geq0 r ≥ 0
sup t ∈ J ∑ ν ∈ N ∣ ∂ t r F ν ( t ) ∣ 2 ≪ r E Λ 2 r . (71) \sup_{t\in J}\sum_{\nu\in\mathcal{N}}|\partial_t^r F_\nu(t)|^2\ll_r E\Lambda^{2r}. \tag*{(71)} t ∈ J sup ν ∈ N ∑ ∣ ∂ t r F ν ( t ) ∣ 2 ≪ r E Λ 2 r . (71) Then F F F
F ν ( t ) = ∑ k ∈ Z c ν , k e ( k ( t − t J ) L ) , ∑ k ∈ Z ∑ ν ∈ N ∣ c ν , k ∣ 2 ≪ E . (72) F_\nu(t)=\sum_{k\in\mathbb{Z}}c_{\nu,k}e\left(\frac{k(t-t_J)}{L}\right),\qquad\sum_{k\in\mathbb{Z}}\sum_{\nu\in\mathcal{N}}|c_{\nu,k}|^2\ll E. \tag*{(72)} F ν ( t ) = k ∈ Z ∑ c ν , k e ( L k ( t − t J ) ) , k ∈ Z ∑ ν ∈ N ∑ ∣ c ν , k ∣ 2 ≪ E . (72) Moreover, for every integer R ≥ 1 R \ge1 R ≥ 1 K ≥ 1 K \ge1 K ≥ 1
∑ ∣ k ∣ > K ∑ ν ∈ N ∣ c ν , k ∣ 2 ≪ R E K − 2 R ( 1 + H Λ ) 2 R , (73) \sum_{|k|>K}\sum_{\nu\in\mathcal{N}} |c_{\nu,k}|^2 \ll_R E K^{-2R}(1+H\Lambda)^{2R}, \tag*{(73)} ∣ k ∣ > K ∑ ν ∈ N ∑ ∣ c ν , k ∣ 2 ≪ R E K − 2 R ( 1 + H Λ ) 2 R , (73) sup t ∈ J ∑ ν ∈ N ∣ ∑ ∣ k ∣ > K c ν , k e ( k ( t − t J ) L ) ∣ 2 ≪ R E K 1 − 2 R ( 1 + H Λ ) 2 R . (74) \sup_{t\in J}\sum_{\nu\in\mathcal{N}}\left|\sum_{|k|>K}c_{\nu,k}e\left(\frac{k(t-t_J)}{L}\right)\right|^2 \ll_R E K^{1-2R}(1+H\Lambda)^{2R}. \tag*{(74)} t ∈ J sup ν ∈ N ∑ ∣ k ∣ > K ∑ c ν , k e ( L k ( t − t J ) ) 2 ≪ R E K 1 − 2 R ( 1 + H Λ ) 2 R . (74) Proof. Write H = ℓ 2 ( N ) \mathcal{H}=\ell^2(\mathcal{N}) H = ℓ 2 ( N ) N \mathcal{N} N H \mathcal{H} H ν ∈ N \nu\in\mathcal{N} ν ∈ N F F F J J J L L L C ∞ ( R / L Z ; H ) C^\infty(\mathbb{R}/L\mathbb{Z};\mathcal{H}) C ∞ ( R / L Z ; H )
Step 1: Fourier coefficients and exact expansion. For k ∈ Z k\in\mathbb{Z} k ∈ Z H \mathcal{H} H
c k = ( c ν , k ) ν ∈ N : = 1 L ∫ J F ( t ) e ( − k ( t − t J ) L ) d t . (75) c_k=(c_{\nu,k})_{\nu\in\mathcal{N}}:=\frac{1}{L}\int_J F(t)e\left(-\frac{k(t-t_J)}{L}\right)\,dt. \tag*{(75)} c k = ( c ν , k ) ν ∈ N := L 1 ∫ J F ( t ) e ( − L k ( t − t J ) ) d t . (75) Since the periodic extension of F F F J J J
F ( t ) = ∑ k ∈ Z c k e ( k ( t − t J ) L ) F(t)=\sum_{k\in\mathbb{Z}}c_ke\left(\frac{k(t-t_J)}{L}\right) F ( t ) = k ∈ Z ∑ c k e ( L k ( t − t J ) ) as an identity in H \mathcal{H} H ν \nu ν
Step 2: zeroth-order Parseval identity. Parseval on the circle of length L L L d t / L dt/L d t / L
∑ k ∈ Z ∥ c k ∥ H 2 = 1 L ∫ J ∥ F ( t ) ∥ H 2 d t = 1 L ∫ J ∑ ν ∈ N ∣ F ν ( t ) ∣ 2 d t ≪ E . \begin{aligned}
\sum_{k\in\mathbb{Z}}\|c_k\|_{\mathcal{H}}^2=\frac{1}{L}\int_J\|F(t)\|_{\mathcal{H}}^2\,dt \\
&=\frac{1}{L}\int_J\sum_{\nu\in\mathcal{N}}|F_\nu(t)|^2\,dt\ll E.
\end{aligned} k ∈ Z ∑ ∥ c k ∥ H 2 = L 1 ∫ J ∥ F ( t ) ∥ H 2 d t = L 1 ∫ J ν ∈ N ∑ ∣ F ν ( t ) ∣ 2 d t ≪ E . The last bound is (5.1) with r = 0 r=0 r = 0 ∥ c k ∥ H 2 = ∑ ν ∈ N ∣ c ν , k ∣ 2 \|c_k\|_{\mathcal{H}}^2=\sum_{\nu\in\mathcal{N}}|c_{\nu,k}|^2 ∥ c k ∥ H 2 = ∑ ν ∈ N ∣ c ν , k ∣ 2
Step 3: weighted Fourier energy. Integrate (5.5) by parts R R R F F F k ∈ Z k\in\mathbb{Z} k ∈ Z
( 2 π i k L ) R c k = 1 L ∫ J ∂ t R F ( t ) e ( − k ( t − t J ) L ) d t . \left(\frac{2\pi i k}{L}\right)^R c_k=\frac{1}{L}\int_J\partial_t^R F(t)e\left(-\frac{k(t-t_J)}{L}\right)\,dt. ( L 2 π ik ) R c k = L 1 ∫ J ∂ t R F ( t ) e ( − L k ( t − t J ) ) d t . Equivalently, the k k k ∂ t R F \partial_t^R F ∂ t R F ( 2 π i k / L ) R c k (2\pi i k/L)^R c_k ( 2 π ik / L ) R c k
∑ k ∈ Z ( 2 π ∣ k ∣ L ) 2 R ∥ c k ∥ H 2 = 1 L ∫ J ∥ ∂ t R F ( t ) ∥ H 2 d t \sum_{k\in\mathbb{Z}}\left(\frac{2\pi|k|}{L}\right)^{2R}\|c_k\|_{\mathcal{H}}^2=\frac{1}{L}\int_J\|\partial_t^R F(t)\|_{\mathcal{H}}^2\,dt k ∈ Z ∑ ( L 2 π ∣ k ∣ ) 2 R ∥ c k ∥ H 2 = L 1 ∫ J ∥ ∂ t R F ( t ) ∥ H 2 d t ≪ R E Λ 2 R . \ll_R E\Lambda^{2R}. ≪ R E Λ 2 R . Consequently,
∑ k ∈ Z ∣ k ∣ 2 R ∥ c k ∥ H 2 ≪ R ( L 2 π ) 2 R E Λ 2 R ≪ R E ( 1 + H Λ ) 2 R . (76) \sum_{k\in\mathbb{Z}}|k|^{2R}\|c_k\|_{\mathcal{H}}^2\ll_R\left(\frac{L}{2\pi}\right)^{2R}E\Lambda^{2R}\ll_R E(1+H\Lambda)^{2R}. \tag*{(76)} k ∈ Z ∑ ∣ k ∣ 2 R ∥ c k ∥ H 2 ≪ R ( 2 π L ) 2 R E Λ 2 R ≪ R E ( 1 + H Λ ) 2 R . (76) because L ≍ H L \asymp H L ≍ H 1 + H Λ 1+H\Lambda 1 + H Λ H Λ < 1 H\Lambda<1 H Λ < 1 J J J Λ \Lambda Λ Λ + H − 1 \Lambda+H^{-1} Λ + H − 1 1 + H Λ 1+H\Lambda 1 + H Λ
Step 4: the L 2 L^2 L 2 For ∣ k ∣ > K |k|>K ∣ k ∣ > K 1 ≤ K − 2 R ∣ k ∣ 2 R 1\le K^{-2R}|k|^{2R} 1 ≤ K − 2 R ∣ k ∣ 2 R
∑ ∣ k ∣ > K ∑ ν ∈ N ∣ c ν , k ∣ 2 = ∑ ∣ k ∣ > K ∥ c k ∥ H 2 ≤ K − 2 R ∑ k ∈ Z ∣ k ∣ 2 R ∥ c k ∥ H 2 \begin{aligned}
\sum_{|k|>K}\sum_{\nu\in\mathcal{N}}|c_{\nu,k}|^2=\sum_{|k|>K}\|c_k\|_{\mathcal{H}}^2 \\
&\le K^{-2R}\sum_{k\in\mathbb{Z}}|k|^{2R}\|c_k\|_{\mathcal{H}}^2
\end{aligned} ∣ k ∣ > K ∑ ν ∈ N ∑ ∣ c ν , k ∣ 2 = ∣ k ∣ > K ∑ ∥ c k ∥ H 2 ≤ K − 2 R k ∈ Z ∑ ∣ k ∣ 2 R ∥ c k ∥ H 2 ≪ R E K − 2 R ( 1 + H Λ ) 2 R . \ll_R EK^{-2R}(1+H\Lambda)^{2R}. ≪ R E K − 2 R ( 1 + H Λ ) 2 R . by (5.6). This proves (5.3).
Step 5: a tail uniform in the spectral parameter. Set
R K ( t ) : = ∑ ∣ k ∣ > K c k e ( k ( t − t J ) L ) ∈ H . \mathcal{R}_K(t):=\sum_{|k|>K}c_k e\left(\frac{k(t-t_J)}{L}\right)\in\mathcal{H}. R K ( t ) := ∣ k ∣ > K ∑ c k e ( L k ( t − t J ) ) ∈ H . For each fixed t ∈ J t\in J t ∈ J ∣ k ∣ − R |k|^{-R} ∣ k ∣ − R ∣ k ∣ R ∥ c k ∥ H |k|^R\|c_k\|_{\mathcal{H}} ∣ k ∣ R ∥ c k ∥ H
∥ R K ( t ) ∥ H 2 ≤ ( ∑ ∣ k ∣ > K ∣ k ∣ − R ∣ k ∣ R ∥ c k ∥ H ) 2 ≤ ( ∑ ∣ k ∣ > K ∣ k ∣ − 2 R ) ( ∑ ∣ k ∣ > K ∣ k ∣ 2 R ∥ c k ∥ H 2 ) . (77) \begin{aligned}
\|\mathcal{R}_K(t)\|_{\mathcal{H}}^2\le\left(\sum_{|k|>K}|k|^{-R}|k|^R\|c_k\|_{\mathcal{H}}\right)^2 \\
&\le\left(\sum_{|k|>K}|k|^{-2R}\right)\left(\sum_{|k|>K}|k|^{2R}\|c_k\|_{\mathcal{H}}^2\right). \tag*{(77)}
\end{aligned} ∥ R K ( t ) ∥ H 2 ≤ ∣ k ∣ > K ∑ ∣ k ∣ − R ∣ k ∣ R ∥ c k ∥ H 2 ≤ ∣ k ∣ > K ∑ ∣ k ∣ − 2 R ∣ k ∣ > K ∑ ∣ k ∣ 2 R ∥ c k ∥ H 2 . (77) For R ≥ 1 R\ge1 R ≥ 1
∑ ∣ k ∣ > K ∣ k ∣ − 2 R ≪ R K 1 − 2 R . \sum_{|k|>K}|k|^{-2R}\ll_R K^{1-2R}. ∣ k ∣ > K ∑ ∣ k ∣ − 2 R ≪ R K 1 − 2 R . Combining this estimate with (5.6) in (5.7), we obtain
sup t ∈ J ∥ R K ( t ) ∥ H 2 ≪ R E K 1 − 2 R ( 1 + H Λ ) 2 R . \sup_{t\in J}\|\mathcal{R}_K(t)\|_{\mathcal{H}}^2\ll_R EK^{1-2R}(1+H\Lambda)^{2R}. t ∈ J sup ∥ R K ( t ) ∥ H 2 ≪ R E K 1 − 2 R ( 1 + H Λ ) 2 R . Finally,
∥ R K ( t ) ∥ H 2 = ∑ ν ∈ N ∣ ∑ ∣ k ∣ > K c ν , k e ( k ( t − t J ) L ) ∣ 2 , \|\mathcal{R}_K(t)\|_{\mathcal{H}}^2=\sum_{\nu\in\mathcal{N}}\left|\sum_{|k|>K}c_{\nu,k}e\left(\frac{k(t-t_J)}{L}\right)\right|^2, ∥ R K ( t ) ∥ H 2 = ν ∈ N ∑ ∣ k ∣ > K ∑ c ν , k e ( L k ( t − t J ) ) 2 , which is (5.4).
Proof of the main theorem
We first prove the following weighted second-moment estimate for the off-diagonal term.
Theorem 6.1. Let S j ( ψ ) S_j(\psi) S j ( ψ ) T ≥ 2 T\ge2 T ≥ 2 1 ≤ U ≤ T 1\le U\le T 1 ≤ U ≤ T ϵ > 0 \epsilon>0 ϵ > 0
∑ j e v e n T < t j ≤ T + U L j ∣ S j ( ψ ) ∣ 2 ≪ ψ , ϵ T 1 + ϵ . (78) \sum_{\substack{j\ \mathrm{even}\\ T<t_j\le T+U}}L_j|S_j(\psi)|^2\ll_{\psi,\epsilon}T^{1+\epsilon}. \tag*{(78)} j even T < t j ≤ T + U ∑ L j ∣ S j ( ψ ) ∣ 2 ≪ ψ , ϵ T 1 + ϵ . (78) Proof. Recall (12) :
S j ( ψ ) = 1 L j ∑ d , a , b ≥ 1 λ j ( a ( a + b ) ) B d a , d ( a + b ) ( t j ) . S_j(\psi)=\frac{1}{L_j}\sum_{d,a,b\geq1}\lambda_j\bigl(a(a+b)\bigr)\mathcal{B}_{da,d(a+b)}(t_j). S j ( ψ ) = L j 1 d , a , b ≥ 1 ∑ λ j ( a ( a + b ) ) B d a , d ( a + b ) ( t j ) . First, we divide the threefold sum into dyadic blocks and estimate the coefficients of the resulting Hecke polynomials. We decompose the retained triples, subject to the Fourier-index and shift truncations from § 3 , into dyadic ranges
d ≍ D , a ≍ X , b ≍ Y . d\asymp D,\qquad a\asymp X,\qquad b\asymp Y. d ≍ D , a ≍ X , b ≍ Y . Call the resulting block B = ( D , X , Y ) \mathcal{B}=(D,X,Y) B = ( D , X , Y )
d a , d ( a + b ) ≤ C ψ T , d 2 b ( 2 a + b ) ≪ T 1 + β . da,\ d(a+b)\leq C_\psi T,\qquad d^2b(2a+b)\ll T^{1+\beta}. d a , d ( a + b ) ≤ C ψ T , d 2 b ( 2 a + b ) ≪ T 1 + β . For each Hecke index A A A T A ( B ) \mathcal{T}_A(\mathcal{B}) T A ( B ) a ( a + b ) = A a(a+b)=A a ( a + b ) = A
C A B ( t ) = ∑ ( d , a , b ) ∈ T A ( B ) B d a , d ( a + b ) ( t ) . \mathcal{C}^{\mathcal{B}}_A(t)=\sum_{(d,a,b)\in\mathcal{T}_A(\mathcal{B})}\mathcal{B}_{da,d(a+b)}(t). C A B ( t ) = ( d , a , b ) ∈ T A ( B ) ∑ B d a , d ( a + b ) ( t ) . The Hecke index has length
A ≍ N : = X ( X + Y ) . A\asymp N:=X(X+Y). A ≍ N := X ( X + Y ) . For every fixed integer r ≥ 0 r\geq0 r ≥ 0
∑ A ∣ ∂ t r C A B ( t ) ∣ 2 ≪ r , ε E B Λ ( Y / X , T ) 2 r , E B = T ε D 2 X Y T 2 . (79) \sum_A\left|\partial_t^r\mathcal{C}^{\mathcal{B}}_A(t)\right|^2\ll_{r,\varepsilon} E_{\mathcal{B}}\Lambda(Y/X,T)^{2r},\qquad E_{\mathcal{B}}=T^\varepsilon\frac{D^2XY}{T^2}. \tag*{(79)} A ∑ ∂ t r C A B ( t ) 2 ≪ r , ε E B Λ ( Y / X , T ) 2 r , E B = T ε T 2 D 2 X Y . (79) We consider only nonempty blocks left after the truncation in § 3 . To keep the bookkeeping transparent, apply Lemma 4.7 and the divisor bound with ε / 10 \varepsilon/10 ε /10 ε \varepsilon ε T ε / 10 T^{\varepsilon/10} T ε /10 T ε T^\varepsilon T ε
Then
∂ t r C A B ( t ) = ∑ ( d , a , b ) ∈ T A ( B ) ∂ t r B d a , d ( a + b ) ( t ) . (80) \partial_t^r\mathcal{C}^{\mathcal{B}}_A(t)=\sum_{(d,a,b)\in\mathcal{T}_A(\mathcal{B})}\partial_t^r\mathcal{B}_{da,d(a+b)}(t). \tag*{(80)} ∂ t r C A B ( t ) = ( d , a , b ) ∈ T A ( B ) ∑ ∂ t r B d a , d ( a + b ) ( t ) . (80) If ( d , a , b ) ∈ T A ( B ) (d,a,b)\in\mathcal{T}_A(\mathcal{B}) ( d , a , b ) ∈ T A ( B )
a ∣ A , b = A / a − a . a\mid A,\qquad b=A/a-a. a ∣ A , b = A / a − a . Thus b b b a a a O ( D ) O(D) O ( D ) d d d
# T A ( B ) ≪ D τ ( A ) . \#\mathcal{T}_A(\mathcal{B})\ll D\tau(A). # T A ( B ) ≪ D τ ( A ) . The truncation in Lemma 2.2 gives d a , d ( a + b ) ≪ ψ T da,d(a+b)\ll_\psi T d a , d ( a + b ) ≪ ψ T A = a ( a + b ) ≪ ψ T 2 A=a(a+b)\ll_\psi T^2 A = a ( a + b ) ≪ ψ T 2 τ ( A ) ≪ ε A ε / 20 \tau(A)\ll_\varepsilon A^{\varepsilon/20} τ ( A ) ≪ ε A ε /20
# T A ( B ) ≪ ψ , ε D T ε / 10 . (81) \#\mathcal{T}_A(\mathcal{B})\ll_{\psi,\varepsilon}DT^{\varepsilon/10}. \tag*{(81)} # T A ( B ) ≪ ψ , ε D T ε /10 . (81) For a ≍ X a\asymp X a ≍ X b ≍ Y b\asymp Y b ≍ Y
b a ≍ Y X . \frac{b}{a}\asymp\frac{Y}{X}. a b ≍ X Y . The piecewise definition (4.43) implies
Λ ( b / a , T ) ≪ Λ ( Y / X , T ) . \Lambda(b/a,T)\ll\Lambda(Y/X,T). Λ ( b / a , T ) ≪ Λ ( Y / X , T ) . Indeed, if Y / X ≤ 1 / 4 Y/X \le1/4 Y / X ≤ 1/4 b / a ≤ 1 b/a \le1 b / a ≤ 1 b / a ≍ Y / X b/a \asymp Y/X b / a ≍ Y / X Y / X ≥ 4 Y/X \ge4 Y / X ≥ 4 b / a > 1 b/a > 1 b / a > 1 log ( 2 + b / a ) ≍ log ( 2 + Y / X ) \log(2+b/a) \asymp\log(2+Y/X) log ( 2 + b / a ) ≍ log ( 2 + Y / X ) 1 / 4 ≤ Y / X ≤ 4 1/4 \le Y/X \le4 1/4 ≤ Y / X ≤ 4
∣ ∂ t r B d a , d ( a + b ) ( t ) ∣ ≪ r , ψ , ε T − 1 ( log T ) Λ ( Y / X , T ) r . (82) \left|\partial_t^r \mathcal{B}_{da,d(a+b)}(t)\right| \ll_{r,\psi,\varepsilon} T^{-1}(\log T)\Lambda(Y/X,T)^r. \tag*{(82)} ∂ t r B d a , d ( a + b ) ( t ) ≪ r , ψ , ε T − 1 ( log T ) Λ ( Y / X , T ) r . (82) Equations (80) and (81) , together with Cauchy’s inequality, give
∣ ∂ t r C A B ( t ) ∣ 2 ≤ # A ( B ) ∑ ( d , a , b ) ∈ A ( B ) ∣ ∂ t r B d a , d ( a + b ) ( t ) ∣ 2 ≪ ψ , ε D T ε / 10 ∑ ( d , a , b ) ∈ A ( B ) ∣ ∂ t r B d a , d ( a + b ) ( t ) ∣ 2 . \left|\partial_t^r \mathcal{C}_{A}^{\mathcal{B}}(t)\right|^2 \le\#\mathcal{A}(\mathcal{B}) \sum_{(d,a,b)\in\mathcal{A}(\mathcal{B})}\left|\partial_t^r \mathcal{B}_{da,d(a+b)}(t)\right|^2
\ll_{\psi,\varepsilon} DT^{\varepsilon/10}\sum_{(d,a,b)\in\mathcal{A}(\mathcal{B})}\left|\partial_t^r \mathcal{B}_{da,d(a+b)}(t)\right|^2. ∂ t r C A B ( t ) 2 ≤ # A ( B ) ( d , a , b ) ∈ A ( B ) ∑ ∂ t r B d a , d ( a + b ) ( t ) 2 ≪ ψ , ε D T ε /10 ( d , a , b ) ∈ A ( B ) ∑ ∂ t r B d a , d ( a + b ) ( t ) 2 . When this is summed over A A A ( d , a , b ) (d,a,b) ( d , a , b ) A = a ( a + b ) A=a(a+b) A = a ( a + b )
∑ A ∣ ∂ t r C A B ( t ) ∣ 2 ≪ ψ , ε D T ε / 10 ∑ d ≍ D a ≍ X b ≍ Y ∣ ∂ t r B d a , d ( a + b ) ( t ) ∣ 2 \sum_A\left|\partial_t^r\mathcal{C}_{A}^{\mathcal{B}}(t)\right|^2\ll_{\psi,\varepsilon}DT^{\varepsilon/10}\sum_{\substack{d\asymp D\\a\asymp X\\b\asymp Y}}\left|\partial_t^r\mathcal{B}_{da,d(a+b)}(t)\right|^2 A ∑ ∂ t r C A B ( t ) 2 ≪ ψ , ε D T ε /10 d ≍ D a ≍ X b ≍ Y ∑ ∂ t r B d a , d ( a + b ) ( t ) 2 ≪ r , ψ , ε D T ε / 10 ( D X Y ) T − 2 + ε / 5 Λ ( Y / X , T ) 2 r . \ll_{r,\psi,\varepsilon}DT^{\varepsilon/10}(DXY)T^{-2+\varepsilon/5}\Lambda(Y/X,T)^{2r}. ≪ r , ψ , ε D T ε /10 ( D X Y ) T − 2 + ε /5 Λ ( Y / X , T ) 2 r . The dyadic box contains O ( D X Y ) O(DXY) O ( D X Y ) (82) . Absorbing the finitely many ε / 10 \varepsilon/10 ε /10 T ε T^\varepsilon T ε
∑ A ∣ ∂ t r C A B ( t ) ∣ 2 ≪ r , ψ , ε T ε D 2 X Y T 2 Λ ( Y / X , T ) 2 r , \sum_A\left|\partial_t^r\mathcal{C}_{A}^{\mathcal{B}}(t)\right|^2\ll_{r,\psi,\varepsilon}T^\varepsilon\frac{D^2XY}{T^2}\Lambda(Y/X,T)^{2r}, A ∑ ∂ t r C A B ( t ) 2 ≪ r , ψ , ε T ε T 2 D 2 X Y Λ ( Y / X , T ) 2 r , which proves (79) .
Now the truncation (25) becomes
D 2 Y ( 2 X + Y ) ≪ T 1 + ε . D^2Y(2X+Y)\ll T^{1+\varepsilon}. D 2 Y ( 2 X + Y ) ≪ T 1 + ε . Next, we fix a non-empty block B = ( D , X , Y ) \mathcal{B}=(D,X,Y) B = ( D , X , Y )
N = X ( X + Y ) , ρ = Y / X , Λ = Λ ( ρ , T ) . N=X(X+Y),\qquad\rho=Y/X,\qquad\Lambda=\Lambda(\rho,T). N = X ( X + Y ) , ρ = Y / X , Λ = Λ ( ρ , T ) . Define
Q B ( T , U ) = ∑ T < t j ≤ T + U j e v e n 1 L j ∣ ∑ A ≍ N λ j ( A ) C A B ( t j ) ∣ 2 . \mathcal{Q}_{\mathcal{B}}(T,U)=\sum_{\substack{T<t_j\le T+U\\j\ {\rm even}}}\frac{1}{L_j}\left|\sum_{A\asymp N}\lambda_j(A)\mathcal{C}_{A}^{\mathcal{B}}(t_j)\right|^2. Q B ( T , U ) = T < t j ≤ T + U j even ∑ L j 1 A ≍ N ∑ λ j ( A ) C A B ( t j ) 2 . Let
H = { min ( T 1 − β , X / Y ) , Y ≤ X , 1 , Y > X , h = min ( U , H ) . H=\begin{cases}
\min(T^{1-\beta},X/Y),&Y\le X,\\
1,&Y>X,
\end{cases}
\qquad h=\min(U,H). H = { min ( T 1 − β , X / Y ) , 1 , Y ≤ X , Y > X , h = min ( U , H ) . We prove
Q B ( T , U ) ≪ T O ( β ) ( T U + N max ( 1 , U H ) ) D 2 X Y T 2 . \mathcal{Q}_{\mathcal{B}}(T,U)\ll T^{O(\beta)}\left(TU+N\max\left(1,\frac{U}{H}\right)\right)\frac{D^2XY}{T^2}. Q B ( T , U ) ≪ T O ( β ) ( T U + N max ( 1 , H U ) ) T 2 D 2 X Y . For this fixed block, choose a smooth square partition of unity { η J } J ∈ J \{\eta_J\}_{J\in\mathcal{J}} { η J } J ∈ J [ T , T + U ] [T,T+U] [ T , T + U ] J = J ( B ) \mathcal{J}=\mathcal{J}(\mathcal{B}) J = J ( B )
∑ J ∈ J η J ( t ) 2 = 1 , sup η J ∈ J , ∣ J ∣ ≍ h , η J ( r ) ( t ) ≪ r h − r . \sum_{J\in\mathcal{J}}\eta_J(t)^2=1,\qquad\sup\eta_J\in\mathcal{J},\qquad|\mathcal{J}|\asymp h,\qquad\eta_J^{(r)}(t)\ll_r h^{-r}. J ∈ J ∑ η J ( t ) 2 = 1 , sup η J ∈ J , ∣ J ∣ ≍ h , η J ( r ) ( t ) ≪ r h − r . The overlap is bounded and the number of intervals is
O ( U / h ) = O ( max ( 1 , U H ) ) . O(U/h)=O\left(\max\left(1,\frac{U}{H}\right)\right). O ( U / h ) = O ( max ( 1 , H U ) ) . Put F A , J ( t ) = η J ( t ) C A B ( t ) F_{A,J}(t)=\eta_J(t)\mathcal{C}^{\mathcal{B}}_A(t) F A , J ( t ) = η J ( t ) C A B ( t )
Q B ( T , U ) = ∑ J ∈ J ∑ j e v e n T < t j ≤ T + U 1 L j ∣ ∑ A λ j ( A ) F A , J ( t j ) ∣ 2 ≤ ∑ J ∈ J ∑ j e v e n 1 L j ∣ ∑ A λ j ( A ) F A , J ( t j ) ∣ 2 . \begin{aligned}
\mathcal{Q}_{\mathcal{B}}(T,U)=\sum_{J\in\mathcal{J}}\sum_{\substack{j\ {\rm even}\\ T<t_j\leq T+U}}\frac{1}{L_j}\left|\sum_A\lambda_j(A)F_{A,J}(t_j)\right|^2 \\
&\leq\sum_{J\in\mathcal{J}}\sum_{j\ {\rm even}}\frac{1}{L_j}\left|\sum_A\lambda_j(A)F_{A,J}(t_j)\right|^2.
\end{aligned} Q B ( T , U ) = J ∈ J ∑ j even T < t j ≤ T + U ∑ L j 1 A ∑ λ j ( A ) F A , J ( t j ) 2 ≤ J ∈ J ∑ j even ∑ L j 1 A ∑ λ j ( A ) F A , J ( t j ) 2 . Leibniz’ rule and (79) give
∑ A ∣ ∂ t r F A , J ( t ) ∣ 2 ≪ r , ψ , β E B ( Λ + h − 1 ) 2 r , E B = T O ( β ) D 2 X Y T 2 . \sum_A|\partial_t^rF_{A,J}(t)|^2\ll_{r,\psi,\beta}\mathcal{E}_{\mathcal{B}}(\Lambda+h^{-1})^{2r},\qquad\mathcal{E}_{\mathcal{B}}=T^{O(\beta)}\frac{D^2XY}{T^2}. A ∑ ∣ ∂ t r F A , J ( t ) ∣ 2 ≪ r , ψ , β E B ( Λ + h − 1 ) 2 r , E B = T O ( β ) T 2 D 2 X Y . Since h ≤ H h\leq H h ≤ H
h Λ ≪ { 1 , Y ≤ X , 1 + log T , Y > X . h\Lambda\ll\begin{cases}1,&Y\leq X,\\1+\log T,&Y>X.\end{cases} h Λ ≪ { 1 , 1 + log T , Y ≤ X , Y > X . Lemma 5.1 , with derivative factor Λ + h − 1 \Lambda+h^{-1} Λ + h − 1 K = T β K=T^\beta K = T β β \beta β
F A , J ( t ) = ∑ ∣ k ∣ ≤ K c A , k , J e ( k ( t − t J ) L J ) + R A , J ( t ) , L J ≪ h , F_{A,J}(t)=\sum_{|k|\leq K}c_{A,k,J}e\left(\frac{k(t-t_J)}{L_J}\right)+R_{A,J}(t),\qquad L_J\ll h, F A , J ( t ) = ∣ k ∣ ≤ K ∑ c A , k , J e ( L J k ( t − t J ) ) + R A , J ( t ) , L J ≪ h , with
∑ k , A ∣ c A , k , J ∣ 2 ≪ E B , \sum_{k,A}|c_{A,k,J}|^2\ll\mathcal{E}_{\mathcal{B}}, k , A ∑ ∣ c A , k , J ∣ 2 ≪ E B , sup t ∈ J ∑ A ∣ R A , J ( t ) ∣ 2 ≪ T − 200 E B . \sup_{t\in J}\sum_A|R_{A,J}(t)|^2\ll T^{-200}\mathcal{E}_{\mathcal{B}}. t ∈ J sup A ∑ ∣ R A , J ( t ) ∣ 2 ≪ T − 200 E B . For each fixed k k k c A , k , J c_{A,k,J} c A , k , J t j t_j t j k k k Lemma 2.3 yield
∑ j e v e n t j ∈ J 1 L j ∣ ∑ ∣ k ∣ ≤ K e ( k ( t j − t J ) L J ) ∑ A c A , k , J λ j ( A ) ∣ 2 ≪ T O ( β ) ( T h + N ) E B . \sum_{\substack{j\ {\rm even}\\t_j\in J}}\frac{1}{L_j}\left|\sum_{|k|\leq K}e\left(\frac{k(t_j-t_J)}{L_J}\right)\sum_Ac_{A,k,J}\lambda_j(A)\right|^2\ll T^{O(\beta)}(Th+N)\mathcal{E}_{\mathcal{B}}. j even t j ∈ J ∑ L j 1 ∣ k ∣ ≤ K ∑ e ( L J k ( t j − t J ) ) A ∑ c A , k , J λ j ( A ) 2 ≪ T O ( β ) ( T h + N ) E B . The uniform tail, the elementary Hecke bound, and Weyl’s upper bound on [ T , 2 T ] [T,2T] [ T , 2 T ] O ( T − 190 ) O(T^{-190}) O ( T − 190 )
Q B ( T , U ) ≪ T O ( β ) U h ( T h + N ) E B + O ( T − 190 ) ≪ T O ( β ) ( T U + N max ( 1 , U H ) ) D 2 X Y T 2 . \begin{aligned}
\mathcal{Q}_{\mathcal{B}}(T,U)\ll T^{O(\beta)}\frac{U}{h}(Th+N)\mathcal{E}_{\mathcal{B}}+O(T^{-190}) \\
&\ll T^{O(\beta)}\left(TU+N\max\left(1,\frac{U}{H}\right)\right)\frac{D^2XY}{T^2}.
\end{aligned} Q B ( T , U ) ≪ T O ( β ) h U ( T h + N ) E B + O ( T − 190 ) ≪ T O ( β ) ( T U + N max ( 1 , H U ) ) T 2 D 2 X Y . Note that the block constraint gives
T U D 2 X Y T 2 ≪ T O ( β ) U , TU\frac{D^2XY}{T^2}\ll T^{O(\beta)}U, T U T 2 D 2 X Y ≪ T O ( β ) U , and
N D 2 X Y T 2 ≪ T O ( β ) X 2 T . N\frac{D^2XY}{T^2}\ll T^{O(\beta)}\frac{X^2}{T}. N T 2 D 2 X Y ≪ T O ( β ) T X 2 . Thus it remains to estimate
X 2 T max ( 1 , U H ) . \frac{X^2}{T}\max\left(1,\frac{U}{H}\right). T X 2 max ( 1 , H U ) . If Y > X Y>X Y > X H = 1 H=1 H = 1 X 2 < Y 2 ≪ T 1 + O ( β ) X^2<Y^2\ll T^{1+O(\beta)} X 2 < Y 2 ≪ T 1 + O ( β ) O ( T O ( β ) U ) O(T^{O(\beta)}U) O ( T O ( β ) U ) Y ≤ X Y\le X Y ≤ X ρ = Y / X \rho=Y/X ρ = Y / X T − 1 + β ≤ ρ ≤ 1 T^{-1+\beta}\le\rho\le1 T − 1 + β ≤ ρ ≤ 1 H = ρ − 1 H=\rho^{-1} H = ρ − 1 ρ X 2 ≪ T 1 + O ( β ) \rho X^2\ll T^{1+O(\beta)} ρ X 2 ≪ T 1 + O ( β ) U ρ ≥ 1 U\rho\ge1 U ρ ≥ 1 O ( T O ( β ) U ) O(T^{O(\beta)}U) O ( T O ( β ) U ) U ρ < 1 U\rho<1 U ρ < 1 O ( T O ( β ) ρ − 1 ) = O ( T 1 + O ( β ) ) O(T^{O(\beta)}\rho^{-1})=O(T^{1+O(\beta)}) O ( T O ( β ) ρ − 1 ) = O ( T 1 + O ( β ) ) 0 < ρ < T − 1 + β 0<\rho<T^{-1+\beta} 0 < ρ < T − 1 + β H = T 1 − β H=T^{1-\beta} H = T 1 − β d ≥ 1 d\ge1 d ≥ 1 d X ≪ ψ T dX\ll_{\psi}T d X ≪ ψ T X ≪ ψ T X\ll_{\psi}T X ≪ ψ T
X 2 T max ( 1 , U T 1 − β ) ≪ ψ T 1 + β . \frac{X^2}{T}\max\left(1,\frac{U}{T^{1-\beta}}\right)\ll_{\psi}T^{1+\beta}. T X 2 max ( 1 , T 1 − β U ) ≪ ψ T 1 + β . In particular, the range U ρ < 1 U\rho<1 U ρ < 1 T 1 + O ( β ) T^{1+O(\beta)} T 1 + O ( β ) T O ( β ) U T^{O(\beta)}U T O ( β ) U 1 ≤ U ≤ T 1\le U\le T 1 ≤ U ≤ T
Q B ( T , U ) ≪ T 1 + O ( β ) . Q_{\mathcal B}(T,U)\ll T^{1+O(\beta)}. Q B ( T , U ) ≪ T 1 + O ( β ) . For each block B \mathcal B B
P j , B = ∑ A λ j ( A ) C A B ( t j ) . P_{j,\mathcal B}=\sum_A\lambda_j(A)C_A^{\mathcal B}(t_j). P j , B = A ∑ λ j ( A ) C A B ( t j ) . Then
S j ( ψ ) = 1 L j ∑ B P j , B + E j f a r + E j s h i f t . S_j(\psi)=\frac{1}{L_j}\sum_{\mathcal B}P_{j,\mathcal B}+\mathcal E_j^{\mathrm{far}}+\mathcal E_j^{\mathrm{shift}}. S j ( ψ ) = L j 1 B ∑ P j , B + E j far + E j shift . There are O ψ ( ( log T ) 3 ) O_{\psi}((\log T)^3) O ψ (( log T ) 3 )
∑ j e v e n T < t j ≤ T + U 1 L j ∣ ∑ B P j , B ∣ 2 ≪ T 1 + ε . (83) \sum_{\substack{j\ \mathrm{even}\\T<t_j\le T+U}}
\frac{1}{L_j}\left|\sum_{\mathcal B}P_{j,\mathcal B}\right|^2\ll T^{1+\varepsilon}. \tag*{(83)} j even T < t j ≤ T + U ∑ L j 1 B ∑ P j , B 2 ≪ T 1 + ε . (83) By Lemmas 2.2 and 3.5 , together with the standard bound L j ≪ ε T ε L_j\ll_{\varepsilon}T^{\varepsilon} L j ≪ ε T ε
∑ j e v e n T < t j ≤ T + U L j ∣ E j f a r + E j s h i f t ∣ 2 ≪ A , ψ , ε T − A , \sum_{\substack{j\ \mathrm{even}\\T<t_j\le T+U}}
L_j\left|\mathcal E_j^{\mathrm{far}}+\mathcal E_j^{\mathrm{shift}}\right|^2\ll_{A,\psi,\varepsilon}T^{-A}, j even T < t j ≤ T + U ∑ L j E j far + E j shift 2 ≪ A , ψ , ε T − A , after choosing the decay exponent in the two tail bounds sufficiently large and using the spectral counting bound. Together with (83) , this proves (78) . □
Proof of Theorem 1.2. Since μ j ( ψ ) − M ( ψ ) = D j ( ψ ) − M ( ψ ) + S j ( ψ ) \mu_j(\psi)-M(\psi)=D_j(\psi)-M(\psi)+S_j(\psi) μ j ( ψ ) − M ( ψ ) = D j ( ψ ) − M ( ψ ) + S j ( ψ )
∣ μ j ( ψ ) − M ( ψ ) ∣ 2 ≤ 2 ∣ D j ( ψ ) − M ( ψ ) ∣ 2 + 2 ∣ S j ( ψ ) ∣ 2 . |\mu_j(\psi)-M(\psi)|^2 \le2|D_j(\psi)-M(\psi)|^2+2|S_j(\psi)|^2. ∣ μ j ( ψ ) − M ( ψ ) ∣ 2 ≤ 2∣ D j ( ψ ) − M ( ψ ) ∣ 2 + 2∣ S j ( ψ ) ∣ 2 . Combining (78) and (10) proves
∑ T < t j ≤ T + U j e v e n L j ∣ μ j ( ψ ) − M ( ψ ) ∣ 2 ≪ ψ , ε T 1 + ε , \sum_{\substack{T<t_j\le T+U\\ j\ \mathrm{even}}}L_j|\mu_j(\psi)-M(\psi)|^2\ll_{\psi,\varepsilon}T^{1+\varepsilon}, T < t j ≤ T + U j even ∑ L j ∣ μ j ( ψ ) − M ( ψ ) ∣ 2 ≪ ψ , ε T 1 + ε , and hence Theorem 1.2 . □
Appendix A. The Weyl law with prescribed parity
For completeness, we record the parity-specific Weyl law used in the proof of Corollary 1.3.
Lemma A.1. Let { u j } \{u_j\} { u j } X = PSL 2 ( Z ) \ H X=\operatorname{PSL}_2(\mathbb{Z})\backslash\mathbb{H} X = PSL 2 ( Z ) \ H R u ( z ) = u ( − z ˉ ) Ru(z)=u(-\bar{z}) R u ( z ) = u ( − z ˉ )
Δ u j = ( 1 4 + t j 2 ) u j , R u j = ε j u j , t j > 0 , ε j ∈ { 1 , − 1 } . \Delta u_j=\left(\frac{1}{4}+t_j^2\right)u_j,\qquad Ru_j=\varepsilon_j u_j,\qquad t_j>0,\quad\varepsilon_j\in\{1,-1\}. Δ u j = ( 4 1 + t j 2 ) u j , R u j = ε j u j , t j > 0 , ε j ∈ { 1 , − 1 } . Then, for T ≥ 2 T\ge2 T ≥ 2
N ± ( T ) : = # { j : 0 < t j < T , ε j = ± 1 } = T 2 24 + O ( T log T ) . N_{\pm}(T):=\#\{j:0<t_j<T,\ \varepsilon_j=\pm1\}=\frac{T^2}{24}+O(T\log T). N ± ( T ) := # { j : 0 < t j < T , ε j = ± 1 } = 24 T 2 + O ( T log T ) . Proof. Put
N ( T ) = ∑ 0 < t j < T 1 , A ( T ) = ∑ 0 < t j < T ε j . N(T)=\sum_{0<t_j<T}1,\qquad A(T)=\sum_{0<t_j<T}\varepsilon_j. N ( T ) = 0 < t j < T ∑ 1 , A ( T ) = 0 < t j < T ∑ ε j . The classical Weyl law for the full cuspidal spectrum gives
N ( T ) = T 2 12 + O ( T log T ) ; (84) N(T)=\frac{T^2}{12}+O(T\log T); \tag*{(84)} N ( T ) = 12 T 2 + O ( T log T ) ; (84) see [5 Theorem 2.28] . Since N − ( T ) = ( N ( T ) − A ( T ) ) / 2 N_-(T)=(N(T)-A(T))/2 N − ( T ) = ( N ( T ) − A ( T )) /2 A ( T ) ≪ T log T A(T)\ll T\log T A ( T ) ≪ T log T
Fix 0 < δ < 2 log ( ( 1 + 5 ) / 2 ) 0<\delta<2\log((1+\sqrt{5})/2) 0 < δ < 2 log (( 1 + 5 ) /2 ) ρ \rho ρ
∫ R ρ ( r ) d r = 1 , supp ρ ^ ⊂ ( − δ , δ ) , ρ ^ ( u ) = ∫ R ρ ( r ) e − i r u d r . \int_{\mathbb{R}}\rho(r)\,dr=1,\qquad\operatorname{supp}\widehat{\rho}\subset(-\delta,\delta),\qquad\widehat{\rho}(u)=\int_{\mathbb{R}}\rho(r)e^{-iru}\,dr. ∫ R ρ ( r ) d r = 1 , supp ρ ⊂ ( − δ , δ ) , ρ ( u ) = ∫ R ρ ( r ) e − i r u d r . Define
h T ( r ) = ∫ − T T ρ ( r − v ) d v , g T ( u ) = 1 2 π ∫ R h T ( r ) e − i r u d r = ρ ^ ( u ) sin ( T u ) π u . h_T(r)=\int_{-T}^{T}\rho(r-v)\,dv,\qquad g_T(u)=\frac{1}{2\pi}\int_{\mathbb{R}}h_T(r)e^{-iru}\,dr=\widehat{\rho}(u)\frac{\sin(Tu)}{\pi u}. h T ( r ) = ∫ − T T ρ ( r − v ) d v , g T ( u ) = 2 π 1 ∫ R h T ( r ) e − i r u d r = ρ ( u ) π u sin ( T u ) . These are admissible test functions for the Selberg trace formula. Here ρ \rho ρ h T h_T h T d = ∣ ∣ r ∣ − T ∣ d=||r|-T| d = ∣∣ r ∣ − T ∣
h T ( r ) = ∫ r − T r + T ρ ( u ) d u . h_T(r)=\int_{r-T}^{r+T}\rho(u)\,du. h T ( r ) = ∫ r − T r + T ρ ( u ) d u . If ∣ r ∣ < T |r|<T ∣ r ∣ < T [ r − T , r + T ] [r-T,r+T] [ r − T , r + T ] [ − d , d ] [-d,d] [ − d , d ] ∣ r ∣ ≥ T |r|\ge T ∣ r ∣ ≥ T { u : ∣ u ∣ ≥ d } \{u:|u|\ge d\} { u : ∣ u ∣ ≥ d } ρ ≥ 0 \rho\ge0 ρ ≥ 0 ∫ R ρ ( u ) d u = 1 \int_{\mathbb{R}}\rho(u)\,du=1 ∫ R ρ ( u ) d u = 1
∣ h T ( r ) − 1 ( − T , T ) ( r ) ∣ ≤ ∫ ∣ u ∣ ≥ d ρ ( u ) d u ≪ B ( 1 + d ) − B , (85) \left|h_T(r)-1_{(-T,T)}(r)\right|\le\int_{|u|\ge d}\rho(u)\,du\ll_B(1+d)^{-B}, \tag*{(85)} h T ( r ) − 1 ( − T , T ) ( r ) ≤ ∫ ∣ u ∣ ≥ d ρ ( u ) d u ≪ B ( 1 + d ) − B , (85) where the last estimate follows from the Schwartz decay of ρ \rho ρ
Apply [1 Theorem 2.1] with N = 1 N = 1 N = 1 χ = 1 \chi= 1 χ = 1 n = − 1 n = -1 n = − 1 T − 1 T_{-1} T − 1 R R R
h T ( i / 2 ) + ∑ j ε j h T ( t j ) = H − ( g T ) − 1 2 π ∫ R h T ( r ) Γ ′ Γ ( 1 + i r ) d r + log ( π 2 ) g T ( 0 ) + 2 ∑ m ≥ 2 Λ ( m ) m g T ( 2 log m ) . (86) h_T(i/2) + \sum_j \varepsilon_j h_T(t_j) = \mathcal{H}_{-}(g_T) - \frac{1}{2\pi}\int_{\mathbb{R}} h_T(r)\frac{\Gamma'}{\Gamma}(1+ir)\,dr
+ \log(\pi\sqrt{2})g_T(0) + 2\sum_{m\geq2}\frac{\Lambda(m)}{m}g_T(2\log m). \tag*{(86)} h T ( i /2 ) + j ∑ ε j h T ( t j ) = H − ( g T ) − 2 π 1 ∫ R h T ( r ) Γ Γ ′ ( 1 + i r ) d r + log ( π 2 ) g T ( 0 ) + 2 m ≥ 2 ∑ m Λ ( m ) g T ( 2 log m ) . (86) Here h T ( i / 2 ) h_T(i/2) h T ( i /2 ) [1 (2.28), (2.34)] into the (NEll + Ell) term of [1 (2.24)] gives
H − ( g T ) = ∑ t ∈ Z ∖ { 0 } c t g T ( ℓ t ) . (87) \mathcal{H}_{-}(g_T) = \sum_{t\in\mathbb{Z}\setminus\{0\}} c_t g_T(\ell_t). \tag*{(87)} H − ( g T ) = t ∈ Z ∖ { 0 } ∑ c t g T ( ℓ t ) . (87) where
ℓ t = 2 log ∣ t ∣ + t 2 + 4 2 , t 2 + 4 = d t f t 2 , \ell_t = 2\log\frac{|t|+\sqrt{t^2+4}}{2}, \qquad t^2+4=d_t f_t^2, ℓ t = 2 log 2 ∣ t ∣ + t 2 + 4 , t 2 + 4 = d t f t 2 , d t d_t d t f t ≥ 1 f_t\geq1 f t ≥ 1
c t = h + ( d t ) log η d t t 2 + 4 ∏ p a ∥ f t ( 1 + ( p − ( d t p ) ) p a − 1 p − 1 ) . c_t = \frac{h^{+}(d_t)\log\eta_{d_t}}{\sqrt{t^2+4}}\prod_{p^a\parallel f_t}\left(1+\left(p-\left(\frac{d_t}{p}\right)\right)\frac{p^a-1}{p-1}\right). c t = t 2 + 4 h + ( d t ) log η d t p a ∥ f t ∏ ( 1 + ( p − ( p d t ) ) p − 1 p a − 1 ) . In this expression, h + ( d ) h^{+}(d) h + ( d ) Q ( d ) \mathbb{Q}(\sqrt{d}) Q ( d ) η d > 1 \eta_d>1 η d > 1 ( d p ) \left(\frac{d}{p}\right) ( p d ) t ≠ 0 t\ne0 t = 0
ℓ t ≥ 2 log 1 + 5 2 > δ . \ell_t \geq2\log\frac{1+\sqrt{5}}{2}>\delta. ℓ t ≥ 2 log 2 1 + 5 > δ . Since supp g T ⊂ ( − δ , δ ) \operatorname{supp}g_T\subset(-\delta,\delta) supp g T ⊂ ( − δ , δ ) H − ( g T ) = 0 \mathcal{H}_{-}(g_T)=0 H − ( g T ) = 0 2 log m ≥ 2 log 2 > δ 2\log m\geq2\log2>\delta 2 log m ≥ 2 log 2 > δ g T ( 0 ) = T / π g_T(0)=T/\pi g T ( 0 ) = T / π h T ( i / 2 ) = O ( 1 ) h_T(i/2)=O(1) h T ( i /2 ) = O ( 1 )
∑ j ε j h T ( t j ) ≪ ∫ R h T ( r ) log ( 2 + ∣ r ∣ ) d r + T + 1 ≪ T log T . (88) \sum_j\varepsilon_j h_T(t_j)\ll\int_{\mathbb{R}}h_T(r)\log(2+|r|)\,dr+T+1\ll T\log T. \tag*{(88)} j ∑ ε j h T ( t j ) ≪ ∫ R h T ( r ) log ( 2 + ∣ r ∣ ) d r + T + 1 ≪ T log T . (88) To remove the smoothing, (A.1) implies
# { j : k ≤ t j < k + 1 } ≪ ( k + 1 ) log ( k + 2 ) ( k ∈ Z ≥ 0 ) . \#\{j:k\leq t_j<k+1\}\ll(k+1)\log(k+2)\qquad(k\in\mathbb{Z}_{\geq0}). # { j : k ≤ t j < k + 1 } ≪ ( k + 1 ) log ( k + 2 ) ( k ∈ Z ≥ 0 ) . The estimate (A.2) with B > 3 B>3 B > 3
∑ j ∣ h T ( t j ) − 1 ( 0 , T ) ( t j ) ∣ ≪ T log T . \sum_j\left|h_T(t_j)-\mathbf{1}_{(0,T)}(t_j)\right|\ll T\log T. j ∑ h T ( t j ) − 1 ( 0 , T ) ( t j ) ≪ T log T . Consequently (A.5) implies A ( T ) ≪ T log T A(T)\ll T\log T A ( T ) ≪ T log T
References
School of Mathematics and Statistics, Shandong University, Weihai, Weihai, Shandong 264209, China
State Key Laboratory of Cryptography and Digital Economy Security, Shandong University, Weihai, 264209, China
Email address: qfsun@sdu.edu.cn
School of Mathematics and Statistics, Shandong University, Weihai, Weihai, Shandong 264209, China
Email address: qzzhang@mail.sdu.edu.cn
