Reducibility of 1-D Quantum Harmonic Oscillator with New Unbounded Oscillatory Perturbations

Abstract

Enlightened by Lemma 1.7 in Liang and Luo (J Differ Equ 270:343–389, 2021), we prove a similar lemma which is based upon oscillatory integrals and Langer’s turning point theory. From it we show that the Schrödinger equation

$$\begin{aligned}&{\mathrm{i}}\partial _t u =-\partial _x^2 u+x^2 u+ \epsilon \langle x\rangle ^\mu \sum _{k\in \Lambda }\left( a_k(\omega t)\sin (k|x|^\beta )+b_k(\omega t )\cos (k|x|^\beta )\right) u, \\&u=u(t,x),~x\in {\mathbb R},~ \beta >1, \end{aligned}$$

can be reduced in ${{\mathcal {H}}}^1({\mathbb R})$ to an autonomous system for most values of the frequency vector $\omega$, where $\Lambda \subset {\mathbb {R}}{\setminus }\{0\}$, $|\Lambda |<\infty$ and $\langle x\rangle :=\sqrt{1+x^2}$. The functions $a_k(\theta )$ and $b_k(\theta )$ are analytic on ${\mathbb {T}}^n_\sigma$ and $\mu \ge 0$ will be chosen according to the value of $\beta$. Comparing with Liang and Luo (J Differ Equ 270:343–389, 2021), the novelty is that the phase functions of oscillatory integral are more degenerate when $\beta >1$.

1 Introduction of the Main Results

1.1 Main Theorem

Following [25] we continue to consider the reducibility for the time dependent Schrödinger equation

$$\begin{aligned}&\mathrm{i}\partial _t u=H_\epsilon (\omega t)u,\quad x\in {\mathbb {R}},\nonumber \\&H_\epsilon :=-\partial _{xx}+x^2+\epsilon X(x,\omega t), \end{aligned}$$

(1.1)

where

$$\begin{aligned} X(x,\theta )=\langle x\rangle ^\mu \sum _{k\in \Lambda }\left( a_k(\theta )\sin (k|x|^\beta )+b_k(\theta )\cos (k|x|^\beta )\right) \end{aligned}$$

with $\Lambda \subset {\mathbb {R}}{\setminus }\{0\}$, $|\Lambda |<\infty$ and $\langle x\rangle :=\sqrt{1+x^2}$. The functions $a_k(\theta )$ and $b_k(\theta )$ are analytic on $\mathbb T^n_\sigma =\{a+b\mathrm{i}\in {\mathbb {C}}^n/2\pi {\mathbb {Z}}^n:|b|<\sigma \}$ with $\sigma >0$ and $\beta >1$ and $\mu \ge 0$ will be chosen in the following. We first introduce some functions and spaces.

Hermite Functions The harmonic oscillator $T=-\partial _{xx}+x^2$ has eigenfunctions $(h_m)_{m\ge 1}$, so called the Hermite functions, namely

$$\begin{aligned} Th_m=(2m-1)h_m,\quad \Vert h_m\Vert _{L^2({\mathbb {R}})}=1,\quad m\ge 1. \end{aligned}$$

(1.2)

Linear Spaces For $s\ge 0$ denote by ${\mathcal {H}}^s$ the domain of $T^{\frac{s}{2}}$ endowed by the graph norm. For $s<0$, the space ${\mathcal {H}}^s$ is the dual of ${\mathcal {H}}^{-s}$. Particularly, for $s\ge 0$ a integer we have

$$\begin{aligned} {\mathcal {H}}^s=\{f\in L^2({\mathbb {R}}):x^\alpha \partial ^\beta f\in L^2({\mathbb {R}}),~\forall ~\alpha ,\beta \in {\mathbb {N}}_0,~\alpha +\beta \le s\}. \end{aligned}$$

We also define the complex weighted-$\ell ^2$-space $\ell _s^2:=\{\xi =(\xi _m\in \mathbb C,m\ge 1):\sum _{m\ge 1}m^s|\xi _m|^2<\infty \}.$ To a function $u\in {\mathcal {H}}^s$ we associate the sequence $\xi$ of its Hermite coefficients by the formula $u=\sum _{m\ge 1}\xi _mh_m(x)$. In the following we will identify the space ${\mathcal {H}}^s$ with $\ell _s^2$ by endowing both space the norm

$$\begin{aligned} \Vert u\Vert _{\mathcal H^s}=\Vert \xi \Vert _s=\left( \sum _{m\ge 1}m^s|\xi _m|^2\right) ^\frac{1}{2}. \end{aligned}$$

Define

$$\begin{aligned} l_*= l(\beta ,\mu )= {\left\{ \begin{array}{ll} \frac{1}{4}\left( \frac{\beta }{6}-\mu \right) ,&{}\quad 1<\beta <2,\\ \frac{1}{4}\left( \frac{2}{9}-\mu \right) ,&{}\quad \beta =2,\\ \frac{1}{4}\left( \frac{\beta -2}{4\beta -2}-\mu \right) ,&{}\quad \beta > 2. \end{array}\right. } \end{aligned}$$

(1.3)

Then we can state our main theorem.

Theorem 1.1

Assume $a_k(\theta )$ and $b_k(\theta )$ are analytic on $\mathbb T^n_\sigma$ with $\sigma >0$ and $\beta >1$ and $\mu$ satisfies

$$\begin{aligned} 0\le \mu< {\left\{ \begin{array}{ll} \frac{\beta }{6},&{}\quad 1<\beta <2,\\ \frac{2}{9},&{}\quad \beta =2,\\ \frac{\beta -2}{4\beta -2},&{}\quad \beta >2. \end{array}\right. } \end{aligned}$$

(1.4)

There exists $\epsilon _*>0$ such that for all $0\le \epsilon <\epsilon _*$ there is a closed set $D_\epsilon \subset D_0=[0,2\pi ]^n$ of asymptotically full measure such that for all $\omega \in D_\epsilon$, the linear Schrödinger equation (1.1) reduces to a linear autonomous equation in ${\mathcal H}^{1}$.

More precisely, for any $\omega \in D_\epsilon$ there exists a linear isomorphism $\Psi _{\omega ,\epsilon }^{\infty }(\theta )\in {\mathfrak {L}}(\mathcal H^{s'})$ with $0\le s'\le 1$, analytically dependent on $\theta \in {\mathbb {T}}^n_{\sigma /2}$ and unitary on $L^2({\mathbb {R}})$, where $\Psi _{\omega ,\epsilon }^{\infty }-\mathrm{Id}\in {\mathfrak {L}}(\mathcal H^0,{\mathcal {H}}^{2l_{*}})\cap {\mathfrak {L}}({\mathcal {H}}^{s'})$ and a bounded Hermitian operator $Q\in {\mathfrak {L}}({\mathcal {H}}^1)$ such that $t\mapsto u(t,\cdot )\in {\mathcal {H}}^1$ satisfies (1.1) if and only if $t\mapsto v(t,\cdot )=\Psi _{\omega ,\epsilon }^{\infty }u(t,\cdot )\in {\mathcal {H}}^1$ satisfies the autonomous equation

$$\begin{aligned} \mathrm{i}\partial _t v=-v_{xx}+x^2v+\epsilon Q(v), \end{aligned}$$

furthermore, there are constants $C,K>0$ such that

$$\begin{aligned}&\mathrm{{Meas}}( D_0{\setminus } D_\epsilon )\le C\epsilon ^\frac{3l_*}{2(2l_*+5)(2l_*+1)},\\&\Vert Q\Vert _{{\mathfrak {L}}({\mathcal {H}}^p,{\mathcal {H}}^{p+4l_*})}+\Vert \partial _\omega Q\Vert _{{\mathfrak {L}}({\mathcal {H}}^p,{\mathcal {H}}^{p+4l_*})}\le K,\quad \omega \in D_\epsilon ,~p\in {\mathbb {N}},\\&\Vert \Psi _{\omega ,\epsilon }^{\infty }(\theta )-\mathrm{Id}\Vert _{{\mathfrak {L}}(\mathcal H^0,\mathcal H^{2l_{*}})},\Vert \Psi _{\omega ,\epsilon }^{\infty }(\theta )-\mathrm{Id}\Vert _{{\mathfrak {L}}(\mathcal H^{s'})}\le C\epsilon ^\frac{1}{3},\quad (\omega ,\theta )\in D_\epsilon \times {\mathbb {T}}^n_{\sigma /2}. \end{aligned}$$

Consequently, Theorem 1.1 follows in the considered range of parameters the ${\mathcal {H}}^1$ norms of the solutions are all bounded forever and the spectrum of the corresponding operator is pure point.

1.2 Related Results and a Critical Lemma

In the following we recall some relevant reducibility results. For 1-D quantum harmonic oscillators(‘QHO’ for short) with periodic or quasi-periodic in time bounded perturbations see [11, 15, 23, 28, 39, 40] .

In [5] Bambusi and Graffi proved the reducibility of 1-D Schrödinger equation with an unbounded time quasiperiodic perturbation in which the potential grows at infinity like $|x|^{2l}$ with a real $l>1$ and the perturbation is bounded by $1+|x|^{\beta }$ with $\beta <l-1$. The reducibility in the limiting case $\beta = l-1$ was proved by Liu and Yuan in [30]. Recently, the results in [5, 30] have been improved by Bambusi in [1, 2], in which he firstly obtained the reducibility results for 1-D QHO with unbounded perturbations.

It seems that the reducibility method in [1, 2] is hard to be applied for 1-D Schrödinger equations with the unbounded oscillatory perturbations(see remark 2.7 in [1]). The authors [25, 27] solved this problem by Langer’s turning point and oscillatory integral estimates. We remark that the critical step in [25] is to build up a decay estimate of the integral $\int _{{\mathbb R}}\langle x\rangle ^{\mu } e^{\mathrm{i} kx} h_m(x) \overline{{h}_n(x)}dx$, in which the phase functions of oscillatory integral are $\phi _{mn}(x): = \zeta _m(x) -\zeta _n(x)+kx,$ where $\zeta _m(x) =\int _{X_m}^x\sqrt{\lambda _m-t^2} dt$ with $X_m=\sqrt{2m-1}=\sqrt{\lambda _m}$.

Comparing with [25], in this paper the phase functions $\Psi _{mn}(x): = \zeta _m(x) -\zeta _n(x)+kx^{\beta }$ with $\beta >1$ are more degenerate. For $1<\beta \le 2$, we use a similar method as [25]. The most difficult part as [25] is the integral $\int _{X_m^{\frac{2}{3}}}^{X_m-X_m^{\nu _2}}\langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx$ where $\nu _2=1-\frac{\beta }{3}$ for $1<\beta <2$ and $\nu _2=\frac{5}{9}$ when $\beta =2$. We have to discuss different cases in order to obtain a suitable lower bound of the derivatives of the phase function. For $\beta >2$ we find a new simple proof which follows from Corollary 3.2 in [24], Lemma 6.1 and a straightforward computation. As Lemma 1.7 in [25] we have the following.

Lemma 1.2

Assume $h_m(x)$ satisfies (1.2). For any $k\ne 0$,

$$\begin{aligned} \left| \int _{{\mathbb R}}\langle x\rangle ^{\mu } e^{\mathrm{i} k|x|^{\beta }}h_m(x) \overline{{h}_n(x)}dx\right| \le C \cdot C_{k,\beta }(mn)^{-l(\beta ,\mu )},\quad m,n\ge 1 \end{aligned}$$

for some absolute constant $C>0$, where $\mu \ge 0, \beta >1$, $l(\beta ,\mu )$ defined in (1.3) and

$$\begin{aligned} C_{k,\beta }= {\left\{ \begin{array}{ll} |\beta (\beta -1)(\beta -2)k|^{-\frac{1}{3}}\vee |k|^{-1}\vee |k|^\frac{1}{4-2\beta },&{}\quad 1<\beta <2,\\ |k|^{-1}\vee 1,&{}\quad \beta =2,\\ |\beta k|^{-1}\vee 1,&{}\quad \beta > 2. \end{array}\right. } \end{aligned}$$

Remark 1.3

In fact $l(1,\mu )= \frac{1}{12}-\frac{\mu }{4}$ has been proved in [25].

In the end we review some relative results. Eliasson–Kuksin [13] initiated to prove the reducibility for PDEs in high dimension. See [22, 26] for higher-dimensional QHO with bounded potential. The first reducibility result for n-D QHO was proved in [7] by Bambusi-Grébert-Maspero-Robert. Towards other PDEs with unbounded perturbations see the reducibility results by Montalto [35] for linear wave equations on ${\mathbb {T}}^d$ and Bambusi, Langella and Montalto [3] for transportation equations [18]. Feola and Grébert [19] set up a reducibility result for a linear Schrödinger equation on the sphere $S^n$ with unbounded potential [20].

The reducibility results usually imply the boundedness of Sobolev norms. Delort [12] constructed a $t^{s/2}$- polynomial growth for 1-D QHO with certain time periodic perturbation [32]. Basing on a Mourre estimate, Maspero [33] proved similar results for 1-D QHO and half - wave equation on ${\mathbb T}$ and the instability is stable in some sense. For a polynomial periodic or quasi-periodic perturbations relative with 1-D QHO we refer to [7, 21, 29, 31]. For 2-D QHO with perturbation which is decaying in t, Faou-Raphaël [17] constructed a solution whose ${\mathcal {H}}^1-$norm presents logarithmic growth with t. For 2-D Schrödinger operator Thomann [38] constructed explicitly a traveling wave whose Sobolev norm presents polynomial growth with t, based on the study in [36] for linear Lowest Landau equations (LLL) with a time-dependent potential. There are also many literatures, e.g. [4, 6, 8,9,10, 16, 34, 41], which are closely relative to the upper growth bound of the solution in Sobolev space.

Our article is organized as follows: in Sect. 2 we state the reducibility theorem, i.e. Theorem 2.1. In Sect. 3, through checking all the assumptions in Theorem 2.1 we prove Theorem 1.1. In Sect. 4 we prove Lemma 1.2 for $1<\beta \le 2$ and the case for $\beta >2$ is delayed in Sect. 5. Some auxiliary lemmas are presented in the “Appendix”.

Notation We use the notations ${\mathbb N}_0=\{0,1,2,\cdots \}$, ${\mathbb N}=\{1,2,\cdots \}$, ${\mathbb {T}}^n={\mathbb {R}}^n/2\pi {\mathbb {Z}}^n$ and ${\mathbb {T}}^n_\sigma =\{a+b \mathrm{i}\in {\mathbb {C}}^n/2\pi \mathbb Z^n:|b|<\sigma \}$. For Hilbert spaces ${\mathcal {H}}_1,{\mathcal {H}}_2$ we denote by ${\mathfrak {L}}({\mathcal {H}}_1,{\mathcal {H}}_2)$ the space of bounded linear operators from ${\mathcal {H}}_1$ to ${\mathcal {H}}_2$ and write ${\mathfrak {L}}({\mathcal {H}}_1,{\mathcal {H}}_1)$ as ${\mathfrak {L}}({\mathcal {H}}_1)$ for simplicity.

2 A KAM Theorem

Following [14, 23] we introduce the KAM Theorem from [26] especially for 1-D case.

2.1 Setting

Linear spaces. For $p\ge 0$ we define $X_p:=\ell _p^2\times \ell _p^2=\{\zeta =(\zeta _a=(\xi _a,\eta _a)\in \mathbb C^2)_{a\in {\mathbb {N}}}, \Vert \zeta \Vert _p<\infty \}$ with $\Vert \zeta \Vert _p^2=\sum \nolimits _{a\in \mathbb N}a^p(|\xi _a|^2+|\eta _a|^2)$. We equip the space with the symplectic structure $\mathrm{i}\sum \nolimits _{a\in \mathbb N}d\xi _a\wedge \eta _a$.

Infinite matrices. Denote by ${\mathcal {M}}_\alpha$ the set of infinite matrices $A:\mathbb N\times {\mathbb {N}}\mapsto {\mathbb {C}}$ with the norm $|A|_\alpha :=\sup \nolimits _{a,b\in \mathbb N}(ab)^\alpha |A_a^b|<\infty$. We also denote ${\mathcal {M}}_\alpha ^+$ be the subspace of ${\mathcal {M}}_\alpha$ satisfying that an infinite matrix $A\in {\mathcal {M}}_\alpha ^+$ if $|A|_{\alpha +}:=\sup \nolimits _{a,b\in \mathbb N}(ab)^\alpha (1+|a-b|)|A_a^b|<\infty$.

In fact one can prove that for all $\alpha >0$, a matrix in ${\mathcal {M}}_\alpha ^+$ defines a bounded operator on $\ell _0^2$. However, when $\alpha \in (0,\frac{1}{2})$, we can’t insure that $\mathcal M_{\alpha }\subset {\mathfrak {L}}(\ell _0^2,\ell _s^2)$ for any $s\in {\mathbb {R}}$. This means that Px makes no sense when the perturbation operator $P\in {\mathcal {M}}_\alpha$ and $x\in \ell _0^2$. Fortunately, from Lemma 2.1 in [22] or Lemma 2.2 in [26] one can show ${\mathcal {M}}_{\alpha }\subset {\mathfrak {L}}(\ell _1^2,\ell _{-1}^2)$ and thus the reducibility in ${\mathcal {H}}^1$ can be built up in Theorem 1.1 instead of $L^2$.

Parameters. In this paper $\omega$ will play the role of a parameter belonging to $D_0=[0,2\pi ]^n$. All the constructed maps will depend on $\omega$ with ${\mathcal {C}}^1$ regularity. When a map is only defined on a Cantor subset of $D_0$ the regularity is understood in Whitney sense.

A class of quadratic Hamiltonians. Let $D\subset D_0,\alpha >0$ and $\sigma >0$. We denote by ${{\mathcal {M}}}_\alpha (D,\sigma )$ the set of mappings as $\mathbb T^n_\sigma \times D\ni (\theta ,\omega )\mapsto Q(\theta ,\omega )\in {\mathcal {M}}_\alpha$ which is real analytic on $\theta \in {\mathbb {T}}^n_\sigma$ and ${\mathcal {C}}^1$ continuous on $\omega \in D$. And we endow this space with the norm $[Q]_\alpha ^{D,\sigma }:=\sup \nolimits _{\begin{array}{c} \omega \in D,|\Im \theta |<\sigma |k|=0,1 \end{array}}|\partial _\omega ^kQ(\theta ,\omega )|_\alpha .$

The subspace of ${\mathcal {M}}_\alpha (D,\sigma )$ formed by $F(\theta ,\omega )$ such that $\partial _\omega ^kF(\theta ,\omega )\in \mathcal M_\alpha ^+,|k|=0,1$, is denoted by ${\mathcal {M}}_\alpha ^+(D,\sigma )$ and endowed with the norm $[F]_{\alpha +}^{D,\sigma }:=\sup \nolimits _{\begin{array}{c} \omega \in D,|\Im \theta |<\sigma |k|=0,1 \end{array}}|\partial _\omega ^k F(\theta ,\omega )|_{\alpha +}$. Besides, the subspace of $M_\alpha (D,\sigma )$ that are independent of $\theta$ will be denoted by ${\mathcal {M}}_\alpha (D)$ and for $N\in {\mathcal {M}}_\alpha (D)$,

$$\begin{aligned}{}[N]_\alpha ^D:=\sup \limits _{\omega \in D,|k|=0,1}|\partial _\omega ^kN(\omega )|_\alpha . \end{aligned}$$

2.2 The Reducibility Theorem

In this section we present an abstract reducibility theorem for a quadratic Hamiltonian quasiperiodic in time of the form

$$\begin{aligned} H(t,\xi ,\eta )=\langle \xi ,N\eta \rangle +\epsilon \langle \xi ,P(\omega t)\eta \rangle ,\quad (\xi ,\eta )\in X_1\subset X_0, \end{aligned}$$

(2.1)

and the corresponding Hamiltonian system

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\dot{\xi }}}=-\mathrm{i}N\xi -\mathrm{i}\epsilon P^T(\omega t)\xi ,\\ {{\dot{\eta }}}=\mathrm{i}N\eta +\mathrm{i}\epsilon P(\omega t)\eta , \end{array}\right. } \end{aligned}$$

(2.2)

where $N=\text {diag}\{\lambda _a,a\in {\mathbb {N}}\}$ satisfies the following spectrum assumptions:

Hypothesis A1-Asymptotics There exist positive constants $c_0,c_1,c_2$ such that

$$\begin{aligned} c_1a\ge \lambda _a\ge c_2a\text { and }|\lambda _a-\lambda _b|\ge c_0|a-b|,~\forall ~a,b\in {\mathbb {N}}. \end{aligned}$$

Hypothesis A2-Second Melnikov Condition in Measure Estimates There exist positive constants $\alpha _1,\alpha _2$ and $c_3$ such that the following holds: for each $0<\kappa <\frac{1}{4}$ and $K>0$ there exists a closed subset $D'=D'(\kappa ,K)\subset D_0$ with $\text {Meas}(D_0{\setminus } D')\le c_3K^{\alpha _1}\kappa ^{\alpha _2}$ such that for all $\omega \in D',k\in {\mathbb {Z}}^n$ with $0<|k|\le K$ and $a,b\in {\mathbb {N}}$ we have $|k\cdot \omega +\lambda _a-\lambda _b|\ge \kappa (1+|a-b|)$.

Then we can state our reducibility results.

Theorem 2.1

Given a nonautonomous Hamiltonian (2.1), we assume that $(\lambda _a)_{a\in {\mathbb {N}}}$ satisfies Hypothesis A1–A2 and $P(\theta )\in {\mathcal {M}}_{\alpha }(D_0,\sigma )$ with $\alpha ,\sigma >0$. Let $\gamma _1=\max \{\alpha _1,n+3\}$ and $\gamma _2=\frac{\alpha \alpha _2}{2\alpha \alpha _2+5}$, then there exists $\epsilon _*>0$ such that for all $0\le \epsilon <\epsilon _*$ there are

• (i) a Cantor set $D_\epsilon \subset D_0$ with $\text {Meas}(D_0{\setminus } D_\epsilon )\le C\epsilon ^{\frac{3\delta \alpha }{2\alpha +1}}$ for a $\delta \in (0,\frac{\gamma _2}{24})$;

• (ii) a ${\mathcal {C}}^1$ family in $\omega \in D_\epsilon$(in Whitney sense), linear, unitary, analytically dependent on $\theta \in {\mathbb {T}}^n_{\sigma /2}$ and symplectic coordinate transformation $\Phi _\omega ^\infty (\theta ): X_0\mapsto X_0,(\omega ,\theta )\in D_\epsilon \times \mathbb T^n_{\sigma /2}$, of the form

  $$\begin{aligned} (\xi _+,\eta _+)\mapsto (\xi ,\eta )=\Phi _\omega ^\infty (\theta )(\xi _+,\eta _+)=(\overline{M}_\omega (\theta )\xi _+,M_\omega (\theta )\eta _+), \end{aligned}$$

  where $\Phi _\omega ^\infty (\theta )-\mathrm{Id}$ satisfies for $0\le s'\le 1$

  $$\begin{aligned} \Vert \Phi _\omega ^\infty (\theta )-\mathrm{Id}\Vert _{{\mathfrak {L}}(X_0,X_{2\alpha })},\Vert \Phi _\omega ^\infty (\theta )-\mathrm{Id}\Vert _{{\mathfrak {L}} (X_{s'})}\le C\epsilon ^{\frac{1}{3}}; \end{aligned}$$

• (iii) a ${\mathcal {C}}^1$ family of autonomous quadratic Hamiltonian in normal forms

  $$\begin{aligned} H_\infty (\xi _+,\eta _+)=\langle \xi _+,N_\infty (\omega )\eta _+\rangle =\sum _{m\ge 1}\lambda _m^\infty \xi _{+,m}\eta _{+,m},\quad \omega \in D_\epsilon , \end{aligned}$$

  where $N_\infty (\omega )=\text {diag}\{\lambda ^{\infty }_m,m\in \mathbb N \}$ is diagonal and is close to N in the sense of

  $$\begin{aligned}{}[N_\infty (\omega )-N]_\alpha ^{D_\epsilon }\le C\epsilon , \end{aligned}$$

  such that

  $$\begin{aligned} H(t,\Phi _\omega ^\infty (\omega t)(\xi _+,\eta _+))=H_\infty (\xi _+,\eta _+),~t\in \mathbb R,~(\xi _+,\eta _+)\in X_1,~\omega \in D_\epsilon . \end{aligned}$$

3 Application to the Quantum Harmonic Oscillator

In this section we will prove Theorem 1.1 by applying Theorem 2.1 to the original Eq. (1.1). Following the strategies in [13], we expand u on the Hermite basis $(h_m)_{m\ge 1}$ as well as ${\bar{u}}$ by the following formula

$$\begin{aligned} u=\sum _{m\ge 1}\xi _mh_m,\quad \bar{u}=\sum _{m\ge 1}\eta _m\bar{h}_m. \end{aligned}$$

Therefore the Eq. (1.1) is equivalent to a nonautonomous Hamiltonian system

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\dot{\xi }}}_m=-\mathrm{i}\frac{\partial H}{\partial \eta _m}=-\mathrm{i}(2m-1)\xi _m- \mathrm{i}\epsilon \left( P^T(\omega t)\xi \right) _m,\\ {{\dot{\eta }}}_m=\mathrm{i}\frac{\partial H}{\partial \xi _m}=\mathrm{i}(2m-1)\eta _m+\mathrm{i}\epsilon \left( P(\omega t)\eta \right) _m, \end{array}\right. }\quad m\ge 1, \end{aligned}$$

(3.1)

where

$$\begin{aligned} H(t,\xi ,\eta )=\langle \xi ,N\eta \rangle +\epsilon \langle \xi ,P(\omega t)\eta \rangle ,\quad (\xi ,\eta )\in X_1\subset X_0, \end{aligned}$$

and $N=\text {diag}\{2m-1,m\ge 1\}$ and

$$\begin{aligned} P_m^n(\omega t)= & {} \sum \limits _{k\in \Lambda } a_k(\omega t) \int _{{\mathbb {R}}} \langle x\rangle ^\mu \sin k|x|^\beta h_m(x)\overline{h_n(x)}dx \nonumber \\&+ \sum \limits _{k\in \Lambda } b_k(\omega t) \int _{{\mathbb {R}}} \langle x\rangle ^\mu \cos k |x|^\beta h_m(x)\overline{h_n(x)}dx, \end{aligned}$$

(3.2)

where the frequencies $\omega \in D_0=[0,2\pi ]^n$ are the external parameters.

The spectrum assumptions can be easily checked by the following two lemmas.

Lemma 3.1

When $\lambda _a=2a-1,a\in {\mathbb {N}}$, Hypothesis A1 holds true with $c_0=c_2=1$ and $c_1=2$.

Lemma 3.2

When $\lambda _a=2a-1,a\in {\mathbb {N}}$, Hypothesis A2 holds true with $\alpha _1=n+1,\alpha _2=1,c_3=c(n)$ and $D_0=[0,2\pi ]^n$,

$$\begin{aligned} D'=\{\omega \in [0,2\pi ]^n:|k\cdot \omega +j|\ge \kappa (1+|j|),~\forall ~j\in \mathbb Z,~k\in {\mathbb {Z}}^n{\setminus }\{0\}\}. \end{aligned}$$

The following lemma is a direct corollary of Lemma 1.2.

Lemma 3.3

Assume that $a_k(\theta )$ and $b_k(\theta )$ are analytic on ${\mathbb {T}}^n_\sigma$ for any nonzero $k\in \Lambda$ with $\sigma >0$ and $\beta >1$ and $\mu$ satisfies (1.4), then there exists $\alpha =l(\beta ,\mu )>0$ such that the matrix function $P(\theta )$ defined by (3.2) is analytic from ${\mathbb {T}}^n_\sigma$ into ${\mathcal {M}}_\alpha$.

Proof of Theorem 1.1

Expanding the Hermite basis $(h_m)_{m\ge 1}$, the Schrödinger equation (1.1) becomes Hamiltonian system (3.1), which is the form of Eq. (2.2) with $\lambda _a=2a-1$. By lemmas above, we can apply Theorem 2.1 to (3.1) with $\gamma _1=n+3,\gamma _2=\frac{\alpha }{2\alpha +5}$ and $\delta =\frac{\gamma _2}{48}$. This follows Theorem 1.1.

More precisely, in new variables given in Theorem 2.1, $(\xi ,\eta )=({\overline{M}}_\omega \xi _+,M_\omega \eta _+)$, system (3.1) is conjugated into an autonomous system of the form:

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\dot{\xi }}}_{+,a}=-\mathrm{i}\lambda _a^\infty (\omega )\xi _{+,a},\\ {{\dot{\eta }}}_{+,a}=\mathrm{i}\lambda _a^\infty (\omega )\eta _{+,a}, \end{array}\right. }\quad a\in {\mathbb {N}}. \end{aligned}$$

Therefore the solution subject to the initial datum $(\xi _+(0),\eta _+(0))$ reads

$$\begin{aligned} (\xi _+(t),\eta _+(t))=(e^{-\mathrm{i}tN_\infty }\xi _+(0),e^{\mathrm{i}tN_\infty }\eta _+(0)),\quad t\in {\mathbb {R}}, \end{aligned}$$

where $N_\infty =\text {diag}\{\lambda ^{\infty }_a, \ a\ge 1\}$. Then the solution of (1.1) with the initial datum $u_0(x)=\sum _{a\ge 1}\xi _a(0)h_a(x)\in {\mathcal {H}}^1$ is formulated by $u(t,x)=\sum _{a\ge 1}\xi _a(t)h_a(x)$ with $\xi (t)=\overline{M}_\omega (\omega t)e^{-\mathrm{i}tN_\infty }M_\omega ^T(0)\xi (0)$, where we use the fact $\left( {\overline{M}}_\omega \right) ^{-1}=M_\omega ^T$ since M is unitary. Now we define the coordinate transformation $\Psi _\omega ^{\infty }(\theta )$ by

$$\begin{aligned} \Psi _\omega ^{\infty }(\theta )\left( \sum _{a\ge 1}\xi _ah_a(x)\right) :=\sum _{a\ge 1}\left( M_\omega ^T(\theta )\xi \right) _ah_a(x)=\sum _{a\ge 1}\xi _{+,a}h_a(x). \end{aligned}$$

Then we have u(t, x) satisfies (1.1) if and only if $v(t,x)=\Psi _\omega ^{\infty }(\omega t)u(t,x)$ satisfies the autonomous equation $\mathrm{i}\partial _tv=-v_{xx}+x^2v+\epsilon Q(v)$, where

$$\begin{aligned} \epsilon Q\left( \sum _{a\ge 1}\xi _ah_a(x)\right) =\sum _{a\ge 1}((N_\infty -N_0)\xi )_ah_a(x)=\sum _{a\ge 1}(\lambda _a^\infty -\lambda _a)\xi _ah_a(x). \end{aligned}$$

The rest estimates are standard (see Lemma 3.4 in [25] for the details). $\square$

4 Proof of Lemma 1.2 When $1<\beta \le 2$

For reader’s convenience, we will use the notations in [25]. In the whole section we will always suppose $\mu \ge 0$ and don’t point it out in the following lemmas.

The eigenfunction of the quantum oscillator operator T is $h_n(x)={(n! 2^n \pi ^{\frac{1}{2}})^{-\frac{1}{2}}}{e^{-\frac{1}{2} x^2} H_n(x)}$, where $H_n(x)$ is the nth Hermite polynomial. Since $h_n(x)$ is an even (or odd) function when n is odd (or even), we only need to estimate

$$\begin{aligned} \int _0^{+\infty } \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx,\quad 1\le m\le n. \end{aligned}$$

(4.1)

By Lemma 4.4 and Remarks 4.5, 4.6 in [25], when $m>m_0$,

$$\begin{aligned} h_m(x)= & {} (\lambda _m-x^2)^{-\frac{1}{4}}\left( \frac{\pi \zeta _m}{2}\right) ^{\frac{1}{2}}H_{\frac{1}{3}}^{(1)}(\zeta _m)\\&+(\lambda _m-x^2)^{-\frac{1}{4}}\left( \frac{\pi \zeta _m}{2}\right) ^{\frac{1}{2}}H_{\frac{1}{3}}^{(1)}(\zeta _m)O \left( \frac{1}{\lambda _m}\right) \nonumber \\:= & {} \psi ^{(m)}_1(x) +\psi ^{(m)}_2(x), \end{aligned}$$

where $\zeta _m(x)=\displaystyle \int _{X_{m}}^x\sqrt{\lambda _m-t^2}dt$ with $X_m^2=\lambda _m(X_m>0)$. Otherwise, when $m\le m_0$, then $h_m(x)=\psi _1^{(m)}(x)+\psi _2^{(m)}(x)$ for $x>2X_{m_0}$, where $\psi _1^{(m)}(x)=(\lambda _m-x^2)^{-\frac{1}{4}}(\frac{\pi \zeta _m}{2})^\frac{1}{2} H_\frac{1}{3}^{(1)}(\zeta _m)$ and $|\psi _2^{(m)}(x)|\le \frac{C}{x^2}|\psi _1^{(m)}(x)|$. Following the same strategies in [25] we distinguish 3 cases to estimate (4.1):

• I. $m,n< C_0: = 2^8m_0^3$;

• II. $m\le m_0$ and $n\ge C_0$;

• III. $m,n>m_0$.

4.1 The Estimates for Case I and Case II

Lemma 4.1

When $n, m<C_0$,

$$\begin{aligned} \left| \int _0^{+\infty } \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{(mn)^{\frac{1}{4}(\frac{\beta }{6}-\mu )}}. \end{aligned}$$

Proof

When $x\le X_0$, from Hölder inequality and $n,m<C_0$, we have

$$\begin{aligned} \left| \int _0^{X_0} \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le X_0^\mu \le \frac{C}{(mn)^{\frac{1}{4}(\frac{\beta }{6}-\mu )}}. \end{aligned}$$

where $X_0$ is a positive constant depending on $C_0$ only. When $x>X_0$, $|X_m^2-x^2|^{-\frac{1}{4}}<1$, we have $\left| \sqrt{\frac{\pi \zeta _m}{2}}H^{(1)}_\frac{1}{3}(\zeta _m)\right| \le e^{-\left| \zeta _m\right| }$ by Lemma 5.4 in [25]. By Lemma 5.5 in [25] we have $\left| \zeta _m\right| \ge \frac{2\sqrt{2}}{3}X_m^\frac{1}{2}(x-X_m)^\frac{3}{2}\ge x-X_0$ for $x> X_0$. Thus

$$\begin{aligned} \left| \int _{X_0}^{+\infty } \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \int _{X_0}^{+\infty } \langle x\rangle ^{\mu }e^{-2(x-X_0)}dx\le Ce^{2X_0} \le \frac{C}{(mn)^{\frac{1}{4}(\frac{\beta }{6}-\mu )}}. \end{aligned}$$

$\square$

Lemma 4.2

For $m\le m_0$ and $n\ge C_0$ and $\mu \ge 0$,

$$\begin{aligned} \left| \int _0^{+\infty } \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{(mn)^{\frac{1}{4}(\frac{\beta }{6}-\mu )}}. \end{aligned}$$

Proof

We divide the integral into two parts.

$$\begin{aligned} \int _0^{+\infty } \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx=\int _0^{X_n^\frac{1}{3}}+\int _{X_n^\frac{1}{3}}^{+\infty }. \end{aligned}$$

Since $x>2X_{m_0}$, we have

$$\begin{aligned} |h_m(x)|\le 2(x^2-X_m^2)^{-\frac{1}{4}}|\sqrt{\frac{\pi \zeta _m}{2}}H^{(1)}_\frac{1}{3}(\zeta _m)|\le 2 e^{-|\zeta _m|}. \end{aligned}$$

On the other hand, for $x\in [0,X_n^\frac{1}{3}]$, one has $|h_n(x)|\le C(X_n^2-x^2)^{-\frac{1}{4}}$. Note $1<\beta <2$, it follows

$$\begin{aligned} \left| \int _0^{X_n^\frac{1}{3}}\langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right|\le & {} C\int _0^{X_n^\frac{1}{3}}\langle x\rangle ^{\mu }(X_n^2-x^2)^{-\frac{1}{4}}dx \le CX_n^{-\frac{1}{6}+\frac{\mu }{3}}\\\le & {} \frac{C}{(mn)^{\frac{1}{4}(\frac{\beta }{6}-\mu )}}. \end{aligned}$$

When $x\ge X_n^\frac{1}{3}\ge 2X_{m_0}$, from Lemma 5.5 in [25], $e^{-|\zeta _m|}\le e^{-C(x-X_m)}$. Note $\Vert h_n(x)\Vert _{L^2}=1$, from Hölder inequality,

$$\begin{aligned} \left| \int _{X_n^\frac{1}{3}}^{+\infty }\langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le C\left( \int _{X_n^\frac{1}{3}}^{+\infty }\langle x\rangle ^{2\mu }e^{-Cx}dx\right) ^\frac{1}{2}\le e^{-CX_n^\frac{1}{3}}. \end{aligned}$$

$\square$

4.2 The Estimate for Case III

In the following we will turn to the complicated case when $m,n>m_0$. We divide the integral into two parts $\int _0^{+\infty } \langle x\rangle ^{\mu } e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx=\int _0^{X_n}+\int _{X_n}^{+\infty }$. We first go to the latter case $\int _{X_n}^{+\infty }$.

4.2.1 The Integral on $[X_n, +\infty )$

Lemma 4.3

For $m_0<m\le n$,

$$\begin{aligned} \displaystyle \left| \int _{X_n}^{+\infty }\langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{12}-\frac{\mu }{4}} n^{\frac{1}{12}-\frac{\mu }{4}}}\le \frac{C}{(mn)^{\frac{1}{4}(\frac{\beta }{6}-\mu )}}. \end{aligned}$$

We first estimate the integral on $[2X_n, +\infty ]$. The following result is clear from [25].

Lemma 4.4

For $m_0<m\le n$,

$$\begin{aligned} \displaystyle \left| \int _{2X_n}^{+\infty }\langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x) \overline{h_n(x)}dx\right| \le e^{-C n}. \end{aligned}$$

For the integral on $[X_n,2X_n]$, we prove that

Lemma 4.5

For $m_0<m\le n$,

$$\begin{aligned} \displaystyle \left| \int ^{2X_n}_{X_n}\langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x) \overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{12}-\frac{\mu }{4}} n^{\frac{1}{12}-\frac{\mu }{4}}}. \end{aligned}$$

Proof

As [25], we only need to estimate the following integral $I: = \int ^{2X_n}_{X_n}\langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx$ since the rest three ones are higher order. I can be divided into two parts as

$$\begin{aligned} |I|&=\displaystyle \left| \Bigg (\int ^{2X_n}_{X_n+X_n^{\frac{1}{3}}}+\int ^{X_n+X_n^{\frac{1}{3}}}_{X_n}\Bigg )\langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\le CX_n^\mu \Big (\int ^{2X_n}_{X_n+X_n^{\frac{1}{3}}}+\int ^{X_n+X_n^{\frac{1}{3}}}_{X_n}\Big )\left| \psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}\right| dx. \end{aligned}$$

From Lemma 5.5 in [25], when $x\ge X_n+X_n^{\frac{1}{3}}$, $\left| \zeta _n\right| \ge \frac{2\sqrt{2}}{3}X_n^\frac{1}{2}(x-X_n)^\frac{3}{2}\ge \frac{2\sqrt{2}}{3}X_n.$ Thus

$$\begin{aligned}&\int ^{2X_n}_{X_n+X_n^{\frac{1}{3}}}\left| \psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}\right| dx \le C\int ^{2X_n}_{X_n+X_n^{\frac{1}{3}}}(x^2-\lambda _m)^{-\frac{1}{4}}(x^2-\lambda _n)^{-\frac{1}{4}} e^{-\left| \zeta _n\right| }dx \\&\quad \le Ce^{-\frac{2\sqrt{2}}{3}X_n} \int ^{2X_n}_{X_n+X_n^{\frac{1}{3}}}(x^2-\lambda _n)^{-\frac{1}{2}}dx\le Ce^{-\frac{2\sqrt{2}}{3}X_n}. \end{aligned}$$

For the second part,

$$\begin{aligned}&\int ^{X_n+X_n^{\frac{1}{3}}}_{X_n}\left| \psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}\right| dx\\&\le C\int ^{X_n+X_n^{\frac{1}{3}}}_{X_n}(x^2-\lambda _m)^{-\frac{1}{4}}(x^2-\lambda _n)^{-\frac{1}{4}}dx \le C\int ^{X_n+X_n^{\frac{1}{3}}}_{X_n} (x^2-\lambda _n)^{-\frac{1}{2}}dx \\&\quad \le CX_n^{-\frac{1}{2}} \int ^{X_n+X_n^{\frac{1}{3}}}_{X_n}(x-X_n)^{-\frac{1}{2}}dx \le CX_n^{-\frac{1}{3}}. \end{aligned}$$

It follows $|I|\le CX_n^{\mu -\frac{1}{3}} \le \frac{C}{m^{\frac{1}{12}-\frac{\mu }{4}}n^{\frac{1}{12}-\frac{\mu }{4}}}$. $\square$

Combining with the above two lemmas we finish Lemma 4.3.

In the following we will estimate the integral on $[0,X_n]$, which is the most complicated case. Note $m_0< m\le n$, the following two cases have to be considered respectively: I. $X_n> 2X_m$; II. $X_m\le X_n \le 2X_m$.

4.2.2 The Integral Estimate on $[0, X_n]$ When $X_n> 2X_m$

Our aim in this part is to build the following

Lemma 4.6

For $k \ne 0$, if $X_n> 2X_m$ and $1<\beta \le 2$, then

$$\begin{aligned} \displaystyle \left| \int ^{X_n}_0 \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C (|k|^\iota \vee 1)}{m^{\frac{1}{8}-\frac{\mu }{4}} n^{\frac{1}{12}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0< m\le n$ and $\iota ={\left\{ \begin{array}{ll} \frac{1}{4-2\beta },&{}\quad 1<\beta <2,\\ 0,&{}\quad \beta =2. \end{array}\right. }$

As [25] we will use the following notation in the remained parts. We denote $f_m(x)=\int _0^{\infty }e^{-t}t^{-\frac{1}{6}}\left( 1+\frac{\mathrm{i}t}{2\zeta _m}\right) ^{-\frac{1}{6}}dt$ and $f_n(x)=\int _0^{\infty }e^{-t}t^{-\frac{1}{6}}\left( 1+\frac{\mathrm{i}t}{2\zeta _n}\right) ^{-\frac{1}{6}}dt$. When $x\in [0,X_{m}]$, from a straightforward computation we have

$$\begin{aligned} \psi _1^{(m)}(x)&= (X_m^2-x^2)^{-\frac{1}{4}}\sqrt{\frac{\pi \zeta _m}{2}} H_\frac{1}{3}^{(1)}(\zeta _m)\\&= (X_m^2-x^2)^{-\frac{1}{4}} \frac{e^{\mathrm{i}\left( \zeta _m-\frac{\pi }{6}-\frac{\pi }{4}\right) }}{\Gamma {\left( \frac{5}{6}\right) }} \int _0^{\infty }e^{-t}t^{-\frac{1}{6}}\left( 1+\frac{\mathrm{i}t}{2\zeta _m}\right) ^{-\frac{1}{6}}dt\\&=C(X_m^2-x^2)^{-\frac{1}{4}}e^{\mathrm{i}\zeta _m(x)}f_m(x). \end{aligned}$$

Similarly, $\overline{\psi _1^{(n)}(x)} =C(X_n^2-x^2)^{-\frac{1}{4}}e^{-\mathrm{i}\zeta _n(x)}\overline{f_n(x)}.$ For $x\in [0,X_m]$, denote $\Psi (x)=(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot f_m(x)\overline{f_n(x)}$ and $g(x)=(\zeta _n(x)-\zeta _m(x)-kx^{\beta })^\prime =\sqrt{X_n^2-x^2}-\sqrt{X_m^2-x^2}-k\beta x^{\beta -1}$,then

$$\begin{aligned} \Psi ^\prime (x)&=\frac{1}{2}x(X_m^2-x^2)^{-\frac{5}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot f_m(x)\overline{f_n(x)}\\&\quad +\frac{1}{2}x(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{5}{4}}\cdot f_m(x)\overline{f_n(x)}\\&\quad +(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot \left( f_m^\prime (x)\overline{f_n(x)}+f_m(x)\overline{f_n^\prime (x)}\right) . \end{aligned}$$

When $x\in [0, X_m]$, $\left| f_m(x)\right| \le \Gamma (\frac{5}{6})$ and $\left| f_n(x)\right| \le \Gamma (\frac{5}{6})$. Thus,

Corollary 4.7

For $x\in [0,X_m)$ and $m\le n$,

$$\begin{aligned} \left| \Psi ^\prime (x)\right|&\le C\Big (x(X_m^2-x^2)^{-\frac{5}{4}}(X_n^2-x^2)^{-\frac{1}{4}} +x(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{5}{4}}\\&\quad +\frac{(X_m^2-x^2)^{\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}}{X_m(X_m-x)^3} + \frac{(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{\frac{1}{4}}}{X_n (X_n-x)^3}\Big )\\&=C\big (J_1+J_2+J_3+J_4\big )\le C(J_1+J_3). \end{aligned}$$

We first estimate the integral on $[0, X_m-X_m^{-\frac{1}{3}}]$.

Lemma 4.8

For $k\ne 0,1<\beta <2$, if $X_n> 2X_m$, then

$$\begin{aligned} \left| \int _0^{X_m-X_m^{-\frac{1}{3}}} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C(|k|^\frac{1}{4-2\beta }\vee 1)}{m^{\frac{1}{8}-\frac{\mu }{4}} n^{\frac{1}{8}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0< m\le n$.

Proof

First we estimate the main term of the integral

$$\begin{aligned} \int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx = C\int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}(\zeta _m-\zeta _n+kx^{\beta })}\Psi (x)dx, \end{aligned}$$

by method of oscillating integral estimate, where $\Psi (x)=(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot f_m(x)\overline{f_n(x)}.$ We discuss two different cases.

Case 1: $k\le \frac{X_n^{2-\beta }}{8}$. In this case, we have

$$\begin{aligned} g(x)\ge \sqrt{X_n^2-x^2}-\sqrt{\frac{X_n^2}{4}-x^2}-\beta kX_n^{\beta -1} \ge \frac{X_n}{2}-2\cdot \frac{X_n^{2-\beta }}{8}X_n^{\beta -1}\ge \frac{X_n}{4}. \end{aligned}$$

Thus, by Lemma 6.1,

$$\begin{aligned}&\left| \int ^{X_m-X_m^{-\frac{1}{3}}}_{0}e^{\mathrm{i}\frac{\zeta _m-\zeta _n+kx^\beta }{X_n}X_n}\langle x\rangle ^\mu \Psi (x)dx\right| \\&\quad \le CX_n^{-1}\left( \left| \left( \langle x\rangle ^\mu \Psi \right) (X_m-X_m^{-\frac{1}{3}})\right| +\int _0^{X_m-X_m^{-\frac{1}{3}}}\left| \left( \langle x\rangle ^\mu \Psi \right) ^\prime (x)\right| dx\right) \\&\quad \le CX_n^{-1}\left( X_m^\mu \left| \Psi (X_m-X_m^{-\frac{1}{3}})\right| +\int _0^{X_m-X_m^{-\frac{1}{3}}}\left( 2\langle x\rangle ^\mu \left( J_1+J_3\right) +\mu \langle x\rangle ^{\mu -1}\frac{ x}{\langle x\rangle }\left| \Psi (x)\right| \right) dx\right) \\&\quad \le CX_n^{-1}X_m^\mu \left( \left| \Psi (X_m-X_m^{-\frac{1}{3}})\right| +\int _0^{X_m-X_m^{-\frac{1}{3}}}\left( J_1+J_3\right) dx\right) \\&\qquad +C\mu X_n^{-1}\int _0^{X_m-X_m^{-\frac{1}{3}}}\langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx. \end{aligned}$$

Clearly,

$$\begin{aligned} X_m^\mu \left| \Psi (X_m-X_m^{-\frac{1}{3}})\right|&\le CX_m^\mu \left( X_m^2-(X_m-X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}} \left( X_n^2-(X_m-X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}}\\&\le C X_m^{-\frac{1}{3}+\mu }, \end{aligned}$$

and

$$\begin{aligned}&\int _0^{X_m-X_m^{-\frac{1}{3}}}\mu \langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx \le C\left( X_m^2-(X_m-X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}}\\&\quad \left( X_n^2-(X_m-X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}}\int _0^{X_m}\mu x^{\mu -1}dx\\&\quad \le C\left( X_m^2-(X_m-X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}} \left( X_n^2-(X_m-X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}}X_m^{\mu }\le CX_m^{-\frac{1}{3}+\mu }, \end{aligned}$$

together with

$$\begin{aligned} \int _0^{X_m-X_m^{-\frac{1}{3}}}J_1dx&\le C \int _0^{X_m-X_m^{-\frac{1}{3}}} x(X_m^2-x^2)^{-\frac{5}{4}}(X_m^2-x^2)^{-\frac{1}{4}}dx \le C X_m^{-\frac{1}{3}}, \end{aligned}$$

and

$$\begin{aligned} \int _0^{X_m-X_m^{-\frac{1}{3}}}J_3dx&\le CX_m^{-1}\int _0^{X_m-X_m^{-\frac{1}{3}}}(X_m-x)^{-3}dx \le C X_m^{-\frac{1}{3}}. \end{aligned}$$

So we obtain $\left| \int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx\right| \le CX_m^{-\frac{1}{3}+\mu }X_n^{-1}.$ Now we turn to remained three terms. Since $m_0< m\le n$,

$$\begin{aligned} \left| \int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _2^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx\right|&\le C\int _0^{X_m-X_m^{-\frac{1}{3}}}X_m^{-2+\mu }(X_m^2-x^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx\\&\le CX_m^{-\frac{3}{2}+\mu }X_n^{-\frac{1}{2}}\le Cn^{-\frac{1}{4}+\frac{\mu }{2}}. \end{aligned}$$

Similarly, when $m_0< m\le n$, we have

$$\begin{aligned} \left| \int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x)\overline{\psi _2^{(n)}(x)}dx\right| \le Cn^{-1+\frac{\mu }{2}} \end{aligned}$$

and

$$\begin{aligned} \left| \int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _2^{(m)}(x)\overline{\psi _2^{(n)}(x)}dx\right| \le Cn^{-1+\frac{\mu }{2}}. \end{aligned}$$

Thus,

$$\begin{aligned} \left| \int ^{X_m-X_m^{-\frac{1}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{n^{\frac{1}{4}-\frac{\mu }{2}}}\le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}}n^{\frac{1}{8}-\frac{\mu }{4}}},\quad m_0<m\le n. \end{aligned}$$

Case 2: $k>\frac{X_n^{2-\beta }}{8}>0$.

Since $m\le n$, we have $2n\le (8k)^{\frac{2}{2-\beta }}+1$. It follows that

$$\begin{aligned} \left| \int _0^{X_m-X_m^{-\frac{1}{3}}} \langle x\rangle ^\mu e^{ \mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le CX_m^\mu \le CX_m^\mu \frac{m^{\frac{1}{8}} n^{\frac{1}{8}}}{m^{\frac{1}{8}}n^{\frac{1}{8}}}\le \frac{Ck^\frac{1}{4-2\beta }}{m^{\frac{1}{8}-\frac{\mu }{4}} n^{\frac{1}{8}-\frac{\mu }{4}}}. \end{aligned}$$

Combining with these two cases we finish the proof. $\square$

Lemma 4.9

For $k\ne 0$, if $X_n>2X_m$, then

$$\begin{aligned} \left| \int _0^{X_m-X_m^{\frac{2}{3}}} \langle x\rangle ^\mu e^{\mathrm{i}kx^{2}}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}} n^{\frac{1}{8}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0< m\le n$.

Proof

We first estimate the main part of the integral. By the oscillating integral estimate,

$$\begin{aligned} \int ^{X_m-X_m^{\frac{2}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^{2}}\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx = C\int ^{X_m-X_m^{\frac{2}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}(\zeta _m-\zeta _n+kx^{2})}\Psi (x)dx, \end{aligned}$$

where $\Psi (x)=(X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot f_m(x)\overline{f_n(x)}.$ Since $g''(x)\ge g''(0)=\frac{1}{X_m}-\frac{1}{X_n}\ge \frac{1}{2}X_m^{-1},$ by Lemma 6.1,

$$\begin{aligned}&\left| \int ^{X_m-X_m^{\frac{2}{3}}}_{0}e^{\mathrm{i}\frac{\zeta _m-\zeta _n+kx^2}{X_n}X_n}\langle x\rangle ^\mu \Psi (x)dx\right| \\&\quad \le C{X_m^\frac{1}{3}}\left( \left| \left( \langle x\rangle ^\mu \Psi \right) (X_m-X_m^{\frac{2}{3}})\right| +\int _0^{X_m-X_m^{\frac{2}{3}}}\left| \left( \langle x\rangle ^\mu \Psi \right) ^\prime (x)\right| dx\right) \\&\quad \le C{X_m^{\frac{1}{3}}}\left( X_m^\mu \left| \Psi (X_m-X_m^{\frac{2}{3}})\right| +\int _0^{X_m-X_m^{\frac{2}{3}}}\left( 2\langle x\rangle ^\mu \left( J_1+J_3\right) +\mu \langle x\rangle ^{\mu -1}\frac{ x}{\langle x\rangle }\left| \Psi (x)\right| \right) dx\right) \\&\quad \le C{X_m^{\frac{1}{3}}}X_m^\mu \left( \left| \Psi (X_m-X_m^{\frac{2}{3}})\right| +\int _0^{X_m-X_m^{\frac{2}{3}}}\left( J_1+J_3\right) dx\right) \\&\qquad +C\mu {X_m^\frac{1}{3}}\int _0^{X_m-X_m^{\frac{2}{3}}}\langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx. \end{aligned}$$

The estimate comes from three terms. Clearly, for $x\in [0,X_m-X_m^\frac{2}{3}]$ we have

$$\begin{aligned} |\Psi (x)|\le & {} C(X_m^2-x^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}\\\le & {} C(X_mX_n)^{-\frac{1}{4}}(X_m-x)^{-\frac{1}{4}}(X_n-x)^{-\frac{1}{4}}\le CX_m^{-\frac{5}{12}}X_n^{-\frac{1}{2}}. \end{aligned}$$

It follows that

$$\begin{aligned} X_m^\mu \left| \Psi (X_m-X_m^{\frac{2}{3}})\right| \le CX_m^{\mu -\frac{5}{12}}X_n^{-\frac{1}{2}}, \end{aligned}$$

and

$$\begin{aligned} \int _0^{X_m-X_m^{\frac{2}{3}}}\mu \langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx \le CX_m^{-\frac{5}{12}}X_n^{-\frac{1}{2}}\int _0^{X_m}\langle x\rangle ^{\mu -1}dx\le CX_m^{\mu -\frac{5}{12}}X_n^{-\frac{1}{2}}, \end{aligned}$$

together with

$$\begin{aligned} \int _0^{X_m-X_m^{\frac{2}{3}}}J_1dx \le C \int _0^{X_m-X_m^{\frac{2}{3}}} x(X_m^2-x^2)^{-\frac{5}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx \le C X_m^{-\frac{5}{12}}X_n^{-\frac{1}{2}}, \end{aligned}$$

and

$$\begin{aligned} \int _0^{X_m-X_m^{\frac{2}{3}}}J_3dx\le CX_m^{-\frac{3}{4}}X_n^{-\frac{1}{2}}\int _0^{X_m-X_m^{\frac{2}{3}}}(X_m-x)^{-\frac{11}{4}}dx \le C X_m^{-\frac{5}{12}}X_n^{-\frac{1}{2}}, \end{aligned}$$

we obtain $\left| \int ^{X_m-X_m^{\frac{2}{3}}}_{0} \langle x\rangle ^\mu e^{\mathrm{i}kx^2}\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx\right| \le CX_m^{-\frac{1}{12}+\mu }X_n^{-\frac{1}{2}}\le C(X_mX_n)^{\frac{\mu }{2}-\frac{1}{4}}.$ The estimate of rest parts of the integral is similar with Lemma 4.8. Thus,

$$\begin{aligned} \left| \int ^{X_m-X_m^{\frac{2}{3}}}_{0} \langle x\rangle ^\mu e^{ \mathrm{i}kx^2}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}}n^{\frac{1}{8}-\frac{\mu }{4}}},\quad m_0<m\le n. \end{aligned}$$

$\square$

Lemma 4.10

If $X_n>2X_m$ and $1<\beta \le 2$, then

$$\begin{aligned} \left| \int _{X_m-X_m^{\nu _1}}^{X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x) \overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{3}{16}-\frac{3}{16}\nu _1-\frac{\mu }{4}} n^{\frac{3}{16}-\frac{3}{16}\nu _1-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0<m\le n$ and

$$\begin{aligned} \nu _1= {\left\{ \begin{array}{ll} -1/3,&{}\quad 1<\beta <2,\\ 2/3, &{}\quad \beta =2. \end{array}\right. } \end{aligned}$$

Proof

First,

$$\begin{aligned}&\left| \int _{X_m-X_m^{\nu _1}}^{X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \le C\int _{X_m-X_m^{\nu _1}}^{X_m} \langle x\rangle ^\mu (X_m^2-x^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx\\&\quad \le C X_m^{-\frac{1}{4}+\mu } (X_n^2-X_m^2)^{-\frac{1}{4}}\int _{X_m-X_m^{\nu _1}}^{X_m}(X_m-x)^{-\frac{1}{4}}dx\\&\quad \le C X_m^{-\frac{1}{4}+\mu } (X_n^2-\frac{X_n^2}{4})^{-\frac{1}{4}}X_m^{\frac{3}{4}\nu _1}\le C X_m^{-\frac{3}{8}+\frac{3}{8}\nu _1+\frac{\mu }{2}} X_n^{-\frac{3}{8}+\frac{3}{8}\nu _1+\frac{\mu }{2}}. \end{aligned}$$

Similarly,

$$\begin{aligned} \left| \int _{X_m-X_m^{\nu _1}}^{X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _{j_1}^{(m)}(x) \overline{\psi _{j_2}^{(n)}(x)}dx\right| \le C X_m^{-\frac{3}{8}+\frac{3}{8}\nu _1+\frac{\mu }{2}} X_n^{-\frac{3}{8}+\frac{3}{8}\nu _1+\frac{\mu }{2}},\ \ j_1,j_2\in \{1,2\}. \end{aligned}$$

Thus we finish the proof. $\square$

Lemma 4.11

When $X_n > 2X_m$ and $1<\beta \le 2$,

$$\begin{aligned} \left| \int _{X_m}^{X_n}\langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}}n^{\frac{1}{12}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0<m\le n$.

Proof

When $X_n>2X_{m_0}$ , $X_m+X_m^{-\frac{1}{3}}\le \frac{X_n}{2}+1\le \frac{3}{4}X_n$. It follows

$$\begin{aligned}&\left| \int ^{X_m+X_m^{-\frac{1}{3}}}_{X_m}\langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le CX_m^\mu \int ^{X_m+X_m^{-\frac{1}{3}}}_{X_m}(x^2-X_m^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}dx\\&\quad \le CX_m^{-\frac{1}{4}+\mu }\left( X_n^2-(X_m+X_m^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}} \int _{X_m}^{X_m+X_m^{-\frac{1}{3}}}(x-X_m)^{-\frac{1}{4}}dx\\&\quad \le CX_m^{-\frac{1}{2}+\frac{\mu }{2}}X_n^{-\frac{1}{2}+\frac{\mu }{2}}. \end{aligned}$$

From $X_n> 2X_m$, we have $X_n-X_n^{-\frac{1}{3}}\ge \frac{3}{2}X_m$, together with Lemma 5.5 in [25], thus

$$\begin{aligned}&\left| \int _{X_m+X_m^{-\frac{1}{3}}}^{\frac{3}{2}X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le CX_m^\mu \int _{X_m+X_m^{-\frac{1}{3}}}^{\frac{3}{2}X_m} (x^2-X_m^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}e^{\mathrm{i}\zeta _m}dx\\&\quad \le C X_m^{-\frac{1}{4}+\mu }\left( X_n^2-(X_n-X_n^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}} \int _{X_m+X_m^{-\frac{1}{3}}}^{\frac{3}{2}X_m} (x-X_m)^{-\frac{1}{4}} e^{-(x-X_m)}dx\\&\quad \le C X_m^{-\frac{1}{4}+\mu }X_n^{-\frac{1}{6}}\int _0^\infty t^{-\frac{1}{4}}e^{-t}dt\le C X_m^{-\frac{1}{4}+\frac{\mu }{2}}X_n^{-\frac{1}{6}+\frac{\mu }{2}}. \end{aligned}$$

When $x\ge \frac{3}{2}X_m$, $x-X_m\ge \frac{1}{3}x$, it follows

$$\begin{aligned}&\left| \int _{\frac{3}{2}X_m}^{X_n-X_n^{-\frac{1}{3}}} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le C\int _{\frac{3}{2}X_m}^{X_n-X_n^{-\frac{1}{3}}} \langle x\rangle ^\mu (x^2-X_m^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}e^{\mathrm{i}\zeta _m}dx\\&\quad \le C X_m^{-\frac{1}{4}}\left( X_n^2-(X_n-X_n^{-\frac{1}{3}})^2\right) ^{-\frac{1}{4}} \int _{\frac{3}{2}X_m}^{X_n-X_n^{-\frac{1}{3}}} (x-X_m)^{-\frac{1}{4}+\mu } e^{-(x-X_m)}dx\\&\quad \le C X_m^{-\frac{1}{4}}X_n^{-\frac{1}{6}}\int _0^\infty t^{-\frac{1}{4}+\mu }e^{-t}dt\le C X_m^{-\frac{1}{4}}X_n^{-\frac{1}{6}}. \end{aligned}$$

Thus,

$$\begin{aligned}&\left| \int _{X_n-X_n^{-\frac{1}{3}}}^{X_n}\langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le C\int _{X_n-X_n^{-\frac{1}{3}}}^{X_n}x^\mu (x^2-X_m^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}e^{\mathrm{i}\zeta _m}dx\\&\quad \le C \left( (X_n-X_n^{-\frac{1}{3}})^2-X_m^2\right) ^{-\frac{1}{4}}X_n^{-\frac{1}{4}} \int _{X_n-X_n^{-\frac{1}{3}}}^{X_n} (X_n-x)^{-\frac{1}{4}}x^\mu e^{-\frac{1}{3}x}dx\\&\quad \le C X_m^{-\frac{1}{2}}X_n^{-\frac{1}{4}} \int _{X_n-X_n^{-\frac{1}{3}}}^{X_n} (X_n-x)^{-\frac{1}{4}}dx\\&\quad \le CX_m^{-\frac{1}{2}}X_n^{-\frac{1}{2}} \le C(X_mX_n)^{-\frac{1}{2}+\frac{\mu }{2}}. \end{aligned}$$

Combining with all the above, we have $\left| \int _{X_m}^{X_n}\langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)} dx \right| \le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}}n^{\frac{1}{12}-\frac{\mu }{4}}}$. The rest estimates are similar as above. $\square$

Combining with Lemmas 4.8, 4.9, 4.10, 4.11, we finish the proof of Lemma 4.6.

4.2.3 The Integral Estimate on $[0, X_n]$ When $X_m\le X_n\le 2X_m$

One can split the integral into

$$\begin{aligned} \int ^{X_n}_0 \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx =\left( \int ^{X_m^\frac{2}{3}}_0+\int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}}+ \int _{X_m-X_m^{\nu _2}}^{X_n}\right) \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx, \end{aligned}$$

and estimate them respectively, where

$$\begin{aligned} \nu _2= {\left\{ \begin{array}{ll} 1-\frac{\beta }{3},&{}\quad 1<\beta <2,\\ \frac{5}{9}, &{}\quad \beta =2. \end{array}\right. } \end{aligned}$$

Our main aim in this part is to build the following two lemmas.

Lemma 4.12

For $X_m \le X_n \le 2X_m$ and $k\ne 0$,

$$\begin{aligned} \displaystyle \left| \int _0^{X_n} \langle x\rangle ^\mu e^{\mathrm{i}kx^2}h_m(x)\overline{h_n(x)}dx\right| \le \frac{ C(|k|^{-1}\vee 1)}{m^{\frac{1}{18}-\frac{\mu }{4}} n^{\frac{1}{18}-\frac{\mu }{4}}}, \end{aligned}$$

where $C>0$, $m_0< m\le n$.

Lemma 4.13

For $X_m \le X_n \le 2X_m$, $k\ne 0$ and $1<\beta <2$,

$$\begin{aligned} \displaystyle \left| \int _0^{X_n} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C(|k|^{-1}\vee |\beta (\beta -1)(\beta -2)k|^{-\frac{1}{3}}\vee 1)}{m^{\frac{\beta }{24}-\frac{\mu }{4}} n^{\frac{\beta }{24}-\frac{\mu }{4}}}, \end{aligned}$$

where $C>0$, $m_0< m\le n$.

From a straightforward computation we have

Lemma 4.14

For $X_m \le X_n \le 2X_m$ and $1<\beta \le 2$,

$$\begin{aligned} \displaystyle \left| \int _0^{X_m^\frac{2}{3}} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{12}-\frac{\mu }{4}} n^{\frac{1}{12}-\frac{\mu }{4}}}, \end{aligned}$$

where $C>0$, $m_0< m\le n$.

Next we estimate the integral on $[X_m^\frac{2}{3},X_m-X_m^{\frac{5}{9}}]$, for which we discuss different cases as the following.

Lemma 4.15

If $X_m\le X_n\le 2X_m$, when $k>0$ and $0\le X_n^2-X_m^2\le kX_m^\frac{4}{3}$, then

$$\begin{aligned} \displaystyle \left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}kx^2}h_m(x)\overline{h_n(x)}dx\right| \le \frac{Ck^{-\frac{1}{2} }}{m^{\frac{7}{36}-\frac{\mu }{4}}n^{\frac{7}{36}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0< m\le n$.

Proof

We first estimate

$$\begin{aligned} \left| \displaystyle \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}kx^2}\psi ^{(m)}_1(x)\overline{\psi ^{(n)}_1(x)}dx\right| = \left| C\displaystyle \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}(\zeta _m-\zeta _n+kx^2)}\Psi (x)dx\right| . \end{aligned}$$

Notice that

$$\begin{aligned} g'(x)&\le \frac{kX_m^\frac{4}{3}X_m}{\sqrt{2X_m^{\frac{14}{9}}-X_m^{\frac{10}{9}}}\sqrt{2X_m^{\frac{14}{9}}-X_m^{\frac{10}{9}}}(\sqrt{2X_m^{\frac{14}{9}}-X_m^{\frac{10}{9}}}+\sqrt{2X_m^{\frac{14}{9}}-X_m^{\frac{10}{9}}})}-2k\\&\le \frac{k X_m^\frac{7}{3}}{2X_m^\frac{7}{3}}-2k= -\frac{3}{2}k, \end{aligned}$$

and by straightforward computation, $g''(x)\ge 0$. It follows $|g'(x)|\ge \frac{3}{2}k$ on $x\in [X_m^\frac{2}{3}, X_m-X_m^\frac{5}{9}]$. By Lemma 6.1,

$$\begin{aligned}&\left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}\frac{2}{3k}(\zeta _m-\zeta _n+kx^2)\cdot \frac{3k}{2}} (X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot f_m(x)\overline{f_n(x)}dx\right| \\&\quad \le Ck^{-\frac{1}{2}}\left[ \left| (\langle x\rangle ^\mu \Psi )(X_m-X_m^{\frac{5}{9}})\right| + \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \left| (\langle x\rangle ^\mu \Psi )^\prime (x)\right| dx\right] \\&\quad \le Ck^{-\frac{1}{2}}\left[ X_m^{\mu }\left| \Psi (X_m-X_m^{\frac{5}{9}})\right| + \langle X_m\rangle ^\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \left| \Psi ^\prime (x)\right| dx\right. \\&\qquad \left. +\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx \right] . \end{aligned}$$

We estimate the above terms one by one. Clearly,

$$\begin{aligned} \left| \Psi (X_m-X_m^{\frac{5}{9}})\right|&\le C\left( X_m^2-(X_m-X_m^{\frac{5}{9}})^2\right) ^{-\frac{1}{4}} \left( X_n^2-(X_m-X_m^{\frac{5}{9}})^2\right) ^{-\frac{1}{4}} \le C X_m^{-\frac{7}{9}}, \end{aligned}$$

and $\int _{X_m^\frac{2}{3}}^{X_m-X_m^{\frac{5}{9}}} \langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx\le CX_m^{-\frac{7}{9}+\mu }$. Besides, we have $|\Psi ^\prime (x)|\le C(J_1+J_3)$ and

$$\begin{aligned} \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}}J_1dx&\le C \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} x(X_m^2-x^2)^{-\frac{5}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx \le C X_m^{-\frac{7}{9}}, \end{aligned}$$

also, $\int _{X_m^\frac{2}{3}}^{X_m-X_m^{\frac{5}{9}}}J_3dx\le C X_m^{-\frac{7}{9}}$. Combining with all the estimates, we obtain

$$\begin{aligned} \left| \displaystyle \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}kx^{2}}\psi ^{(m)}_1(x)\overline{\psi ^{(n)}_1(x)}dx\right| \le Ck^{-\frac{1}{2}} X_m^{-\frac{7}{9}+\mu }. \end{aligned}$$

The estimates for the rest three terms are easier. In fact, by $m > m_0$,

$$\begin{aligned} \left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}kx^2}\psi ^{(m)}_2(x)\overline{\psi ^{(n)}_1(x)}dx\right|&\le CX_m^{-2+\mu }\int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}}(X_m^2-x^2)^{-\frac{1}{2}}dx\\&\le CX_m^{-\frac{5}{2}+\mu } \int _{0}^{X_m-X_m^\frac{5}{9}}(X_m-x)^{-\frac{1}{2}}dx \le CX_m^{-\frac{7}{9}+\mu }. \end{aligned}$$

The estimates for the other two are similar. Thus,

$$\begin{aligned} \left| \displaystyle \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}}\langle x\rangle ^\mu e^{\mathrm{i}kx^2}h_m(x)\overline{h_n(x)}dx\right| \le&\frac{Ck^{-\frac{1}{2} }}{m^{\frac{7}{36}-\frac{\mu }{4}}n^{\frac{7}{36}-\frac{\mu }{4}}}. \end{aligned}$$

$\square$

Lemma 4.16

For $k>0$, $X_m\le X_n\le 2X_m$, if $kX_m^\frac{4}{3}\le X_n^2-X_m^2$, then

$$\begin{aligned} \displaystyle \left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^\frac{5}{9}} \langle x\rangle ^\mu e^{\mathrm{i}kx^2}h_m(x)\overline{h_n(x)}dx\right| \le \frac{Ck^{-\frac{1}{3}}}{m^{\frac{1}{18}-\frac{\mu }{4}} n^{\frac{1}{18}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0< m\le n$.

Proof

For $kX_m^\frac{4}{3}\le X_n^2-X_m^2$, straightforward computation shows that $g'''(x)\ge 0$. So

$$\begin{aligned} g''(x)\ge g''(0)=\frac{1}{X_m}-\frac{1}{X_n} =\frac{X_n^2-X_m^2}{X_mX_n(X_m+X_n)}\ge \frac{kX_m^\frac{4}{3}}{3X_mX_mX_n}\ge \frac{k}{6}X_m^{-\frac{5}{3}}. \end{aligned}$$

Then by Lemma 6.1,

$$\begin{aligned}&\Big |\int _{X_m^\frac{2}{3}}^{X_m-X_m^{\frac{5}{9}}}\langle x\rangle ^\mu e^{\mathrm{i}\frac{\zeta _m-\zeta _n+kx^2}{kX_m^{-\frac{5}{3}}} kX_m^{-\frac{5}{3}}}(X_n^2-x^2)^{-\frac{1}{4}}(X_m^2-x^2)^{-\frac{1}{4}}f_m(x)\overline{f_n(x)}dx\Big | \nonumber \\&\quad \le Ck^{-\frac{1}{3}}X_m^{\frac{5}{9}}\left[ \left| (\langle x\rangle ^\mu \Psi )(X_m-X_m^{\frac{5}{9}})\right| + \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\frac{5}{9}}} \left| (\langle x\rangle ^\mu \Psi )^\prime (x)\right| dx\right] \nonumber \\&\quad \le Ck^{-\frac{1}{3}}X_m^{\frac{5}{9}}\bigg [X_m^\mu \left| \Psi (X_m-X_m^{\frac{5}{9}})\right| + X_m^\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\frac{5}{9}}}\left| \Psi ^\prime (x)\right| dx \nonumber \\&\qquad +\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\frac{5}{9}}} \langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx\bigg ]. \end{aligned}$$

The rest part of the proof is similar with Lemma 4.15. $\square$

Lemma 4.17

For $\forall k < 0, X_m\le X_n\le 2X_m$, we have

$$\begin{aligned} \displaystyle \left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C(|k|^{-1}\vee 1)}{m^{\frac{1}{6}\beta -\frac{1}{24}+\frac{\nu _2}{8}-\frac{\mu }{4} } n^{\frac{1}{6}\beta -\frac{1}{24}+\frac{\nu _2}{8}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0<m\le n$.

Proof

We first estimate

$$\begin{aligned} \left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i} kx^\beta }\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx\right| =\left| C\int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}}\langle x\rangle ^\mu e^{\mathrm{i}(\zeta _m-\zeta _n+kx^\beta )}\Psi (x)dx\right| . \end{aligned}$$

Notice that $g(x)=\sqrt{X_n^2-x^2}-\sqrt{X_m^2-x^2}-\beta kx^{\beta -1} \ge - kX_m^{\frac{2}{3}(\beta -1)}$ and $g^\prime (x)>0$, then by Lemma 6.1,

$$\begin{aligned}&\left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}}\langle x\rangle ^\mu e^{\mathrm{i}\frac{\zeta _m-\zeta _n+kx^\beta }{|k|X_m^{\frac{2}{3}(\beta -1)}}|k|X_m^{\frac{2}{3}(\beta -1)}} (X_m^2-x^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}f_m(x)\overline{f_n(x)}dx\right| \\&\quad \le \frac{C}{|k|}X_m^{-\frac{2}{3}(\beta -1)}\left[ \left| (\langle x\rangle ^\mu \Psi )(X_m-X_m^{\nu _2})\right| +\int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}}|(\langle x\rangle ^\mu \Psi )^\prime (x)|dx\right] \\&\quad \le C|k|^{-1}X_m^{-\frac{2}{3}(\beta -1)}\bigg [X_m^\mu \left| \Psi (X_m-X_m^{\nu _2})\right| +X_m^\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}}|\Psi ^\prime (x)|dx \\&\qquad +\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}}\langle x\rangle ^{\mu -1}|\Psi (x)|dx\bigg ]. \end{aligned}$$

The rest part of the proof is similar with Lemma 4.15. $\square$

Finally, we have

Lemma 4.18

For $k>0, 1<\beta <2, X_m\le X_n\le 2X_m$, we have

$$\begin{aligned} \displaystyle \left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C|\beta (\beta -1)(\beta -2)k|^{-\frac{1}{3}} }{m^{\frac{\beta }{24}-\frac{\mu }{4}} n^{\frac{\beta }{24}-\frac{\mu }{4}}}, \end{aligned}$$

where $m_0<m\le n$.

Proof

First we estimate

$$\begin{aligned} \left| \displaystyle \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi ^{(m)}_1(x)\overline{\psi ^{(n)}_1(x)}dx\right| = \left| C\displaystyle \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i}(\zeta _m-\zeta _n+kx^\beta )}\Psi (x)dx\right| . \end{aligned}$$

Since

$$\begin{aligned} g''(x)\ge -\beta (\beta -1)(\beta -2)kx^{\beta -3}\ge -\beta (\beta -1)(\beta -2)kX_m^{\beta -3}, \end{aligned}$$

then by Lemma 6.1,

$$\begin{aligned}&\left| \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i}(\zeta _m-\zeta _n+kx^\beta )} (X_m^2-x^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}\cdot f_m(x)\overline{f_n(x)}dx\right| \\&\quad \le C|\beta (\beta -1)(\beta -2)k|^{-\frac{1}{3}}X_m^{1-\frac{\beta }{3}}\left[ \left| (\langle x\rangle ^\mu \Psi )(X_m-X_m^{\nu _2})\right| + \int _{X_m^\frac{2}{3}}^{X_m-X_m^\nu } \left| (\langle x\rangle ^\mu \Psi )^\prime (x)\right| dx\right] \\&\quad \le C|\beta (\beta -1)(\beta -2)k|^{-\frac{1}{3}}X_m^{1-\frac{\beta }{3}}\bigg [X_m^\mu \left| \Psi (X_m-X_m^{\nu _2})\right| + X_m^\mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu }}\left| \Psi ^\prime (x)\right| dx\\&\qquad + \mu \int _{X_m^\frac{2}{3}}^{X_m-X_m^{\nu _2}} \langle x\rangle ^{\mu -1}\left| \Psi (x)\right| dx\bigg ]. \end{aligned}$$

The rest part of the proof is similar with Lemma 4.15. $\square$

For the last part of the integral, we have

Lemma 4.19

For $\forall k \ne 0, X_m\le X_n\le 2X_m$, $1<\beta \le 2$,

$$\begin{aligned} \displaystyle \left| \int _{X_m-X_m^{\nu _2}}^{X_m}\langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{8}-\frac{\nu _2}{8}-\frac{\mu }{4}} n^{\frac{1}{8}-\frac{\nu _2}{8}-\frac{\mu }{4}}}. \end{aligned}$$

Here $m_0< m\le n$.

Proof

First,

$$\begin{aligned} \left| \int _{X_m-X_m^{\nu _2}}^{X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x)\overline{\psi _1^{(n)}(x)}dx\right|&\le C\int _{X_m-X_m^{\nu _2}}^{X_m}\langle x\rangle ^\mu (X_m^2-x^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx\\&\le CX_m^\mu \int _{X_m-X_m^{\nu _2}}^{X_m}(X_m^2-x^2)^{-\frac{1}{2}}dx\\&\le CX_m^{-\frac{1}{2}+\mu }\int _{X_m-X_m^{\nu _2}}^{X_m}(X_m-x)^{-\frac{1}{2}}dx\\&\le CX_m^{-\frac{1}{2}+\mu }X_m^{\frac{\nu _2}{2}}\le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}-\frac{\nu _2}{8}}n^{\frac{1}{8}-\frac{\mu }{4}-\frac{\nu _2}{8}}}. \end{aligned}$$

It follows $\displaystyle \left| \int _{X_m-X_m^{\nu _2}}^{X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{8}-\frac{\mu }{4}-\frac{\nu _2}{8}}n^{\frac{1}{8}-\frac{\mu }{4}-\frac{\nu _2}{8}}}$. $\square$

Lemma 4.20

For $k\ne 0,\ X_m\le X_n\le 2X_m$ and $1<\beta \le 2$, we have

$$\begin{aligned} \Big |\int _{X_m}^{X_n} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\Big |\le \frac{C}{m^{-\frac{\nu _2}{16}+\frac{1}{8}-\frac{\mu }{4}} n^{-\frac{\nu _2}{16}+\frac{1}{8}-\frac{\mu }{4}}}. \end{aligned}$$

Proof

For the integral on $[X_m,X_n]$, we discuss two different cases:

Case 1.$X_n - X_n^{-\nu _2} \ge X_m + X_m^{-\nu _2}$. We split the integral into three parts. First,

$$\begin{aligned}&\left| \int ^{X_m+X_m^{-\nu _2}}_{X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le \int ^{X_m+X_m^{-\nu _2}}_{X_m} \langle x\rangle ^\mu (x^2-X_m^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx\\&\quad \le C X_m^\mu X_m^{-\frac{1}{4}} X_n^{-\frac{1}{4}} (X_n-X_m-X_m^{-\nu _2})^{-\frac{1}{4}}\int ^{X_m+X_m^{-\nu _2}}_{X_m}(x-X_m)^{-\frac{1}{4}}dx\\&\quad \le C X_m^\mu X_m^{-\frac{1}{4}} X_n^{-\frac{1}{4}} X_n^{\frac{\nu _2}{4}}X_m^{-\frac{3}{4}\nu _2} \le C n^{-\frac{1}{8}-\frac{\nu _2}{8}+\frac{\mu }{4}}m^{-\frac{1}{8}-\frac{\nu _2}{8}+\frac{\mu }{4}}. \end{aligned}$$

When $x \ge X_m+X_m^{-\nu _2}$, we have $\mathrm{i}\zeta _m \le -(x-X_m)$. It follows

$$\begin{aligned}&\left| \int _{X_m+X_m^{-\nu _2}}^{X_n-X_n^{-\nu _2}} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le C\int _{X_m+X_m^{-\nu _2}}^{X_n-X_n^{-\nu _2}} \langle x\rangle ^\mu (x^2-X_m^2)^{-\frac{1}{4}} (X_n^2-x^2)^{-\frac{1}{4}}e^{ \mathrm{i}\zeta _m}dx\\&\quad \le C (2X_m)^\mu X_m^{-\frac{1}{4}}\left( X_n^2-(X_n-X_n^{-\nu _2})^2\right) ^{-\frac{1}{4}} \int _{X_m+X_m^{-\nu _2}}^{X_n-X_n^{-\nu _2}} (x-X_m)^{-\frac{1}{4}} e^{ \mathrm{i}\zeta _m}dx\\&\quad \le C (2X_m)^\mu X_m^{-\frac{1}{4}}X_n^{\frac{\nu _2-1}{4}}\int _0^\infty t^{-\frac{1}{4}}e^{-t}dt\le C n^{-\frac{1}{8}+\frac{\nu _2}{16}+\frac{\mu }{4}}m^{-\frac{1}{8}+\frac{\nu _2}{16}+\frac{\mu }{4}}. \end{aligned}$$

Finally, from

$$\begin{aligned}&\left| \int _{X_n-X_n^{-\nu _2}}^{X_n}\langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le \int _{X_n-X_n^{-\nu _2}}^{X_n} \langle x\rangle ^\mu (x^2-X_m^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx\\&\quad \le C (2X_m)^\mu \left( (X_n-X_n^{-\nu _2})^2-X_m^2\right) ^{-\frac{1}{4}}X_n^{-\frac{1}{4}} \int _{X_n-X_n^{-\nu _2}}^{X_n} (X_n-x)^{-\frac{1}{4}}dx\\&\quad \le C n^{-\frac{1}{8}-\frac{\nu _2}{8}+\frac{\mu }{4}}m^{-\frac{1}{8}-\frac{\nu _2}{8}+\frac{\mu }{4}}, \end{aligned}$$

it follows $\Big |\int _{X_m}^{X_n} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\Big |\le \frac{C}{(nm)^{\frac{1}{8}-\frac{\mu }{4}-\frac{\nu _2}{16}}}.$

Case 2. $X_n - X_n^{-\nu _2} < X_m + X_m^{-\nu _2}$. In fact, notice that the function $(x-X_m)^{-\frac{1}{4}}(X_n-x)^{-\frac{1}{4}}$ is symmetric on $[X_m,X_n]$, we obtain

$$\begin{aligned}&\left| \int _{X_m}^{X_n} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }\psi _1^{(m)}(x) \overline{\psi _1^{(n)}(x)}dx\right| \\&\quad \le \int _{X_m}^{X_n} \langle x\rangle ^\mu (x^2-X_m^2)^{-\frac{1}{4}}(X_n^2-x^2)^{-\frac{1}{4}}dx\\&\quad \le C (2X_m^\mu )X_m^{-\frac{1}{4}}X_n^{-\frac{1}{4}}\int _{X_m}^{X_n} (x-X_m)^{-\frac{1}{4}}(X_n-x)^{-\frac{1}{4}}dx\\&\quad \le C (2X_m^\mu )X_m^{-\frac{1}{4}}X_n^{-\frac{1}{4}}\int _{X_m}^{X_n} (x-X_m)^{-\frac{1}{2}}dx\\&\quad \le C X_m^\mu X_m^{-\frac{1}{4}}X_n^{-\frac{1}{4}}(X_n-X_m)^\frac{1}{2}\le (nm)^{\frac{1}{4} \mu -\frac{1}{8}-\frac{1}{4}\nu _2}. \end{aligned}$$

Thus, $\displaystyle \left| \int _{X_m}^{X_n} \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{m^{\frac{1}{8}+\frac{\nu _2}{8}-\frac{\mu }{4}}n^{\frac{1}{8}+\frac{\nu _2}{8}-\frac{\mu }{4}}}.$ Combining with the above two cases, we finish the proof. $\square$

Lemmas 4.12 and 4.13 follow directly by the lemmas in Sect. 4.2.3. Combining with all the lemmas in this section we finish the proof of Lemma 1.2 for $1<\beta \le 2$.

5 Proof of Lemma 1.2 When $\beta >2$

In the following we will suppose that $m\le n$ without losing the generality. As the case $1<\beta \le 2$ we only need to estimate the integral on $[0,\infty ]$. We first apply Theorem 3.1 in [24] to obtain the integral estimates on $[2X_m,\infty )$ as follows.

Lemma 5.1

For $\mu \ge 0$, then we have

$$\begin{aligned} \left| \int _{2X_m}^\infty \langle x\rangle ^\mu e^{\mathrm{i}kx^\beta }h_m(x)\overline{{h}_n(x)}dx\right| \le \frac{C}{(mn)^{\frac{1}{4}\left( \frac{1}{3}-\mu \right) }}. \end{aligned}$$

Proof

By Theorem 3.1 in [24] we have

$$\begin{aligned}&\left| \int _{2X_m}^\infty \langle x\rangle ^\mu e^{\mathrm{i kx^\beta }}h_m(x)\overline{{h}_n(x)}dx\right| \\&\quad \le C\int _{2X_m}^\infty \langle x\rangle ^{-1} \langle x\rangle ^{\mu +1}|h_m(x)||h_n(x)|dx\\&\quad \le C\Vert \langle x\rangle ^{-1}\Vert _{L^2({\mathbb {R}})}^\frac{1}{2}\cdot \Vert \langle x\rangle ^{\mu +1} h_m(x)\Vert _{L^2(\{x:|x|\ge 2X_m\})}^\frac{1}{2}\cdot \Vert h_n\Vert _{L^\infty ({\mathbb {R}})}\\&\quad \le CX_m^{-\frac{1}{6}}X_n^{-\frac{1}{6}}\le C(mn)^{\frac{1}{4}\left( \mu -\frac{1}{3}\right) }. \end{aligned}$$

$\square$

Next we consider the integral on $[0,2X_m]$. Define $\nu _3=\frac{5\beta -4}{2(\beta -1)(2\beta -1)}\in (0,1)$ when $\beta >2$. In the following we will discuss two different cases depending on whether $X_n^{\nu _3}\ge 2X_m$ or not.

The Case $X_n^{\nu _3}\ge 2X_m$ In this case we directly estimate the rest integral on $[0,2X_m]$.

Lemma 5.2

For $X_n^{\nu _3}\ge 2X_m$ and $\beta >2$, then

$$\begin{aligned} \displaystyle \left| \int _{0}^{2X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{(mn)^{\frac{1}{4}\left( \frac{\beta -2}{4\beta -2}-\mu \right) }}. \end{aligned}$$

Proof

From Lemma 6.2,

$$\begin{aligned}&\left| \int _{0}^{2X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x) \overline{h_n(x)}dx\right| \\&\quad \le \int _{0}^{2X_m} \langle x\rangle ^\mu |h_m(x)||h_n(x)|dx \le C X_m^\mu \int _{0}^{X_n^{\nu _3}} |h_m(x)||h_n(x)|dx\\&\quad \le C X_m^\mu \Vert h_m(x)\Vert _{L^2}\Vert h_n(x)\Vert _{L^p}\left( \int _0^{X_n^{\nu _3}}dx\right) ^{\frac{1}{q}}\\&\quad \le C X_m^{\mu }X_n^{-\left( \frac{1}{2}-\frac{1}{p}\right) }X_n^{\frac{\nu _3}{q}} \le C (mn)^{\frac{1}{4}\left( \mu -\frac{\beta -2}{4\beta -2}\right) }, \end{aligned}$$

where $\frac{1}{p}+\frac{1}{q}=\frac{1}{2},q=\frac{4\beta -3}{\beta -1}>4.$ $\square$

The Case $X_n^{\nu _3}<2X_m$ In this case we split the rest integral into two parts as follows.

Lemma 5.3

For $k\ne 0, X_n^{\nu _3}< 2X_m$ and $\beta >2$, then

$$\begin{aligned} \displaystyle \left| \int _{X_n^{\nu _3}}^{2X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C(|k\beta |^{-1}\vee 1)}{(mn)^{\frac{1}{4}\left( \frac{\beta -2}{4\beta -2}-\mu \right) }}. \end{aligned}$$

Proof

Denote $\psi (x)=\langle x\rangle ^{\mu }h_m(x)h_n(x),\phi (x)=x^\beta$. Notice that when $x\in [X_n^{\nu _3},2X_m]$, $|\phi '(x)|\ge \beta X_n^{\nu _3(\beta -1)}$. So by Lemma 6.1,

$$\begin{aligned} \left| \int _{X_n^{\nu _3}}^{2X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x) \overline{h_n(x)}dx\right| \le C|k\beta |^{-1}X_n^{-\nu _3(\beta -1)}\left[ \left| \psi (2X_m)\right| + \int _{X_n^{\nu _3}}^{2X_m} \left| \psi ^\prime (x)\right| dx\right] , \end{aligned}$$

where $\left| \psi (2X_m)\right| \le C X_m^{\mu }$, and $|\int _{X_n^{\nu _3}}^{2X_m} \langle x\rangle ^{\mu -1}h_m(x)\overline{h_n(x)}dx|\le CX_m^{\mu -1}$. By Lemma 6.2, $\Vert h_m'(x)\Vert _{L^2}\le CX_m$. Thus,

$$\begin{aligned} \left| \int _{X_n^{\nu _3}}^{2X_m} \langle x\rangle ^{\mu }h_m'(x)\overline{h_n(x)}dx\right| \le CX_m^{\mu }X_m=CX_m^{\mu +1},\\ \left| \int _{X_n^{\nu _3}}^{2X_m} \langle x\rangle ^{\mu }h_m(x)\overline{h_n'(x)}dx\right| \le CX_m^{\mu }X_n\le CX_m^{\mu }X_n. \end{aligned}$$

Combining with all the conclusions we have

$$\begin{aligned} \left| \int _{X_n^{\nu _3}}^{2X_m} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x) \overline{h_n(x)}dx\right|&\le C|k\beta |^{-1}X_n^{-\nu _3(\beta -1)+1} X_m^{\mu }\le C |k\beta |^{-1} (X_mX_n)^\frac{\mu }{2}X_n^{-\frac{\beta -2}{4\beta -2}}\\&\le C |k\beta |^{-1}(mn)^{\frac{1}{4}\left( \mu -\frac{\beta -2}{4\beta -2}\right) }. \end{aligned}$$

$\square$

Lemma 5.4

For $X_n^{\nu _3}< 2X_m$ and $\beta >2$, then

$$\begin{aligned} \left| \int _{0}^{X_n^{\nu _3}} \langle x\rangle ^\mu e^{\mathrm{i}kx^{\beta }}h_m(x)\overline{h_n(x)}dx\right| \le \frac{C}{(mn)^{\frac{1}{4}\left( \frac{\beta -2}{4\beta -2}-\mu \right) }}. \end{aligned}$$

Proof

The proof is similar as Lemma 5.2, we omit it. $\square$

Hence, combining the above four Lemmas we obtain Lemma 1.2 when $\beta >2$.

Appendix

In the following we will introduce two technical lemmas without proof. The first lemma provides an estimate of oscillatory integral. For more details, see [37].

Lemma 6.1

[37] Suppose $\phi$ is real-valued and smooth in (A, B), $\psi$ is complex-valued, and that $|\phi ^{(k)}(x)|\ge 1$ for all $x\in (A,B)$. Then

$$\begin{aligned} \left| \int _A^B e^{\mathrm{i\lambda \phi (x)}}\psi (x)dx\right| \le c_k\lambda ^{-1/k}\left[ |\psi (B)|+\int _A^B|\psi ^\prime (x)|dx\right] \end{aligned}$$

holds when:

• (i) $k\ge 2$, or

• (ii) $k=1$ and $\phi ^\prime (x)$ is monotonic. The bound $c_k$ is independent of $\phi$, $\psi$ and $\lambda$.

The second lemma shows that the $L^p$-norm of the eigenfunction of harmonic oscillating operator can be controlled by its $L^2$-norm.

Lemma 6.2

[24] Sppose that h(x) is the eigenfunction of Herimite operator with the corresponding eigenvalue $\mu ^2$. Then

$$\begin{aligned} \Vert h\Vert _{L^p}\le \mu ^{\rho (p)}\Vert h\Vert _{L^2}, \end{aligned}$$

where

$$\begin{aligned} \rho (p)= {\left\{ \begin{array}{ll} -(\frac{1}{2}-\frac{1}{p}), &{}\quad 2\le p<4,\\ -\frac{1}{3}+\frac{1}{3}(\frac{1}{2}-\frac{1}{p}),&{}\quad 4<p\le \infty . \end{array}\right. } \end{aligned}$$