A Mechanical Model of Brownian Motion for One Massive Particle Including Slow Light Particles

Abstract

We provide a connection between Brownian motion and a classical mechanical system. Precisely, we consider a system of one massive particle interacting with an ideal gas, evolved according to non-random mechanical principles, via interaction potentials, without any assumption requiring that the initial velocities of the environmental particles should be restricted to be “fast enough”. We prove the convergence of the (position, velocity)-process of the massive particle under a certain scaling limit, such that the mass of the environmental particles converges to 0 while the density and the velocities of them go to infinity, and give the precise expression of the limiting process, a diffusion process.

1 Introduction

In this paper, we consider a mechanical model consisting of a massive particle in an ideal gas (i.e., a Rayleigh gas model), and present a mathematical proof that under certain conditions, when the mass m of a gas particle converges to 0, while the number density and velocity distribution of the gas scale like $m^{-\frac{1}{2}}$, the motion of the massive particle converges to a Brownian motion (i.e., a Langevin process).

1.1 Preliminary

Brownian motion, first observed by Brown in 1827, is a well-known physical phenomenon concerning the dynamics of a small particle immersed into a fluid in equilibrium, e.g., a grain of pollen in a glass of water [14]. It is an interesting problem in mathematical physics to describe the Brownian motion phenomenology by classical mechanical models.

The first physical explanation of Brownian motion was given by Einstein: the motion is coming about as a result of the repeated collisions of the massive particle with the numerous much smaller but faster fluid atoms. In more mathematical terms the explanation is often presented in the following rough way: since the massive particle is collided by a big number of very light water particles, if we could assume that the interactions from each light particle at each time are independent, then by the central limit theorem for the sum of i.i.d. random variables, this will give in a suitable limit the Brownian motion.

However, this assumption of independence can hardly be justified, even in a model where only interactions through collisions are considered, since there exists the possibility of re-collisions. This becomes a more evident and significant drawback when considering the model of interactions caused by potentials. Therefore, the actual motion of the massive particle can not be explained as resulting from a sum of i.i.d. random variables, it is not even a Markov process.

So in order to study this phenomenon more precisely, one needs to construct some model which is consistent with the mentioned dependence on the past. In such a model, a massive particle interacts with a gas of infinitely many light particles, with the dynamics fully deterministic and Newtonian, as long as the initial condition is given. The only source of randomness is from the initial configuration of the light particles. The problem we will be concerned with is to describe the motion of the massive particle in the Brownian limit, where the mass m of the light particles goes to 0, while the density and the velocities of them have order $m^{-1/2}$. The scaling is done in such a way that the variance of the momentum transfer stays of order 1. See the introductions of [12, 13] for a more detailed explanation with respect to the reason of this scaling. We notice that one expects that the non-Markovian character of the dynamics disappears when $m \rightarrow 0$, since at least when the initial velocities of the light particles are fast enough, the interactions should be short enough.

This type of model, called a mechanical model of Brownian motion, was first introduced and studied by Holley [9], for the case where the whole system is in dimension $d=1$, and the interactions are given by collisions. This model was later extended by, e.g., Dürr et al. [6,7,8], Calderoni et al. [2], to the case of higher dimensional spaces. Szász and Tóth [15] also considered some related problem. We notice that in all these papers, the interactions were just of the collisions type.

Kusuoka and Liang [12] and Liang [13] considered this type of problem with interactions given by compactly supported smooth potentials, under certain conditions. In particular, [12, 13] assumed that all light particles are sufficiently fast, precisely, all light particles have initial velocities not less than $m^{-1/2} (2C_0+1)$, with $C_0$ a positive constant determined by the potential functions. See (1.2) for the definition of $C_0$. This technical condition was used in an essential way in [12, 13]. In detail, this assumption ensured that light particles cross the valid range of interaction in a bounded time, and never reenter the valid range. Therefore, when considering the behavior of the light particles, we could use the approximation that the massive particle is frozen. This is our freezing approximation which will be explained later in details.

However, in a physically more relevant model, there also exist light particles with initial kinetic energies less than $2C_0+1$, equivalently, with initial velocities less than $m^{-1/2} (2C_0+1)$. As explained, this could not be covered by [12, 13]. Indeed, in a continuous interaction potential model with possibly not sufficiently fast light particles, the effective interaction time durations between particles might be unbounded. This is heuristically clear by considering the much simpler model with the massive particle frozen—if the initial relative position $x-X$ of a light particle (with the massive particle frozen) is parallel to its initial velocity v but with opposite directions, and with its energy $\frac{m}{2} |v|^2 + U (x-X)$ equal to the maximum of the potential U (which is assumed to be finite in the present paper), then after its first hit, the light particle will stop at the position that attains the maximum of the potential, hence the effective interaction time duration would be infinity.

Certainly, in a repulsive interaction potential model as discussed in the present paper, as long as $x-X$ is not totally parallel to v, the particle could leave the valid range at some finite time. However, the observation above suggests that the effective interaction duration could nevertheless be very long. In our model the situation is even more complicated since the massive particle is also evolving.

In this paper, we consider this continuous repulsive interaction potential model with possibly not “sufficiently fast” light particles, which, as discussed above, means that the effective interaction durations are not bounded. We notice that our model does not include the Lennard-Jones potential model, especially, does not include the Weeks–Chandler–Andersen potential model (see, for example, [5] for its formulation, the potential of which is repulsive and compactly supported), in which the potential diverges to $\infty$ as the inter-particle distance converges to 0. On the other hand, we notice that the difficulties of our present model and the Lennard-Jones model are different—as long as we assume that the initial energies of the light particles are bounded below (which is assumed in the present paper), in the Lennard–Jones model, the effective interaction time duration is always finite (but it has its own difficulty that the second derivative of the potential function is unbounded, which means that the nearness of the positions does not imply the nearness of the forces, see also Problem 3 of Sect. 4); whereas in the present model, the difficulty is the possibility of unbounded interaction time durations.

There are a lot of further papers related to our topic, in the sense of “deriving Brownian motion from dynamics involving a dependence on the past” (or “re-collisions” for the collisional interactions). For example, Chernov and Dolgopyat [4] considered a model with only one massive particle and one light particle but with full re-collisions, Caprino et al. [3] considered a model with the mean-field approximation from the beginning, and with a different scaling. See also the references therein. Simulation results on this topic have also been discussed, for example, Kim and Karniadakis [11] and the references therein.

However, in the literature, to the best knowledge of the author, there are not so many papers concerning our problem of “deriving Brownian motion from a Hamiltonian dynamics consisting of massive particle(s) with infinitely many ideal gas light particles”, except the ones [2, 6,7,8,9, 12, 13] quoted before. Especially, for the case where interactions are given by potentials, [12, 13] are the only ones that the author knows. As for a potential model with possibly slow light particles (hence possibly long interaction durations as explained), we believe that the present paper is the first progress. We notice that the Markov approximation method used in, e.g. [6, 15], is not applicable to smooth potential interaction models. We prove our convergence in this paper with the help of the martingale problem theory. This framework of proof was also used in [2, 12, 13]. Precisely, we first prove the tightness of the considered family of probabilities, and then prove that any cluster point of it must be the unique diffusion described. See also Sect. 2.1 for more detailed explanations.

We remark that our model (i.e., with smooth potential interactions and possibly low initial energy light particles) has the following evident difficulty when compared with hard core models or with potential models with sufficiently fast light particles: the system is strongly non-Markovian, due to the extensions in times of the interactions. Since we are not assuming that the initial velocities of light particles are at least $m^{-1/2} (2C_0+1)$, which was the case for [12, 13], the interacting times could be arbitrarily long, depending on the initial states of the considered light particles. For hard core models, although still non-Markovian because of the possibility of re-collisions, each interaction happens in an instant; while for smooth potential interaction models, if the initial velocities of the light particles are at least $m^{-1/2} (2C_0+1)$ as in [12, 13], the interacting times are bounded. However, in our model, as explained, this could not be the case.

One more evident difference between these models concerns the velocities of the light particles after interactions: in the hard core model, after each collision, the gas particle changes its velocity a lot—almost reflecting—, since the masses of the light particles and of the massive particle(s) are too different; in the model of smooth potential interaction but with light particles which are fast enough, as discussed in [12, 13], each light particle simply “almost passes through” (this was also one of the main heuristic ideas of [12, 13]); whereas in our model, the velocities of light particles after interactions could be in any direction varying from “almost passing through” to “almost reflecting”.

Finally, we remark that, for the case where there is only one massive particle (as in this paper and in part of [12, 13]), our limiting process for smooth potential interaction model coincides with the one for the hard core model, which was given by [6]. See [13, Remark 1] for a detailed explanation. (For the case where there are at least two massive particles, we could not make the comparison since the limiting process for the hard core model is unknown).

1.2 Description of the Model and Statement of the Result

Let us now describe our model in details. We consider a dynamical system that consists of one massive particle immersed into an environment of infinitely many light particles with mass $m > 0$ (we will take the limit $m \rightarrow 0$ later on). Without loss of generality, we assume that the mass of the massive particle is equal to 1. The initial condition of the environment is given by ${\widetilde{\omega }} \in Conf (\mathbf{R}^d \times \mathbf{R}^d)$. Here $Conf ( * )$ stands for the set of all non-empty closed subsets of $*$ which have no cluster point. Also, $(x, v) \in {\widetilde{\omega }}$ means that there exists an environmental particle with position x and velocity v at time 0. The distribution of ${\widetilde{\omega }}$ will be given later. As soon as the initial condition of the system is given, our system is totally deterministic, Newtonian, with the Hamiltonian given by $\frac{1}{2} |V|^2 + \sum _{(x, v)} \frac{m}{2} |v|^2 + \sum _{(x, v)} U (X-x)$. Here (X, V) is the state (i.e., the position and the velocity) of the massive particles, and (x, v) is the state of each light particle. So the interaction between the massive particle and light particles is given by a potential function U. We assume that $U \in C_0^{\infty } (\mathbf{R}^d)$, the set of smooth functions on $\mathbf{R}^d$ with compact supports, and concentrate ourselves to the case where U is spherical-symmetric and gives us a repulsive force, so we are assuming the following:

• U1. $d > 1$, $U \in C_0^{\infty } (\mathbf{R}^d)$, and there exists a constant $R_U > 0$ and a smooth function $h: [0, \infty ) \rightarrow [0, \infty )$ such that $U (x) = h(|x|)$ for any $x \in \mathbf{R}^d$, $U(x) = 0$ if $|x| \ge R_U$, and $h'(a) < 0$ for any $a \in (0, R_U)$. Also, we assume that $h''(0) < 0$.

For any $a, b \in \mathbf{R}^d$, let (a, b) and $a \cdot b$ denote their inner product, and when $a \ne \mathbf{0}$, let $\pi ^{\perp }_a b$ denote the component of b that is perpendicular to a, i.e., $\pi ^{\perp }_a b := b - (b, \frac{a}{|a|}) \frac{a}{|a|}$. The spherical-symmetry of U in the assumption U1 ensures that the freezing-approximation particles can evolve only in the directions of v and $\pi ^{\perp }_v (x-X)$, so helps us to estimate the interacting time durations (see, for example, the proof of Lemma 3.13), see also Lemma 4.2 for the benefit of the spherical-symmetry of U; and the repulsive property in the assumption U1 ensures that at least the freezing-approximation particles leave the valid range in a finite time for sure as long as $\pi ^{\perp }_v (x-X) \ne 0$.

For any initial condition ${\tilde{\omega }}$ and time $t \in [0, \infty )$, let $(X^{(m)}(t, {\tilde{\omega }}), V^{(m)}(t, {\tilde{\omega }}))$ denote the state of the massive particle at time t, and for any $(x, v) \in {\tilde{\omega }}$, let $(x^{(m)}(t, x, v, {\tilde{\omega }}), v^{(m)}(t, x, v, {\tilde{\omega }}))$ denote the state at time t of the light particle which had state (x, v) at time 0. So our dynamical system is given by the following infinite system of ordinary differential equations:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle { \frac{d}{dt} X^{(m)} (t, {\widetilde{\omega }}) = V^{(m)} (t, {\widetilde{\omega }}), }\\ \displaystyle { \frac{d}{dt} V^{(m)} (t, {\widetilde{\omega }}) = - \int _{\mathbf{R}^d \times \mathbf{R}^d} \nabla U \big (X^{(m)} (t, {\widetilde{\omega }}) - x^{(m)} (t, x, v, {\widetilde{\omega }})\big ) \mu _{{\widetilde{\omega }}} (dx, dv), } \\ \displaystyle { \big (X^{(m)} (0, {\widetilde{\omega }}), V^{(m)} (0, {\widetilde{\omega }})\big ) = (X_0, V_0), } \\ \displaystyle { \frac{d}{dt} x^{(m)} (t, x, v, {\widetilde{\omega }}) = v^{(m)} (t, x, v, {\widetilde{\omega }}), } \\ \displaystyle { m \frac{d}{dt} v^{(m)} (t, x, v, {\widetilde{\omega }}) = - \nabla U \big (x^{(m)} (t, x, v, {\widetilde{\omega }}) - X^{(m)} (t, {\widetilde{\omega }})\big ), } \\ \displaystyle { \big (x^{(m)} (0, x, v, {\widetilde{\omega }}), v^{(m)} (0, x, v, {\widetilde{\omega }})\big ) = (x, v), \qquad (x, v) \in {\widetilde{\omega }}. } \end{array} \right. \end{aligned}$$

(1.1)

Here $\mu _{{\widetilde{\omega }}} (\ \cdot \ )$ is the counting measure determined by ${\widetilde{\omega }}$: $\mu _{{\widetilde{\omega }}} (A) = \sharp ({\widetilde{\omega }} \cap A)$ for any $A \in \mathcal {B} (\mathbf{R}^d \times \mathbf{R}^d)$, the set of Borel subsets of $\mathbf{R}^d \times \mathbf{R}^d$. ($\sharp ( \ \cdot \ )$ thus denoting the number of points in the argument). When there is no risk of confusion, we will omit the superscript (m) and the parameter ${\widetilde{\omega }}$. So X(t) stands for $X^{(m)} (t, {\widetilde{\omega }})$, etc. .

The only randomness of our model comes from the distribution of the environmental initial condition ${\widetilde{\omega }}$, which is given by the following. Let $\rho : [0, \infty ) \times \mathbf{R}^d \rightarrow [0, \infty )$ be a measurable function such that $\sup _{z \in \mathbf{R}^d} \rho (u, z) \rightarrow 0$ rapidly as $u \rightarrow \infty$ (see conditions A1, A2 and A3 below for detailed assumptions with respect to $\rho$). Let ${\widetilde{\lambda _m}}$ be the non-atomic Radon measure on $\mathbf{R}^d \times \mathbf{R}^d$ given by

$$\begin{aligned} {\widetilde{\lambda _m}} (dx, dv) = m^{\frac{d-1}{2}} \rho \left( \frac{m}{2} |v|^2, x - X_0 \right) dx dv, \end{aligned}$$

and let ${\widetilde{P_m}} (d {\widetilde{\omega }})$ be the Poisson point process with the intensity measure ${\widetilde{\lambda _m}}$. So ${\widetilde{P_m}}$ is a probability measure on ${\widetilde{\Omega }} = Conf (\mathbf{R}^d \times \mathbf{R}^d)$. We assume that the distribution of ${\widetilde{\omega }}$ is given by ${\widetilde{P_m}}$. (See, e.g., [10] for more details with respect to Poisson point processes).

• A1. There exists a constant ${\overline{v}} > 0$ such that $\rho (u, z) = 0$ for any $u < \frac{1}{2} {\overline{v}}^2$ and $z \in \mathbf{R}^d$.

• A2. $\rho (u, -z) = \rho (u, z)$ for any $z \in \mathbf{R}^d$ and $u \in [0, \infty )$. Also, there exist a function $\rho _0: [0, \infty ) \rightarrow [0, \infty )$ and a constant $R_1>0$ such that $\rho (u, z) = \rho _0 (u)$ as long as $|z| \ge R_1$ for any $u \in [0, \infty )$.

• A3. $\int _{\mathbf{R}^d} (1+|v|^3) \rho _{max} (\frac{1}{2} |v|^2) dv < \infty .$ Here $\rho _{max} (u) := \sup _{z \in \mathbf{R}^d} \rho (u, z)$, $u \in [0, \infty )$.

(A1) means that all light particles have initial velocities not smaller than $m^{-1/2} {\overline{v}}$, equivalently, we are assuming that the kinetic energies of all light particles are bounded below. This enables us to estimate the interacting time durations of those light particles with initial skews $|\pi ^{\perp }_v (x-X(\cdot ))|$ which are not too small (see Proposition 2.1). We emphasize again that ${\overline{v}}$ could be any positive number in our model. (The case with ${\overline{v}} = 0$ is more complicated, see Problem 1 of Sect. 4).

(A2) is satisfied, for example, if $\rho (u, z)$ is a function of u and U(z) , i.e., if the initial distribution of the light particles depends on the kinetic energies and the potentials of the light particles. In particular, different from [12, 13], our model is also consistent with a model such that the initial distribution of the light particles is not affected by the existence of the massive particle (i.e., the case where $\rho (u, z)$ does not depend on z). The second half of (A2) implies that even if the initial position of the massive particle does affect the initial distribution of light particles, it affects only a bounded neighborhood of the massive particle, so since the velocities of all light particles are of order $m^{-\frac{1}{2}}$, after a time period that is short enough (precisely, for $t \ge m^{\frac{1}{2}} a_m$, where $a_m$ is defined in Sect. 2.5), the distribution of the incoming light particles is almost independent of the massive particle. Also, since we have only one massive particle in our model, the first half of (A2) (i.e., the even property of $\rho (u, z)$ with respect to z) is enough to ensure that the mean of the forces of our freezing-approximation $\varphi$ is 0 (see Lemma 3.33). We remark that, in the case where there are more than one massive particles as in [12, 13], to ensure that the mean of the forces of the freezing-approximation $\varphi$ is 0, one needed to assume that $\rho$ is a function of $\frac{m}{2} |v|^2 + \sum _{i=1}^N U_i(x-X_{i, 0})$, i.e., the initial distribution of the light particles depends on the total energy (the summation of kinetic energy and the potentials) of the light particles.

(A3) assumes that $\rho (u, z)$ decreases rapidly enough when $u \rightarrow \infty$, uniformly with respect to z. So we are assuming that there are not too many extremely fast light particles.

We notice that our intensity function of the present model includes that of [12, 13] as a special case. To see this, write the function $\rho$ of [12, 13] as ${\overline{\rho }}$ to distinguish the notations. Then the model of [12, 13] is given by $\rho (u, z) := {\overline{\rho }} (u + U(z-X_0))$. Now our (A2) is satisfied trivially, and our (A3) is a direct consequence of (A2) of [12, 13]. Also, the assumption (A1) of [12, 13] ensures that (A1) of our present paper is satisfied with ${\overline{v}}$ given by $2 C_0 + 1$. Here

$$\begin{aligned} C_0 := \sqrt{2 R_U \Vert \nabla U \Vert _{\infty }}. \end{aligned}$$

(1.2)

We emphasize again that the constant $2C_0+1$ was used in an essential way in [12, 13]: they proved the fact that if the initial speed of a light particle is not smaller than $(2 C_0 + 1) m^{-1/2}$, then until any given $T>0$, the period of time that it could be in the valid range of the interaction with the massive particle is bounded by $3 C_0^{-1} (R_U + |X_0| + n T + 1) m^{1/2}$ (see, for example, [12, Proposition 3.6.5]). In other words, [12, 13] assumed from the beginning that the initial velocities of light particles are fast enough such that the interactions with the massive particle could not “stop” these light particles. In particular, when describing the behavior of light particles, [12, 13] could use the so-called freezing-approximation $\varphi (t, x, v; X)$ and $\psi (t, x, v; X)$ (see (1.3) and (1.4) for their definitions, and see Lemma 3.20 for the meaning of our expression “approximation”). See also Sect. 2 for more explanations with respect to this freezing approximation.

This idea of freezing-approximation is also used in this paper, with more precise estimates – as explained, since we are not assuming in this paper that the initial velocities are that fast, it does not hold in our model that the interacting times are bounded. Our idea to tackle this problem will be explained in Sect. 2.

Let

$$\begin{aligned} E= & {} \big \{ (x, v) \in \mathbf{R}^d \times (\mathbf{R}^d {\setminus } \{ 0 \}); \ x \cdot v = 0 \big \}, \\ E_v= & {} \big \{ x \in \mathbf{R}^d; x \cdot v = 0 \big \}, \qquad v \in \mathbf{R}^d {\setminus } \{ 0 \}, \end{aligned}$$

and let $\nu (dx, dv)$ be the measure on E given by $\nu (dx, dv) = |v| {\widetilde{\nu }} (dx; v) dv$, where ${\widetilde{\nu }} (dx; v)$ is the Lebesgue measure on $E_v$. E is used in the domain of the ray representation $\Psi$ defined in (2.1) later.

By a slight modification of [12, Sect. 3.3, pp. 751–758] (the only modification needed is the definition of $G_n$ – let $G_n := \{ (t, x, v) \in \mathbf{R} \times E; |x|< R_0, |t| < T + m^{1/2} {\overline{v}}^{-1} R_0 \}$ now. Here $R_0 \ge R_U + |X_0| + nT + 1$ as defined in (2.4) ), we have that under (U1) (A1) and (A3), for any $m > 0$, (1.1) has a unique solution for ${\widetilde{P_m}}$-almost every ${\widetilde{\omega }}$.

Remark 1.1

If we define $G_n := \{ (t, x, v) \in \mathbf{R} \times E; |x|< R_0, |t| < T + m^{1/2} |v|^{-1} R_0 \}$, and let $\lambda$ be as defined in (2.2), then with $S_{d-1} := \int _{\mathbf{R}^{d-1}} 1_{ \{ |x| \le 1 \} } dx$, we have by (A3) that

$$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{G_n} (u, x, v) (1+|v|^2) \lambda (du, dx, dv) \\&\quad \le m^{-1} \int _{\mathbf{R}^d} (1+|v|^2) |v| \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) dv \int _{E_v} 1_{ \{ |x|< R_0 \} } {\widetilde{\nu }} (dx; v) \int _\mathbf{R} 1_{ \left\{ |u|< T + m^{1/2} |v|^{-1} R_0 \right\} } d u \\&\quad = m^{-1} (2 R_0)^{d-1} S_{d-1} \int _{\mathbf{R}^d} (1+|v|^2) \big (T|v| + m^{1/2} R_0\big ) \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) dv \\&\quad < \infty \end{aligned}$$

for any fixed m, T and n. Therefore, by checking the proof of [12, Sect. 3.3, pp. 751–758] carefully, we can get the uniqueness of the solution of (1.1) for ${\widetilde{P_m}}$-almost every ${\widetilde{\omega }}$ and any $m > 0$ by assuming (A3) and (U1) (with $U \in C_0^2 (\mathbf{R}^d)$ instead of $U \in C_0^{\infty } (\mathbf{R}^d)$) only. This ensures the existence of the solution of our Problem 1 presented in Sect. 4. $\square$

We are interested in the limit behavior of the massive particle when $m \rightarrow 0$. In order to formulate our main result, we first prepare several notations, which are the same as in [12]. First, for any $X \in \mathbf{R}^{d}$ and $(x, v) \in \mathbf{R}^{2d}$, let $\varphi (t, x, v; X) = (\varphi ^0 (t, x, v; X), \varphi ^1 (t, x, v; X))$ denote the solution of the following system of ordinary differential equations:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle {\frac{d}{dt} \varphi ^0 (t, x, v; X) = \varphi ^1 (t, x, v; X)} \\ \displaystyle {\frac{d}{dt} \varphi ^1 (t, x, v; X) = - \nabla U \big (\varphi ^0 (t, x, v; X) - X\big ) } \\ \displaystyle {(\varphi ^0 \big (0, x, v; X), \varphi ^1 (0, x, v; X) \big ) = (x, v)}. \end{array} \right. \end{aligned}$$

(1.3)

We notice that (1.3) is the same as the second half of (1.1) with $m = 1$, except that the quantity $X^{(1)}(t)$ of (1.1) is substituted by X in (1.3).

As in [12, 13], for any $X \in \mathbf{R}^d$ and $(x, v) \in E$, we have that

$$\begin{aligned} \psi (t, x, v; X) := (\psi ^0 (t, x, v; X), \psi ^1 (t, x, v; X)) := \lim _{s \rightarrow \infty } \varphi (t+s, x - s v, v; X) \end{aligned}$$

(1.4)

is well-defined. Indeed, $\psi (t, x, v; X) = \varphi (t+s, x - sv, v; X)$ for any $s \ge (-t) \vee (\frac{R_U+|X|+1}{|v|})$. $\psi (t, x, v; X)$ with proper X is our freezing-approximation mentioned above.

Also, for any $(x, v) \in E$, $X, V \in \mathbf{R}^d$ and $a \in \mathbf{R}$, let z(t; x, v, X, V, a) denote the solution of

$$\begin{aligned} \left\{ \begin{array}{l} \frac{d^2}{dt^2} z(t) = - \nabla ^2 U (\psi ^0 (t, x, v, X) - X) \Big ( z(t) - (t+a) V \Big ), \\ \lim \nolimits _{t \rightarrow - \infty } z(t) = \lim \nolimits _{t \rightarrow - \infty } \frac{d}{dt} z(t) = 0. \end{array} \right. \end{aligned}$$

(1.5)

So z(t; x, v, X, V, a) is a linear function of V.

Our limiting diffusion generator L on function over $\mathbf{R}^{2d}$ is given by the following:

$$\begin{aligned} L = \frac{1}{2} \sum _{k, l = 1}^d a_{kl} \frac{\partial ^2}{\partial V_k \partial V_l} + \sum _{k, l= 1}^d b_{kl} V_l \frac{\partial }{\partial V_k} + \sum _{k = 1}^d V_k \frac{\partial }{\partial X_k}, \end{aligned}$$

(1.6)

with

$$\begin{aligned} a_{kl}= & {} \int _E \left( \int _{- \infty }^{\infty } \nabla _k U \big (\psi ^0 (t, x, v; X) - X\big ) dt \right) \\&\times \left( \int _{- \infty }^{\infty } \nabla _l U \big (\psi ^0 (t, x, v; X) - X\big ) dt \right) \rho _0 \Big (\frac{1}{2} |v|^2\Big ) \nu (dx, dv), \end{aligned}$$

and $b_{kl}: \mathbf{R}^d \rightarrow R$, $k, l = 1, \cdots , d$, are $C^{\infty }$-functions determined by the following relation:

$$\begin{aligned}&- \int _E \left( \int _{- \infty }^{\infty } \nabla ^2 U \big (\psi ^0 (t, x, v, X) - X\big ) z(t, x, v, X, V, -t) dt \right) \rho _0 \Big (\frac{1}{2} |v|^2\Big ) \nu (dx, dv) \\&\quad = \sum _{l=1}^d b_{kl} V_l^{\ell }. \end{aligned}$$

The coefficients a and b correspond to the 0-order and the 1-order approximations, respectively, of our freezing-approximation. We notice that as proved in [13, pp. 248–249], a and b are indeed independent of X, since there is only one massive particle in our model. We express them in the present way, since the heuristic meanings of the present formulations are more clear. One more advantage of this formulation is to maintain consistency of the notations with [12, 13], which discussed the case with more than one massive particles. We also remark that the integrals with respect to t in the definitions of $a_{kl}$ and $b_{kl}$ are finite by Proposition 2.1 (2).

Finally, our metric on $C ([0, \infty ); \mathbf{R}^{2d})$ is given by

$$\begin{aligned} dist (w_1, w_2) := \sum _{k = 1}^{\infty } 2^{-k} \Big ( 1 \wedge \max _{t \in [0, k]} |w_1 (t) - w_2 (t)| \Big ), \qquad w_1, w_2 \in \mathbf{R}^{2d}. \end{aligned}$$

Our main result is the following.

Theorem 1.2

Assume (U1) and (A1)–(A3). Also, assume that $d > 2 (1+\Vert h''\Vert _{\infty })(- h''(0))^{-1/2} + 1$. Then the distribution of $\{ (X^{(m)} (t), V^{(m)} (t)); t \ge 0 \}$ under ${\widetilde{P_m}}$ converges weakly as $m \rightarrow 0$ to the diffusion process with generator L in $(C ([0, \infty ); \mathbf{R}^{2d}), dist)$.

Remark 1.3

We remark that the assumption $d > 2 (1+\Vert h''\Vert _{\infty })(- h''(0))^{-1/2} + 1$ in Theorem 1.2 implies that $d > 5$. This assumption is closely related to our estimate of the interacting time durations (Proposition 2.1). It might be possible to weaken this condition if one could estimate the effective interaction time durations more accurately. On the other hand, as explained below, we need at least $d > 3$ to apply our method of this paper.

On the one hand, we need $\alpha < \frac{\sqrt{\varepsilon _1}}{2 (1+\Vert h''\Vert _{\infty })}$ (here $\varepsilon _1$ is a lower bound of $- \frac{h'(y)}{y}$ in a neighbor of 0, which is approximately equal to $-h''(0)$, and $\alpha$ is a constant that we introduce in Sect. 2 to handle the singularity) in our estimate of the effective interaction durations of the light particles satisfying $| x - \pi ^{\perp }_v X ({\tilde{r}})| \ge m^{\alpha }$ (see Proposition 2.1 (4)), since we need our approximation error, which is dominated by $m^{\frac{1}{2}} | x - \pi ^{\perp }_v X ({\tilde{r}})|^{- (1+\Vert h''\Vert _{\infty }) \varepsilon _1^{-\frac{1}{2}}}$, to be of order o(1) as long as $| x - \pi ^{\perp }_v X ({\tilde{r}})| \ge m^{\alpha }$ (see the proof of Lemma 3.24 for details). On the other hand, we need $\alpha (d-1) - \frac{1}{2} > 0$ to ensure that the total force from singular light particles (i.e., those light particles with their initial states satisfying $| x - \pi ^{\perp }_v X (\cdot ) | < m^{\alpha }$) to be negligible: since the initial velocities are of order $m^{-\frac{1}{2}}$, and the density of light particles is of order $m^{-\frac{1}{2}}$, we get that the number of light particles satisfying this initial condition is of order $m^{\alpha (d-1) - 1}$; also, the force from each of these light particles is of order $m^{\frac{1}{2}}$, so the total force from singular light particles is of order $m^{\alpha (d-1) - \frac{1}{2}}$.

Combining these, we need $\frac{\sqrt{\varepsilon _1}}{2 (1+\Vert h''\Vert _{\infty })} > \frac{1}{2 (d-1)}$, equivalently, at least $d > (1+\Vert h''\Vert _{\infty })(- h''(0))^{-1/2} + 1$ to apply our method of this paper.

Moreover, for the case $d \le 3$, as the following calculation shows, although we can prove that the diffusion coefficient $a_{kl}$ of our limiting generator L is finite, we are not even able to prove that the drift coefficient is finite.

By Proposition 2.1 (2) below, we have that $| \int _{- \infty }^{\infty } \nabla _k U (\psi ^0 (t, x, v; X) - X) dt | = | \psi ^1 (-\infty , x, v, X) - \psi ^1 (\infty , x, v, X) | 1_{ \{ |x| \le R_0 \} } \le 2 |v| 1_{ \{ |x| \le R_0 \} }$. So $| a_{kl} | \le \int _E 4 |v|^2 1_{ \{ |x| \le R_0 \} } \rho _0 (\frac{1}{2} |v|^2) \nu (dx, dv) \le 4 (2 R_0)^{d-1} \int _{\mathbf{R}^d} |v|^3 \rho _0 (\frac{1}{2} |v|^2) dv < \infty$.

However, for the drift coefficient $b_{kl}$, our estimation of this paper (see Lemma 3.38 and Proposition 2.1 (2) below) only ensures that the term $\int _{- \infty }^{\infty } \nabla ^2 U (\psi ^0 (t, x, v, X) - X) z(t, x, v, X, V, -t) dt$ in the definition of $b_{kl}$ is dominated by a constant multiplies $\Big ( \log \frac{1}{| x - \pi ^{\perp }_v X|} \Big )^2 | x - \pi ^{\perp }_v X|^{- (1+C_1) \varepsilon _1^{-\frac{1}{2}}}$, which is integrable with respect to $\rho _0 (\frac{1}{2} |v|^2) \nu (dx, dv)$ on E only if $d > (1+C_1) \varepsilon _1^{-\frac{1}{2}} + 1 ( \ge 3 )$.

We close this remark by emphasizing that we do not mean that our limiting generator is not well-defined for $d \le 3$, we just mean that we are not sure whether it is well-defined for $d \le 3$ by our estimate in this paper. $\square$

The paper is organized as follows. In Sect. 2, we explain the main ideas of this paper. In Sect. 3, we present the proof of Theorem 1.2. In Sect. 4, we give a brief summary and several concluding remarks.

2 Several Notations and Basic Ideas

In this section, we define several notations and explain the main ideas of this paper. In Sect. 2.1, we try to explain the main ideas with minimum mathematics. The mathematical formulations of these ideas are given in Sects. 2.2–2.5. Finally, in Sect. 2.6, we give a brief summary.

2.1 Explanation of the Main Ideas

For any $n \in \mathbf{N}$, let $\sigma _n := \inf \{ t > 0; |V_t| \ge n \}$. We notice that in order to prove Theorem 1.2, a result with respect to $t \in [0, \infty )$, it suffices to prove the assertion for $t \in [0, T \wedge \sigma _n]$ for any $T>0$ and $n \in \mathbf{N}$. Choose and fix any $T>0$ and $n \in \mathbf{N}$ from now on.

Since most of the basic ideas of [12, 13] except the boundedness of the interacting time durations (which, as explained in Sect. 1, does not hold in our model) is also used in the present paper, let us start with the explanation of these common ideas.

The first idea that needs to be mentioned is the ray representation defined in (2.1) below. Since until time $T \wedge \sigma _n$, the massive particle is in $\{ X \in \mathbf{R}^d; |X| \le |X_0| + nT \}$, a bounded domain, the effective interaction range for light particles is also bounded. Therefore, a light particle with its initial position far enough from the origin keeps a uniform motion until its first entrance to this bounded effective range. This is the base of our idea of the ray representation : if the effective interaction time durations are bounded, then for any $t \in [0, T \wedge \sigma _n ]$ and $(r, x, v) \in \mathbf{R} \times E$, a light particle with initial state (position, velocity) $\Psi (r, x, - m^{\frac{1}{2}} v) = (x - m^{\frac{1}{2}} r v, m^{-\frac{1}{2}} v)$ is in the effective interaction range if and only if $r \approx t$; in other words, at each time t, the massive particle get forces from only those light particles with initial states $\Psi (r, x, - m^{\frac{1}{2}} v)$ with $r \approx t$. Although the effective interaction time durations are not bounded in the present paper, this idea of ray representation is still useful in order to give the entrance time of a light particle to the valid interaction range. The model after application of this ray representation is given by (2.3) below.

Freezing approximation is also an important idea. As explained, for the case with bounded effective interaction durations as in [12, 13], when considering the force caused by a marked light particle, the movement of the massive particle during the effective interaction time duration of this light particle is very small, so the evolution of this light particle can be approximated by the evolution of a particle with the massive particle frozen. This is our freezing approximation $\varphi$ and $\psi$. Also, the first order approximation error (i.e., $z (\cdot )$ defined by (1.5), see Lemma 3.21 below) gives us the drift term in our limiting process. In the present paper, although the effective interaction durations are not bounded, by presenting accurate estimates of the effective interaction time durations (see $t_1$ defined in (2.11)), we prove that this idea of freezing approximation is still applicable.

The common framework of the present paper and [12, 13] is as follows: with the help of the explained ray representation and freezing approximation, we prove that the velocity of the massive particle can be re-expressed as the sum of a martingale term, a smooth term and a negligible term, with each of these terms tight. Precisely, the families of the distributions of these stochastic processes are tight in $\wp (D([0, T]; \mathbf{R}^d))$, the set of probability measures on the Skorohod space $D([0, T]; \mathbf{R}^d)$ (see Proposition 3.31 for the formulation). See Sect. 3.4.1 for a brief review of the Skorohod space and the basic facts that are used to get the tightness, and see [1] for more details. In particular, we get that the family of the distributions of position/velocity processes is tight in $\wp (C([0, T]; \mathbf{R}^d))$, the set of probability measures on $C([0, T]; \mathbf{R}^d)$ (the set of continuous $\mathbf{R}^d$-valued functions on [0, T] ). We recall the well-known fact that on a separable metric space, the tightness of a family of probabilities is equivalent to the sequentially compactness of its closure, i.e., any infinite countable subset of it has a subsequence that converges weakly. We then prove the convergence of the drift term (see Sect. 3.5) and the convergence of the quadratic variation of the martingale term (see (3.24)) as $m \rightarrow 0$. This implies that any of the cluster point(s) of the considered distributions of position/velocity processes as $m \rightarrow 0$ is the unique solution of the martingale problem L, i.e., for any $f \in C_0^{\infty } (\mathbf{R}^d \times \mathbf{R}^d)$, after taking the limit $m \rightarrow 0$, the distribution of $\Big \{ f (X(t \wedge \sigma _n), V(t \wedge \sigma _n)) - f (X_0, V_0) - \int _0^{t \wedge \sigma _n} L f ( X(s), V(s) ) ds; \ t \in [0, T] \Big \}$ under $P_m$ is a martingale. So by the one-to-one correspondence between diffusion and solution of the corresponding martingale problem, we get the expected convergence of the position/velocity processes. See [12, Sect. 5] for the detailed calculation.

We would like to emphasize that, although both our stochastic process $\{ V^{(m)}(t); t \in [0, T] \}$ before taking limit and the corresponding expected limiting process are continuous, its martingale part (M(t) defined by (3.20)) is not continuous.

We now explain the original main ideas of the present paper. As explained, the biggest difference between the present paper and [12, 13], which is also the biggest difficulty of the present paper, is that the effective interaction time durations are not bounded in our model, even for the freezing approximation. However, we prove in this paper (see Proposition 2.1 (2) and (2.11)) that this unboundedness happens only in a order of $\log | \pi ^{\perp }_v (x-X)|$, the logarithm of the initial skew between the particles. This log order estimate of the effective interaction time durations is one of the main ideas of this paper, and is used in an essential way; when applying Gronwall’s inequality to estimate the approximation error between the states of the light particle and its freezing approximation, we get an estimate of exponential order with respect to the effective interaction time duration (see, e.g., Lemma 3.20). So a log-order estimate of the effective interaction time is necessary. Also, we remark that as mentioned in Remark 1.3, opposite to [12, 13], since the effective interaction time durations are not bounded in the present model, the approximation errors of our freezing approximation (for example, the quantity $\kappa _1 (t)$ defined in (3.10)) could not be of order $m^{\frac{1}{2}}$ uniformly with respect to $(x, v) \in E$. Nevertheless, we get the expected convergence by an accurate calculation (see also Sect. 2.4, especially (2.12)).

The estimation of the effective interaction time duration for a light particle is more complicated, and the method used by [12, 13] is not applicable in our model. In [12, 13], as in the case for the freezing approximation, since the initial velocity of a light particle were not less than $m^{-\frac{1}{2}} (2C_0+1)$, the light particle keeps a speed not less than $m^{-\frac{1}{2}} C_0$ in the direction of its initial velocity, so the light particle passes the valid interaction range in a short time (see [12, Propositions 3.6.1 and 3.6.5], which is formulated as Proposition 3.22 (1) in this paper). However, as explained, for a light particle with not sufficiently high initial energy, this could not be the case – even for a freezing approximation, the velocity of the particle might be 0 at some point. Therefore, since the massive particle is also evolving, it is hopeless to track the evolution of the light particle during its effective interaction duration directly.

This difficulty is solved in the following way, with the help of the freezing approximation: since the effective interaction time duration of the freezing approximation is of log order as explained (see Sect. 3.1), as long as the light particle is not too singular (i.e., if $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, where $\alpha$ is a constant restricted by (2.9) and (2.10), see also the paragraph following it), the approximation error of our freezing-approximation (which is discussed in Sect. 3.2) could be small enough, so with the help of the information with respect to the behavior of the freezing approximation particle after its exiting time from the valid interaction range, we get that the light particle could not be in the valid interaction range after this time, too (see Sect. 3.3).

The introduction of $\alpha$ is also one of our main ideas of this paper: instead of trying to “track” all of the light particles, we use the fact that the total effect from those “singular” light particles (i.e., those with $| x - \pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha }$) is small enough.

Also, as explained right after the assumption A2, the system needs a short time before the density of the incoming light particles getting almost no affect from the massive particle. Since the efficient interaction time durations are not bounded in the present paper, the definition of this short time is certainly different from that of [12, 13], and is valid for only those non-singular light particles. The explicit definition is given in Sect. 2.5.

In Sects. 2.2–2.5, we present the mathematical formulations of the ideas that we just explained: we formulate the ray representation and present our model after having applied the ray representation in Sect. 2.2; we give the explicit definition of the necessary constants in Sects. 2.3 and 2.5; and we formulate our estimations of the effective interaction durations of the light particles and the freezing approximation particles in Sect. 2.4.

2.2 Ray Representation

We first formulate the ray representation explained in Sect. 2.1. Our ray representation $\Psi$ is defined as follows:

$$\begin{aligned} \Psi :&\mathbf{R} \times E \rightarrow \mathbf{R}^d \times (\mathbf{R}^d {\setminus } \{ 0 \}), \nonumber \\&(s, (x, v)) \mapsto \Psi (s, (x, v)) = (x-sv, v). \end{aligned}$$

(2.1)

We remark that in this new space $\mathbf{R} \times E$, v is still the initial velocity of the light particle, while x is not the initial position of it anymore: now x is only the component of its initial position that is perpendicular to the velocity. Also, s plays an important role as explained: it is approximately the time that the light particle with initial condition $\Psi (s, x, v)$ interacts with the massive particle.

We now apply the ray representation (2.1) to our model. See [12] for detailed calculation. Let $\Omega = Conf (\mathbf{R} \times E)$, let $\lambda (dr, dx, dv)$ be the measure on $\Omega$ given by

$$\begin{aligned} \lambda (dr, dx, dv) = \lambda _m (dr, dx, dv) = m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v - X_0 \right) dr \nu (dx, dv), \end{aligned}$$

(2.2)

(with $\rho$ and $\nu$ as in Sect. 1), and let $P_m (d \omega ) = P_{\lambda _m} (d \omega )$ be the Poisson point process on $Conf (\mathbf{R} \times E)$ with intensity function $\lambda _m (dr, dx, dv)$. Then we can convert our problem with respect to $Conf (\mathbf{R}^d \times \mathbf{R}^d)$ to a problem with respect to $Conf (\mathbf{R} \times E)$. Our $\omega \in \Omega$ has distribution $P_m$, and for each initial condition $\omega$, we are considering the following system of infinite ODEs (we omit the superscription (m) for the sake of simplicity):

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle { \frac{d}{dt} X (t, \omega ) = X (t, \omega ), } \\ \displaystyle { \frac{d}{dt} V (t, \omega ) = - \int _{\mathbf{R} \times E} \nabla U \left( X (t, \omega ) - x\left( t, \Psi \left( r, x, m^{- \frac{1}{2}} v\right) , \omega \right) \right) \mu _{\omega } (dr, dx, dv), } \\ \displaystyle { \big (X (0, \omega ), V (0, \omega )\big ) = (X_0, V_0), } \\ \displaystyle { \frac{d}{dt} x \big (t, \Psi (r, x, v), \omega \big ) = v \big (t, \Psi (r, x, v), \omega \big ), } \\ \displaystyle { m \frac{d}{dt} v (t, \Psi (r, x, v), \omega ) = - \nabla U \big (x (t, \Psi (r, x, v), \omega ) - X (t, \omega )\big ), } \\ \displaystyle { \big (x (0, \Psi (r, x, v), \omega ), v (0, \Psi (r, x, v), \omega )\big ) = \Psi (r, x, v), \qquad (r, x, v) \in \omega . } \end{array} \right. \end{aligned}$$

(2.3)

2.3 Time for Freezing and Definition of the “Singularity”

As explained, we are going to approximate $x (s, \Psi (r, x, m^{-1/2} v))$ by $\varphi ^0 (m^{-\frac{1}{2}} s, \Psi (m^{-\frac{1}{2}} r, x, v; X))$ or $\psi ^0 (m^{-\frac{1}{2}} (s-r), x, v; X)$ with some proper X. In this section, let us first explain a little bit more about the freezing time of our freezing approximation: we are going to use different X in our freezing approximations (equivalently, different time to freeze the massive particle) for different purposes. To be precise, we take $X = X_0$ when proving that the force during the very first time duration (i.e., the duration $s \in [0, m^{\frac{1}{2}} a_m ]$, where $a_m$ is defined in Sect. 2.5 below) is negligible; we take $X = X(s)$ when proving the convergence in the last step of our proof; and we take $X = X({\tilde{r}})$ with ${\tilde{r}}$ given by (2.5) below for estimating the effective interaction duration of the light particle (see Proposition 2.1 (4)) and in order to get a measurable approximation (see Lemma 3.19 below. The measurability is necessary for estimating the variation of the corresponding term under Poisson point process measure). See (3.18) for the concrete usage of these freezing-times in the decomposition of $V(t \wedge \sigma _n)$.

Let

$$\begin{aligned}&R_0 := (R_U \vee R_1) + |X_0| + nT + 1, \nonumber \\&\tau := ({\overline{v}} \wedge C_0)^{-1} R_0. \end{aligned}$$

(2.4)

As will be proved later (Proposition 2.1 (3) or Proposition 3.22 (3)), if $r \ge t + m^{1/2} \tau$, then $| x (u, \Psi (r, x, m^{-1/2}v)) - X(u)| \ge R_U$ for all $u \in [0, t]$, which means that this light particle does not enter the valid range until time t. So the behavior of the massive particle at time t is $\mathcal {F}_{(-\infty , t + m^{1/2} \tau ] \times E}$-measurable (Lemma 3.19). For any $r \in \mathbf{R}$, ${\tilde{r}}$ is defined by

$$\begin{aligned} {\tilde{r}} := ((r - m^{1/2} \tau ) \vee 0) \wedge T \wedge \sigma _n. \end{aligned}$$

(2.5)

Next, let us present our definition of singularity. Write $C_1 := \Vert h'' \Vert _{\infty }$. Then

$$\begin{aligned}&| \nabla ^2 U (x) y | \le C_1 |y|, \qquad \hbox { for any } x, y \in \mathbf{R}^d, \end{aligned}$$

(2.6)

$$\begin{aligned}&| \nabla U (y_1) - \nabla U (y_2) | \le C_1 |y_1 - y_2|, \qquad \hbox { for any } y_1, y_2 \in \mathbf{R}^d. \end{aligned}$$

(2.7)

The proof is easy, and is given in Appendix.

Recall that by assumption, $h''(0) < 0$ and $d > 2 (1+C_1)(- h''(0))^{-1/2} + 1$. So there exists a constant $\varepsilon _1 \in (0, -h''(0))$ such that

$$\begin{aligned} d > 2 (1+C_1) \varepsilon _1^{-1/2} + 1. \end{aligned}$$

(2.8)

Therefore, there exists a constant $\alpha > 0$ such that

$$\begin{aligned}&\alpha> \frac{1}{d-1}, \qquad ( d>1\hbox { by assumption (U1))} \end{aligned}$$

(2.9)

$$\begin{aligned}&\alpha < \frac{\varepsilon _1^{1/2}}{2(1+C_1)}. \end{aligned}$$

(2.10)

(so, in particular, $\alpha \le \frac{1}{4}$). These conditions are chosen such that $\alpha$ satisfies the following conditions: On the one hand, $\alpha$ is big enough ((2.9)) such that the number of “singular” light particles (i.e., with $| x - \pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha }$) converges to 0. See, for example, Lemmas 3.41, 3.42 and 3.45. On the other hand, $\alpha$ is small enough ((2.10)) such that those light particles with $| x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ leave the valid interaction range in a certain time and the error of our freezing-approximation is small enough (Proposition 2.1 (4) and Lemma 3.20).

2.4 Estimation of the Effective Interaction Time Duration

Since our interaction force is repulsive, for a freezing approximation, $| x - \pi ^{\perp }_v X |$ is a lower bound of the inter-particle distance (see, e.g., Lemma 3.11), and plays an important role when estimating the effective interaction time duration.

Let $\varepsilon _2 > 0$ be a constant such that $h(y) \le \frac{1}{36} {\overline{v}}^2$ for any $y \ge R_U - \varepsilon _2$. Since $h'(0) = 0$ by (U1) and $- \varepsilon _1 > h''(0)$ by definition, there exists a constant $\varepsilon _3 > 0$ such that $h'(y) \le - \varepsilon _1 y$ for any $y \in (0, \varepsilon _3)$. Let $\varepsilon _4 > 0$ be a constant such that $- h'(y) \ge \varepsilon _4$ for any $y \in [\varepsilon _3, \frac{3}{4} R_U \vee (R_U - \varepsilon _2)]$, and let $C_2 := 9 \tau + 2 \varepsilon _4^{-1} (2C_0 + \sqrt{2 \Vert U\Vert _{\infty }})$. Finally, for any $y \ge 0$ and $v \in \mathbf{R}^d$ satisfying $|v| \ge {\overline{v}}$, let

$$\begin{aligned} t_1 (v, y) := \left\{ \begin{array}{ll} 2 \tau , &{} \hbox { if } \quad |v| \ge 2 C_0 + 1, \\ C_2 + 2 \varepsilon _1^{-1/2} \Big ( \log \frac{2 \varepsilon _3}{y} \Big ) 1_{ \left\{ y \le \frac{R_U}{2} \wedge (2 \varepsilon _3) \right\} }, &{} \hbox { if } \quad |v| \in [ {\overline{v}}, 2 C_0 + 1). \end{array} \right. \end{aligned}$$

(2.11)

Here $\log \frac{1}{y}$ is understood to be $\infty$ for $y = 0$, i.e., $t_1 (v, y) = \infty$ if $|v| \in [ {\overline{v}}, 2 C_0 + 1)$ and $y = 0$.

We shall prove in Proposition 3.22 that if $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, then the light particle with initial condition $\Psi (r, x, m^{-1/2} v)$ could be in the valid range at time $m^{1/2} r + t$ only if $t \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)]$. As claimed, this is also one of our main ideas of this paper: although the effective interaction time durations of these light particles (with $| x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$) are not of order $m^{1/2}$, which was the case for $|v| \ge 2 C_0 + 1$ as proven by [12], they are dominated by $m^{1/2} (t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + \tau )$. See Proposition 2.1 for the precise statement. On the other hand, since $t_1$ is at most of order $\log$, with the help of the general result

$$\begin{aligned} \int _{(0, R]} |\log r|^k r^a dr < \infty , \qquad \hbox { for any } R> 0, k \in \mathbf{N} \hbox { and } a > -1, \end{aligned}$$

(2.12)

this does not cause any essential problem after taking the integral. Several more estimates of this type are given in Lemma 4.5.

We summarize our key estimates for the efficient interaction time durations as follows:

Proposition 2.1

For any $(x, v) \in E$, $r \in \mathbf{R}$, $s \in [0, T \wedge \sigma _n]$ and $X \in \mathbf{R}^d$ satisfying $|X| \le |X_0| + nT$, we have the following:

• 1. $| \nabla U (\varphi ^0 (s, x-rv, v; X) - X) | \le \Vert \nabla U \Vert _{\infty } 1_{ \{ |x-\pi ^{\perp }_v X| \le R_U + 1 \} } 1_{ \{ r \in [-\tau , s + \tau ] \} }.$

• 2. $| \nabla ^k U (\psi ^0 (s, x, v; X) - X) | \le \Vert \nabla ^k U \Vert _{\infty } 1_{ \{ |x-\pi ^{\perp }_v X| \le R_U + 1 \} } 1_{[-\tau , t_1 (v, |x-\pi ^{\perp }_v X|)]} (s)$ for any $k \in \mathbf{N}$.

• 3. $| \nabla U ( x(s, \Psi (r, x, m^{-1/2} v)) - X(s) ) | \le \Vert \nabla U \Vert _{\infty } 1_{ \{ r \in [- m^{1/2} \tau , s + m^{1/2} \tau ] \} } 1_{ \{ |x-\pi ^{\perp }_v X({\tilde{r}})| \le R_U + 1 \} }.$

• 4. Assume that $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ and $\alpha$ satisfies (2.10) in addition, then

  $$\begin{aligned}&| \nabla U \left( x\big (s, \Psi (r, x, m^{-1/2} v)\big ) - X(s) \right) | \\&\le \Vert \nabla U \Vert _{\infty } 1_{ \left\{ |x-\pi ^{\perp }_v X({\tilde{r}})| \le R_U + 1 \right\} } 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \right\} } 1_{ \{ r \ge - m^{1/2} \tau \} }. \end{aligned}$$

Proposition 2.1 is proved in Sects. 3.1–3.3. Precisely, since $|X| \le |X_0| + n T$, by (3.2), we get (1) as a direct consequence of Lemma 3.1 (1) (2) and Proposition 3.3 (1); and we get (2) as a direct consequence of Lemma 3.2 and Proposition 3.4. Finally, (3) and (4) are direct consequences of Lemma 3.18 and Proposition 3.22.

2.5 A Short Time Duration Right After Starting

Let

$$\begin{aligned} a_m := C_2 + 2 \varepsilon _1^{-1/2} \log \frac{2\varepsilon _3}{m^{\alpha }} + (\tau \vee 1). \end{aligned}$$

(2.13)

We remark that as long as $m \le (2 \varepsilon _3)^{1/\alpha }$, we have that

$$\begin{aligned} y \in \mathbf{R}^d, |y| \ge m^{\alpha } \Rightarrow t_1 (v, |y|) \le a_m - \tau . \end{aligned}$$

(2.14)

$a_m$ is used in the decomposition of V(t) , since the situations for $s \in (0, m^{1/2} a_m]$ and $s > m^{1/2} a_m$ are different (see (3.18)). We notice that $m a_m^6 e^{(1+C_1) a_m} \rightarrow 0$ as $m \rightarrow 0$ by (2.10). As will be proved in Lemmas 3.35 and 3.36, this ensures that the forces during the time interval $[0, m^{1/2} a_m]$ are negligible when $m \rightarrow 0$. Also, by Lemma 2.2 below, we get that our freezing time ${\tilde{r}}$ defined by (2.5) is given by ${\tilde{r}} = r - m^{1/2} \tau$ if $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, $s \in [0, T \wedge \sigma _n ]$ and $s \ge m^{1/2} a_m$.

Lemma 2.2

Assume that $\nabla U \Big ( x(s, \Psi (r, x, m^{-1/2} v)) - X(s) \Big ) \ne 0$ or $\nabla U (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))) \ne 0$. Also, assume that $m \le (2 \varepsilon _3)^{1/\alpha }$, $s \in [0, T \wedge \sigma _n]$, $s \ge m^{1/2} a_m$ and $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$. Then $r \in [m^{1/2} \tau , s + m^{1/2} \tau ]$ and ${\tilde{r}} = r - m^{1/2} \tau$.

Proof

Since $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ by assumption, the first assumption combined with Proposition 2.1 (2) and (4), implies that $m^{-1/2} (s-r) \in [-\tau , t_1(v, |x-\pi ^{\perp }_v X({\tilde{r}})|)]$, equivalently, $r \in [s - m^{1/2} t_1(v, |x-\pi ^{\perp }_v X({\tilde{r}})|), s + m^{1/2} \tau ]$. This combined with (2.14) and our assumption $s \in [m^{1/2} a_m, T \wedge \sigma _n]$ implies that for any $m \le (2 \varepsilon _3)^{1/\alpha }$, we have $r \in [m^{1/2} \tau , s + m^{1/2} \tau ]$, so $r - m^{1/2} \tau \in [0, s] \subset [0, T \wedge \sigma _n]$, hence ${\tilde{r}} = r - m^{1/2} \tau$. $\square$

2.6 Summary

In summary, we first apply the ray representation to specify the first entrance of each light particle to its valid interaction range, then use $a_m$, $\alpha$ and freezing-approximation to decompose V(t) in the following way: first we ignore the time period $[0, m^{1/2} a_m]$, since the density of the incoming light particles during this period might depend on the massive particle; then we ignore those light particles without enough skew (i.e., those with $|x - \pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha }$), since the behavior of this type of light particles is too difficult to be tracked; finally, after these approximations, we apply our freezing-approximation. The first approximation explained above is fine heuristically because the time period is short enough, and the second one is fine because the total number of such “singular” light particles is small enough. As a result, we are able to re-express V(t) as a sum of a martingale term, a smooth term and a negligible term, and get the desired tightness (see Sect. 3.4.2 for the concrete decomposition, and see Proposition 3.31 for the result). Finally, with the help of this decomposition, we prove the convergence of the smooth term (see Sect. 3.5). As explained, by the well-known martingale problem theorem, this implies our Theorem 1.2.

The concrete proof of Theorem 1.2 is given in Sect. 3.

3 Proof of Theorem 1.2

We present the proof of Theorem 1.2 is this section. In Sect. 3.1, we discuss the behaviors of $\varphi$ and $\psi$. In Sect. 3.2, we estimate the approximation error of our freezing approximation. With the help of these results, we discuss the behavior of the light particles in Sect. 3.3. In detail, $\varphi$ is used for the discussion with respect to $x (t, \Psi (r, x, m^{-1/2}v))$ with $|r| \le m^{1/2} \tau$, and $\psi$ is used for those with $r \ge m^{1/2} \tau$. In Sect. 3.4, we prove our key decomposition: we rewrite V(t) as a sum of a martingale term, a smooth term and a negligible term, with the concrete expressions of the martingale term and the smooth term given. In particular, we get the tightness. Finally, in Sect. 3.5, with the help of the results of Sect. 3.4, we prove the desired convergence, by proving that a certain part of the smooth term is actually also negligible.

3.1 Some Discussion for $\varphi$ and $\psi$

In this section, we discuss the behaviors of $\varphi$ and $\psi$. Precisely, we prove that they are far from the valid range after certain times (see Propositions 3.3 and 3.4).

To begin with, Lemmas 3.1 and 3.2 consider the problem of until when does the particle keep a uniform motion. For any $X \in \mathbf{R}^d$, let $R_X := R_U + |X| + 1$ and $s_1 := s_1 (v, X) := |v|^{-1} R_X$. So $s_1 \le \tau$ if $|v| \ge {\overline{v}}$ and $|X| \le |X_0| + nT$.

Lemma 3.1

For any $r \in \mathbf{R}$, $t \ge 0$ and $(x, v) \in E$, we have that $\varphi (t, x-rv, v; X) = (x-rv+tv, v)$ and $| \varphi ^0 (t, x-rv, v; X) - X | \ge R_U + 1$ if (1) $|x-\pi ^{\perp }_v X| \ge R_U + 1$ or (2) $r \ge t + s_1$.

Proof

First we consider the case $|x-\pi ^{\perp }_v X| \ge R_U + 1$. Since $x \cdot v = 0$, we have that $|x-rv+sv-X| \ge |x-\pi ^{\perp }_v X| \ge R_U + 1$ for any $s \ge 0$, so the particle keeps a uniform motion. Therefore, $\varphi (t, x-rv, v; X) = (x-rv+tv, v)$, hence $|\varphi ^0 (t, x-rv, v; X) - X| = |x-rv+tv-X| \ge |x-\pi ^{\perp }_v X| \ge R_U + 1$ for any $t \ge 0$.

The assertion for $r \ge t + s_1$ is similar. In this case, for any $s \in [0, t]$, we have that $r - s \ge s_1$, hence $r - s + (X, \frac{v}{|v|}) \frac{1}{|v|} \ge s_1 - \frac{|X|}{|v|} = \frac{R_U+1}{|v|}$, so $|x - r v + s v - X| \ge |r - s + (X, \frac{v}{|v|}) \frac{1}{|v|}| \cdot |v| \ge R_U + 1$. Therefore, the particle keeps a uniform motion during the time interval [0, t] . So $\varphi (t, x-rv, v; X) = (x-rv+tv, v)$, and $| \varphi ^0 (t, x-rv, v; X) - X | = |x - r v + t v - X| \ge R_U + 1$. $\square$

Since $\psi ^0 (t, x, v; X) = \lim _{s \rightarrow \infty } \varphi (t+s, x - s v, v; X)$ by definition, we get the following as a corollary of Lemma 3.1.

Lemma 3.2

For any $(x, v) \in E$, we have that $\psi (t, x, v; X) = (x+tv, v)$ and $| \psi ^0 (t, x, v; X) - X | \ge R_U + 1$ if (1) $|x-\pi ^{\perp }_v X| \ge R_U + 1$ or (2) $t \le - s_1$. $\square$

We next consider the case where the conditions of Lemmas 3.1 or 3.2 are not satisfied. For any $X \in \mathbf{R}^d$ and $(x, v) \in \mathbf{R}^{2d}$, let

$$\begin{aligned} d^{\varphi } (t, x, v; X):= & {} | \varphi ^0 (t, x, v; X)-X |, \\ v^{d, \varphi } (t, x, v; X):= & {} \frac{d}{dt} d^{\varphi } (t, x, v; X). \end{aligned}$$

Finally, let $C_3 := 5 s_1 + 2 \varepsilon _4^{-1} (2 C_0 + \sqrt{2 \Vert U\Vert _{\infty }})$, where $C_0$ is as defined in (1.2), and for any $X \in \mathbf{R}^d$ and $(x, v) \in E$ satisfying $|v| \ge {\overline{v}}$, let

$$\begin{aligned} t_2 (x, v, X) := \left\{ \begin{array}{ll} 2 C_0^{-1} R_X, &{} \hbox { if } \quad |v| \ge 2 C_0, \\ 8 s_1, &{} \hbox { if } \quad |v| \in [{\overline{v}}, 2 C_0) \hbox { and } |x-\pi ^{\perp }_v X| \ge \frac{R_U}{2}, \\ C_3 + 2 \varepsilon _1^{-1/2} \log ^+ \frac{2 \varepsilon _3}{|x-\pi ^{\perp }_v X|}, &{} \hbox { if } \quad |v| \in [{\overline{v}}, 2 C_0) \hbox { and } 0 \le |x-\pi ^{\perp }_v X| < \frac{R_U}{2}. \end{array} \right. \end{aligned}$$

(3.1)

Here $\log ^+ y := (\log y) \vee 0$ for any $y>0$. The definition of $t_2 (x, v, X)$ is similar to that of $t_1 (v, *)$. We introduce this notation to make our statements of this section easier to be understood. We remark that

$$\begin{aligned} t_1 (v, |x-\pi ^{\perp }_v X|) \ge t_2 (x, v, X) \qquad \hbox { if } \quad |X| \le |X_0| + n T, \end{aligned}$$

(3.2)

and

$$\begin{aligned} t_1 (v, |x-\pi ^{\perp }_v X|) \ge t_2 (x, v, X) + \tau , \quad \hbox { if } \quad |v| \in [{\overline{v}}, 2C_0+1), |X| \le |X_0| + n T. \end{aligned}$$

(3.3)

Our main result of this section is the following:

Proposition 3.3

For any $X \in \mathbf{R}^d$, $r \in \mathbf{R}$ and $(x, v) \in E$ satisfying $|v| \ge {\overline{v}}$, we have that $d^{\varphi } (t, x-rv, v; X) \ge R_U + 1$ and $v^{d, \varphi } (t, x-rv, v; X) > 0$ if at least one of the following conditions is satisfied:

• 1. $r \le - s_1$ and $t \ge 0$,

• 2. $|r| \le s_1$ and $t \ge t_2 (x, v, X)$,

• 3. $r \ge s_1$ and $t + s_1 - r \ge t_2 (x, v, X)$.

Similarly, for any $X \in \mathbf{R}^d$ and $(x, v) \in E$, let

$$\begin{aligned} d^{\psi } (t, x, v; X):= & {} | \psi ^0 (t, x, v; X)-X |, \\ v^{d, \psi } (t, x, v; X):= & {} \frac{d}{dt} d^{\psi } (t, x, v; X). \end{aligned}$$

Then we have the following.

Proposition 3.4

For any $X \in \mathbf{R}^d$ and $(x, v) \in E$ satisfying $|v| \ge {\overline{v}}$, we have that $| \psi ^0 (t, x, v; X) - X | \ge R_U + 1$ and $v^{d, \psi } (t, x, v; X) > 0$ for any $t \ge t_2 (x, v, X) - s_1$.

Proof

This follows easily by Proposition 3.3 (2) or (3) since $\psi (t, x, v; X) = \varphi (t+s_1, x-s_1 v, v; X)$ for any $t \ge - s_1$. $\square$

We prove Proposition 3.3 in the rest of Sect. 3.1. First, we have the following by a direct calculation:

Lemma 3.5

For any $x, v, X \in \mathbf{R}^d$, we have the following:

• (1) $\frac{d}{dt} \Big ( \frac{1}{2} |\varphi ^1 (t, x, v; X)|^2 + h (d^{\varphi } (t, x, v; X)) \Big ) = 0$,

• (2) $\frac{d}{dt} \Big ( (\varphi ^0 (t, x, v; X)-X, \varphi ^1 (t, x, v; X) )^2 - d^{\varphi } (t, x, v; X)^2 |\varphi ^1 (t, x, v; X)|^2 \Big ) = 0$.

$\square$

Lemma 3.6

For any $x, v, X \in \mathbf{R}^d$, we have the following:

• (1) $v^{d, \varphi } (t, x, v; X) = \Big ( \varphi ^1 (t, x, v; X), \frac{\varphi ^0 (t, x, v; X)-X}{d^{\varphi } (t, x, v; X)} \Big )$ for any $t \ge 0$,

• (2) $\frac{d}{dt} v^{d, \varphi } (t, x, v; X) = \frac{1}{d^{\varphi } (t, x, v; X)} \Big | \pi ^{\perp }_{\varphi ^0 (t, x, v; X)-X} \varphi ^1 (t, x, v; X) \Big |^2 - h' (d^{\varphi } (t, x, v; X))$ for any $t \ge 0$,

• (3) $v^{d, \varphi } (t, x, v; X)$ is monotone non-decreasing with respect to $t \ge 0$,

• (4) $| v^{d, \varphi } (t, x, v; X) |^2 = - \frac{ |v|^2 |\pi ^{\perp }_v (x-X)|^2}{d^{\varphi } (t, x, v; X)^2} - 2 h (d^{\varphi } (t, x, v; X)) + |v|^2 + 2 h (|x-X|)$ for any $t \ge 0$.

Proof

(1) and (2) are easy by direct calculation. (3) is a direct consequence of (2) since $-h' \ge 0$. For the last assertion, we have that

$$\begin{aligned}&d^{\varphi } (t, x, v; X)^2 \Big ( | v^{d, \varphi } (t, x, v; X) |^2 - |\varphi ^1 (t, x, v; X)|^2 \Big ) \\&\quad = \Big ( \varphi ^1 (t, x, v; X), \varphi ^0 (t, x, v; X) - X \Big )^2 - d^{\varphi } (t, x, v; X)^2 |\varphi ^1 (t, x, v; X)|^2 \\&\quad = \Big ( \varphi ^1 (0, x, v; X), \varphi ^0 (0, x, v; X) - X \Big )^2 - d^{\varphi } (0, x, v; X)^2 |\varphi ^1 (0, x, v; X)|^2 \\&\quad = (v, x-X)^2 - |x-X|^2 |v|^2 \\&\quad = - |v|^2 |\pi ^{\perp }_v (x-X)|^2, \end{aligned}$$

where we used (1) of this lemma when passing to the second line, we used Lemma 3.5 (2) when passing to the third line, and we used the fact that $(a, b)^2 - |a|^2 |b|^2 = - |b|^2 |\pi ^{\perp }_b a|^2$ for any $a, b \in \mathbf{R}^d$ when passing to the last line. Since $|\varphi ^1 (t, x, v; X)|^2 = |v|^2 + 2 h (|x-X|) - 2 h (d^{\varphi } (t, x, v; X))$ by Lemma 3.5 (1), we get our assertion (4). $\square$

Also, by the definition of $\psi$, we get the following properties with respect to $\psi$, as a corollary of Lemmas 3.5 and 3.6. Lemma 3.7 will be used later.

Lemma 3.7

For any $(x, v) \in E$ and $X \in \mathbf{R}^d$, we have the following:

• 1. $\frac{d}{dt} \Big ( \frac{1}{2} |\psi ^1 (t, x, v; X)|^2 + h (d^{\psi } (t, x, v; X)) \Big ) = 0$,

• 2. $v^{d, \psi } (t, x, v; X) = \Big ( \psi ^1 (t, x, v; X), \frac{\psi ^0 (t, x, v; X)-X}{d^{\psi } (t, x, v; X)} \Big )$ for any $t \in \mathbf{R}$,

• 3. $| v^{d, \psi } (t, x, v; X) |^2 = - \frac{ |v|^2 |x-\pi ^{\perp }_v X|^2}{d^{\psi } (t, x, v; X)^2} - 2 h (d^{\psi } (t, x, v; X)) + |v|^2$ for any $t \in \mathbf{R}$.

Proof of Proposition 3.3 (1) and (3)

We first prove the first assertion of Proposition 3.3. Suppose $r \le - s_1$. Then we have that $|x-rv+sv-X| \ge (s-r) |v| - |X| \ge s_1 |v| - |X| \ge R_U + 1$ for any $s \ge 0$, so the particle keeps a uniform motion. Therefore, for any $t \ge 0$, we have that $\varphi (t, x-rv, v; X) = (x - rv + tv, v)$. So $|d^{\varphi } (t, x-rv, v; X) = |x-rv+tv-X| \ge R_U + 1$, and by Lemma 3.6 (1),

$$\begin{aligned}&v^{d, \varphi } (t, x-rv, v; X) \\&\quad = \left( v, \frac{x-rv+tv-X}{|x-rv+tv-X|} \right) = \frac{1}{|x-rv+tv-X|} \Big ( (t-r) |v|^2 - (v, X) \Big ) \\&\quad \ge \frac{1}{|x-rv+tv-X|} \left( \frac{R_X}{|v|} |v|^2 - |X| \cdot |v| \right) \ge \frac{|v|}{|x-rv+tv-X|} (R_X-|X|) > 0. \end{aligned}$$

This completes the proof of Proposition 3.3 (1).

For the third assertion of Proposition 3.3, we just notice that if $r \ge s_1$, then $\varphi (t, x-rv, v; X) = \varphi (t-r+s_1, x-s_1 v, v; X)$ for any $t \ge r - s_1$. Indeed, suppose $r \ge s_1$, then for any $s \in [0, r-s_1]$, we have that $|x-rv+sv-X| \ge |s-r| \cdot |v| - |X| \ge R_U + 1$, so the particle keeps a uniform motion during the time interval $[0, r-s_1]$, hence $\varphi (r-s_1, x-rv, v; X) = (x-rv+(r-s_1)v, v) = (x-s_1 v, v)$. Therefore, for any $t \ge r - s_1$, we have that $\varphi (t, x-rv, v; X) = \varphi (t-r+s_1, \varphi (r-s_1, x-rv, v; X); X) = \varphi (t-r+s_1, x-s_1 v, v; X)$.

Hence the third assertion of Proposition 3.3 is an easy corollary of the second assertion. $\square$

We prove the second assertion of Proposition 3.3 in the rest of Sect. 3.1. First, we notice the following.

Lemma 3.8

Let $|r| \le s_1$. Then $\xi _1 := \inf \{ t \ge 0; d^{\varphi } (t, x-rv, v; X) < R_U + \frac{1}{2} \} \in [0, 2 s_1] \cup \{ \infty \}$.

Proof

Suppose that $\xi _1 \notin [0, 2 s_1]$. Then $d^{\varphi } (t, x-rv, v; X) \ge R_U + \frac{1}{2}$ for all $t \in [0, 2 s_1]$, hence the particle keeps a uniform motion during this time period. So for any $t \in [0, 2 s_1]$, we have that $\varphi (t, x-rv, v; X) = (x - rv + tv, v)$, hence

$$\begin{aligned} \Big ( R_U + \frac{1}{2} \Big )^2 \le |x-rv+tv-X|^2 = |x-\pi ^{\perp }_v X|^2 + \left( t - r - \Big ( X, \frac{v}{|v|} \Big ) \frac{1}{|v|} \right) ^2 |v|^2.\nonumber \\ \end{aligned}$$

(3.4)

If $r + \Big ( X, \frac{v}{|v|} \Big ) \frac{1}{|v|} \ge 0$, then $r + \Big ( X, \frac{v}{|v|} \Big ) \frac{1}{|v|} \in [0, 2 s_1]$, so applying (3.4) to $t = r + \Big ( X, \frac{v}{|v|} \Big ) \frac{1}{|v|}$, we get that $|x-\pi ^{\perp }_v X| \ge R_U + \frac{1}{2}$, hence $|x-rv+tv-X| \ge R_U + \frac{1}{2}$ for any $t \ge 0$, which means that $\xi _1 = \infty$. If $r + \Big ( X, \frac{v}{|v|} \Big ) \frac{1}{|v|} < 0$, then applying (3.4) to $t = 0$, we get that $|x-\pi ^{\perp }_v X|^2 + \Big ( 0 - r - \Big ( X, \frac{v}{|v|} \Big ) \frac{1}{|v|} \Big )^2 |v|^2 \ge (R_U + \frac{1}{2})^2$, hence again, $|x-rv+tv-X| > R_U + \frac{1}{2}$ for any $t \ge 0$, which means that $\xi _1 = \infty$. $\square$

As a corollary of Lemma 3.8, we prove in Lemma 3.9 that in order to prove Proposition 3.3 (2), it suffices to prove that

$$\begin{aligned} d^{\varphi } (t, x-rv, v; X) \ge R_U + 1, \qquad \hbox { if } t \ge t_2 (x, v, X). \end{aligned}$$

(3.5)

Lemma 3.9

Assume the same conditions as in Proposition 3.3 (2), i.e., we assume that $r \in \mathbf{R}$, $(x, v) \in E$, $|v| \ge {\overline{v}}$ and $|r| \le s_1$. Also, assume that $d^{\varphi } (t_2 (x, v, X), x-rv, v; X) \ge R_U + 1$. Then we get that $v^{d, \varphi } (t, x-rv, v; X) > 0$ for any $t \ge t_2 (x, v, X)$.

Proof

First we notice that $t_2 (x, v; X) \ge 2 s_1$ by the definition of $t_2 (x, v; X)$. Let $\xi _1$ be as in Lemma 3.8. Then by Lemma 3.8, we get that $\xi _1 \in [0, 2 s_1] \cup \{ \infty \}$.

If $\xi _1 \in [0, 2 s_1]$, then $\xi _1 \le 2 s_1 \le t_2 (x, v; X)$. Also, $d^{\varphi } (\xi _1, x-rv, v, X) \le R_U + \frac{1}{2}$ by definition, and $d^{\varphi } (t_2 (x, v, X), x-rv, v, X) \ge R_U + 1$ by assumption. So by the mean-value theorem, there exists a ${\tilde{t}} \in [\xi _1, t_2 (x, v, X)]$ such that $v^{d, \varphi } ({\tilde{t}}, x-rv, v; X) > 0$. This combined with Lemma 3.6 (3) implies our assertion.

If $\xi _1 = \infty$, then $d^{\varphi } (t, x-rv, v; X) \ge R_U + \frac{1}{2}$ for all $t \ge 0$, so the particle keeps a uniform motion, hence $\varphi (t, x-rv, v; X) = (x-rv+tv, v)$ for $t \ge 0$. Therefore, for any $t \ge t_2 (x, v, X)$, since $|r| \le s_1$, we get that $t - r \ge s_1$, hence

$$\begin{aligned}&\Big ( \varphi ^1 (t, x-rv, v; X), \varphi ^0 (t, x-rv, v; X) - X \Big ) \\&\quad = ( v, x - rv + tv - X ) = (t-r) |v|^2 - (v, X) \ge |v|^{-1} R_X |v|^2 - |v| \cdot |X| \\&\quad = |v| (R_X - |X|) > 0, \end{aligned}$$

so by Lemma 3.6 (1), we get that $v^{d, \varphi } (t, x-rv, v; X) > 0$. $\square$

By Lemma 3.9, in order to prove Proposition 3.3 (2), it suffices to prove (3.5) under the conditions of Proposition 3.3 (2). We prove this from now on. We first prepare the following general result for later use.

Lemma 3.10

Let g be a function satisfying the following conditions: $g \in C^2 ([0, \infty ))$, $g(0) > 0$, $g'(0) \ge 0$, and $g''(t) = l(t) g (t)$, for some $l(t) \ge 0$ for any $t \ge 0$. Then g(t) is monotone non-decreasing with respect to $t \ge 0$.

Proof

Let $\xi _2 := \inf \{ t > 0; g'(t) < 0 \}$. It suffices to prove that $\xi _2 = + \infty$. Suppose $\xi _2 < \infty$. Then for any $n \in \mathbf{N}$, there exists a $t_n \in [\xi _2, \xi _2 + \frac{1}{n}]$ such that $g'(t_n) < 0$. On the other hand, we have that $g'(t) \ge 0$ for any $t \in [0, \xi _2]$, so for any $t \in [0, \xi _2]$, we have that $g(t) \ge g(0) > 0$. By the continuity of g, this implies that there exists a $N \in \mathbf{N}$ such that for any $t \in [0, \xi _2 + \frac{1}{N}]$, we have $g(t) > 0$, hence $g''(t) = l(t) g(t) \ge 0$. Therefore,

$$\begin{aligned} 0 > g' (t_N) = \int _0^{t_N} g''(s) ds + g'(0) \ge 0, \end{aligned}$$

which yields a contradiction. So $\xi _2 = + \infty$. $\square$

Lemma 3.11

Suppose that $(x, v) \in E$ and $X \in \mathbf{R}^d$ satisfy $|x - \pi ^{\perp }_v X| \ne 0$. Then for any $r \in \mathbf{R}$, we have that $( \varphi ^0 (t, x-rv, v; X)-X ) \cdot \frac{x - \pi ^{\perp }_v X}{|x - \pi ^{\perp }_v X|}$ is monotone non-decreasing with respect to $t \ge 0$.

Proof

Let $g(t) := ( \varphi ^0 (t, x-rv, v; X)-X ) \cdot \frac{x - \pi ^{\perp }_v X}{|x - \pi ^{\perp }_v X|}$. Then $g(0) = (x - r v - X) \cdot \frac{x - \pi ^{\perp }_v X}{|x - \pi ^{\perp }_v X|} = |x - \pi ^{\perp }_v X| > 0$, $g'(0) = v \cdot \frac{x - \pi ^{\perp }_v X}{|x - \pi ^{\perp }_v X|} = 0$. Also, $g''(t) = - \frac{h'(d^{\varphi } (t, x-rv, v; X))}{d^{\varphi } (t, x-rv, v; X)} g(t)$ and $- \frac{h'(d^{\varphi } (t, x-rv, v; X))}{d^{\varphi } (t, x-rv, v; X)} \ge 0$ for any $t \ge 0$. So we get our assertion by Lemma 3.10. $\square$

Now we are ready to prove that (3.5) holds under the conditions of Proposition 3.3 (2). We first prove it for the case $|v| \ge 2 C_0$. For this case, we have the following as a consequence of [12, Proposition 3.2.2].

Lemma 3.12

For any $(x, v) \in E$ with $|v| \ge 2 C_0$, we have that

$$\begin{aligned} d^{\varphi } (t, x-rv, v; X) \ge R_U + 1 \end{aligned}$$

if $t \ge 2 C_0^{-1} R_X$ and $|r| \le s_1$.

Proof

Suppose that $(x, v) \in E$ and $|v| \ge 2 C_0$. Then by [12, Proposition 3.2.2], we have for any $r \in \mathbf{R}$ and $s \ge 0$ that

$$\begin{aligned} \varphi ^1 (s, x-rv, v; X) \cdot \frac{v}{|v|} \ge C_0. \end{aligned}$$

Therefore, for any $t \ge 2 C_0^{-1} R_X$, since $|r| \le s_1 = |v|^{-1} R_X$ by assumption, we get that

$$\begin{aligned}&d^{\varphi } (t, x-rv, v; X) = \Big | \varphi ^0 (t, x-rv, v; X) - X \Big | \\&\quad \ge \Big ( \varphi ^0 (t, x-rv, v; X) - X \Big ) \cdot \frac{v}{|v|} \\&\quad = \int _0^t \varphi ^1 (s, x-rv, v; X) \cdot \frac{v}{|v|} ds + \Big ( \varphi ^0 (0, x-rv, v; X) - X \Big ) \cdot \frac{v}{|v|} \\&\quad \ge C_0 t + (x-rv-X) \cdot \frac{v}{|v|} \ge 2 R_X - |r| \cdot |v| - |X| \\&\quad \ge 2 R_X - R_X - |X| = R_U + 1. \end{aligned}$$

$\square$

From now on, we consider those particles with $|v| < 2 C_0$ and $|r| \le s_1$. In Lemma 3.13 below, we consider those particles that are skewed enough from the beginning (i.e., with $|x-\pi ^{\perp }_v X| \ge \frac{1}{2} R_U$), and prove that they leave the valid interacting range quickly.

Lemma 3.13

Suppose that $r \in \mathbf{R}$, $(x, v) \in E$, $|x-\pi ^{\perp }_v X| \ge \frac{1}{2} R_U$ and $|r| \le s_1$. Then

$$\begin{aligned} | \varphi ^0 (t, x-rv, v; X) - X | \ge R_U + 1 \end{aligned}$$

for any $t \ge 8 s_1$.

Proof

Let $e_1 := \frac{x-\pi ^{\perp }_v X}{|x-\pi ^{\perp }_v X|}$ and $e_2 := \frac{v}{|v|}$. Define

$$\begin{aligned} \xi _3 := \inf \left\{ t \ge 0; \varphi ^1 (t, x-rv, v; X) \cdot e_2 \le \frac{|v|}{2} \right\} . \end{aligned}$$

If $\xi _3 \ge 4 s_1$, then $\varphi ^1 (4 s_1, x-rv, v; X) \cdot e_2 \ge \frac{|v|}{2}$, and

$$\begin{aligned}&\big ( \varphi ^0 (4 s_1, x-rv, v; X) - X \big ) \cdot e_2 \\&= \big ( \varphi ^0 (0, x-rv, v; X) - X \big ) \cdot e_2 + \int _0^{4 s_1} \varphi ^1 (s, x-rv, v; X) \cdot e_2 ds \\&\ge (x - r v - X) \cdot \frac{v}{|v|} + \frac{|v|}{2} \cdot 4 |v|^{-1} R_X \\&\ge - s_1 |v| - |X| + 2 R_X = R_U + 1. \end{aligned}$$

In particular, by applying Lemma 3.10 to $g(t) := \big ( \varphi ^0 (t + 4 s_1, x-rv, v; X) - X \big ) \cdot e_2$, we get that $\big ( \varphi ^0 (t, x-rv, v; X) - X \big ) \cdot e_2$ is monotone nondecreasing with respect to $t \ge 4 s_1$. Combining these, we get that $| \varphi ^0 (t, x-rv, v; X) - X | \ge R_U + 1$ for any $t \ge 4 s_1$.

We next deal with the case $\xi _3 < 4 s_1$. We notice that

$$\begin{aligned} - \nabla U \big (\varphi ^0 (t, x-rv, v; X) - X\big ) \cdot e_1 \ge \frac{1}{2} \nabla U \big (\varphi ^0 (t, x-rv, v; X) - X\big ) \cdot e_2, \quad \hbox { for any } t \ge 0. \end{aligned}$$

(3.6)

Indeed, if $\nabla U (\varphi ^0 (t, x-rv, v; X) - X) = 0$, then (3.6) is trivial, so it suffices to prove it for the case $\nabla U (\varphi ^0 (t, x-rv, v; X) - X) \ne 0$. In this case, we have that $| \varphi ^0 (t, x-rv, v; X) - X | \le R_U$. On the other hand, we have by Lemma 3.11 and our assumption that $( \varphi ^0 (t, x-rv, v; X) - X ) \cdot e_1 \ge (x-rv-X) \cdot e_1 = |x-\pi ^{\perp }_v X| \ge \frac{R_U}{2}$. Therefore,

$$\begin{aligned} \big ( \varphi ^0 (t, x-rv, v; X) - X \big ) \cdot e_1 \ge - \frac{1}{2} \big ( \varphi ^0 (t, x-rv, v; X) - X \big ) \cdot e_2. \end{aligned}$$

Since $- \frac{h'(d^{\varphi } (t, x-rv, v; X))}{d^{\varphi } (t, x-rv, v; X)}$ is always non-negative, this implies that

$$\begin{aligned}&- \nabla U \big (\varphi ^0 (t, x-rv, v; X) - X\big ) \cdot e_1 \\&\quad = - \frac{h'\big (d^{\varphi } (t, x-rv, v; X)\big )}{d^{\varphi } (t, x-rv, v; X)} \big ( \varphi ^0 (t, x-rv, v; X) - X \big ) \cdot e_1 \\&\quad \ge \frac{1}{2} \frac{h'\big (d^{\varphi } (t, x-rv, v; X)\big )}{d^{\varphi } (t, x-rv, v; X)} \big ( \varphi ^0 (t, x-rv, v; X) - X \big ) \cdot e_2 \\&\quad = \frac{1}{2} \nabla U \big (\varphi ^0 (t, x-rv, v; X) - X\big ) \cdot e_2. \end{aligned}$$

This completes the proof of (3.6).

We notice that $\varphi ^1 (0, x-rv, v; X) \cdot e_1 = 0$ and $\varphi ^1 (0, x-rv, v; X) \cdot e_2 = |v|$. Also, $\varphi ^1 (\xi _3, x-rv, v; X) \cdot e_2 = \frac{|v|}{2}$. Therefore, by (3.6), we have that

$$\begin{aligned} \varphi ^1 (\xi _3, x-rv, v; X) \cdot e_1= & {} \int _0^{\xi _3} - \nabla U \big (\varphi ^0 (t, x-rv, v; X) - X\big ) \cdot e_1 dt \nonumber \\\ge & {} \frac{1}{2} \int _0^{\xi _3} \nabla U \big (\varphi ^0 (t, x-rv, v; X) - X\big ) \cdot e_2 dt \nonumber \\= & {} \frac{1}{2} \Big ( \varphi ^1 (0, x-rv, v; X) \cdot e_2 - \varphi ^1 (\xi _3, x-rv, v; X) \cdot e_2 \Big ) \nonumber \\= & {} \frac{1}{2} \Big ( |v| - \frac{|v|}{2} \Big ) = \frac{|v|}{4}. \end{aligned}$$

(3.7)

Also, for any $s \ge \xi _3$, since $(\varphi ^0 (s, x-rv, v; X) - X) \cdot e_1 > 0$ by Lemma 3.11, we have that $- \nabla U (\varphi ^0 (s, x-rv, v; X) - X) \cdot e_1 \ge 0$. Therefore, for any $t \ge \xi _3$, we have by (3.7) that

$$\begin{aligned}&\varphi ^1 (t, x-rv, v; X) \cdot e_1 \nonumber \\&\quad = \varphi ^1 (\xi _3, x-rv, v; X) \cdot e_1 + \int _{\xi _3}^t \Big ( - \nabla U (\varphi ^0 (s, x-rv, v; X) - X) \cdot e_1 \Big ) ds \nonumber \\&\quad \ge \varphi ^1 (\xi _3, x-rv, v; X) \cdot e_1 \ge \frac{|v|}{4}. \end{aligned}$$

(3.8)

Also, we have $\Big ( \varphi ^0 (\xi _3, x-rv, v; X) - X \Big ) \cdot e_1 \ge 0$ by Lemma 3.11. Combining this with (3.8), we get that

$$\begin{aligned} \Big ( \varphi ^0 (t, x-rv, v; X) - X \Big ) \cdot e_1\ge & {} \Big ( \varphi ^0 (\xi _3, x-rv, v; X) - X \Big ) \cdot e_1 + \int _{\xi _3}^t \frac{|v|}{4} ds \\\ge & {} \frac{|v|}{4} (t-\xi _3) \end{aligned}$$

for any $t \ge \xi _3$. Also, recall that we are assuming $\xi _3 < 4 s_1$ now. Therefore, for any $t \ge 8 s_1$, we have that $|\varphi ^0 (t, x-rv, v; X)-X| \ge \Big ( \varphi ^0 (t, x-rv, v; X) - X \Big ) \cdot e_1 \ge \frac{|v|}{4} (t-\xi _3) \ge \frac{|v|}{4} \cdot 4|v|^{-1} R_X = R_X \ge R_U + 1$. $\square$

Finally, we deal with the case $|x - \pi ^{\perp }_v X| \le \frac{R_U}{2}$ and $|v| \in [{\overline{v}}, 2C_0]$. Divide $d^{\varphi } (t, x-rv, v; X) < R_U+1$ into three parts: $d^{\varphi } (t, x-rv, v; X) \in [\frac{3}{4} R_U \vee (R_U - \varepsilon _2), R_U+1)$, $d^{\varphi } (t, x-rv, v; X) \in [\varepsilon _3, \frac{3}{4} R_U \vee (R_U - \varepsilon _2))$ and $d^{\varphi } (t, x-rv, v; X) < \varepsilon _3$. We notice that $\frac{d^2}{dt^2} d^{\varphi } (t, x-rv, v; X) \ge 0$ by Lemma 3.6 (2), hence $d^{\varphi }$ is convex, so the particle passes through the first two domains at most twice, and could be in the last domain at most once. Therefore, by Lemmas 3.8 and 3.9, in order to prove Proposition 3.3 (2) for $|x - \pi ^{\perp }_v X| \le \frac{R_U}{2}$ and $|v| \in [{\overline{v}}, 2C_0]$, it suffices to prove the following lemma:

Lemma 3.14

For any $r \in \mathbf{R}$, $(x, v) \in E$ and $X \in \mathbf{R}^d$, we have the following:

• 1. If $|x - \pi ^{\perp }_v X| \le \frac{R_U}{2}$ and $|v| \ge {\overline{v}}$, then

  $$\begin{aligned} \int _{[0, \infty )} 1_{\left\{ d^{\varphi } (t, x-rv, v; X) \in \left[ \frac{3}{4} R_U \vee (R_U - \varepsilon _2), R_U+1\right) \right\} } dt \le 2 \sqrt{2} |v|^{-1} (R_U+1). \end{aligned}$$

• 2. If $|x - \pi ^{\perp }_v X| \le \frac{R_U}{2}$, then

  $$\begin{aligned} \int _{[0, \infty )} 1_{ \left\{ d^{\varphi } (t, x-rv, v; X) \in \left[ \varepsilon _3, \frac{3}{4} R_U \vee (R_U - \varepsilon _2)\right) \right\} } dt \le 2 \varepsilon _4^{-1} \Big ( |v| + \sqrt{2 \Vert U\Vert _{\infty }} \Big ). \end{aligned}$$

• 3. If $0 < |x - \pi ^{\perp }_v X| \le \frac{R_U}{2}$, then

  $$\begin{aligned} \int _{[0, \infty )} 1_{ \{ d^{\varphi } (t, x-rv, v; X) < \varepsilon _3 \} } dt \le 2 {\varepsilon _1}^{-1/2} \log ^+ \Big ( \frac{2 \varepsilon _3}{|x - \pi ^{\perp }_v X|} \Big ). \end{aligned}$$

Proof

(1) For any $t \ge 0$ satisfying $d^{\varphi } (t, x-rv, v; X) \ge \frac{3}{4} R_U \vee (R_U - \varepsilon _2)$, we have by Lemma 3.6 (4) that

$$\begin{aligned}&|v^{d, \varphi } (t, x-rv, v; X)|^2 \\&\quad = - |v|^2 \frac{|x - \pi ^{\perp }_v X|^2}{d^{\varphi } (t, x-rv, v; X)^2} - 2 h(d^{\varphi }(t, x-rv, v; X)) + |v|^2 + 2 h(|x-rv-X|) \\&\quad \ge - |v|^2 \frac{(R_U/2)^2}{(3R_U/4)^2} - \frac{1}{18} {\overline{v}}^2 + |v|^2 = \frac{5}{9} |v|^2 - \frac{1}{18} {\overline{v}}^2 \ge \frac{1}{2} |v|^2. \end{aligned}$$

Therefore, $\int _{[0, \infty )} 1_{ \{ d^{\varphi } (t, x-rv, v; X) \in [\frac{3}{4} R_U \vee (R_U - \varepsilon _2), R_U+1) \} } dt \le 2 \frac{R_U + 1}{\frac{1}{\sqrt{2}} |v|} = 2 \sqrt{2} |v|^{-1} (R_U+1).$

(2) For any $t \ge 0$ satisfying $d^{\varphi } (t, x-rv, v; X) \in (\varepsilon _3, \frac{3}{4} R_U \vee (R_U - \varepsilon _2))$, we have by Lemma 3.6 (2) and the definition of $\varepsilon _4$ that

$$\begin{aligned}&\frac{d}{dt} v^{d, \varphi } (t, x-rv, v; X) \\&= \frac{1}{d^{\varphi } (t, x-rv, v; X)} \Big | \pi ^{\perp }_{\varphi ^0 (t, x-rv, v; X)\,-\,X} \varphi ^1 (t, x-rv, v; X) \Big |^2 - h' (d^{\varphi } (t, x-rv, v; X)) \\&\ge \varepsilon _4. \end{aligned}$$

Since $|v^{d, \varphi } (t, x-rv, v; X)| \le |v| + \sqrt{2 \Vert U\Vert _{\infty }}$ by Lemma 3.6 (4), we get our assertion (2).

(3) Write the minimum of $\{ d^{\varphi } (t, x-rv, v; X); t \ge 0 \}$ as ${\overline{r}}$. It suffices to consider the case where ${\overline{r}} < \varepsilon _3$. By Lemma 3.6 (2), we have that $\frac{d}{dt} v^{d, \varphi } (t, x-rv, v; X) \ge - h'(d^{\varphi } (t, x-rv, v; X))$, so if $d^{\varphi } (t, x-rv, v; X) < \varepsilon _3$, then $\frac{d}{dt} v^{d, \varphi } (t, x-rv, v; X) > \varepsilon _1 d^{\varphi } (t, x-rv, v; X)$. Since

$$\begin{aligned}&\frac{d}{dt} \left( |v^{d, \varphi } (t, x-rv, v; X)|^2 - \varepsilon _1 d^{\varphi } (t, x-rv, v; X)^2 \right) \\&\quad = 2 v^{d, \varphi } (t, x-rv, v; X) \left( \frac{d}{dt} v^{d, \varphi } (t, x-rv, v; X) - \varepsilon _1 d^{\varphi } (t, x-rv, v; X) \right) , \end{aligned}$$

this implies that the derivatives of $|v^{d, \varphi } (t, x-rv, v; X)|^2 - \varepsilon _1 d^{\varphi } (t, x-rv, v; X)^2$ and $d^{\varphi } (t, x-rv, v; X)$ have the same sign in $\Big \{ d^{\varphi } (t, x-rv, v; X) < \varepsilon _3 \Big \}$. So $|v^{d, \varphi } (t, x-rv, v; X)|^2 - \varepsilon _1 d^{\varphi } (t, x-rv, v; X)^2$ attains its minimum in $\Big \{ d^{\varphi } (t, x-rv, v; X) < \varepsilon _3 \Big \}$ at ${\overline{r}}$, too. So in this domain, we have that

$$\begin{aligned} |v^{d, \varphi } (t, x-rv, v; X)|^2 - \varepsilon _1 d^{\varphi } (t, x-rv, v; X)^2 \ge - \varepsilon _1 {\overline{r}}^2, \end{aligned}$$

hence

$$\begin{aligned} |v^{d, \varphi } (t, x-rv, v; X)| \ge \sqrt{\varepsilon _1 \left( d^{\varphi } (t, x-rv, v; X)^2 - {\overline{r}}^2\right) }. \end{aligned}$$

Therefore, if ${\overline{r}} \le \varepsilon _3$, we get that

$$\begin{aligned}&\int _{[0, \infty )} 1_{ \{ d^{\varphi } (t, x-rv, v; X) \le \varepsilon _3 \} } dt \le 2 \int _{{\overline{r}}}^{\varepsilon _3} \frac{1}{\sqrt{\varepsilon _1} \sqrt{a^2 - {\overline{r}}^2}} da \\&\quad = \frac{2}{\sqrt{\varepsilon _1}} \log \left( \varepsilon _3 + \sqrt{\varepsilon _3^2 - {\overline{r}}^2} \right) - \frac{2}{\sqrt{\varepsilon _1}} \log {\overline{r}} \le \frac{2}{\sqrt{\varepsilon _1}} \log \Big (\frac{2 \varepsilon _3}{\overline{r}}\Big ). \end{aligned}$$

Since ${\overline{r}} \ge |x-\pi ^{\perp }_v X|$ by Lemma 3.11, this implies our assertion. $\square$

Proof of Proposition 3.3 (2)

We first check that (3.5) holds: for the case with $|v| \ge 2 C_0$ this is proved by Lemma 3.12; for the case with $|v| \in [{\overline{v}}, 2 C_0)$ and $|x-\pi ^{\perp }_v X| \ge \frac{1}{2} R_U$ it is proved by Lemma 3.13; and for the case with $|v| \in [{\overline{v}}, 2 C_0)$ and $|x-\pi ^{\perp }_v X| \in (0, \frac{1}{2} R_U]$ it is proved by Lemmas 3.8 and 3.14.

This combined with Lemma 3.9 completes the proof of Proposition 3.3 (2). $\square$

3.2 Error Estimate of the Freezing-Approximation

We use $\psi ^0 (m^{-1/2} (t-r), x, v; X)$ with proper X as an approximation of $x(t, \Psi (r, x, m^{-1/2} v))$ for r large enough, and use $\varphi ^0 (m^{-1/2} t, \Psi (m^{-1/2} r, x, v); X_0)$ as an approximation of it for |r| small. In this section, we discuss the error estimates of these approximations. In Sect. 3.3, we will use the first order estimates of both of them, combined with the results of Sect. 3.1, to prove that the light particles with $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ are out of the valid range after certain times. Also, we discuss the second order estimate for the approximation by $\psi (\cdot , x, v; X)$, which is necessary for formulating the limiting process.

We first quote the well-known Gronwall’s Lemma in the following form:

Lemma 3.15

(Gronwall) Suppose that a $\mathbf{R}$-valued continuous function $g(\cdot )$ satisfies

$$\begin{aligned} 0 \le g(t) \le \beta _1 (t) + \beta _2 \int _0^t g(s) ds, \quad 0 \le t \le {\overline{t}}, \end{aligned}$$

with ${\overline{t}} > 0$, $\beta _2 \ge 0$ and $\beta _1: [0, {\overline{t}}] \rightarrow \mathbf{R}$ integrable. Then

$$\begin{aligned} g(t) \le \beta _1 (t) + \beta _2 \int _0^t \beta _1 (s) e^{\beta _2 (t-s)} ds, \quad 0 \le t \le {\overline{t}}. \end{aligned}$$

In particular, if $\beta _1 (t) = {\tilde{\beta }}_1 (t) t$ with some non-decreasing ${\tilde{\beta }}_1 ( \cdot )$, then

$$\begin{aligned} g(t) \le \frac{{\tilde{\beta }}_1 (t)}{\beta _2} (e^{\beta _2 t} - 1), \quad 0 \le t \le {\overline{t}}. \end{aligned}$$

Proof

The first assertion is the well-known Gronwall’s Lemma itself. For the second half, it suffices to notice that ${\tilde{\beta }}_1 ( \cdot )$ is non-decreasing and that $\int _0^t s e^{\beta _2 (t-s)} ds = e^{\beta _2 t} \Big [ \Big ( - \frac{1}{\beta _2} t e^{- \beta _2 t} - \frac{1}{\beta _2^2} e^{- \beta _2 t} \Big ) + \frac{1}{\beta _2^2} \Big ]$. $\square$

As an easy corollary of Lemma 3.15, we get the following variation of Gronwall’s Lemma, which is used several times in this paper.

Lemma 3.16

Let y(t) be a $\mathbf{R}^d$-valued function defined on $\mathbf{R}$ satisfying the following: there exist a non-decreasing non-negative function $\beta _1 ( \cdot )$ and constants $\beta _2, {\overline{t}}_1, {\overline{t}}_2 > 0$ such that

$$\begin{aligned} \Big | \frac{d^2}{dt^2} y(t) \Big | \le \beta _1 (t) + \beta _2 |y(t)|, \qquad \hbox { for all } t \in [-{\overline{t}}_1, {\overline{t}}_2], \end{aligned}$$

and $y (-{\overline{t}}_1) = \frac{d}{dt}y (-{\overline{t}}_1) = 0$. Then we have that

$$\begin{aligned} |y(t)| \vee \Big | \frac{d}{dt} y(t) \Big | \le \frac{2 \beta _1 (t)}{1+\beta _2} \Big ( e^{\frac{1}{2} (1+\beta _2)(t+{\overline{t}}_1)} - 1 \Big ), \qquad \hbox { for all } t \in [-{\overline{t}}_1, {\overline{t}}_2]. \end{aligned}$$

Proof

Let $g_1 (t) := \Big ( |y(t)|^2 + |y'(t)|^2 \Big )^{1/2}$. Then by assumption, we get that

$$\begin{aligned} |g_1'(t)|= & {} \frac{|y(t) \cdot y'(t) + y'(t) \cdot y''(t)|}{( |y(t)|^2 + |y'(t)|^2 )^{1/2}} \le \frac{|y'(t)| \Big ( |y(t)| + \beta _1 (t) + \beta _2 |y(t)| \Big )}{( |y(t)|^2 + |y'(t)|^2 )^{1/2}} \\\le & {} \beta _1 (t) + (1+\beta _2) \frac{1}{2} ( |y(t)|^2 + |y'(t)|^2 )^{1/2} \\= & {} \beta _1 (t) + \frac{1}{2} (1+\beta _2) g_1 (t), \quad t \in [-{\overline{t}}_1, {\overline{t}}_2]. \end{aligned}$$

Here when passing to the second line, we used the general fact that $\frac{ab}{(a^2+b^2)^{1/2}} \le \frac{1}{2} (a^2+b^2)^{1/2}$ for any $a, b \in \mathbf{R}$. Let $g_2 (t) := g_1 (t-{\overline{t}}_1)$. Then $g_2 (0) = 0$ and $|g_2 '(t)| \le \beta _1 (t-{\overline{t}}_1) + \frac{1}{2} (1+\beta _2) g_2 (t)$ for any $t \in [0, {\overline{t}}_1+{\overline{t}}_2]$. Therefore, since $\beta _1 ( \cdot )$ is non-decreasing, we get that

$$\begin{aligned} g_2 (t) \le \beta _1 (t-{\overline{t}}_1) t + \frac{1}{2} (1+\beta _2) \int _0^t g_2 (s) ds, \qquad t \in [0, {\overline{t}}_1+{\overline{t}}_2]. \end{aligned}$$

By Lemma 3.15, this implies that

$$\begin{aligned} g_2(t) \le \frac{2 \beta _1 (t-{\overline{t}}_1)}{1+\beta _2} \Big ( e^{\frac{1}{2} (1+\beta _2)t} - 1 \Big ), \qquad t \in [0, {\overline{t}}_1+{\overline{t}}_2], \end{aligned}$$

which implies our assertion. $\square$

Now we are ready to consider the difference between $x (t, \Psi (r, x, m^{-1/2} v))$ and $\varphi ^0 (m^{-1/2} t, \Psi (m^{-1/2} r, x, v); X_0)$.

Lemma 3.17

Assume that $0 \le m^{1/2} s \le T \wedge \sigma _n$. Then we have that

$$\begin{aligned}&\Big | x(m^{1/2} s, \Psi (r, x, m^{-1/2}v)) - \varphi ^0 (s, \Psi (m^{-1/2} r, x, v); X_0) \Big | \\&\qquad \vee \Big | m^{1/2} v(m^{1/2} s, \Psi (r, x, m^{-1/2}v)) - \varphi ^1 (s, \Psi (m^{-1/2} r, x, v); X_0) \Big | \\&\le \frac{2 C_1 n m^{1/2} s}{1+C_1} \Big ( e^{\frac{1}{2} (1+C_1) s} - 1 \Big ). \end{aligned}$$

Proof

Let $y (s) := x(m^{1/2} s, \Psi (r, x, m^{-1/2}v)) - \varphi ^0 (s, \Psi (m^{-1/2} r, x, v); X_0)$. Then we have that $y(0) = (x-r m^{-1/2} v) - (x-r m^{-1/2} v) = 0$, and $y'(0) = m^{1/2} v(0, \Psi (r, x, m^{-1/2}v)) - \varphi ^1 (0, \Psi (m^{-1/2} r, x, v); X_0) = v - v = 0$. Also, $\frac{d^2}{ds^2} y(s) = - \nabla U (x(m^{1/2} s, \Psi (r, x, m^{-1/2}v)) - X(m^{1/2} s)) + \nabla U (\varphi ^0 (s, \Psi (m^{-1/2} r, x, v); X_0) - X_0),$ so by (2.7), we get that

$$\begin{aligned} \Big | \frac{d^2}{ds^2} y(s) \Big | \le C_1 \Big ( |y(s)| + |X(m^{1/2} s) - X_0| \Big ) \le C_1 |y(s)| + C_1 n m^{1/2} s. \end{aligned}$$

By Lemma 3.16, this implies our assertion. $\square$

For r large enough, we are going to approximate $x (t, \Psi (r, x, m^{-1/2} v))$ by the scattering $\psi ^0 (\cdot , x, v, X(\cdot ))$ with some proper $X(\cdot )$ (see (3.10) and Lemma 3.20 below). In order to apply Lemma 3.16 to estimate the approximation error of this approximation, we need to prove first that they have the same value for t small enough. We do this in Lemma 3.18 below.

Lemma 3.18

For any $(x, v) \in E$ with $|v| \ge {\overline{v}}$, we have that

$$\begin{aligned} \left( x \big (s, \Psi (r, x, m^{-1/2}v)\big ), v (s, \Psi (r, x, m^{-1/2}v)) \right) = \Psi (r-s, x, m^{-1/2} v) \end{aligned}$$

(3.9)

and

$$\begin{aligned} |x \big (s, \Psi (r, x, m^{-1/2}v)\big ) - X(s)| \ge R_U + 1 \end{aligned}$$

if $s \in [0, T \wedge \sigma _n]$ and $r \in (-\infty , -m^{1/2} \tau ] \cup [s + m^{1/2} \tau , \infty )$.

Proof

Since $s \in [0, T \wedge \sigma _n]$, we have that $|X(u)| \le |X_0| + n T$ for any $u \in [0, s]$. If $r \ge s + m^{1/2} \tau$ or $r \le - m^{1/2} \tau$, then for any $u \in [0, s]$, we have that $|u-r| \ge m^{1/2} \tau \ge m^{1/2} {\overline{v}}^{-1} R_0$, hence

$$\begin{aligned}&| x - r m^{-1/2} v + u m^{-1/2} v - X(u) | \\&\quad \ge m^{-1/2} |u-r| |v| - |X_0| - n T \ge m^{-1/2} m^{1/2} {\overline{v}}^{-1} R_0 {\overline{v}} - |X_0| - n T \\&\quad = R_0 - |X_0| - n T = R_U + 1, \end{aligned}$$

so the light particle keeps a uniform motion during the time interval [0, s] . Therefore, (3.9) holds, hence $|x (s, \Psi (r, x, m^{-1/2} v)) - X(s)| = |x - m^{-1/2} (r-s) v - X(s)| \ge R_U + 1$. $\square$

For any $t \ge 0$, let $\mathcal {F}_t := \mathcal {F}_{(-\infty , t + m^{1/2} \tau ] \times E}$. Then we get the following as a corollary of Lemma 3.18. This is used later.

Lemma 3.19

For any $t \in [0, T \wedge \sigma _n]$, we have that X(t) is $\mathcal {F}_t$-measurable. $\square$

Choose any $c_m \rightarrow 0$ (when $m \rightarrow 0$). We are going to approximate $x (m^{1/2} t + s, \Psi (s, x, m^{-1/2} v))$ by the scattering $\psi ^0 (t, x, v; X(s-c_m))$ (see (3.10) and Lemma 3.20 below). Let

$$\begin{aligned} \kappa _1 (t) := x (m^{1/2} t + s, \Psi (s, x, m^{-1/2} v)) - \psi ^0 (t, x, v; X(s-c_m)). \end{aligned}$$

(3.10)

We prove in Lemma 3.20 below that the first order error $\kappa _1 (t)$ is small enough. Lemma 3.20 with $c_m = \tau m^{1/2}$ is enough for proving that $x (m^{1/2} t + r, \Psi (r, x, m^{-1/2} v))$ is out of the valid interaction range for $t \notin [-\tau , t_1(v, |x-\pi ^{\perp }_v X({\tilde{r}})|)]$ and $r \ge m^{1/2} \tau$. However, in Sect. 3.4, Lemma 3.20 with general $c_m$ as in our present form is necessary in order to estimate the second order error of our freezing-approximation.

Lemma 3.20

Assume that $(x, v) \in E$, $|v| \ge {\overline{v}}$ and $s - c_m, s - m^{1/2} \tau \in [0, T \wedge \sigma _n]$. Then we have the following:

• 1. $\kappa _1 (t) = 0$ if $0 \le m^{1/2} t + s \le T \wedge \sigma _n$ and $t \le - \tau$.

• 2.
  $$\begin{aligned} \frac{d^2}{dt^2} \kappa _1 (t)= & {} - \nabla U (\kappa _1 (t) + \psi ^0 (t, x, v; X(s-c_m)) - X(m^{1/2} t + s)) \nonumber \\&+ \nabla U (\psi ^0 (t, x, v; X(s-c_m)) - X(s-c_m)). \end{aligned}$$

  (3.11)

• 3. for any $b_m \ge \tau$, we have that

  $$\begin{aligned} |\kappa _1(t)| \vee |\kappa _1'(t)| \le \frac{2 C_1 n}{1+C_1} (m^{1/2} b_m + |c_m|) \Big ( e^{\frac{1}{2} (1+C_1) (t+\tau )} - 1 \Big ) \end{aligned}$$

  for any $t \in [-\tau , b_m]$ satisfying $0 \le m^{1/2} t + s \le T \wedge \sigma _n$.

We remark that $b_m$ and $c_m$ in Lemmas 3.20 and 3.21 may depend on (x, v) and s.

Proof

Under the given assumption, we notice that $|X(s-c_m)| \le |X_0| + nT$, so for any $t \le - \tau$ satisfying $0 \le m^{1/2} t + s \le T \wedge \sigma _n$, we have by Lemma 3.18 and Lemma 3.2 (2) that $x (m^{1/2} t + s, \Psi (s, x, m^{-1/2} v)) = \psi ^0 (t, x, v; X(s-c_m)) = x + tv$, hence $\kappa _1 (t) = 0$.

The second assertion is trivial by definition.

For the last assertion, we notice that since $t \in [-\tau , b_m]$, $s - c_m, m^{1/2} t + s \in [0, T \wedge \sigma _n ]$ and $b_m \ge \tau$, we have that

$$\begin{aligned} |X(m^{1/2}t+s) - X(s-c_m)| \le n |m^{1/2} t + c_m| \le n (m^{1/2} b_m + |c_m|). \end{aligned}$$

Therefore, by (3.11) and (2.7), we get that

$$\begin{aligned} \Big | \frac{d^2}{dt^2} \kappa _1 (t) \Big | \le C_1 \left( |\kappa _1(t)| + n (m^{1/2} b_m + |c_m|)\right) \end{aligned}$$

for any $t \in [-\tau , b_m]$ satisfying $0 \le m^{1/2} t + s \le T \wedge \sigma _n$. Also, $\kappa _1 (-\tau ) = \kappa _1' (-\tau ) = 0$ by the first assertion of this lemma. These combined with Lemma 3.16 imply our third assertion. $\square$

Lemma 3.20 is enough for our proof of the tightness. However, to find the explicit expression of the limiting process (see Sect. 3.5 for details), we need the second order error estimate, too. Let

$$\begin{aligned} \kappa _2 (t) := \kappa _1 (t) - m^{1/2} z \left( t, x, v; X(s-c_m), V(s-c_m), m^{-1/2} c_m\right) . \end{aligned}$$

Lemma 3.21 below gives an estimate of $\kappa _2 (t)$.

The following estimate, which will be used several times in this paper, is trivial by a direct calculation.

$$\begin{aligned} \frac{2 C_1}{1+C_1} (e^b - 1) + 1 \le 2 e^b, \qquad \hbox { for all } b > 0. \end{aligned}$$

(3.12)

Using this we shall prove the following:

Lemma 3.21

Assume that $(x, v) \in E$, $|v| \ge {\overline{v}}$, $b_m \ge \tau$, $- b_m \le m^{-1/2} c_m \le \tau$, $- \tau \le t \le b_m$ and $s - c_m, s + m^{1/2} t, s - m^{1/2} \tau \in [0, T \wedge \sigma _n]$. Then for any $a \in [0, 1]$, we have that

$$\begin{aligned} |\kappa _2 (t)|\le & {} \frac{8 n^2}{1+C_1} (2C_1 \vee 1) (\Vert \nabla ^3 U \Vert _{\infty } \vee 1) e^{\left( 1+\frac{a}{2}\right) (1 + C_1) (t + \tau )} \Big ( m^{1/2} b_m + |c_m| \Big )^{1+a} \\&+ \frac{2 C_1}{1+C_1} \Big ( e^{\frac{1}{2} (1 + C_1) (t + \tau )} - 1 \Big ) \int _{s - m^{1/2} \tau }^{(s + m^{1/2} t) \vee (s-c_m)} |V(u) - V(s-c_m)| du. \end{aligned}$$

Proof

First we notice that $\kappa _2 (-\tau ) = \frac{d}{dt} \kappa _2 (-\tau ) = 0$ by Lemmas 3.20 (1), 3.2 (2) and the definition of z. Also, by definition and a simple calculation, we get that

$$\begin{aligned} \frac{d^2}{dt^2} \kappa _2 (t)= & {} - \int _0^1 d \theta \Big \{ \nabla ^2 U \Big ( \psi ^0 (t, x, v; X(s-c_m)) - X(s-c_m) \\&+ \theta \big [ \kappa _1(t) - (X(m^{1/2}t+s) - X(s-c_m)) \big ] \Big ) \\&- \nabla ^2 U (\psi ^0 (t, x, v; X(s-c_m)) - X(s-c_m)) \Big \} \\&\cdot \left[ \kappa _1 (t) - X(m^{1/2}t+s) + X(s-c_m) \right] \\&- \nabla ^2 U (\psi ^0 (t, x, v; X(s-c_m)) - X(s-c_m)) \\&\cdot \left\{ \kappa _2 (t) - \big [X(m^{1/2}t+s) - X(s-c_m) - (m^{1/2} t+c_m) V (s-c_m)\big ] \right\} . \end{aligned}$$

We have that $|X(m^{1/2}t+s) - X(s-c_m) | \le n |m^{1/2} t + c_m|$ and

$$\begin{aligned}&\Big | X(m^{1/2}t+s) - X(s-c_m) - (m^{1/2} t+c_m) V (s-c_m) \Big | \\&= \Big | \int _{s-c_m}^{m^{1/2}t+s} (V(u) - V (s-c_m)) du \Big | \le \int _{s - m^{1/2} \tau }^{(s + m^{1/2} t) \vee (s-c_m)} |V(u) - V(s-c_m)| du. \end{aligned}$$

Combining these with (2.6), for any $a \in [0, 1]$, we get that

$$\begin{aligned} \Big | \frac{d^2}{dt^2} \kappa _2 (t) \Big |\le & {} (2C_1)^{1-a} \Vert \nabla ^3 U \Vert _{\infty }^a \Big ( |\kappa _1(t)| + n |m^{1/2} t + c_m| \Big )^{1+a} \\&+ C_1 \left( |\kappa _2(t)| + \int _{s - m^{1/2} \tau }^{(s + m^{1/2} t) \vee (s-c_m)} |V(u) - V(s-c_m)| du \right) \end{aligned}$$

for any $t \in [-\tau , b_m]$. This combined with Lemma 3.20 and (3.12) implies that

$$\begin{aligned} \Big | \frac{d^2}{dt^2} \kappa _2 (t) \Big |\le & {} (2C_1 \vee 1) (\Vert \nabla ^3 U \Vert _{\infty } \vee 1) \Big ( 2 n ( m^{1/2} b_m + |c_m| ) e^{\frac{1}{2} (1+C_1) (t+\tau )} \Big )^{1+a} \\&+ C_1 \int _{s - m^{1/2} \tau }^{(s + m^{1/2} t) \vee (s-c_m)} |V(r) - V(s-c_m)| dr + C_1 |\kappa _2(t)|. \end{aligned}$$

This combined with Lemma 3.16 implies our assertion. $\square$

3.3 Behavior of x(t, x, v)

In this section, we study the sojourn time of the light particle in the valid interaction range. Precisely, we consider the question about when does $\nabla U (x (m^{1/2} t + r, \Psi (r, x, m^{-1/2} v)) - X(m^{1/2} t + r)) \ne 0$ hold. Our main result of this section is the following.

Proposition 3.22

There exists a $m_1 \ge 0$ such that for any $m \in (0, m_1]$, $(x, v) \in E$ satisfying $|v| \ge {\overline{v}}$ and $r, t \in \mathbf{R}$ satisfying $m^{1/2} t + r \in [0, T \wedge \sigma _n]$, we have that

$$\begin{aligned} \Big | x (m^{1/2} t + r, \Psi (r, x, m^{-1/2} v)) - X(m^{1/2} t + r) \Big | \ge R_U \end{aligned}$$

if at least one of the following conditions is satisfied:

• 1. $|v| \ge 2 C_0 + 1$ and $t \notin [-\tau , 2\tau ]$,

• 2. $r \le - m^{1/2} \tau$,

• 3. $t \le - \tau$,

• 4. $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge R_U + \frac{1}{2}$,

• 5. $\alpha$ satisfies (2.10), $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ and $t \ge t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)$.

We prove Proposition 3.22 in Sect. 3.3. Since the notation $\tau$ in [12] is equal to $C_0^{-1} R_0$, which is dominated by our $\tau$ in this paper, we get the first assertion of Proposition 3.22 as a trivial corollary of [12, Proposition 3.6.5]. The assertions (2) and (3) are trivial by Lemma 3.18. We prove the assertions (4) and (5) of Proposition 3.22 in the rest of Sect. 3.3.

Proposition 3.22 (4) is included in Lemma 3.23 below.

Lemma 3.23

Assume that $r, t \in \mathbf{R}$ satisfy $m^{1/2} t + r \in [0, T \wedge \sigma _n]$. Let $(x, v) \in E$ with $|v| \ge {\overline{v}}$. Also, assume that $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge R_U + \frac{1}{2}$. Then as long as $m \le (4n\tau )^{-2} \wedge (n^{-2} {\overline{v}}^2)$, we have that

$$\begin{aligned} (x(m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)), v(m^{1/2} t + r, \Psi (r, x, m^{-1/2}v))) = (x+tv, m^{-1/2}v) \end{aligned}$$

and

$$\begin{aligned} |x\left( m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)\right) - X (m^{1/2} t + r) | \ge R_U. \end{aligned}$$

We remark that by Lemma 3.23, at least until $T \wedge \sigma _n$, those light particles with $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge R_U + \frac{1}{2}$ keep uniform motions, and never enter the valid interaction range.

Proof of Lemma 3.23

If $r \le -m^{1/2} \tau$ or $t \le - \tau$, then our assertion is nothing but Lemma 3.18. From now on, we assume that $r \ge -m^{1/2} \tau$ and $t \ge - \tau$. Therefore, $r - m^{1/2} \tau \le r + m^{1/2} t \le T \wedge \sigma _n$, hence ${\tilde{r}} = (r - m^{1/2} \tau ) \vee 0$ and $m^{1/2} t + r \ge (r - m^{1/2} \tau ) \vee 0 = {\tilde{r}}$.

We first deal with the case $t \in [-\tau , \tau ]$. First we notice that

$$\begin{aligned} \left( x\big ({\tilde{r}}, \Psi (r, x, m^{-1/2} v)\big ), v\big ({\tilde{r}}, \Psi (r, x, m^{-1/2} v)\big ) \right) = \big (x - (\tau \wedge m^{-1/2} r) v, m^{-1/2} v\big ). \end{aligned}$$

(3.13)

Indeed, if $r \le m^{1/2} \tau$, then ${\tilde{r}} = 0$, hence $\Big ( x({\tilde{r}}, \Psi (r, x, m^{-1/2} v)), v({\tilde{r}}, \Psi (r, x, m^{-1/2} v)) \Big ) = (x - m^{-1/2} r v, m^{-1/2} v)$; if $r > m^{1/2} \tau$, then ${\tilde{r}} = r - m^{1/2} \tau$, hence by Lemma 3.18, we get that $\Big ( x({\tilde{r}}, \Psi (r, x, m^{-1/2} v)), v({\tilde{r}}, \Psi (r, x, m^{-1/2} v)) \Big ) = \Psi (r-{\tilde{r}}, x, m^{-1/2} v) = (x-\tau v, m^{-1/2} v)$.

Also, for any $u \in [{\tilde{r}}, m^{1/2} t + r]$, since $t \le \tau$ by assumption and ${\tilde{r}}, u \in [0, T \wedge \sigma _n]$, we get that $|X({\tilde{r}}) - X(u)| \le n (u-{\tilde{r}}) \le n (m^{1/2} t + r - (r - m^{1/2} \tau )) \le 2 n m^{1/2} \tau$. Therefore, since $m \le (4n\tau )^{-2}$ by assumption, we get that

$$\begin{aligned}&|x - m^{-1/2} r v + u m^{-1/2} v - X(u)| \ge | x - \pi ^{\perp }_v X({\tilde{r}}) | - |X({\tilde{r}}) - X(u)| \nonumber \\&\ge R_U + \frac{1}{2} - 2 n m^{1/2} \tau \ge R_U, \qquad \qquad u \in [{\tilde{r}}, m^{1/2} t + r]. \end{aligned}$$

(3.14)

So the particle keeps a uniform motion during the time interval $u \in [{\tilde{r}}, m^{1/2} t + r]$. Combining this with (3.13), we get that

$$\begin{aligned}&v\big (m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)\big ) = v\big ({\tilde{r}}, \Psi (r, x, m^{-1/2} v)\big ) = m^{-1/2} v, \\&x\big (m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)\big ) \\&\qquad = x - (\tau \wedge m^{-1/2} r) v + (m^{1/2} t + r - {\tilde{r}}) m^{-1/2} v \\&\qquad = x + \big (m^{1/2} t + r - {\tilde{r}} - (m^{1/2} \tau ) \wedge r \big ) m^{-1/2} v = x + t v. \end{aligned}$$

This combined with (3.14) (with $u = m^{1/2} t + r$) implies our assertion, and completes our proof for the case $t \in [-\tau , \tau ]$.

In particular, if $r + m^{1/2} \tau \le T \wedge \sigma _n$, then $(x(m^{1/2} \tau + r, \Psi (r, x, m^{-1/2}v)), v(m^{1/2} \tau + r, \Psi (r, x, m^{-1/2}v))) = (x + \tau v, m^{-1/2}v)$. We prove in the following that this implies our assertions for $t \ge \tau$. Indeed, let $\xi _4 := \inf \{ t > \tau ; |x(m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)) - X (m^{1/2} t + r) | < R_U \}$. Then $|x(m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)) - X (m^{1/2} t + r) | \ge R_U$ for any $t \in [\tau , \xi _4]$, hence $(x(m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)), v(m^{1/2} t + r, \Psi (r, x, m^{-1/2}v))) = (x + t v, m^{-1/2}v)$ for any $t \in [\tau , \xi _4]$. Therefore, it suffices to prove that $m^{1/2} \xi _4 + r \ge T \wedge \sigma _n$. Suppose not. Then for any $t \in [\tau , \xi _4]$, since $m \le n^{-2} {\overline{v}}^2$ by assumption, we have that $(m^{-1/2} v - V(m^{1/2} t + r)) \cdot \frac{v}{|v|} \ge m^{-1/2} |v| - n \ge 0$. Therefore,

$$\begin{aligned} R_U= & {} \Big | x\left( m^{1/2} \xi _4 + r, \Psi (r, x, m^{-1/2}v)\right) - X (m^{1/2} \xi _4 + r) \Big | \\\ge & {} \Big ( x(m^{1/2} \xi _4 + r, \Psi (r, x, m^{-1/2}v)) - X (m^{1/2} \xi _4 + r) \Big ) \cdot \frac{v}{|v|} \\= & {} \Big \{ x\left( m^{1/2} \tau + r, \Psi (r, x, m^{-1/2}v)\right) - X (m^{1/2} \tau + r) \\&\quad + \,m^{1/2} \int _{\tau }^{\xi _4} \Big ( v\left( m^{1/2} t + r, \Psi (r, x, m^{-1/2}v)\right) - V (m^{1/2} t + r) \Big ) dt \Big \} \cdot \frac{v}{|v|} \\= & {} \Big \{ x + \tau v - X (m^{1/2} \tau + r) + m^{1/2} \int _{\tau }^{\xi _4} \Big ( m^{-1/2} v - V(m^{1/2} t + r) \Big ) dt \Big \} \cdot \frac{v}{|v|} \\\ge & {} \Big \{ x + \tau v - X (m^{1/2} \tau + r) \Big \} \cdot \frac{v}{|v|} \ge \tau |v| - (|X_0| + nT) \\\ge & {} R_0 - (|X_0| + nT) = R_U + 1, \end{aligned}$$

which yields a contradiction. $\square$

Finally, we prove the assertion (5) of Proposition 3.22. We notice that $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) \ge 2 \tau$ by definition, so by (1) (2) and (4) of Proposition 3.22, we only need to consider those light particles with $|v| \in [{\overline{v}}, 2C_0+1)$, $r \ge - m^{1/2} \tau$ and $| x - \pi ^{\perp }_v X({\tilde{r}})| \in [m^{\alpha }, R_U + \frac{1}{2}]$. In other words, it suffices to prove the following lemma:

Lemma 3.24

Under (2.10), there exists a $m_2 > 0$ such that for any $m \in (0, m_2]$, we have that

$$\begin{aligned} |x\left( m^{1/2} t + r, \Psi (r, x, m^{-1/2} v)\right) - X(m^{1/2} t + r) | \ge R_U \end{aligned}$$

as long as $(x, v) \in E$, $|v| \in [{\overline{v}}, 2C_0+1)$, $|x-\pi ^{\perp }_v X({\tilde{r}})| \in [ m^{\alpha }, R_U + \frac{1}{2}]$, $t \ge t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)$, $r \ge - m^{1/2} \tau$ and $m^{1/2} t + r \in [0, T \wedge \sigma _n]$.

We prove Lemma 3.24 by using the results of Sects. 3.1 and 3.2. As explained before, we approximate $x (m^{1/2} t + r, \Psi (r, x, m^{-1/2} v))$ by $\varphi (t + m^{-1/2} r, \Psi (m^{-1/2} r, x, v); X_0)$ if $|r| \le m^{1/2} \tau$, and by $\psi (t, x, v; X({\tilde{r}}))$ if $r > m^{1/2} \tau$. Let a(t, r, x, v) denote this approximation, i.e., for any $(x, v) \in E$, $r \ge - m^{1/2} \tau$ and $t \in \mathbf{R}$, define $a (t, r, x, v) = (a^0 (t, r, x, v), a^1 (t, r, x, v))$ by

$$\begin{aligned} a (t, r, x, v) := \left\{ \begin{array}{ll} \varphi \left( t + m^{-1/2} r, \Psi (m^{-1/2} r, x, v); X_0\right) , &{} \hbox { if } \quad |r| \le m^{1/2} \tau , \\ \psi \left( t, x, v; X({\tilde{r}})\right) , &{} \hbox { if } \quad r > m^{1/2} \tau . \end{array} \right. \end{aligned}$$

Also, let $d^a (t, r, x, v) := |a^0 (t, r, x, v) - X({\tilde{r}})|$ and $v^{d, a} (t, r, x, v) := \frac{d}{dt} d^a (t, r, x, v) = \Big ( a^1 (t, r, x, v), \frac{a^0 (t, r, x, v) - X({\tilde{r}})}{|a^0 (t, r, x, v) - X({\tilde{r}})|} \Big )$. Then we have the following.

Lemma 3.25

Suppose that $(x, v) \in E$, $r \ge - m^{1/2} \tau$ and $|v| \ge {\overline{v}}$. Then we have the following:

• 1. $| a^0 \left( t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|), r, x, v\right) - X({\tilde{r}}) | \ge R_U + 1$,

• 2. $v^{d, a} \left( t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|), r, x, v\right) > 0$.

Proof

Let $s_1 := |v|^{-1} R_{X({\tilde{r}})}$. Then $s_1 \le \tau$. With the help of (3.3), we get our assertions for $r \in [-m^{1/2} \tau , - m^{1/2} s_1]$ by Proposition 3.3 (1); for $r \in (- m^{1/2} s_1, m^{1/2} s_1]$ by Proposition 3.3 (2); for $r \in (m^{1/2} s_1, m^{1/2} \tau ]$ by Proposition 3.3 (3); and for $r > m^{1/2} \tau$ by Proposition 3.4. $\square$

In order to prove Lemma 3.24, we first prove that the light particle is out of the valid interaction range at time $m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r$.

Lemma 3.26

Assume (2.10). Then there exists a $m_3 > 0$ such that for any $m \in (0, m_3]$, we have that

$$\begin{aligned}&\Big | x\left( m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r, \Psi (r, x, m^{-1/2} v)\right) \\&\quad - X\left( m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r\right) \Big | \ge R_U + \frac{1}{2} \end{aligned}$$

as long as $(x, v) \in E$, $|v| \in [{\overline{v}}, 2C_0+1)$, $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, $r \ge - m^{1/2} \tau$ and $m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r \in [0, T \wedge \sigma _n]$.

Proof

To simplify notations, write $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)$ as $t_0$ in the proof of this lemma. By Lemma 3.17 and Lemma 3.20 (3) for $c_m = m^{1/2} \tau$, we get that

$$\begin{aligned}&| x\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - a^0 (t_0, r, x, v) | \\&\quad \le \frac{2 C_1}{1+C_1} n m^{1/2} (t_0 + \tau ) \Big ( e^{\frac{1}{2} (1+C_1) (t_0+\tau )} - 1 \Big ). \end{aligned}$$

Also, since $m^{1/2} t_0 + r, {\tilde{r}} \in [0, T \wedge \sigma _n]$, we have that $| X(m^{1/2} t_0 + r) - X({\tilde{r}}) | \le n |m^{1/2} t_0 + r - {\tilde{r}}| \le n m^{1/2} (t_0 + \tau )$. Combining these with Lemma 3.25 (1), we get that

$$\begin{aligned}&| x\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - X(m^{1/2} t_0 + r) | \\&\ge | a^0 (t_0, r, x, v) - X({\tilde{r}}) | - | x\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - a^0 (t_0, r, x, v) | \\&\qquad - | X(m^{1/2} t_0 + r) - X({\tilde{r}}) | \\&\ge R_U + 1 - 2 n m^{1/2} (t_0 + \tau ) e^{\frac{1}{2} (1+C_1) (t_0+\tau )}. \end{aligned}$$

So it suffices to choose $m_3 > 0$ such that $2 n m^{1/2} (t_0 + \tau ) e^{\frac{1}{2} (1+C_1) (t_0+\tau )} \le \frac{1}{2}$ for any $m \in (0, m_3]$. We have that $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ by assumption, so as long as $m \le (2 \varepsilon _3)^{1/\alpha }$, we have that $t_0 = t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) \le C_2 + 2 \varepsilon _1^{-1/2} \log \frac{2 \varepsilon _3}{m^{\alpha }}$. Therefore,

$$\begin{aligned}&2 n m^{1/2} (t_0 + \tau ) e^{\frac{1}{2} (1+C_1) (t_0+\tau )} \\&\quad \le 2 n e^{\frac{1}{2} (1+C_1) (C_2 + \tau + 2 \varepsilon _1^{-1/2} \log (2 \varepsilon _3 ) ) } \left( C_2 + \tau + 2 \varepsilon _1^{-1/2} \log \frac{2 \varepsilon _3}{m^{\alpha }}\right) m^{\frac{1}{2} - \frac{1}{2} (1+C_1) 2 \varepsilon _1^{-1/2} \alpha }. \end{aligned}$$

Since the right hand side above converges to 0 as long as $(1+C_1) \varepsilon _1^{-1/2} \alpha < \frac{1}{2}$, we get our assertion. $\square$

Notice: we do not need $|x-\pi ^{\perp }_v X(s - a m^{1/2})| \le R_U + \frac{1}{2}$ in Lemma 3.26. However, this condition is necessary for Lemma 3.27 below.

In order to prove that the particle could not enter the valid interaction range after $m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r$, either, we also need to discuss its velocity at this time. We first notice the following general estimate:

$$\begin{aligned} \Big | v_1 \cdot \frac{x_1}{|x_1|} - v_2 \cdot \frac{x_2}{|x_2|} \Big | \le |v_1 - v_2| + 2 \frac{|v_2|}{|x_2|} |x_1 - x_2|, \quad \hbox { for any } x_1, x_2, v_1, v_2 \in \mathbf{R}^d.\nonumber \\ \end{aligned}$$

(3.15)

Indeed, since $\Big | \frac{x_1}{|x_1|} - \frac{x_2}{|x_2|} \Big | = \Big | \frac{x_1 - x_2}{|x_2|} + \Big ( \frac{1}{|x_1|} - \frac{1}{|x_2|} \Big ) x_1 \Big | \le \frac{2 |x_1-x_2|}{|x_2|},$ we get that

$$\begin{aligned} \Big | v_1 \cdot \frac{x_1}{|x_1|} - v_2 \cdot \frac{x_2}{|x_2|} \Big |= & {} \Big | (v_1 - v_2) \cdot \frac{x_1}{|x_1|} + v_2 \cdot \Big ( \frac{x_1}{|x_1|} - \frac{x_2}{|x_2|} \Big ) \Big | \\\le & {} |v_1 - v_2| + |v_2| \Big | \frac{x_1}{|x_1|} - \frac{x_2}{|x_2|} \Big | \le |v_1 - v_2| + \frac{2 |v_2|}{|x_2|} |x_1-x_2|. \end{aligned}$$

Let

$$\begin{aligned} v^{d, 2} (t, \Psi (r, x, m^{-1/2} v)) := \Big ( v(t, \Psi (r, x, m^{-1/2} v)), \frac{x(t, \Psi (r, x, m^{-1/2} v)) - X(t)}{|x(t, \Psi (r, x, m^{-1/2} v)) - X(t)|} \Big ). \end{aligned}$$

We notice that $v^{d, 2} (t, \Psi (r, x, m^{-1/2} v))$ is not equal to $\frac{d}{dt} |x(t, \Psi (r, x, m^{-1/2} v)) - X(t)|$. Then we have the following.

Lemma 3.27

Assume (2.10). Then there exist constants $m_4, C_4 > 0$ such that for any $m \in (0, m_4]$, we have that

$$\begin{aligned} v^{d, 2} (m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r, \Psi (r, x, m^{-1/2} v)) \ge C_4 m^{-1/2} \end{aligned}$$

as long as $(x, v) \in E$, $|x-\pi ^{\perp }_v X({\tilde{r}})| \in [ m^{\alpha }, R_U + \frac{1}{2}]$, $r \ge - m^{1/2} \tau$ and $m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r \in [0, T \wedge \sigma _n]$.

Proof

To simplify notations, we write $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)$ as $t_0$ in the proof of this lemma, too. By Lemmas 3.17 and 3.20 (3) for $c_m = m^{1/2} \tau$, we have that

$$\begin{aligned}&| m^{1/2} v\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - a^1 (t_0, r, x, v) | \\&\quad \vee | x\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - a^0 (t_0, r, x, v) | \\&\le \frac{2 C_1 n}{1 + C_1} m^{1/2} (t_0 + \tau ) \Big ( e^{\frac{1}{2} (1+C_1) (t_0+\tau )} - 1 \Big ). \end{aligned}$$

Also, by Lemma 3.5 (1) and Lemma 3.7 (1), we have that $|a^1 (t_0, r, x, v)| \le |v| + \sqrt{2 \Vert U\Vert _{\infty }}$. Since $m^{1/2} t_0 + r, {\tilde{r}} \in [0, T \wedge \sigma _n]$, we have $| X(m^{1/2} t_0 + r) - X({\tilde{r}}) | \le n |m^{1/2} t_0 + r - {\tilde{r}}| \le n m^{1/2} (t_0 + \tau )$. These combined with (3.15) and Lemma 3.25 (1) imply that

$$\begin{aligned}&| m^{1/2} v^{d, 2} \left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - v^{d, a} (t_0, r, x, v) | \nonumber \\&\le | m^{1/2} v\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - a^1 (t_0, r, x, v) | \nonumber \\&\qquad + \frac{2(|v| + \sqrt{2 \Vert U\Vert _{\infty }})}{R_U + 1} \Big ( | x\left( m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\right) - a^0 (t_0, r, x, v) | \nonumber \\&\qquad + |X(m^{1/2} t_0 + r) - X({\tilde{r}})| \Big ) \nonumber \\&\le \left( 1 + \frac{2(|v| + \sqrt{2 \Vert U\Vert _{\infty }})}{R_U + 1} \right) \frac{2 C_1 n}{1 + C_1} m^{1/2} (t_0 + \tau ) \Big ( e^{\frac{1}{2} (1+C_1) (t_0+\tau )} - 1 \Big ) \nonumber \\&\qquad + \frac{2(|v| + \sqrt{2 \Vert U\Vert _{\infty }})}{R_U + 1} n m^{1/2} (t_0 + \tau ) \nonumber \\&\le 4 (2|v| + 2 \sqrt{2 \Vert U\Vert _{\infty }} + 1) n m^{1/2} t_0 e^{\frac{1}{2} (1+C_1) (t_0+\tau )}. \end{aligned}$$

(3.16)

On the other hand, let $C_5 := \Big ( 1 - \frac{(R_U+\frac{1}{2})^2}{(R_U+1)^2} \Big )^{1/2}$, then by Lemmas 3.6 (4), 3.7 (3) and 3.25 (1), we get that

$$\begin{aligned} |v^{d, a} (t_0, r, x, v)|^2 \ge - \frac{ |v|^2 |x - \pi ^{\perp }_v X({\tilde{r}})|^2}{d^a (t_0, r, x, v)^2} + |v|^2 \ge \left( 1 - \frac{\left( R_U+\frac{1}{2}\right) ^2}{(R_U+1)^2} \right) |v|^2 = C_5^2 |v|^2. \end{aligned}$$

By Lemma 3.25 (2), this implies that $v^{d, a} (t_0, r, x, v) \ge C_5 |v|.$ Combining this with (3.16), we get that

$$\begin{aligned}&m^{1/2} v^{d, 2} \big (m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\big ) \nonumber \\&\ge C_5 |v| - 4 \big (2|v| + 2 \sqrt{2 \Vert U\Vert _{\infty }} + 1\big ) n m^{1/2} t_0 e^{\frac{1}{2} (1+C_1) (t_0+\tau )} \nonumber \\&= \big ( C_5 - 8 n m^{1/2} t_0 e^{\frac{1}{2} (1+C_1) (t_0+\tau )} \big ) |v| - 4 \big (2\sqrt{2 \Vert U\Vert _{\infty }} + 1\big ) n m^{1/2} t_0 e^{\frac{1}{2} (1+C_1) (t_0+\tau )}. \nonumber \\ \end{aligned}$$

(3.17)

Since $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }$ by assumption, we have that $t_0 = t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) \le C_2 + 2 \varepsilon _1^{-1/2} \log \frac{2 \varepsilon _3}{m^{\alpha }}$. Since $\frac{1}{2} - (1+C_1) \varepsilon _1^{-1/2} \alpha > 0$ by assumption, we get that $m^{1/2} t_0 e^{\frac{1}{2} (1+C_1) (t_0+\tau )} \rightarrow 0$ as $m \rightarrow 0$. Also, $|v| \ge {\overline{v}}$. Therefore, there exists a $m_4 \in (0, 1]$ such that the right hand side of (3.17) with $m \in (0, m_4]$ is dominated below by a strictly positive constant. This gives us our assertion. $\square$

Now, we are ready to prove that those light particles with $|x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$ could never enter the valid range after $m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + r$.

Proof of Lemma 3.24

As we did before, we shall write $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)$ as $t_0$ to simplify notations. Let $\xi _5 := \inf \{ t > t_0: |x(m^{1/2} t + r, \Psi (r, x, m^{-1/2} v)) - X(m^{1/2} t + r) | < R_U \}$. It suffices to prove that $m^{1/2} \xi _5 + r > T \wedge \sigma _n$.

Let $e_3 := \frac{x(m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)) - X(m^{1/2} t_0 + r)}{|x(m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)) - X(m^{1/2} t_0 + r)|}$. Then by Lemma 3.27 we have that $v(m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)) \cdot e_3 \ge C_4 m^{-1/2}$. We notice that the particle keeps a uniform motion during the time interval $[ m^{1/2} t_0 + r, m^{1/2} \xi _5 + r ]$, hence $v(m^{1/2} t + r, \Psi (r, x, m^{-1/2} v)) = v(m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v))$ for any $t \in [t_0, \xi _5]$. Suppose that $m^{1/2} \xi _5 + r \le T \wedge \sigma _n$. Then $0 \le m^{1/2} t_0 + r \le m^{1/2} \xi _5 + r \le T \wedge \sigma _n$, hence $|X(m^{1/2} t_0 + r) - X(m^{1/2} \xi _5 + r)| \le n m^{1/2} (\xi _5-t_0)$. Combining these with Lemma 3.26, we get that if $m \le m_3 \wedge m_4$, then

$$\begin{aligned} R_U= & {} | x\big (m^{1/2} \xi _5 + r, \Psi (r, x, m^{-1/2} v)\big ) - X\big (m^{1/2} \xi _5 + r\big ) | \\\ge & {} \Big ( x\big (m^{1/2} \xi _5 + r, \Psi (r, x, m^{-1/2} v)\big ) - X(m^{1/2} \xi _5 + r) \Big ) \cdot e_3 \\= & {} |x\big (m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\big ) - X(m^{1/2} t_0 + r)| \\&+ m^{1/2} (\xi _5 - t_0) v\big (m^{1/2} t_0 + r, \Psi (r, x, m^{-1/2} v)\big ) \cdot e_3 \\&+ \Big ( X(m^{1/2} t_0 + r) - X(m^{1/2} \xi _5 + r) \Big ) \cdot e_3 \\\ge & {} R_U + \frac{1}{2} + m^{1/2} \big (\xi _5 - t_0) (C_4 m^{-1/2} - n\big ). \end{aligned}$$

If $m \le n^{-2} C_4^2$ in addition, the right hand side above is greater than or equal to $R_U + \frac{1}{2}$, which yields a contradiction. $\square$

3.4 Tightness and Decomposition of V(t)

In Sects. 3.4 and 3.5, we re-express V(t) as a sum of a martingale term, a smooth term and a negligible term. The heuristic meanings of these quantities are the same as those discussed in [12, 13]. Precisely, the martingale term is approximately the variance of the force after our freezing-approximation (see the term $I^3 (t)$ defined below). We remark that, when there are more than one massive particles, the mean of the force after our freezing-approximation (the term $I^5 (t)$ defined below) gives us approximately the resulting interaction between the massive particles – since there is only one massive particle in our present model, this term is approximately 0 in our case (see [12, Lemma 4.3.3] and the proof of Lemma 3.37). The smooth part (the term $I^4 (t)$ defined below) is given approximately by the first order of our approximation error.

Also, the discussion with respect to $I^6 (t)$ and $I^9 (t)$ defined below consists of two steps: we first prove in Sect. 3.4 that they can also be regarded as a part of the smooth term; and with the help of the results of Sect. 3.4, we prove in Sect. 3.5 that they are actually negligible.

Necessary estimates of these quantities are also provided in Sect. 3.4. In particular, we get the desired tightness with the help of the Skorohod space (see Sect. 3.4.1 for a brief review of Skorohod space).

Section 3.4 is organized as follows. In Sect. 3.4.1, we review several basic facts with respect to the Skorohod space and the tightness of the probability measures on it; in Sect. 3.4.2, we present our key decomposition of V(t) and the tightness (Proposition 3.31), the main result of Sect. 3.4; Sects.  3.4.3 and 3.4.4 give the proof of Proposition 3.31 except for the part with respect to the martingale term, which will be presented in Sect. 3.4.5.

3.4.1 Basic Facts for Tightness

We briefly review several basic facts with respect to tightness. As claimed in Sect. 2, the martingale part of V(t) is not continuous, so we consider the problem in the framework of the Skorohod space. We begin by recalling some basic facts with respect to the Skorohod space $(D ([0, T]; \mathbf{R}^d), d^0)$ and the tightness of the probability measures on it (see Billingsley [1] for more details).

For any $T > 0$, $D ([0, T]; \mathbf{R}^d)$ denotes the Skorohod space:

$$\begin{aligned} D ([0, T]; \mathbf{R}^d)= & {} \Big \{ w: [0, T] \rightarrow \mathbf{R}^d; \quad w(t) = w(t+) := \lim _{s \downarrow t} w(s), t \in [0, T), \\&\qquad \qquad \hbox { and } w(t-) := \lim _{s \uparrow t} w(s) \hbox { exists}, t \in (0, T] \Big \}, \end{aligned}$$

with the metric $d^0 = d_T^0$ given by $d^0 (w, {\widetilde{w}}) = \inf _{\lambda \in \Lambda } \Big \{ \Vert \lambda \Vert ^0 \vee \Vert w - {\widetilde{w}} \circ \lambda \Vert _{\infty } \Big \}$ for any $w, {\widetilde{w}} \in D ([0, T]; \mathbf{R}^d)$, where $\Lambda = \Big \{ \lambda : [0, T] \rightarrow [0, T]; \hbox { continuous, non-decreasing}, \lambda (0) = 0, \lambda (T) = T \Big \},$ $\Vert w \Vert _{\infty } = \sup _{0 \le t \le T} |w(t)|$, and $\Vert \lambda \Vert ^0 = \sup _{0 \le s < t \le T} \Big | \log \frac{\lambda (t) - \lambda (s)}{t-s} \Big |$ for any $\lambda \in \Lambda$.

It is well-known that $(D ([0, T]; \mathbf{R}^d), d^0)$ is a complete metric space, $C ([0, T]; \mathbf{R}^d)$ is closed in $(D ([0, T]; \mathbf{R}^d), d^0)$, and the Skorohod topology relativized to $C ([0, T]; \mathbf{R}^d)$ coincides with the uniform topology there (see, e.g., [1]). Let $\wp (D ([0, T]; \mathbf{R}^d))$ denote the space of all probabilities on $D ([0, T]; \mathbf{R}^d)$.

Our base for the proof of tightness is Theorem 3.28 below, which is essentially a corollary of [1, Theorem 13.2]. See [12, Theorem3.4.1] for its proof.

Theorem 3.28

([12]) Let $(\Omega _n, {{\mathcal {F}}}_n, Q_n)$, $n \in \mathbf{N}$, be probability spaces, and let $X_n: \Omega _n \rightarrow D ([0, T]; \mathbf{R}^d)$, $n \in \mathbf{N}$, be measurable. Let $\mu _{X_n} = Q_n \circ X_n^{-1}$. Suppose that there exist constants $\varepsilon , \beta , \gamma , C > 0$ such that

• (1) $E^{Q_n} \big [ \Vert X_n ( \ \cdot \ ) \Vert _{\infty }^{\varepsilon } \big ] \le C$,

• (2) $E^{Q_n} \Big [ |X_n (r) - X_n (s)|^{\beta } |X_n (s) - X_n (t)|^{\beta } \Big ] \le C |t-r|^{1 + \varepsilon }$ for any $0 \le r \le s \le t \le 1$,

• (3) $E^{Q_n} \Big [ |X_n (s) - X_n (t)|^{\varepsilon } \Big ] \le C |t-s|^{\gamma }$ for any $0 \le s \le t \le 1$,

for any $n \in \mathbf{N}$. Then $\big \{ \mu _{X_n} \big \}_{n=1}^{\infty }$ is tight in $\wp (D ([0, T]; \mathbf{R}^d))$. $\square$

As easy consequences of Theorem 3.28, we have Lemmas 3.29 and 3.30 below, which were also used in [12, 13].

Lemma 3.29

([12]) Let $(\Omega _n, \{ {{\mathcal {F}}}_n (t) \}_{t \in [0, T]}, Q_n)$, $n \in \mathbf{N}$, be filtered probability spaces, and let $f_n: [0, T] \times \Omega _n \rightarrow \mathbf{R}$ be $\{ {{\mathcal {F}}}_n (t) \}_{t \in [0, T]}$-adapted, $n \in \mathbf{N}$. If

$$\begin{aligned} \sup _{n \in \mathbf{N}} \sup _{s \in [0, T]} E^{Q_n} \Big [ |f_n (s)|^2 \Big ] < \infty , \end{aligned}$$

then $\Big \{ \hbox {the distribution of } \{ \int _0^t f_n (s) ds \}_{t \in [0, T]} \hbox { under } Q_n; n \in \mathbf{N} \Big \}$ is tight in $\wp (C([0, T]; \mathbf{R}^d))$.

$\square$

Lemma 3.30

([12]) Let $(\Omega _n, \{ {{\mathcal {F}}}_n (t) \}_{t \in [0, T]}, Q_n)$, $n \in \mathbf{N}$, be filtered probability spaces, and let $\{ M_n (t) \}_t$ be $(\{ {{\mathcal {F}}}_n (t) \}_{t \in [0, T]}, Q_n)$-martingales. If there exists a constant $C>0$ such that

$$\begin{aligned} E^{Q_n} \Big [ |M_n(t)-M_n(s)|^2 \Big | {{\mathcal {F}}}_s \Big ] \le C (t-s), \qquad 0 \le s < t \le T, \end{aligned}$$

then $\Big \{ \hbox {the distribution of } \{ M_n (t) \}_{t \in [0, T]} \hbox { under } Q_n; n \in \mathbf{N} \Big \}$ is tight in $\wp (D([0, T]; \mathbf{R}^d))$. $\square$

3.4.2 First Decomposition—The Result

We present our key decomposition of V(t) in this section (see Proposition 3.31). Rewrite $V(t \wedge \sigma _n) = - \int _0^{t \wedge \sigma _n} ds \int _{\mathbf{R} \times E} \nabla U (X(s) - x (s, \Psi (r, x, m^{-1/2} v)) \mu _{\omega } (dr, dx, dv)$ as

$$\begin{aligned} V(t \wedge \sigma _n) = I^1 (t) + \cdots I^{11} (t), \end{aligned}$$

(3.18)

with

$$\begin{aligned} I^1 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{[0, m^{1/2} a_m]} (s) ds \int _{\mathbf{R} \times E} \Big \{ \nabla U \big (X(s) - x (s, \Psi (r, x, m^{-1/2} v))\big ) \\&\qquad - \nabla U \big (X_0 - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big )\big ) \Big \} \mu _{\omega } (dr, dx, dv), \\ I^2 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{[0, m^{1/2} a_m]} (s) ds \int _{\mathbf{R} \times E} \nabla U \big (X_0 - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big )\big ) \mu _{\omega } (dr, dx, dv), \\ I^3 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} \nabla U \big (X({\tilde{r}}) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X({\tilde{r}})\big )\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\ I^4 (t)= & {} - \int _0^{t \wedge \sigma _n} ds \int _{\mathbf{R} \times E} f_1 (s, r, x, v) {\overline{\lambda }} (dr, dx, dv), \\ I^5 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} \nabla U (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v; X(s)) \lambda (dr, dx, dv), \\ I^6 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }} \Big ( f_3 (s, r, x, v) - f_2 (s, r, x, v)\Big ) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\ I^7 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }} f_2 (s, r, x, v) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\ I^8 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha }} f_3 (s, r, x, v) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\ I^9 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} \Big ( f_4 (s, r, x, v) - f_1 (s, r, x, v) \Big ) {\overline{\lambda }} (dr, dx, dv), \\ I^{10} (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} f_4 (s, r, x, v) \Big ( \lambda (dr, dx, dv) - {\overline{\lambda }} (dr, dx, dv) \Big ), \\ I^{11} (t)= & {} \int _0^{t \wedge \sigma _n} 1_{[0, m^{1/2} a_m]} (s) ds \int _{\mathbf{R} \times E} f_1 (s, r, x, v) {\overline{\lambda }} (dr, dx, dv). \\ \end{aligned}$$

Here

$$\begin{aligned} {\overline{\lambda }} (dr, dx, dv) := m^{-1} \rho _0 \left( \frac{1}{2} |v|^2\right) dr \nu (dx, dv), \end{aligned}$$

$a_m$ is as defined in (2.13), and

$$\begin{aligned}&f_1 (s, r, x, v) \\&:= - \nabla ^2 U \big (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v; X(s)) \\&\qquad \cdot m^{1/2} z (m^{-1/2}(s-r), x, v, X(s), V(s), -m^{-1/2} (s-r)), \\&f_2 (s, r, x, v) \\&:= \nabla ^2 U (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))) \\&\qquad \Big ( - m^{1/2} z(m^{-1/2}(s-r), x, v; X({\tilde{r}}), V({\tilde{r}}), - m^{-1/2}({\tilde{r}}-r)) + (s-{\tilde{r}}) V({\tilde{r}}) \Big ), \\&f_3 (s, r, x, v) \\&:= \nabla U (X(s) - x (s, \Psi (r, x, m^{-1/2} v))) - \nabla U (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))), \\&f_4 (s, r, x, v) \\&:= \nabla U (X(s) - x (s, \Psi (r, x, m^{-1/2} v))) - \nabla U (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v; X(s)). \end{aligned}$$

$I^1 (t)$ and $I^2 (t)$ in the decomposition above correspond to the force during the time interval right after starting (i.e., $s \le m^{1/2} a_m$). They are negligible by virtue of the shortness of the time (see Lemmas 3.35 and 3.36). This blank of time, after necessary treatment, is refilled at the last step (see the definition of $I^4(t)$ and $I^{11} (t)$). As explained, the term $I^3 (t)$ gives us approximately the martingale term. The discussion with respect to $I^3 (t)$ is a little bit complicated, and will be given in Sect. 3.4.5. The term $I^4 (t)$ gives us our drift term. Indeed, it is trivial that

$$\begin{aligned}&I^4 (t) = - \int _0^{t \wedge \sigma _n} ds \int _{\mathbf{R} \times E} \nabla ^2 U (X(s) - \psi ^0 (u, x, v; X(s)) z (u, x, v, X(s), V(s), -u) \nonumber \\&\qquad \qquad \times \rho _0 \Big ( \frac{1}{2}|v|^2 \Big ) du \nu (dx, dv), \end{aligned}$$

(3.19)

which depends on m only via $X(\cdot )$. The necessary estimate for the tightness of $I^4 (t)$ is essentially the same as that for the fact that $I^{11}$ is negligible (see Lemma 3.43 (1) and (2)). The others, as will be proven in Sects. 3.4.3, 3.4.4 and 3.5, are approximately 0. To be precise, we prove in Sect. 3.4.3 that $I^1 (t)$, $I^2 (t)$, $I^5 (t)$, $I^7 (t)$, $I^8 (t)$, $I^{10} (t)$ and $I^{11} (t)$ converge to 0 fast enough. Also, as announced, the terms $I^6 (t)$ and $I^9 (t)$ are discussed in two steps: we prove in Sect. 3.4.4 that they can also be considered as a part of our smooth part, which is enough to prove our tightness of the distribution of $\{ (X (t \wedge \sigma _n), V (t \wedge \sigma _n)); t \in [0, T] \}$ under $P_m$ for $m \in (0, 1]$; then, in Sect. 3.5, with the help of these results, we prove that they are also negligible.

In order to formulate our main result of Sect. 3.4, we prepare several notations. Let

$$\begin{aligned} N ( (0, t] \times A ) := \mu _{\omega } ( (m^{1/2} \tau , m^{1/2} \tau + t] \times A), \qquad A \subset E. \end{aligned}$$

Since $U (x-rv-X_0) = 0$ for all $r \ge m^{1/2} \tau$, we get that N is a Poisson point process with intensity ${\overline{\lambda }} (dt, dx, dv)$. Let

$$\begin{aligned} {\overline{N}} (dt, dx, dv) := N (dt, dx, dv) - {\overline{\lambda }} (dt, dx, dv). \end{aligned}$$

Finally, let

$$\begin{aligned}&M (t) := - \int _{(0, t] \times E} {\overline{N}} (dr, dx, dv) \int _{\left[ 0, m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| ) + \tau \big )\right] } \nonumber \\&\qquad \times \nabla U \left( X(r \wedge \sigma _n) - \psi ^0 \big (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n)\big )\right) ds. \end{aligned}$$

(3.20)

M(t) is the martingale part of our decomposition, which is, as explained before, approximately equal to $I^3 (t)$.

Now we are ready to formulate our main result of Sect. 3.4.

Proposition 3.31

There exists an $\{ \mathcal {F}_t \}_t$-adapted process $\eta (t)$ and a constant $C_6$ such that

• 1.
  $$\begin{aligned} V ( t \wedge \sigma _n ) - V(0) = M ( t \wedge \sigma _n ) + I^4 (t) + I^6 (t) + I^9 (t) + \eta (t), \end{aligned}$$

  (3.21)

• 2. $I^4 (t)$, $I^6 (t)$ and $I^9 (t)$ are $\{ \mathcal {F}_t \}_t$-adapted, $C^1$-class with respect to t, and

  $$\begin{aligned} \sup _{m \in (0, 1] } \sup _{t \in [0, T]} E^{P_m} \Big [ \Big | \frac{d}{dt} I^k (t) \Big |^2 \Big ] \le C_6, \qquad k \in \{ 4, 6, 9 \}, \end{aligned}$$

  (3.22)

• 3. $M(\cdot )$ is a càdlàg $\{ \mathcal {F}_t \}_t$-martingale with its jumps satisfying $| \Delta M(t) | \le C_6 m^{1/2}$, and

  $$\begin{aligned} E^{P_m} \Big [ |M(t) - M(s)|^2 \Big | \mathcal {F}_s \Big ] \le C_6 |t-s| \end{aligned}$$

  (3.23)

  for any $0 \le s < t \le T$ and $m \in (0, 1]$, also, for any $k, l \in \{ 1, \cdots , d \}$ and any bounded $g: [0, \infty ) \times \Omega \rightarrow \mathbf{R}$ such that $g (s, \cdot )$ is $\mathcal {F}_s$-measurable, we have that

  $$\begin{aligned} E^{P_m} \left[ \sup _{t \in [0, T] } \Big | \int _{(0, t \wedge \sigma _n]} g (s) \Big ( d [ M^k, M^l ]_s - a_{kl} ds \Big ) \Big |^2 \right] \le C_6 \Vert g\Vert _{\infty }^2 m. \end{aligned}$$

  (3.24)
• 4.
  $$\begin{aligned} E \Big [ \sup _{u \in [0, T] } |\eta (u)|^2 \Big ] \rightarrow 0, \end{aligned}$$

  and there exists a constant $\varepsilon > 0$ such that

  $$\begin{aligned} E \Big [ \sup _{u \in [s - m^{1/2} a_m, s] } | \eta (u) - \eta (s - m^{1/2} a_m) |^2 \Big ] \le C_6 m^{\varepsilon } \end{aligned}$$

  (3.25)

  for any $s \in [m^{1/2} a_m, T]$.

In particular, the distributions of $\{ M (t) + \eta (t); t \in [0, t] \}$ and $\{ I^k (t); t \in [0, T] \}$ with $k \in \{ 4, 6, 9 \}$ under $P_m$ are tight in $\wp (C([0, T]; \mathbf{R}^d))$.

The tightness in Proposition 3.31 is a direct consequence of the other assertions of the same Proposition and Lemmas 3.29, 3.30. So it suffices to prove the first half of Proposition 3.31.

The quantity $\eta (t)$ of Proposition 3.31 is given by the sum of $I^1 (t)$, $I^2 (t)$, $I^5 (t)$, $I^7 (t)$, $I^8 (t)$, $I^{10} (t)$, $I^{11} (t)$ and a part of $I^3 (t)$. Lemma 3.32 below proves that there exists a constant $\varepsilon > 0$ such that when $m \rightarrow 0$, all these terms except the last one are of order $o (m^{\varepsilon })$. Precisely, we prove the following:

Lemma 3.32

There exists a constant $\varepsilon > 0$ such that when $m \rightarrow 0$,

$$\begin{aligned} E^{P_m} \Big [ \sup _{0 \le t \le T} |I (t)|^2 \Big ]^{1/2} = o(m^{\varepsilon }), \end{aligned}$$

(3.26)

with I given by $I^1 (t)$, $I^2 (t)$, $I^5 (t)$, $I^7 (t)$, $I^8 (t)$, $I^{10} (t)$, $I^{11} (t)$.

Lemma 3.32 certainly implies that the contribution from these terms in $\eta (t)$ satisfies the estimates in Proposition 3.31 (4).

We prove Lemma 3.32 in Sect. 3.4.3. (3.22) for $k = 4$ is formulated as a part of Lemma 3.43. The proof of (3.22) for $k \in \{ 6, 9 \}$ is given in Sect. 3.4.4. The discussion with respect to $I^3 (t)$ is presented in Sect. 3.4.5.

3.4.3 Proof of Lemma 3.32

We start our proof by noticing the following two facts, which hold by virtue of the translation property and the symmetry of $\varphi (t, \cdot , \cdot ; X)$ and $\psi (t, \cdot , \cdot ; X)$ with respect to X. See Appendix (Sect. A.2) for their proofs.

Lemma 3.33

• 1. The following holds:

  $$\begin{aligned} \int _{\mathbf{R} \times E} \nabla U \left( X_0 - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big )\right) \lambda (dr, dx, dv) = 0. \end{aligned}$$

  (3.27)

• 2. For any $X \in \mathbf{R}^d$, the following holds:

  $$\begin{aligned} \int _{\mathbf{R} \times E} \nabla U \big (X - \psi ^0 (u, x, v; X)\big ) {\overline{\lambda }} (du, dx, dv) = 0. \end{aligned}$$

  (3.28)

The proof of the following general result with respect to the Poisson point process follows easily from the definitions. We formulate it here without proof.

Lemma 3.34

Let $f: \mathbf{R} \times E \times \Omega \rightarrow \mathbf{R}$ be a function such that $f (r, x, v, \cdot )$ is $\mathcal {F}_{(-\infty , r] \times E}$-measurable for any $(r, x, v) \in \mathbf{R} \times E$. Then we have that

$$\begin{aligned} E \Big [ \Big ( \int f d (\mu _{\omega } - \lambda ) \Big )^2 \Big ] = E \Big [ \int f^2 d \lambda \Big ]. \end{aligned}$$

$\square$

As a consequence of Lemma 3.34, we get that

$$\begin{aligned}&E \Big [ \Big ( \int f d \mu _{\omega } \Big )^2 \Big ] \le 2 E \Big [ \int f^2 d \lambda \Big ] + 2 E \Big [ \Big (\int f d \lambda \Big )^2 \Big ], \end{aligned}$$

(3.29)

$$\begin{aligned}&E \Big [ \Big ( \int f d (\mu _{\omega } + \lambda ) \Big )^2 \Big ] \le 2 E \Big [\int f^2 d \lambda \Big ] + 8 E \Big [ \Big ( \int f d \lambda \Big )^2 \Big ]. \end{aligned}$$

(3.30)

Lemmas 3.35 and 3.36 below imply that $I^1 (t)$ and $I^2 (t)$ satisfy (3.26), by the virtue of (2.10).

Lemma 3.35

There exists a constant $C_7 > 0$ such that

$$\begin{aligned} E^{P_m} \Big [ \sup _{0 \le t \le T} |I^1 (t)|^2 \Big ] \le C_7 m a_m^6 e^{(1+C_1) a_m}, \qquad m \in (0, 1]. \end{aligned}$$

Proof

Since $|X({\tilde{r}})| \le |X_0| + nT$, we have that $1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| \le R_U + 1 \} } \le 1_{ \{ |x| \le R_0 \} }$ and $1_{ \{ | x - \pi ^{\perp }_v X_0| \le R_U + 1 \} } \le 1_{ \{ |x| \le R_0 \} }$. So by (2.7), Proposition 2.1 (1) (3), Lemma 3.17 and (3.12), we get for any $s \in [0, T \wedge \sigma _n \wedge m^{1/2} a_m]$ that

$$\begin{aligned}&\Big | \nabla U \big (X(s) - x (s, \Psi (r, x, m^{-1/2} v))\big ) - \nabla U \big (X_0 - \varphi ^0 (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0))\big ) \Big | \\&\le 1_{[0, R_0]} (|x|) 1_{[-m^{1/2} \tau , s + m^{1/2} \tau ]} (r) C_1 \\&\qquad \cdot \left( n s + \Big | x (s, \Psi \big (r, x, m^{-1/2} v)\big ) - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big ) \Big | \right) \\&\le 1_{[0, R_0]} (|x|) 1_{\big [-m^{1/2} \tau , m^{1/2} a_m + m^{1/2} \tau \big ]} (r) C_1 \cdot \frac{2 C_1 n}{1 + C_1} m^{1/2} a_m e^{\frac{1}{2} (1+C_1) a_m}. \end{aligned}$$

Therefore,

$$\begin{aligned} |I^1 (t)|\le & {} m^{1/2} a_m 2 C_1 n m^{1/2} a_m e^{\frac{1}{2} (1+C_1) a_m} \\&\qquad \cdot \int _{\mathbf{R} \times E} 1_{[0, R_0]} (|x|) 1_{\big [-m^{1/2} \tau , m^{1/2} a_m + m^{1/2} \tau \big ]} (r) \mu _{\omega } (dr, dx, dv). \end{aligned}$$

Since $a_m \ge \tau \vee 1$ by definition, this combined with Lemma 4.4 (2) implies that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{0 \le t \le T} |I^1 (t)|^2 \Big ] \\&\le \Big ( 2 C_1 n m a_m^2 e^{\frac{1}{2} (1+C_1) a_m} \Big )^2 C \Big ( m^{-2} \Big ( m^{1/2} a_m + m^{1/2} \tau \Big )^2 + 1 \Big ) \\&\le 20 C_1^2 C n^2 m a_m^6 e^{(1+C_1) a_m}. \end{aligned}$$

$\square$

Lemma 3.36

There exists a constant $C_8 > 0$ such that

$$\begin{aligned} E^{P_m} \Big [ \sup _{0 \le t \le T} |I^2 (t)|^2 \Big ] \le C_8 m^{1/2} a_m^3, \qquad m \in (0, 1]. \end{aligned}$$

Proof

By Lemma 3.33 (1), we get that

$$\begin{aligned} I^2 (t)= & {} - \int _0^{t \wedge \sigma _n} 1_{[0, m^{1/2} a_m]} (s) ds \int _{\mathbf{R} \times E} \nabla U \big (X_0 - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big )\big )\\&(\mu _{\omega } - \lambda ) (dr, dx, dv). \end{aligned}$$

For any $s \in [0, m^{1/2} a_m]$, by Lemma 3.34, Proposition 2.1 (1) and Lemma 4.4 (1), we get that

$$\begin{aligned}&E \left[ \Big | \int _{\mathbf{R} \times E} \nabla U \big (X_0 - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big )\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \right] \\&\quad = \int _{\mathbf{R} \times E} |\nabla U \big (X_0 - \varphi ^0 \big (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0)\big )\big ) |^2 \lambda (dr, dx, dv) \\&\quad \le \Vert \nabla U \Vert _{\infty }^2 \int _{\mathbf{R} \times E} 1_{[0, R_0]} (|x|) 1_{[-m^{1/2} \tau , m^{1/2} a_m + m^{1/2} \tau ]} (r) \lambda (dr, dx, dv) \\&\quad \le \Vert \nabla U \Vert _{\infty }^2 C m^{-1} \big (m^{1/2} a_m + m^{1/2} \tau \big ) \le 2 \Vert \nabla U \Vert _{\infty }^2 C m^{-1/2} a_m. \end{aligned}$$

Therefore,

$$\begin{aligned}&E^{P_m} \Big [ \sup _{0 \le t \le T} |I^2 (t)|^2 \Big ] \\&\quad \le E^{P_m} \Big [ m^{1/2} a_m \int _0^T 1_{[0, m^{1/2} a_m]} (s) ds \\&\qquad \Big | \int _{\mathbf{R} \times E} \nabla U (X_0 - \varphi ^0 (m^{-1/2} s, \Psi (m^{-1/2} r, x, v; X_0))) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \Big ] \\&\quad \le m a_m^2 2 \Vert \nabla U \Vert _{\infty }^2 C m^{-1/2} a_m. \end{aligned}$$

So we get our assertion with $C_8 := 2 \Vert \nabla U \Vert _{\infty }^2 C$. $\square$

Since $\alpha > \frac{1}{2(d-1)}$ by (2.9), Lemma 3.37 below implies that $I^5 (t)$ also satisfies (3.26).

Lemma 3.37

There exists a constant $C_9 > 0$ such that

$$\begin{aligned} |I^5 (t)| \le C_9 m^{\alpha (d-1) - \frac{1}{2}} \end{aligned}$$

for any $t \in [0, T]$ and $m \in (0, 1]$.

Proof

First, for any $s \in [0, T \wedge \sigma _n]$, we have by Lemma 3.33 (2) that

$$\begin{aligned} \int _{\mathbf{R} \times E} \nabla U \big (X(s) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X(s)\big )\big ) {\overline{\lambda }} (dr, dx, dv) = 0. \end{aligned}$$

Assume $|x-\pi ^{\perp }_v X(s)| \ge m^{\alpha }$ for a while. Then by Proposition 2.1 (2) and (2.14), for any $m \le (2 \varepsilon _3)^{1/\alpha }$, we have that $\nabla U (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v; X(s))) \ne 0$ implies $r \ge s - m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X(s)|) \ge s - m^{1/2} (a_m - \tau )$. This combined with $s \ge m^{1/2} a_m$ implies that $r \ge m^{1/2} \tau$, so $| x - m^{-1/2} r v - X_0 | \ge m^{-1/2} r |v| - |X_0| \ge R_1$, hence $\lambda (dr, dx, dv) = {\overline{\lambda }} (dr, dx, dv)$ by (A2).

Therefore,

$$\begin{aligned} I^5 (t)= & {} I^5 (t) + \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \nonumber \\&\int _{\mathbf{R} \times E} \nabla U \big (X(s) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X(s)\big )\big ) {\overline{\lambda }} (dr, dx, dv) \nonumber \\= & {} \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \nonumber \\&\int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X(s)|< m^{\alpha }} \nabla U \big (X(s) - \psi ^0 \big (m^{-1/2} (s-r),x, v; X(s)\big )\big ) {\overline{\lambda }} (dr, dx, dv) \nonumber \\&- \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds\nonumber \\&\int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X(s)| < m^{\alpha }} \nabla U \big (X(s) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X(s)\big )\big ) \lambda (dr, dx, dv). \nonumber \\ \end{aligned}$$

(3.31)

We notice that for any $(x, v) \in E$ satisfying $|x-\pi ^{\perp }_v X(s)| \ne 0$, we have $|\psi ^1 (-\infty , x, v; X(s))| = |\psi ^1 (+\infty , x, v; X(s))| = |v|$, so

$$\begin{aligned}&\Big | \int _{\mathbf{R}} \nabla U \big (X(s) - \psi ^0 (u, x, v; X(s))\big ) du \Big |\\&\quad = |\psi ^1 (\infty , x, v; X(s)) - \psi ^1 (-\infty , x, v; X(s))| \le 2 |v|. \end{aligned}$$

Therefore, we have for any $s \in [0, T \wedge \sigma _n]$ that

$$\begin{aligned}&\Big | \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X(s)|< m^{\alpha }} \nabla U \big (X(s) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X(s)\big )\big ) \lambda (dr, dx, dv) \Big | \\&\quad \le \int _{E, |x-\pi ^{\perp }_v X(s)| < m^{\alpha }} 2 |v| m^{-1/2} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) \nu (dx dv) \\&\quad \le 2 m^{-1/2} (2m^{\alpha })^{d-1} \int _{\mathbf{R}^d} |v|^2 \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) dv. \end{aligned}$$

The same estimate holds for the assertion with $\lambda (dr, dx, dv)$ substituted by ${\overline{\lambda }} (dr, dx, dv)$, too. Therefore, we get our assertion with $C_9 := 2T \cdot 2^d \int _{\mathbf{R}^d} |v|^2 \rho _{max} (\frac{1}{2} |v|^2) dv$. $\square$

We next deal with the term $I^7(t)$. Lemma 3.39 below is more than enough to prove that $I^7(t)$ satisfies (3.26) under (2.8). We prove it in the present form since it is also used in the discussion with respect to $I^6(t)$ (see the proof of (3.22) for $k = 6$ in Sect. 3.4.4).

We first prepare the following estimate with respect to $z(\cdot ; x, v, X, V, a)$.

Lemma 3.38

For any ${\overline{t}} \ge \tau$ and $c_m \ge \tau$ ($m \in (0, 1]$), we have for any $|a| \le c_m$, $|V| \le n$, $X \in \mathbf{R}^d$ and $(x, v) \in E$ that

$$\begin{aligned}&|z(t; x, v, X, V, a)| \vee \Big | \frac{d}{dt} z(t; x, v, X, V, a) \Big | \\&\quad \le 2 ({\overline{t}} + c_m) n \Big ( e^{\frac{1}{2} (1+C_1)(t+\tau )} - 1 \Big ), \quad t \in [-\tau , {\overline{t}}]. \end{aligned}$$

Proof

We have by the definition of z, (2.6) and our assumption that

$$\begin{aligned} \left| \frac{d^2}{dt^2} z(t)\right| \le C_1 |z(t)| + C_1 ({\overline{t}} + c_m) n, \quad t \in [-\tau , {\overline{t}}]. \end{aligned}$$

Also, $z(-\tau ) = \frac{d}{dt} z (-\tau ) = 0$. Therefore, we get our assertion by Lemma 3.16. $\square$

Lemma 3.39

Under (2.8), there exists a constant $C_{10} > 0$ (which may depend on $\alpha$) such that

• 1. for any $t \in [0, T]$, we have that

  $$\begin{aligned}&E^{P_m} \Big [ 1_{ [m^{1/2} a_m, T \wedge \sigma _n ] } (t) \Big | \int _{\mathbf{R} \times E, | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha } } f_2 (t, r, x, v) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \Big ] \\&\quad < C_{10} m^{\frac{1}{2}}. \end{aligned}$$

• 2. $E^{P_m} \Big [ \sup _{t \in [0, T]} |I^7 (t)|^2 \Big ] \le C_{10} m^{1/2}.$

Proof

By re-choosing $C_{10}$ if necessary, the second assertion is a direct consequence of the first assertion. We prove the first assertion in the following.

For any $t \in \mathbf{R}$, if $f_2 (t, r, x, v) \ne 0$, then by Proposition 2.1 (2), we get that $m^{-1/2} (t-r) \in [- \tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)]$. So if $t \in [m^{1/2} a_m, T \wedge \sigma _n]$ and $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }$ in addition, then by (2.14), for any $m \le (2 \varepsilon _3)^{1/\alpha }$, we have $r - m^{1/2} \tau \in [0, T \wedge \sigma _n]$. Also, we notice that $1_{ \{ r - m^{1/2} \tau \in [0, T \wedge \sigma _n] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha } \} } f_2 (t, r, x, v)$ is $\mathcal {F}_{(-\infty , r] \times E}$-measurable for any r by Lemma 3.19. Therefore, with the help of Lemma 3.34, we get that

$$\begin{aligned}&E^{P_m} \Big [ 1_{ [m^{1/2} a_m, T \wedge \sigma _n ] } (t) \Big | \int _{\mathbf{R} \times E, | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha } } f_2 (t, r, x, v) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \Big ] \\&\quad \le E^{P_m} \Big [ \Big | \int _{\mathbf{R} \times E} 1_{ \{ r - m^{1/2} \tau \in [0, T \wedge \sigma _n] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha } \} } f_2 (t, r, x, v) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \Big ] \\&\quad = \int _{\mathbf{R} \times E} E^{P_m} \Big [ 1_{ \{ r - m^{1/2} \tau \in [0, T \wedge \sigma _n] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha } \} } f_2 (t, r, x, v)^2 \Big ] \lambda (dr, dx, dv). \end{aligned}$$

So it suffices to prove that the right hand side above is dominated by $C_{10} m^{1/2}$ with some proper constant $C_{10}$.

On the other hand, if $f_2 (t, r, x, v) \ne 0$, then by Proposition 2.1 (2), we get that $t - r \in [-m^{1/2} \tau , m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)]$. This combined with $r - m^{1/2} \tau \in [0, T \wedge \sigma _n]$ implies that $t - {\tilde{r}} = t - r + m^{1/2} \tau \in [0, m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + m^{1/2} \tau ]$, hence $| (t-{\tilde{r}}) V({\tilde{r}}) | \le n m^{1/2} (t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + \tau )$. So by Lemma 3.38 (with $c_m$ and ${\overline{t}}$ given by $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)$), we get that

$$\begin{aligned}&\Big | z\big (m^{-1/2}(t-r), x, v; X({\tilde{r}}), V({\tilde{r}}), - m^{-1/2}({\tilde{r}}-r)\big ) - m^{-1/2} (t-{\tilde{r}}) V({\tilde{r}}) \Big | \\&\quad \le 4 n t_1 \big (v, |x-\pi ^{\perp }_v X({\tilde{r}})|\big ) \exp \left( \frac{1}{2} (1+C_1) \big (t_1\big (v, |x-\pi ^{\perp }_v X({\tilde{r}})|\big ) + \tau \big ) \right) . \end{aligned}$$

Therefore, by Proposition 2.1 (2),

$$\begin{aligned}&\int _{\mathbf{R} \times E} E^{P_m} \Big [ 1_{ \{ r - m^{1/2} \tau \in [0, T \wedge \sigma _n] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha } \} } f_2 (t, r, x, v)^2 \Big ] \lambda (dr, dx, dv) \\&\quad \le \int _{\mathbf{R} \times E} \lambda (dr, dx, dv) 1_{ \{ |x| \le R_0 \} } 1_{ \{ t - r \in [-m^{1/2} \tau , m^{1/2} t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \} } \Vert \nabla ^2 U \Vert _{\infty }^2 \\&\qquad \times m \left( 4 n t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) \exp \Big ( \frac{1}{2} (1+C_1) (t_1(v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + \tau ) \Big ) \right) ^2. \end{aligned}$$

Since $(1+C_1) 2 \varepsilon _1^{-1/2} < d - 1$ by (2.8), by Lemma 4.5 (1), we get that the right hand side above is dominated by $\Vert \nabla ^2 U \Vert _{\infty }^2 16 n^2 C m^{1/2}$. This completes our proof. $\square$

In order to prove that $I^8 (t)$ also satisfies (3.26), we first prepare the following.

Lemma 3.40

There exists a constant $C_{11} > 0$ such that for any $(x, v) \in \mathbf{R}^{2d}$ and $t \in [0, T \wedge \sigma _n]$, we have

$$\begin{aligned} |v(t, x, v)| \le |v| + C_{11} m^{-1/2}. \end{aligned}$$

Proof

For any $t \in [0, \sigma _n]$, we have that

$$\begin{aligned}&\Big | \frac{d}{dt} \Big ( \frac{m}{2} |v(t, x, v)|^2 + U (x(t, x, v) - X(t)) \Big ) \Big | \\&\quad = \Big | - \nabla U (x(t, x, v) - X(t)) \cdot V(t) \Big | \le n \Vert \nabla U \Vert _{\infty }, \end{aligned}$$

so

$$\begin{aligned} \Big | \Big ( \frac{m}{2} |v(t, x, v)|^2 + U (x(t, x, v) - X(t)) \Big ) - \Big ( \frac{m}{2} |v|^2 + U(x-X_0) \Big ) \Big | \le n \Vert \nabla U \Vert _{\infty } t. \end{aligned}$$

Therefore, for any $t \in [0, T \wedge \sigma _n]$, we have

$$\begin{aligned} |v(t, x, v)|^2\le & {} \sqrt{|v|^2 + 2 m^{-1} (n T \Vert \nabla U \Vert _{\infty } + \Vert U \Vert _{\infty } )} \\\le & {} |v| + m^{-1/2} \sqrt{2 (n T \Vert \nabla U \Vert _{\infty } + \Vert U \Vert _{\infty })}. \end{aligned}$$

$\square$

$I^8 (t)$ satisfies (3.26) by Lemma 3.41 below, since $\alpha > \frac{1}{2(d-1)}$ by (2.9).

Lemma 3.41

There exists a constant $C_{12} > 0$ such that

$$\begin{aligned} E^{P_m} \Big [ \sup _{t \in [0, T]} |I^8 (t)|^2 \Big ] \le C_{12} (m^{\alpha (d-1)} + m^{2 \alpha (d-1)-1}), \qquad m \in (0, 1]. \end{aligned}$$

Proof

For any $(x, v) \in E$, $t \in [0, T]$ and $r \in \mathbf{R}$, by Lemmas 3.40 and 3.18, we have

$$\begin{aligned}&\Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \nabla U (X(s) - x(s, \Psi (r, x, m^{-1/2}v))) \Big | \nonumber \\&\quad = \Big | m \Big ( v (t \wedge \sigma _n, \Psi (r, x, m^{-1/2}v)) - v (t \wedge \sigma _n \wedge m^{1/2} a_m, \Psi (r, x, m^{-1/2}v)) \Big ) \Big | \nonumber \\&\quad \le 2 m^{1/2} (|v| + C_{11}) 1_{[-m^{1/2} \tau , T + m^{1/2} \tau ]} (r). \end{aligned}$$

(3.32)

This estimate is also used in the proof of Lemma 3.60. Similarly,

$$\begin{aligned}&\Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \nabla U (X({\widetilde{r}}) - \psi ^0(m^{-1/2} (s-r)x, v; X({\widetilde{r}})) \Big | \\&\quad \le 2 m^{1/2} |v| 1_{[-m^{1/2} \tau , T + m^{1/2} \tau ]} (r). \end{aligned}$$

Therefore,

$$\begin{aligned} |I^8 (t)| \le \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha }} 4 m^{1/2} (|v| + C_{11}) 1_{[-m^{1/2} \tau , T + m^{1/2} \tau ]} (r) (\mu _{\omega } + \lambda ) (dr, dx, dv). \end{aligned}$$

Combining this with (3.30) and Lemma 4.5 (2), we get that

$$\begin{aligned} E^{P_m} \Big [ \sup _{t \in [0, T]} |I^8 (t)|^2 \Big ] \le 32 m \cdot C m^{\alpha (d-1) -1} (T+2\tau ) + 8 \cdot 16 m \cdot \Big ( C m^{\alpha (d-1) -1} (T+2\tau ) \Big )^2, \end{aligned}$$

which implies our assertion. $\square$

Lemma 3.42 below implies that $I^{10} (t)$ also satisfies (3.26), since $\alpha > \frac{1}{2(d-1)}$ by (2.9).

Lemma 3.42

There exists a constant $C_{13}>0$ such that

$$\begin{aligned} \sup _{t \in [0, T], m \in (0, 1], \omega \in \Omega } |I^{10} (t)| \le C_{13} m^{\alpha (d-1) - \frac{1}{2}}. \end{aligned}$$

Proof

First we notice that if $|r| \ge m^{1/2} \tau$, then $|x - m^{-1/2} r v - X_0| \ge \tau |v| - |X_0| \ge R_1$, hence $\lambda (dr, dx, dv) = {\overline{\lambda }} (dr, dx, dv)$ by (A2).

Also, if $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }$, $s \ge m^{1/2} a_m$, $s \in [0, T \wedge \sigma _n]$ and $\nabla U (X(s) - x (s, \Psi (r, x, m^{-1/2} v))) \ne 0$, then by Lemma 2.2, we get that $r \ge m^{1/2} \tau$. As discussed above, this implies in turn that $\lambda (dr, dx, dv) = {\overline{\lambda }} (dr, dx, dv)$. Therefore,

$$\begin{aligned}&\Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} \nabla U (X(s) - x (s, \Psi (r, x, m^{-1/2} v))) ( \lambda - {\overline{\lambda }} ) (dr, dx, dv) \Big | \nonumber \\&\quad \le \int _0^T ds \int _{\mathbf{R} \times E, | x - \pi ^{\perp }_v X({\tilde{r}}) | < m^{\alpha }} 1_{ \{ |r| \le m^{1/2} \tau \} } \Vert \nabla U \Vert _{\infty } ( \lambda + {\overline{\lambda }} ) (dr, dx, dv) \nonumber \\&\quad \le \Vert \nabla U \Vert _{\infty } T m^{-1} (2 m^{\alpha })^{d-1} 2 m^{1/2} \tau \int _{\mathbf{R}^d} 2 \rho _{max} \Big ( \frac{1}{2}|v|^2 \Big ) |v| dv. \end{aligned}$$

(3.33)

Similarly, if $| x - \pi ^{\perp }_v X(s) | \ge m^{1/2} \tau$, $s \ge m^{1/2} a_m$, $s \in [0, T \wedge \sigma _n]$ and $\nabla U (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v, X(s)) \ne 0$, then by Proposition 2.1 (2) and (2.14), $m^{-1/2} (s-r) \le t_1 (v, |x - \pi ^{\perp }_v X(s)|) \le a_m - \tau$, so for any $m \le (2 \varepsilon _3)^{1/\alpha }$, we have $r \ge s - m^{1/2} a_m + m^{1/2} \tau \ge m^{1/2} \tau$, hence $\lambda (dr, dx, dv) = {\overline{\lambda }} (dr, dx, dv)$. Therefore,

$$\begin{aligned}&\Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} \nabla U (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v, X(s)) ( \lambda - {\overline{\lambda }} ) (dr, dx, dv) \Big | \nonumber \\&\quad \le \int _0^T ds \int _{\mathbf{R} \times E, | x - \pi ^{\perp }_v X(s) | < m^{\alpha }} 1_{ \{ |r| \le m^{1/2} \tau \} } \Vert \nabla U \Vert _{\infty } ( \lambda + {\overline{\lambda }} ) (dr, dx, dv) \nonumber \\&\quad \le \Vert \nabla U \Vert _{\infty } T m^{-1} (2 m^{\alpha })^{d-1} 2 m^{1/2} \tau \int _{\mathbf{R}^d} 2 \rho _{max} \Big ( \frac{1}{2}|v|^2 \Big ) |v| dv. \end{aligned}$$

(3.34)

Combining (3.33) and (3.34), we get our assertion. $\square$

We prove in Lemma 3.43 below that $I^{11}(t)$ satisfies (3.26) and that $I^4 (t)$ satisfies the estimate in (3.22). Lemma 3.43 is also used in the proof of (3.22) for $k = 9$ (see Sect. 3.4.4).

Lemma 3.43

There exists a constant $C_{14} > 0$ such that

• 1. $\sup _{m \in (0, 1] } \sup _{t \in [0, T \wedge \sigma _n]} \Big | \int _{\mathbf{R} \times E} f_1 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \le C_{14}.$

• 2. $\sup _{m \in (0, 1] } \sup _{t \in [0, T]} E^{P_m} \Big [ \Big | \frac{d}{dt} I^4 (t) \Big |^2 \Big ] \le C_{14}.$

• 3. $|I^{11} (t)| \le C_{14} m^{1/2} a_m.$

Proof

By the definitions of $I^4 (t)$ and $I^{11} (t)$, the second and the third assertions are trivial by the first assertion. We prove the first assertion in the following. By a simple change of variable, $\Big | \int _{\mathbf{R} \times E} f_1 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big |$ is equal to

$$\begin{aligned} \Big | \int _{\mathbf{R} \times E} \nabla ^2 U \big (X(t) - \psi ^0 (u, x, v; X(t)\big ) z (u, x, v, X(t), V(t), -u) \rho _0 \Big ( \frac{1}{2}|v|^2 \Big ) du \nu (dx, dv) \Big |. \end{aligned}$$

So for any $t \in [0, T \wedge \sigma _n]$, we get by Proposition 2.1 (2) and Lemma 3.38 that

$$\begin{aligned}&\Big | \int _{\mathbf{R} \times E} f_1 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \\&\quad \le \Big | \int _{\mathbf{R} \times E} \Vert \nabla ^2 U \Vert _{\infty } 1_{ \{ |x| \le R_0 \} } 1_{ \{ u \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(t) )] \} } \\&\qquad \times 4 t_1 (v, |x-\pi ^{\perp }_v X(t)| ) n e^{\frac{1}{2} (1+C_1) (u+\tau )} \rho _0 \Big ( \frac{1}{2}|v|^2 \Big ) du \nu (dx, dv), \end{aligned}$$

which, since $d > (1+C_1) \varepsilon _1^{-1/2} + 1$ by (2.8), is bounded for $m \in (0, 1]$ by Lemma 4.5 (1). This completes our proof. $\square$

3.4.4 Smoothness for $I^6 (t)$ and $I^9 (t)$

In this section, we prove that $I^6(t)$ and $I^9 (t)$ satisfy (3.22).

Proof of (3.22) with $k=6$

By Lemma 3.39 (1), it suffices to prove that

$$\begin{aligned} E^{P_m} \Big [ 1_{ \{ m^{1/2} a_m \le t \le \sigma _n \} } \Big ( \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha } } |f_3 (t, r, x, v)| (\mu _{\omega } + \lambda ) (dr, dx, dv) \Big )^2 \Big ] \end{aligned}$$

is bounded for $m \in (0, 1]$ and $t \in [0, T]$. Since $|x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, we get by Proposition 2.1 (2), (4), (2.7), Lemma 3.20 and (3.12) that

$$\begin{aligned}&|f_3 (t, r, x, v)| \\&\quad \le 1_{ \{ |x-\pi ^{\perp }_v X({\tilde{r}})| \le R_U + 1 \} } 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_1(v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \} } \\&\qquad \times C_1 \Big ( |X(t) - X({\tilde{r}})| + |x (t, \Psi (r, x, m^{-1/2} v)) - \psi ^0 (m^{-1/2} (t-r), x, v; X({\tilde{r}}))| \Big ) \\&\quad \le 1_{ \{ |x-\pi ^{\perp }_v X({\tilde{r}})| \le R_U + 1 \} } 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_1(v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \} } \\&\qquad \times C_1 2 n m^{1/2} (t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + \tau ) e^{\frac{1}{2} (1+C_1) (m^{-1/2} (t-r) + \tau )}. \end{aligned}$$

This combined with (3.30) and Lemma 4.5 (1) implies our assertion, since $d > 2 (1+C_1) \varepsilon _1^{-1/2} + 1$ by (2.8). $\square$

Before proving (3.22) for $k=9$, we first prepare several estimates.

Lemma 3.44

There exists a constant $m_5 \in (0, 1]$ such that for any $m \in (0, m_5]$, if $t \in [0, T \wedge \sigma _n]$, $| x - \pi ^{\perp }_v X({\tilde{r}}) | \le m^{\alpha }$ and $m^{-1/2} (t-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X(t)|)]$, then $| x - \pi ^{\perp }_v X(t)| \le 2 m^{\alpha }$.

Proof

First, we get by assumption that

$$\begin{aligned} t-{\tilde{r}} \in [0, m^{1/2} (t_1 (v, | x - \pi ^{\perp }_v X(t)|) + \tau )]. \end{aligned}$$

(3.35)

Indeed, since $t \in [0, T \wedge \sigma _n]$ and $t - r \ge - m^{1/2} \tau$, we get that $r - m^{1/2} \tau \le T \wedge \sigma _n$, hence ${\tilde{r}} = (r - m^{1/2} \tau ) \vee 0 \ge r - m^{1/2} \tau$. So $t-{\tilde{r}} \le t - r + m^{1/2} \tau \le m^{1/2} t_1 (v, | x - \pi ^{\perp }_v X(t)|) + m^{1/2} \tau$, and $t-{\tilde{r}} = (t - (r - m^{1/2} \tau )) \wedge t \ge 0$.

By (3.35) and the assumption $| x - \pi ^{\perp }_v X({\tilde{r}}) | \le m^{\alpha }$, we get that

$$\begin{aligned}&|x - \pi ^{\perp }_v X(t)| \le | x - \pi ^{\perp }_v X({\tilde{r}}) | + |X(t) - X({\tilde{r}})| \nonumber \\&\quad \le m^{\alpha } + n m^{1/2} (t_1 (v, | x - \pi ^{\perp }_v X(t)|) + \tau ). \end{aligned}$$

(3.36)

If $|v| \ge 2 C_0 + 1$, then $t_1 (v, | x - \pi ^{\perp }_v X(t)|) = 2 \tau$, so (3.36) becomes $|x - \pi ^{\perp }_v X(t)| \le m^{\alpha } + n m^{1/2} 3 \tau$. Since $\alpha < \frac{1}{2}$ as a consequence of (2.10), as long as $m \le (3 n \tau )^{-1/({\frac{1}{2}-\alpha })}$, this implies that $|x - \pi ^{\perp }_v X(t)| \le 2 m^{\alpha }$. We next consider the case $|v| \in [{\overline{v}}, 2C_0+1)$. To simplify notations, let $a := |x - \pi ^{\perp }_v X(t)|$. So (3.36) becomes

$$\begin{aligned} a \le m^{\alpha } + n m^{1/2} \left( C_2 + \tau + 2 \varepsilon _1^{-1/2} \left( \log \frac{2 \varepsilon _3}{a}\right) 1_{ \left\{ a \le \frac{R_U}{2} \wedge (2 \varepsilon _3) \right\} } \right) . \end{aligned}$$

If $a > \frac{R_U}{2} \wedge (2 \varepsilon _3)$, then we get $\frac{R_U}{2} \wedge (2 \varepsilon _3) < a \le m^{\alpha } + n m^{1/2} ( C_2 + \tau )$, which is impossible for $m>0$ small enough. If $a \le \frac{R_U}{2} \wedge (2 \varepsilon _3)$, then we get that $a \le m^{\alpha } + n m^{1/2} \Big ( C_2 + \tau + 2 \varepsilon _1^{-1/2} \log \frac{2\varepsilon }{a} \Big )$. If $a \ge 2 m^{\alpha }$ in addition, then this implies that $m^{\alpha } \le n m^{1/2} \Big ( C_2 + \tau + 2 \varepsilon _1^{-1/2} \log \frac{2\varepsilon }{a} \Big ) \le n m^{1/2} \Big ( C_2 + \tau + 2 \varepsilon _1^{-1/2} \log \frac{2\varepsilon }{2 m^{\alpha }} \Big )$, which, again, is impossible if $m>0$ is small enough, since $\alpha < \frac{1}{2}$. $\square$

For any $t \in [0, T \wedge \sigma _n]$, divide $\mathbf{R} \times E$ into $\mathbf{R} \times E = B_1 \cup B_2 (t) \cup B_3 (t)$, with

$$\begin{aligned}&B_1 := \{ (r, x, v) \in \mathbf{R} \times E; | x - \pi ^{\perp }_v X({\tilde{r}}) |< m^{\alpha } \}, \nonumber \\&B_2 (t) := \{ (r, x, v) \in \mathbf{R} \times E; | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }, | x - \pi ^{\perp }_v X(t) | < m^{\alpha } \}, \nonumber \\&B_3 (t) := \{ (r, x, v) \in \mathbf{R} \times E; | x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }, | x - \pi ^{\perp }_v X(t) | \ge m^{\alpha } \}. \end{aligned}$$

(3.37)

This decomposition is also used in the proof of Lemma 3.60.

By the virtue of (2.9), Lemma 3.45 implies that the integrals of $f_4$ on $B_1$ and $B_2$ converge to 0 as $m \rightarrow 0$.

Lemma 3.45

There exists a constant $C_{15}>0$ such that for any $t \in [0, T]$, we have

• 1. ${\displaystyle { \Big | \int _{B_1} f_4 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \le C_{15} \Big ( m^{\alpha (d-1) - 1} + m^{\alpha (d-1) - \frac{1}{2}} \log (m^{-1}) \Big ), }}$

• 2. ${\displaystyle { \Big | \int _{B_2(t)} f_4 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \le C_{15} \Big ( m^{\alpha (d-1) - \frac{1}{2}} a_m + m^{\alpha (d-1) - \frac{1}{2}} \log (m^{-1}) \Big ). }}$

Proof

(1) By Proposition 2.1 (3), we have $\Big | \nabla U (X(t) - x (t, \Psi (r, x, m^{-1/2} v))) \Big | \le \Vert \nabla U \Vert _{\infty } 1_{ \{ r \in [- m^{1/2} \tau , T + m^{1/2} \tau ] \} }$; also, by Proposition 2.1 (2) and Lemma 3.44, we have on $B_1$ that $| \nabla U (X(t) - \psi ^0 (m^{-1/2} (t-r), x, v; X(t))) | \le \Vert \nabla U \Vert _{\infty } 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X(t)|)] \}} 1_{ \{ | x - \pi ^{\perp }_v X(t)| \le 2 m^{\alpha } \} }$. Therefore, with the help of Lemma 4.5 (2) (3), we get for any $t \in [0, T]$ that

$$\begin{aligned}&\Big | \int _{B_1} f_4 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \nonumber \\&\quad \le \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \le m^{\alpha } \} } 1_{ \{ r \in [-m^{1/2} \tau , T + m^{1/2} \tau ] \} } \Vert \nabla U \Vert _{\infty } {\overline{\lambda }} (dr, dx, dv) \\&\qquad + \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X(t)| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X(t)|)] \} } \Vert \nabla U \Vert _{\infty } {\overline{\lambda }} (dr, dx, dv) \\&\le C \Vert \nabla U \Vert _{\infty } \Big ( (T + 2 \tau ) m^{\alpha (d-1) - 1} + m^{\alpha (d-1) - \frac{1}{2}} \log (m^{-1}) \Big ). \end{aligned}$$

(2) It suffices to consider the case where $m \le (2 \varepsilon _3)^{1/\alpha }$.

On $B_2(t)$, since $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }$, we get by Proposition 2.1 (4) and (2.14) that $\Big | \nabla U (X(t) - x (t, \Psi (r, x, m^{-1/2} v))) \Big | \le \Vert \nabla U \Vert _{\infty } 1_{ \{ m^{-1/2} (t-r) \in [- \tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|) ] \} } \le \Vert \nabla U \Vert _{\infty } 1_{ \{ m^{-1/2} (t-r) \in [- \tau , a_m - \tau ] \} }$; also, by Proposition 2.1 (2), we have that $| \nabla U (X(t) - \psi ^0 (m^{-1/2} (t-r), x, v; X(t))) | \le \Vert \nabla U \Vert _{\infty } 1_{ \{ t-r \in [-m^{1/2} \tau , m^{1/2} t_1 (v, | x - \pi ^{\perp }_v X(t)|)] \} }$. So with the help of Lemma 4.5 (2) (3), we get that

$$\begin{aligned}&\Big | \int _{B_2 (t)} f_4 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \\&\quad \le \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X(t) | \le m^{\alpha } \} } 1_{ \{ m^{-1/2} (t-r) \in [- \tau , a_m - \tau ] \} } \Vert \nabla U \Vert _{\infty } {\overline{\lambda }} (dr, dx, dv) \\&\qquad + \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X(t) | \le m^{\alpha } \} } 1_{ \{ m^{-1/2} (t-r) \in [- \tau , t_1 (v, | x - \pi ^{\perp }_v X(t)|)] \} } \Vert \nabla U \Vert _{\infty } {\overline{\lambda }} (dr, dx, dv) \\&\quad \le C \Vert \nabla U \Vert _{\infty } \Big ( m^{\alpha (d-1) - \frac{1}{2}} a_m + m^{\alpha (d-1) - \frac{1}{2}} \log (m^{-1}) \Big ). \end{aligned}$$

$\square$

Now we are ready to prove (3.22) for $k=9$.

Proof of (3.22) for $k=9$. By definition and Lemma 3.43 (1), it suffices to prove that

$$\begin{aligned} \sup _{m \in (0, 1] } \sup _{t \in [0, T]} 1_{ \{ m^{1/2} a_m \le t \le \sigma _n \} } \Big | \int _{\mathbf{R} \times E} f_4 (t, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | < \infty . \end{aligned}$$

The integrals on $B_1$ and $B_2 (t)$ converge to 0 by Lemma 3.45. We prove in the following that the integral on $B_3 (t)$ is bounded. Let $t_3 (r, t, x, v) = t_1 (v, | x - \pi ^{\perp }_v X(t)|) \vee t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|)$. Then on $B_3 (t)$, for any $m \le (2 \varepsilon _3)^{1/\alpha }$, we have by (2.14) that

$$\begin{aligned} t_3 (r, t, x, v) \le a_m - \tau . \end{aligned}$$

(3.38)

If $f_4 (t, r, x, v) \ne 0$, then by Proposition 2.1 (2) (4), we have on $B_3 (t)$ that $|x| \le R_0$ and $m^{-1/2} (t-r) \in [-\tau , t_3 (r, t, x, v)]$, which combined with $t \ge m^{1/2} a_m$ and (3.38) implies that $r - m^{1/2} \tau \ge 0$, hence Lemma 3.20 (3) is applicable (with s, t, $c_m$ and $b_m$ given by r, $m^{-1/2} (t-r)$, $r-t$ and $t_3 (r, t, x, v)$, respectively). Combining these, we get by Lemma 3.20 (3) that

$$\begin{aligned}&\Big | f_4 (t, r, x, v) \Big | \\&\quad \le C_1 \Big | x \big (t, \Psi (r, x, m^{-1/2} v)\big ) - \psi ^0 \big (m^{-1/2} (t-r), x, v; X(t)\big ) \Big | \\&\qquad \cdot 1_{ \{ |x| \le R_0 \} } 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_3 (r, t, x, v)] \} } \\&\quad \le C_1 1_{ \{ |x| \le R_0 \} } 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_3 (r, t, x, v)] \} } 4 n m^{1/2} t_3 (r, t, x, v) e^{\frac{1}{2} ( 1+C_1 ) (m^{-1/2} (t-r) + \tau )} \\&\quad \le C_1 1_{ \{ |x| \le R_0 \} } 4 n m^{1/2} \exp \left( \frac{1}{2} ( 1+C_1 ) (m^{-1/2} (t-r) + \tau ) \right) \\&\qquad \times \Big ( 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X(t)|)] \} } t_1 (v, | x - \pi ^{\perp }_v X(t)|) \\&\qquad \qquad + 1_{ \{ m^{-1/2} (t-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|) ] \} } t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|) \Big ). \end{aligned}$$

Since $(1+C_1) \varepsilon _1^{-1/2} < d - 1$ by (2.8), this combined with Lemma 4.5 (1) implies that the considered integral on $B_3 (t)$ is bounded for $m \in (0, 1]$, hence completes our proof. $\square$

3.4.5 The Martingale Part

We deal with the term $I^3 (t)$ in this section. Precisely, we prove that $I^3 (t) - M(t)$ satisfies the same estimates as in Proposition 3.31 (4) with respect to $\eta (t)$, and that M(t) satisfies Proposition 3.31 (3). This will complete the proof of Proposition 3.31.

The proof is similar to that of [12, Section 4.4] with slight modification, so we give only the sketch.

We notice that if $\nabla U (X(r - m^{1/2} \tau ) - \psi ^0 (m^{-1/2} (s-r), x, v; X(r - m^{1/2} \tau ))) \ne 0$, $s \ge m^{1/2} a_m$ and $| x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, then by Proposition 2.1 (2) and (2.14), for any $m \le (2 \varepsilon _3)^{1/\alpha }$, we get that $m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|)]$, hence $r \in [s - m^{1/2} t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|), s + m^{1/2} \tau ] \subset [m^{1/2} \tau , s + m^{1/2} \tau ]$, so if $s \in [0, T \wedge \sigma _n]$ in addition, then ${\tilde{r}} = r - m^{1/2} \tau$. So we have the following decomposition of $I^3 (t)$ by definition:

$$\begin{aligned} I^3 (t) = I^{3, 1} (t) + I^{3, 2} (t) + I^{3, 3} (t) + I^{3, 4} (t), \qquad t \in [0, T], \end{aligned}$$

with

$$\begin{aligned}&I^{3, 1} (t) = - \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})|< m^{\alpha } \} } \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\&I^{3, 2} (t) = - \int _0^{t \wedge \sigma _n} ds \int _{\mathbf{R} \times E} 1_{ \{ r \in [m^{1/2} \tau , s + m^{1/2} \tau ] \} } \\&\qquad \nabla U \big (X(r - m^{1/2} \tau ) - \psi ^0 (m^{-1/2} (s-r), x, v; X(r - m^{1/2} \tau ))\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\&I^{3, 3} (t) = \int _0^{t \wedge \sigma _n} ds \int _{\mathbf{R} \times E} 1_{ \{ r \in [m^{1/2} \tau , s + m^{1/2} \tau ] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha } \} } \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv), \\&I^{3, 4} (t) = \int _0^{t \wedge \sigma _n} 1_{(0, m^{1/2} a_m]} (s) ds \int _{\mathbf{R} \times E} 1_{ \{ r \in [m^{1/2} \tau , s + m^{1/2} \tau ] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha } \} } \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv). \end{aligned}$$

We prove in Lemmas 3.46 and 3.47 below that $I^{3, 1} (t)$, $I^{3, 3} (t)$ and $I^{3, 4} (t)$ satisfy (3.26), which implies that they satisfy the same estimates as in Proposition 3.31 (4).

Lemma 3.46

There exists a constant $C_{16} > 0$ such that

$$\begin{aligned} E^{P_m} \Big [ \sup _{t \in [0, T]} |I^{3, k} (t)|^2 \Big ] \le C_{16} m^{\alpha (d-1) - \frac{1}{2}} \log (m^{-1}), \qquad k \in \{ 1, 3 \}, m \in (0, 1]. \end{aligned}$$

Proof

$X({\tilde{r}})$ is $\mathcal {F}_{(-\infty , r] \times E}$-measurable by Lemma 3.19. So for $k \in \{ 1, 3 \}$, we have by Lemma 3.34 and Proposition 2.1 (2) that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{t \in [0, T]} |I^{3, k} (t)|^2 \Big ] \\&\quad \le T \int _0^T ds E \Big [ \Big | \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})|< m^{\alpha } \} } \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \Big ] \\&\quad = T \int _0^T ds E \Big [ \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})|< m^{\alpha } \} } \\&\qquad \Big | \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big ) \Big |^2 \lambda (dr, dx, dv) \Big ] \\&\quad \le T \Vert \nabla U \Vert _{\infty }^2 \int _0^T ds E \Big [ \int _{\mathbf{R} \times E} 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| < m^{\alpha } \} } \\&\qquad 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \} } \lambda (dr, dx, dv) \Big ]. \end{aligned}$$

This combined with Lemma 4.5 (3) implies our assertion. $\square$

Lemma 3.47

There exists a constant $C_{17} > 0$ such that

$$\begin{aligned} E^{P_m} \Big [ \sup _{t \in [0, T]} |I^{3, 4} (t)|^2 \Big ] \le C_{17} m^{\frac{1}{2}} a_m^2, \qquad m \in (0, 1]. \end{aligned}$$

Proof

We have by Lemma 3.34 that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{t \in [0, T]} |I^{3, 4} (t)|^2 \Big ] \nonumber \\&\quad \le m^{\frac{1}{2}} a_m E^{P_m} \Big [ \int _0^T 1_{(0, m^{1/2} a_m]} (s) ds \Big | \int _{\mathbf{R} \times E} 1_{ \{ r \in [m^{1/2} \tau , s + m^{1/2} \tau ] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha } \} } \nonumber \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big ) (\mu _{\omega } - \lambda ) (dr, dx, dv) \Big |^2 \Big ] \nonumber \\&\quad = m^{\frac{1}{2}} a_m \int _0^T 1_{(0, m^{1/2} a_m]} (s) ds E^{P_m} \Big [ \int _{\mathbf{R} \times E} 1_{ \{ r \in [m^{1/2} \tau , s + m^{1/2} \tau ] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha } \} } \nonumber \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big )^2 \lambda (dr, dx, dv) \Big ]. \end{aligned}$$

(3.39)

By Proposition 2.1 (2) and Lemma 4.5 (1), we have that

$$\begin{aligned}&E^{P_m} \Big [ \int _{\mathbf{R} \times E} 1_{ \{ r \in [m^{1/2} \tau , s + m^{1/2} \tau ] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha } \} } \\&\qquad \nabla U \big (X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}}))\big )^2 \lambda (dr, dx, dv) \Big ] \\&\quad \le \Vert \nabla U \Vert _{\infty }^2 E^{P_m} \Big [ \int _{\mathbf{R} \times E} 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \} } 1_{ \{ |x - \pi ^{\perp }_v X({\tilde{r}})| \le R_U + 1 \} } \lambda (dr, dx, dv) \Big ] \\&\quad \le \Vert \nabla U \Vert _{\infty }^2 C m^{-1/2}. \end{aligned}$$

This combined with (3.39) implies our assertion. $\square$

Finally, we deal with the term $I^{3, 2}$. Let

$$\begin{aligned} {\widetilde{I^{3, 2}}} (t):= & {} - \int _0^t ds \int _{(0, s] \times E} \nabla U (X(r \wedge \sigma _n) \\&- \psi ^0 (m^{-1/2} (s-r) - \tau , x, v; X(r \wedge \sigma _n))) {\overline{N}} (dr, dx, dv). \end{aligned}$$

As discussed before, $r \ge m^{1/2} \tau$ implies $\lambda (dr, dx, dv) = {\overline{\lambda }} (dr, dx, dv)$. Therefore, by a trivial change of variable, we get that

$$\begin{aligned} I^{3, 2} (t) = {\widetilde{I^{3, 2}}} (t \wedge \sigma _n). \end{aligned}$$

We first notice the following estimate with respect to $\frac{d}{dt} {\widetilde{I^{3, 2}}} (t)$.

Lemma 3.48

There exists a constant $C_{18} > 0$ such that

$$\begin{aligned} E^{P_m} \Big [ \Big | \frac{d}{dt} {\widetilde{I^{3, 2}}} (t) \Big |^2 \Big ] \le C_{18} m^{-1/2}, \qquad \hbox { for all } t \in [0, T], m \in (0, 1]. \end{aligned}$$

Proof

We have by definition, Lemma 3.34 and Proposition 2.1 (2) that

$$\begin{aligned}&E^{P_m} \Big [ \Big | \frac{d}{dt} {\widetilde{I^{3, 2}}} (t) \Big |^2 \Big ] \\&\quad = E^{P_m} \Big [ \int _{(0, t] \times E} \Big | \nabla U (X(r \wedge \sigma _n) - \psi ^0 (m^{-1/2} (t-r) - \tau , x, v; X(r \wedge \sigma _n))) \Big |^2 {\overline{\lambda }} (dr, dx, dv) \Big ] \\&\quad \le \Vert \nabla U \Vert _{\infty }^2 E^{P_m} \Big [ \int _{(0, t] \times E} 1_{ \{ |x-\pi ^{\perp }_v X(r \wedge \sigma _n)| \le R_U + 1 \} } \\&\qquad 1_{ \{ m^{-1/2} (t-r) - \tau \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(r \wedge \sigma _n)|) ] \} } {\overline{\lambda }} (dr, dx, dv) \Big ]. \end{aligned}$$

This combined with Lemma 4.5 (1) implies our assertion. $\square$

Also, by a simple calculation, we have that

$$\begin{aligned} {\widetilde{I^{3, 2}}} (t)= & {} - \int _{(0, t] \times E} {\overline{N}} (dr, dx, dv) \\&\int _{[0, t-r]} \nabla U (X(r \wedge \sigma _n) - \psi ^0 (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n))) ds. \end{aligned}$$

Since by Proposition 2.1 (2), $s \ge m^{1/2} \Big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n)| ) + \tau \Big )$ implies $\nabla U (X(r \wedge \sigma _n) - \psi ^0 (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n))) = 0$, we get that ${\widetilde{I^{3, 2}}} (t)$ can be rewritten as

$$\begin{aligned} {\widetilde{I^{3, 2}}} (t) = M (t) + \eta _1 (t), \end{aligned}$$

(3.40)

with M(t) as defined in (3.20), and

$$\begin{aligned}&\eta _1 (t) := \int _{(0, t] \times E} {\overline{N}} (dr, dx, dv)\\&\qquad \int _{[(t-r) \wedge \{ m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )|) + \tau \big ) \}, m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n)|) + \tau \big ) ]} \\&\qquad \nabla U \big (X(r \wedge \sigma _n) - \psi ^0 \big (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n)\big )\big ) ds. \end{aligned}$$

Summarizing our result up to now, we get that in order to complete our proof of Proposition 3.31, it suffices to prove that M(t) satisfies Proposition 3.31 (3), and that $\eta _1 (t)$ satisfies the same estimates as in Proposition 3.31 (4). Let $C_{19} := \Big ( 3 \tau \Vert \nabla U \Vert _{\infty } \Big ) \vee (4C_0 + 2)$. We first prove the following:

Lemma 3.49

$\{ M (t) \}_{t \in [0, T]}$ is a càdlàg martingale, and the jumps of $\{ M (t) \}_{t \in [0, T]}$ and $\{ \eta _1 (t) \}_{t \in [0, T]}$ satisfy $| \Delta M (t) | \le C_{19} m^{1/2}$, $| \Delta \eta _1 (t) | \le C_{19} m^{1/2}$.

Proof

The fact that $\{ M (t) \}_{t \in [0, T]}$ is a càdlàg martingale is trivial.

We estimate $| \Delta M (t) |$ in the following. If $|v| \ge 2 C_0 + 1$, then $t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n)|) = 2 \tau$ by definition, so $\Big | \int _{[0, m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n)|) + \tau \big )]} \nabla U (X(r \wedge \sigma _n) - \psi ^0 (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n))) ds \Big | \le m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n)|) + \tau \big ) \Vert \nabla U \Vert _{\infty } = m^{1/2} 3 \tau \Vert \nabla U \Vert _{\infty }$. If $|v| \le 2 C_0 + 1$, then

$$\begin{aligned}&\Big | \int _{[0, m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )|) + \tau \big )]} \nabla U (X (r \wedge \sigma _n ) - \psi ^0 (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n ))) ds \Big | \\&\quad = m^{1/2} \Big | \psi ^1 (t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )|), x, v; X(r \wedge \sigma _n )) - \psi ^1 (-\tau , x, v; X(r \wedge \sigma _n )) \Big | \\&\quad \le m^{1/2} 2 |v| \le m^{1/2} 2 (2 C_0 + 1). \end{aligned}$$

The estimation with respect to $| \Delta \eta _1 (t) |$ is gotten in exactly the same way, and we omit the proof here. $\square$

Proof of Proposition 3.31 (3)

By Lemma 3.49, if suffices to prove (3.23) and (3.24).

We prove (3.23) by using the following general result: for any $f: \mathbf{R} \times E \times \Omega \rightarrow \mathbf{R}$ such that $f (r, x, v, \cdot )$ is $\mathcal {F}_r$-measurable for any $(r, x, v) \in \mathbf{R} \times E$, we have for any $s, t \in \mathbf{R}$ satisfying $s < t$ that

$$\begin{aligned}&E \left[ \Big ( \int _{[s, t] \times E} f (r, x, v, \cdot ) {\overline{N}} (dr, dx, dv) \Big )^2 \Big | \mathcal {F}_s \right] \nonumber \\&\quad = E \left[ \int _{[s, t] \times E} f^2 (r, x, v, \cdot ) {\overline{\lambda }} (dr, dx, dv) \Big | \mathcal {F}_s \right] . \end{aligned}$$

(3.41)

For any $s, t \in [0, T]$ satisfying $s < t$, we have by definition, (3.41) and Proposition 2.1 (2) that

$$\begin{aligned}&E^{P_m} \Big [ |M(t) - M(s)|^2 \Big | \mathcal {F}_s \Big ] \\&\quad = E^{P_m} \Big [ \int _{(s, t] \times E} \Big | \int _{[0, m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| ) + \tau \big )]} \\&\qquad \nabla U (X(r \wedge \sigma _n) - \psi ^0 (m^{-1/2} u - \tau , x, v; X(r \wedge \sigma _n))) du \Big |^2 {\overline{\lambda }} (dr, dx, dv) \Big | \mathcal {F}_s \Big ] \\&\quad \le m \Vert \nabla U \Vert _{\infty }^2 E^{P_m} \Big [ \int _{(s, t] \times E} (t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| ) + \tau )^2 \\&\qquad \qquad \times 1_{ \{ |x - \pi ^{\perp }_v X(r \wedge \sigma _n)| \le R_U + 1 \} } {\overline{\lambda }} (dr, dx, dv) \Big | \mathcal {F}_s \Big ]. \end{aligned}$$

Combining this with Lemma 4.5 (4), we get (3.23).

We next prove (3.24). Let $A_k (s, x, v) := m^{-\frac{1}{2}} \int _\mathbf{R} \nabla _k U (X(s) - \psi ^0 (m^{-\frac{1}{2}} u - \tau , x, v; X(s))) du$. Then $|A_k (s, x, v) | \le \Vert \nabla U \Vert _{\infty } (t_1 (v, | x - \pi ^{\perp }_v X(s) | ) + \tau ) 1_{ \{ | x - \pi ^{\perp }_v X(s) | \le R_U + 1 \} }$ by Proposition 2.1 (2). Also, for any $s \le \sigma _n$, we have by definition that

$$\begin{aligned} a_{kl} d s = m \int _E A_k (s, x, v) A_l (s, x, v) {\overline{\lambda }} (ds, dx, dv) \end{aligned}$$

and

$$\begin{aligned}{}[M^k, M^l]_s = m \int _{(0, s] \times E} A_k (s, x, v) A_l (s, x, v) N (d s, d x, d v). \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{(0, t \wedge \sigma _n]} g (s) \Big ( d [ M^k, M^l ]_s - a_{kl} ds \Big ) = m \int _{(0, t \wedge \sigma _n]} g (s) A_k (s, x, v) A_l (s, x, v) {\overline{N}} (ds, dx, dv), \end{aligned}$$

which is a martingale. So by Doob’s inequality and (3.41), we get that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{t \in [0, T] } \Big | \int _{(0, t \wedge \sigma _n]} g (s) \Big ( d [ M^k, M^l ]_s - a_{kl} ds \Big ) \Big |^2 \Big ] \\&\quad \le 4 E^{P_m} \Big [ \Big | m \int _{(0, T \wedge \sigma _n] \times E} g (s) A_k (s, x, v) A_l (s, x, v) {\overline{N}} (ds, dx, dv) \Big |^2 \Big ] \\&\quad = 4 m^2 E^{P_m} \Big [ \int _{(0, T \wedge \sigma _n] \times E} g (s)^2 A_k (s, x, v)^2 A_l (s, x, v)^2 {\overline{\lambda }} (ds, dx, dv) \Big ] \\&\quad \le 4 m E^{P_m} \Big [ \int _{(0, Ts\wedge \sigma _n ]} ds \int _E \Vert g\Vert _{\infty }^2 \Vert \nabla U \Vert _{\infty }^4 (t_1 (v, | x - \pi ^{\perp }_v X(s) | ) + \tau )^4 \\&\qquad \times 1_{ \{ | x - \pi ^{\perp }_v X(s) | \le R_U + 1 \} } \rho _0 \Big ( \frac{1}{2} |v|^2 \Big ) \nu (dx, dv) \Big ], \end{aligned}$$

which, by Lemma 4.5 (3), is bounded by $4 \Vert g\Vert _{\infty }^2 \Vert \nabla U \Vert _{\infty }^4 C T m$. So (3.24) holds. $\square$

By Proposition 3.31 (3) and Lemma 3.30, we get easily the following.

Lemma 3.50

$\{ \hbox {the distribution of } \big \{ M (t) \big \}_{t \in [0, T]} \hbox { under } P_m \}_{m \in (0, 1]}$ is tight in $\wp (D([0, T]; \mathbf{R}^d))$. $\square$

With Lemmas 3.49 and 3.50 in hand, we get Lemmas 3.51 and 3.52 below by exactly the same argument as in [12] (see the proof of [12, Lemmas 4.4.2–4.4.4]).

Lemma 3.51

Any cluster point of $\{ \big \{ M (t) \big \}_{t \in [0, T]} \hbox { under } P_m \}_{m \rightarrow 0}$ in $\wp (D([0, T]; \mathbf{R}^d))$ constitutes a continuous canonical process. $\square$

Lemma 3.52

For any $\varepsilon > 0$, we have that

$$\begin{aligned} \limsup _{\delta \rightarrow 0} \limsup _{m \rightarrow 0} P_m \Big ( \sup _{0 \le s \le t \le T, |s-t| \le \delta } |M (t) - M (s)| > \varepsilon \Big ) = 0. \end{aligned}$$

(3.42)

$\square$

Also, since

$$\begin{aligned}&\sup _{\omega \in \Omega , m \in (0, 1] } \int _{(0, T] \times E} \Big ( m^{1/2} t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n)|) \Big )^k \\&\qquad 1_{ \{ |x - \pi ^{\perp }_v X(r \wedge \sigma _n)| \le R_U + 1 \} } {\overline{\lambda }} (dr, dx, dv) < \infty \end{aligned}$$

for $k \in \{ 2, 4 \}$, by exactly the same argument as in [12, Lemmas 4.4.5], we get the following.

Lemma 3.53

There exists a constant $C_{20}>0$ such that

$$\begin{aligned} \sup _{m \in (0, 1]} E^{P_m} \Big [ \sup _{t \in [0, T]} |M (t)|^4 \Big ] \le C_{20}. \end{aligned}$$

$\square$

Similarly, by the same argument as in the proof of [12, (4.18)], we get the following.

Lemma 3.54

There exists a constant $C_{21} > 0$ such that

$$\begin{aligned} E^{P_m} [ |\eta _1 (t)|^6] \le C_{21} m^{3/2}, \qquad t \in [0, T], m \in (0, 1]. \end{aligned}$$

Proof

The quantity A in the proof of [12, (4.18)] is now given by $A = \Big | \int _{(t-r) \wedge \{ m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| )+ \tau \big ) \} } ^{m^{1/2} \big ( t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| ) + \tau \big ) } \nabla U (X(r \wedge \sigma _n) - \psi ^0 (m^{-1/2} s - \tau , x, v; X(r \wedge \sigma _n))) ds \Big |$. So

$$\begin{aligned}&A \le 1_{ \{ r \ge t - m^{1/2} t_1 (v, | x - \pi ^{\perp }_v X (r \wedge \sigma _n)|) - m^{1/2} \tau \} } \\&\qquad \times m^{1/2} (t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| ) + \tau ) \Vert \nabla U \Vert _{\infty } 1_{ \{ | x - \pi ^{\perp }_v X (r \wedge \sigma _n)| \le R_U + 1 \} }. \end{aligned}$$

So for any $k \in \mathbf{N}$, we get that

$$\begin{aligned}&\int _{(0, t] \times E} A^k {\overline{\lambda }} (dr, dx, dv) \\&\quad \le m^{k/2} \Vert \nabla U \Vert _{\infty }^k \int _{(0, t] \times E} 1_{ \{ r \ge t - m^{1/2} (t_1 (v, |x - \pi ^{\perp }_v X(r \wedge \sigma _n )| ) + \tau ) \} } \\&\qquad ( t_1 (v, | x - \pi ^{\perp }_v X (r \wedge \sigma _n )|) + \tau )^k 1_{ \{ | x - \pi ^{\perp }_v X (r \wedge \sigma _n)| \le R_U + 1 \} } {\overline{\lambda }} (dr, dx, dv). \end{aligned}$$

Hence by Lemma 4.5 (4), we get that

$$\begin{aligned} \int _{(0, t] \times E} A^k {\overline{\lambda }} (dr, dx, dv) \le \Vert \nabla U \Vert _{\infty }^k C m^{\frac{k-1}{2}}, \qquad k \in \{ 0, 1, \cdots , 6 \}. \end{aligned}$$

So by exactly the same argument as in the proof of [12, (4.18)], with a proper new C, we get that

$$\begin{aligned} E^{P_m} [ |\eta _1 (t)|^6] \le C \Big ( \Big (m^{\frac{2-1}{2}}\Big )^3 + \Big (m^{\frac{3-1}{2}}\Big )^2 + \Big (m^{\frac{2-1}{2}}\Big ) \cdot \Big (m^{\frac{4-1}{2}}\Big ) + m^{\frac{6-1}{2}} \Big ), \end{aligned}$$

which implies our assertion. $\square$

Recall that the jumps of $\eta _1 (t)$ are dominated by $C_{19} m^{1/2}$ by Lemma 3.49. So by exactly the same argument as in the proof of [12, Lemma 4.4.6], we get the following result with the help of (3.40), Lemmas 3.48, 3.52 and 3.53:

Lemma 3.55

$\displaystyle { \lim _{m \rightarrow 0} E^{P_m} \Big [ \sup _{0 \le t \le T} \Big | \eta _1 (t) \Big |^2 \Big ] = 0. }$ $\square$

Finally, we prove the following result with respect to the variations of M(t) and $\eta _1 (t)$:

Lemma 3.56

There exists a constant $C_{22}>0$ such that for any $s \in [m^{1/2} a_m, T]$,

• 1. $E \Big [ \sup _{u \in [s - m^{1/2} a_m, s] } |M (u) - M (s - m^{1/2} a_m)|^2 \Big ] \le C_{22} m^{1/2} a_m.$

• 2. $E \Big [ \sup _{u \in [s - m^{1/2} a_m, s] } | \eta _1 (u) - \eta _1 (s - m^{1/2} a_m) |^2 \Big ] \le C_{22} m^{1/2} a_m^2.$

Proof

Since $M(\cdot )$ is a càdlàg martingale, we have by Doob’s inequality and (3.23) that $E \Big [ \sup _{u \in [s - m^{1/2} a_m, s] } |M (u) - M (s - m^{1/2} a_m)|^2 \Big ] \le 4 E \Big [ |M (s) - M (s - m^{1/2} a_m)|^2 \Big ] \le 4 C_6 m^{1/2} a_m.$ So we get our first assertion.

For the second assertion, we have by definition that $| \eta _1 (u) - \eta _1 (s - m^{1/2} a_m) |^2 \le 2 | {\widetilde{I^{3, 2}}} (u) - {\widetilde{I^{3, 2}}} (s - m^{1/2} a_m) |^2 + 2 |M (u) - M (s - m^{1/2} a_m)|^2$. By Lemma 3.48, we have that

$$\begin{aligned}&E \Big [ \sup _{u \in [s - m^{1/2} a_m, s] } | {\widetilde{I^{3, 2}}} (u) - {\widetilde{I^{3, 2}}} (s - m^{1/2} a_m) |^2 \Big ] \\&\quad \le E \Big [ \Big ( \int _{s - m^{1/2} a_m}^{s} \Big | \frac{d}{du} {\widetilde{I^{3, 2}}} (u) \Big | du \Big )^2 \Big ] \\&\quad \le (m^{1/2} a_m)^2 \sup _{u \in [s - m^{1/2} a_m, s] } E \Big [ \Big | \frac{d}{du} {\widetilde{I^{3, 2}}} (u) \Big |^2 \Big ] \\&\quad \le (m^{1/2} a_m)^2 C_{18} m^{-1/2} = C_{18} m^{1/2} a_m^2. \end{aligned}$$

This combined with our first assertion implies our second assertion. $\square$

Summarizing our result up to now, we have completed the proof of Proposition 3.31 with $\eta (t)$ given by $\eta (t) := I^1 (t) + I^2 (t) + I^5 (t) + I^7 (t) + I^8 (t) + I^{10} (t) + I^{11} (t) + I^{3, 1} (t) + I^{3, 3} (t) + I^{3, 4} (t) + \eta _1 (t)$.

3.5 Decaying of $I^6(t)$ and $I^9(t)$

We completed the proof of Proposition 3.31 in Sect. 3.4.5. In this section, we prove that the terms $I^6(t)$ and $I^9(t)$ are actually negligible, by using Proposition 3.31. See Lemmas 3.59 and 3.60 below for the explicit expression. As explained in Sect. 2.1, by exactly the same argument of the martingale problem theory as was used in [12, Sect. 5], which was also briefly summarized in [13, page 317], this will complete the proof of Theorem 1.2.

We start our proof by the following observation.

Lemma 3.57

Assume that $(y, v) \in E$ and $s, r \in \mathbf{R}$ satisfy $|y| \ge m^{\alpha }$, $s \in [m^{\frac{1}{2}} a_m, T \wedge \sigma _n]$ and $- \tau \le m^{-1/2} (s-r) \le t_1 (v, |y|)$. Then we have for any $m \le (2 \varepsilon _3)^{1/\alpha }$ that

$$\begin{aligned} \int _{r - m^{1/2} \tau }^s |V(u) - V({\tilde{r}})| du \le 2 m^{1/2} a_m \sup _{u \in [s - m^{1/2} a_m, s]} |V (s-m^{1/2} a_m) - V(u)|. \end{aligned}$$

Proof

Since $|y| \ge m^{\alpha }$, by (2.14) we have for any $m \le (2 \varepsilon _3)^{1/\alpha }$ that $t_1 (v, |y|) \le a_m - \tau$. This combined with $m^{-1/2} (s-r) \le t_1 (v, |y|)$ implies that $r - m^{1/2} \tau \ge s - m^{1/2} t_1 (v, |y|) - m^{1/2} \tau \ge s-m^{1/2} a_m$. Our assertion is now trivial. $\square$

Lemma 3.58 below is easy by Proposition 3.31.

Lemma 3.58

There exist constants $\varepsilon , C_{23} > 0$ such that for any $s \in [m^{1/2} a_m, T]$, we have that

$$\begin{aligned} E^{P_m} \Big [ 1_{ \{ s \le \sigma _n \} } \sup _{u \in [s - m^{1/2} a_m, s]} |V (s-m^{1/2} a_m) - V(u)|^2 \Big ] \le C_{23} m^{\varepsilon }. \end{aligned}$$

Proof

To simplify notation, let $P (t) := I^4 (t) + I^6 (t) + I^9 (t)$. Then by Proposition 3.31 (1), we have that

$$\begin{aligned}&E^{P_m} \Big [ 1_{ \{ s \le \sigma _n \} } \sup _{u \in [s - m^{1/2} a_m, s]} |V (s-m^{1/2} a_m) - V(u)|^2 \Big ] \nonumber \\&\quad \le 3 E^{P_m} \Big [ \sup _{u \in [s - m^{1/2} a_m, s]} |\eta (s-m^{1/2} a_m) - \eta (u)|^2 \Big ] \nonumber \\&\qquad + 3 E^{P_m} \Big [ \sup _{u \in [s - m^{1/2} a_m, s]} |M (s-m^{1/2} a_m) - M(u)|^2 \Big ] \nonumber \\&\qquad + 3 E^{P_m} \Big [ \sup _{u \in [s - m^{1/2} a_m, s]} |P (s-m^{1/2} a_m) - P (u)|^2 \Big ]. \end{aligned}$$

(3.43)

The first expectation on the right hand side of (3.43) is dominated by $3 C_6 m^{\varepsilon }$ by (3.25) of Proposition 3.31. Also, as proved in Lemma 3.56 (1), Proposition 3.31 (3) implies that the second expectation on the right hand side of (3.43) is dominated by $12 C_6 m^{1/2} a_m$.

Finally, for the third term on the right hand side of (3.43), we have by Proposition 3.31 (2) that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{u \in [s - m^{1/2} a_m, s]} |P (s-m^{1/2} a_m) - P(u)|^2 \Big ] \\&\quad \le (m^{1/2} a_m)^2 \sup _{t \in [0, T]} E^{P_m} \Big [ \Big | \frac{d}{dt} P(t) \Big |^2 \Big ] \le 3 C_6 m a_m^2. \end{aligned}$$

$\square$

Lemma 3.59

We have that

$$\begin{aligned} \lim _{m \rightarrow 0} E \Big [ \sup _{t \in [0, T]} |I^6 (t)| \Big ] = 0. \end{aligned}$$

Proof

Let

$$\begin{aligned}&F_1 (s, r, x, v) := |X(s) - X({\tilde{r}})| + | x (s, \Psi (r, x, m^{-1/2} v)) \\&\qquad - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) |, \\&F_2 (s, r, x, v) := F_1 (s, r, x, v)^2, \\&F_3 (s, r, x, v) := |X(s) - X({\tilde{r}}) - (s-{\tilde{r}}) V({\tilde{r}})| \\&\quad + | x (s, \Psi (r, x, m^{-1/2} v)) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) \\&\qquad \qquad - m^{1/2} z (m^{-1/2} (s-r), x, v; X({\tilde{r}}), V({\tilde{r}}), - m^{-1/2}({\tilde{r}}-r)) |. \end{aligned}$$

Then by an easy calculation, we get the following:

$$\begin{aligned} | (f_3 - f_2) (s, r, x, v) | \le \frac{1}{2} \Vert \nabla ^3 U \Vert _{\infty } F_2 (s, r, x, v) + \Vert \nabla ^2 U \Vert _{\infty } F_3 (s, r, x, v). \end{aligned}$$

(3.44)

Indeed, by definition and a simple calculation, we have that

$$\begin{aligned}&(f_3 - f_2) (s, r, x, v) \\&= \int _0^1 \Big \{ \nabla ^2 U \Big ( X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) \\&\qquad +\, \theta [ X(s) - X({\tilde{r}}) - x (s, \Psi (r, x, m^{-1/2} v)) + \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) ] \Big ) \\&\qquad - \nabla ^2 U \Big ( X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) \Big ) \Big \} \\&\qquad \cdot \Big ( X(s) - X({\tilde{r}}) - x (s, \Psi (r, x, m^{-1/2} v)) + \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) \Big ) d \theta \\&\qquad +\, \nabla ^2 U \Big ( X({\tilde{r}}) - \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) \Big ) \cdot \Big ( X(s) - X({\tilde{r}}) - (s-{\tilde{r}}) V({\tilde{r}}) \\&\qquad - \,x (s, \Psi (r, x, m^{-1/2} v)) + \psi ^0 (m^{-1/2} (s-r), x, v; X({\tilde{r}})) \\&\qquad +\, m^{1/2} z (m^{-1/2} (s-r), x, v; X({\tilde{r}}), V({\tilde{r}}), - m^{-1/2}({\tilde{r}}-r)) \Big ). \end{aligned}$$

This gives us (3.44).

Also, by Proposition 2.1 (2) (4), we get that if $| x - \pi ^{\perp }_v X({\tilde{r}}) | \ge m^{\alpha }$, then $| (f_3 - f_2) (s, r, x, v) | \ne 0$ implies that $m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)]$ and $| x - \pi ^{\perp }_v X({\tilde{r}}) | \le R_U + 1$. So in order to prove Lemma 3.59, it suffices to prove the following:

$$\begin{aligned}&E^{P_m} \Big [ \sup _{t \in [0, T]} \Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }} ( \mu _{\omega } + \lambda ) (dr, dx, dv) F_k (s, r, x, v) \nonumber \\&\quad \times 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)] \} } 1_{ \{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \le R_U + 1 \} } \Big | \Big ] \rightarrow 0, \quad k = 2, 3. \end{aligned}$$

(3.45)

We prove (3.45) in the following.

As same as in the proof of Lemma 2.2, if $m \le (2 \varepsilon _3)^{1/\alpha }$, $m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)]$, $s \ge m^{1/2} a_m$ and $| x - \pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }$, then ${\tilde{r}} = r - m^{1/2} \tau$, hence $s - {\tilde{r}} = s - r + m^{1/2} \tau \in [0, m^{1/2} t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}})|) + m^{1/2} \tau ]$. This combined with Lemma 3.20 (with s, t and $b_m$ given by r, $m^{-1/2} (s-r)$ and $t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)$, respectively) and (3.12) implies that

$$\begin{aligned} F_1 (s, r, x, v) \le 4 n m^{1/2} t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |) e^{ \frac{1}{2} (1+C_1) (m^{-1/2} (s-r) + \tau ) }. \end{aligned}$$

Therefore, since $(1+C_1) 2 \varepsilon _1^{-1/2} < d - 1$, by Lemma 4.5 (1), we get that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{t \in [0, T]} \Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }} ( \mu _{\omega } + \lambda ) (dr, dx, dv) \\&\qquad \times F_2 (s, r, x, v)1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)] \right\} } 1_{ \left\{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \le R_U + 1 \right\} } \Big | \Big ] \\&\quad \le 2 \int _0^T ds E^{P_m} \Big [ \int _{\mathbf{R} \times E} 16 n^2 m t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)^2 e^{ (1+C_1) (m^{-1/2} (s-r) + \tau ) } \\&\qquad \times 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)] \right\} } 1_{ \left\{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \le R_U + 1 \right\} } \lambda (dr, dx, dv) \Big ] \\&\quad \le C m^{1/2}. \end{aligned}$$

So (3.45) with $k = 2$ holds.

We next prove (3.45) with $k = 3$. In the considered domain, we have that $|X(s) - X({\tilde{r}}) - (s-{\tilde{r}}) V({\tilde{r}})| \le \int _{r - m^{1/2} \tau }^s |V(u) - V({\tilde{r}})| du$. Combining this with Lemma 3.21 (with s, t, $c_m$ and $b_m$ given by r, $m^{-1/2} (s-r)$, $m^{1/2} \tau$ and $t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)$, respectively) and (3.12), we get for $m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)]$ that

$$\begin{aligned}&F_3 (s, r, x, v) \\&\le 8 n^2 ( \Vert \nabla ^3 U \Vert _{\infty } \vee 1 ) ( 2 C_1 \vee 1 ) e^{ \left( 1+\frac{a}{2}\right) (1+C_1) (t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |) + \tau ) } \\&\qquad \times m^{\frac{1+a}{2}} (t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |) + \tau )^{1+a} \\&\qquad + 2 e^{ \frac{1}{2} (1+C_1) (t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |) + \tau ) } \int _{r - m^{1/2} \tau }^s |V(u) - V({\tilde{r}})| du. \end{aligned}$$

Choose $a \in (0, 1)$ such that $(1+\frac{a}{2}) (1+C_1) 2 \varepsilon _1^{-1/2} < d - 1$. Then by Lemma 4.5 (1), the integral with respect to the first term on the right hand side above has order $O (m^{a/2})$, hence converges to 0 as $m \rightarrow 0$. We deal with the second term on the right hand side above in the following.

By Lemma 3.57 and Schwarz inequality, we have that

$$\begin{aligned}&E^{P_m} \Big [ \sup _{t \in [0, T]} \Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) \int _{\mathbf{R} \times E, |x-\pi ^{\perp }_v X({\tilde{r}})| \ge m^{\alpha }} e^{ \frac{1}{2} (1+C_1) (t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |) + \tau ) }\\&\qquad \times \int _{r - m^{1/2} \tau }^s |V(u) - V({\tilde{r}})| du 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)] \right\} }\\&\qquad \times 1_{ \left\{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \le R_U + 1 \right\} } ( \mu _{\omega } + \lambda ) (dr, dx, dv) \Big | \Big ] \\&\quad \le 2 m^{1/2} a_m \int _0^T ds E^{P_m} \Big [ 1_{ \{ m^{1/2} a_m \le s \le \sigma _n \} } \sup _{u \in [s - m^{1/2} a_m, s]} |V (s-m^{1/2} a_m) - V(u)|^2 \Big ]^{1/2} \\&\qquad \times E^{P_m} \Big [ \Big ( \int _{\mathbf{R} \times E} e^{ \frac{1}{2} (1+C_1) (t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |) + \tau ) } \\&\qquad \times 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, | x - \pi ^{\perp }_v X({\tilde{r}}) |)] \right\} } 1_{ \left\{ | x - \pi ^{\perp }_v X({\tilde{r}}) | \le R_U + 1 \right\} } ( \mu _{\omega } + \lambda ) (dr, dx, dv) \Big )^2 \Big ]^{1/2}. \end{aligned}$$

Since $(1+C_1) 2 \varepsilon _1^{-1/2} < d - 1$, by Lemma 3.58, (3.30) and Lemma 4.5 (1), the right hand side above is dominated by $2 m^{1/2} a_m T (C_{23} m^{\varepsilon })^{1/2} \Big ( 2 C m^{-1/2} + 8 (C m^{-1/2})^2 \Big )^{1/2}$, which converges to 0 as $m \rightarrow 0$. This completes the proof of (3.45) with $k = 3$, and in turn completes the proof of Lemma 3.59. $\square$

In the second half of this section, we prove that $I^9$ is also negligible. This will complete our proof of Theorem 1.2.

Lemma 3.60

We have that

$$\begin{aligned} \lim _{m \rightarrow 0} E^{P_m} \Big [ \sup _{t \in [0, T]} |I^9 (t)| \Big ] = 0. \end{aligned}$$

Proof

Let $B_1$, $B_2 ( \cdot )$ and $B_3 ( \cdot )$ be as defined in (3.37). By Lemma 3.45, it suffices to estimate the integrals of $f_1$ on $B_1$, $B_2 (s)$, and the integral of $f_4 - f_1$ on $B_3 (s)$. Precisely, we prove the following (3.46) and (3.47):

$$\begin{aligned}&E^{P_m} \left[ \sup _{t \in [0, T]} \Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{B_1 \cup B_2 (s)} f_1 (s, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \right] \rightarrow 0, \nonumber \\\end{aligned}$$

(3.46)

$$\begin{aligned}&E^{P_m} \left[ \sup _{t \in [0, T]} \Big | \int _0^{t \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{B_3 (s) } ( f_4 - f_1 ) (s, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \right] \rightarrow 0.\nonumber \\ \end{aligned}$$

(3.47)

We first prove (3.46). If $\nabla U^2 (X(s) - \psi ^0 (m^{-1/2} (s-r), x, v; X(s)) \ne 0$, then by Proposition 2.1 (2), we have that $m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(s)|)]$. By Lemma 3.44, this combined with $|x - \pi ^{\perp }_v X({\tilde{r}})| \le m^{\alpha }$ implies $|x - \pi ^{\perp }_v X(s)| \le 2 m^{\alpha }$. Therefore, applying Lemma 3.38 (with ${\overline{t}}$ and $c_m$ given by $t_1 (v, |x - \pi ^{\perp }_v X(s)|)$), with the help of (3.12), we get the following:

$$\begin{aligned}&\Big | \int _{\mathbf{R} \times E} 1_{B_1 \cup B_2 (s)} (r, x, v) f_1 (s, r, x, v) {\overline{\lambda }} (dr, dx, dv) \Big | \\&\quad \le m^{1/2} C_1 4 n \int _{\mathbf{R} \times E} 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(s)|)] \} } t_1 (v, |x-\pi ^{\perp }_v X(s)|) \\&\qquad e^{\frac{1}{2} (1+C_1) (t_1 (v, |x-\pi ^{\perp }_v X(s)|) + \tau ) } 1_{ \{ |x - \pi ^{\perp }_v X(s)| \le 2 m^{\alpha } \} } {\overline{\lambda }} (dr, dx, dv), \end{aligned}$$

which, by Lemma 4.5 (3), is dominated by $C_1 4 n \cdot C m^{\alpha (d - 1 - (1+C_1) \varepsilon _1^{-1/2}) } (\log (m^{-1}) )^2$. Since $d - 1 > (1+C_1) \varepsilon _1^{-1/2}$, this converges to 0 as $m \rightarrow 0$. This implies (3.46).

Finally, we prove (3.47). Let $t_3 := t_3 (s, r, x, v) := t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) \vee t_1 (v, |x-\pi ^{\perp }_v X(s)|)$. Then on $B_3 (s)$, by (2.14) we have for any $m \le (2 \varepsilon _3)^{1/\alpha }$ that $t_3 \le a_m - \tau$. If $(f_4 - f_1) (s, r, x, v) \ne 0$, then by Proposition 2.1 (2) (4), we get that $|x| \le R_0$ and $m^{-1/2} (s-r) \le t_3$, which combined with $s \ge m^{1/2} a_m$ implies that $r - m^{1/2} \tau \ge 0$. Also, by exactly the same argument as that we used for (3.44), we get that

$$\begin{aligned}&| (f_4 - f_1) (s, r, x, v) | \\&\quad \le \frac{1}{2} \Vert \nabla ^3 U \Vert _{\infty } \Big | x \big (s, \Psi (r, x, m^{-1/2} v)\big ) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X(s)\big ) \Big |^2 \\&\qquad + \Vert \nabla ^2 U \Vert _{\infty } \Big | x \big (s, \Psi (r, x, m^{-1/2} v)\big ) - \psi ^0 \big (m^{-1/2} (s-r), x, v; X(s)\big ) \\&\qquad - m^{1/2} z \big (m^{-1/2} (s-r), x, v; X(s), V(s), - m^{-1/2}(s-r)\big ) \Big |. \end{aligned}$$

As same as in the proof of Lemma 3.59, choose $a \in (0, 1)$ such that $(1+\frac{a}{2}) (1+C_1) 2 \varepsilon _1^{-1/2} < d - 1$. By Lemmas 3.20 and 3.21, we get that

$$\begin{aligned}&| (f_4 - f_1) (s, r, x, v) | \\&\quad \le 1_{ \{ |x| \le R_0 \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_3] \} } \\&\qquad \times \Big ( \frac{1}{2} \Vert \nabla ^3 U \Vert _{\infty } \Big \{ \frac{2 C_1 n}{1+C_1} m^{1/2} 2 t_3 \Big ( e^{\frac{1}{2} (1+C_1) (t_3 + \tau )} - 1 \Big ) \Big \}^2 \\&\qquad + \Vert \nabla ^2 U \Vert _{\infty } \Big \{ \frac{8 n^2}{1+C_1} (2C_1 \vee 1) (\Vert \nabla ^3 U \Vert _{\infty } \vee 1) e^{\left( 1+\frac{a}{2}\right) (1 + C_1) ( t_3 + \tau )} \Big ( m^{1/2} 2 t_3 \Big )^{1+a} \\&\qquad + \frac{2 C_1}{1+C_1} \Big ( e^{\frac{1}{2} (1 + C_1) (t_3 + \tau )} - 1 \Big ) \int _{r - m^{1/2} \tau }^s |V(u) - V(s)| du \Big \} \Big ). \end{aligned}$$

So with some proper constant $C_{24}$, we have that

$$\begin{aligned}&| (f_4 - f_1) (s, r, x, v) | \le 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_3] \} } 1_{ \{ |x| \le R_0 \} } C_{24} \\&\qquad \times \left( m t_3^2 e^{(1+C_1) t_3 } + m^{\frac{1+a}{2}} t_3^2 e^{(1+\frac{a}{2}) (1+C_1) t_3 } + e^{\frac{1}{2} (1+C_1) t_3 } \int _{r - m^{1/2} \tau }^s |V(u) - V(s)| du \right) . \end{aligned}$$

The integral corresponding to $m t_3^2 e^{(1+C_1) t_3 }$ converges to 0 by Lemma 4.5 (1), by the virtue of (2.8). Therefore, in order to prove (3.47), it suffices to prove the following (3.48) $\sim$ (3.51).

$$\begin{aligned}&m^{\frac{1+a}{2}} E^{P_m} \Big [ \int _0^{T \wedge \sigma _n} ds \int _{\mathbf{R} \times E} 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \right\} } 1_{ \{ |x| \le R_0 \} } \nonumber \\&\qquad \times t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)^2 e^{\left( 1+\frac{a}{2}\right) (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) } {\overline{\lambda }} (dr, dx, dv) \Big ] \rightarrow 0, \end{aligned}$$

(3.48)

$$\begin{aligned}&m^{\frac{1+a}{2}} E^{P_m} \Big [ \int _0^{T \wedge \sigma _n} ds \int _{\mathbf{R} \times E} 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(s)|)] \right\} } 1_{ \{ |x| \le R_0 \} } \nonumber \\&\qquad \times t_1 (v, |x-\pi ^{\perp }_v X(s)|)^2 e^{\left( 1+\frac{a}{2}\right) (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X(s)|) } {\overline{\lambda }} (dr, dx, dv) \Big ] \rightarrow 0, \end{aligned}$$

(3.49)

$$\begin{aligned}&E^{P_m} \Big [ \int _0^{T \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} 1_{ \left\{ |x - \pi ^{\perp }_v X({\tilde{r}})| > m^{\alpha } \right\} } \nonumber \\&\qquad \times 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|)] \right\} } 1_{ \{ |x| \le R_0 \} } e^{\frac{1}{2} (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) } \nonumber \\&\qquad \times \int _{r - m^{1/2} \tau }^s |V(u) - V(s)| du {\overline{\lambda }} (dr, dx, dv) \Big ] \rightarrow 0, \end{aligned}$$

(3.50)

$$\begin{aligned}&E^{P_m} \Big [ \int _0^{T \wedge \sigma _n} 1_{(m^{1/2} a_m, \infty )} (s) ds \int _{\mathbf{R} \times E} 1_{ \{ |x - \pi ^{\perp }_v X(s)| > m^{\alpha } \} } \nonumber \\&\qquad \times 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(s)|)] \right\} } 1_{ \{ |x| \le R_0 \} } e^{\frac{1}{2} (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X(s)|) } \nonumber \\&\qquad \times \int _{r - m^{1/2} \tau }^s |V(u) - V(s)| du {\overline{\lambda }} (dr, dx, dv) \Big ] \rightarrow 0. \end{aligned}$$

(3.51)

Since $(1+\frac{a}{2}) (1+C_1) 2 \varepsilon _1^{-1/2} < d - 1$, (3.48) and (3.49) are direct consequences of Lemma 4.5 (1).

As for (3.50) and (3.51), by 2.14, we have for any $m \le (2 \varepsilon _3)^{1/\alpha }$ that $|x - \pi ^{\perp }_v X({\tilde{r}})| > m^{\alpha }$ implies $t_1 (v, |x-\pi ^{\perp }_v X(s)| \le a_m - \tau$, so if $m^{-1/2} (s-r) \le t_1 (v, |x-\pi ^{\perp }_v X(s)|$ in addition, then $r - m^{1/2} \tau \ge s - m^{1/2} a_m$. Also, we have that $(1+C_1) \varepsilon _1^{-1/2} < d - 1$. Therefore, by Lemmas 3.57 and 4.5 (1), both of the left hand sides of (3.50) and (3.51) are dominated by

$$\begin{aligned} C E^{P_m} \left[ \int _0^T 1_{ \{ m^{1/2} a_m \le s \le \sigma _n \} } a_m \sup _{ u \in [s - m^{1/2} a_m, s] } |V(u) - V(s - m^{1/2} \tau )| ds \right] . \end{aligned}$$

This combined with Lemma 3.58 implies (3.50) and (3.51). $\square$

4 Concluding Remarks

For a Rayleigh gas model with a single massive particle and a repulsive potential, we presented a mathematical proof that the motion of the massive particle converges to a Brownian motion when the mass m of the light particles converges to 0, while their number density and velocity distribution scale like $m^{-\frac{1}{2}}$, under the assumption that the initial kinetic energies of the light particles are bounded below. We do not need any assumption requiring that the initial velocities of the environmental particles should be restricted to be “fast enough”. As a result, the interaction time durations between the massive particle and the light particles are unbounded.

Here are several possible generalizations that are more physically relevant:

• 1. A model with no minimum constraint on the initial kinetic energies of the gas particles. In such a model, besides the possibly small initial skews between particles as discussed in this paper, the possibly slow initial velocities of the light particles is also a source of singularity. In other words, even for a freezing approximation particle, when the initial velocity v converges to 0, the effective interaction time duration diverges to infinity. Moreover, a careful calculation suggests that this divergence is much worse than a log order. So it is hopeless to remove this singularity by simply applying the method that we used in this paper (i.e., the introduction of $\alpha$ in this paper). Some new estimation is necessary. A generalization in this direction is now in progress by the author.

• 2. Lowering the dimensionality $d \le 5$. As explained in Remark 1.3, the method of this paper is not applicable to the low dimensional case. When $d \le 3$, we are even not able to prove that the generator L of the limiting process is well-defined. A more accurate estimate will be necessary.

• 3. A model with interaction potential that diverges to infinity as the inter-particle distance converges to 0. For example, the Weeks-Chandler-Andersen potential or the Lennard-Jones potential. In such a model, the second derivative of the potential function is not bounded, so estimations with respect to the effective interaction time durations do not imply directly estimations with respect to the distance between the light particles and our freezing approximations; and estimations with respect to the distances between the particles do not imply directly the estimations with respect to the differences between the corresponding forces. It is expected that a method similar to the one used for low initial kinetic energies (i.e., the first problem formulated in this remark) might be effective. This is also planned as a forthcoming topic of our research.

Appendix

We give the proof of (2.6) and (2.7) in Sect. A.1; give the proof of Lemma 3.33 in Sect. A.2; present several estimates that are used to prove that $I^1 (t)$ and $I^2 (t)$ converge to 0 fast enough in Sect. A.3; finally, in Sect. A.4, we present several necessary estimates with respect to integrals involving $t_1 (v; *)$.

1.1 A.1 Proof of (2.6) and (2.7)

By a simple calculation, we have that $\nabla _i \nabla _j U (x) = \Big ( h''(|x|) - \frac{h'(|x|)}{|x|} \Big ) \frac{x_i}{|x|} \frac{x_j}{|x|} + \delta _{ij} \frac{h'(|x|)}{|x|}$ for any $i, j \in \{ 1, \cdots , d \}$, so $\nabla ^2 U(x) y = \Big ( h''(|x|) - \frac{h'(|x|)}{|x|} \Big ) (y, \frac{x}{|x|}) \frac{x}{|x|} + \frac{h'(|x|)}{|x|} y = h''(|x|) \pi _x y + \frac{h'(|x|)}{|x|} \pi ^{\perp }_x y$. Since $|\frac{h'(|x|)}{|x|}| \le \Vert h'' \Vert _{\infty }$, this implies (2.6).

Also, (2.7) is an easy consequence of (2.6) since $| \nabla U (y_1) - \nabla U (y_2) | = \int _0^1 \nabla ^2 U(y_2 + \theta (y_1-y_2)) (y_1-y_2) d \theta$.

1.2 A.2 Proof of Lemma 3.33

We give the proof of Lemma 3.33 in this section. We first notice the following translation property (Lemma 4.1) and symmetry (Lemma 4.2) of $\varphi$, which are clear heuristically.

Lemma 4.1

For any $(x, v) \in \mathbf{R}^{2d}$, $t \ge 0$ and $X \in \mathbf{R}^d$, we have that

$$\begin{aligned} \varphi ^0 (t, x, v; X) - X = \varphi ^0 (t, x-X, v; \mathbf{0}). \end{aligned}$$

Proof

Just notice that both sides above satisfy the same ODE with the same initial conditions. $\square$

Lemma 4.2

For any $X \in \mathbf{R}^d$ and $(x, v) \in E$, let $\iota (x, v; X) := \Big ( \iota ^0 (x, v; X), \iota ^1 (x, v; X) \Big ) := (2X-x, -v)$. Then we have the following.

$$\begin{aligned} \varphi \big (t, \iota (x, v; X); X\big ) = \iota \big ( \varphi (t, x, v; X); X\big ), \qquad t \ge 0. \end{aligned}$$

Proof

First we notice that $\frac{d}{dt} \varphi ^0 (t, \iota (x, v; X); X) = \varphi ^1 (t, \iota (x, v; X); X)$ and $\frac{d}{dt} \iota ^0 ( \varphi (t, x, v; X); X) = \frac{d}{dt} \Big ( 2X - \varphi ^0 (t, x, v; X) \Big ) = - \varphi ^1 (t, x, v; X) = \iota ^1 ( \varphi (t, x, v; X); X)$. We have by definition that $(\varphi ^0 (0, \iota (x, v; X); X), \varphi ^1 (0, \iota (x, v; X); X) ) = (2X-x, -v) = \iota ( \varphi (0, x, v; X); X).$ So $\varphi (t, \iota (x, v; X); X)$ and $\iota ( \varphi (t, x, v; X); X)$ satisfy the same initial condition. Also,

$$\begin{aligned} \frac{d^2}{dt^2} \varphi ^0 (t, \iota (x, v; X); X) = - \nabla U ( \varphi ^0 (t, \iota (x, v; X); X) - X ), \end{aligned}$$

and

$$\begin{aligned}&\frac{d^2}{dt^2} \iota ^0 ( \varphi (t, x, v; X); X) = \frac{d^2}{dt^2} \Big ( 2X - \varphi ^0 (t, x, v; X) \Big ) \\&\quad = \nabla U \big ( \varphi ^0 (t, x, v; X) - X \big ) = - \nabla U \big ( (2X - \varphi ^0 (t, x, v; X) ) - X \big ) \\&\quad = - \nabla U \big ( \iota ^0 ( \varphi (t, x, v; X); X) - X \big ). \end{aligned}$$

So $\varphi ^0 (t, \iota (x, v; X); X)$ and $\iota ^0 ( \varphi (t, x, v; X); X)$ satisfy the same ODE, too. Therefore, $\varphi ^0 (t, \iota (x, v; X); X) = \iota ^0 ( \varphi (t, x, v; X); X)$, which implies our assertion. $\square$

Now we are ready to prove the first assertion of Lemma 3.33.

Proof of Lemma 3.33 (1)

We first notice that the left hand side of (3.27) does not depend on $X_0$. Indeed, by definition and Lemma 4.1, we have that

$$\begin{aligned}&\hbox {(LHS) of }(3.27) \\&\quad = \int _{\mathbf{R}^d} |v| dv \int _\mathbf{R} dr \int _{E_v} \nabla U \left( - \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r v - X_0, v, \mathbf{0})\right) \\&\qquad \times m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v - X_0\right) {\tilde{\nu }} (dx; v). \end{aligned}$$

Writing $x - m^{-1/2} r v - X_0$ as $x - \pi ^{\perp }_v X_0 - m^{-1/2} ( r + m^{1/2} \frac{(X_0, v)}{|v|^2} ) v$, by change of variable $x - \pi ^{\perp }_v X_0 \rightarrow x$, $r + m^{1/2} \frac{(X_0, v)}{|v|^2} \rightarrow r$ for any fix v, this gives us that

$$\begin{aligned}&\hbox {(LHS) of } (3.27) \nonumber \\&\quad = \int _{\mathbf{R}^d} |v| dv \int _\mathbf{R} dr \int _{E_v} \nabla U \left( - \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r v, v, \mathbf{0})\right) \nonumber \\&\qquad \times m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v\right) {\tilde{\nu }} (dx; v), \end{aligned}$$

(5.1)

the right hand side of which does not depend on $X_0$.

Let J denote the right hand side of (5.1). So it suffices to prove that $J=0$. By Lemma 4.2 with $X=0$, we have that $- \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r, v, \mathbf{0}) = \varphi ^0 (m^{-1/2} s, -x + m^{-1/2} r, -v, \mathbf{0})$. Substituting this into the definition of J, since $\rho (u, - z) = \rho (u, z)$ for any $(u, z) \in [0, \infty ) \times \mathbf{R}^d$ by (A2), we get by change of variable $x \rightarrow -x$ and $v \rightarrow -v$ that

$$\begin{aligned} J= & {} \int _{\mathbf{R}^d} |v| dv \int _\mathbf{R} dr \int _{E_v} \nabla U ( \varphi ^0 (m^{-1/2} s, x - m^{-1/2} r v, v, \mathbf{0})) \\&\qquad \times m^{-1} \rho \left( \frac{1}{2} |v|^2, x - m^{-1/2} r v\right) {\tilde{\nu }} (dx; v) \\= & {} - J, \end{aligned}$$

hence $J=0$. $\square$

The second assertion of Lemma 3.33 is proved similarly. We first notice the following property of $\psi$.

Lemma 4.3

For any $(x, v) \in E$, $u \in \mathbf{R}$ and $X \in \mathbf{R}^d$,

• 1. $\psi ^0 (u, x, v; X) - X = \psi ^0 (u - \frac{(X, v)}{|v|^2}, x - \pi ^{\perp }_v X, v; \mathbf{0})$.

• 2. $\psi ^0 (u, -x, -v; \mathbf{0}) = - \psi ^0 (u, x, v; \mathbf{0})$.

Proof

(1) By definition and Lemma 4.1, we have that

$$\begin{aligned}&\psi ^0 (u, x, v; X) - X = \lim _{s \rightarrow \infty } \varphi ^0 (u+s, x - s v, v; X) - X \\&\quad = \lim _{s \rightarrow \infty } \varphi ^0 (u+s, x - s v - X, v; \mathbf{0}) \\&\quad = \lim _{s \rightarrow \infty } \varphi ^0 \left( \Big ( u - \frac{(X, v)}{|v|^2} \Big ) + \Big ( s + \frac{(X, v)}{|v|^2} \Big ), x - \pi ^{\perp }_v X - \Big ( s + \frac{(X, v)}{|v|^2} \Big ) v, v; \mathbf{0}\right) \\&\quad = \psi ^0 \left( u - \frac{(X, v)}{|v|^2}, x - \pi ^{\perp }_v X, v; \mathbf{0} \right) . \end{aligned}$$

(2) By definition and Lemma 4.2, we have that

$$\begin{aligned}&\psi ^0 (u, -x, -v; \mathbf{0}) = \lim _{s \rightarrow \infty } \varphi ^0 (u+s, -x + s v, - v; \mathbf{0}) \\&\quad = - \lim _{s \rightarrow \infty } \varphi ^0 (u+s, x - s v, v; \mathbf{0}) = - \psi ^0 (u, x, v; \mathbf{0}). \end{aligned}$$

$\square$

Proof of Lemma 3.33 (2)

By Lemma 4.3 (1) and change of variable $u - \frac{(X, v)}{|v|^2} \rightarrow u$, $x - \pi ^{\perp }_v X \rightarrow x$ for any fixed $v \in \mathbf{R}^d$, we have that

$$\begin{aligned}&\hbox {(LHS) of }(3.28) \\&\quad = \int _{\mathbf{R} \times E} \nabla U \left( - \psi ^0 \Big ( u - \frac{(X, v)}{|v|^2}, x - \pi ^{\perp }_v X, v; \mathbf{0} \Big ) \right) m^{-1} \rho _0 \Big (\frac{1}{2} |v|^2\Big ) du \nu (dx, dv) \\&\quad = \int _{\mathbf{R} \times E} \nabla U \big (- \psi ^0 (u, x, v; \mathbf{0}) \big ) m^{-1} \rho _0 \Big (\frac{1}{2} |v|^2\Big ) du \nu (dx, dv). \end{aligned}$$

Let J denote the right hand side above. Then by Lemma 4.3 (2) and change of variable $x \rightarrow -x$, $v \rightarrow - v$, we get that $J = - J$, hence $J=0$. $\square$

1.3 A.3 Estimates for $I^1 (t)$ and $I^2 (t)$

Lemma 4.4 is used in the estimates with respect to $I^1 (t)$ and $I^2 (t)$ (see Lemmas 3.35 and 3.36).

Lemma 4.4

For any $A>0$, there exists a constant $C>0$ such that for any $B_m \ge 0$ ($m \in (0, 1]$), we have that

• 1.
  $$\begin{aligned} \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|)^2 \lambda _m (dr, dx, dv) \le C_9 m^{-1} B_m, \end{aligned}$$
• 2.
  $$\begin{aligned} E^{P_m} \left[ \Big ( \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|) \mu _{\omega } (dr, dx, dv) \Big )^2 \right] \le C_9 (m^{-2} B_m^2 + 1). \end{aligned}$$

Proof

Let $C := 2 (2 A)^{d-1} \int _{\mathbf{R}^d} (1+|v|^2) \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv$. Then C is finite by assumption. We have that

$$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|^2) \lambda _m (dr, dx, dv) \\&\quad \le \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|^2) m^{-1} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dr {\tilde{\nu }} (dx; v) dv \\&\quad \le C m^{-1} B_m, \end{aligned}$$

which gives us our first assertion.

Therefore, by (3.29), we get that

$$\begin{aligned}&E^{P_m} \left[ \Big ( \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|) \mu _{\omega } (dr, dx, dv) \Big )^2 \right] \\&\le 2 \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|)^2 \lambda _m (dr, dx, dv) \\&\qquad + 2 \left( \int _{\mathbf{R} \times E} 1_{[0, A]} (|x|) 1_{ \{ |r| \le B_m \} } (1+|v|) \lambda _m (dr, dx, dv) \right) ^2 \\&\le 2 C m^{-1} B_m + 2 (C m^{-1} B_m)^2. \end{aligned}$$

So we get our second assertion by re-choosing C in an appropriate way. $\square$

1.4 A.4 Several Estimates with Respect to Integrals Involving $t_1 (v, *)$

As seen, the valid interaction time of each light particle is $t_1 (v, |x-\pi ^{\perp }_v X({\tilde{r}})|) + \tau$, which is not bounded with respect to (x, v). However, by Lemma 4.5, this does not cause any problem after taking integrals.

Lemma 4.5

Fix any $T>0$ and $n \in \mathbf{N}$. Then there exists a constant C such that

• 1. for any $\beta > 0$ such that $2 \varepsilon _1^{-1/2} \beta < d - 1$, $u_0 \in \mathbf{R}$, $k \in \{ 0, 1, 2 \}$, $s \in [0, T \wedge \sigma _n]$ and any $u (r, s) \in [0, T \wedge \sigma _n]$,

  $$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{ \{ |x| \le R_0 \} } 1_{ \left\{ m^{-1/2} (s-r) - u_0 \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \right\} } t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|)^k \\&\qquad \exp \Big ( \beta t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \Big ) (\lambda + {\overline{\lambda }}) (dr, dx, dv) \le C m^{-1/2}, \end{aligned}$$

• 2. for any $-\infty< c_1< c_2 < \infty$ (which may depend on m),

  $$\begin{aligned} \int _{\mathbf{R} \times E} 1_{ \left\{ |x-\pi ^{\perp }_v X({\tilde{r}})| \le m^{\alpha } \right\} } 1_{ \left\{ r \in [c_1, c_2] \right\} } (|v|+1)^2 (\lambda + {\overline{\lambda }}) (dr, dx, dv) \le C m^{\alpha (d-1) - 1} (c_2-c_1), \end{aligned}$$

• 3. for any $s \in [0, T \wedge \sigma _n]$ and $u (r, s) \in [0, T \wedge \sigma _n]$,

  $$\begin{aligned}&\int _{\mathbf{R} \times E} 1_{ \left\{ |x-\pi ^{\perp }_v X(u (r, s))| \le 2 m^{\alpha } \right\} } 1_{ \left\{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \right\} } (\lambda + {\overline{\lambda }}) (dr, dx, dv) \\&\qquad \times t_1 \big (v, |x-\pi ^{\perp }_v X(u (r, s))|\big )^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \ell \right) \\&\qquad \le C m^{\alpha \Big ( (d-1 - \ell (1+C_1) \epsilon _1^{-1/2} \Big ) - \frac{1}{2}} \Big ( \log m^{-1} \Big )^{k+1}, \quad k, \ell \in \{ 0, 1 \}, \end{aligned}$$
• 4.
  $$\begin{aligned} \int _E t_1 (v, |x-\pi ^{\perp }_v X|)^k 1_{ \left\{ |x-\pi ^{\perp }_v X| \le R_U + 1 \right\} } \rho \left( \frac{1}{2} |v|^2\right) \nu (dx, dv) \le C, \quad k \in \{ 0, 1, \cdots , 7 \}. \end{aligned}$$

Proof

The proofs of these assertions are similar – first apply the change of variable $y = x - \pi ^{\perp }_v X (\cdot )$ for any fixed v and r, then apply Fubini’s Theorem to switch the order of integrations with respect to y and r if necessary, and finally estimate the integral with respect to x. We present the proofs of (1) and (3) in the following. The proofs of (2) and (4) are similar but simpler, since we do not need Fubini’s Theorem in these proofs, and we omit them here.

(1) By definition and (A3), we get that the left hand side of (1) is dominated by

$$\begin{aligned}&m^{-1} \int _{\mathbf{R}^d} 2 \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dx; v) \\&\qquad \times 1_{ \left\{ |x-\pi ^{\perp }_v X(u (r, s))| \le 2 R_0 \right\} } 1_{ \left\{ m^{-1/2} (s-r) - u_0 \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \right\} } \\&\qquad \times t_1 \left( v, |x-\pi ^{\perp }_v X(u (r, s))|\right) ^k \exp \left( \beta t_1 \big (v, |x-\pi ^{\perp }_v X(u (r, s))|\big ) \right) . \end{aligned}$$

Applying the change of variable $y = x-\pi ^{\perp }_v X(u (r, s))$ for any fixed v and r, this is equal to

$$\begin{aligned}&2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dy; v) \\&\qquad 1_{ \{ |y| \le 2 R_0 \} } 1_{ \{ m^{-1/2} (s-r) - u_0 \in [-\tau , t_1 (v, |y|) \} } t_1 (v, |y|)^k \exp \Big ( \beta t_1 (v, |y|) \Big ). \end{aligned}$$

Applying Fubini’s Theorem to switch the order of integrations with respect to x and r, since $t_1 (v, *) \ge \tau$, this is dominated by

$$\begin{aligned} 4 m^{-1/2} \int _{\mathbf{R}^d} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _{E_v} 1_{ \{ |y| \le 2 R_0 \} } t_1 (v, |y|)^{k+1} \exp \Big ( \beta t_1 (v, |y|) \Big ) {\tilde{\nu }} (dy; v). \end{aligned}$$

So in order to prove (1), it suffices to prove that

$$\begin{aligned} \sup _{v \in \mathbf{R}^d} \int _{E_v} 1_{ \{ |y| \le 2 R_0 \} } t_1 (v, |y|)^{k+1} \exp \Big ( \beta t_1 (v, |y|) \Big ) {\tilde{\nu }} (dy; v) < \infty . \end{aligned}$$

(5.2)

By the definition of $t_1 (v, |y|)$ and the help of polar coordinates, it in turn suffice to prove that

$$\begin{aligned} \int _{\left[ 0, (2 R_0) \wedge \frac{R_U}{2} \wedge (2 \varepsilon _3)\right] } r^{- \beta 2 \varepsilon _1^{-1/2} + d - 2} \Big ( \log \frac{1}{r} \Big )^{k+1} dr < \infty . \end{aligned}$$

(5.3)

On the other hand, by (2.12), we have that (5.3) holds as long as $- \beta 2 \varepsilon _1^{-1/2} + d - 2 > -1$. This completes our proof of (1).

(3) By definition and (A3), we get that the left hand side of (3) is dominated by

$$\begin{aligned}&2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \Big ( \frac{1}{2} |v|^2 \Big ) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dx; v) \\&\qquad \times 1_{ \{ |x-\pi ^{\perp }_v X(u (r, s))| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \} } \\&\qquad \times t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|)^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |x-\pi ^{\perp }_v X(u (r, s))|) \ell \right) . \end{aligned}$$

Applying the change of variable $y = x-\pi ^{\perp }_v X(u (r, s))$ for any fixed v and r, this is equal to

$$\begin{aligned}&2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \left( \frac{1}{2} |v|^2 \right) |v| dv \int _\mathbf{R} dr \int _{E_v} {\tilde{\nu }} (dy; v) \\&\qquad \times 1_{ \{ |y| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |y|) \} } t_1 (v, |y|)^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |y|) \ell \right) \\&\quad = 2 m^{-1} \int _{\mathbf{R}^d} \rho _{max} \left( \frac{1}{2} |v|^2 \right) |v| dv \int _{E_v} {\tilde{\nu }} (dy; v) \int _\mathbf{R} dr \\&\qquad \times 1_{ \{ |y| \le 2 m^{\alpha } \} } 1_{ \{ m^{-1/2} (s-r) \in [-\tau , t_1 (v, |y|) \} } t_1 (v, |y|)^k \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |y|) \ell \right) \\&\le 4 m^{-\frac{1}{2}} \int _{\mathbf{R}^d} \rho _{max} \left( \frac{1}{2} |v|^2 \right) |v| dv \int _{E_v} {\tilde{\nu }} (dy; v) \\&\qquad \times 1_{ \{ |y| \le 2 m^{\alpha } \} } t_1 (v, |y|)^{k+1} \exp \left( \frac{1}{2} (1+C_1) t_1 (v, |y|) \ell \right) . \end{aligned}$$

It suffices to consider the case where $m>0$ is small enough such that $2 m^{\alpha } \le \frac{R_U}{2} \wedge (2 \varepsilon _3)$. So by the definition of $t_1 (v, |y|)$, it suffices to prove the following:

$$\begin{aligned}&\sup _{v \in \mathbf{R}^d} \int _{E_v} {\tilde{\nu }} (dy; v) 1_{ \{ |y| \le 2 m^{\alpha } \} } (\log |y|^{-1})^{k+1} \exp \left( \frac{1}{2} \ell (1+C_1) 2 \epsilon _1^{-1/2} \log \frac{2 \varepsilon _3}{|y|} \right) \nonumber \\&\le C m^{\alpha \Big ( (d-1 - \ell (1+C_1) \epsilon _1^{-1/2} \Big )} \Big ( \log m^{-1} \Big )^{k+1}. \end{aligned}$$

(5.4)

On the other hand, by changing variables to polar coordinates, the left hand side of (5.4) is equal to

$$\begin{aligned} C \int _0^{2 m^{\alpha }} r^{d-2} r^{- \ell (1+C_1) \epsilon _1^{-1/2}} \Big ( \log r^{-1} \Big )^{k+1} dr. \end{aligned}$$

For any fixed $\ell \in \{ 0, 1 \}$, we have that $d - 1 - \ell (1+C_1) \epsilon _1^{-1/2} > 0$. Now we get our assertion by the well-known result that for any $p \ne -1$,

$$\begin{aligned}&\int r^p \log r d r = \frac{r^{p+1}}{p+1} \Big ( \log r - \frac{1}{p+1} \Big ), \\&\int r^p \Big ( \log r \Big )^2 d r = \frac{r^{p+1}}{p+1} \left( \Big ( \log r \Big )^2 - \frac{2}{p+1} \log r + \frac{2}{(p+1)^2} \right) . \end{aligned}$$

$\square$