An invariance principle for the stochastic heat equation

Abstract

We approximate the white-noise driven stochastic heat equation by replacing the fractional Laplacian by the generator of a discrete time random walk on the one dimensional lattice, and approximating white noise by a collection of i.i.d. mean zero random variables. As a consequence, we give an alternative proof of the weak convergence of the scaled partition function of directed polymers in the intermediate disorder regime, to the stochastic heat equation; an advantage of the proof is that it gives the convergence of all moments.

Appendices

Appendix: A local limit theorem

The following local limit theorem is a modification of Proposition 3.3 in [13]. Below $p_t$ is the density for the Stable($\alpha$) process with generator $-\nu (-\Delta )^{\alpha /2}$.

Theorem A.1

Suppose Assumption 1.1 holds. Then for any $0\leqslant b\leqslant 1$ and any $c>0$

$$\begin{aligned}&\sup _{k \in \mathbf {Z}} \;\sup _{|x-(k-\mu [nt])|\leqslant cn^{(1-b)/\alpha }}\left| n^{\frac{1}{\alpha }} \mathrm {P}_{[nt]}(k) - p_{\frac{[nt]}{n}}\left( x n^{-\frac{1}{\alpha }}\right) \right| \nonumber \\&\quad \lesssim \frac{1}{n^{a/\alpha } t^{(1+a)/\alpha }} + \frac{1}{n^{b/\alpha }t^{2/\alpha }}, \end{aligned}$$

(A.1)

uniformly for $1/n\leqslant t\leqslant T$.

Proof

To simplify notation we denote $\tilde{t}=[nt]/n$. We first bound the expression on the left for $x=k-\mu [nt]$. Fourier inversion gives

$$\begin{aligned} \begin{aligned} p_{\tilde{t}}(x) = \frac{1}{2\pi } \int _{\mathbf {R}} e^{-\mathrm {i} xz} e^{-\nu \tilde{t}|z|^\alpha } \mathrm{d}z, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \mathrm {P}_{[nt]}(k) = \frac{1}{2\pi } \int _{-\pi }^{\pi } e^{-\mathrm {i} k z} \cdot \left[ \phi (z)\right] ^{[nt]}. \end{aligned}$$

Therefore

$$\begin{aligned}\begin{aligned}&(2\pi )\left| n^{\frac{1}{\alpha }} \mathrm {P}_{[nt]}(k) - p_{\tilde{t}}\left( (k-\mu [nt])n^{-\frac{1}{\alpha }}\right) \right| \\&\quad \leqslant \int _{ \left[ -\pi n^{1/\alpha },\pi n^{1/\alpha }\right] ^c}e^{-\nu \tilde{t}|z|^\alpha } \mathrm{d}z + \int _{-\pi n^{1/\alpha }}^{\pi n^{1/\alpha }} \left| e^{-\nu \tilde{t}|z|^\alpha }- \tilde{\phi }\left( \frac{z}{n^{1/\alpha }}\right) ^{[nt]}\right| \mathrm{d}z. \end{aligned} \end{aligned}$$

The first term can be bound just as term $I_2$ in Proposition 3.3 of [13], and we get

$$\begin{aligned} \int _{ \left[ -\pi n^{1/\alpha },\pi n^{1/\alpha }\right] ^c}e^{-\nu \tilde{t}|z|^\alpha } \mathrm{d}z \lesssim \frac{1}{n^{a/\alpha } t^{(1+a)/\alpha }}. \end{aligned}$$

For the second term we split the region of integration depending on whether or not z is in

$$\begin{aligned} A_{t,n} :=\Big \{z\in \mathbf {R}; \, |z|\leqslant \frac{n^{a/\{\alpha (a+\alpha )\}}}{t^{1/(a+\alpha )}} \Big \}. \end{aligned}$$

Our assumptions on the characteristic function $\phi$ imply

$$\begin{aligned} \Big \vert \tilde{\phi }\big (\frac{z}{n^{1/\alpha }}\big )\Big \vert \leqslant 1- c_1 \frac{|z|^\alpha }{n} \end{aligned}$$

on $|z|\leqslant \pi n^{1/\alpha }$, and so it follows as in [13] that

$$\begin{aligned} \int _{A_{t,n}^c\cap [-\pi n^{1/\alpha },-\pi n^{1/\alpha }]} \left| e^{-\nu \tilde{t}|z|^\alpha }- \tilde{\phi }\left( \frac{z}{n^{1/\alpha }}\right) ^{[nt]}\right| \, \mathrm{d}z \lesssim \frac{1}{n^{a/\alpha } t^{(1+a)/\alpha }}. \end{aligned}$$

We next need to consider the above integrand over the region $z \in A_{t,n}$. First observe

$$\begin{aligned} \begin{aligned} e^{-\nu \tilde{t}|z|^\alpha }- \tilde{\phi }\left( \frac{z}{n^{1/\alpha }}\right) ^{[nt]}&= e^{-\nu t|z|^\alpha } - \exp \left[ [nt] \log \tilde{\phi }\left( \frac{z}{n^{1/\alpha }}\right) \right] \\&=e^{-\nu t|z|^\alpha } \left\{ 1-\exp \left[ [nt] \log \tilde{\phi }\left( \frac{z}{n^{1/\alpha }}\right) +\nu \tilde{t}|z|^\alpha \right] \right\} \\&= e^{-\nu t|z|^\alpha } \left\{ 1-\exp \left[ [nt] \log \left( 1-\nu \frac{|z|^\alpha }{n}+\mathcal {D}\left( \frac{z}{n^{1/\alpha }}\right) \right) \right. \right. \\&\quad \left. \left. +\nu \tilde{t}|z|^\alpha \right] \right\} . \end{aligned} \end{aligned}$$

It is easy to see that $[nt]\mathcal {D}\big (z/n^{1/\alpha }\big )$ is bounded on $A_{t,n}$, and so

$$\begin{aligned}\begin{aligned} \int _{A_{t,n}\cap [-\pi n^{1/\alpha },-\pi n^{1/\alpha }]} \left| e^{-\nu \tilde{t}|z|^\alpha }- \tilde{\phi }\left( \frac{z}{n^{1/\alpha }}\right) ^{[nt]}\right| \, \mathrm{d}z&\lesssim \int _{\mathbf {R}} e^{-\nu \tilde{t}|z|^\alpha } \cdot (nt) \left| \frac{z}{n^{1/\alpha }}\right| ^{a+\alpha } \\&\lesssim \frac{1}{n^{a/\alpha } t^{(1+a)/\alpha }}. \end{aligned} \end{aligned}$$

We thus have the required bound in the case $x=k-\mu [nt]$. In the case of a general x such that $|x-(k-\mu [nt])|\leqslant n^{(1-b)/\alpha }$ we have

$$\begin{aligned}\begin{aligned}&\left| p_{\tilde{t}}(xn^{-\frac{1}{\alpha }})-p_{\tilde{t}}((k-\mu [nt])n^{-\frac{1}{\alpha }})\right| \\&\quad \leqslant \int _{\mathbf {R}} \left| e^{-\mathrm {i}z(k-\mu [nt])n^{-\frac{1}{\alpha }}}- e^{-\mathrm {i}zxn^{-\frac{1}{\alpha }} } \right| e^{-\nu \tilde{t}|z|^\alpha }\, \mathrm{d}z \\&\quad \lesssim \int _{\mathbf {R}} 1\wedge \frac{|z|}{n^{b/\alpha }} \cdot e^{-\nu \tilde{t}|z|^\alpha } \mathrm{d}z \\&\quad \leqslant \frac{1}{n^{b/\alpha }t^{2/\alpha }}\int _{|w|\leqslant n^{b/\alpha }t^{1/\alpha }} |w| e^{-\nu |w|^\alpha } \mathrm{d}w +\frac{1}{t^{1/\alpha }} \int _{|w|> n^{b/\alpha }t^{1/\alpha }} e^{-\nu |w|^\alpha }\mathrm{d}w\\&\quad \lesssim \frac{1}{n^{b/\alpha }t^{2/\alpha }}. \end{aligned} \end{aligned}$$

This completes the proof. $\square$

The local limit theorem has the following consequence.

Corollary A.2

Let Assumption 1.1 hold. Suppose $a_n$ is an integer valued sequence such that

$$\begin{aligned} \frac{a_n-\mu [nt]}{n^{1/\alpha }} \rightarrow a, \end{aligned}$$

as $n \rightarrow \infty$ for some constant a. Then for fixed $t>0$

$$\begin{aligned} \frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i=0}^{[nt]} \mathrm {P}_i(a_n) \rightarrow \int _0^t\frac{\mathrm{d}s}{s^{1/\alpha }}\, p_1\left( \frac{a}{s^{1/\alpha }}\right) \; \text {as } n \rightarrow \infty . \end{aligned}$$

Proof

We write

$$\begin{aligned} \begin{aligned} \frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i=1}^{[nt]} \mathrm {P}_i(a_n)&= \frac{1}{n}\sum _{i=1}^{[nt]} n^{\frac{1}{\alpha }}\mathrm {P}_i(a_n) \\&= \frac{1}{n} \sum _{i=1}^{[nt]} \left[ p_{\frac{i}{n}}\left( \frac{a_n-\mu [nt]}{n^{1/\alpha }}\right) + \frac{n^{1/\alpha }}{i^{(1+a)/\alpha }}\right] \end{aligned} \end{aligned}$$

Use the scaling property: for any $c>0$

$$\begin{aligned} p_t(x) = c p_{c^\alpha t}(cx) \end{aligned}$$

(A.2)

to write the above as

$$\begin{aligned} \frac{1}{n} \sum _{i=1}^{[nt]} \frac{1}{(i/n)^{1/\alpha }}\cdot p_1\left( \left( \frac{n}{i}\right) ^{1/\alpha } \cdot \frac{a_n-\mu [nt]}{n^{1/\alpha }}\right) +o(1). \end{aligned}$$

The rest is a Riemann sum approximation. $\square$

Appendix: Bounds required for Proposition 3.5

The main results of this section are Lemmas B.4 and B.5 which were needed in the proof of Proposition 3.5.

Lemma B.1

The following holds

$$\begin{aligned} \sup _{n}\sup _{w\in \mathbf {Z}} \frac{[n^\theta ]}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} P\big (Y_{i[n^\theta ]}=w\big ) \leqslant \sup _{n} \frac{[n^\theta ]}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} P\big (Y_{i[n^\theta ]}=0\big ) <\infty . \end{aligned}$$

Proof

The first inequality is a simple consequence of the fact that

$$\begin{aligned} P(Y_j=w) = \frac{1}{2\pi }\int _{-\pi }^{\pi } e^{-\mathrm i w z} \vert \phi (z)\vert ^{2j} \mathrm{d}w \leqslant P(Y_j=0) \end{aligned}$$

for any w, since the characteristic function of $Y_j$ is $\vert \phi (z)\vert ^{2j}$. As for the finiteness of the sum observe that $P(Y_j=0)$ is decreasing with j and therefore

$$\begin{aligned} {[}n^\theta ]\sum _{i=1}^{r-1} P\big (Y_{i[n^\theta ]}=0\big ) \leqslant \sum _{j=0}^n P(Y_j=0) \lesssim n^{(\alpha -1)/\alpha }, \end{aligned}$$

the last inequality following from (2.7). $\square$

It follows from Assumption 1.1 on the characteristic function that the distributions function F(x) of $X_1$ belongs to the domain of normal attraction of the symmetric stable law with exponent $\alpha$ (see section 35 of [17]). This means that one has to scale the centered $X_n-\mu n$ by a constant multiple of $n^{1/\alpha }$. One can characterize such distribution functions.

Lemma B.2

([17]) A necessary and sufficient condition for F to be in the domain of normal attraction of a Stable($\alpha$) law, $0<\alpha <2$, is the existence of contants $c_1,c_2\geqslant 0,\, c_1+c_2>0$ such that

$$\begin{aligned} F(x)&= \big (c_1 + f_1(x)\big ) \frac{1}{|x|^\alpha }, \qquad \text {for } x<0, \\ F(x)&= 1-\big (c_2+ f_2(x)\big ) \frac{1}{|x|^\alpha }, \;\;\text {for } x>0, \end{aligned}$$

where the functions $f_1$ and $f_2$ satisfy

$$\begin{aligned} \lim _{x \rightarrow -\infty } f_1(x) = \lim _{x\rightarrow \infty } f_2(x)=0. \end{aligned}$$

We use the above lemma to deduce the following large deviation estimate.

Lemma B.3

There exists a constant $c_3>0$ such that

$$\begin{aligned} P \big (\vert X_{[nt]-r[n^\theta ]}\vert \geqslant c_3n^\theta \big )\lesssim \frac{1}{n^{(\alpha -1)\theta }}. \end{aligned}$$

Proof

For $1<\alpha <2$ we use the result in [20]. This gives for $c_3>|\mu |$

$$\begin{aligned} P(|X_{[nt]-r[n^\theta ]}| \geqslant c_3n^\theta )&\lesssim n^\theta P(|X_1| \geqslant c_3 n^\theta )\\&\lesssim \frac{n^\theta }{n^{\alpha \theta }}, \end{aligned}$$

from the previous lemma. For $\alpha =2$, one can use Exercise 3.3.19 in [11] to conclude that $X_1$ has second moments. One then uses Chebyschev’s inequality to prove the lemma. $\square$

For the rest of this section we provide the proofs of the bounds required for the two terms in (3.14). Below is the bound on the first term.

Lemma B.4

The first term in (3.14) has the bound

$$\begin{aligned} \frac{1}{n^{(\alpha -1)/\alpha }} \Big \vert \sum _{i,j} \sum _{k,l} \mathrm {P}_j(l) \cdot \Big [ \mathrm {P}_j(l)-\mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )\Big ]\Big \vert \lesssim \frac{1}{n^{(\alpha -1)\theta }} + \frac{n^{\theta +o(1)}}{n^{(\alpha -1)/\alpha }}, \end{aligned}$$

where the limits of the summation are as in (3.12).

Proof

We write

$$\begin{aligned} \begin{aligned}&\frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i,j} \sum _{k,l} \mathrm {P}_j(l) \cdot \Big [ \mathrm {P}_j(l)-\mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )\Big ] \\&\quad = \frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i,j} P(X_j=\tilde{X}_j) \\&\qquad - \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i,j}\sum _{k,l}\sum _{w\in \mathbf {Z}} P\left( X_{i[n^\theta ]}=l-w,\, \tilde{X}_{[in^\theta ]}=k[n^\gamma ]\right) \cdot \mathrm {P}_{j-[in^\theta ]}(w), \end{aligned} \end{aligned}$$

(B.1)

the equality holding by an application of the Markov property. We focus on the second term for now, we will return to the first term later. We split the sum according to whether $i=0$ or not. The $i=0$ term is of order $n^{\theta }n^{-(\alpha -1)/\alpha }$. For the term corresponding to $i\geqslant 1$ replace $\tilde{X}_{i[n^\theta ]} = k[n^\gamma ]$ by $\tilde{X}_{i[n^\theta ]}=l$ by using Theorem A.1 with $b=1-\alpha \gamma$ to obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i,j}\sum _{k,l} \sum _w P\left( X_{i[n^\theta ]}=l-w,\, \tilde{X}_{[in^\theta ]}=k[n^\gamma ]\right) \cdot \mathrm {P}_{j-[in^\theta ]}(w) \\&\quad =O\left( \frac{n^\theta }{n^{(\alpha -1)/\alpha }}\right) + \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\sum _{j}\left[ O\left( \frac{1}{(in^\theta )^{(1+a)/\alpha }} \right) + O\left( \frac{n^\gamma }{(in^\theta )^{2/\alpha }}\right) \right] \\&\qquad +\frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\sum _{j}\sum _{k,l} \sum _{w} P\left( X_{i[n^\theta ]}=l-w, \tilde{X}_{[in^\theta ]}=l\right) \cdot \mathrm {P}_{j-[in^\theta ]}(w) \\&\quad =O\left( \frac{n^\theta }{n^{(\alpha -1)/\alpha }}\right) +O\left( \frac{n^{o(1)}}{n^{\min (a, \alpha -1)/\alpha }}\right) \\&\qquad +\frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\sum _{j}\sum _{w} P\left( X_{i[n^\theta ]}-\tilde{X}_{i[n^\theta ]}=w\right) \cdot \mathrm {P}_{j-[in^\theta ]}(w). \end{aligned} \end{aligned}$$

Returning to our expression (B.1) we now can write it as

$$\begin{aligned} \begin{aligned}&\frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i,j} P(X_j=\tilde{X}_j) -\frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} \sum _j\sum _{|w|\leqslant c_3n^\theta } P\left( X_{i[n^\theta ]}-\tilde{X}_{i[n^\theta ]}=w\right) \\&\quad \cdot \mathrm {P}_{j-[in^\theta ]}(w) \\&\quad +O\left( \frac{1}{n^{(\alpha -1)\theta }}\right) +O\left( \frac{n^\theta }{n^{(\alpha -1)/\alpha }}\right) +O\left( \frac{n^{o(1)}}{n^{\min (a, \alpha -1)/\alpha }}\right) , \end{aligned} \end{aligned}$$

where we have used Lemma B.1 and Lemma B.3 with the constant $c_3$ from there. Let us now consider each of the first two terms above. Using Theorem A.1 as well as (A.2) one gets for the first term

$$\begin{aligned} \frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i,j} P(X_j=\tilde{X}_j) = \frac{\alpha \tilde{p}_1(0) }{\alpha -1} \cdot \left( \frac{r[n^\theta ]-1}{n}\right) ^{(\alpha -1)/\alpha } + O\left( \frac{n^{o(1)}}{n^{\min (a, \,\alpha -1)/\alpha }}\right) , \end{aligned}$$

where $\tilde{p}_1(\cdot )$ is the transition kernel of $-2\nu (-\Delta )^{\alpha /2}$. Using Theorem A.1 again with $b=1-\alpha \theta$, we get for $|w|\leqslant c_3n^\theta$

$$\begin{aligned}&\frac{[n^\theta ]}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\sum _{|w|\leqslant c_3[n^\theta ]} P\left( X_{i[n^\theta ]}-\tilde{X}_{i[n^\theta ]}=w\right) \cdot \mathrm {P}_{j-i[n^\theta ]}(w) \\&\quad =\frac{[n^\theta ]}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\left[ \frac{\tilde{p}_1(0)}{(i[n^\theta ])^{1/\alpha }}+ O\left( \frac{1}{(in^\theta )^{(1+a)/\alpha }}\right) +O\left( \frac{n^\theta }{(in^\theta )^{2/\alpha }}\right) \right] \\&\quad = \frac{\alpha \tilde{p}_1(0) }{\alpha -1} \cdot \left( \frac{r[n^\theta ]-1}{n}\right) ^{(\alpha -1)/\alpha } +O\left( \frac{n^{\theta (\alpha -1)/\alpha }}{n^{(\alpha -1)/\alpha }}\right) \\&\qquad + O\left( \frac{n^{o(1)}}{n^{\min (a,\, \alpha -1)/\alpha }}\right) + O\left( \frac{n^{\theta +o(1)}}{n^{(\alpha -1)/\alpha }}\right) . \end{aligned}$$

Collecting all our estimates and recalling our conditions (3.2) on $\gamma$ and $\theta$ completes the proof. $\square$

We next bound the second term in (3.14).

Lemma B.5

The second term in (3.14) has the bound

$$\begin{aligned} \frac{1}{n^{(\alpha -1)/\alpha }}\Big \vert \sum _{i,j} \sum _{k,l} \mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )\cdot \Big [ \mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )-\mathrm {P}_j(l)\Big ]\Big \vert \lesssim \frac{n^{\gamma }}{n^{(\alpha -1)\theta }}+ \frac{n^{\theta +\gamma +o(1)}}{n^{(\alpha -1)/\alpha }} \end{aligned}$$

where the limits in the summation are as in (3.14).

Proof

We separate out the $i=0$ term and obtain

$$\begin{aligned} \begin{aligned}&\frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i,j} \sum _{k,l} \mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )\cdot \Big [ \mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )-\mathrm {P}_j(l)\Big ] \\&\quad = O \left( \frac{n^{\theta +\gamma }}{n^{(\alpha -1)/\alpha }}\right) + \frac{1}{n^{(\alpha -1)/\alpha }} \sum _{i=1}^{r-1} \sum _{j, k,l} \sum _w \mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )\\&\qquad \cdot \Big [ \mathrm {P}_{i[n^\theta ]}\big (k[n^\gamma ]\big )-\mathrm {P}_{i[n^\theta ]}(l-w)\Big ]\cdot \mathrm {P}_{j-i[n^\theta ]}(w) \end{aligned} \end{aligned}$$

The reason for the error bound for the first term is that restricting $i=0$ forces $k=0$, and the number of terms in the summation over j and l is of order $n^{\theta +\gamma }$. As in Lemma B.4 we split the sum over w according to whether $|w|\leqslant c_3[n^\theta ]$ or not. In the case $|w|\leqslant c_3[n^\theta ]$ we use Theorem A.1 with $b=1-\alpha \theta$. Thus the above is

$$\begin{aligned} \begin{aligned}&=O \left( \frac{n^{\theta +\gamma }}{n^{(\alpha -1)/\alpha }}\right) +\frac{n^{\gamma +\theta }}{n^{(\alpha -1)/\alpha }} \sum _{i=1}^{r-1} \left[ O\left( \frac{1}{(in^\theta )^{(1+a)/\alpha }}\right) + O\left( \frac{n^\theta }{(in^\theta )^{2/\alpha }}\right) \right] \\&\quad + \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} \sum _{j,k,l}\sum _{|w|\geqslant c_3n^\theta } \mathrm {P}_{i[n^\theta ]}(k[n^\gamma ])\cdot \left[ \mathrm {P}_{i[n^\theta ]}(k[n^\gamma ]) -\mathrm {P}_{i[n^\theta ]}(l-w)\right] \\&\quad \cdot \mathrm {P}_{j-i[n^\theta ]}(w) \\&= O\left( \frac{n^{\gamma +o(1)}}{n^{\min (a, \alpha -1)/\alpha }}\right) + O \left( \frac{n^{\gamma +\theta +o(1)}}{n^{(\alpha -1)/\alpha }}\right) \\&\quad + \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} \sum _{j,k,l}\sum _{|w|\geqslant c_3n^\theta } \mathrm {P}_{i[n^\theta ]}(k[n^\gamma ])\cdot \left[ \mathrm {P}_{i[n^\theta ]}(k[n^\gamma ]) -\mathrm {P}_{i[n^\theta ]}(l-w)\right] \\&\quad \cdot \mathrm {P}_{j-i[n^\theta ]}(w) \end{aligned} \end{aligned}$$

To bound the last term we ignore the difference in the expression and instead bound the sum. By Lemmas B.1 and B.3

$$\begin{aligned} \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} \sum _{j,k,l}\sum _{|w|\geqslant c_3n^\theta } \mathrm {P}^2_{i[n^\theta ]}(k[n^\gamma ])\cdot \mathrm {P}_{j-i[n^\theta ]}(w) \lesssim \frac{n^{\gamma }}{n^{(\alpha -1)\theta }}, \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1} \sum _{j,k,l}\sum _{|w|\geqslant c_3n^\theta } \mathrm {P}_{i[n^\theta ]}(k[n^\gamma ])\cdot \mathrm {P}_{i[n^\theta ]}(l-w)\cdot \mathrm {P}_{j-i[n^\theta ]}(w) \\&\quad \lesssim \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\sum _j\sum _{y=0}^{[n^\gamma ]-1}\sum _{|w|\geqslant c_3n^\theta } P\left( \tilde{X}_{i[n^\theta ]}- X_{i[n^\theta ]} = y-w\right) \cdot \mathrm {P}_{j-i[n^\theta ]}(w) \\&\quad \lesssim \frac{1}{n^{(\alpha -1)/\alpha }}\sum _{i=1}^{r-1}\sum _j\sum _{y=0}^{[n^\gamma ]-1}\sum _{|w|\geqslant c_3n^\theta } P\left( \tilde{X}_{i[n^\theta ]}- X_{i[n^\theta ]} = 0\right) \cdot \mathrm {P}_{j-i[n^\theta ]}(w) \\&\quad \lesssim \frac{n^{\gamma }}{n^{(\alpha -1)\theta }}, \end{aligned} \end{aligned}$$

where we used Lemma B.1 in the last step. Collecting our estimates completes the proof. $\square$