Density fluctuations for exclusion processes with long jumps

Abstract

We show that the stationary density fluctuations of exclusion processes with long jumps, whose rates are of the form $c^\pm |y-x|^{-(1+\alpha )}$ where $c\pm$ depends on the sign of $y-x$, are given by a fractional Ornstein–Uhlenbeck process for $\alpha \in (0,\frac{3}{2})$. When $\alpha =\frac{3}{2}$ we show that the density fluctuations are tight, in a suitable topology, and that any limit point is an energy solution of the fractional Burgers equation, previously introduced in Gubinelli and Jara (Stoch Partial Differ Equ Anal Comput 1(2):325–350, 2013) in the finite volume setting.

Appendices

Appendix A: The spectral gap inequality

A classical problem in the theory of Markov chains is the study of the time that the chain needs to reach the equilibrium. In the case of a (continuous time) finite state ergodic Markov chain it is known that the convergence to equilibrium is exponentially fast. Therefore the relevant question is the exponential rate at which this happens. Let $\{x(t); t \ge 0\}$ be an ergodic Markov chain on a finite state space E. Let $\mu$ be its unique invariant measure. For $f: E \rightarrow {\mathbb R}$ and $x \in E$, let $P_t f(x) = {\mathbb E} [ f(x(t))| x(0)=x]$. Let $\langle \cdot \rangle _\mu$ denote the expectation with respect to $\mu$. Then we define

$$\begin{aligned} \lambda = - \sup _{f:E \rightarrow {\mathbb R}} \limsup _{t \rightarrow \infty } \frac{1}{t} \log (\big \Vert P_t f- \langle f\rangle _\mu \big \Vert _{L^2(\mu )}). \end{aligned}$$

The number $1/\lambda$ is known as the relaxation time of the chain $\{x(t); t \ge 0\}$. In the case on which the chain $\{x(t); t \ge 0\}$ is reversible with respect to $\mu$, the number $\lambda$ is equal to the spectral gap of the generator A of the chain $\{x(t); t \ge 0\}$, that is, the absolute value of the largest non-zero eigenvalue of A. In that case, we have the variational formula

$$\begin{aligned} \lambda ^{-1} = \sup _{\langle f\rangle _\mu =0} \frac{\langle f^2\rangle _\mu }{\langle -fAf\rangle _{\mu }}. \end{aligned}$$

(A.1)

When the chain is not reversible, this variational formula provides an upper bound for $\lambda ^{-1}$. A natural question in the theory of Markov chains is to estimate the spectral gap of a Markov chain, or of a family of Markov chains of increasing complexity.

For the symmetric simple random walk restricted to $\{1,\dots ,n\}$ it is well known that $\lambda ^{-1} = {\mathscr {O}}(n^2)$. It turns out that this property of the random walk over finite intervals, by means of a computation of Nash type, allows one to show that in the case of the symmetric simple random walk on ${\mathbb Z}$,

$$\begin{aligned} \big \Vert P_t f - \langle f\rangle _\mu \big \Vert _{L^2(\mu )} = o\big (\tfrac{1}{t^a}\big ) \text { for any } a < \tfrac{1}{2}, \end{aligned}$$

where $\mu$ is the counting measure on ${\mathbb Z}$. Therefore a sharp upper bound on the spectral gap of finite-state Markov chains gives valuable information even in the case of chains on infinite state spaces.

Consider the symmetric random walk restricted to the set $\Lambda _\ell = \{1,\dots ,\ell \}$. In the case where the random walks have long jumps we have the following result:

Proposition A.1

Let $p(\cdot )$ be given by (2.2). There exists $\kappa >0$ such that

$$\begin{aligned} \sum _{x \in \Lambda _\ell } f(x)^2 \le \kappa \ell ^\alpha \sum _{x,y \in \Lambda _\ell } p(y-x) \big (f(y)-f(x)\big )^2 \end{aligned}$$

(A.2)

for any $\ell \in {\mathbb N}$ and any $f: \Lambda _\ell \rightarrow {\mathbb R}$ such that

$$\begin{aligned} \sum _{x \in \Lambda _\ell } f(x) =0. \end{aligned}$$

(A.3)

Remark A.2

This proposition is telling us that the spectral gap of a Markov chain with jump rates given in (2.2) and defined on the interval $\Lambda _\ell$, is bounded from below by $\frac{1}{\kappa \ell ^\alpha }$. In addition, pairing together the two terms involving x and y, we see that only the behavior of the symmetric part of $p(\cdot )$ is relevant for this proposition.

The proof of this proposition is in fact very simple. For that purpose note that

$$\begin{aligned} \sum _{x,y \in \Lambda _\ell } p(y-x) \big (f(y)-f(x)\big )^2=\sum _{x,y \in \Lambda _\ell }2s(y-x)\big ( f(y) -f(x) \big )^2. \end{aligned}$$

(A.4)

To conclude, use the fact that for f satisfying (A.3), it holds:

$$\begin{aligned} \sum _{x,y \in \Lambda _\ell } \big ( f(y) -f(x) \big )^2 = 2 \ell \sum _{x \in \Lambda _\ell } f(x)^2, \end{aligned}$$

together with $s(y-x) \ge \frac{c^++c^-}{2 \ell ^{1+\alpha }}$ for any $x,y \in \Lambda _\ell$.

As a corollary of Proposition A.1 we can obtain a lower bound for the spectral gap of the exclusion process with transition rate $p(\cdot )$:

Corollary A.3

Let $p(\cdot )$ be defined by (2.2). Let $f: \Omega \rightarrow {\mathbb R}$ be a local function with ${{\mathrm{supp}}}(f) \subseteq \Lambda _\ell$. Assume that $\int f d\mu _\sigma = 0$ for any $\sigma \in [0,1]$. Then,

$$\begin{aligned} \int f^2 d\mu _\sigma \le \kappa \ell ^\alpha \sum _{x,y \in \Lambda _\ell } p(y-x) \int \big (\nabla _{x,y} f\big )^2 d\mu _\sigma \end{aligned}$$

for any $\sigma \in [0,1]$.

The simplest way to prove this corollary is by means of the Aldous’ conjecture, proved in [8], which says that the spectral gap of an exclusion process with symmetric rates is equal to the spectral gap of the random walk with the same rates. Another proof using a comparison principle can be found in [25].

Appendix B: Proof of Proposition 2.2

We do the proof for the case $\alpha >1$, the others being analogous. The proof of the proposition is elementary, but very tedious. Recall the definition of ${\mathscr {L}}_n$ and ${\mathscr {L}}$ from (2.6) and (2.4), respectively. First note that by the definition of $m_n^\alpha$ in (2.3), for this regime of $\alpha$ we rewrite ${\mathscr {L}}_n f(\tfrac{x}{n}) = n^\alpha \sum _{y} p(y)\psi _{x/n}(\tfrac{y}{n})$ and ${\mathscr {L}} f(x) = \int _{\mathbb R} p(y) \psi _x(y)dy$ where $\psi _{u}(v) = f(u+v)-f(u)-vf'(u),$ for $f\in {\mathscr {C}}^2({\mathbb R})$. Second, note that ${\mathscr {L}}_n^+ f(\tfrac{x}{n}) = n^\alpha \sum _{y \ge 1} p(y)\psi _{x/n}(\tfrac{y}{n})$ and ${\mathscr {L}}^+ f(x) = \int _0^\infty \frac{c^+}{y^{1+\alpha }} \psi _x(y)dy$ are well-defined and that it is enough to show (2.7) for ${\mathscr {L}}_n^+$ and ${\mathscr {L}}^+$. For $x \in {\mathbb N}$, define $P(x) = \sum _{y \ge x} p(y)$ and $a(x) = x^\alpha P(x)-\frac{c^+}{\alpha }$. Note that a(x) tends to 0, as $x\rightarrow \infty$. The idea is to perform an integration by parts in the formula of ${\mathscr {L}}_n^+f$ in order to work with the more regular object $P(\cdot )$. By writing $p(y)=P(y)-P(y+1)$, performing a summation by parts and a Taylor expansion on $\psi$, we see that ${\mathscr {L}}_n^+f(\tfrac{x}{n}) = n^{\alpha -1} \sum _{y\ge 1} P(y)\psi _{x/n}'(\tfrac{y}{n})+R_1^n(x)$, where $R_1^n(x)$ is an error term which satisfies $|R_1^n(x)| \le \Vert \psi ''_{x/n}\Vert _\infty n^{\alpha -2} \sum _{y \ge 1} P(y).$ Note that $\Vert \psi _{x/n}'\Vert _\infty {\le 2\Vert f'\Vert _\infty }$, which does not depend on x. This last sum is equal to $\sum _{y \ge 1} y p(y)<+\infty$ and since $\alpha <2$, $R^1_n(x)$ vanishes, as $n\rightarrow \infty$. For $y\ge 1$, let $A(y) = \sup _{z \ge y} |a(z)|$. We have that

$$\begin{aligned} {\mathscr {L}}^+_nf(\tfrac{x}{n})= n^{\alpha -1} \sum _{y\ge 1} \frac{c^+}{\alpha y^\alpha }\psi _{x/n}'(\tfrac{y}{n})+n^{\alpha -1} \sum _{y\ge 1} \frac{a(y)}{y^\alpha }\psi _{x/n}'(\tfrac{y}{n}). \end{aligned}$$

(B.1)

Note that $\psi _u'$ is bounded and that $\psi _u(0)=\psi '_u(0)=0$. Therefore, there exists a constant K such that $|\psi _u'(v)| \le Kv$ for any $v\in [0,1]$ and such that $|\psi _u'(v)| \le K$ for any $v >1$. In fact, we can choose $K=\max \{2\Vert f'\Vert _\infty , \Vert f''\Vert _\infty \}$. Therefore, the second term on the right hand side of (B.1) is bounded by

$$\begin{aligned}&\Big |n^{\alpha -1}\sum _{y=1}^n \frac{a(y)}{y^\alpha }\psi _{x/n}'(\tfrac{y}{n})\Big | +\Big |n^{\alpha -1}\sum _{y\ge n+1}\frac{a(y)}{y^\alpha }\psi _{x/n}'(\tfrac{y}{n})\Big | \\&\quad \le Kn^{\alpha -2} \Big (A(1)\sum _{y=1}^k y^{1-\alpha } + A(k+1)\sum _{y=k+1}^n y^{1-\alpha }\Big )+K n^{\alpha -1} A(n)\sum _{y\ge n+1} \frac{1}{y^\alpha }\\&\quad \le \frac{K}{2-\alpha }\Big (A(1) (\tfrac{k}{n})^{2-\alpha }+A(k+1)\Big )+\frac{K A(n+1)}{\alpha -1} \end{aligned}$$

for any $k<n$. Choosing, for example, $k = \sqrt{n}$ , the last sums vanish, as $n \rightarrow \infty$. Note that

$$\begin{aligned} {\mathscr {L}}^+ f\big (\tfrac{x}{n}\big ) = -\frac{c^+ \psi _x(y)}{\alpha y^\alpha }\Big |_{y=0}^\infty + \int _0^\infty \frac{c^+ \psi _x'(y)}{y^\alpha } dy = c^+\int _0^\infty \frac{ \psi _x'(y)}{\alpha y^\alpha } dy, \end{aligned}$$

since $\psi _x(y)$ is quadratic around $y =0$ and linear for $y \gg 1$. Moreover, the first sum on the right hand side of (B.1) is just a Riemann sum for this last integral. Since the function $\frac{\psi _x'(y)}{y^\alpha }$ is continuous at $y=0$ and it decays like $\frac{1}{y^\alpha }$, this Riemann sum converges to the corresponding integral. Note that this convegence is uniform in x, since $\frac{\psi _x'(y)}{y^\alpha }$ is equicontinuous in x. Finally, since for $0<y<z$, we have that $\Big |\frac{\psi '(z)}{z^\alpha }-\frac{\psi '(y)}{y^\alpha }\Big | \le \frac{C(z-y)}{y^\alpha }$ for some constant C which depends only on $\Vert f''\Vert _\infty$, we conclude that

$$\begin{aligned} \Big | n^{\alpha -1} \sum _{y \ge 1} \frac{c^+}{\alpha y^\alpha } \psi _{x/n}'(\tfrac{y}{n}) - \int _{\tfrac{1}{n}}^\infty \frac{c^+ \psi '(y)}{y^\alpha }dy\Big | \le C n^{\alpha -2} \sum _{y\ge 1} \frac{1}{y^\alpha }. \end{aligned}$$

Since the last sum is finite and $\alpha <2$, we have just shown that

$$\begin{aligned} \lim _{n \rightarrow \infty } | {\mathscr {L}}^+_nf(\tfrac{x}{n}) - {\mathscr {L}}^+f(\tfrac{x}{n})| =0. \end{aligned}$$

Moreover, since all the constants above do not depend on x, the limit is uniform in x, showing the first half of the proposition. The second half can be proved in a similar way.

Appendix C: Proof of Proposition 2.7

For the reader’s convenience, we repeat here various definitions introduced in Sect. 2. Let ${\mathscr {L}}$ be a generator of a Lévy process in ${\mathbb R}$. Let $\{{\mathscr {W}}_t; t \ge 0\}$ be a Brownian motion on $L^2({\mathbb R})$ and let ${\mathscr {S}} = \frac{1}{2}({\mathscr {L}} + {\mathscr {L}}^*)$ be the symmetric part of the operator ${\mathscr {L}}$. We say that a stochastic process $\{{\mathscr {Y}}_t; t \ge 0\}$ is a stationary solution of the infinite-dimensional Ornstein–Uhlenbeck equation

$$\begin{aligned} d {\mathscr {Y}}_t = {\mathscr {L}}^* {\mathscr {Y}}_t dt + \sqrt{-2 \chi } {\mathscr {S}} d {\mathscr {W}}_t \end{aligned}$$

(C.1)

if for each $t \in [0,T]$ the random variable ${\mathscr {Y}}_t$ is a white noise of variance $\chi$ and for any differentiable function $f: [0,T] \rightarrow {\mathbb S}({\mathbb R})$ the process

$$\begin{aligned} {\mathscr {Y}}_t(f_t) - {\mathscr {Y}}_0(f_0) - \int _0^t {\mathscr {Y}}_s( (\partial _s+{\mathscr {L}}) f_s) ds \end{aligned}$$

is a martingale of quadratic variation $2 \chi \int _0^t \langle f_s , -{\mathscr {S}} f_s \rangle ds.$ We will prove following result:

Proposition C.1

Two stationary solutions of (C.1) have the same law.

Proof

Let f be a function in ${\mathbb S}({\mathbb R})$ and take $t \ge 0$. Let $\{P_t; t \ge 0\}$ the semigroup associated to the generator ${\mathscr {L}}$, that is, $P_t = e^{t {\mathscr {L}}}$ for any $t \ge 0$. Since ${\mathscr {L}}$ is the generator of a Lévy process, $\{P_t; t \ge 0\}$ is a strongly continuous, contraction semigroup on ${\mathscr {C}}_b({\mathbb R})$. In particular $f_s = P_{t-s} t$ is a differentiable trajectory on ${\mathscr {C}}_b({\mathbb R})$ satisfying $\frac{d}{ds} f_s = - {\mathscr {L}} f_s$ for any $s \le t$ and $f_t = f$. Since $\{P_t; t \ge 0\}$ is also a contraction in $L^1({\mathbb R})$, it is a contraction in $L^2({\mathbb R})$. Note that $\{f_s; s \le t\}$ is not a legitimate test function, since although $P_{t-s} f$ is infinitely differentiable, it does not satisfy the decay properties needed to be in ${\mathbb S}({\mathbb R})$. However, $\{f_s; s \le t\}$ can be approximated in $L^2$ by differentiable functions $f_\epsilon : [0,t] \rightarrow {\mathbb S}({\mathbb R})$, justifying the use of $\{f_s; s \le t\}$ as a test function. Since $(\partial _s + {\mathscr {L}}) f_s =0$, we conclude that

$$\begin{aligned} {\mathscr {M}}_{s,t}(f) =: {\mathscr {Y}}_s(P_{t-s} f) - {\mathscr {Y}}_0(P_t f) \end{aligned}$$

is a martingale of quadratic variation (with respect to s)

$$\begin{aligned} 2 \chi \int _0^t\langle P_s f, -{\mathscr {S}} P_s f\rangle ds. \end{aligned}$$

Since the quadratic variation of ${\mathscr {M}}_{s,t}(f)$ is deterministic, $\{{\mathscr {M}}_{s,t}(f); s\in [0, t]\}$ and in consequence $\{{\mathscr {Y}}_t; t\in [0, T]\}$ are Gaussian processes. Note that

$$\begin{aligned} \tfrac{d}{dt} \langle P_t f, P_t f\rangle = 2 \langle P_t f, {\mathscr {L}} P_t f\rangle = -2 \langle P_t f, -{\mathscr {S}} P_t f\rangle . \end{aligned}$$

Therefore

$$\begin{aligned} 2 \chi \int _0^t \langle P_s f, -{\mathscr {S}} P_s f\rangle = \chi \big (\langle f,f\rangle - \langle P_t f, P_t f\rangle \big ). \end{aligned}$$

We conclude that ${\mathscr {Y}}_t(f)$ can be written as the sum of two independent Gaussian variables: ${\mathscr {Y}}_0(P_t f)$, which depends only on the initial law and ${\mathscr {M}}_{t,t}(f)$, which is independent of ${\mathscr {Y}}_0$. Since $\{{\mathscr {Y}}_t; t\in [0, T]\}$ is a Gaussian process, it is characterized by its covariance structure. By stationarity, the computations above show that for any $0 \le s \le t \le T$ and any $f,g \in {\mathbb S}({\mathbb R})$,

$$\begin{aligned} E[ {\mathscr {Y}}_t(f) {\mathscr {Y}}_s(g)]&= E[ {\mathscr {Y}}_{t-s}(f) {\mathscr {Y}}_0(g)] = E[({\mathscr {Y}}_0(P_{t-s}f)+{\mathscr {M}}_{t-s,t-s}(f)) {\mathscr {Y}}_0(g)] \\ {}&= \chi \langle P_{t-s} f,g\rangle , \end{aligned}$$

which shows uniqueness in law of the process $\{{\mathscr {Y}}_t; t\in [0, T]\}$. $\square$

Appendix D: Auxiliary computations

In this section we collect the proofs of some estimates that are needed in Sect. 4.

1.1 D.1 Bounds on Taylor expansions of test functions

In the computations below, it will be useful to control the decay at infinity of the error terms of the Taylor expansion for test functions. For $g:{\mathbb R} \rightarrow {\mathbb R}$, $x \in {\mathbb R}$ and $M >0$, define

$$\begin{aligned} \Vert g(x)\Vert _{M,\infty } = \sup _{|y-x| \le M} | g(y) |. \end{aligned}$$

We have the following:

Lemma D.1

Let $g \in {\mathbb S}({\mathbb R})$. Then, for any $\ell \in {\mathbb N}$,

$$\begin{aligned} \lim _{x \rightarrow \pm \infty } |x|^\ell \Vert g(x)\Vert _{M,\infty } =0. \end{aligned}$$

The proof of this lemma is elementary, so we omit it. We can apply this lemma to obtain the aforementioned bounds on Taylor expansions of test functions: let $f \in {\mathbb S}({\mathbb R})$, $x \in {\mathbb R}$ and $\ell \in {\mathbb N}$. For any $M >0$ and any $y \in {\mathbb R}$ such that $|y-x| \le M$, we have that

$$\begin{aligned} \Big | f(y) - \sum _{i=0}^{\ell -1} \frac{f^{(i)}(x)}{i !} (y-x)^\ell \Big | \le \frac{|y-x|^\ell }{\ell !} \Vert f^{(\ell )}(x)\Vert _{M,\infty }. \end{aligned}$$

(D.1)

This is an immediate consequence of the Taylor formula with Lagrange error and Lemma D.1.

1.2 D.2 Estimation of (4.1)

In order to estimate the first term of (4.1) we first make a change of variables $z=y-x$, we fix M and write it as

$$\begin{aligned}&c_1(\rho ) n^{2\alpha -2} \sum _{x\in {\mathbb Z}}\sum _{|z|=1}^{Mn} \frac{1}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big ) - f\big (\tfrac{x}{n}\big ) \big )^4\nonumber \\&\quad +\,c_1(\rho ) n^{2\alpha -2} \sum _{x\in {\mathbb Z}}\sum _{|z|\ge M n} \frac{1}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big ) - f\big (\tfrac{x}{n}\big ) \big )^4. \end{aligned}$$

(D.2)

The second term in (D.2) can be bounded from above by

$$\begin{aligned} C n^{2\alpha -2} \sum _{x\in {\mathbb Z}}\sum _{|z|\ge M n} \frac{1}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big )^4+C n^{2\alpha -2} \sum _{x\in {\mathbb Z}}\sum _{|z|\ge M n} \frac{1}{|z|^{2+2\alpha }} \big (f\big (\tfrac{x}{n}\big ) \big )^4. \end{aligned}$$

By making a change of variables, using the fact that $f\in { {\mathbb S}}({\mathbb R})$ and since

$$\begin{aligned} n^{1+2\alpha }\sum _{|z|> Mn}\frac{1}{|z|^{2+2\alpha }} <\infty \end{aligned}$$

we conclude that the two previous sums are of order $O(n^{-2})$. To compute the first term in (D.2), we use Lemma D.1 to conclude that

$$\begin{aligned} n^{2\alpha -2} \sum _{x \in {\mathbb Z}}\sum _{|z|=1}^{Mn} \frac{1}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big ) - f\big (\tfrac{x}{n}\big ) \big )^4\le n^{2\alpha -6}\sum _{x \in {\mathbb Z}}\Vert f'(x/n)\Vert _{M/n,\infty }^4 \sum _{|z|=1}^{Mn}|z|^{2-2\alpha }. \end{aligned}$$

Now we use (D.1) and since

$$\begin{aligned} n^{2\alpha -3}\sum _{|z|=1}^{Mn}\frac{1}{|z|^{2\alpha -2}}=O(n^{2\alpha -3}), \end{aligned}$$

we obtain that the second term in (D.2) is of order $O(n^{2\alpha -5})$ and therefore vanishes as $n\rightarrow \infty .$ The second term of (4.1), by a change of variables $z=y-x$ and fixing M, can be written as

$$\begin{aligned} \begin{aligned}&c_2(\rho ) \frac{1}{n} \sum _{x\in {\mathbb Z}} \left( \frac{1}{n}\sum _{|z|=1}^{Mn} \frac{n^{1+\alpha }}{|z|^{1+\alpha }}\big (f\big (\tfrac{z+x}{n}\big )-f\big (\tfrac{x}{n}\big )\big )^2\right) ^2\\&\quad + c_2(\rho ) \frac{1}{n} \sum _{x\in {\mathbb Z}} \left( \frac{1}{n}\sum _{|z|\ge Mn} \frac{n^{1+\alpha }}{|z|^{1+\alpha }}\big (f\big (\tfrac{z+x}{n}\big )-f\big (\tfrac{x}{n}\big )\big )^2\right) ^2. \end{aligned} \end{aligned}$$

By repeating exactly the same computations as above, the first term of last expression is of order $O(n^{\alpha -3})$ while the second one is of order $O(n^{-1})$. This shows that the second term of (4.1) vanishes, as $n\rightarrow \infty$.

1.3 D.3 Estimation of (4.3)

The term (4.3), by a change of variables $z=y-x$ and fixing M, can be written as

$$\begin{aligned} \begin{aligned}&c_3(\rho ) \frac{1}{n^2} \sum _{x\in {\mathbb Z}}\sum _{|z|=1}^{Mn} \frac{n^{2+2\alpha }}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big ) -f \big ( \tfrac{x}{n}\big )\big )^2\\&\quad + c_3(\rho ) \frac{1}{n^2} \sum _{x\in {\mathbb Z}} \sum _{|z|\ge Mn}\frac{n^{2+2\alpha }}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big ) -f \big ( \tfrac{x}{n}\big )\big )^2. \end{aligned} \end{aligned}$$

Repeating the same procedure as above we can see that the second term above is of order $O(n^{-2})$, while the first one can be bounded from above by

$$\begin{aligned} Cn^{2\alpha -2}\sum _{|z|=1}^{Mn}{|z|^{-2\alpha }}. \end{aligned}$$

Last sum is convergent if $\alpha <1/2$ and if $\alpha \ge 1/2$, it is divergent. When $\alpha =1/2$, the order of divergence is $\log (n)$. Therefore, we can conclude that (4.3) is of order o(1) for $\alpha <1$ and bounded for $\alpha =1$.

1.4 D.4 Proof of Lemma 4.5

The expectation in (4.4) is bounded from above by

$$\begin{aligned} \begin{aligned} c_4(\rho ) n^{2\alpha -1} \sum _{|x-y|\ge K_n} a(y-x)^2 \big (f\big (\tfrac{y}{n}\big ) -f \big ( \tfrac{x}{n}\big )\big )^2, \end{aligned} \end{aligned}$$

(D.3)

and, by the change of variables $z=y-x$ and by fixing M, we can write it as

$$\begin{aligned} \begin{aligned}&c_4(\rho ) n^{2\alpha -1} \sum _{x\in {\mathbb Z}}\sum _{K_n\le |z|\le Mn} \frac{1}{|z|^{2+2\alpha }} \big (f\big (\tfrac{z+x}{n}\big ) -f \big ( \tfrac{x}{n}\big )\big )^2\\&\quad + c_4(\rho ) n^{2\alpha -1} \sum _{x\in {\mathbb Z}}\sum _{|z|\ge Mn}\frac{1}{|z|^{2+2\alpha }}\big (f\big (\tfrac{z+x}{n}\big ) -f \big ( \tfrac{x}{n}\big )\big )^2. \end{aligned} \end{aligned}$$

Now, doing the same arguments as above, we see that the first term in last expression if of order $O(n^{-1})$, while the second one can be bounded from above by

$$\begin{aligned} \begin{aligned} C n^{2\alpha -2}\sum _{ |z|=K_n}^{Mn}|z|^{-2\alpha } \end{aligned} \end{aligned}$$

which is of order $\frac{n^{2\alpha -2}}{K_n^{2\alpha -1}}$ and vanishes if $K_n \gg n^{\frac{2\alpha -2}{2\alpha -1}}$.

1.5 D.5 Proof of Lemma 4.6

Note that the expectation in (4.5) is bounded from above by

$$\begin{aligned} c_4(\rho ) n^{2\alpha -1} \sum _{|y-x| \le K} a^2(y-x) \frac{(y-x)^4}{n^4} \Vert f''((y-x)/n)\Vert ^2_{K/n,\infty }. \end{aligned}$$

As above, by making the change of variables $z=y-x$ and using the fact that $f\in {\mathbb S}({\mathbb R})$ we can bound the previous expression by

$$\begin{aligned} Cn^{2\alpha -4}\sum _{|z| \le K}z^{2-2\alpha } \end{aligned}$$

which is of order $\frac{K^{3-2\alpha }}{n^{4-2\alpha }}.$ Since $K \gg n^{\frac{2\alpha -2}{2\alpha -1}}$, the expectation vanishes as $n\rightarrow \infty$.

1.6 D.6 Proofs of Lemmas 4.10 and 4.11

We start with the proof of Lemma 4.10. By the Cauchy–Schwarz inequality and by (4.8), the expectation in (4.9) is bounded from above by

$$\begin{aligned} Kt^2 n^{2\alpha -3} \left( \sum _{y=1}^{K-1} y a(y)\right) ^2\sum _x f'\big (\tfrac{x}{n}\big )^2 \int \psi _x^{K} (\eta )^2 \mu _\rho (d\eta ) \le C(f,\rho )\frac{ t^2 n^{2\alpha -2}}{K}, \end{aligned}$$

(D.4)

which vanishes, as $n\rightarrow \infty$, if $K \gg n^{2\alpha -2}$. Above we used the fact that $\sum _{y=1}^{K-1} y a(y)<\infty$, since we are in the regime $\alpha >1$. Now we prove Lemma 4.11. For each $j=1,\ldots , K$, we use Proposition 4.9 with $F(\eta )=\sum _{x \in {\mathscr {Z}}_j}F_x(\eta )$ and

$$\begin{aligned} F_x(\eta ):=n^{\alpha -3/2} \sum _{y=1}^{K-1}y a(y) \big \{ {\bar{\eta }}(x) {\bar{\eta }}(x+y) - \psi _x^{K} (\eta )\big \} f'\big (\tfrac{x}{n} \big ). \end{aligned}$$

Therefore, the expectation in (4.9) is bounded from above by

$$\begin{aligned} c_4(\rho ) t K \frac{K^{\alpha }}{n^\alpha } n^{2\alpha -3} \sum _{x} f'\big ( \tfrac{x}{n}\big )^2 \sum _{y=1}^{K-1} y^2 a(y)^2 \le C(f,\rho ) \frac{ K^{1+\alpha } t}{n^{2-\alpha }}. \end{aligned}$$

(D.5)

We note that above we used the fact that $\sum _{y=1}^{K-1} y^2 a(y)^2<\infty$. We conclude that last term vanishes if $K \ll n^{\frac{2-\alpha }{1+\alpha }}$.

1.7 D.7 Proof of Lemma 4.14

By the Minkowski’s inequality and the estimate in (4.12), the expectation in (4.13) is bounded from above by

$$\begin{aligned} tC(f,\rho )\left( \sum _{i=1}^\ell \sqrt{\frac{n^{\gamma _i (1+\alpha )}}{n^{\gamma _{i-1}(2\alpha -1)+2-\alpha }}}\right) ^2=C(f,\rho ) \frac{t \ell ^2}{n^\delta }. \end{aligned}$$

1.8 D.8 Proof of Lemma 4.15

At at a first glance we note that by Proposition 4.9 and by (4.8), the expectation in (4.15) is bounded by

$$\begin{aligned} C(\rho ,f) t n^{2\alpha -2} \frac{(K^{j+1})^{1+\alpha }}{n^\alpha } \int (\Psi _x^j)^2 d\mu _\rho \le C(f,\rho ) t\frac{ (K^{j+1})^{\alpha +1}}{(K^j)^2n^{2-\alpha }}. \end{aligned}$$

(D.6)

But this bound will not be sufficient for us. Therefore, in order to prove the lemma we use the multiscale structure introduced in [16]. For that purpose, let j be fixed and take the sequence of boxes $\ell _0=K^j$, $\ell _1=2\ell _0$ and for $p\ge 2$, $\ell _p=2^p\ell _0$. Suppose that there exists P sufficiently big such that $2^P \ell _0=K^{j+1}$. Performing a telescopic sum and using the Minkowski’s inequality together with (D.6), the expectation on the left hand side of (4.15) is bounded from above by

$$\begin{aligned} C(f,\rho )t\left( \sum _{p=1}^P{\frac{(2^{p+1}\ell _0)^{\alpha -1}}{n^{2-\alpha }(2^p\ell _0)^{2}}}\right) ^2\le C(f,\rho )t \frac{(2^P\ell _0)^{\alpha -1}}{n^{2-\alpha }}. \end{aligned}$$

(D.7)

This proves (4.15) for the case $K^{j+1}=2^P \ell _0$. For the other cases, we take P sufficiently big such that $2^P\ell _0\le K^{j+1}\le 2^{P+1}\ell _0$. Let ${\tilde{\Psi }}_x^i=\Psi _x^{2^i\ell _0}$. Then, by using the inequality $(x+y)^2\le 2x^2+y^2$, the expectation on the left hand side of (4.15) is bounded from above by

$$\begin{aligned}&2{\mathbb E}_n\left[ \left( \int _0^t n^{\alpha -3/2} \sum _{x} \big ( \Psi _x^j(\eta _s^n) - {\tilde{\Psi }}_x^{j}(\eta _s^n)\big ) f' \big (\tfrac{x}{n} \big ) ds\right) ^2\right] \nonumber \\&\quad +\,2{\mathbb E}_n\left[ \left( \int _0^t n^{\alpha -3/2} \sum _{x} \big ( {\tilde{\Psi }}_x^{j}(\eta _s^n)- {\Psi }_x^{j+1}(\eta _s^n)\big ) f' \big (\tfrac{x}{n} \big ) ds\right) ^2\right] . \end{aligned}$$

(D.8)

From (D.7), the first expectation in (D.8) is bounded from above by

$$\begin{aligned} C(f,\rho )t \frac{(2^P\ell _0)^{\alpha -1}}{n^{2-\alpha }}, \end{aligned}$$

while from (D.6) the second expectation is bounded from above by

$$\begin{aligned} C(f,\rho ) t\frac{ (K^{j+1})^{\alpha +1}}{(2^P\ell _0)^2n^{2-\alpha }}\le C(f,\rho ) t\frac{ (2^{P+1}\ell _0)^{\alpha +1}}{(2^P\ell _0)^2n^{2-\alpha }}. \end{aligned}$$

Putting together the two previous estimates, the proof of (4.15) ends.

1.9 D.9 Proof of (4.18)

First we compute the price to double the size the box. For that purpose, let M be given. By Proposition 4.9 and (4.8) we have that

$$\begin{aligned} {\mathbb E}_n \left[ \left( \int _0^t \sum _x \big ( \psi _x^M(\eta _s^n) -\psi _x^{2M}(\eta _s^n)\big ) f'\big (\tfrac{x}{n} \big ) ds \right) ^2 \right] \le C(f,\rho ) t \sqrt{\frac{M}{n}}. \end{aligned}$$

Define $M_0=M$ and $M_i = 2^i M$ for $i \in {\mathbb N}$. By writing a telescopic sum, using Minkowski’s inequality and the previous estimate, we see that

$$\begin{aligned} \begin{aligned}&{\mathbb E}_n \left[ \left( \int _0^t \sum _x \big ( \psi _x^M(\eta _s^n) -\psi _x^{M_\ell }(\eta _s^n)\big ) f'\big (\tfrac{x}{n} \big ) ds \right) ^2 \right] \\&\quad ={\mathbb E}_n \left[ \left( \int _0^t \sum _x \sum _{i=0}{\ell -1}\big ( \psi _x^{2^iM}(\eta _s^n) -\psi _x^{2^{i+1}M}(\eta _s^n)\big ) f'\big (\tfrac{x}{n} \big ) ds \right) ^2 \right] \\&\quad \le \frac{C(f,\rho ) t}{\sqrt{n}} \left( \sum _{i=0}^{\ell -1} M_i^{1/4}\right) ^2\le C(f,\rho )t \sqrt{\frac{M_\ell }{n}}. \end{aligned} \end{aligned}$$

Taking $M= K_n$ and $M_\ell = \varepsilon n$ the proof ends.