The Scaling Limit of the Directed Polymer with Power-Law Tail Disorder

Abstract

In this paper, we study the so-called intermediate disorder regime for a directed polymer in a random environment with heavy-tail. Consider a simple symmetric random walk $(S_n)_{n\ge 0}$ on $\mathbb {Z}^d$, with $d\ge 1$, and modify its law using Gibbs weights in the product form $\prod _{n=1}^{N} (1+\beta \eta _{n,S_n})$, where $(\eta _{n,x})_{n\ge 0, x\in {\mathbb {Z}}^d}$ is a field of i.i.d. random variables whose distribution satisfies ${\mathbb {P}}(\eta >z) \sim z^{-\alpha }$ as $z\rightarrow \infty$, for some $\alpha \in (0,2)$. We prove that if $\alpha < \min (1 + \frac{2}{d} ,2)$, when sending N to infinity and rescaling the disorder intensity by taking $\beta =\beta _N \sim N^{-\gamma }$ with $\gamma =\frac{d}{2\alpha }(1+\frac{2}{d}-\alpha )$, the distribution of the trajectory under diffusive scaling converges in law towards a random limit, which is the continuum polymer with Lévy $\alpha$-stable noise constructed in the companion paper (Berger and Lacoin in The continuum directed polymer in Lévy Noise, 2020. arXiv:2007.06484v2).

Appendices

Appendix A: Stochastic comparison: expectation versus integrals

We present here a technical result which allows to replace some expectations with respect to a random variable whose law $\mu$ satisfies $\mu ([u,+\infty ))= \varphi (u)u^{-\alpha }$ by integrals with respect to the measure with density $\alpha u^{-(1+\alpha )}\varphi (u) \mathrm{d}u$ (which is not necessarily a probability).

Proposition A.1

Let $\mu$ be a probability measure on ${\mathbb {R}}_+$ that satisfies $\mu \big ( [t,+\infty ) \big )=\varphi (t)t^{-\alpha }$ for some slowly varying $\varphi$. There exist constants C and $B_0$ (depending on $\varphi$ and $\alpha$) such that for all $k\in {\mathbb {N}}$ and for all non-decreasing function $f : {\mathbb {R}}_+^k \rightarrow {\mathbb {R}}_+$ with $f(0)=0$, and all $B\ge B_0$, we have

$$\begin{aligned} \int _{[0,B)^k} f(u_1,\ldots ,u_k) \prod _{i=1}^k \mu (\mathrm{d}u_k)&\le C^k \int _{[0, 2B)^k} f(u_1,\ldots ,u_k) \prod _{i=1}^k u_i^{-(1+\alpha )} \varphi (u_i)\mathrm{d}u_i \, . \end{aligned}$$

Proof

The first remark is that we need to show the result only in the case $k=1$. Integrating successively the functions $u_i \mapsto f(u_1, \ldots , u_k)$ then yields the result. Also, it is sufficient to check the result for a function f that is differentiable and bounded (the other cases can be obtained by monotone convergence). We set $\overline{F}(u) = \mu ([u,\infty ))$. Using an integration by parts, and applying these inequalities (recall that $f'(u)\ge 0$) we get

$$\begin{aligned} \int _{[0,B)} f(u) \mu (\mathrm{d}u) \, = \int _{[0,B)} f'(u) \overline{F}(u) \mathrm{d}u - f(B) \overline{F}(B) \end{aligned}$$

(A.1)

Now we set

$$\begin{aligned} \overline{\varphi }(u):= u^{\alpha } \int ^{\infty }_{u} \alpha v^{-(1+\alpha )} \varphi (v) \mathrm{d}v. \end{aligned}$$

Since $\overline{\varphi }$ is asymptotically equivalent to $\varphi$ at $\infty$, and to $\alpha u^{\alpha }\log u$ at 0, we have $\varphi \le C \overline{\varphi }$.

$$\begin{aligned} \int _{[0,B)} f'(u) \overline{F}(u) \mathrm{d}u&\le C \int _{[0,B)} f'(u) u^{-\alpha } \overline{\varphi }(u) \mathrm{d}u \\&=C\alpha \left( \int _{[0,B)} f(u) u^{-(1-\alpha )} \varphi (u) \mathrm{d}u + f(B) B^{-\alpha }\overline{\varphi }(B)\right) \,, \end{aligned}$$

where we used another integration by parts for the last identity. Hence we obtain that

$$\begin{aligned} \int _{[0,B)} f(u) \mu (\mathrm{d}u) \le C \alpha \int _{[0,B)} f(u) u^{-(1-\alpha )} \varphi (u) \mathrm{d}u + C\alpha f(B) B^{-\alpha } \overline{\varphi }(B) \,. \end{aligned}$$

(A.2)

Now, to conclude with use that f is non-decreasing and that $\overline{\varphi }$ and $\varphi$ are asymptotically equivalent to obtain that for B sufficiently large

$$\begin{aligned} f(B) B^{-\alpha }\overline{\varphi }(B) \le C' \int _{[B,2B)} f(u) u^{-(1+\alpha )} \varphi (u) \mathrm{d}u , \end{aligned}$$

so that the second term in (A.2) can be absorbed into the first one. $\quad \square$

Proposition A.2

Let $\mu$ be a probability measure on ${\mathbb {R}}_+$ such that $\mu \big ([t,\infty ) \big ) \, \leqslant \,t^{-\alpha } \varphi (t)$ for all $t\ge 0$, with $\alpha \in (0,2)$. Then there is some constant C such that for all $k\in {\mathbb {N}}$ and for all non-increasing function $f : {\mathbb {R}}_+^k \rightarrow {\mathbb {R}}_+$ with bounded support, we have

$$\begin{aligned} \int _{{\mathbb {R}}^k_+} f(u_1,\ldots ,u_k) \prod _{i=1}^k u_i^2 \mu (\mathrm{d}u_i) \le C^k \int _{{\mathbb {R}}^k_+} \ f(u_1,\ldots ,u_k) \prod _{i=1}^k u_i^{1-\alpha } \varphi (u_i)\mathrm{d}u. \end{aligned}$$

(A.3)

Proof

As for Proposition A.1, we only need to prove the result in the case $k=1$, for a differentiable function f. Let T be such that the support of f is included in [0, T) and that of $f'$ is included in (0, T). We define $\widetilde{\varphi }$ by

$$\begin{aligned} \widetilde{\varphi }(u)= (2-\alpha )u^{\alpha -2}\int ^t_0 \varphi (u) u^{1-\alpha } \mathrm{d}u. \end{aligned}$$

(A.4)

Let also $\widetilde{\mu }$ denote the measure on [0, T] defined by $\widetilde{\mu }(\mathrm{d}t)= t^2 \mu (\mathrm{d}t)$. By an integration by parts, we get, using our assumption on $\mu$, that for all $t\ge 0$

$$\begin{aligned} \widetilde{\mu }([0,t)) = \int _0^t s^2 \mu (\mathrm{d}s)&= - t^2 \mu ((t,\infty )) + 2 \int _0^t s \mu ((s,\infty )) \mathrm{d}s\\&\, \leqslant \,2 \int _0^t s^{1-\alpha } \varphi (s) \mathrm{d}s = \frac{2}{2-\alpha } t^{2-\alpha } \widetilde{\varphi }(t) \,. \end{aligned}$$

Therefore, thanks to an integration by parts (using that $f(T)=0$), we get that

$$\begin{aligned} \int _{[0,T]} f(u) u^2 \mu (\mathrm{d}u)&= -\int _{[0,T]} f'(u) \widetilde{\mu }([0,u])\mathrm{d}u \le C\int _{[0,T]} (-f'(u)) u^{2-\alpha } \widetilde{\varphi }(u)\mathrm{d}u \, , \end{aligned}$$

where we have used that $-f'(u) \ge 0$ so the inequality goes in the right direction. We conclude the proof by another integration by parts. $\quad \square$

Let us conclude this section with the proof of a useful (and standard) identity, used repeatedly in the paper.

Lemma A.3

For any $t >0$, $k\ge 0$ and $\zeta _1,\dots ,\zeta _{k+1} >0$, using the convention $s_0=0$ and $s_{k+1}=t$ we have, for all $k\ge 1$,

$$\begin{aligned} \int _{0<s_1< \cdots<s_k < t } \prod _{i=1}^{k+1} (s_i-s_{i-1})^{\zeta _i-1} \mathrm{d}s_i = t^{\sum _{i=1}^{k+1}\zeta _i-1} \frac{\prod _{i=1}^{k+1}\Gamma (\zeta _i)}{\Gamma (\sum _{i=1}^{k+1}\zeta _i)} \,. \end{aligned}$$

Let us stress that in this paper we use this identity with $\zeta _i=\zeta$ for all $i=1,\dots , k$ and either $\zeta _{k+1}=\zeta$ or $\zeta _{k+1}=1$.

Proof

By scaling it is sufficient to prove the identity for $t=1$. We have

$$\begin{aligned} \prod _{i=1}^{k+1}\Gamma (\zeta _i) =\int _{(0,\infty )^{k+1}} \prod _{i=1}^{k+1} u^{\zeta _i-1}_i e^{-\sum _{i=1}^{k+1} u_i}\mathrm{d}u_i \,. \end{aligned}$$

(A.5)

Using the change of variables $(u_1,\dots ,u_{k+1})\rightarrow (s_1,\dots ,s_k,v)$ where $v:=\sum _{i=1}^{k+1} u_i$ and $s_j:=\frac{\sum _{i=1}^j u_i}{v}$ for $j\in \llbracket 0,k\rrbracket$, we obtain

$$\begin{aligned} \prod _{i=1}^{k+1}\Gamma (\zeta _i) = \int _{(0,\infty )} v^{\sum _{i=1}^{k+1}\zeta _i-1} e^{-v} \mathrm{d}v \int _{0<s_1< \cdots<s_k < 1 } \prod _{i=1}^{k+1} (s_i-s_{i-1})^{\zeta _i-1} \prod _{i=1}^k \mathrm{d}s_i \, , \end{aligned}$$

(A.6)

which yields the result. $\quad \square$

Appendix B: Tightness for $\xi ^{\eta }_N$

First of all, let us recall the definition of the functional space $H_\mathrm{loc}^s({\mathbb {R}}^{d+1})$. Given $s\in {\mathbb {R}}$, let $H^s({\mathbb {R}}^{d+1})$ be defined as the topological closure of the space of smooth and compactly supported functions, with respect to the norm

$$\begin{aligned} \Vert f\Vert _{H^s} =\Big ( \int _{{\mathbb {R}}^{d+1}} (1+|z|^2)^s |\widehat{f} (z)|^2 \mathrm{d}z \Big )^{1/2} \, , \end{aligned}$$

where $\widehat{f} (z) = \int _{{\mathbb {R}}^{d+1}} f(x) e^{ -i x \cdot z} \mathrm{d}x$ is the Fourier transform of f. The associated local Sobolev space is given by

$$\begin{aligned} H_\mathrm{loc}^{s} ({\mathbb {R}}^{d+1}) := \big \{ f :f\psi \in H^{s} \text { for every compactly supported } \psi \in C^{\infty } \big \} \end{aligned}$$

with the topology induced by the family of semi-norms $(\Vert f \psi \Vert _{H^s} )_{\psi }$.

Proof of Lemma 3.7

First of all, let us notice that we can write

$$\begin{aligned} \xi _{N,\eta } - \xi _{N,\eta }^{ (a)} := \frac{1}{V_N} \sum _{(n,x)\in \mathbb {H}_d} \overline{\eta }^{ (a)}_{n,x}\, \delta _{(\frac{n}{N}, \frac{x}{\sqrt{N/d}})} \,, \end{aligned}$$

(B.1)

where

$$\begin{aligned} \overline{\eta }^{ (a)}_{n,x} := \left( \eta _{n,x} - {\mathbb {E}}\big [\eta \, \big | \, 1+\eta< a V_N \big ] \right) \mathbf{1}_{\{ 1+ \eta _{n,x} < a V_N \}}\, . \end{aligned}$$

(B.2)

Notice that ${\mathbb {E}}[ \overline{\eta }^{ (a)}_{n,x}] =0$. Now using (4.3), the monotonicity in a and continuity at 0, there exists a function $\varepsilon : (0,1)\rightarrow {\mathbb {R}}_+$ with $\lim _{a\rightarrow 0} \varepsilon (a)=0,$ such that for every $N\ge 1$

$$\begin{aligned} V^{-2}_N{\mathbb {E}}[ (\overline{\eta }^{ (a)}_{n,x} )^2] \, \leqslant \,\varepsilon (a) N^{ -(\frac{d}{2} +1)}. \end{aligned}$$

Hence we have

$$\begin{aligned} {\mathbb {E}}\left[ \langle \xi _{N,\eta } - \xi _{N,\eta }^{ (a)}, \psi \rangle ^2 \right] \le \varepsilon (a) N^{ - (\frac{d}{2} +1)} \sum _{(n,x)\in {\mathbb {H}}^d} \psi \left( \frac{n}{N}, \frac{x}{\sqrt{N/d}} \right) ^2. \end{aligned}$$

(B.3)

Since the Riemann sum in the r.h.s. converges, we have

$$\begin{aligned} \limsup _{N \rightarrow +\infty } {\mathbb {E}}\left[ \langle \xi _{N,\eta } - \xi _{N,\eta }^{(a)}, \psi \rangle ^2 \right] \, \leqslant \,C_{\psi } \, a^{2-\alpha } \,, \end{aligned}$$

which concludes the proof. $\quad \square$

Proof of Lemma 3.2

We have to show that for every smooth $\psi$ with compact support, the sequence $\xi ^{\eta ,\psi }_N:=\psi \times \xi ^{\eta }_N$ is tight in $H^{s}({\mathbb {R}}^{d+1})$. This corresponds to showing that $\widehat{\xi }^{\eta ,\psi }_N$ is tight in $L^2(\mu ^s)$ for $\mu ^s=(1+ |z|^2)^{-s}\mathrm{d}z$.

We are going to show that with large probability $\widehat{\xi }^{\eta ,\psi }_N\in K_{R}$ where $K_R$ is defined (for a fixed $s'>s$)

$$\begin{aligned} K_R:=&\bigg \{ f \ : \ \int |f(z)|^2 (1+ |z|^2)^{-s'} \mathrm{d}z \le R \nonumber \\&\qquad \text { and } \forall a\in {\mathbb {R}}^{d+1}, \ \int |f(z+a)-f(z)|^2 (1+ |z|^2)^{-s} \mathrm{d}z \le R |a| \bigg \}. \end{aligned}$$

(B.4)

Since $K_R$ is compact (by Frechet–Kolmogorov criterion) this is sufficient to conclude that the distribution of $\xi ^{\eta ,\psi }_N$ is tight.

To see that $\widehat{\xi }^{\eta ,\psi }_N\in K_R$ with large probability, we first observe that $\xi ^{\eta ,\psi }_N$ coincides with large probability with $\xi ^{\eta ,\psi ,[0,b)}_N$ (constructed from the environment $\eta ^{[0,b)}$, recall (3.8)). Then we have by a computation similar to (B.3), for all N sufficiently large

$$\begin{aligned} {\mathbb {E}}\left[ \big |\widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z)\big |^2 \right] \le C_b \Big (\int |\psi |^2\Big ) \end{aligned}$$

(B.5)

so that

$$\begin{aligned} {\mathbb {P}}\left[ \int \left| \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z)\right| ^2 (1+ |z|^2)^{-s'} \mathrm{d}z \ge R \right] \le \frac{1}{R} C_{b,\psi }\, . \end{aligned}$$

(B.6)

For the second point we observe that

$$\begin{aligned} {\mathbb {E}}\left[ \left| \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z+a)- \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(a)\right| ^2 \right] \le C'_{b,\psi } |a|^2 \int |x|^2 |\psi (x)|^2 \mathrm{d}x. \end{aligned}$$

(B.7)

(Note that $\widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z+a)- \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(a)$ is the Fourier transform of the map $x\mapsto (e^{i a.x}-1) \psi \times \xi ^{\eta ,\psi ,[0,b)}_N$, so we are simply bounding the first factor by |a||x|.) We therefore have that

$$\begin{aligned} {\mathbb {P}}\left( \int \left| \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z+a)- \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(a)\right| ^2(1+ |z|^2)^{-s} \mathrm{d}z \ge |a| \right) \le C_{b,\psi }'' |a|. \end{aligned}$$

(B.8)

Hence, using a union bound, we obtain that

$$\begin{aligned}&{\mathbb {P}}\left( \exists k\ge k_0, \exists i\in \llbracket 1,d\rrbracket \,\, \int \left| \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z+2^{-k}e_i)- \widehat{\xi }^{\eta ,\psi ,[0,b)}_N(z)\right| ^2(1+ |z|^2)^{-s} \mathrm{d}z \ge 2^{-k} \right) \\&\quad \le \varepsilon (k_0), \end{aligned}$$

with $\lim _{k_0\rightarrow \infty } \varepsilon (k_0)=0$. This is sufficient to conclude that $\widehat{\xi }^{\eta ,\psi ,[0,b)}_N\in K_R$ with probability close to one, and thus so is $\widehat{\xi }^{\eta ,\psi }_N$. $\quad \square$