Recurrence for vertex-reinforced random walks on Z with weak reinforcements

We prove that any vertex-reinforced random walk on the integer lattice with non-decreasing reinforcement sequence $w$ satisfying $w(k) = o(k^{\alpha})$ for some $\alpha<1/2$ is recurrent. This improves on previous results of Volkov (2006) and Schapira (2012).


Introduction
In this paper, we consider a one-dimensional vertex-reinforced random walk (VRRW) with non-decreasing weight sequence w : N → (0, ∞), that is a stochastic process X = (X n ) n≥0 on Z, starting from X 0 = 0, with transition probabilities: P{X n+1 = X n ± 1 | F n } = w(Z n (X n ± 1)) w(Z n (X n + 1)) + w(Z n (X n − 1)) where F n def = σ(X 1 , . . . , X n ) is the natural filtration of the process and Z n (x) def = #{0 ≤ k ≤ n, X k = x} is the local time of X on site x at time n. This process was first introduced by Pemantle in [3] and then studied in the linear case w(k) = k + 1 by Pemantle and Volkov in [5]. They proved the surprising fact that the walk visits only finitely many sites. This result was subsequently improved by Tarrès [7,8] who showed that the walk eventually gets stuck on exactly 5 consecutive sites almost surely. When the reinforcement sequence grows faster than linearly, the walk still gets stuck on a finite set but whose cardinality may be smaller than 5, see [1,9] for details. On the other hand, Volkov [9] proved that for sub-linearly growing weight sequences of order n α with α < 1, the walk necessarily visits infinitely many sites almost-surely. Later, Schapira [6] improved this result showing that, when α < 1/2, the VRRW is either transient or recurrent. The main result of this paper is to show that the walk is, indeed, recurrent. Theorem 1.1. Assume that the weight sequence satisfies w(k) = o(k α ) for some α < 1/2. Then X is recurrent i.e. it visits every site infinitely often almost-surely.
Let us mention that, simultaneously with the writing of this paper, a similar result was independently obtained by Chen and Kozma [2] who proved recurrence for the VRRW with weights of order n α , α < 1/2, using a clever martingale argument combined with previous local time estimates from Schapira [6]. The argument in this paper, while also making use of a martingale, is self-contained and does not rely upon previous results of Volkov [9] or Schapira [6]. In particular, we do not require any assumption on the regular variation of the weight function w.

A martingale
Obviously, multiplying the weight function by a positive constant does not change the process X. Thus, we now assume without loss of generality that w(0) = 1. We define the two-sided sequence (a x ) x∈Z by where ε > 0 will be chosen later during the proof of the theorem. Define also We construct from X two processes (M n ) n≥0 and (∆ n (z), z < X n ) n≥0 in the following way: 2. By induction, M n and (∆ n (z), z < X n ) having been constructed, Note that the quantities ∆ have a simple interpretation: for any n and z < X n , the value ∆ n (z) is positive and corresponds to the increments of M n the last time before n that the walk X jumped from site z to site z + 1 (with the convention ∆ n (z) = 1 for negative z if no such jumps occurred yet). By extension, we also define ∆ n def = ∆ n (X n ) at the current position as the "would be" increment of M n if X makes its next jumps to the right (at time n + 1) i.e.
We will also use the notation τ y to denote the hitting time of site y, Proposition 2.1. The process M is an F n -martingale and, for n ≥ 0, we have In particular, for y = 1, 2, . . . , the process M n∧τ−y is bounded below by −y/2, hence it converges a.s.
Proof. Since ∆ n (·) and Z n (·) are F n -measurable, by definition of M , thus M is indeed a martingale. Furthermore, by construction, at each time i when the process X crosses an edge {x, x + 1} from left to right, the process M increases by ∆ i = ∆ i+1 (x) > 0. If at some later time, say j > i, X crosses this edge again (and thus in the other direction), the martingale decreases by a x+1 ∆ j (x) = a x+1 ∆ i . Moreover, by convention ∆ 0 (z) = 1 and a z = 1 2 for z < 0 so that M decreases by 1 2 each time it crosses a new edge of the negative half line for the first time. Putting these facts together, we get exactly (1). Finally, since a z < 1 for any z ∈ Z, each term in the sum (1) is positive, hence M n∧τ−y is bounded below by 1 2 inf i≤n∧τ−y X i ≥ −y/2.
Proof. We prove by induction on n that for n ≤ τ −y , Recalling that w(k) ≥ 1 and a k ≤ 1 for any k, it is straightforward that (3) holds for n = 0. Now, assume the result for n and consider the two cases: • If X n+1 = X n −1. Then for any −y ≤ z ≤ X n+1 , we have ∆ n+1 (z) = ∆ n (z) whereas w(Z n+1 (z)) ≥ w(Z n (z)). Thus (3) holds for n + 1.
We can now recover, with our assumptions on w, Volkov's result [9] stating that the walk does not get stuck on any finite interval. Proof. On {τ −y = ∞}, the combination of (1) and Proposition 2.2 give Denoting by e n = (s n , s n + 1) the edge which has been most visited at time n, we deduce that on the event {τ −y = ∞}, where Z n (e n ) denotes the number of times the edge e n has been crossed from left to right before time n.
Using that max(Z n (s n ), Z n (s n + 1)) ≤ 2Z n (e n ) and that w(k) = o( √ k) and that M n∧τ−y converges, we conclude that on {τ −y = ∞}, either Z n (e n ) remains bounded or a sn+1 takes values arbitrarily close to 1. In any case, this means that X goes arbitrarily far to the right hence lim sup n X n = +∞.
3 Proof of theorem 1.1 Fix y > 0 and consider the event E y = {inf n X n = −y + 1}. Pick v > 0 and define N z to be the number of jumps of X from site z to site z + 1 before time τ v (according to the previous proposition τ v is finite on E y so all the N z are finite). From (4), grouping together the contributions to M of each edge (z, z + 1), we get, on E y , where C > 0 and α < 1/2 only depend on the weight function w. Finally, lemma 4.1 below states that if we choose ε > 0 small enough, the sum above becomes arbitrarily large as v tends to infinity. On the other hand, we also know that M converges on this event so necessarily P{E y } = 0. Since this result holds for any y > 0, we get inf X n = −∞ a.s. By symmetry, sup X n = +∞ a.s. which implies that the walk visits every site of the integer lattice infinitely often almost surely.

An analytic lemma
Proof. The idea is to group the b i 's into packets with respect to their value. Consider a reordering of the b i 's:b 0 ≥b 2 ≥ . . . ≥b K . Fix a positive integer l and group these numbers into l + 1 packets We can choose the K i 's growing geometrically such that the sizes of the packets satisfy We now regroup each term of the sum (5) according to which packet the central b i (the one appearing in the numerator) belongs. Assume by contradiction that the sum (5) is bounded, say by A.
We first consider only the terms corresponding to packet P 1 . Since there are at least K 4 l terms, we obtain the inequality where the constant C does not depend on K or on the sequence (b i ). We now deal with packets k = 2, . . . , l. Thanks to (6) and since every denominator in (5) involves only two b j other than the one appearing in the numerator, there are least one third of the terms belonging to packet P k that do not contain any b j from a packet with smaller index (i.e. with larger value). So, there is at least K 4 l such terms for which we can get a lower bound the same way we did for packet P 1 . We deduce that, summing over the terms corresponding to packet P k , with the convention K 0 = 0 and where C again does not depend on K or (b i ). Finally, we obtain a lower bound for the sum of the terms belonging to the last packet P l+1 by taking 1 2 as the lower bound for the numerator, and considering only the terms for which no b i 's from any other packet appear in the denominator (again, there are at least K 4 l such terms). This give the inequality Combining (7) and (8), we get by induction that for some constant C (depending on l), .
For ε small enough such that ε 1−2α ≤ 1 2 we obtaiñ Recalling that the sum (5) contains the term corresponding tob 0 but is also, by assumption, bounded above by A, we find Finally, we choose l large enough such that 1−2α 2(2α) l+1 − 1 − ε > 0 and we get a contradiction by letting K tends to infinity.