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Comptes RendusMathématiqueQiaonan Gu, Kan Jiang and Lifeng XiBedford–McMullen carpets meet nonlinear curvesVolume 363 (2025), p. 845-851Online since: 15 July 2025https://doi.org/10.5802/crmath.752This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/C EN T R EMER S ENN EThe Comptes Rendus. Mathématique are a member of theMersenne Center for open scientific publishingwww.centre-mersenne.org — e-ISSN : 1778-3569Comptes Rendus. Mathématique2025, Vol. 363, p. 845-851https://doi.org/10.5802/crmath.752Research article / Article de rechercheDynamical systems / Systèmes dynamiquesBedford–McMullen carpets meet nonlinearcurvesLes tapis de Bedford–McMullen intersectent des courbesnon linéairesQiaonan Gu a, Kan Jiang a and Lifeng Xi aa Department of Mathematics, Ningbo University, Ningbo, People’s Republic of ChinaE-mails: 13586815362@163.com (Qiaonan Gu), jiangkan@nbu.edu.cn, xilifeng@nbu.edu.cnAbstract. Let f be a regular function (the precise definition will be provided in the paper). Then, for anyα ∈ (0,2) and any ϵ> 0, there exists a Bedford–McMullen carpet K with convex hull [0,1]2 such thatα−ϵ< dimH (K ) <α and{(x, y) : y = f (x)}∩K = {(0,1)},where dimH denotes the Hausdorff dimension. In particular, this result holds for y = cos(x). Unlike classicalmethods for analyzing slicing sets, our proofs rely exclusively on mathematical induction.Résumé. Soit f une fonction régulière (la définition précise sera donnée dans l’article). Alors, pour toutα ∈ (0,2) et tout ϵ> 0, il existe un tapis de Bedford–McMullen K d’enveloppe convexe [0,1]2 tel queα−ϵ< dimH (K ) <α et{(x, y) : y = f (x)}∩K = {(0,1)},où dimH désigne la dimension de Hausdorff. En particulier, ce résultat est valable pour y = cos(x). Contrai-rement aux méthodes classiques d’analyse des ensembles de coupe, notre démonstration repose exclusive-ment sur un raisonnement par récurrence.2020 Mathematics Subject Classification. 28A80, 28A78.Funding. Kan Jiang was supported by Zhejiang Provincial Natural Science Foundation of China under GrantNo. LMS25A010009. Lifeng Xi was supported by National Natural Science Foundation of China under GrantNos. 12171259, 11831007, and 11771226.Manuscript received 13 February 2025, revised 9 April 2025, accepted 18 April 2025.1. IntroductionLet K ⊂ R2 be a fractal set and Γ ⊂ R2 be a curve. In geometric measure theory, determining theHausdorff dimension of K ∩Γ is a challenging problem [2]. Classical results, such as Marstrand’stheorem [2], provide upper bounds for the dimension of linear slices. Roughly speaking, fora given fractal set K ⊂ R2 and a line Γ ⊂ R2, generically, the Hausdorff dimension of K ∩ Γ isdimH (K )−1. The reader may refer to Mattila’s book [2, Theorem 10.11]. However, the situationbecomes significantly more complicated when dealing with nonlinear slicing sets. For instance,when K is a Bedford–McMullen carpet and Γ is a nonlinear curve, the analysis of K ∩ Γ isgenerally non-trivial. To date, there has been limited references in this area. In this paper, weISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/846 Qiaonan Gu, Kan Jiang and Lifeng Xifocus on the case where K is a Bedford–McMullen carpet and Γ is a nonlinear curve satisfyingspecific conditions. We prove that although some Bedford–McMullen carpet has large Hausdorffdimension, it simultaneously intersects many smooth curves with only finite points.We begin by introducing the following definition.Definition 1. Let f ∈C∞(R) be a function. Suppose that s is the smallest positive integer such thatf (s)(0) ̸= 0, where f (s) denotes the s-th order of the derivative and s ∈ N≥2. Let m = ns ,n ≥ 10100.We say that f is a regular function if it satisfies the following properties:(1) f (0) = 1, 1−1/m > f (1−1/n) > f (1) > 1/m;(2) f is strictly decreasing on [0,1];(3) (n −1)−s ≤ − f (s)(x)s! ≤ 1−1/m, ∀ x ∈ [0,1/3].Remark 2. We give some remarks on the condition:1−1/m > f (1−1/n) > f (1) > 1/m.Since f is strictly decreasing, f (1−1/n) > f (1) is clear. If we let n →∞, then the above inequalitiesbecomef (0) = 1 ≥ f (1) ≥ 0.The first inequality is clear as f is strictly decreasing. If0 ≤ a ≤ n/3−1, 0 ≤ b ≤ m/3,then the third condition implies that∣∣∣∣ f (s)(x)s! (a +1)sns∣∣∣∣≤ ∣∣∣∣ f (s)(x)s! 13s∣∣∣∣≤ m −b −1m , x ∈ [0,1/3].We shall use this inequality in the proof of Theorem 5.We present some essential definitions that will be used throughout the paper. Let N+ denotethe set of all positive integers. For a given integer k ∈ N+, we define N≥k ={x ∈ N+ : x ≥ k}. Forany positive real number α ∈R+, we denote by [α] the integer part of α. For instance, [2.025] = 2.We recall the definition of Bedford–McMullen carpet. Let n,m ∈ N≥2, m > n, and let A be asubset of {0,1, . . . ,n−1}×{0,1, . . . ,m−1}. Without loss of generality, we assume that the cardinalityof A is at least 2, and call A the digit set. We define a class of self-affine functions as follows:ga(x, y) =(x +xin,y + yim), a = (xi , yi ) ∈A .The collection of {ga}a∈A is called an iterated function system (IFS). Hutchinson [1] proved thatthere exists a unique non-empty compact set K satisfyingK = ⋃a∈Aga(K ).Such a set K is referred to as a Bedford–McMullen carpet. For any k ∈N+ and ai ∈A with 1 ≤ i ≤ k,we definega1a2···ak([0,1]2)= ga1 ◦ ga2 ◦ · · · ◦ gak ([0,1]2)as a basic rectangle at the k-th level. If the curve y = f (x) intersects a basic rectangle I , we denoteby (x, y) f ∈ I .The following result, due to McMullen [3], provides the Hausdorff dimension of a Bedford–McMullen carpet.Theorem 3. Let K be a Bedford–McMullen carpet associated with the IFS {gai }ai∈A . Then theHausdorff dimension of K is given bydimH (K ) =ln(∑n−1j=0 tlogm nj)lnn,Qiaonan Gu, Kan Jiang and Lifeng Xi 847where t j denotes the number of indices i such that ( j , i ) ∈A .We state the main result of this paper.Theorem 4. Let f be a regular function defined in Definition 1. Then for any α ∈ (0,2) and anyϵ> 0, there exists a Bedford–McMullen carpet K with the convex hull [0,1]2 such thatα−ϵ< dimH (K ) <α,and that {(x, y) : y = f (x)}∩K = {(0,1)},where dimH stands for the Hausdorff dimension.Theorem 4 immediately follows from the following result.Theorem 5. Let f be a regular function defined in Definition 1. Let K be a Bedford–McMullencarpet defined by the digit setD = {(0,0), (0,m −1), (n −1,0), (n −1,m −1), (ai ,bi )}, 1 ≤ i ≤ u, u ∈N+.If m = ns with n ≥ 10100, and 0 ≤ ai ≤ n/3−1, 0 ≤ bi ≤ m/3, ai ,bi ∈N+∪ {0}, then{(x, y) : y = f (x)}∩K = {(0,1)}.Applying Theorem 4, we obtain the following example.Example 6. For anyα ∈ (0,2), ϵ> 0 and any closed interval [κ1,κ2] ⊂ (0,1), there exists a Bedford–McMullen carpet K such thatα−ϵ< dimH (K ) <α,( ⋃β∈[0.9,1]{(x, y) : y = cos(βx)})∩K = {(0,1)},and ( ⋃γ∈[κ1,κ2]{(x, y) : y = 1−γx2})∩K = {(0,1)}.Let u(x) = cos(βx), β ∈ [0.9,1] and v(x) = 1−γx2, γ ∈ [κ1,κ2]. It is easy to check that for thesetwo functions we have s = 2. Hence, when we construct some Bedford–McMullen carpet, we canassume m = n2. Note that κ1 and κ2 are two fixed numbers. Hence, if n, m = n2 are sufficientlylarge, v(x) and u(x) are regular functions. So, Example 6 follows from Theorem 4 and the proof ofTheorem 5.By Theorem 4, numerous other functions can be readily identified. We do not give moreexamples. The key insight of this example lies in the existence of a uniform Bedford–McMullenset that intersects a variety of distinct functions at a single unique point.This paper is organized as follows. In Section 2, we prove Theorem 4 via Theorem 5. Section 3is entirely devoted to the detailed proof of Theorem 5.2. Proof of Theorem 4 by using Theorem 5Proof. For any α ∈ (0,1), if m is sufficiently large, then [mα] ∈ (0,m/3). Thus, we may select [mα]basic rectangles within the vertical strip with the form [0,1/n]×[i /m, (i +1)/m], 0 ≤ i ≤ [m/3]−1,m = ns . Additionally, we include the four basic rectangles at the corners of the unit square at thefirst level. We denote the resulting Bedford–McMullen carpet by K1. In light of Theorem 3, theHausdorff dimension of K1 is given bylog(([mα]+1)1/s +21/s)logn.848 Qiaonan Gu, Kan Jiang and Lifeng XiSince m = ns , the above dimension converges to α as n → ∞. Furthermore, by means ofTheorem 5, the curve y = f (x) meets the Bedford–McMullen carpet K1 with a unique point, i.e.(0,1).For α ∈ (1,2), we have α−1 ∈ (0,1). Thus, for sufficiently large m,[mα−1] ∈ [0,m/3].We can then choose [n/3] distinct vertical strips (each of the form[i /n, (i +1)/n]× [0,1], where0 ≤ i ≤ [ n3 ]−2) in the unit square at the first level. For each strip, we select [mα−1] basic rectanglesof the form: [i /n, (i +1)/n]× [ j /m, ( j +1)/m], 0 ≤ i ≤ [n/3]−2, 0 ≤ j ≤ [m/3]−1.We also include the four-cornered basic rectangles at the first level. Let K2 be the resultingBedford–McMullen carpet. Applying McMullen’s result once again, the Hausdorff dimension ofK2 islog(([mα−1]+1)1/s + ([mα−1])1/s × ([ n3 ]−2)+21/s)logn,which converges to α as n →∞. By virtue of Theorem 5, we conclude thatK2 ∩{(x, y) : y = f (x)}= {(0,1)}.3. Proof of Theorem 5Proof. The digit set D consists of four digits{(0,0), (0,m −1), (n −1,0), (n −1,m −1)}.Define the following functions:f1(x, y) = (x/n, y/m),f3(x, y) =((x +n −1)/n, y/m),f4(x, y) =((x +n −1)/n, (y +m −1)/m),f5(x, y) =(x/n, (y +m −1)/m).To represent u digits (ai ,bi ), 1 ≤ i ≤ u, we use a uniform function:f2(x, y) =((x +a)/n, (y +b)/m), 0 ≤ a ≤ n/3−1, 0 ≤ b ≤ m/3, a,b ∈N+∪ {0}.For instance, let n = 5, m = 25, a = 1, b = 11, y = cos(x), the reader may refer to Figure 1 for thefirst iteration.In what follows, we let F (x1, x2) = x2 − f (x1). Suppose y = f (x) and (x, y) ∈ K . Since f (0) = 1, itfollows that the curve y = f (x) intersects f5([0,1]2). Moreover, it is straightforward to verify that:F(f1(1,1))≤ F ( f3(1,1))= 1/m − f (1) < 0,as f (1) > 1/m. Since f is strictly decreasing and∂x1 F =− f ′(x1) ≥ 0, ∂x2 F = 1 > 0,by the above inequalities, it follows that the four corners of the rectangles, i.e. f3([0,1]2)andf1([0,1]2), are lying below the curve y = f (x). Thus, the curve y = f (x) does not intersectf1([0,1]2)or f3([0,1]2). We shall use similar ideas to prove other cases. Next, observe that:F(f2(1,1))= b +1m− f(a +1n).Qiaonan Gu, Kan Jiang and Lifeng Xi 849Figure 1. y = cos(x) intersects basic rectangles at the first level.By Taylor’s formula and the definition of s, there exists ξ ∈ (0,(a +1)/n)) such thatF(f2(1,1))= b +1m− f (0)− f(s)(ξ)s!(a +1)sns.To proveF(f2(1,1))< 0,it suffices to show ∣∣∣∣ f (s)(ξ)s! (a +1)sns∣∣∣∣≤ m −b −1m .Since 0 ≤ a ≤ n/3−1, 0 ≤ b ≤ m/3, we only need:∣∣∣∣ f (s)(ξ)s! (a +1)sns∣∣∣∣≤ ∣∣∣∣ f (s)(ξ)s! 13s∣∣∣∣≤ m −b −1m .The last inequality holds due to Remark 2. Hence, the curve y = f (x) does not intersect f2([0,1]2).Using the conditions of the regular function, we haveF(f4(0,0))= 1−1/m − f (1−1/n) > 0,and1/m < f (1) < 1−1/m.Hence, at the first level, the curve y = f (x) intersects only f5([0,1]2).If (x, y) f ∈ f5([0,1]2), we can show via induction that for any k ∈N+,(x, y) f ∈ f5k([0,1]2).Taking k → ∞, we obtain (x, y) = (0,1). To prove this, assume that at the k-th level, (x, y) f ∈f5k([0,1]2), we aim to show (x, y) f ∈ f5k+1([0,1]2). Since f is strictly decreasing, it suffices to verify:F(f5k 3(1,1))< 0, F ( f5k 4(0,0))> 0, F ( f5k 2(1,1))< 0,and1− (m −1)/mk+1 < f (1/nk ) < 1−1/mk+1.850 Qiaonan Gu, Kan Jiang and Lifeng XiHere f5k i (x, y) = f5 ◦ · · · ◦ f5︸ ︷︷ ︸k◦ fi (x, y), i = 2,3,4. Note thatF(f5k 3(1,1))= 1−1/mk +1/mk+1 − f ( 1nk).By Taylor’s theorem, there exists some ς ∈ (0,1/nk ) such thatF(f5k 3(1,1))= 1−1/mk +1/mk+1 −1− f (s)(ς)s!1nks.Hence, to show F(f5k 3(1,1))< 0, it suffices to prove− f(s)(ς)s!< 1−1/m, ς ∈ (0,1/nk ) ⊂ (0,1/3).However, this is one of the conditions for the regular function.For the second inequality, we haveF(f5k 4(0,0))= 1−1/mk+1 − f (n −1nk+1).Employing Taylor’s theorem:F(f5k 4(0,0))= 1−1/mk+1 −(1+ f (s)(η)s!(n −1nk+1)s),where η ∈ (0, n−1nk+1). To show F(f5k 4(0,0))> 0, it suffices to have− f (s)(x)/s! > (n −1)−sfor any x ∈ [0,1/3]. This is guaranteed by the conditions of the regular function. For the thirdinequality:F(f5k 2(1,1))= b +1mk+1+1− 1mk− f(a +1nk+1).Utilizing Taylor’s theorem:b +1mk+1+1− 1mk− f(a +1nk+1)= b +1mk+1+1− 1mk−(1+ f(s)(κ)s!(a +1nk+1)s),where κ ∈ (0,(a +1)/nk+1). However, by the following conditions, i.e.m = ns , 0 ≤ a ≤ n/3−1 and 0 ≤ b ≤ m/3,the inequality F(f5k 2(1,1))< 0 holds from∣∣ f (s)(x)/s!∣∣≤ 3s((m −b −1)/m), ∀ x ∈ [0,1/3].This is a consequence of the regular function, as noted in Remark 2. Finally, we verify:1− (m −1)/mk+1 < f (1/nk ) < 1−1/mk+1.In terms of Taylor’s theorem again,f (1/nk ) = f (0)+ f(s)(ξ)s!1nks= 1+ f(s)(ξ)s!1nks,where ξ ∈ (0,1/nk ). Assuming:1ns= 1/m < − f(s)(x)s!< 1−1/mfor any x ∈ [0,1/3], the inequalities hold. This is ensured by the conditions of the regular function.Qiaonan Gu, Kan Jiang and Lifeng Xi 851AcknowledgmentsThe authors would like to express their gratitude to the referee for the helpful comments.Declaration of interestsThe authors do not work for, advise, own shares in, or receive funds from any organizationthat could benefit from this article, and have declared no affiliations other than their researchorganizations.References[1] J. E. Hutchinson, “Fractals and self-similarity”, Indiana Univ. Math. J. 30 (1981), no. 5,pp. 713–747.[2] P. Mattila, Geometry of sets and measures in Euclidean spaces. Fractals and rectifiability.,1st paperback edition, Cambridge Studies in Advanced Mathematics, Cambridge UniversityPress, 1999, xii+343 pages.[3] C. McMullen, “The Hausdorff dimension of general Sierpiński carpets”, Nagoya Math. J. 96(1984), pp. 1–9.

Bedford–McMullen carpets meet nonlinear curves

Idriss Mazari-Fouquer

Another look at qualitative properties of eigenvalues using effective Hamiltonians

Asaf Nachmias

Yuval Peres

Comptes RendusMathématiqueAsaf Nachmias and Yuval PeresEvery recurrent network has a potential tending to infinityVolume 363 (2025), p. 793-798Online since: 4 July 2025https://doi.org/10.5802/crmath.758This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/C EN T R EMER S ENN EThe Comptes Rendus. Mathématique are a member of theMersenne Center for open scientific publishingwww.centre-mersenne.org — e-ISSN : 1778-3569Comptes Rendus. Mathématique2025, Vol. 363, p. 793-798https://doi.org/10.5802/crmath.758Research article / Article de rechercheProbability theory / ProbabilitésEvery recurrent network has a potentialtending to infinityTout réseau récurrent a un potentiel tendant vers l’infiniAsaf Nachmias a and Yuval Peres ba Department of Mathematical Sciences, Tel Aviv University, Israelb Beijing Institute of Mathematical Sciences and Applications, Beijing, ChinaE-mails: asafnach@tauex.tau.ac.il, yperes@gmail.comAbstract. A rooted network consists of a connected, locally finite graph G , equipped with edge conductancesand a distinguished vertex o. A nonnegative function on the vertices of G which vanishes at o, has Laplacian 1at o, and is harmonic at all other vertices is called a potential. We prove that every infinite recurrent rootednetwork admits a potential tending to infinity. This is an analogue of classical theorems due to Evans andNakai in the settings of Euclidean domains and Riemannian surfaces.Résumé. Un réseau enraciné est un graphe G localement fini et connexe, doté de conductances sur les arêteset d’un sommet distingué o. Un potentiel est une fonction sur les sommets de G qui est positive, s’annuleen o, a un laplacien 1 en o et est harmonique en tous les autres sommets. Nous prouvons que tout réseauenraciné, récurrent et infini admet un potentiel tendant vers l’infini. Il s’agit d’un analogue de théorèmesclassiques de Evans et Nakai dans le cadre des domaines euclidiens et des surfaces riemanniennes.Funding. This work was supported by ERC consolidator grant 101001124 (UniversalMap) as well as ISFgrants 1294/19 and 898/23.Manuscript received 27 February 2025, revised and accepted 6 May 2025.1. IntroductionA rooted network (G ,c,o) consists of a connected, locally finite graph G = (V ,E), endowed withpositive edge weights c : E → (0,∞) and a distinguished vertex o ∈V . The corresponding networkrandom walk with transition probabilities p(x, y) = cx y /cx is reversible, with reversing measurecx =∑z∼x cxz . Define the network Laplacian of a function f : V →R by∆ f (v) = ∑x∼vcv x(f (x)− f (v)).A potential on (G ,c,o) is a function h : V → [0,∞) that is harmonic off o (i.e., ∆h(v) = 0 forevery vertex v ̸= o), normalized so that h(o) = 0 and ∆h(o) = 1. In every infinite rooted networkthere is at least one potential, see Claim 5. Our focus is on recurrent networks.We say that a function h : V → [0,∞) tends to infinity and write h → ∞, if for all M > 0 theset{v ∈ V : h(v) ≤ M} is finite. Observe that on Z, the potentials h1(n) = max(n,0) for n ∈ Z andh2(n) = max(−n,0) do not tend to infinity, but any non-trivial convex combination of them does.ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/794 Asaf Nachmias and Yuval PeresOn Z2, on the other hand, it is well known that there is a unique potential h with h(v) =Θ(log|v |)as v → ∞, see [6, Sections 12 and 31]. Recently there has been renewed interest in recurrentpotential theory, motivated by close connections to unimodular random networks and uniformspanning trees in work of Berestycki, van Engelenburg and Hutchcroft, see [1,2].Theorem 1. On any infinite recurrent rooted network there exists a potential which tends toinfinity.The earliest theorem of this type is due to G. C. Evans [3] who showed that every compactpolar set K in Rd supports a probability measure such that its Newtonian potential is infiniteon K . Using Evans’ approach, Nakai [4] proved that every parabolic noncompact Riemanniansurface admits a function which is harmonic off a compact set and tends to infinity (i.e., its levelsets are compact). In Nakai’s proof, the set K is replaced by the Stone–Čech boundary of thesurface. Nakai’s argument can be adapted to prove Theorem 1, but the use of the Stone–Čechcompactification makes this argument non-constructive. Here we present a more constructiveprobabilistic approach, using the minimax theorem.Remark 2. Theorem 1 implies that if a rooted network admits a unique potential, then it musttend to infinity. This answers a question raised following [1, Proposition 5.1].Remark 3. It is well known that the existence of a potential tending to infinity implies recurrence.Indeed, let h be a potential, let {X t } be the network random walk and consider the hitting timeτo = min{n ≥ 0 : Xn = o}. Then h(Xn∧τo ) is a nonnegative martingale, whence it almost surelyconverges; this contradicts transience if h →∞.2. Preliminaries2.1. Dipoles and potentialsThe Green kernel Go(x, y) of the random walk {Xn}n≥0 killed at o is defined byGo(x, y) :=∞∑n=0Px(Xn = y, τo > n).The corresponding Green density,go(x, y) = Go(x, y)cy,is symmetric by reversibility. A dipole from o to another vertex y in a network (G ,c) is a functionf : V → [0,∞), for which f (o) = 0 and∀ v ∈V ∆ f (v) = 1o(v)−1y (v).Claim 4. In a recurrent network, the Green density go( · , y) for the killed random walk is a dipolefrom o to y. Moreover, supy go(x, y) <∞ for each x ∈V .Proof. First observe that go( · , y) is harmonic off {o, y} and satisfies go(o, y) = 0. Second,∆go( · , y)(y) =−1, since∑x∼ycx y(go(y, y)−go(x, y))=Go(y, y)− ∑x∼ycx ycyGo(x, y) = 1. (1)Also,∆go( · , y)(o) =∑x∼ocox go(x, y) =∑x∼oGo(y, x)coxcx= 1, (2)where the last equality is obtained by summing, over n ≥ 0, the identity∑x∼oPy(Xn = x, τo > n)coxcx=Py(Xn+1 = o, τo > n)=Py (τo = n +1)and using recurrence. Thus, go( · , y) is a dipole.Asaf Nachmias and Yuval Peres 795Lastly, the identity (2) implies that go(x, y) ≤ c−1ox for every x ∼ o. More generally, since go( · , y)is superharmonic off o, for any x ̸= o we have go(x, y) ≤ ∏ℓi=1 c−1xi−1xi , where x0, . . . , xℓ is a simplepath from o to x. □Claim 5. The space of potentials on a recurrent infinite rooted (G ,c,o) is a nonempty, compact,convex subset of RV endowed with the product topology.Proof. By Claim 4, for any sequence of vertices yn →∞, Cantor diagonalization yields a subse-quential pointwise limit of go( · , yn), and any such limit is a potential. Next, for any potential hwe have h(x) ≤ ∏ℓi=1 c−1xi−1xi , where x0, . . . , xℓ is a simple path from o to x, by the argument in theproof of Claim 4, so the set of potentials is a closed subset of a compact product space. Finally,convexity is obvious. □2.2. The Doob transform determined by a potentialAny potential h defines a Markov chain on the state space Sh ={v : h(v) > 0} via the Doob h-transform that has transition probabilitiesph(x, y) = p(x, y)h(y)h(x).The corresponding Markov chain {Yn}n≥0 is called the h-process. Write Phµ for the law of the h-process with initial distribution µ and Ehµ for the corresponding expectation.For any finite sequenceγ= (γ1, . . . ,γk ) in V , we writePγ1 (γ) :=∏ki=2 p(γi−1,γi ). Ifγ is containedin Sh , we define Phγ1(γ) similarly using ph(γi−1,γi ). We have the telescoping productPhγ1 (γ) =Pγ1 (γ)h(γk )h(γ1). (3)We will use the initial distribution µh defined byµh(v) = cov h(v) for all v ∼ o. (4)Observe thatPhµh (γ) = coγ1 h(γ1)Phγ1 (γ) = coγ1Pγ1 (γ)h(γk ) = coPo(oγ)h(γk ). (5)There is a standard relation between the Green kernels of the h-process and the networkrandom walk killed at o.Claim 6. Let h be a potential on a recurrent rooted network (G ,c,o). Then the h-process {Yn}n≥0is transient. Moreover, its Green kernel Gh(x, y) =∑k≥0Phx (Yk = y), defined for x, y ∈ Sh , satisfiesGh(x, y) =Go(x, y) h(y)h(x), (6)and for all v ∈ ShGh(µh , v) = h(v)cv , (7)where Gh(µ, v) :=∑x µ(x)Gh(x, v).Proof. Transience will follow once we show Gh(x, y) <∞. To prove (6), we sum (3) over all pathsfrom x to y that avoid o. Note that by harmonicity of h off o, any such path is contained in Sh .Next, (7) follows from (6) byGh(µh , y) =∑vcov h(v)Gh(v, y) = h(y)∑vcov Go(v, y) = h(y)cy ,using (2) and the symmetry of go( · , · ) in the last step. □796 Asaf Nachmias and Yuval PeresClaim 7. Let h be a potential for a recurrent rooted network (G ,c,o) and consider the h-pro-cess {Yn} started at v ∈ Sh . Then for every M > 0,Phv(h(Yℓ) > M)−→ 1 as ℓ→∞.This was also observed in [2, Lemma 5].Proof. Recall that {Xn} is the network random walk. By (3), we havePhv(h(Yℓ) ≤ M)= Ev [h(Xℓ)h(v)1{h(Xℓ)≤M and τo≥ℓ}]≤ Mh(v)Pv (τo ≥ ℓ),which tends to 0 as ℓ→∞ by recurrence. □3. Proof of main theoremLet d( · , · ) denote graph distance. For any integer r > 0, we setB(o,r ) = {w : d(o, w) < r } and ∂B(o,r ) = {w : d(o, w) = r },Next we state a formally weaker version of Theorem 1, where we can choose a different potentialfor each radius, and show that it implies the theorem.Lemma 8. LetΨ be the set of potentials on a recurrent infinite rooted network (G ,c,o). ThenM(r ) := maxψ∈Ψminv∈∂B(o,r )ψ(v) −→∞ as r →∞. (8)Remark 9. For every integer r > 0 and ψ ∈ Ψ, the minimum of ψ on V \ B(o,r ) is attained on∂B(o,r ). Indeed, let {X t } be the network random walk started at w ̸∈ B(o,r ) and let τr be thehitting time of ∂B(o,r ). Then ψ(X t∧τr ) is a positive martingale, so by recurrence and optionalstopping,ψ(w) ≥ Ew[ψ(Xτr )]≥ minv∈∂B(o,r )ψ(v).Consequently, M(r ) = maxψ∈Ψminw∈V \B(o,r )ψ(w) is non-decreasing in r .Proof of Theorem 1 given Lemma 8. For each integer r , let ψr be a potential where the maxi-mum in the definition (8) of M(r ) is attained. For every n ≥ 1, choose rn such that M(rn) ≥ n2n .By Claim 5,h :=∞∑n=12−nψrnis a potential. Next, if w ∉ B(o,rn), then h(w) ≥ 2−nψrn (w) ≥ n. Thus, for each n, the set{v ∈V : h(v) < n} is finite. □3.1. Preliminary lemmasThe following lemma shows that if the dipoles go( · , vn) converge to a potential h, then thenetwork random walks conditioned on τvn < τo converge weakly to the h-process. We actuallyneed this for dipole mixtures. Given a probability measure η on V , writego( · ,η) :=∑vη(v)go( · , v).Lemma 10. Let (G ,c,o) be a recurrent rooted network. Let rn →∞ be a sequence of integers and foreach n let ηrn be a probability measure supported on ∂B(o,rn). Assume that the sequence go( · ,ηrn )converges pointwise to a potential h as n → ∞. Then for any finite path γ in Sh starting at aneighbor of o, we have ∑vηrn (v)Po(oγ∣∣τv < τ+o )−→Phµh (γ) as n →∞. (9)Asaf Nachmias and Yuval Peres 797Proof. First note that for v, w different from o, we have go(w, v) = Pw (τv < τo)go(v, v) and bypath reversal,coPo(τv < τ+o ) = cvPv (τo < τ+v ) = cv Go(v, v)−1 = go(v, v)−1.Therefore, if γ= (γ1, . . . ,γℓ) does not contain the vertex v , thenPo(oγ∣∣τv < τ+o )=Po(oγ)Pγℓ (τv < τo)Po(τv < τ+o )= coPo(oγ)go(γℓ, v). (10)Integrating this with respect to ηrn gives∑vηrn (v)Po(oγ∣∣τv < τ+o )= coPo(oγ)go(γℓ,ηrn ) −→ coPo(oγ)h(γℓ),as n →∞ by our hypothesis. This yields the desired result by (5). □Let h be a potential and set TM = inf{t > 0 : h(X t ) ≥ M}where {X t } is the network randomwalk. Optional stopping for the nonnegative martingale h(X t∧τo ) gives Pv (TM < τo) ≤ h(v)/M .Careful path reversal and conditioning yields the following.Lemma 11. Let h be a potential for a recurrent rooted network (G ,c,o) and let v ̸= o be a vertex.Then the network random walk {X t } satisfiesPo(h(Xℓ) ≥ M∣∣τv < τ+o )≤ h(v)M ,for all M > 0 and integer 0 < ℓ< d(o, v), where τ+o = min{t ≥ 1 : X t = o}.Proof. We may assume that h(v) < M . Since ℓ< d(o, v) we havePo(h(Xℓ) ≥ M∣∣τv < τ+o )≤Po(TM < τv ∣∣τv < τ+o ). (11)Recall that coPo(τv < τ+o ) = cvPv (τo < τ+v ) by path reversal. Similarly,coPo(TM < τv < τ+o ) = cvPv (TM < τo < τ+v ),so Po(TM < τv∣∣τv < τ+o )=Pv (TM < τo ∣∣τo < τ+v ). (12)Consider the random walk started at v and write L = max{t < τo : X t = v}. Then XL , . . . , Xτo isdistributed as a random walk started at v , conditioned on τo < τ+v , and stopped at o. HencePv(TM < τo∣∣τo < τ+v )=Pv (∃ s ∈ (L,τo) : h(Xs ) ≥ M)≤Pv (TM < τo). (13)To bound the last probability we note that St = h(X t∧τo ) is a non-negative martingale, so byoptional stoppingh(v) = Ev S0 ≥ Ev STM ≥ MPv (TM < τo).Combining this with (11), (12) and (13) concludes the proof. □3.2. Proof of Lemma 8Recall that Ψ denotes the set of potentials on (G ,c,o). Given an integer r > 0, let Υr be thecompact space of probability measures on ∂B(o,r ). Observe that for every ψ : V →Rminv∈∂B(o,r )ψ(v) = minη∈Υr∑vη(v)ψ(v). (14)Next, we will apply Sion’s minimax theorem [5] to the function fr : Ψ×Υr →R, defined byfr (ψ,η) =∑v∈∂B(o,r )η(v)ψ(v).As noted in Claim 5, the setΨ is a convex subset of the linear space RV endowed with the producttopology. Additionally, it is clear that fr is continuous and bilinear and that Υr is a compact798 Asaf Nachmias and Yuval Peresconvex subset of the linear topological space R∂B(o,r ). Hence the assumptions of Sion’s theoremhold, and we obtain, using (14), that M(r ) defined in (8) satisfiesM(r ) = maxψ∈Ψminη∈Υrfr (ψ,η) = minη∈Υrmaxψ∈Ψfr (ψ,η). (15)For each integer r > 0, let ηr ∈ Υr denote a minimizer of the right-hand side of (15). ByClaim 4, there exists a sequence rn → ∞ such that go( · ,ηrn ) converges pointwise to a functionh : V → [0,∞). Clearly h is a potential.Fix m > 0. By Claim 7, there exists an integer ℓ such that Phµh(h(Yℓ) ≥ 2m)≥ 1/2. We now applyLemma 10 and sum (9) over all paths γ= (γ1, . . . ,γℓ) ⊂ Sh such that h(γℓ) ≥ 2m. We infer thatlimn→∞∑vηrn (v)Po(h(Xℓ) ≥ 2m∣∣τv < τ+o )=Phµh (h(Yℓ) ≥ 2m)≥ 1/2.Lemma 11 states that if ℓ< rn and v ∈ ∂B(o,rn), thenPo(h(Xℓ) ≥ 2m∣∣τv < τ+o )≤ h(v)2m .Combining the last two inequalities yields thatliminfn→∞ frn (h,ηrn ) = liminfn→∞∑vηrn (v)h(v) ≥ m.In conjunction with (15), this gives that liminfn M(rn) ≥ m for every m > 0. Since M(r ) is non-decreasing (Remark 9), it must tend to infinity as r →∞. □AcknowledgmentsWe thank Omer Angel, Ori Gurel-Gurevich and Fedja Nazarov for helpful discussions.Declaration of interestsThe authors do not work for, advise, own shares in, or receive funds from any organizationthat could benefit from this article, and have declared no affiliations other than their researchorganizations.References[1] N. Berestycki and D. van Engelenburg, “Harnack inequality and one-endedness of UST onreversible random graphs”, Probab. Theory Relat. Fields 188 (2024), no. 1-2, pp. 487–548.[2] D. van Engelenburg and T. Hutchcroft, “The number of ends in the uniform spanning treefor recurrent unimodular random graphs”, Ann. Probab. 52 (2024), no. 6, pp. 2079–2103.[3] G. C. Evans, “Potentials and positively infinite singularities of harmonic functions”,Monatsh. Math. Phys. 43 (1936), no. 1, pp. 419–424.[4] M. Nakai, “On Evans potential”, Proc. Japan Acad. 38 (1962), pp. 624–629.[5] M. Sion, “On general minimax theorems”, Pac. J. Math. 8 (1958), pp. 171–176.[6] F. Spitzer, Principles of random walk, Second edition, Graduate Texts in Mathematics,Springer, 1976, pp. xiii+408.

Every recurrent network has a potential tending to infinity

Karthik Elamvazhuthi

A linear test for global nonlinear controllability

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