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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.

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