Dimension estimates on circular (s,t)-Furstenberg sets
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This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY-NC 4.0 https://creativecommons.org/licenses/by-nc/4.0/ Dimension estimates on circular (s,t)-Furstenberg sets © 2022 Annales Fennici Mathematici Published version Liu, Jiayin Liu, J. (2023). Dimension estimates on circular (s,t)-Furstenberg sets. Annales Fennici Mathematici, 48(1), 299-324. https://doi.org/10.54330/afm.128073 2023
Annales Fennici Mathematici Volumen 48, 2023, 299–324 Dimension estimates on circular (s, t)-Furstenberg sets Jiayin Liu Abstract. In this paper, we show that circular (s, t)-Furstenberg sets in R2have Hausdorff dimension at least max{t 3+s, (2t+ 1)s−t}for all 0< s, t ≤1. This result extends the previous dimension estimates on circular Kakeya sets by Wolff. Furstenbergin (s, t)-ympyräjoukkojen ulottuvuuden arvioita Tiivistelmä. Tässä työssä osoitetaan, että tason R2Furstenbergin (s, t)-ympyräjoukkojen Hausdorffin ulottuvuus on vähintään max{t 3+s, (2t+ 1)s−t}kaikilla 0< s, t ≤1. Tämä tulos yleistää Wolffin aiemmin todistamia Kakeyan ympyräjoukkojen ulottuvuusarvioita. 1. Introduction Let Fbe a circular (s, t)-Furstenberg set in R2. That is, there exists a parameter set K⊂R3 +with Hausdorff dimension dimHK≥t such that for every (x, r)∈K, (1.1) dimH(F∩S(x, r)) ≥s where R3 +:= {(x, r)=(x1, x2, r)|r > 0}and S(x, r)is the circle centered at x∈R2 with radius r. A special class of circular (1,1)-Furstenberg sets is the family of circular Kakeya sets, that is, Borel sets in R2that contain circles of every radius. The study on the Hausdorff dimension of Furstenberg sets was initiated from their linear version. In this paper, we call a set F⊂R2a linear (s, t)-Furstenberg set if there exists a parameter set Kin A(2,1) with dimHK≥t such that for every L∈K, dimH(F∩L)≥s where A(n, k)denotes the family of k-dimensional affine subspaces in Rn. In 1999, Wolff [16] showed that linear (s, 1)-Furstenberg sets with parameter set Kcontaining lines in every direction have Hausdorff dimension at least (1.2) max{1 2+s, 2s}for all 0< s ≤1. https://doi.org/10.54330/afm.128073 2020 Mathematics Subject Classification: Primary 28A75; Secondary 28A78, 28A80. Key words: Furstenberg set, circular Furstenberg set, Hausdorff dimension. J. L. is supported by the Academy of Finland via the projects: Quantitative rectifiability in Euclidean and non-Euclidean spaces, Grant No. 314172, and Singular integrals, harmonic functions, and boundary regularity in Heisenberg groups, Grant No. 328846. c 2023 The Finnish Mathematical Society
300 Jiayin Liu In the sequel, there is a series of works improving the above lower bound and providing the one for linear (s, t)-Furstenberg sets with some of them only considering special values of s, t. We refer the readers to [9, 1, 11, 10, 12, 7, 3, 2, 13] and references therein. Moreover, in higher dimensions, one can similarly define linear (s, t)-Furstenberg sets with parameter set Kin A(n, k). See [5, 6] for some recent progress. It is not clear whether the above lower bound estimates on the Hausdorff dimension for linear (s, t)-Furstenberg sets in R2are sharp for any value of sand t except s= 1. Hence determining the sharp lower bound remains open for Hausdorff dimension of linear (s, t)-Furstenberg sets. In terms of circular (s, t)-Furstenberg sets in R2, Wolff in [17, Corollary 3] showed that circular Kakeya sets in R2have full dimension 2employing techniques from harmonic analysis. Also, in [15, Corollary 3], Wolff proved that Borel sets in R2 consisting of circles with t-dimensional set of centers have Hausdorff dimension at least 1 + t. Later, in [8], as an application of their techniques to prove a Marstrandtype restricted projection theorem, Käenmäki–Orponen–Venieri were able to show that the above lower bound 1 + tin [15] holds true for analytic t-dimensional family of circles. Hence they provide an alternative method showing the dimension of sets containing full circles. Since the above results concern special cases of circular (1, t)- Furstenberg sets, these bounds are sharp. To the best of the author’s knowledge, these works and earlier results on families of full circles are the only ones concerning the Hausdorff dimension for circular Furstenberg sets. In this paper, we extend the existing result to general 0< s, t ≤1. We show the following: Theorem 1.1. For any 0< s ≤1and 0< t ≤1, the Hausdorff dimension of a circular (s, t)-Furstenberg set Fin R2is at least (1.3) max{t 3+s, (2s−1)t+s}. We remark that for any 0< t ≤1, if 0< s ≤2 3, then the maximum in (1.3) is attained by t 3+s. Otherwise, it is achieved by (2s−1)t+s. Indeed, these two bounds are obtained by different approaches. Hence Theorem 1.1 is a combination of the following two theorems. Theorem 1.2. For any 0< s ≤1and 0< t ≤1, the Hausdorff dimension of a circular (s, t)-Furstenberg set Fin R2is at least t 3+s. Theorem 1.3. For any 1 2< s ≤1and 0< t ≤1, the Hausdorff dimension of a circular (s, t)-Furstenberg set Fin R2is at least (2s−1)t+s. Below, we briefly outline our ideas of the proof of Theorem 1.2 and Theorem 1.3, which will imply Theorem 1.1. Here, we will focus on explaining some informal ideas on obtaining the Minkowski dimension lower bounds for circular Furstenberg sets. Then we can derive the Hausdorff dimension lower bounds from the Minkowski dimension lower bounds in a standard way. To this end, in the proof, we will work with a discretized version of the circular (s, t)-Furstenberg set Fin the following sense. That is, instead of studying the tdimensional parameter set K, we will concentrate on a finite subset V⊂Kwhich is a (δ, t)-set (See Definition 2.2). In brief, Vis a
Dimension estimates on circular (s, t)-Furstenberg sets 301 δ-separated set with cardinality δ−tand satisfies a t-dimensional non-concentration condition. With this discretized circular Furstenberg set Sz∈VS(z)∩F, we consider an arbitrary cover U={B(xi, ri)}i∈Ik1of this set by balls of radii between δ/2and δ where δ= 2−k1(k1∈N) is sufficiently small. We will give a lower bound of #Ik1 independent of the choice of the cover U. Recall that the desired lower bound is t 3+s in Theorem 1.2 and (2t+ 1)s−tin Theorem 1.3, so we need to show that (1.4) #Ik1&1 δt 3+s in Theorem 1.2 and (1.5) #Ik1&1 δ(2t+1)s−t if 1 2< s ≤1in Theorem 1.3. Indeed, this will imply X i∈Ik1 r t 3+s i&1 δt 3+s δt 3+s&1in Theorem 1.2 and X i∈Ik1 r(2t+1)s−t i&1 δ(2t+1)s−t δ(2t+1)s−t&1if 1 2< s ≤1in Theorem 1.3, which further imply that the t 3+s(resp. (2t+1)s−t) dimensional Hausdorff measure of Fis positive and therefore the Hausdorff dimension of Fis at least t 3+s(resp. (2t+ 1)s−t). To show (1.4), we adapt the approach for showing the lower bound for the Hausdorff dimension of linear (s, 1)-Furstenberg sets used by Wolff in [16] together with some geometric observations from planar geometry. The heuristic idea is that, since three points determine a unique circle in the plane provided they are not collinear, we can show that three well-separated δ-balls Bi, Bj, Bkdetermine a “unique” circle S(z)(not necessarily unique in reality, see the statement before (3.22)), z∈V, with the help of Lemma 2.5, which intuitively means that there exists a unique circle S(z)with z∈Vsuch that S(z)∩Bl6=∅for l=i, j, k. This further enables us to identify the circle S(z)with the triple (i, j, k). Indeed, the above manipulations are motivated by Wolff [16] to show the lower bound 1/2 + sin (1.2) for the Hausdorff dimension of linear (s, 1)-Furstenberg sets where 1/2appears from the fact that two points determine a unique line in the plane. For circular (s, 1)- Furstenberg sets, we can only get the lower bound 1/3+ ssince we need three points to determine a circle. On the other hand, since S(z)∩Fhas Hausdorff dimension no less than s, we need, roughly speaking, at least ∼δ−sδ-balls in Uto cover S(z)∩F. Hence we can identify each S(z)∩Fby the triples (i, j, k)∈ Ik1×Ik1×Ik1 (or equivalently, (Bi, Bj, Bk)∈ U × U × U) where S(z)∩Bl6=∅for l=i, j, k. Then each S(z)∩Fgives rise to δ−s(δ−s−1)(δ−s−2) ∼δ−3smany distinct triples (i, j, k)∈ Ik1×Ik1×Ik1representing three distinct δ-balls in Uand therefore we obtain a total number #V×δ−3s=δ−3s−tmany distinct triples. Finally, since all these triples are contained in Ik1×Ik1×Ik1, we deduce that (#Ik1)3&δ−3s−t, which gives (1.4). This is the rough idea behind the proof of the Minkowski dimesion version of Theorem 1.2.
302 Jiayin Liu On the other hand, inequality (1.5) is obtained by applying the result from Käenmäki–Orponen–Venieri in [8] utilised to find the Hausdorff dimension of t-dimensional analytic sets of circles. Heuristically, as discussed above, since one needs at least ∼δ−sδ-balls in Uto cover S(z)∩Ffor each z∈V, if each δ-ball in Uonly intersects one S(z)∩Ffor some z∈V, then Uconsists of at least δ−s#V∼δ−s−t many δ-balls. However, this may not be the case. In general, if each δ-ball in U intersects no more than δ−ξ(0< ξ ≤t) many sets from the family {S(z)∩F}z∈V, then we can deduce that Uconsists of at least δ−s#V δ−ξ∼δ−s−t δ−ξmany δ-balls. Actually, by applying [8, Lemma 5.1], we can show that for more than half of points zin V, there exists S0(z)⊂S(z)∩Fwith dimHS0(z) = dimH[S(z)∩F]≥ssuch that each δball in Uintersects no more than δt(2s−2) many sets from the family {S0(z)}z∈Vwhere t(2s−2) arises from the choice of the parameter λwhen applying Lemma 5.1 in [8] to guarantee (4.15) holds. We refer readers to the discussion around (4.17) in Section 4 for details. This fact will imply that there exist at least δ−s#V δt(2s−2) ∼δ−[(2t+1)s−t]many δ-balls in U, which is equivalent to say #Ik1&δ−[(2t+1)s−t]. Hence (1.5) holds and this concludes a heuristic discussion regarding Theorem 1.3. Finally, we remark that we do not know if the bound max{t 3+s, (2s−1)t+s}in Theorem 1.1 is sharp and we here make a conjecture that the sharp lower bound for Hausdorff dimension of circular (s, 1)-Furstenberg sets is 1 2+3 2sfor 0< s ≤1. Indeed, in the following example, based on the example in [16], we construct a circular (s, 1)- Furstenberg set whose Hausdorff dimension does not exceed 1 2+3 2sfor all 0< s ≤1. Example 1.4. Due to the construction in [16, Section 1] by Wolff, for all 0< s ≤ 1, there exists a linear (s, 1)-Furstenberg set F⊂B(0,4) \B(0,1) whose Hausdorff dimension does not exceed 1 2+3 2s. Now considering R2as the complex plane C, using the map ω:C→C,z7→ 1 z, all lines in Care mapped to circles through (0,0). Also noticing that ω|B(0,4)\B(0,1) is a biLipschitz homeomorphism, we deduce that F0:= ω(F)is a circular (s, 1)-Furstenberg set with same dimension as F. That is, dimH(F0)≤1 2+3 2s. The paper is organised as follows. In Section 2, we clarify our notations and symbols, as well as introduce definitions and results employed in the proof. Sections 3 and 4 are devoted to showing the proof of Theorem 1.2 and 1.3 respectively. In the last section, Section 5, we complete the proof of some auxiliary lemmas needed in the proof of Theorem 1.2 using planar geometry. Acknowledgement. J. L. would like to thank K. Fässler and T. Orponen for many motivating discussions and their constant support. J. L. would also like to convey his gratitude to the anonymous referee for pointing out a mistake in the proof of Theorem 1.2 and for providing many valuable suggestions which significantly improved the final presentation of the paper. 2. Preliminaries In this paper, we denote by Sδ(x, r)the δ-neighbourhood of S(x, r), i.e. Sδ(x, r) := B(x, r +δ)\B(x, r −δ). We also use the notation z= (x, r)∈R3. Moreover, we use the notation f.g(resp. f.hg) for f≤kg (resp. f≤k(h)g) where kis a constant that depends only on the ambient space (resp. the parameter h), and may change from line to line. Likewise, f&gand f∼gare understood correspondingly.
Dimension estimates on circular (s, t)-Furstenberg sets 303 The notation Hsstands for the s-dimensional Hausdorff measure, and Hs ∞stands for s-dimensional Hausdorff content. The notation | · | and k · k will denote the Lebesgue measure and the Euclidean distance respectively in R2or R3. We also use dist(A, B)to denote Euclidean distance between Aand Bwhere Aand Bcan be either points or sets. #Awill denote the cardinality of a set A. We have the following observation which makes it possible to restrict ourselves to circular Furstenberg sets with bounded parameter set. Remark 2.1. (i) Since we are concerned with the Hausdorff dimension of the circular Furstenberg set F, we claim that it is enough to consider the case that F has parameter set K⊂B0where (2.1) B0={(x, r)∈R3|x∈B(0,1 4)and 1 2≤r≤2}. To see this, consider the following covering of the parameter space R3 +. For k, l, m ∈ Z, let Dk,l,m := {(x, r)∈R3|x∈B((22m−2k, 22m−2l),22m−2)and 22m−1≤r≤22m+1}. Then R3 +=[ k,l,m Dk,l,m and B0=D0,0,0. Hence for each > 0sufficiently small, there exists k, l, msuch that (2.2) dimH(K)−dimH(K∩Dk,l,m)< . Let Fbe the circular Furstenberg set with parameter set K∩Dk,l,m. Denote by Sy:R2→R2,Sy(x) := x−yfor any y∈R2and by Dλ:R2→R2,Dλ(x) := λx for any λ > 0. Then, letting y= (22m−2k,22m−2l)and λ= 2−2m, we observe that e F:= D2−2m◦S(22m−2k,22m−2l)(F) is a circular Furstenberg set the parameter set e Kcontained in B0and satisfying (2.3) dimH(e K) = dimH(K∩Dk,l,m). If Fis a circular (s, t)-Furstenberg set, then by (2.2) and (2.3), for 0< < t, we know e Fis a circular (s, t −)-Furstenberg set and (2.4) dimHF≥dimHe Ffor every 0< < t. Now, assume Theorem 1.1 holds for circular Furstenberg sets with parameter set contained in B0, then (2.5) dimHe F≥max{t− 3+s, (2s−1)(t−) + s}for every 0< < t. Combining (2.4) and (2.5), we deduce that dimHF≥lim →0max{t− 3+s, (2s−1)(t−) + s}= max{t 3+s, (2s−1)t+s}. Hence to show Theorem 1.1, we only need to consider the case that Fhas parameter set K⊂B0. (ii) Note that |Sδ(x, r)| ≤ c0δfor all (x, r)∈B0where c0is an absolute constant. We introduce the following:
304 Jiayin Liu Definition 2.2. ((δ, q)-sets) Let δ∈(0,1), q > 0, and let P⊂Rnbe a finite δ-separated set. We say that Pis a (δ, q)-set, if it satisfies the estimate (2.6) #{P∩B(x, r)}.r δq, x ∈Rn, r > δ. We recall from [4, Lemma 3.13] the following Lemma 2.3. Let δ, q > 0, and let Q⊂Rnbe any set with Hq ∞(Q) =: β > 0. Then there exists a (δ, q)-set P⊂Qwith cardinality #P&β·δ−q. Remark 2.4. If Q⊂B0and Hq ∞(Q) = β, by Lemma 2.3, we know that for any δ > 0, there exists a (δ, q)-set P⊂Qwith cardinality #P&βδ−q. Furthermore, letting r= diam B0in (2.6), we know #P.δ−q, if δ < diam B0. We conclude that βδ−q.#P.δ−q. To show Theorem 1.2, we need to establish the following result from planar geometry. Since the proof relies on two more auxiliary lemmas, we postpone it to the last section. Lemma 2.5. Let A, B, C ∈R2such that min{kA−Bk,kA−Ck,kB−Ck,2} ≥ 2c. For a > 0such that a < 1 20c2, define (2.7) W:= b−a≤ kx−Ak ≤ b+a, (x, b)∈R2×[1 2,2] : b−a≤ kx−Bk ≤ b+a, b−a≤ kx−Ck ≤ b+a . Then (2.8) diam W.a c2. It is worth mentioning that Lemma 2.5 shares a very similar conclusion with the one in [16, Lemma 3.2 (Mastrand’s 3-circle lemma)]. Indeed, if we let =δ=a, r=b,λ=c,t= 1/2−aand r1=r2=r3=atherein, then the set Win Lemma 2.5 will be contained in Ωtλ defined in [16, Lemma 3.2]. And the conclusion of [16, Lemma 3.2] says that Ωtλ is contained in the union of two ellipsoids in R3with diam Ωtλ .a c2. Since we only consider the case r1=r2=r3=a(that is, Cδ(xi, ri) become balls B(xi,2a)for i= 1,2,3in [16, Lemma 3.2]), we can deduce that Wlies in one cuboid in R3based on an approach which differs completely from the one of [16, Lemma 3.2]. Now, we start the preparation for the proof of Theorem 1.3. Let P⊂R3be a (δ, q)-set. For any p∈P, let ∆pbe the Dirac measure centered at p. Then (2.9) µP:= 1 #PX p∈P ∆p is a probability measure satisfying the Frostman condition µP(B(z, r)) .rqfor all z∈R3and r > δ. Indeed, for any ball B(z, r)with r > δ we have µP(B(z, r)) = 1 #P X p∈P ∆p!(B(z, r)) = 1 #PX p∈P ∆p(B(z, r)) =1 #P#(P∩B(z, r)) .rq. Below in Section 3 and 4, thanks to Remark 2.1(i), we will assume the circular (s, t)-Furstenberg set Fhas parameter set K⊂B0.
Dimension estimates on circular (s, t)-Furstenberg sets 305 3. Proof of Theorem 1.2 Proof of Theorem 1.2. Let Fbe a circular (s, t)-Furstenberg set with parameter set K⊂B0. It suffices to show, for any > 0,0< s0< s and 0< t0< t, dimH(F)≥t0 3+s0−. Hence in the following, we fix s0, t0and 0< < t0 3+s0. We notice that there exists α > 0and K1⊂Ksuch that Ht0 ∞(K1)> α, where (3.1) K1:= {z∈K| Hs0 ∞(F∩S(z)) > α}. Indeed, by the subadditivity of Hausdorff content, and the fact K=[ n{z∈K| Hs0 ∞(F∩S(z)) >1 n}, we deduce the existence of αsuch that Ht0 ∞(K1)> α for K1defined as in (3.1). Next, since > 0, we can find δ0=δ0(, s0)>0sufficiently small such that for any 0< δ < δ0, we have (3.2) δ−log 1 δ−(8 3+12 s0) >1. and (3.3) √640δ < τ =τ(δ) := π−11 161/s01 log 1 δ2/s0 <1. Then we choose k0to be an integer larger than log( 1 δ0)also satisfying (3.4) α > ∞ X k=k0 1 k2. Now, we outline the main steps of the proof. We start with an arbitrary cover U={B(xi, ri)}i∈I of Fby balls of radius less than 2−k0. In the sequel, we will derive a lower bound X i∈I rσ i&,t0,s01 with σ=t0/3 + s0−independent of the choice of the particular cover. This will imply Hσ(F)>0. To this end, we divide the proof into 4 steps. Let Ik:= {i∈ I | 2−(k+1) < ri≤2−k}, Fk:= n[B(xi, ri)|i∈ Iko. First, in Step 1, we will deduce that there exists k1≥k0and a (δ, t0)-set V⊂K with δ= 2−k1such that for every circle z= (x, r)∈V, we have (3.5) Hs0 ∞(S(z)∩Fk1)> k−2 1. Then, in Step 2, we modify Wolff’s approach for linear (s, 1)-Furstenberg sets to fit our circular case. For each circle S(z)with z∈V, we will extract from S(z)three τ-separated arcs h+ z, h− z, h× zsuch that (3.6) Hs0 ∞(h+ z∩Fk1)&k−2 1,Hs0 ∞(h− z∩Fk1)&k−2 1,Hs0 ∞(h× z∩Fk1)&k−2 1.
306 Jiayin Liu These arcs enable us to define an index set T ⊂ Ik1× Ik1× Ik1×Vwhose cardinality will be estimated in the following steps and will imply the lower bound for #Ik1. Next, in Step 3, we will deduce that the cardinality of Tis upper bounded by the cardinality of Ik1with the help of Lemma 2.5. Indeed, we will show #T.(#Ik1)3τ−6. Finally, in Step 4, we will estimate the lower bound of #Twhich also serves as the one of #Ik1, hence #Iwith the aid of (3.6). This will enable us to conclude the proof. Step 1. Let αbe as in (3.4). Hence by pigeonhole principle we deduce that for each S(z)∈K1, there exists k(z)≥k0such that Hs0 ∞(S(z)∩F∩Fk(z))> k(z)−2. Moreover, by applying pigeonhole principle again we obtain that there exists k1≥k0such that (3.7) Ht0 ∞(K2)> k−2 1 where K2:= {z∈K1:k(z) = k1}. We remark that for every circle z∈K2, we have (3.8) ∞>Hs0 ∞(S(z)∩Fk1)≥ Hs0 ∞(S(z)∩F∩Fk1)> k−2 1. By letting δ= 2−k1,q=t0and Q=K2in Lemma 2.3, we know that there exists a(δ, t0)-set V⊂K2with cardinality (3.9) #V&Ht0 ∞(K2)·δ−t0. Hence for every z∈V, (3.8) implies (3.5), which concludes Step 1. Step 2. We start the procedure of extracting three disjoint arcs for any S(z), z = (x, r)∈V, which is illustrated in Figures 1, 2 and 3. Let η:= η(z) = Hs0 ∞(S(z)∩Fk1). Also let γ= ( η 16)1/s0. Divide S(z)into Narcs I1,··· , INsuch that •the length of I1,··· , IN−1is γ, •the length of INis at most γ, •and Nγ ≥2πr. Since γ= ( η 16)1/s0≤1 16 and z= (x, r)∈B0implies r > 1 2, we know N≥2πr γ≥π 1 16 ≥16. Note that if Iis an arc in S(z), then (3.10) Hs0 ∞(I)≤(diam I)s0≤(H1(I))s0. This implies for all l= 1,··· , N, (3.11) Hs0 ∞(Il∩Fk1)≤ Hs0 ∞(Il)≤γs0=η 16. See Figure 1 for Narcs.
Dimension estimates on circular (s, t)-Furstenberg sets 313 Now, we outline the main steps of the proof. We start with an arbitrary cover U={B(xi, ri)}i∈I of Fby balls of radius less than 2−k0. In the sequel, we will derive a lower bound X i∈I rσ i&,t0,s01 with σ= (2t0+ 1)s0−t0−independent of the choice of the particular cover. This will imply Hσ(F)>0. To this end, we divide the proof into 3 steps. Let Ik:= {i∈ I | 2−(k+1) < ri≤2−k}, Fk:= {SB(xi, ri)|i∈ Ik}. First, in Step 1, we will deduce that there exists k1≥k0and a (δ, t0)-set V⊂K with δ= 2−k1such that (4.7) 1 k2 1·δ−t0.#V.δ−t0, and for every circle z= (x, r)∈V, we have (4.8) Hs0 ∞(S(z)∩Fk1)> k−2 1. Next, in Step 2, we associate a finite measure µsupported on Vusing (2.9). Then we apply Lemma 4.1 to obtain that there exists G⊂Vand Sδ 2(z)contained in the δ-neighbourhood of S(z)∩Fk1, such that for every z∈Gand w∈Sδ 2(z), (4.9) #{z0∈G|w∈Sδ 2(z0)}.t0Cη,C,t0,s0δt0(2s0−2−η)(log 1 δ)4t0+2. Finally, in Step 3, we will provide a lower bound of the cardinality #Ik1by combining the upper bound in Step 2 as well as the lower bounds on the cardinality #Gand the Lebesgue measure |Sδ 2(z)|. Explicitly, we have #Ik1&,t0,s0 1 δ(2t0+1)s0−t0(1+η) 1 (log 1 δ)6+4t0. This will enable us to conclude the proof. Step 1. Employing the same arguments as in Step 1 in the proof of Theorem 1.2, we can deduce the existence of k1and V⊂K2⊂K1satisfying (4.8) and the first inequality in (4.7). The second inequality in (4.7) is derived from Remark 2.4. Here, we omit the details. Step 2. Define µVas in (2.9) applied to P=V. Then we know µVis a probability measure satisfying the Frostman condition µV(B(z, r)) ≤CHt0 ∞(K2)−1rt0< Ck2 1rt0=C(log 1 δ)2rt0 for all z∈R3and r > δ. Hence by setting µ:= µV (log 1 δ)2, we know that µhas total measure (log 1 δ)−2<1, sptµ=V⊂B0and µ(B(z, r)) ≤Crt0=: Crt0 for all z∈R3and r > δ. Let mδ µbe the corresponding multiplicity function with respect to µdefined as in (4.1).
314 Jiayin Liu Applying Lemma 4.1 with t=t0,δ= 2−k1,ηas in (4.2), µ= (log 1 δ)−2µV,D=V and (4.10) λ= (2c04s0k2 1)−1δ1−s0, we obtain that for A=Cη,C,t0·δ−η, there is a set G=G(k1, s0, t0, )⊂Vwith (4.11) µ(V\G)< A−t0/3 such that the following holds for all z∈G: (4.12) |Sδ(z)∩{w|mµ δ(w)≥At0λ−2t0δt0}| ≤ λ|Sδ(z)|. Because |Sδ(z)| ≤ c0δfor all z∈B0, (4.12) becomes (4.13) |Sδ(z)∩{w|mµ δ(w)≥At0λ−2t0δt0}| ≤ c0λδ. Moreover, recalling that δ= 2−k1and k1≥k0, we know that 0< δ < δ0. Hence by Cη,C,t0≥1, (4.4) and the choice of ηin (4.2) we deduce A−t0 3≤δηt0 3≤1 4 1 (log 1 δ)2=1 4µ(V). Hence (4.11) becomes (4.14) µ(V\G)<1 4µ(V). For z∈G, let S1(z) := S(z)∩Fk1and Sδ 1(z)be the δ-neighbourhood of S1(z). Our next goal is to substitute the right hand side term λ|Sδ(z)|in (4.12) by the term 1 2|Sδ 1(z)|with the help of the proper choice of λas in (4.10). This means, in the sense of 2-dimensional Lebesgue measure, more than half of the points in Sδ 1(z)have low multiplicity. To this end, we claim that (4.15) |Sδ 1(z)| ≥ 1 4s0k2 1 δ2−s0. To see (4.15), let P(z)be a maximal 2δ-separated set in S1(z). Then Sp∈P(z)B(p, 2δ) forms a cover of S1(z). Hence Hs0 ∞(S1(z)) ≤#P(z)(4δ)s0. which, combined with (4.8), implies #P(z)≥ Hs0 ∞(S1(z)) 1 (4δ)s0≥1 4s0k2 1 1 δs0. On the other hand, we have Sp∈P(z)B(p, δ)⊂Sδ 1(z). Hence by {B(p, δ)}p∈P(z)being mutually disjoint, we deduce |Sδ 1(z)|≥|Sp∈P(z)B(p, δ)|= #Pδ2π≥π 4s0k2 1 δ2−s0>1 4s0k2 1 δ2−s0, which gives (4.15). Noticing that |Sδ(z)| ≤ c0δ,S1(z)⊂S(z)and combining (4.13) as well as (4.15), we arrive at (4.16) |Sδ 1(z)∩{w|mµ δ(w)≥At0λ−2t0δt0}| ≤ λc0δ≤λc04s0k2 1δs0−1|Sδ 1(z)|. Now recall A=Cη,C,t0·δ−ηand λ= (2c04s0k2 1)−1δ1−s0= (2c04s0)−1(log 1 δ)−2δ1−s0. Then (4.16) becomes |Sδ 1(z)∩{w|mµ δ(w)≥Cη,C,t0,s0δt0(2s0−1−η)(log 1 δ)4t0}| ≤ 1 2|Sδ 1(z)|
Dimension estimates on circular (s, t)-Furstenberg sets 315 where we recall Cη,C,t0,s0defined in (4.5). For each z∈G, define the low-multiplicity set Sδ 2(z) := {w∈Sδ 1(z)|mµ δ(w)< Cη,C,t0,s0δt0(2s0−1−η)(log 1 δ)4t0}. Then we have (4.17) |Sδ 2(z)| ≥ 1 2|Sδ 1(z)|. See Figure 4 for an illustration of S1(z),Sδ 1(z)and Sδ 2(z). Figure 4. An illustration of S1(z),Sδ 1(z)and Sδ 2(z). Notice that mµ δ(w)< Cη,C,t0,s0δt0(2s0−1−η)(log 1 δ)4t0is equivalent to µ({z0∈R3|w∈Sδ(z0)})< Cη,C,t0,s0δt0(2s0−1−η)(log 1 δ)4t0, which, combined with (4.7), indicates that for w∈Sδ 2(z), it holds #{z0∈V|w∈Sδ(z0)} ≤ #V·Cη,C,t0,s0δt0(2s0−1−η)(log 1 δ)4t0+2 .t0Cη,C,t0,s0δt0(2s0−2−η)(log 1 δ)4t0+2. Furthermore, by the inclusions G⊂Vand Sδ 2(z)⊂Sδ(z), we conclude (4.9), which finishes Step 2. Step 3. We will lower bound #Ik1in the following. First notice that if {Sδ(z)}z∈G were mutually disjoint, we could lower bound #Ik1by summing up the number of balls Bi(i∈ Ik1)needed to cover each Sδ 2(z)since no ball could simultaneously intersect two of these sets. However, in general, {Sδ(z)}z∈Gmay not be mutually disjoint, which needs a bit more efforts to get the lower bound of #Ik1. Let e Fk1:= [ i∈Ik1 B(xi,4ri).
316 Jiayin Liu We deduce that (4.18) [ z∈G Sδ 2(z)⊂e Fk1. Indeed, for any w∈Sδ 2(z), there exists w0∈S(z)∩Fk1such that kw−w0k< δ. On the other hand, we know that w0∈B(xi, ri)for some xi∈ Ik1and ri>2−(k1+1) = δ/2, which implies kw0−xik< ri and hence kw−xik< δ +ri<3ri. In addition, by (4.7) and (4.14), we can infer that (4.19) #G&#V&1 δt0 1 (log 1 δ)2. Moreover by recalling (4.9) we obtain that for every w∈Sz∈GSδ 2(z), N(w) := #{z0∈G|w∈Sδ 2(z0)}.t0Cη,C,t0,s0δt0(2s0−2−η)(log 1 δ)4t0+2 and hence combining (4.19), we can estimate [ z∈G Sδ 2(z)=X z∈GˆχSδ 2(z)(w)1 N(w)dw &t0(Cη,C,t0,s0δt0(2s0−2−η)(log 1 δ)4t0+2)−1X z∈GSδ 2(z) &t0(Cη,C,t0,s0)−11 δt0(2s0−2−η) 1 (log 1 δ)4t0+2 1 δt0 1 (log 1 δ)2min z∈G{|Sδ 2(z)|} &η,t0,s0 1 δt0(2s0−2−η) 1 (log 1 δ)4t0+2 1 δt0 1 (log 1 δ)2δ2−s01 (log 1 δ)2 (4.20) where in the last inequality we employ (4.15) and (4.17). Therefore, combining (4.18) and (4.20) we arrive at #Ik1δ2&e Fk1≥[ z∈G Sδ 2(z)&η,t0,s0 1 δt0(2s0−2−η) 1 (log 1 δ)4t0+2 1 δt0 1 (log 1 δ)2δ2−s01 (log 1 δ)2, which implies #Ik1&η,t0,s0 1 δ(2t0+1)s0−t0(1+η) 1 (log 1 δ)6+4t0. Since Ik1⊂ I, we deduce that X i∈I r(2t0+1)s0−t0− i≥X i∈Ik1 r(2t0+1)s0−t0− i &η(,t0,s0),t0,s02−k1((2t0+1)s0−t0−)1 δ(2t0+1)s0−t0(1+η) 1 (log 1 δ)6+4t0 &η(,t0,s0),t0,s0δt0η−1 (log 1 δ)6+4t0 &,t0,s0δ−/21 (log 1 δ)6+4t0>1,
Dimension estimates on circular (s, t)-Furstenberg sets 317 where in the third inequality we recall δ= 2−k1and in the fourth as well as the last inequality we recall (4.3). This enables us to deduce dimH(F)≥(2t0+ 1)s0−t0− for any 1 2< s0< s,0< t0< t and > 0. Therefore, dimH(F)≥(2t+ 1)s−t= (2s−1)t+s. We conclude the proof. 5. Proof of Lemma 2.5 This section is devoted to the proof of Lemma 2.5. For the readers’ convenience, we restate Lemma 2.5 in the following. Lemma 5.1. Let A, B, C ∈R2such that min{kA−Bk,kA−Ck,kB−Ck} ≥ 2c with c < 1. For a > 0such that a < 1 20c2, define W:= b−a≤ kx−Ak ≤ b+a, (x, b)∈R2×[1 2,2] : b−a≤ kx−Bk ≤ b+a, b−a≤ kx−Ck ≤ b+a . Then diam W.a c2. We briefly explain the approach. We will decompose Was W=[ b∈I⊂[1/2,2] W(b)×{b}. Then for each fixed b, W(b) := b−a≤ kx−Ak ≤ b+a, x∈R2:b−a≤ kx−Bk ≤ b+a, b−a≤ kx−Ck ≤ b+a =Sa(A, b)∩Sa(B, b)∩Sa(C, b) is a subset in R2formed by the intersection of three annuli. We will show that W(b)6=∅only for branging in a set Iwith diameter .a c2. Moreover, if W(b)6= ∅, then A, B, C form a non-degenerate 4ABC with circumcenter Mand W(b)is contained in a rhombus centered at Mwith diameter .a c2. This will imply diam W.a c2. The above justification is contained in next two auxiliary lemmas. In what follows, given A, B ∈R2and 0< a < c2 20, we denote by Ra,c AB the rectangle centered at the middle point of AB whose short sides have length 9a cand long sides have length 6 parallel to the bisector of AB. Lemma 5.2. Let A, B ∈R2and b∈[1 2,2]. If c < min{1,kA−Bk 2}and 0< a < c2 20 <1, then Sa(A, b)∩Sa(B, b)⊂ Ra,c AB.
318 Jiayin Liu Proof. Let kA−Bk= 2u. Without loss of generality, we assume A= (−u, 0) and B= (u, 0). It is easy to see that Sa(A, b)∩Sa(B, b) ={x∈R2|b−a≤ kx−Ak ≤ b+a, b −a≤ kx−Bk ≤ b+a} ⊂U:= {x= (x1, x2)∈R2|max{kx−Ak,kx−Bk} ≤ 3, −2a≤ kx−Ak−kx−Bk ≤ 2a}. Since u=kA−Bk 2> c > a, from planar geometry we know that the set {x∈R2| kx−Ak−kx−Bk=±2a} consisting of points, whose absolute difference of distances to the two fixed points A and Bis the constant 2a, is a hyperbola in R2determined by the equation y(x) = y(x1, x2) = 1 where y:R2→Ris defined by y(x) = y(x1, x2)7→ x2 1 a2−x2 2 u2−a2. Then we observe that {x∈R2| −2a≤ kx−Ak−kx−Bk ≤ 2a}={x∈R2|y(x1, x2)≤1} and hence U= [B((−u, 0),3) ∩B((u, 0),3)] ∩{x∈R2|y(x1, x2)≤1}, which implies U⊂ {x∈R2| |x2| ≤ 3, y(x1, x2)≤1}. Figure 5 shows the case that u= 2 and a= 0.75. Figure 5. The case u= 2 and a= 0.75.
Dimension estimates on circular (s, t)-Furstenberg sets 319 Letting |x2|= 3 in the equation y(x1, x2) = 1, we have |x1|=aq1 + 9 u2−a2. Since 20a<c2<1and u>c, it holds (5.1) ar1 + 9 u2−a2< ar1 + 9 c2−a2< as1 + 9 8 9c2< ar81 4c2=9 2 a c<3. This implies that the rectangle with four vertices (±9 2 a c,±3) has short side length 9a c and long side length 6. By recalling the definition of Ra,c AB, we have Sa(A, b)∩Sa(B, b)⊂U⊂ Ra,c AB ={x∈R2| |x1| ≤ 9 2 a c,|x2| ≤ 3}, which concludes the proof. Lemma 5.3. Let A, B, C ∈R2such that min{kA−Bk,kA−Ck,kB−Ck,2} ≥ 2c. Let b∈[1 2,2]. Then for a > 0such that a < 1 20c2< b, define (5.2) W(b) := b−a≤ kx−Ak ≤ b+a, x∈R2:b−a≤ kx−Bk ≤ b+a, b−a≤ kx−Ck ≤ b+a If the triangle 4ABC is degenerate, then (5.3) W(b) = ∅for all b∈[1 2,2]. If 4ABC is non-degenerate, let Mbe the circumcenter of 4ABC and h:= kM−Ak=kM−Bk=kM−Ck. Then, we have (5.4) W(b)⊂BM, K a c2for all b∈1 2,2. In addition, if W(b)6=∅, then (5.5) b∈h−Ka c2, h +Ka c2∩1 2,2. Here in (5.4) and (5.5),Kis an absolute constant. Proof. Without loss of generality, we assume the side BC of 4ABC has maximal length. Then ∠A:= ∠BAC ≥π/3. Since W(b) = Sa(A, b)∩Sa(B, b)∩Sa(C, b), from Lemma 5.2 we know (5.6) W(b)⊂ Ra,c AB ∩Ra,c AC. Below we estimate diam(Ra,c AB ∩Ra,c AC)from above. Denote by L1and L2the bisector of AB and AC respectively. Hence D:= L1∩AB is the middle point of AB and E:= L2∩AC is the middle point of AC. See Figure 6 for an illustration. Let d=9 2 a c. Since 20a < c2, we have (5.7) d=9 2 a c<9 40c < 1 4c. Case 1. ∠A=π. That is, 4ABC degenerates. By (5.7), it is easy to see Ra,c AB ∩Ra,c AC =∅, which, with help of (5.6), implies W(b) = ∅for all b∈[1 2,2]. That is, (5.3) holds.
320 Jiayin Liu Figure 6. An illustration for L1,L2,Ra,c AB and Ra,c AC . Case 2. ∠A∈(π−arctan(2c/9), π). We will show that (5.8) Ra,c AB ∩Ra,c AC =∅. Denote kA−Bk= 2uand kA−Ck= 2v. Since Mis the circumcenter of 4ABC, it is the intersection of lines L1and L2. Then the line L3passing through Aand Mdivides R2into two connected components. Since the center Dof Ra,c AB and the center Eof Ra,c AC are contained in different connected components above and d < 1 4c by (5.7), a sufficient condition for Ra,c AB ∩Ra,c AC =∅is that (5.9) Ra,c AB ∩L3=∅and Ra,c AC ∩L3=∅. See Figure 7 for an illustration. Figure 7. An illustration for Case 2. Recall that half of the length of the short sides of Ra,c AB and Ra,c AC is d=9 2 a c. By assumption ∠A∈(π−arctan(2c/9), π), this implies ∠DMA+∠EMA ≤arctan(2c/9). Hence (5.10) tan ∠DMA < 2c 9≤c−d 3≤u−d 3and tan ∠EMA < 2c 9≤c−d 3≤v−d 3 where in the second inequality we apply d < c 3from (5.7). Now we explain how (5.10) implies (5.9). Let D0be the intersection of the line segment AD and the long side of the triangle Ra,c AB. Also, let L0 1:= L1+ (D0−D). That is, line L0 1is the translation
Dimension estimates on circular (s, t)-Furstenberg sets 321 of line L1by the vector D0−Din R2. Denote the intersection of L0 1and L3by M0. See Figure 8 for an illustration. Figure 8. An illustration for D0,M0and L0 1. We observe that (5.11) ∠D0M0A=∠DMA and tan ∠D0M0A=kA−D0k kD0−M0k=u−d kD0−M0k where in the last inequality we recall that kA−D0k=kA−Dk−kD−D0k,kA−Dk=u and kD−D0k=d. Combining (5.10) and (5.11), we deduce that u−d kD0−M0k (5.11) = tan ∠D0M0A(5.10) <u−d 3, which implies kD0−M0k>3. This, combined with the fact that half of the length of the long sides of Ra,c AB is 3, shows that Ra,c AB ∩L3=∅. By a similar argument, we also have Ra,c AC ∩L3=∅with the help of (5.10). This shows that (5.9) is true and hence (5.8) holds. Case 3. ∠A∈[π/3, π −arctan(2c/9)]. In this case, W(b)may not be empty. Now, we assume that W(b)6=∅, which implies that Ra,c AB ∩ Ra,c AC 6=∅. Moreover, denote by Vd Lithe closed d-neighbourhood of lines Li,i= 1,2. Then Vd L1∩Vd L2is a rhombus TMcentered at Msatisfying Ra,c AB ∩Ra,c AC ⊂ TM. We will show that (5.12) diam TM≤324 a c2. See Figure 9 for an illustration. Denote the length of two diagonals of TMby d1and d2and the the length of four sides of TMby l. We have diam TM= max{d1, d2},(5.13) d2 1+d2 2= 4l2 (5.14) and (5.15) l=2d sin ∠A.
322 Jiayin Liu Figure 9. An illustration for the estimate ky−Mk. Since ∠A∈[π/3, π −arctan(2c/9)], we have (5.16) sin ∠A≥sin arctan 2c 9≥sin c 9≥c 18 where in the second last inequality we use the fact that arctan y > y 2if 0< y < 1 and in the last inequality we use the fact that sin y > y 2if 0< y < 1. Combining (5.13), (5.14), (5.15) and (5.16), we obtain (5.17) diam TM≤2l≤72d c= 324 a c2 where in the last equality we recall d=9 2 a c. Therefore, we conclude (5.12). Combining Case 2 and Case 3, we conclude (5.4). Finally, we show (5.5). Let x∈W(b). By (5.2) and (5.13), we have |b−h|=|b−kM−Ak| ≤ |b−kx−Ak|+kx−Mk.a+a c2.a c2. The proof is complete. Now, we are in a position to show: Proof of Lemma 2.5. For b∈[1 2,2], define f W(b) := W(b)×{b} = b−a≤ kx−Ak ≤ b+a, (x1, x2, x3)=(x, x3)∈R3:b−a≤ kx−Bk ≤ b+a, x3=b b−a≤ kx−Ck ≤ b+a, . First we assume 4ABC degenerates. Then by (5.3), we know f W(b) = ∅for all b∈[1 2,2]. Hence the lemma holds for this case. Next, we assume 4ABC is non-degenerate. Then by (5.4), we have (5.18) f W(b)⊂B((M, b), K a c2)∩{x3=b} ⊂ R3for all b∈[1 2,2], where Mis the circumcenter of the triangle 4ABC.