Research article

Function space properties of the Cauchy transform on the Sierpinski gasket

  • Received: 23 October 2022 Revised: 06 December 2022 Accepted: 19 December 2022 Published: 29 December 2022
  • MSC : 28A80, 30C55, 30E20

  • Let Sj(z)=εj+(zεj)/2 be an iterated function system, where εj=e2jπi/3 for j=0,1,2. Then, there exists a uniform self-similar measure μ supported on a compact set K, which is the attractor of {Sj}2j=0. The Hausdorff dimension of the attractor K is α=log3/log2. Let F(z)=K(zω)1dμ(ω) be the Cauchy transform of μ. In this paper, we consider the Hardy space and the multiplier property of F. We prove that F belongs to Hp for 0<p<1/(2α) and that F is a multiplier of some class of function space.

    Citation: Songran Wang, Zhinmin Wang. Function space properties of the Cauchy transform on the Sierpinski gasket[J]. AIMS Mathematics, 2023, 8(3): 6064-6073. doi: 10.3934/math.2023306

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  • Let Sj(z)=εj+(zεj)/2 be an iterated function system, where εj=e2jπi/3 for j=0,1,2. Then, there exists a uniform self-similar measure μ supported on a compact set K, which is the attractor of {Sj}2j=0. The Hausdorff dimension of the attractor K is α=log3/log2. Let F(z)=K(zω)1dμ(ω) be the Cauchy transform of μ. In this paper, we consider the Hardy space and the multiplier property of F. We prove that F belongs to Hp for 0<p<1/(2α) and that F is a multiplier of some class of function space.



    The Cauchy transform of a measure in the plane is a useful tool for geometric measure theory [1,2,3], and it has also important applications in solving integral equations [4,5]. If the measure is a self-similar measure, the Cauchy transform of it has very rich fractal behavior. Stricharz et. al. [6] initiated the study of the Cauchy transform F(z)=K(zω)1dμ(ω) of a self-similar measure μ with compact support K, and they proved that F has a Holder continuous extension over K and showed how to compute the Laurent expansion of F in the complement of a disk containing K. Soon afterwards, more analytic and geometric properties of F were given by Dong and Lau [7,8,9,10,11,12]: for example, the asymptotic behavior of the Laurent coefficients of F and the region of starlikeness of F. They also gave estimates for the Taylor coefficients of the Cauchy transforms of some special Hausdorff measures [13,14]. For the special case that K is the Sierpinski gasket, and μ is the normalized Hausdorff measure on K, Dong and Lau [8,9,10,11] carried out a detailed study of the properties of the mapping of the Cauchy transform on K and investigated some open problems proposed in [6]. Away from K, F is well-behaved, but the image of F is chaotic near the boundary of K and is difficult to catch [see 6, 7, 12]. In this paper, we will consider the properties of the function spaces of F(z) near the Sierpinski gasket.

    Let Sj(z)=εj+(zεj)/2 be an iterated function system, where εj=e2jπi/3 for j=0,1,2. The attractor K of {Sj}2j=0 is just the Sierpinski gasket (Figure 1). It is well known that K is a compact set, CK is a multiply connected domain, and the Hausdorff dimension of K is α=log3/log2. We denote unbounded connected region of CK by 0 and the triangular connected region of CK by n(n1). Then, CK=n=0n.

    Figure 1.  Sierpinski gasket.

    Let μ be the uniform self-similar measure on K, i.e., μ is the restriction of the α -Hausdorff measure on K normalized to a probability measure. With slight abusing of notation, we let Hα be the Hausdorff measure normalized on K. From the basic property of the Hausdorff measure [15], for EC, we have Hα(ϕ(E))=Hα(E), where ϕ can be the complex conjugation or the rotation of eiθ. Also, for any nZ,Hα(2nE)=2αnH(E). The Cauchy transform of μ=Hα|K is

    F(z)=KdHα(w)zω. (1.1)

    Our main consideration is on the dyadic points of 0. With fixed k, for 1m2k1, let

    zk,m=m2kε1+(1m2k)ε2=12+m2k12k3i.

    These are the dyadic points on the line segment joining the two vertices ε1 and ε2. The dyadic points on the other two sides of 0 can be obtained by zk,m multiplied by εj,j=1,2. It suffices to consider zk,m since εjF(εjz)=F(z),j=0,1,2.

    The paper is organized as follows. In Section 2, we introduce some necessary results and notations. In Section 3, we give an Hp space property of F(1/z) on |z|<1. In the final section, we study the multiplier property of F(1/z) on |z|<1.

    In this section, we first give some necessary notations and propositions firstly. Let T=eπi(K1) be a relocation of the Sierpinski gasket K. The new vertices are at 0,3eπi/6,3eπi/6. Set SjK=Kj,j=0,1,2. Let Tj=eπi(Kj1),j=0,1,2, denote the three triangular components of T containing the respective vertices. We define the "Sierpinski cones" of T (Figure 2) as A0=nZ2n(T1T2). For =1,,5, let A=eπi/3A0, and

    H(z)=AdHα(ω)(zω)2.
    Figure 2.  Sierpinski cones.

    It is easy to check that H(2z)=2α2H(z) by the scaling property of Hausdorff measure. In the sequel, we need the following propositions.

    Proposition 2.1. [9] There exists some constant C>0 such that,

    maxdist(z,K)t|F(z)|Ctα2,t>0.

    Proposition 2.2. [9] For 0<ρ<1, there exists some constant C>0 which depends on ρ such that for |argz|<5π/6 and 0<|z|ρ3,

    |F(1+z)+H3(z)|C.

    For the details of the proof of the above two propositions, we can see [10].

    In this section, we consider the function space property of F(1z) on D={z:|z|<1}. The Hardy space Hp consists of analytic functions f in D such that

    fp=sup0r<1(12π2π0|f(reiθ)|pdθ)1/p<+.

    Theorem 3.1. Let g(z)=F(1z) for zD. Then, g(z)Hp for 0<p<12α and gHp for p12α, where α is the Hausdorff dimension of K.

    Remark Similarly, we may prove that g(k)(z)Hp for 0<p<1k+1α and g(k)(z)Hp for p1k+1α.

    Proof. Note that g(z) is analytic in D, and g(eiθ) exists for θ{0,2π/3,4π/3}. By Theorem 2.6 in [16, p. 21], we only need to prove g(eiθ)Lp for 0<p<1/(2α), and g(eiθ)Lp for p1/(2α).

    For π/3θ<0, let z=eiθ and z=ρeiθ0, where ρ>0. By the sine rule, we have

    dist(eiθ,K)=sin(π6θ)|eiθz|=sinθ2(3cosθ2+sinθ2)sinθ2|. (3.1)

    From Proposition 2.1, there exists some constant C>0 such that

    |g(z)|Cdist(eiθ,K)α2C|θ|α2,π3θ<0.

    Notice that Hα and K are symmetric with respect to the real-axis. Then, g(ˉz)=¯g(z), and π/3π/3|g(eiθ)|pdθ=20π/3|g(eiθ)|pdθ. Hence, for 0<p<1/(2α),

    ππ|g(eiθ)|pdθ=60π/3|g(eiθ)|pdθCπ/30θp(α2)dθ<+.

    The above inequality gives g(eiθ)Lp for 0<p<1/(2α).

    Next, we will prove g(eiθ)L12α. For 0<t3/2 and |θ∣<5π/6, from Proposition 2.2, we obtain

    |F(1+teiθ)+2(2α)NH3(2Nteiθ)|C1, (3.2)

    where the positive integer N satisfies 1/22Nt<1. For 0<t<1, let 1+teiθ=eiφ. Then,

    φ=φ(t)=arctant1t2/41t2/2andθ=θ(t)=π2+arcsint2. (3.3)

    Since F(eiφ)=e2iφg(eiφ)) and H3(2z)=2α2H3(z), we have

    |e2iφg(eiφ)2(2α)NH3(2Nteiθ)|C1 (3.4)

    by using (3.2). Define β=arcsin(t/2) and b=b(t):=2Nt. Noting that b=b(t)[12,1) and i(eiβ1)=2sin(β/2)eiβ/2, we see that

    H3(bieiβ)=A3dHα(w)(biw+bi(eiβ1))2=A3dHα(w)(biw)2+k=1(k+1)A3(2bsin(β2))kekβi2dHα(w)(biw)k+2:=H3(bi)+ε(t). (3.5)

    To estimate ε(t), we set E1=T1,E2=T2. Then,

    |ε(t)|k=1(k+1)A3(2bsin(β/2))kdHα(w)|biwk+2=k=1(k+1)n=3nE1E2(2bsin(β/2))kdHα(w)|bi2nwk+2k=1(k+1)n=0(38)nE1E2(2bsin(β/2))kdHα(w)|2nbiwk+2+k=1(k+1)n=1(13)nE1E2(2bsin(β/2))kdHα(w)|bi2nwk+2.

    With consideration of geometric factors, for b[1/2,1), n1 and wE1E2, the two inequalities |w2nbi∣≥3/4 and |bi2nw∣≥3b/2 hold. Hence,

    |ε(t)|1615k=1(k+1)(43)k+2(2bsin(β2))k+13k=1(k+1)(233b)k+2(2bsin(β2))k.

    By sinβ=t/2, it is easy to check that sin(β/2)=11t2/4/2<t/3 for small t>0. This shows that we can find constants C2>0 and δ>0 such that

    |ε(t)|C2t,0<tδ. (3.6)

    From (3.4)–(3.6), we know that

    |g(eiφ)|2(2α)N(|H3(bi)|C2t)C12(2α)N|H3(bi)|C,

    where C is a positive constant. This implies that, for 0<tδ, we have

    (C+|g(eiφ)|)12α2N|H3(bi)|12α. (3.7)

    Let the positive integer N0 satisfy 2Nδ for all NN0. Note that φ(t)c1>0 for 0<tδ. According to (3.7), we obtain

    φ(2N)φ(2N1)(C+|g(eiφ)|)12αdφ2Nφ(2N)φ(2N1)|H3(bi)|12αdφc12N2N2N1|H3(2Nti)|12αdt=c111/2|H3(xi)|12αdx:=c2.

    We can check that H3(z) is non-constant analytic in |argz|<5π/6. This gives c2>0. Noting that φ(2N1)0+ as N, we have

    φ(2N0)0(C+|g(eiφ)|)12αdφ=+N=N0φ(2N)φ(2N1)(C+|g(eiφ)|)12αdφ=+.

    By using (a+b)p2p(ap+bp) for a>0,b>0 and p>0, we have

    φ(2N0)0|g(eiφ)|12αdφ=+,

    which implies that g(z)Hp for p1/(2α).

    In this section, we consider the multiplier property of g(z). Let Λ denote the set of complex-valued Borel measures on T={z:|z|=1}, let kλ(z)=(1z)λ for λ>0, and kλ(z)=log11z+1 for λ=0. Here, we choose the branch of kλ(z) which equals 1 when z=0. Let Fλ denote the family of functions h for which there exists μΛ such that

    h(z)=Tkλ(ζz)dμ(ζ),|z|<1. (4.1)

    Each Fλ is a Banach space with respect to the norm defined by

    hFλ=inf{μ∥:μΛsuchthat(4.1)holds},

    where μ denotes the total variation of the measure μ. The spaces Fλ were introduced in [17,18], and some roperties of functions in Fλ were obtained in [19,20].

    An analytic function υ(z) in D is called a multiplier of Fλ provided that υ(z)h(z)Fλ for all hFλ. Let Mλ denote the set of all multipliers of Fλ. Mλ is a Banach space with respect to the norm defined by

    υMλ=sup{υhFλ:hFλ,hFλ1}.

    The family Mλ has been studied in [19,20,21]. In this section, we will consider the multiplier property of g(z)=F(1/z) with respect to Fλ.

    Theorem 4.1. For each β0, g(z)Mβ. For any small ε>0, g(z)F2α+ε and g(z)F2αε, where α is the Hausdorff dimension of K.

    Proof. Since gHp for some p>1, we have gMβ for each β0 by Theorem 3.1 in [21, p. 621]. It follows from the remark of Theorem 3.1 that g(z)H1/(3α)H1/(3α+ε). Together with Theorem 3 in [17, p. 116], we see that H1/pFp for p1. Hence, gF3α+ε. Note that fFλ if and only if fFλ+1 [17, p. 112]. Consequently g(z)F2α+ε follows. From [7, p. 70], we obtain that g(z)=z+n=1a3n+1z3n+1 for |z|<1, and c1nαa3n+1c2nα for n1, with constants c1>0 and c2>0. Assume that g(z)F2αε. Since every complex measure on T is of bounded variation, it follows easily that there exists some constant c>0 such that |(3n+1)a3n+1|cn1αε,n1. This is a contradiction. Then, the result follows.

    In view of Theorem 4.1, an interesting question is to determine if g(z)F2α, which is equivalent to ([17, p. 115]).

    h(z)=n=0n+1dn(2α)KωndHα(ω)znF1, (4.2)

    where dn(λ)=Γ(n+λ)Γ(n+1)Γ(λ) is defined by (1x)λ=n=0dn(λ)xn. It follows from Stirling's formula that dn(λ)(n+1)1λ=Γ(λ)1+c(λ)(n+1)1+O((n+1)2). Then,

    n+1dn(2α)dn(α+1)=(n+1)1α(n+1)αdn(2α)dn(α+1)=c0+c1n+1+cn,

    where |cn|C(n+1)2. If we substitute this into (4.2), then we have

    h(z)=c0KdHα(ω)(1zω)α+1+c1n=0dn(α+1)n+1KωndHα(ω)zn+n=0cndn(α+1)KωndHα(ω)zn:=c0h1(z)+c1h2(z)+h3(z).

    Since |dn(α+1)KωndHα(ω)|C by [7], it follows that h2(z)H2 and h3(z)H, which imply c1h2(z)+h3(z)F1 as HH2H1F1. Consequently,

    g(z)F2αh1(z)=KdHα(ω)(1zω)α+1F1. (4.3)

    In view of (4.3), by [18], we know that g(z)F2α if and only if z0h1(t)dtF0. This leads us to consider

    fε(z)=KdHα(ω)(1ωz)αε,|z|<1. (4.4)

    Theorem 4.2. fεMβ for each β0 if ε>0, and f0(z)Mβ for each β0.

    Proof. For the first assertion, we only need to show fε(z)Hp for some p>1 by Theorem 3.1 in [21, p. 621]. Noting that (1x)λ=n=0dn(λ)xn for |x|<1, it follows easily by the H¨older inequality that

    12π2π0|fε(reiθ)|pdθ12π2π0(K|ω||1reiθω|(α+1ε)dHα(ω))pdθCK12π2π0dθ|1reiθω|p(α+1ε)dHα(ω)=CK12π2π0|n=0dn(p2(α+1ε))rneinθωn|2dθdHα(ω)=Cn=0d2n(p2(α+1ε))K|ω|2ndHα(ω)r2n.

    It follows from Proposition 4.2 in [7] that K|ω|ndHα(ω)Cnα. Combining this with dn(λ)Γ(λ)1nλ1(n), we get that

    12π2π0|fε(reiθ)|pdθCn=1np(α+1ε)2αr2n.

    Notice that p(α+1ε)2αε1 as p1, we can choose p>1 such that p(α+1ε)2α<1ε/2. Hence, fεHp for p>1.

    For the second assertion, it is sufficient to prove that f0(z) is unbounded in D. By Theorem 5.2 in [7], we get that KωndHα(ω)=0 for n3k, and there exists some constant c1>0 such that Kω3kdHα(ω)c1kα for all k1. Note that dn(α)c2nα1 for some constant c2>0 and all n1. It follows that there exists some constant c3>0 such that

    f0(x)=n=0dn(α)KωndHα(ω)xn=1+n=1d3n(α)Kω3ndHα(ω)x3nc3n=1x3nn,x1.

    Although we can not prove f0(z)=z0h1(t)dtF0(or g(z)F2α), yet we can prove f0(z)BMOA, which consists of all functions fH1 satisfying

    fBMOA=supIT1|I|I|f(ζ)fI||dζ|<,

    where the supremum is taken over all arcs IT with |I|=I|dζ| and fI=|I|1If(ζ)|dζ|. It should be noted that F0BMOAHp for all p>0 [21, p. 617].

    Theorem 4.3. f0(z)BMOA.

    Proof. We first prove that there exists some positive constant C such that

    |f0(z)|C|1z3|,|z|<1. (4.5)

    It is equivalent to prove that p(z):=(1z3)f0(z) is bounded for |z|<1. It is easy check that p(z) is continue on {z:|z|1/2}. Hence, max|z|1/2|p(z)|<. Next, we prove p(z) is bounded for 1/2<|z|<1. Let Ω={reiθ:1/2<r<1,π/3θ0}. For zΩ, let d=dist(z1,K). Obviously, d>0 as 1<|z|1<2. Noting that p(e2πi/3z)=p(z), |p(ˉz)|=|p(z)|, and we can check that there exists some positive constant C1 such that

    |f0(z)|C1KdHα(ω)|1zω|α+1C1d.

    With consideration of geometry, we find that there exists some constant C2>0 such that d=dist(z1,K)C2|1z| for zΩ. Hence,

    |p(z)|C1C12|1z3||1z|1C3,zΩ.

    Note that |p(e2πi/3z)|=|p(z)|, |p(ˉz)|=|p(z)|. We obtain that p(z) is bounded for 1/2<|z|<1, and (4.5) follows.

    It is known that an analytic function ψ(z) on D belongs to BMOA if and only if |ψ(z)|2(1|z|2)dxdy/π is a Carleson measure [22, p. 240]. By using (4.5), we have |f0(z)|C|1z3|1. Hence, for any small sector Sh(θ0)={reiθ:1hr<1,|θθ0|h},

    suph>01hSh(θ0)|f0(z)|2(1|z|2)dxdyπCsuph>01hSh(0)1|z|2|1z3|2dxdyC.

    This shows that |f0(z)|2(1|z|2)dxdy/π is a Carleson measure, and the result follows.

    This work was supported by the NNSF of China (Grant No. 12101219) and the Hunan Provincial NSF (Grant No. 2022JJ40141). Also, the authors are grateful to Professor Xin-Han Dong for his guidance to complete this paper.

    The authors declare no conflicts of interest.



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    1. Songran Wang, Zhimin Wang, Bergman Space Properties of Fractional Derivatives of the Cauchy Transform of a Certain Self-Similar Measure, 2024, 13, 2075-1680, 268, 10.3390/axioms13040268
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