
Let Sj(z)=εj+(z−εj)/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(z−ω)−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/(2−α) 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+(z−εj)/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(z−ω)−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/(2−α) 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(z−ω)−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+(z−εj)/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, C∖K is a multiply connected domain, and the Hausdorff dimension of K is α=log3/log2. We denote unbounded connected region of C∖K by △0 and the triangular connected region of C∖K by △n(n≥1). Then, C∖K=∪∞n=0△n.
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 E⊂C, we have Hα(ϕ(E))=Hα(E), where ϕ can be the complex conjugation or the rotation of eiθ. Also, for any n∈Z,Hα(2nE)=2αnH(E). The Cauchy transform of μ=Hα|K is
F(z)=∫KdHα(w)z−ω. | (1.1) |
Our main consideration is on the dyadic points of ∂△0. With fixed k, for 1≤m≤2k−1, let
zk,m=m2kε1+(1−m2k)ε2=−12+m−2k−12k√3i. |
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(K−1) be a relocation of the Sierpinski gasket K. The new vertices are at 0,√3eπi/6,√3e−πi/6. Set SjK=Kj,j=0,1,2. Let Tj=eπi(Kj−1),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=⋃n∈Z2n(T1∪T2). For ℓ=1,⋯,5, let Aℓ=eℓπi/3A0, and
Hℓ(z)=∫AℓdHα(ω)(z−ω)2. |
It is easy to check that Hℓ(2z)=2α−2Hℓ(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)|≤Ctα−2,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|≤ρ√3,
|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
∥f∥p=sup0≤r<1(12π∫2π0|f(reiθ)|pdθ)1/p<+∞. |
Theorem 3.1. Let g(z)=F(1z) for z∈D. Then, g′(z)∈Hp for 0<p<12−α and g′∉Hp for p≥12−α, where α is the Hausdorff dimension of K.
Remark Similarly, we may prove that g(k)(z)∈Hp for 0<p<1k+1−α and g(k)(z)∉Hp for p≥1k+1−α.
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/(2−α), and g′(eiθ)∉Lp for p≥1/(2−α).
For −π/3≤θ<0, let z=eiθ and z∗=ρe−iθ∈∂△0, where ρ>0. By the sine rule, we have
dist(e−iθ,K)=sin(π6−θ)|e−iθ−z∗|=−sinθ2(√3cosθ2+sinθ2)≥∣sinθ2|. | (3.1) |
From Proposition 2.1, there exists some constant C>0 such that
|g′(z)|≤Cdist(e−iθ,K)α−2≤C|θ|α−2,−π3≤θ<0. |
Notice that Hα and K are symmetric with respect to the real-axis. Then, g′(ˉz)=¯g′(z), and ∫π/3−π/3|g′(eiθ)|pdθ=2∫0−π/3|g′(eiθ)|pdθ. Hence, for 0<p<1/(2−α),
∫π−π|g′(eiθ)|pdθ=6∫0−π/3|g′(eiθ)|pdθ≤C∫π/30θp(α−2)dθ<+∞. |
The above inequality gives g′(eiθ)∈Lp for 0<p<1/(2−α).
Next, we will prove g′(eiθ)∉L12−α. For 0<t≤√3/2 and |θ∣<5π/6, from Proposition 2.2, we obtain
|F′(1+teiθ)+2(2−α)NH3(2Nteiθ)|≤C1, | (3.2) |
where the positive integer N satisfies 1/2≤2Nt<1. For 0<t<1, let 1+teiθ=eiφ. Then,
φ=φ(t)=arctant√1−t2/41−t2/2andθ=θ(t)=π2+arcsint2. | (3.3) |
Since F′(eiφ)=−e−2iφg′(e−iφ)) and H3(2z)=2α−2H3(z), we have
|e−2iφg′(e−iφ)−2(2−α)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(eiβ−1)=−2sin(β/2)eiβ/2, we see that
H3(bieiβ)=∫A3dHα(w)(bi−w+bi(eiβ−1))2=∫A3dHα(w)(bi−w)2+∞∑k=1(k+1)∫A3(2bsin(β2))kekβi2dHα(w)(bi−w)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)|bi−w∣k+2=∞∑k=1(k+1)∞∑n=−∞3n∫E1∪E2(2bsin(β/2))kdHα(w)|bi−2nw∣k+2≤∞∑k=1(k+1)∞∑n=0(38)n∫E1∪E2(2bsin(β/2))kdHα(w)|2−nbi−w∣k+2+∞∑k=1(k+1)∞∑n=1(13)n∫E1∪E2(2bsin(β/2))kdHα(w)|bi−2−nw∣k+2. |
With consideration of geometric factors, for b∈[1/2,1), n≥1 and w∈E1∪E2, the two inequalities |w−2−nbi∣≥3/4 and |bi−2−nw∣≥√3b/2 hold. Hence,
|ε(t)|≤1615∞∑k=1(k+1)(43)k+2(2bsin(β2))k+13∞∑k=1(k+1)(2√33b)k+2(2bsin(β2))k. |
By sinβ=t/2, it is easy to check that sin(β/2)=√1−√1−t2/4/√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<t≤δ. | (3.6) |
From (3.4)–(3.6), we know that
|g′(e−iφ)|≥2(2−α)N(|H3(bi)|−C2t)−C1≥2(2−α)N|H3(bi)|−C, |
where C is a positive constant. This implies that, for 0<t≤δ, we have
(C+|g′(e−iφ)|)12−α≥2N|H3(bi)|12−α. | (3.7) |
Let the positive integer N0 satisfy 2−N≤δ for all N≥N0. Note that φ′(t)≥c1>0 for 0<t≤δ. According to (3.7), we obtain
∫φ(2−N)φ(2−N−1)(C+|g′(e−iφ)|)12−αdφ≥2N∫φ(2−N)φ(2−N−1)|H3(bi)|12−αdφ≥c12N∫2−N2−N−1|H3(2Nti)|12−αdt=c1∫11/2|H3(xi)|12−αdx:=c2. |
We can check that H3(z) is non-constant analytic in |argz|<5π/6. This gives c2>0. Noting that φ(2−N−1)→0+ as N→∞, we have
∫φ(2−N0)0(C+|g′(e−iφ)|)12−αdφ=+∞∑N=N0∫φ(2−N)φ(2−N−1)(C+|g′(e−iφ)|)12−αdφ=+∞. |
By using (a+b)p≤2p(ap+bp) for a>0,b>0 and p>0, we have
∫φ(2−N0)0|g′(e−iφ)|12−αdφ=+∞, |
which implies that g′(z)∉Hp for p≥1/(2−α).
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)=(1−z)−λ for λ>0, and kλ(z)=log11−z+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 μ∈Λ 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
∥h∥Fλ=inf{∥μ∥:μ∈Λsuchthat(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 h∈Fλ. Let Mλ denote the set of all multipliers of Fλ. Mλ is a Banach space with respect to the norm defined by
∥υ∥Mλ=sup{∥υh∥Fλ:h∈Fλ,∥h∥Fλ≤1}. |
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 β≥0, g(z)∈Mβ. For any small ε>0, g′(z)∈F2−α+ε and g′(z)∉F2−α−ε, where α is the Hausdorff dimension of K.
Proof. Since g′∈Hp for some p>1, we have g∈Mβ for each β≥0 by Theorem 3.1 in [21, p. 621]. It follows from the remark of Theorem 3.1 that g″(z)∈H1/(3−α)⊂H1/(3−α+ε). Together with Theorem 3 in [17, p. 116], we see that H1/p⊂Fp for p≥1. Hence, g″∈F3−α+ε. Note that f∈Fλ if and only if f′∈Fλ+1 [17, p. 112]. Consequently g′(z)∈F2−α+ε follows. From [7, p. 70], we obtain that g(z)=z+∑∞n=1a3n+1z3n+1 for |z|<1, and c1n−α≤a3n+1≤c2n−α for n≥1, with constants c1>0 and c2>0. Assume that g′(z)∈F2−α−ε. 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|≤cn1−α−ε,n≥1. This is a contradiction. Then, the result follows.
In view of Theorem 4.1, an interesting question is to determine if g′(z)∈F2−α, which is equivalent to ([17, p. 115]).
h(z)=∞∑n=0n+1dn(2−α)∫KωndHα(ω)zn∈F1, | (4.2) |
where dn(λ)=Γ(n+λ)Γ(n+1)Γ(λ) is defined by (1−x)−λ=∑∞n=0dn(λ)xn. It follows from Stirling's formula that dn(λ)(n+1)1−λ=Γ(λ)−1+c(λ)(n+1)−1+O((n+1)−2). Then,
n+1dn(2−α)dn(α+1)=(n+1)1−α(n+1)αdn(2−α)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)=c0∫KdHα(ω)(1−zω)α+1+c1∞∑n=0dn(α+1)n+1∫KωndHα(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 H∞⊂H2⊂H1⊂F1. Consequently,
g′(z)∈F2−α⟺h1(z)=∫KdHα(ω)(1−zω)α+1∈F1. | (4.3) |
In view of (4.3), by [18], we know that g′(z)∈F2−α if and only if ∫z0h1(t)dt∈F0. This leads us to consider
fε(z)=∫KdHα(ω)(1−ωz)α−ε,|z|<1. | (4.4) |
Theorem 4.2. fε∈Mβ for each β≥0 if ε>0, and f0(z)∉Mβ for each β≥0.
Proof. For the first assertion, we only need to show f′ε(z)∈Hp for some p>1 by Theorem 3.1 in [21, p. 621]. Noting that (1−x)−λ=∑∞n=0dn(λ)xn for |x|<1, it follows easily by the H¨older inequality that
12π∫2π0|f′ε(reiθ)|pdθ≤12π∫2π0(∫K|ω||1−reiθω|(α+1−ε)dHα(ω))pdθ≤C∫K12π∫2π0dθ|1−reiθω|p(α+1−ε)dHα(ω)=C∫K12π∫2π0|∞∑n=0dn(p2(α+1−ε))rneinθωn|2dθdHα(ω)=C∞∑n=0d2n(p2(α+1−ε))∫K|ω|2ndHα(ω)r2n. |
It follows from Proposition 4.2 in [7] that ∫K|ω|ndHα(ω)≤Cn−α. Combining this with dn(λ)∼Γ(λ)−1nλ−1(n→∞), we get that
12π∫2π0|f′ε(reiθ)|pdθ≤C∞∑n=1np(α+1−ε)−2−αr2n. |
Notice that p(α+1−ε)−2−α→−ε−1 as p→1, we can choose p>1 such that p(α+1−ε)−2−α<−1−ε/2. Hence, f′ε∈Hp 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 n≠3k, and there exists some constant c1>0 such that ∫Kω3kdHα(ω)≥c1k−α for all k≥1. Note that dn(α)≥c2nα−1 for some constant c2>0 and all n≥1. 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)x3n≥c3∞∑n=1x3nn→∞,x→1−. |
Although we can not prove f0(z)=∫z0h1(t)dt∈F0(or g′(z)∈F2−α), yet we can prove f0(z)∈BMOA, which consists of all functions f∈H1 satisfying
∥f∥BMOA=supI⊂T1|I|∫I|f(ζ)−fI||dζ|<∞, |
where the supremum is taken over all arcs I⊂T with |I|=∫I|dζ| and fI=|I|−1∫If(ζ)|dζ|. It should be noted that F0⊂BMOA⊂Hp 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
|f′0(z)|≤C|1−z3|,|z|<1. | (4.5) |
It is equivalent to prove that p(z):=(1−z3)f′0(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/3≤θ≤0}. For z∈Ω, let d=dist(z−1,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
|f′0(z)|≤C1∫KdHα(ω)|1z−ω|α+1≤C1d. |
With consideration of geometry, we find that there exists some constant C2>0 such that d=dist(z−1,K)≥C2|1−z| for z∈Ω. Hence,
|p(z)|≤C1C−12|1−z3||1−z|−1≤C3,z∈Ω. |
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 |f′0(z)|≤C|1−z3|−1. Hence, for any small sector Sh(θ0)={reiθ:1−h≤r<1,|θ−θ0|≤h},
suph>01h∫Sh(θ0)|f′0(z)|2(1−|z|2)dxdyπ≤Csuph>01h∫Sh(0)1−|z|2|1−z3|2dxdy≤C′. |
This shows that |f′0(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 |