MODULUS OF CONTINUITY OF THE DIRICHLET SOLUTIONS

Transcription

1 MODULUS OF CONTINUITY OF THE DIRICHLET SOLUTIONS HIROAKI AIKAWA Abstract. Let D be a bounded domain in R n with n 2. For a function f on D we denote by H D f the Dirichlet solution, for the Laplacian, of f over D. It is classical that if D is regular, then H D maps the family of continuous boundary functions to the family of harmonic functions in D continuous up to the boundary D. We show that improved continuity of a boundary function f ensures improved continuity of H D f in the context of general modulus of continuity. 1. Introduction Let D be a bounded domain in R n with n 2. For a function f on D we denote by H D f the Dirichlet solution, for the Laplacian, of f over D, i.e., H D f is harmonic in D and H D f = f on D. It is well known that if D is regular, then H D maps the family of continuous boundary functions to the family of harmonic functions in D continuous up to the boundary D. It may be natural to think that improved continuity of a boundary function f ensures improved continuity of H D f. Let 0 < β 1. We say that f is β-hölder continuous if f (x) f (y) C x y β. Here, C stands for an absolute positive constant whose value is unimportant and may change from one occurrence to the next. In the previous paper [1] we characterized the family of domains for which β-hölder continuity is preserved by H D. In particular, we showed that there is no domain which preserves 1-Hölder (or Lipschitz) continuity. See [2] for further generalizations to p-harmonic functions in metric measure spaces. This paper concerns similar problems in the context of a general modulus of continuity. Let M be the family of positive nondecreasing concave functions ψ on (0, ) with ψ(0) = lim t 0 ψ(t) = 0. The concavity assumption on ψ is natural in view of the Hölder continuity case. See Section 5 for more on these assumptions. Obviously, the case of Hölder continuity is covered by the functions ψ(t) = min{t β, 1}, which belongs to M for 0 < β 1. For 2000 Mathematics Subject Classification. 31A05, 31A20, 31B05, 31B25. Key words and phrases. Modulus of continuity, Dirichlet solution, concave function. This work was supported in part by Kakenhi No and No

2 2 HIROAKI AIKAWA α > 0 let ( log t) α for 0 < t < 1/e α+1, ψ α (t) = (α + 1) α for t 1/e α+1. Then ψ α M. In fact, if 0 < t < 1/e α+1, then ψ α(t) = α t ( log t) α 1 > 0 and ψ α( log t) α 2 α(t) = (α log t) < 0. We see that ψ t 2 α is a modulus of continuity weaker than Hölder continuity. We observe that M includes many functions besides these two types of functions. See Proposition 2.1. Let ψ M. For an arbitrary bounded set E R n, n 2, we consider the family Λ ψ (E) of all bounded continuous functions f on E with f ψ,e = sup x E f (x) f (y) f (x) + sup x,y E ψ( x y ) x y Take another φ M and define the operator norm H D ψ φ = H D f φ,d sup. f Λ ψ ( D) f ψ, D f ψ, D 0 <. The finiteness of H D ψ φ is of interest. Hinkkanen [3] considered the problem mainly for planar domains. However, the most interesting case ψ = φ was treated only for Hölder continuity. Our previous papers [1] and [2] also dealt only with Hölder continuity. The case ψ = φ = ψ α with the above ψ α appeared (at least tacitly) in connection with Kleinian groups. Shiga [5] and [6] proved a Hardy-Littlewood type theorem for holomorphic functions with respect to ψ α. Let us state his result [6, Theorem 2.2], albeit slightly changed to suit our purposes. Theorem A. Let f be a holomorphic function on the unit disk and continuous on =. Suppose that there exists α > 0 such that f (e iθ 1 ) f (e iθ 2 ) Cψ α ( θ 1 θ 2 ) for 0 θ 1, θ 2 < 2π. Then f (z) Cδ (z) 1 ψ α (δ (z)) for z, where δ (z) = 1 z = dist(z, ). This results suggests a harmonic Hardy-Littlewood type result. Theorem B. Let h be a harmonic function on and continuous on. Suppose that there exists α > 0 such that Then h(e iθ 1 ) h(e iθ 2 ) Cψ α ( θ 1 θ 2 ) for 0 θ 1, θ 2 < 2π. h(z) Cδ (z) 1 ψ α (δ (z)) for z.

3 MODULUS OF CONTINUITY 3 Obviously Theorem B yields Theorem A immediately. Hardy-Littlewood type theorems and the modulus of continuity of the Dirichlet solution are closely related. Proposition 2.3 below in Section 2 shows that Theorem B follows from Theorem C. H ψ ψ < for every ψ(t) = ψ α (t) with α > 0. Shiga [6] proved Theorem A by means of the explicit form of Cauchy s integral formula for the unit disk with the aid of very smart calculations. Our motivation is to establish a counterpart of Theorem C for a general domain and a general modulus of continuity. Since there is no explicit form of the Poisson integral for general domains, we need an approach different from Shiga s. One might think of Widman s estimates for the Green function [8]; these are not applicable because of the general nature of the domains considered here and the subtleness of modulus of continuity. Instead, we shall apply a barrier and harmonic measure method, as in [1]. This method enables us to deal not only with ψ α but also a general modulus of continuity ψ M. It also reveals that the smoothness of the domain has no significance. Without loss of generality, we may assume that D is a bounded regular domain (see [1, Proposition 1]). For each a D we define a test function τ a,ψ on D by τ a,ψ (ξ) = ψ( ξ a ) for ξ D. We shall see in Lemma 2.4 that τ a,ψ Λ ψ ( D). Theorem 1.1. Let ψ M. Then the following are equivalent: (i) H D ψ ψ <. (ii) There is a constant C 1 such that whenever a D. H D τ a,ψ (x) Cψ( x a ) for x D, The second condition of Theorem 1.1 can be verified by using harmonic measure. By ω(x, E, U) we denote the harmonic measure evaluated at x of E in U. We write B(x, r) and S (x, r) for the open ball and the sphere of center at x and radius r, respectively. Definition 1.2 (Global Harmonic Measure Decay property). We say that D enjoys the Global Harmonic Measure Decay property with ψ (abbreviated to the GHMD(ψ) property) if ψ( x a ) (1.1) ω(x, D \ B(a, r), D) C ψ(r) whenever a D and r > 0. for x D B(a, r),

4 4 HIROAKI AIKAWA Remark 1.3. If r > diam(d), then D \ B(a, r) =, so that (1.1) trivially holds. Note that if (1.1) holds only for 0 < r < R with some R > 0 independent of a D, then so does for all r > 0 with different C. Theorem 1.4. Let ψ M. If H D ψ ψ <, then D satisfies the GHMD(ψ) property. Conversely, let Ψ M and suppose that there is R > 0 such that (1.2) R r dψ(t) Ψ(t) C ψ(r) Ψ(r) for 0 < r < R. If D satisfies the GHMD(Ψ) property, then H D ψ ψ <. Corollary 1.5. Let 0 < α < α. If D satisfies the GHMD(ψ α ) property, then H D ψ ψ < with ψ(t) = ψ α (t). More geometrically we have the following corollaries. Corollary 1.6. Let D R n, n 2, be a Lipschitz domain. Then H D ψ ψ < for every ψ(t) = ψ α (t) with α > 0. Corollary 1.7. Let D R 2 be a finitely connected plane domain with no puncture. Then H D ψ ψ < for every ψ(t) = ψ α (t) with α > 0. Remark 1.8. The above corollary gives an extension of Theorem C and hence an alternative proof of Theorem A. The plan of this paper is as follows. In the next section we shall prove Theorem 1.1 after preliminary observations for M. In Section 3 we shall prove Theorem 1.4 and Corollary 1.5. Further remarks on the harmonic measure decay property in Section 4 yield Corollary 1.6 and Corollary 1.7. Finally, we shall present an observation, due to S. T. Kuroda [4], for the relevance of the class M, which may be of independent interest. Acknowledgments. The author would like to thank Shiga for showing his preprints [5] and [6]. These give a starting point of this paper. The author also would like to thank Kuroda for showing his preprint [4]. The author is grateful to the referee for many valuable comments. 2. Proof of Theorem 1.1 Let us begin with elementary observations. Proposition 2.1. Let φ, ψ M. Then φ ψ M. Proof. It is easy to see that φ ψ is a nondecreasing function and that φ ψ(0) = 0. For the concavity, let s, t > 0 and 0 < λ < 1. Then λ(φ ψ)(s) + (1 λ)(φ ψ)(t) = λφ(ψ(s)) + (1 λ)φ(ψ(t)) φ(λψ(s) + (1 λ)ψ(t)) φ(ψ(λs + (1 λ)t)) = (φ ψ)(λs + (1 λ)t)),

5 MODULUS OF CONTINUITY 5 where the concavity of φ is used in the first inequality and, for the second inequality, the monotonicity of φ and the concavity of ψ are used. Thus the concavity of φ ψ follows, and hence φ ψ M. Lemma 2.2. Let ψ M. Then (i) ψ(t)/t is nonincreasing, i.e., if 0 < s < t, then ψ(s) ψ(t). s t (ii) ψ is subadditive, i.e. if s, t > 0, then ψ(s + t) ψ(s) + ψ(t). (iii) ψ(t) is doubling, i.e., If 0 < s t 2s, then ψ(s) ψ(t) 2ψ(s). Proof. (i) Since ψ(0) = 0, the concavity of ψ proves (i). (ii) Let f (t) = ψ(t)/t. This is a nonincreasing function by (i). If s, t > 0, then ψ(s) + ψ(t) ψ(s + t) = s f (s) + t f (t) (s + t) f (s + t) = s( f (s) f (s + t)) + t( f (t) f (s + t)) 0. Hence (ii) follows. (iii) Since ψ is monotonic, we need only show that ψ(2s) 2ψ(s) which is, in turn, an immediate consequence of (i). Let δ D (x) = dist(x, D). Proposition 2.3. Let ψ M. Suppose h is a harmonic function on D. If then h(x) h(y) Cψ( x y ) for x, y D, (2.1) h(x) Cδ D (x) 1 ψ(δ D (x)) for x D. Proof. Let x D. By assumption (2.2) h(x) h(y) Cψ(δ D (x)) for y B(x, 2 1 δ D (x)), where the doubling property of ψ is used. Let r = δ D (x)/2. By the Poisson integral formula h(y) = P(y, z)h(z)dσ(z) for y B(x, r), S (x,r) where P(y, z) = r2 y x 2 σ n r y z is the Poisson kernel for B(x, r) with σ n n being the surface measure of a unit sphere. Subtract h(x) from the both sides and differentiate them with respect to y. Then h(y) = y P(y, z)(h(z) h(x))dσ(z) for y B(x, r). S (x,r)

6 6 HIROAKI AIKAWA If y B(x, r/2) = B(x, 4 1 δ D (x)), then y P(y, z) Cr n for z S (x, r), so that (2.2) yields Hence (2.1) holds. h(y) Cr 1 ψ(δ D (x)) for y B(x, 4 1 δ D (x)). Lemma 2.4. Let ψ M and let τ a,ψ be as above. Then τ a,ψ Λ ψ ( D) and there is C independent of a D such that Proof. By definition τ a,ψ ψ, D C. sup τ a,ψ (ξ) ψ(diam(d)) <. ξ D Let ξ, ξ D. Let us estimate τ a,ψ (ξ) τ a,ψ (ξ ) = ψ( ξ a ) ψ( ξ a ). Put t = ξ a and s = ξ a. Without loss of generality we may assume that 0 < s < t. It follows from Lemma 2.2 (ii) and the monotonicity that τ a,ψ (ξ) τ a,ψ (ξ ) ψ(t) ψ(s) ψ(t s)+ψ(s) ψ(s) = ψ(t s) ψ( ξ ξ ), since t s ξ ξ. Hence τ a,ψ (ξ) τ a,ψ (ξ ) ψ( ξ ξ ) 1. Thus τ a,ψ Λ ψ ( D) and τ a,ψ ψ, D ψ(diam(d)) + 1 <. Proof of Theorem 1.1. (i) = (ii). Suppose H D ψ ψ <. Then Lemma 2.4 gives Hence H D τ a,ψ ψ,d H D ψ ψ τ a,ψ ψ, D C H D ψ ψ <. H D τ a,ψ (x) H D τ a,ψ (y) Cψ( x y ) for x, y D. Let x D be a point such that δ D (x) = x x. Letting y x gives Then H D τ a,ψ (x) τ a,ψ (x ) Cψ( x x ). H D τ a,ψ (x) Cψ( x x ) + τ a,ψ (x ) = Cψ( x x ) + ψ( x a ). Now x x = δ D (x) x a. Also, x a x x + x a 2 x a, so that ψ( x a ) ψ(2 x a ) 2ψ( x a ) by the doubling property of ψ. Hence as required. Thus (ii) follows. H D τ a,ψ (x) Cψ( x a )

7 MODULUS OF CONTINUITY 7 (ii) = (i). Suppose (ii) holds. Let f Λ ψ ( D) and let x, y D. It is sufficient to show that H D f (x) H D f (y) C f ψ, D ψ( x y ). Without loss of generality, we may assume that 0 < δ D (y) δ D (x). Let x, y D be points such that δ D (x) = x x and δ D (y) = y y. Let f 0 (ξ) = f (ξ) f (x ). It is easy to see that f 0 ψ, D 2 f ψ, D < and (2.3) f 0 (ξ) f ψ, D ψ( ξ x ) = f ψ, D τ x,ψ(ξ). Let us consider two cases. Case 1. x y 1 2 δ D(x). Let us estimate H D f = H D f 0 on B(x, 2 3 δ D(x)). If z B(x, 2 3 δ D(x)), then (ii) and (2.3) give H D f 0 (z) f ψ, D H D τ x,ψ(z) C f ψ, D ψ( z x ) C f ψ, D ψ(δ D (x)), where the doubling property of ψ is used in the last inequality. Representing H D f 0 as the Poisson integral and differentiating under the integral sign, as in the proof of Proposition 2.3, we obtain H D f (z) = H D ψ(δ D (x)) f 0 (z) C f ψ, D for z B(x, 1 δ D (x) 2 δ D(x)). The mean value theorem, Lemma 2.2 (i) and (iii) give H D f (x) H D ψ(δ D (x)) f (y) C f ψ, D x y C f ψ, D ψ( x y ). δ D (x) Thus (i) follows in this case. Case 2. x y > 1δ 2 D(x). By assumption we have x y > 1δ 2 D(x) 1 δ 2 D(y). It is easy to see that x y x y + δ D (x) + δ D (y) 5 x y. Since D is regular, it follows from (ii) and (2.3) that H D f (x) f (x ) = H D f 0 (x) C f ψ, D H D τ x,ψ(x) C f ψ, D ψ(δ D (x)) C f ψ, D ψ( x y ), where the doubling property of ψ is used in the last inequality. Similarly, we have H D f (y) f (y ) C f ψ, D ψ( x y ). The doubling property of ψ gives again f (x ) f (y ) f ψ, D ψ( x y ) C f ψ, D ψ( x y ). Collecting the above inequalities, we obtain H D f (x) H D f (y) C f ψ, D ψ( x y ). Thus (i) follows.

8 8 HIROAKI AIKAWA 3. Proofs of Theorem 1.4 and Corollary 1.5 Proof of Theorem 1.4. Suppose H D ψ ψ <. Let a D. Since χ D\B(a,r) (ξ) ψ(r) 1 τ a,ψ (ξ) for ξ D, it follows from the maximum principle and Theorem 1.1 that ω(x, D \ B(a, r), D) ψ(r) 1 H D τ a,ψ (x) Cψ(r) 1 ψ( x a ) for x D, so that D enjoys the GHMD(ψ) property. Thus the first assertion follows. Let us prove the second assertion. Because of the local nature we may assume that R = in (1.2), i.e., we may assume that (3.1) r dψ(t) Ψ(t) C ψ(r) Ψ(r) for 0 < r <. Now let a D and x D. Let r = r 0 = x a. Define an increasing sequence {r j } by ψ(r j ) = 2 j ψ(r) for j 1. Then (3.2) r dψ(t) Ψ(t) = = r j+1 r j dψ(t) Ψ(t) 1 Ψ(r j+1 ) ψ(r j+1 ) ψ(r j ) Ψ(r j+1 ) = r j+1 r j ψ(r j ) Ψ(r j+1 ) = 1 4 dψ(t) j=1 ψ(r j+1 ) Ψ(r j ), where the definition of r j is used in the last two equalities. By definition τ a,ψ (ξ) ψ(r)χ D B(a,r) (ξ) + ψ(r j+1 )χ D\B(a,r j )(ξ) for ξ D. Hence the maximum principle yields H D τ a,ψ (x) H D (ψ(r)χ D B(a,r) )(x) + ψ(r) + ψ(r) + ψ(r j+1 )H D (χ D\B(a,r j ))(x) ψ(r j+1 )ω(x, D \ B(a, r j ), D) ψ(r j+1 ) Ψ(r) Ψ(r j ) = ψ(r) + ψ(r 1 ) Ψ(r) Ψ(r 0 ) + ψ(r j+1 ) Ψ(r) Ψ(r j=1 j ) dψ(t) 3ψ(r) + 4Ψ(r) r Ψ(t),

9 MODULUS OF CONTINUITY 9 where the last inequality follows from r = r 0, ψ(r 1 ) = 2ψ(r) and (3.2). By (3.1) we have H D τ a,ψ (x) 3ψ(r) + CΨ(r) ψ(r) Ψ(r) Cψ(r). Thus (ii) in Theorem 1.1 holds, and hence H D ψ ψ <. Proof of Corollary 1.5. Let 0 < α < α. Let 0 < r < 1/e α+1. Then 1/e α+1 1/e α+1 1/e α+1 dψ α (t) r ψ α (t) = α (log 1/t) α 1 dt ( 1) α α 1 dt = α log r (log 1/t) α t r t t. Letting s = log 1/t, we obtain that this integral is equal to α log 1/r α+1 s α α 1 ds = α [ ] s α log 1/r α α α α+1 α ( 1) α α α log = α α r α α Thus (1.2) holds. Hence Theorem 1.4 proves H D ψ ψ <. 4. Proofs of Corollary 1.6 and Corollary 1.7 ψ α (r) ψ α (r). Let us begin with observations on the harmonic measure decay property. Definition 4.1 (Local Harmonic Measure Decay property). We say that D enjoys the Local Harmonic Measure Decay property with ψ (abbreviated to the LHMD(ψ) property) if ψ( x a ) (4.1) ω(x, D S (a, r), D B(a, r)) C for x D B(a, r), ψ(r) whenever a D and r > 0. Remark 4.2. In the above definition we may assume that (4.1) holds only for 0 < r < R with some R > 0. See Remark 1.3. Let ψ M. If the LHMD(ψ) property holds for D, then so does the GHMD(ψ) property. If ψ(t) = t α, then the converse is true as well (see [1]). For general ψ M, we do not know whether the converse holds or not. Remark 4.3. See Sugawa [7] for the plane case. Remark 4.4. If ψ(t) = t β, then we can replace ψ( x a )/ψ(r) by ψ( x a /r) in (1.1) and (4.1). For general ψ M, however, such a replacement is not valid because of the inhomogeneity of ψ. It is convenient to introduce a certain order among functions in M. Definition 4.5. Let φ, ψ M. We say that φ ψ if there are r 0 > 0 and C > 0 such that φ(s) φ(r) C ψ(s) for 0 < s < r r 0. ψ(r)

10 10 HIROAKI AIKAWA The definition of the GHMD property readily gives Proposition 4.6. Let φ, ψ M and let φ ψ. If the GHMD(φ) property holds for D, then so does the GHMD(ψ) property. The order and the composition have the following relationship. Proposition 4.7. Let φ, ψ M. Then ψ φ ψ. Proof. Let 0 < s < r. Put S = ψ(s) and R = ψ(r). Since 0 < S R from the monotonicity of ψ, it follows from Lemma 2.2 (i) that Hence φ(ψ(r)) ψ(r) = φ(r) R φ(s ) S = φ(ψ(s)). ψ(s) ψ(s) ψ(r) φ(ψ(s)) φ(ψ(r)) = φ ψ(s) φ ψ(r). Let α > 0 and β > 0. Then ψ α (t β ) = β α ψ α (t), so that t β ψ α (t) by Proposition 4.7. Hence Corollary 1.5 and Proposition 4.6 give Corollary 4.8. Let β > 0. If D satisfies the GHMD(t β ) property, then H D ψ ψ < for every ψ(t) = ψ α (t) with α > 0. Proof of Corollary 1.6. It is easy to see that every Lipschitz domain D satisfies the LHMD(t β ) property for some β > 0, and so the GHMD(t β ) property. Hence Corollary 4.8 gives Corollary 1.6. Proof of Corollary 1.7. We observe that a finitely connected plane domain with no puncture satisfies the GHMD(t β ) property for 0 < β < 1/2. Hence Corollary 4.8 gives Corollary Remark In Section 1 we have assumed that ψ(t) is concave. We observe that this assumption can essentially be relaxed to the nonincreasing property of ψ(t)/t. This observation is due to S. T. Kuroda [4]. We include it with his permission. Proposition 5.1. Let ψ(t) be a positive nondecreasing function on (0, ) with ψ(0) = lim t 0 ψ(t) = 0. Suppose that ψ(t)/t is nonincreasing. By U ψ we denote the family of concave functions φ on (0, ) such that ψ φ. Let Ψ(t) = inf{φ(t) : φ U ψ } for t > 0 be the concave upper envelope of ψ. Then (i) Ψ is a concave nondecreasing function on (0, ) with Ψ(0) = 0. (ii) 1 Ψ(t) ψ(t) Ψ(t) for t > 0. 2

11 MODULUS OF CONTINUITY 11 Proof. (i) We prove only the concavity of Ψ as the rest is easy. Let s > 0 and t > 0 and let 0 < λ < 1. Then for every φ U ψ we have λφ(s) + (1 λ)φ(t) φ(λs + (1 λ)t)). Taking the infimum for φ U ψ, we obtain λψ(s) + (1 λ)ψ(t) = λ inf φ U ψ φ(s) + (1 λ) inf φ U ψ φ(t) inf φ U ψ ( λφ(s) + (1 λ)φ(t) ) inf φ U ψ φ(λs + (1 λ)t) = Ψ(λs + (1 λ)t). Thus Ψ is a concave function. (ii) Let us prove the first inequality since the second is obvious. Let t 0 > 0. If ψ(t 0 ) = Ψ(t 0 ), then there is nothing to prove. Suppose ψ(t 0 ) < Ψ(t 0 ). By definition we infer that Ψ(t) is a linear function in a neighborhood of t 0. Let (t 1, t 2 ) be the largest interval including t 0 and on which ψ(t) < Ψ(t). Then Ψ(t) is linear on [t 1, t 2 ], Ψ(t 1 ) = ψ(t 1 ) and Ψ(t 2 ) = ψ(t 2 ). Let p(t) be the linear function whose graph is the line connecting the points (t 1, ψ(t 1 )) and (t 2, ψ(t 2 )). Let q(t) be the linear function whose graph is the line connecting the points (0, ψ(t 0 )) and (t 0, Ψ(t 0 )). We observe that p(t 0 ) = q(t 0 ) = Ψ(t 0 ) and p(0) ψ(t 0 ) = q(0), and hence q(t) p(t) for t t 0. See Figure 1. Evaluate p(t) and q(t) at t = t 2. Then the explicit form q(0)=ψ(t 0 ) p(t) q(t) p(0) ψ(t) t 1 t 0 t 2 Figure 1. The concave upper envelope. q(t) = t 1 0 (p(t 0) ψ(t 0 ))(t t 0 ) + p(t 0 ) gives p(t 0 ) ψ(t 0 ) (t 2 t 0 ) + p(t 0 ) p(t 2 ) = ψ(t 2 ), t 0 so that t 2 p(t 0 ) = ( t 2 t ) p(t 0 ) t 2 t 0 ψ(t 0 ) + ψ(t 2 ). t 0 t 0 t 0

12 12 HIROAKI AIKAWA Since ψ(t)/t is nonincreasing, it follows that Ψ(t 0 ) = p(t 0 ) t 2 t 0 ψ(t 0 ) + t 0 ψ(t 0 ) ψ(t 2 ) ψ(t 0 ) + t 0 = 2ψ(t 0 ). t 2 t 2 t 0 Thus the first inequality of (ii) follows. Remark 5.2. The nonincreasing property of ψ(t)/t in the assumptions of the proposition can be replaced by the subadditivity of ψ. See Lemma 2.2 (ii). References [1] H. Aikawa, Hölder continuity of the Dirichlet solution for a general domain, Bull. London Math. Soc. 34 (2002), no. 6, article [2] H. Aikawa and N. Shanmugalingam, Hölder estimates of p-harmonic extension operators, J. Differential Equations 220 (2006), no. 1, article [3] A. Hinkkanen, Modulus of continuity of harmonic functions, J. Analyse Math. 51 (1988), [4] S. T. Kuroda, Diagonalization modulo norm ideals; spectral method and modulus of continuity, RIMS Kôkyûroku Bessatsu, vol. B16, Res. Inst. Math. Sci. (RIMS), Kyoto, [5] H. Shiga, Riemann mappings of invariant components of Kleinian groups, J. London Math. Soc. (2) 80 (2009), [6], Modulus of continuity, a Hardy-Littlewood theorem and its application, Infinite dimensional Teichmüller space and moduli space, RIMS Kôkyûroku Bessatsu, Res. Inst. Math. Sci. (RIMS), Kyoto, to appear. [7] T. Sugawa, Uniformly perfect sets: analytic and geometric aspects, Sugaku Expositions 16 (2003), no. 2, [8] K.-O. Widman, Inequalities for the Green function and boundary continuity of the gradient of solutions of elliptic differential equations, Math. Scand. 21 (1967), (1968). Department of Mathematics, Hokkaido University, Sapporo , Japan address: aik@math.sci.hokudai.ac.jp