Examples of non-Dini domains with large singular sets

1 Introduction

We consider the following question, which is inspired by a classical question asked by Bers (see the Introduction in [9]):

When D is a C 1 , 1 domain, Lin proved that S has zero ( d − 1 ) -dimensional Hausdorff measure, and that S is a ( d − 2 ) -dimensional set, see [11, Theorem 2.3]. Adolfsson et al. [2] (see also Kenig and Wang [5] for an alternative proof) extended the result to convex domains. This was then followed by the works of Adolfsson and Escauriaza [1] and Kukavica and Nyström [10], who proved (using different methods) the result for C 1 -Dini domains (see Definition 2.1). Recently, Tolsa [14] proved that for all C 1 domains (or Lipschitz domains with sufficiently small Lipschitz constant), the set S has zero ( d − 1 ) -dimensional Hausdorff measure.

In a recent work, we proved the following theorem:

In short, the theorem says that when D is a C 1 -Dini domain, the singular set at the interior and boundary, S ( u ) ∩ B R ( 0 ) , is ( d − 2 ) -rectifiable, and its ( d − 2 ) -dimensional Hausdorff measure is finite. We remark that a similar result for convex domains can be found in [12].

It is natural to ask whether such a fine estimate (i.e., ℋ d − 2 estimate) of the singular set can be extended to more general domains, for example, Lipschitz domains with small constants. In that setting, recall Tolsa showed that S ( u ) ∩ ∂ D has surface measure zero in B R ( 0 ) (see [14]). The answer is no in general because if the domain is less regular than C 1 -Dini, the gradient of harmonic functions that vanish on an open subset of the boundary may not exist everywhere in that open set, and thus it does not make sense to talk about its ℋ d − 2 -measure. The goal of this article is to provide counterexamples demonstrating that C 1 -Dini domains are indeed the optimal class of domains for which Theorem 1.1 holds. More precisely, if D is less regular than C 1 -Dini and a harmonic function u vanishes on an open subset of ∂ D , we cannot make sense of ∇ u at the boundary in general. However, in the special case when u is a nonnegative harmonic function in D that vanishes in ∂ D ∩ B 2 R ( 0 ) with 0 ∈ ∂ D , by the comparison principle (see Lemma 2.3) u is comparable to the Green’s function G in D ∩ B R ( 0 ) . Hence, for σ -almost every x ∈ ∂ D ∩ B R ( 0 ) (where σ ≔ ℋ d − 1 ∣ ∂ D denotes the boundary surface measure of D ), we have

where ∂ n G denotes the normal derivative of G at the boundary and d ω / d σ denotes the Poisson kernel of the harmonic measure ω (whose pole is the same as that of the Green’s function G ). Since the upper and lower densities of the Radon measure ω are defined everywhere (and take values in [ 0 , + ∞ ] ), we can use the following set^[1]

in place of the boundary singular set of u , namely { p ∈ ∂ D : ∇ u ( p ) = 0 } . Roughly speaking, we showed that when the domain D barely fails to be C 1 -Dini, the ℋ d − 2 -measure of the above set could be infinite.

We remark that when D is a C 1 -Dini domain and the harmonic function u in consideration is nonnegative, the Hopf maximum principle implies that

where ∂ ν u denotes the normal derivative of u with the normal vector ν pointing inward, see [3]. (The Hopf maximum principle in [3] was proven for solutions to parabolic equations, but by taking a slice at a fixed positive time, the elliptic analog follows.) In particular, this implies that for C 1 -Dini domains, the singular set

In a related work [4, Section 9], the author constructed the following example (credited to Tolsa): there exist Lipschitz domains D ⊂ R 2 with small constants such that the singular set at the boundary

has Hausdorff dimension as close to 1 as we want. In particular, it indicates that for Lipschitz domains, one cannot expect a better answer to (Q) than saying that the singular set has zero surface measure. (For comparison, our examples show that in order to obtain the sharp ( d − 2 ) -dimensional estimate for (Q) as in [9], the assumption of C 1 -Dini domains cannot be weakened.) These Lipschitz domains are built by taking the union of cones with vertices at a fat Cantor set, whose Hausdorff dimension can be chosen sufficiently close to 1. The purpose of their example is similar to ours, but the constructions are completely different. Besides, our example of domains is better than C 1 -regular, instead of just Lipschitz, but the singular set is 0-dimensional (albeit infinite) rather than ( 1 − ε ) -dimensional.

The proof of Theorem 1.2 is inspired by the work of the first-named author [6]. It was demonstrated there that one can construct a Lipschitz domain in R 2 with prescribed tangent vectors on its boundary such that its harmonic measure is given by the exponential of the Hilbert transform of that prescribed function, see, e.g., [6, Lemma 1.11]. This article is organized as follows. We recall some definitions and preliminary results in Section 2. Then, to fix ideas, we first construct Lipschitz domains with the desired properties in Section 3. These domains are explicit and not difficult to visualize. In Section 4, we construct the desired C 1 domains for every modulus of continuity θ satisfying (1.1).

2 Preliminaries

The following integral will be used repeatedly in the computation of Hilbert transforms, so we state it as a lemma here.

We recall the following lemmas about positive solutions to elliptic partial differential equations (for a reference, see [7]).

3 Lipschitz domains

Let H : R → R be the Heaviside step function, namely

Let { x k } be a sequence of distinct points in R ⧹ { 0 } such that x k → 0 . (Then, in particular, { x k } is bounded, say ∣ x k ∣ ≤ 1 .) Let c be a positive real number and { a k } be a sequence in R + such that

We define a function f : R → R as follows:

Clearly,

Let K be the Hilbert transform operator, modulo a constant, defined as follows:

whenever the limit on the right-hand side exists. Here, χ E denotes the characteristic function of the set E . Recall that the Hilbert transform maps L ∞ ( R ) functions into functions in the bounded mean oscillations (BMO) space. Simple computations show that

and hence, formally, we have

In fact, by the assumption (3.1), we have that

Hence,

where the limit is taken in the BMO space.

The function on the right-hand side of (3.3) is well-defined and continuous in R ⧹ { 0 } . In fact, assume that x ≠ 0 and x ≠ x k for every k . Since log ∣ ⋅ ∣ is a continuous function in R ⧹ { 0 } , it is uniformly continuous on compact subset of R ⧹ { 0 } . Let E be a compact subset of R ⧹ { 0 } containing x such that E ∩ { x k : k ∈ N } = ∅ . We have that, for any y ∈ E ,

and the convergence is uniform. In particular, there exists k 0 ∈ N depending on E such that for any k ≥ k 0 and y ∈ E , we have that

Hence, for any m ≥ ℓ ≥ k 0 , we have

Therefore,

is well-defined and continuous in E . Therefore, we have shown that K f ( x ) is well-defined and continuous on R ⧹ ( { x k } ∪ { 0 } ) . In addition, we have

Moreover, near each x k , we have

Since x k ′ ≠ x k for every k ′ ≠ k , and x k ′ → 0 ≠ x k , we have that

Then, as long as 0 < ∣ x − x k ∣ < δ k / 2 ,

In particular, the second term in (3.6), e k ( ⋅ ) , is bounded near x k , and thus

Let V ( x , t ) and W ( x , t ) be the Poisson integrals of f ( x ) and K f ( x ) , respectively, in the upper half plane R + 2 . (The Poisson integral of K f ( x ) is well-defined because K f ( x ) has logarithmic growth at infinity according to (3.5).) Since f ( x ) is continuous and bounded in R ⧹ ( { x k } ∪ { 0 } ) , by the classical theory^[2] for every x ∈ R ⧹ ( { x k } ∪ { 0 } ) , we have that V ( z ) → f ( x ) as z → x . Since K f ( x ) is continuous in  R ⧹ ( { x k } ∪ { 0 } ) , we also have that W ( z ) → K f ( x ) as z → x for every x ∈ R ⧹ ( { x k } ∪ { 0 } ) . Moreover, let

and recall that F ℓ → K f in BMO ( R ) and pointwise in R ⧹ ( { x k } ∪ { 0 } ) . We claim that for any ( x , t ) ∈ R + 2 ,

where P t denotes the Poisson kernel in R + 2 with P t ( ξ ) = 1 π t ξ 2 + t 2 . Then, it immediately follows that

In the second equality, we use the fact that the Poisson integral of log ∣ x ∣ in R + 2 is just log ∣ ( x , t ) ∣ . To prove the claim, let

We let

for any allowable function h on R . For the first term, we have

which converges to 0 as ℓ → + ∞ . However, since ∣ x k ∣ ≤ 1 and log ∣ ⋅ ∣ is square-integrable near the origin, for any E that is a neighborhood of the origin, we have by the Minkowski integral inequality that

as ℓ → + ∞ . Hence,

which also converges to 0 as ℓ → + ∞ . This finishes the proof of the claim (3.9). In particular, (3.10) implies that

with logarithmic decay.

Let z = x + i y , and we define the function

One can verify that − W and V satisfy the Cauchy-Riemann equations and thus g is analytic in R + 2 . Let G ( z ) ≔ exp g ( z ) . Then, G is analytic in R + 2 and has a nontangential limit toward the boundary for almost every x ∈ ∂ R + 2 , since G is nontangentially bounded at almost every boundary point.

and

where we used the maximum principle for the Poisson integral.

Let Φ denote the antiderivative of G in R + 2 . More precisely, for any z ∈ R + 2 , let γ z denote any rectifiable curve from i to z and let

This function is well-defined (i.e., independent of the choice of curve) since R + 2 is simply connected. Besides, for any z = ( x , t ) ∈ R + 2 , by choosing γ z to be the line segment connecting i to z , we can easily show that

namely Φ ( z ) ∈ C . Since ∣ Φ ′ ( z ) ∣ = ∣ G ( z ) ∣ ≠ 0 , Φ is locally a conformal mapping. We claim that Φ is injective. Assume there are two distinct points z 1 , z 2 ∈ R + 2 such that Φ ( z 1 ) = Φ ( z 2 ) . Let γ 0 denote the line segment in R + 2 connecting z 1 to z 2 . More precisely, we consider the parametrization γ 0 ( t ) = z 1 + t ( z 2 − z 1 ) with t ∈ [ 0 , 1 ] . We have that

and

Hence, it follows that

In particular, the real part of the above integral also vanishes, i.e.,

However, by (3.14), we have that

Combined with G ( z ) ≠ 0 , this is a contradiction with (3.15). Therefore, Φ is injective.

For any z ∈ ∂ R + 2 and z ≠ x k , z ≠ 0 , by the properties of the Poisson integrals V and W , it is easy to see that

and moreover, it is independent of the choice of the curve γ z ⊂ R + 2 ¯ . Next, we show that Φ ( z ) is well-defined as z → x k and z → 0 and is independent of the choice of the curve. For fixed k , let z and z ′ be arbitrary points in R + 2 ¯ ⧹ { 0 } with z , z ′ ≠ x k ′ for any k ′ ∈ N such that δ ≔ max { ∣ z − ( x k , 0 ) ∣ and ∣ z ′ − ( x k , 0 ) ∣ } is sufficiently small. Let γ denote a rectifiable curve in R + 2 ¯ connecting z and z ′ such that γ does not intersect the origin or x k ′ for any k ′ ∈ N . Then, by (3.13),

Since c π a k < 1 2 by the assumption (3.1), by carefully choosing the curve γ (e.g., by taking γ to be the union of an arc on ∂ B δ ( x k ) and a line segment on a ray from x k ), we can guarantee that

Therefore, Φ ( z ) is continuous and finite as z → x k . To show Φ ( z ) is continuous at the origin, let z = ( x 0 , t 0 ) and z ′ = ( x 0 ′ , t 0 ′ ) be arbitrary points in R + 2 ¯ ⧹ { 0 } that are sufficiently close to the origin. Let δ ≔ max { ∣ z ∣ , ∣ z ′ ∣ } , and clearly ∣ x 0 ∣ , ∣ x 0 ′ ∣ , t 0 , t 0 ′ < δ . Let t ∗ ≔ max { t 0 + δ , t 0 ′ + δ } . Let γ 1 denote the vertical line segment between z and ( x 0 , t ∗ ) , γ 2 denote the horizontal line segment between ( x 0 , t ∗ ) and ( x 0 ′ , t ∗ ) , and γ 3 denote the vertical line segment between ( x 0 ′ , t ∗ ) and z ′ , each parametrized by unit length. For each i ∈ { 1 , 2 , 3 } , we have

Since we always have that

it follows that