A Peak Set of Hausdorff Dimension 2n − 1 for the Algebra A(D) in the Boundary of a Domain D with C⌃2 Boundary

Abstract

We consider a bounded strictly pseudoconvex domain $\Omega \subset \mathbb {C}^{n}$ with $C^{2}$ boundary. Then, we show that any compact Ahlfors–David regular subset of $\partial \Omega$ of Hausdorff dimension $\beta \in (0,2n-1]$ contains a peak set E of Hausdorff dimension equal to $\beta$.

1 Introduction

1.1 Historical Background

Let $\Omega \subset \subset \mathbb {C}^{n}$ be a domain. If we have a compact set $K\subset \partial \Omega$ and $f\in A(\Omega )$ such that $|f|<1$ on $\overline{\Omega }{\setminus } K$ and $f=1$ on K we say that K is a peak set for¹$A(\Omega )$ and f is a peak function for K. It is possible to generalize this concept to a peak interpolation set. A compact set $K\subset \partial \Omega$ is said to be a peak interpolation set for $A(\Omega )$ if every $g\in C(K)$ extends to an $f\in A(\Omega )$ that satisfies

$$\begin{aligned} |f(z)|<\max \left\{ |g(w)|:w\in K\right\} \end{aligned}$$

for all $z\in \overline{\Omega }{\setminus } K$.

Crucial sources of information about peak sets are [15, 16]. Peak sets in the boundary of the unit disc $\mathbb {D}$ have been extensively studied. The most important result in this area is the Fatou–Rudin–Carleson theorem (see [1, 3, 18]) which states that the classes of peak and peak interpolation sets for $A(\mathbb {D})$ coincide and are precisely the compact subsets of one-dimensional Lebesgue measure zero in $\partial \mathbb {D}$. However, as one can expect, the situation in higher dimensions is more complicated. Before we give examples of peak sets of maximal Hausdorff measure, we summarize the current state of knowledge in the regular case. Rudin generalized the Carleson-Fatou-Rudin theorem to higher dimensions in the following way:

Theorem

[17] Let $\Omega$ be a bounded strictly pseudoconvex domain with $C^{2}$-boundary in $\mathbb {C}^{n}$. Moreover, let M be $C^{1}$-submanifold of $\partial \Omega$, such that the tangent space $T_{w}(M$) lies in the maximal complex subspace of $T_{w}(\partial \Omega )$ for every $w\in M$. Then, every compact subset of M is a peak interpolation set for $A(\Omega )$.

Nagel and Rudin (see [13, 14]) studied the intersection of a peak set for $A(\Omega )$ for any $C^{1}$-domain $\Omega$ with a transverse curve; i.e. a curve in $\partial \Omega$ whose tangent space at every point is not contained in the maximal complex of the tangent space of $\partial \Omega$ at that point. They showed that if $\gamma$ is a transverse curve of class $C^{2}$ in $\partial \Omega$, then any bounded holomorphic function on $\Omega$ has nontangential boundary values $\{\mu _{\gamma }\}$-almost everywhere, where $\mu _{\gamma }$ is the measure on $\partial \Omega$ defined by²$\int fd\mu _{\gamma }:=\int _{0}^{1}f(\varphi (t))dt$ for all $f\in C(\partial \Omega )$ and a parametrization $\varphi$. It is easy to notice that if $G\in A(\Omega )$ is a peak function for some set K, then $F=\exp (i\log (1-G))$ is a bounded holomorphic function on $\Omega$ with no limit along any curve in $\Omega$ that ends at a point of K. Hence, a peak interpolation set of a strictly pseudoconvex domain intersects any transverse curve $\gamma$ in a set of $\{\mu _{\gamma }\}$-measure zero.

Topologically, peak sets and maximum modulus sets are small in strictly pseudoconvex domains. The real topological dimension of a maximum modulus set is not bigger than n (see [4]). Moreover, as Stout (see [19]) observed, for a peak set it is not bigger than $n-1$. In particular, a peak set and a maximum modulus set must have empty interiors. However we might ask if every peak interpolation set for $A(\Omega )$ is contained in some ($n-1$)-dimensional complex tangential submanifold of $\partial \Omega$? The answer to this question is no: from the measure-theoretic point of view, peak sets and maximum modulus sets no longer have to be so small. Tumanov constructed a peak set of Hausdorff dimension 2.5 in the unit sphere $\partial \mathbb {B}\subset C^{3}$. Henriksen proved (see [7]) that every strictly pseudoconvex domain with $C^{\infty }$ boundary in $\mathbb {C}^{n}$ has a peak set with a Hausdorff dimension $2n-1$. Moreover, if $\Omega$ is a circular, bounded, strictly convex domain with $C^{2}$ boundary it is possible to construct a peak set $K\subset \partial \Omega$ which intersects all the circles in $\partial \Omega$ centered at the origin (see [8, 10]). These examples indicate that the characterization of all peak sets for $A(\Omega )$ in the case of a strictly pseudoconvex region $\Omega$ is not trivial. It is known that every peak set K is also a peak interpolation set, which implies that any compact $T\subset K$ is also a peak set (see [16, p. 206]). Moreover, any compact subset of the Euclidean space of Hausdorff dimension $m>0$ contains a compact subset of Hausdorff dimension $\beta$ for each $0\le \beta \le m$. (see [5, Theorem 2.10.47]). Therefore, the peak sets mentioned in [7, 8, 10, 20] contain peak sets of any lower Hausdorff dimension. However, if we choose a compact set $K\subset \partial \Omega$ then there is no universal way to construct a peak set as in [7, 8, 10, 20] inside K.

1.2 Motivations

Our inspiration is a Henriksen’s result from [7]. One can ask whether this is the best possible result in terms of Hausdorff dimension. Henriksen’s method is based on the $\overline{\partial }$ problem and requires $C^{\infty }$ boundary assumption. Our methods do not require the use of a theory related to the $\overline{\partial }$-equation. We use the so-called Holomorphic Support Function (HSF). The idea of HSF was successfully realized by Löw in his construction of inner functions for strictly pseudoconvex domains with $C^{2}$ boundary or just pseudoconvex with $C^{\infty }$ boundary (see [11]). We generalize this idea in the paper [9], where we only use a function $\Phi (\cdot ,w)\in A(\Omega )$ with

$$\begin{aligned} \exp \left( -c_{2}\left\| z-w\right\| ^{2}\right) \le \left| \Phi (z,w)\right| \le \exp \left( -c_{1}\left\| z-w\right\| ^{2}\right) \end{aligned}$$

(1.1)

for $z,w\in \partial \Omega$. In particular, strictly pseudoconvex domains with $C^{2}$ boundary have the above property. Using HSF when constructing functions with given boundary properties can sometimes bring better results than the methods based on $\overline{\partial }$-problem. It turns out that a construction of peak sets with maximal possible Hausdorff dimension is an example of such a problem.

1.3 Main Result

The following fact is the main result of this paper:

Theorem

(see Theorem 8) Assume that $\Omega \subset \mathbb {C}^{n}$ is a bounded strictly pseudoconvex domain with $C^{2}$ boundary. If $X\subset \partial \Omega$ is a compact Ahlfors–David regular set of Hausdorff dimension $d\in (0,2n-1]$ then there exists a compact set $E\subset X$ of Hausdorff dimension equal to d which is a peak set for $A(\Omega )$ i.e. there exists a function $F\in A(\Omega )$ such that $E=\{z\in \partial \Omega :F(z)=1\}$ and $|F|<1$on $\overline{\Omega }{\setminus } E$.

1.4 Applications

Assume that $K\subset \partial \mathbb {B}^{n}$, then the following properties are equivalent (see [16, Theorem 10.1.2]):

• (1) K is a zero set i.e., there exists $f\in A(\mathbb {B}^{n})$ such that $f(\xi )=0$ for every $\xi \in K$ and $f(z)\ne 0$ for $z\in \mathbb {\overline{B}}^{n}{\setminus } K$.

• (2) K is a peak set.

• (3) K is a peak interpolation set i.e., to every $g\in C(K)$ ($g\not \equiv 0$) corresponds an $f\in A(\mathbb {B}^{n})$ such that $f(\xi )=g(\xi )$ for $\xi \in K$ and $|f(z)|<\left\| g\right\| _{K}$ for every $z\in \overline{\mathbb {B}}^{n}{\setminus } K$.

• (4) K is a null set for every $v\in A^{\bot }$. More explicitly, the requirement is that $|v|(K)=0$ whenever v is a regular complex Borel measure on $\partial \mathbb {B}^{n}$ satisfying $\int fdv=0$ for $f\in A(\mathbb {B}^{n})$.

Valskii (see [21]) proved the above equivalence for smoothly bounded, star-shaped, strictly pseudoconvex domains in $\mathbb {C}^{n}$. Chollet (see [2]) showed the same for arbitrary strictly pseudoconvex domains.

1.5 Organization of the Paper

We are going to start our construction by selecting a compact $X\subset \partial \Omega$ which is Ahlfors–David regular and has Hausdorff dimension equal to d. We shall present an example of such a set (see Example 2). Actually, any compact manifold in $\partial \Omega$ or self-similar Cantor set, satisfying the open set condition is an example of Ahlfors–David regular set, see Example 2 and [12, p. 92]. Moreover, any image of Ahlfors–David regular set by a $C^{1}$ diffeomorphism is also Ahlfors–David regular. Therefore it is enough to construct the appropriate peak set only for strictly convex domains and use Fornæss Embedding Theorem (see [6]).

We prove (see Lemma 1) the multidimensional version of Tumanov’s lemma (see [20, Lemma 3]) and use it as the basic tool for determining the Hausdorff dimension.

It is hard to estimate the real part of a peak function from below in the whole ball K(x, r) where $x\in \partial \Omega$ is a peak point. However, we present estimations of this sort in a big part (in measure-theoretic sense) of this ball. This part is denoted as an (x, r)-segment. We introduce the notion of $\rho$-separatedness to control the distances between the centers of the balls. To apply Lemma 1 it is necessary to choose a proper number of $\rho$-separated segments contained in a master segment. The Lemma 3 guarantees that such a choice is possible.

Theorem 7 describes some useful properties of peak functions which we use to obtain a peak set with required properties. In particular, for a given n and $r_{n}>0$ we construct a compact set $E_{n}$ as a sum of $(\cdot ,r_{n})$-segments. Our peak set (see Theorem 8) can be now defined as $E=\bigcap _{n\in \mathbb {N}}E_{n}$.

1.6 Notations

For $\alpha ,\delta >0$ and countable family $\mathfrak {U}$ of open sets U with diameter ${\text {diam}(U)}$ we define:

$$\begin{aligned} h^{\alpha }(\mathfrak {U}):=\sum _{U\in \mathfrak {U}}\text {diam}(U)^{\alpha } \end{aligned}$$

Now the Hausdorf measure is:

$$\begin{aligned} H_{\delta }^{\alpha }(E):=\inf _{\mathfrak {U}}\left\{ h^{\alpha }(\mathfrak {U}):E\subset \bigcup _{U\in \mathfrak {U},d(U)<\delta }U\right\} , \end{aligned}$$

where the infimum is taken over all countable covers $\mathfrak {U}$ of E by open sets. We define Hausdorff dimension:

$$\begin{aligned} \dim _{H}(E):=\sup \{\alpha \ge 0:H^{\alpha }(E)>0\}=\inf \{\alpha \ge 0:H^{\alpha }(E)=0\}, \end{aligned}$$

where $H^{\alpha }(E)=\lim _{\delta \rightarrow 0}H_{\delta }^{\alpha }(E)$.

For a given set S we define a ball:

$$\begin{aligned} K_{S}(x,r)&:=\{y\in S:\left\| x-y\right\| <r\} \end{aligned}$$

Let $d>0$. We say that $X\subset \partial \Omega$ is d-dimensional Ahlfors–David³ regular set iff there exist $a,b>0$ such that:

$$\begin{aligned} ar^{d}\le H^{d}(K_{X}(x,r))\le br^{d} \end{aligned}$$

(1.2)

for any $x\in X$ and $r\in (0,1]$.

Moreover for $r>0$ and a given point $x\in X$ a compact $Q=Q(x,r)$ subset of X is (x, r)-segment iff $Q\subset K_{X}(x,r)$.

A set $A\subset X$ is $\rho$-separated iff $\left\| x-y\right\| \ge \rho$ for $x\ne y\in A$.

2 Preliminary Estimates

Our essential tool for calculating the Hausdorff dimension of a set is (compare with the Tumanov’s Lemm 3 [20]).

Lemma 1

Let $d>\alpha >0$ and $c>0.$ Assume that X is a d-dimensional Ahlfors–David regular set. Let $\{r_{j}\}_{j=0}^{\infty }$ be a sequence of positive numbers decreasing to zero. For a given $j\in \mathbb {N}$ let $\omega _{j}$ be a finite subset of X. Let us define $E_{j}=\bigcup _{x\in \omega _{j}}Q(x,r_{j})$ where $Q(x,r_{j})$ denotes a $(x,r_{j})$-segment.

Suppose that

• (1) In each segment $Q(x,r_{j-1})$ of $E_{j-1}$ there are at least $\left( \frac{r_{j-1}}{r_{j}}\right) ^{\alpha }$ segments of $E_{j}$.

• (2) The distance between balls $K_{X}(x,r_{j})$ and $K_{X}(y,r_{j})$ is at least

  $$\begin{aligned} \rho _{j}:=cr_{j-1}\left( \frac{r_{j}}{r_{j-1}}\right) ^{\nicefrac {\alpha }{d}} \end{aligned}$$

  for $x\ne y\in \omega _{j}$.

Then, $H^{\alpha }(E)>0$ where $E=\bigcap _{j\in \mathbb {N}}E_{j}$.

Proof

Without loss of generality we assume that $E_{j}\subset E_{j-1}$, $\omega _{0}=\{x_{0}\}$ and $r_{0}=\frac{1}{2}$. Let us define

$$\begin{aligned} \mathfrak {U}:=\left\{ K_{X}(x,r)\right\} _{j\in \mathbb {N},r=r_{j},x\in \omega _{j}}. \end{aligned}$$

Observe that $E:=\bigcap _{j\in \mathbb {N}}E_{j}$ is a compact set. We prove the Lemma in two steps.

Case (1):

If $\mathfrak {F}\subset \mathfrak {U}$ is a finite cover of E then $1\le h^{\alpha }(\mathfrak {F}).$

Suppose that $h^{\alpha }(\mathfrak {F})<1$. Let N be the number of balls in $\mathfrak {F}$. Assume that N is as small as possible. If $N=1$ then $\mathfrak {F}=\left\{ K_{X}(x_{0},\frac{1}{2})\right\}$ and $h^{\alpha }(\mathfrak {F})=1$ which is impossible, so $N>1$. Let $K_{X}(x,r)\in \mathfrak {F}$. There exists j such that $K_{X}(x,r)=K_{X}(x_{j},r_{j})$ and $Q_{j}:=Q(x_{j},r_{j})$ is a $(x_{j},r_{j})$-segment in $E_{j}$ for some $x_{j}\in \omega _{j}$. Assume that index j is the maximal possible. There exists $Q_{j-1}:=Q(x_{j-1},r_{j-1})$ for some $x_{j-1}\in \omega _{j-1}$ such that $Q_{j}\subset Q_{j-1}\subset K(x_{j-1},r_{j-1})$. Since N is minimal $\mathfrak {F}{\setminus } K_{X}(x_{j},r_{j})$ will not cover E and $\mathfrak {F}$ does not contain a ball covering $Q_{j-1}$. Since j is maximal $\mathfrak {F}$ does not contain a ball with radius $r_{k}$ for $k>j$. Therefore $Q_{j-1}\cap E$ is covered by balls of radii equal to $r_{j}$. Let

$$\begin{aligned} S=\left\{ K_{X}(x,r_{j})\right\} _{x\in \omega _{j},K_{X}(x,r_{j})\cap Q_{j-1}\cap E\ne \emptyset } \end{aligned}$$

be a family of such balls. In particular, $S\subset \mathfrak {F}$. Now, we define the following cover $\mathfrak {F}_{2}:=\mathfrak {F}{\setminus } S\cup \{K(x_{j-1},r_{j-1})\}$ of E. Since S contains no fewer than $\left( \frac{r_{j-1}}{r_{j}}\right) ^{\alpha }$ balls with radius $r_{j}$ we estimate

$$\begin{aligned} h^{\alpha }(\mathfrak {F}_{2})&\le 2^{\alpha }r_{j-1}^{\alpha }+\sum _{K_{X}(x,r_{k})\in \mathfrak {F}{\setminus } S}2^{\alpha }r_{k}^{\alpha }\le 2^{\alpha }r_{j}^{\alpha }\left( \frac{r_{j-1}}{r_{j}}\right) ^{\alpha }+\sum _{K_{X}(x,r_{k})\in \mathfrak {F}{\setminus } S}2^{\alpha }r_{k}^{\alpha }\\&\le \sum _{K_{X}(x,r_{j})\in S}2^{\alpha }r_{j}^{\alpha }+\sum _{K_{X}(x,r_{k})\in \mathfrak {F}{\setminus } S}2^{\alpha }r_{k}^{\alpha }=h^{\alpha }(\mathfrak {F})<1. \end{aligned}$$

Observe that S contains at least$\Big \lceil \left( \frac{r_{j-1}}{r_{j}}\right) ^{\alpha }\Big \rceil \ge 2$ balls, so $\mathfrak {F}_{2}$ contains less balls than $\mathfrak {F}$, which is impossible.

Case (2):

Let $\gamma =\frac{b2^{2d+\alpha }\left( 2+\frac{1}{c}\right) ^{d}}{ac^{d}}>0$. For a given $\mathfrak {G}$ a countable open cover of E it is possible to choose another finite cover $\mathfrak {F}\subset \mathfrak {U}$ by balls as in the Case (1) which additionally fulfills $h^{\alpha }(\mathfrak {F})\le \gamma h^{\alpha }(\mathfrak {G}).$

Since E is a compact set there exists a finite cover $\tilde{\mathfrak {G}}$ of E such that $\tilde{\mathfrak {G}}\subset \mathfrak {G}$. Let $U\in \tilde{\mathfrak {G}}$. There exists an index $j=j(U)$ such that U intersects precisely one segment $Q_{j-1}$ in $E_{j-1}$ and $n(U)>1$ segments $Q(x_{1},r_{j}), \ldots ,Q(x_{n(U)},r_{j})$ in $E_{j}$. Let us set:

$$\begin{aligned} \mathfrak {F}_{U}&=\left\{ K_{X}(x_{k},r_{j(U)})\right\} _{k\in \{1, \ldots ,n(U)\}}\\ \mathfrak {F}&=\left\{ V\right\} _{U\in \tilde{\mathfrak {G}},V\in \mathfrak {F}_{U}}. \end{aligned}$$

Note that $\mathfrak {F}\subset \mathfrak {U}$ is a finite cover of E. We prove that $h^{\alpha }(\mathfrak {F})\le \gamma h^{\alpha }(\mathfrak {G}).$

Let $\delta _{j-1}=\text {diam}(U\cap Q_{j-1})$ denote the diameter of $U\cap Q_{j-1}$. In particular, $\delta _{j-1}\le \min \{\text {diam}(U),2r_{j-1})$. Since $n(U)\ge 2$ we observe $\rho _{j}\le \delta _{j-1}$. If $x\in U\cap Q_{j-1}$ then $U\cap Q_{j-1}\subset \overline{K_{X}(x,\delta _{j-1})}$. In particular, $\overline{K_{X}(x,\delta _{j-1}+r_{j})}$ contains $\{x_{1}, \ldots ,x_{n(U)}\}\subset \omega _{j}$. Observe that

$$\begin{aligned} \rho _{j}&=cr_{j-1}\left( \frac{r_{j}}{r_{j-1}}\right) ^{\nicefrac {\alpha }{d}}>cr_{j}\\ \rho _{j}^{d}&=c^{d}r_{j-1}^{d-\alpha }r_{j}^{\alpha }. \end{aligned}$$

We have

$$\begin{aligned} \bigcup _{i=1}^{n(U)}K_{X}\left( x_{i},\nicefrac {\rho _{j}}{2}\right) \subset K_{X}\left( x,\delta _{j-1}+r_{j}+\nicefrac {\rho _{j}}{2}\right) . \end{aligned}$$

Since all the balls $K_{X}\left( x_{i},\nicefrac {\rho _{j}}{2}\right)$ are disjoint, we can use (1.2) to obtain

$$\begin{aligned} n(U)a\left( \frac{\rho _{j}}{2}\right) ^{d}\le b\left( \delta _{j-1}+r_{j}+\frac{\rho _{j}}{2}\right) ^{d}. \end{aligned}$$

(2.1)

We estimate (the last equality defines $\gamma$)

$$\begin{aligned} n(U)\le \frac{b2^{d}\left( \delta _{j-1}+\left( \frac{1}{c}+\frac{1}{2}\right) \rho _{j}\right) ^{d}}{a\rho _{j}^{d}}\le \frac{b2^{d}\left( 2+\frac{1}{c}\right) ^{d}\delta _{j-1}^{d}}{a\rho _{j}^{d}}=:\frac{\gamma c^{d}\delta _{j-1}^{d}}{2^{\alpha +d}\rho _{j}^{d}}. \end{aligned}$$

Now, using $\delta _{j-1}\le \min \{\text {diam}(U),2r_{j-1})$ we calculate

$$\begin{aligned} h^{\alpha }(\mathfrak {F})&=\sum _{U\in \tilde{\mathfrak {G}}}h^{\alpha }(\mathfrak {F}_{U})\le \sum _{U\in \mathfrak {G}}h^{\alpha }(\mathfrak {F}_{U})\le \sum _{U\in \mathfrak {G}}n(U)2^{\alpha }r_{j(U)}^{\alpha }\le \sum _{U\in \mathfrak {G}}\frac{\gamma c^{d}\delta _{j(U)-1}^{d}r_{j(U)}^{\alpha }}{2^{d}\rho _{j(U)}^{d}}\\&\le \sum _{U\in \mathfrak {G}}\frac{\gamma c^{d}\text {diam}(U)^{\alpha }\delta _{j(U)-1}^{d-\alpha }r_{j(U)}^{\alpha }}{2^{d}\rho _{j(U)}^{d}}\le \sum _{U\in \mathfrak {G}}\frac{\gamma c^{d}2^{d}\text {diam}(U)^{\alpha }r_{j(U)-1}^{d-\alpha }r_{j(U)}^{\alpha }}{2^{d}c^{d}r_{j(U)-1}^{d-\alpha }r_{j(U)}^{\alpha }}\\&\le \sum _{U\in \mathfrak {G}}\gamma \text {diam}(U)^{\alpha }=\gamma h^{\alpha }(\mathfrak {G}). \end{aligned}$$

Since $h^{\alpha }(\mathfrak {F})\ge 1$ we have $h^{\alpha }(\mathfrak {G})\ge \frac{1}{\gamma }$, which implies that $H^{\alpha }(E)\ge \frac{1}{\gamma }$. $\square$

We use Lemma 1 to obtain an example of a compact Ahlfors–David regular set of given Hausdorff dimension.

Example 2

Let m be a natural number and $\beta \in (0,m]$. Then there exists a compact set $X\subset [0,1]^{m}$ and $a,b>0$ such that

$$\begin{aligned} ar^{\beta }\le H^{\beta }(K_{X}(x,r))\le br^{\beta } \end{aligned}$$

for any $x\in X$ and $0<r<1$.

Proof

For the purposes of this example, let’s define $\left\| x-y\right\| :=2\max _{j}\left| x_{j}-y_{j}\right|$. Since $H^{m}$ coincides with the Lebesgue measure on $\mathbb {R}^{m}$ we assume that $\beta <m$. Let $s\in (0,\frac{1}{2})$ be such $1=2^{m}s^{\beta }$ and $\gamma =1-2s$. Assume that:

$$\begin{aligned} A:=\left\{ 0,\frac{1-s}{s}\right\} ^{m}. \end{aligned}$$

Let us define the following similitudes:

$$\begin{aligned} \varphi _{a}:x\rightarrow s(x+a) \end{aligned}$$

for $a\in A$. Let $E_{0}=[0,1]^{m}$ and

$$\begin{aligned} E_{n+1}:=\bigcup _{a\in A}\varphi _{a}(E_{n}). \end{aligned}$$

Let us define the Cantor set: $X=\bigcap _{k\in \mathbb {N}}E_{k}$. In particular, $E_{k}$ is the sum of balls with radius $s^{k}$ and the minimal distance between them is equal $2\gamma s^{k-1}$. Each ball in $E_{j-1}$ contains precisely

$$\begin{aligned} 2^{m}=s^{-\beta }=\left( \frac{s^{j-1}}{s^{j}}\right) ^{\beta } \end{aligned}$$

balls from $E_{j}$. The distance between any two balls in $E_{j}$ is not smaller than

$$\begin{aligned} 2\gamma s^{j-1}2^{-1}s^{\frac{\beta }{m}}=\gamma s^{j-1}\left( \frac{s^{j}}{s^{j-1}}\right) ^{\frac{\beta }{m}}. \end{aligned}$$

By Lemma 1 we have $H^{\beta }(X)>0$ (balls are treated as segments).

Let $\delta >0$ and j be such a large natural number that $s^{j}<\delta$. Then $E_{j}$ is a cover of X with

$$\begin{aligned} h^{\beta }(E_{j})=\sum _{Q\in E_{j}}\left( 2s^{j}\right) ^{\beta }=2^{mj}2^{\beta }s^{j\beta }=2^{\beta }. \end{aligned}$$

In particular, $H_{\delta }^{\beta }(X)\le 2^{\beta }$ and $H^{\beta }(X)\le 2^{\beta }$. Let u be the smallest integer for which $2s^{u+1}<\gamma$. We prove that

$$\begin{aligned} a&=s^{\beta (1+u)}H^{\beta }(X)\\ b&=\gamma ^{-\beta }H^{\beta }(X). \end{aligned}$$

Now, consider $x\in X$ and a ball $K_{X}(x,r)$ with $r<1$. There exists an index $j=j(x,r)$ such that $K_{X}(x,r)$ intersects precisely one of the balls $Q_{j-1}$ in $E_{j-1}$ and $k>1$ balls in $E_{j}$. In particular, $K_{X}(x,r)\subset Q_{j-1}$ which implies:

$$\begin{aligned} r\le s^{j-1}. \end{aligned}$$

Since the distance between any two balls in $E_{j}$ is not smaller than $2\gamma s^{j-1}$ we have

$$\begin{aligned} 2\gamma s^{j-1}\le 2r. \end{aligned}$$

Now, we estimate

$$\begin{aligned} H^{\beta }(K_{X}(x,r))&\le H^{\beta }(Q_{j-1}\cap X)=H^{\beta }(s^{j-1}X)=s^{\beta (j-1)}H^{\beta }(X)\\&\le \left( \frac{r}{\gamma }\right) ^{\beta }H^{\beta }(X)=br^{\beta }. \end{aligned}$$

Since $2s^{u+1}<\gamma$ we have the following inequality

$$\begin{aligned} 2s^{j+u}<s^{j}\gamma s^{-1}=\gamma s^{j-1}\le r. \end{aligned}$$

There exists a ball $Q_{j+u}$ from $E_{j+u}$ such that $x\in Q_{j+u}$. But $Q_{j+u}$ has a diameter $2s^{j+u}<r$, which implies that

$$\begin{aligned} Q_{j+u}\cap X\subset K_{X}(x,r). \end{aligned}$$

Now, we estimate

$$\begin{aligned} H^{\beta }(K_{X}(x,r))&\ge H^{\beta }(Q_{j+u}\cap X)=H^{\beta }(s^{j+u}X)=s^{\beta (j+u)}H^{\beta }(X)\\&\ge s^{\beta (j-1)}s^{\beta (1+u)}H^{\beta }(X)\ge r{}^{\beta }s^{\beta (1+u)}H^{\beta }(X)=ar^{\beta }. \end{aligned}$$

$\square$

The following observation enables us to combine the properties of natural peak functions in strictly convex domains with the structure of separated sets, which yields a construction of a Cantor set with a large Hausdorff dimension.

Lemma 3

Assume that X is a bounded d-dimensional Ahlfors–David regular set:

$$\begin{aligned} ar^{d}\le H^{d}(K_{X}(x,r))\le br^{d} \end{aligned}$$

for any $x\in X$ and $r\in (0,1]$. For a given $\theta \in (0,1)$ we can choose a constant $c_{\theta }>0$ such that for a given $\alpha \in (0,d)$, a relatively open set⁴U in X with $U\subset K_{X}(x,r)$ and $H^{d}(U)>\theta r^{d}$ there exists $c_{\theta }r\eta ^{\nicefrac {\alpha }{d}}$-separated set $A\subset U$ with properties

• $\#A\ge \eta ^{-\alpha }$ ($\#A$ denotes the number of elements in A),

• $\overline{K_{X}(\xi ,r\eta )}\subset U$ for $\xi \in A$.

for all small enough $\eta >0$.

Proof

Let us denote $V:=\left\{ z\in U:\inf _{w\in \partial U}\left\| z-w\right\| >r\eta \right\}$. We will show that it is enough to choose the constant $c_{\theta }$ so that the following equality is satisfied $\frac{a\theta }{b^{2}2^{d}c_{\theta }^{d}}=1$.

Let B be a maximal possible $r\eta$-separated subset of V. Note that B is finite. Let us select a maximal $c_{\theta }r\eta ^{\nicefrac {\alpha }{d}}$-separated subset $B_{1}$ from B. Next, we select a maximal $c_{\theta }r\eta ^{\nicefrac {\alpha }{d}}$-separated subset $B_{2}$ from $B{\setminus } B_{1}$. We proceed this way till we exhaust B. Let $B_{N}$ be the last non-empty set in this procedure. Let $\xi \in B_{N}$. Observe that $K_{X}(\xi ,c_{\theta }r\eta ^{\nicefrac {\alpha }{d}})\cap B_{k}\ne \emptyset$ for $k=1, \ldots ,N-1$. In particular, $K_{X}(\xi ,c_{\theta }r\eta ^{\nicefrac {\alpha }{\beta }})$ contains at least N different elements $\left\{ \xi _{1}, \ldots ,\xi _{N}\right\}$ from B. Since $K_{X}(\xi _{j},r\eta )\cap K_{X}(\xi _{k},r\eta )=\emptyset$ for $j\ne k$ and $K_{X}(\xi _{j};r\eta )\subset K_{X}(\xi ;r\eta +c_{\theta }r\eta ^{\nicefrac {\alpha }{d}})$ therefore we have

$$\begin{aligned} \bigcup _{k=1}^{N}K_{X}(\xi _{k},r\eta )\subset K_{X}(\xi ;r\eta +c_{\theta }r\eta ^{\nicefrac {\alpha }{d}})\subset K_{X}(\xi ;2c_{\theta }r\eta ^{\nicefrac {\alpha }{d}}) \end{aligned}$$

and

$$\begin{aligned} Nar^{d}\eta ^{d}\le b\left( 2c_{\theta }r\eta ^{\nicefrac {\alpha }{d}}\right) ^{d} \end{aligned}$$

for all small enough $\eta >0$. We observe

$$\begin{aligned} V\subset \bigcup _{\xi \in B}\overline{K_{X}}\left( \xi ,r\eta \right) \subset U, \end{aligned}$$

which implies:

$$\begin{aligned} H^{d}(V)\le \sum _{j=1}^{N}\sum _{\xi \in B_{j}}H^{d}\left( K_{X}\left( \xi ,r\eta \right) \right) . \end{aligned}$$

In particular, there exists index $j_{0}$ such that

$$\begin{aligned} H^{d}(V)\le N\sum _{\xi \in B_{j_{0}}}H^{d}\left( K_{X}\left( \xi ,r\eta \right) \right) . \end{aligned}$$

We choose $A=B_{j_{0}}.$ Now, we have:

$$\begin{aligned} \theta r^{d}\le N(\#A)br^{d}\eta ^{d}\le (\#A)br^{d}\eta ^{d}\frac{b2^{d}c_{\theta }^{d}}{a\eta ^{d-\alpha }} \end{aligned}$$

which implies:

$$\begin{aligned} \#A\ge \frac{a\theta \eta ^{-\alpha }}{b^{2}2^{d}c_{\theta }^{d}}=\eta ^{-\alpha }. \end{aligned}$$

$\square$

3 Strictly Convex Domain

Let $\Omega \subset \mathbb {C}^{n}$ be a bounded strictly convex $C^{2}$ smooth domain. We consider $X\subset \partial \Omega$ a compact d-dimensional Ahlfors–David regular set of diameter less than 1. Let us denote by $v_{z}$ the unit normal vector to $\partial \Omega$ at the point $z\in \partial \Omega$ which is directed outwards from $\Omega$. There exist constants $c_{1},c_{2}>0$ such that (see [9]):

$$\begin{aligned} -c_{1}\left\| z-\xi \right\| ^{2}\le \Re \left\langle z-\xi ,v_{\xi }\right\rangle \le -c_{2}\left\| z-\xi \right\| ^{2} \end{aligned}$$

(3.1)

for $z,\xi \in \partial \Omega$. We decrease the constant $c_{2}>0$ if necessary and assume that $c_{2}\in (0,1)$. The right inequality can be improved:

Lemma 4

There exists a constant $c_{2}\in (0,1)$ such that

$$\begin{aligned} \Re \left\langle w-\xi ,v_{\xi }\right\rangle \le -c_{2}\left\| w-\xi \right\| ^{2} \end{aligned}$$

(3.2)

for $\xi \in \partial \Omega$ and $w\in \overline{\Omega }$.

Proof

Near the boundary we have points in the form $w=w_{z,t}=z-tv_{z}$ where $z\in \partial \Omega$ and $t\in [0,\delta ]$. We assume that $\delta <1$ is so small that $\Re \left\langle v_{z},v_{\xi }\right\rangle >c_{2}$ for $z,\xi \in \partial \Omega$ with $\left\| z-\xi \right\| \le \delta$. Since the angle $\angle (w{}_{z,t},z,\xi )$ is less than $\frac{\pi }{2}$ and $t\le 1$, we have

$$\begin{aligned} \left\| w_{z,t}-\xi \right\| ^{2}\le \left\| w_{z,t}-z\right\| ^{2}+\left\| z-\xi \right\| ^{2}\le \left\| w_{z,t}-z\right\| +\left\| z-\xi \right\| ^{2}. \end{aligned}$$

Now, using (3.1) we estimate

$$\begin{aligned} \Re \left\langle w_{z,t}-\xi ,v_{\xi }\right\rangle&=\Re \left\langle z-tv_{z}-\xi ,v_{\xi }\right\rangle =-t\Re \left\langle v_{z},v_{\xi }\right\rangle +\Re \left\langle z-\xi ,v_{\xi }\right\rangle \\ {}&\le -tc_{2}-c_{2}\left\| z-\xi \right\| ^{2}=-c_{2}\left( \left\| w_{z,t}-z\right\| +\left\| z-\xi \right\| ^{2}\right) \\ {}&\le -c_{2}\left\| w_{z,t}-\xi \right\| ^{2}. \end{aligned}$$

In particular, we show the required property with additional assumptions $\left\| z-\xi \right\| \le \delta$. To obtain (3.2) it is enough to observe that $\overline{\Omega }$ is compact domain and $\Re \left\langle z-\xi ,v_{\xi }\right\rangle <0$ for all $z\in \overline{\Omega }{\setminus }\{\xi \}$, so it is enough to decrease the constant $c_{2}$. $\square$

Lemma 5

If $\rho \in (0,1)$ and A is $\rho$-separated subset of X with

$$\begin{aligned} A_{k}(z)=\left\{ x\in A:\frac{\rho k}{2}\le \left\| x-z\right\| \le \frac{\rho (k+1)}{2}\right\} \end{aligned}$$

for $z\in \overline{\Omega }$, then $\#A_{0}(z)\le 1$, $\#A_{k}(z)\le \frac{b}{a}(k+2)^{d}$ and $\#A\le \frac{b}{a}3^{d}\rho ^{-d}$.

Proof

Observe that $K_{X}(x,\nicefrac {\rho }{2})\cap K_{X}(y,\nicefrac {\rho }{2})=\emptyset$ for $x\ne y\in A$. This implies that $A_{0}(z)$ contains at most one element. Moreover,

$$\begin{aligned} \bigcup _{x\in A_{k}(z)}K_{X}\left( x,\frac{\rho }{2}\right) \subset K_{X}\left( z,\frac{\rho (k+2)}{2}\right) . \end{aligned}$$

Now, we estimate:

$$\begin{aligned} \#A_{k}(z)a\frac{\rho ^{d}}{2^{d}}&\le \sum _{x\in A_{k}(z)}H^{d}\left( K_{X}\left( x,\frac{\rho }{2}\right) \right) =H^{d}\left( \bigcup _{x\in A_{k}(z)}K_{X}\left( x,\frac{\rho }{2}\right) \right) \\ {}&\le H^{d}\left( K_{X}\left( z,\frac{\rho (k+2)}{2}\right) \right) \le b\frac{\rho ^{d}(k+2)^{d}}{2^{d}}. \end{aligned}$$

Let $x_{0}\in A$. Since $\text {diam}(X)\le 1$ we have

$$\begin{aligned} \bigcup _{x\in A}K_{X}\left( x,\frac{\rho }{2}\right) \subset K_{X}\left( x_{0},\text {diam}(X)+\frac{\rho }{2}\right) \subset K_{X}\left( x_{0},\frac{3}{2}\right) , \end{aligned}$$

which implies

$$\begin{aligned} \#Aa\left( \frac{\rho }{2}\right) ^{d}\le b\left( \frac{3}{2}\right) ^{d}. \end{aligned}$$

$\square$

The $(\xi ,r)$-segment $K_{X}(\xi ,r)\cap \{z:\Re g_{\xi ,r}(z)>\frac{1}{6}\}$ is comparable to the whole ball $K_{X}(\xi ,r)$ in the following way

Lemma 6

Let $\xi \in X$, $r\in (0,1)$. There exists $\varphi \in [0,2\pi ]$ such that the function $g_{\xi ,r}(z):=\exp \left( c_{1}^{-1}r^{-2}\left\langle z-\xi ,v_{\xi }\right\rangle +i\varphi \right)$ has the following properties

• (1) $3|g_{\xi ,r}(z)|>1$ if $z\in \partial \Omega$ and $\left\| z-\xi \right\| <r$,

• (2) $3H^{d}(\{z\in K_{X}(\xi ,r):6\Re g_{\xi ,r}(z)>1\})\ge H^{d}(K_{X}(\xi ,r))$.

Proof

Let $z\in \partial \Omega$ be such that $\left\| z-\xi \right\| <r$. Then, using (3.1) we estimate

$$\begin{aligned} |g_{\xi ,r}(z)|=\exp \left( c_{1}^{-1}r^{-2}\Re \left\langle z-\xi ,v_{\xi }\right\rangle \right) \ge \exp \left( -r^{-2}\left\| z-\xi \right\| ^{2}\right)>\exp (-1)>\frac{1}{3}. \end{aligned}$$

Let $g_{\xi ,r,k}(z):=\exp \left( c_{1}^{-1}r^{-2}\left\langle z-\xi ,v_{\xi }\right\rangle +\frac{2\pi ik}{3}\right)$. For a given $z\in \overline{\Omega }{\setminus }\{\xi \}$ there exists an index $k_{z}\in \{0,1,2\}$ such that

$$\begin{aligned} \arg \left( c_{1}^{-1}r^{-2}\left\langle z-\xi ,v_{\xi }\right\rangle +\frac{2\pi ik_{z}}{3}\right) \in \left[ \frac{-\pi }{3},\frac{\pi }{3}\right] . \end{aligned}$$

In particular,

$$\begin{aligned} \max _{k=0,1,2}\Re g_{\xi ,r,k}>|g_{\xi ,r,k}(z)|\cos (\nicefrac {\pi }{3})=\frac{1}{2}|g_{\xi ,r}(z)|>\nicefrac {1}{6} \end{aligned}$$

for $z\in K_{X}(\xi ,r)$, so there exists index k such that the required properties are fulfilled for $g_{\xi ,r}=g_{\xi ,r,k}$. $\square$

The following fact is essential for our construction.

Theorem 7

There exists constants $C>2$, $\beta \in (0,1)$ such that for each small enough $r>0$ and Cr-separated subset A of X there exists a holomorphic function $f\in \mathcal {O}(\mathbb {C}^{n})$ with the following properties

• (1) $|f(z)|>3$ if $z\in \partial \Omega$ and $\left\| z-\xi \right\| <r$ for some $\xi \in A$,

• (2) $3H^{d}(\{z\in K_{X}(\xi ,r):\Re f(z)>1\})\ge H^{d}(K_{X}(\xi ,r))$ for $\xi \in A$,

• (3) $\left| f\right| \le 13$ on $\overline{\Omega }$,

• (4) $\left| f\right| \le \beta ^{1/r}$ on $\overline{\Omega }{\setminus }\bigcup _{x\in A}K_{\overline{\Omega }}(x,\sqrt{r})$,

• (5) $\left| f(z)-f(w)\right| \le Cr^{-2}\left\| z-w\right\|$ for $z,w\in \overline{\Omega }$.

Proof

Let us consider

$$\begin{aligned} f(z):=12\sum _{\xi \in A}\exp \left( c_{1}^{-1}r^{-2}\left\langle z-\xi ,v_{\xi }\right\rangle +i\varphi _{\xi }\right) . \end{aligned}$$

We choose the set $\{\varphi _{\xi }\}_{\xi \in A}\subset [0,2\pi ]$ such that the function

$$\begin{aligned} g_{\xi ,r}(z):=\exp \left( c_{1}^{-1}r^{-2}\left\langle z-\xi ,v_{\xi }\right\rangle +i\varphi _{\xi }\right) \end{aligned}$$

fulfills the properties from Lemma 6 for $\xi \in A$, $r\in (0,1)$. There exists $C>2$ so large that $\frac{13}{c_{1}}<C$ and

$$\begin{aligned} 12\sum _{k=1}^{\infty }\frac{b}{a}(k+2)^{d}\exp \left( \frac{-c_{2}C^{2}k^{2}}{4c_{1}}\right) \le 1. \end{aligned}$$

(3.3)

For a given $z\in \overline{\Omega }$ let us define

$$\begin{aligned} A_{k}(z):=\left\{ \xi \in A:\frac{Ckr}{2}\le \left\| z-\xi \right\| \le \frac{C(k+1)r}{2}\right\} . \end{aligned}$$

If $\xi \in A_{k}(z)$ then we estimate

$$\begin{aligned} r^{-2}\Re \left\langle z-\xi ,v_{\xi }\right\rangle \le -r^{-2}c_{2}\left\| z-\xi \right\| ^{2}\le -r^{-2}c_{2}\left( \frac{Ckr}{2}\right) ^{2}\le \frac{-c_{2}C^{2}k^{2}}{4}. \end{aligned}$$

By Lemmas 4 and 5 for a given $z\in \overline{\Omega }$, we have the following estimation

$$\begin{aligned} \sum _{\eta \in A{\setminus } A_{0}(z)}\left| g_{\eta ,r}(z)\right|&=\sum _{\eta \in A{\setminus } A_{0}(z)}\exp \left( c_{1}^{-1}r^{-2}\Re \left\langle z-\eta ,v_{\eta }\right\rangle \right) \nonumber \\ {}&\le \sum _{k=1}^{\infty }\sum _{\eta \in A_{k}(z)}\exp \left( c_{1}^{-1}r^{-2}\Re \left\langle z-\eta ,v_{\eta }\right\rangle \right) \nonumber \\ {}&\le \sum _{k=1}^{\infty }\sum _{\eta \in A_{k}(z)}\exp \left( \frac{-c_{2}C^{2}k^{2}}{4c_{1}}\right) \nonumber \\ {}&\le \sum _{k=1}^{\infty }\frac{b}{a}(k+2)^{d}\exp \left( \frac{-c_{2}C^{2}k^{2}}{4c_{1}}\right) \nonumber \\ {}&\le \frac{1}{12}. \end{aligned}$$

(3.4)

By Lemma 5 and above inequality, we observe the property (3) holds:

$$\begin{aligned} \left| f(z)\right|&\le 12\sum _{\xi \in A}\left| g_{\xi ,r}(z)\right| \le 12+12\sum _{\eta \in A{\setminus } A_{0}(z)}\left| g_{\eta ,r}(z)\right| \le 13 \end{aligned}$$

for $z\in \overline{\Omega }.$

Set $\xi \in A$. Let $z\in \partial \Omega$ be such that $\left\| z-\xi \right\|<r<\frac{Cr}{2}$ and

$$\begin{aligned} \Re g_{\xi ,r}(z)>\frac{1}{6}. \end{aligned}$$

We observe that $A_{0}(z)=\{\xi \}$, which due to the inequality (3.4) implies

$$\begin{aligned} \Re f(z)\ge&12\Re g_{\xi ,r}(z)-12\sum _{\eta \in A{\setminus } A_{0}(z)}\left| g_{\eta ,r}(z)\right| >2-1=1, \end{aligned}$$

and gives the property (2)

$$\begin{aligned} H^{d}\left( \left\{ z\in K_{X}(\xi ,r):\Re f(z)>1\right\} \right)&\ge H^{d}\left( \left\{ z\in K_{X}(\xi ,r):\Re g_{\xi ,r}(z)>\frac{1}{6}\right\} \right) \\ {}&\ge \frac{1}{3}H^{d}\left( K_{X}(\xi ,r)\right) . \end{aligned}$$

Moreover, by Lemma 6 and (3.4), we obtain the property (1)

$$\begin{aligned} |f(z)|\ge&12\left| g_{\xi ,r}(z)\right| -12\sum _{\eta \in A{\setminus } A_{0}(z)}\left| g_{\eta ,r}(z)\right| >4-1=3 \end{aligned}$$

for $z\in \partial \Omega$ with $\left\| z-\xi \right\| <r$ for some $\xi \in A$.

Let $z\in \overline{\Omega }{\setminus }\bigcup _{\xi \in A}K_{\overline{\Omega }}\left( \xi ,\sqrt{r}\right)$. By Lemma 5, we have at most $\frac{b3^{d}}{aC^{d}r^{d}}$ elements in A. Now, using Lemma 4, we obtain the property (4) for $\beta =\exp \left( -\frac{c_{2}}{2c_{1}}\right)$

$$\begin{aligned} |f(z)|&\le 12\sum _{\xi \in A}\left| g_{\xi ,r}(z)\right| \le 12\sum _{\xi \in A}\exp \left( -c_{1}^{-1}r^{-2}c_{2}\left\| z-\xi \right\| ^{2}\right) \\&{\le 12\sum _{\xi \in A}\exp \left( \nicefrac {-c_{1}^{-1}c_{2}}{r}\right) \le \frac{12b3^{d}}{aC^{d}r^{d}}\exp { \left( -\frac{c_{2}}{c_{1}r}\right) {<\beta ^{1/r}}}} \end{aligned}$$

for all $r>0$ small enough.

If $z_{t}=z+t(w-z)$ and $\phi (t):=f(z_{t})$ then

$$\begin{aligned} \phi '(t)=12\sum _{\xi \in A}\frac{\left\langle w-z,v_{\xi }\right\rangle }{c_{1}r^{2}}g_{\xi ,r}(z_{t}), \end{aligned}$$

so due to (3.4) we have property (5)

$$\begin{aligned} \left| f(z)-f(w)\right|&=\left| \phi (1)-\phi (0)\right| \le \max _{t\in [0,1]}|\phi '(t)|\\&\le \frac{12\left\| w-z\right\| }{c_{1}r^{2}}\max _{t\in [0,1]}\left( 1+\sum _{\xi \in A{\setminus } A_{0}(z_{t})}\left| g_{\xi ,r}(z_{t})\right| \right) \le \frac{13\left\| w-z\right\| }{c_{1}r^{2}}\\&<Cr^{-2}\left\| w-z\right\| \end{aligned}$$

for $z,w\in \overline{\Omega }.$ $\square$

4 Peak Set in Strictly Pseudoconvex Domain

Now, we are ready to finish our construction.

Theorem 8

Assume that $\Omega \subset \mathbb {C}^{n}$ is a bounded strictly pseudoconvex domain with $C^{2}$ boundary. If $X\subset \partial \Omega$ is a compact Ahlfors–David regular set of Hausdorff dimension $d\in (0,2n-1]$ then there exists a compact set $E\subset X$ of Hausdorff dimension equal to d which is a peak set for $A(\Omega )$ i.e., there exists a function $F\in A(\Omega )$ such that $E=\{z\in \partial \Omega :F(z)=1\}$ and $|F|<1$ on $\overline{\Omega }{\setminus } E$.

Proof

First, we observe that the boundary of a strictly pseudoconvex domain embedded in the boundary of a strictly convex domain is Ahlfors–David regular. In particular, by Fornæss Embedding Theorem (see [6]) it is enough to prove our result if $\Omega$ is a bounded strictly convex domain with $C^{2}$ boundary.

Assume that $\alpha \in (0,d)$ and $\theta =2^{-d-2}\min \{a,1\}$. We choose the constant $c_{\theta }$ from Lemma 3 and the constants $C,\beta$ from Theorem 7. On X we consider the topology induced from $\mathbb {C}^{n}$. First we construct a Cantor set $E\subset X\subset \partial \Omega$ of Hausdorff dimension at least $\alpha$ and not bigger than d. Since X has Hausdorff dimension equal to d, it is enough to prove that E has Hausdorff dimension at least $\alpha$.

By induction, we construct numbers $r_{j}\in (0,1/(8C)]$, finite sets $A_{j}\subset X$, compact sets $E_{j}\subset X$, and holomorphic functions $f_{j}$ on $\mathbb {C}^{n}$ meeting the following properties (for $j\ge 1$):

• (1) $E_{j}=\bigcup _{x\in A_{j}}Q(x,r_{j})$ where

  • (a) $Q(x,r_{j})\subset \overline{K_{X}\left( x,\nicefrac {r_{j}}{2}\right) }$,

  • (b) $H^{d}(\mathring{Q}(x,r_{j}))>\theta r_{j}^{d}$, where $\mathring{Q}(x,r_{j})$ denotes interior⁵ of $(x,r_{j})$-segment $Q(x,r_{j})$ in X.

• (2) in each segment $Q(x,r_{j-1})$ in $E_{j-1}$ there are at least $\left( \frac{r_{j-1}}{r_{j}}\right) ^{\alpha }$ segments of $E_{j}$,

• (3) the distance between any two balls $K_{X}(x,r_{j})$ and $K_{X}(y,r_{j})$ for $x\ne y\in A_{j}$ is not smaller than

  $$\begin{aligned} \rho _{j}:=\frac{c_{\theta }}{2}r_{j-1}\left( \frac{r_{j}}{r_{j-1}}\right) ^{\nicefrac {\alpha }{d}}, \end{aligned}$$

• (4) $r_{j-1}^{3}>r_{j}$,

• (5) $\Re f_{j}\ge 0.5$ on $\bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j}^{3})$,

• (6) $\left| f_{j}\right| <13$ on $\overline{\Omega }$,

• (7) $\left| f_{j}\right| <2^{-j}$ on $\overline{\Omega }{\setminus }\bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j-1})$.

Let $r_{0}=1$, $E_{0}=\overline{K_{X}(x_{0},\nicefrac {1}{2})}$, $A_{0}=\{x_{0}\}$ and $f_{0}\equiv 1$. Now suppose we have just constructed the numbers $r_{0}, \ldots ,r_{j-1}$, finite sets of centers $A_{0}, \ldots ,A_{j-1}$, compact sets $E_{0},\dots ,E_{j-1}$ and holomorphic functions $f_{0}, \ldots ,f_{j-1}$.

We use Lemma 3 to each interior of segment $\mathring{Q}(\xi ,r_{j-1})$ for $\xi \in A_{j-1}$ and get universal $\eta _{0}\in (0,1)$. This is possible because $A_{j-1}$ is a finite set. Now, for a given $\eta \in (0,\eta _{0})$ we consider the sets $A_{j,\xi }\subset Q(\xi ,r_{j-1})$ from Lemma 3 which are $c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}$-separated, has at least $\eta ^{-\alpha }$ elements and $\bigcup _{\zeta \in A_{j,\xi }}\overline{K_{X}(\zeta ,r_{j-1}\eta )}\subset \mathring{Q}(\xi ,r_{j-1})$. Let us define $A_{j}:=\bigcup _{\xi \in A_{j-1}}A_{j,\xi }$. Without loss of generality we assume that $\eta$ is so small that $A_{j}$ is $c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}$-separated and

(i):

$c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}>\frac{1}{2}Cr_{j-1}\eta$,

(ii):

$\frac{52r_{j-1}\eta }{c_{1}}\le \frac{1}{2}$,

(iii):

$r_{j-1}\ge \sqrt{\frac{1}{2}r_{j-1}\eta }$,

(iv):

$\beta ^{2/\left( r_{j-1}\eta \right) }\le 2^{-j}$ ($\beta \in (0,1)$ is from Theorem 7).

Since $A_{j}$ is now also $\frac{1}{2}Cr_{j-1}\eta$-separated, we can use Theorem 7 with $r:=\frac{1}{2}r_{j-1}\eta$ to obtain $f_{j}(z)$ with the following properties

• $r_{j-1}^{3}>r_{j-1}\eta$ and $c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}>4r_{j-1}\eta$,

• $3\,H^{d}(\{z\in K_{X}(\xi ,\frac{1}{2}r_{j-1}\eta ):\Re f_{j}(z)>1\})\ge H^{d}(K_{X}(\xi ,\frac{1}{2}r_{j-1}\eta ))$ for $\xi \in A_{j}$,

• $\left| f_{j}\right| \le 13$ on $\overline{\Omega }$,

• $\left| f_{j}\right| \le \beta ^{2/(r_{j-1}\eta )}\le 2^{-j}$ on $\overline{\Omega }{{\setminus }}\bigcup _{x\in A_{j}}K_{\overline{\Omega }}\left( x,\sqrt{\frac{1}{2}r_{j-1}\eta }\right)$,

• $\left| f_{j}(z)-f_{j}(w)\right| \le \frac{C}{\left( \frac{1}{2}r_{j-1}\eta \right) }\left\| z-w\right\|$ for $z,w\in \overline{\Omega }$.

We define $r_{j}:=r_{j-1}\eta$ and

$$\begin{aligned} Q(\zeta ,r_{j})&:=\overline{\left\{ z\in K_{X}\left( \zeta ,\nicefrac {r_{j}}{2}\right) :\Re f_{j}(z)>1\right\} }\\ E_{j}&:=\bigcup _{\zeta \in A_{j}}Q(\zeta ,r_{j}). \end{aligned}$$

The properties (4), (6) are now obvious.

We observe for $\zeta \in A_{j}$ the property (1)

$$\begin{aligned} H^{d}(\mathring{Q}(\zeta ,r_{j}))\ge \frac{1}{3}H^{d}\left( K_{X}(\xi ,\frac{1}{2}r_{j})\right) \ge \frac{a}{3}\left( \frac{r_{j}}{2}\right) ^{d}>\theta r_{j}^{d}. \end{aligned}$$

Let $\zeta \in A_{j-1}$. In the segment $Q(\zeta ,r_{j-1})$ there are contained precisely $\#A_{j,\zeta }$ segments of $E_{j}$, but $\#A_{j,\zeta }\ge \eta ^{-\alpha }=\left( \frac{r_{j-1}}{r_{j}}\right) ^{\alpha }$ which gives the property (2).

We estimate the distance between any two balls $K_{X}(x,r_{j})$ and $K_{X}(y,r_{j})$ for $x\ne y\in A_{j}$

$$\begin{aligned} c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}-2r_{j}\ge c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}-2r_{j-1}\eta >\frac{1}{2}c_{\theta }r_{j-1}\eta ^{\nicefrac {\alpha }{d}}=\frac{c_{\theta }}{2}r_{j-1}\left( \frac{r_{j}}{r_{j-1}}\right) ^{\nicefrac {\alpha }{d}}, \end{aligned}$$

which implies the property (3).

Let $w\in E_{j}$ and $z\in \bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j}^{3})$. Since $\Re f_{j}(w)\ge 1$ and $\left\| z-w\right\| <r_{j}^{3}$ we have

$$\begin{aligned} \left| \Re f_{j}(z)-\Re f_{j}(w)\right| \le \left| f_{j}(z)-f_{j}(w)\right| \le \frac{C}{\left( \frac{1}{2}r_{j-1}\eta \right) ^{2}}\left\| z-w\right\| \le 4Cr_{j}\le \frac{1}{2} \end{aligned}$$

which implies $\Re f_{j}(z)\ge 0.5$ i.e., the property (5).

If $z\in \overline{\Omega }{\setminus }\bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j-1})$ then $z\in \overline{\Omega }{\setminus }\bigcup _{x\in A_{j}}K_{\overline{\Omega }}(x,r_{j}+r_{j-1})\subset \overline{\Omega }{\setminus }\bigcup _{x\in A_{j}}K_{\overline{\Omega }}(x,\sqrt{\frac{1}{2}r_{j-1}\eta })$, so we obtain the property (7)

$$\begin{aligned} \left| f_{j}(z)\right| \le \beta ^{2/\left( r_{j-1}\eta \right) }\le 2^{-j}. \end{aligned}$$

Let us define $E:=\bigcap _{j=0}^{\infty }E_{j}$ and $f:=\sum _{j=0}^{\infty }f_{j}$. We will now present some observations that will gradually lead us to the end of the proof.

(a):

By Lemma 1, the compact set E has Hausdorff dimension at least $\alpha$.

(b):

Since $r_{j}\searrow 0$ and $E_{j+1}\subset E_{j}$ we easily observe that

$$\begin{aligned} \bigcup _{x\in E_{j+1}}K_{\overline{\Omega }}(x,r_{j+1})\subset \bigcup _{x\in E_{j}}K_{\overline{\Omega }}(E_{j},r_{j}) \end{aligned}$$

and $\bigcap _{j=0}^{\infty }\bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j})=\bigcap _{j=0}^{\infty }\bigcup _{x\in E_{j}}K_{X}(x,r_{j})=E$. Moreover, f is holomorphic on $\Omega$ and continuous on

$$\begin{aligned} \overline{\Omega }{\setminus }\bigcap _{j=0}^{\infty }\bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j})=\overline{\Omega }{\setminus } E. \end{aligned}$$

(c):

Let $z\in \bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j}){\setminus }\bigcup _{x\in E_{j+1}}K_{\overline{\Omega }}(x,r_{j+1}),$ for $j\ge 1$. Then, $z\in \bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j})\subset \bigcup _{x\in E_{j}}K_{\overline{\Omega }}(x,r_{j-1}^{3})\subset \bigcup _{x\in E_{j-k}}K_{\overline{\Omega }}(x,r_{j-k}^{3})$ and $\Re f_{j-k}\ge 0.5$ for $k\in \{1, \ldots ,j\}$. Moreover, $z\notin \bigcup _{x\in E_{j+1}}K_{\overline{\Omega }}(x,r_{j+1})$ which implies that $z\notin \bigcup _{x\in E_{j+2}}K_{\overline{\Omega }}(x,r_{j+1})$. Moreover we have $z\notin \bigcup _{x\in E_{j+k}}K_{\overline{\Omega }}(x,r_{j+k-1})$ for $k\ge 2$, which implies $\left| f_{j+k}(z)\right| <2^{-j-k}$ for $k\ge 2$. Now, we estimate

$$\begin{aligned} \Re f(z)&\ge \sum _{k=0}^{\infty }\Re f_{k}(z)\ge \sum _{k=0}^{j-1}\Re f_{k}-\left| f_{j}\right| -\left| f_{j+1}\right| -\sum _{k=j+2}^{\infty }\left| f_{k}(z)\right| \\&\ge \frac{j-1}{2}-26-\sum _{k=j+2}^{\infty }2^{-k}\ge \frac{j-1}{2}-27. \end{aligned}$$

(d):

If $z\in \overline{\Omega }{\setminus }\bigcup _{x\in E_{1}}K_{\overline{\Omega }}(x,r_{1})$ then $z\in \overline{\Omega }{\setminus }\bigcup _{x\in E_{2}}K_{\overline{\Omega }}(x,r_{1})\subset \overline{\Omega }{\setminus }\bigcup _{x\in E_{k+1}}K_{\overline{\Omega }}(x,r_{k})$ for $k\ge 1$ which implies $\left| f_{k}(z)\right| <2^{-k}$ for $k\ge 2$ and

$$\begin{aligned} \Re f(z)&\ge \sum _{k=0}^{\infty }\Re f_{k}(z)\ge \Re f_{0}(z)-|f_{1}(z)|-\sum _{k=2}^{\infty }\left| f_{k}(z)\right| \\ {}&\ge 1-13-\sum _{k=2}^{\infty }2^{-k}\ge -13 \end{aligned}$$

(e):

In particular, $\lim _{z\rightarrow w\in E}\Re f(z)=\infty$ and $\Re f\ge -27$ on $\overline{\Omega }$.

Let us define

$$\begin{aligned} F_{\alpha }(z):=\exp \left( \frac{-1}{f+28}\right) . \end{aligned}$$

We easily observe that $F_{\alpha }$ is holomorphic on $\Omega$ and continuous on $\overline{\Omega }$. Moreover, $F_{\alpha }=1$ on E and $\left| F_{\alpha }\right| <1$ on $\overline{\Omega }{\setminus } E$. We have proved that it is possible to choose a compact peak set $Q_{k}$ of Hausdorff dimension at least $\alpha _{k}=d-\frac{1}{k}$ and diameter at most $\frac{1}{k}$. The sets $Q_{k}$ can be chosen so that $Q_{j}\cap Q_{t}=\emptyset$ for all $t\ne j$ and there exists additional point $P\in X$ such that $Q=\{P\}\cup \bigcup _{k=1}^{\infty }Q_{k}$ is a compact set. Then, $H^{\alpha _{k}}(Q)\ge H^{\alpha _{k}}(Q_{k})$ so Q has Hausdorff dimension at least $\alpha _{k}$. In particular Q has the Hausdorff dimension equal to d. $\square$

Remark

Assume that $\Omega _{1}$ is a bounded pseudoconvex domain with $C^{2}$ boundary. If $P\in \partial \Omega$ is a strictly pseudoconvex point then there exists a strictly pseudoconvex domain $\Omega _{2}$ with $C^{2}$ boundary such that $\Omega _{1}\subset \Omega _{2}$ and $\partial \Omega _{1}\cap U=\partial \Omega _{2}\cap U$ for some neighborhood U of P. Now due to Theorem 8 there exists a peak set $K\subset U$ for $A(\Omega _{2})$ with the properties as in Theorem 8. Such a set K is a peak set for $A(\Omega _{1})$.