2·2. Hyperbolic metric

The hyperbolic metrics $2|dx|/(1-|x|^2)$ on ${{\mathbb{B}}^n}$ and $|dx|/x_n$ on ${{\mathbb{H}}^n}$ induce the hyperbolic distances $h_{{{\mathbb{B}}^n}}(x,y)$ and $h_{{{\mathbb{H}}^n}}(x,y)$ respectively. When $n=2,$ any domain G of $\overline{\mathbb{R}}^2={\overline{\mathbb{C}}}$ with $\mathrm{card\,}(\partial G)\geqslant 3$ is known to have a holomorphic universal covering projection p of the unit disk $\mathbb{B}^2$ onto G. Thus the hyperbolic distance $h_G$ of G can be defined by

\[h_G(z_1,z_2)=\min_{\zeta_1\in p^{-1}(z_1), \zeta_2\in p^{-1}(z_2)} h_{\mathbb{B}^2}(\zeta_1,\zeta_2)=\inf_{\gamma\in\Gamma} \int_\gamma \rho_G(z)|dz|,\]

where $\Gamma$ is the set of all rectifiable curves joining $z_1$ and $z_2$ in G and $\rho_G(z)$ denotes the hyperbolic density determined by the relation $2/(1-|\zeta|^2)=\rho(p(\zeta))|p'(\zeta)|,\; \zeta\in\mathbb{B}^2$ (see [ Reference Beardon and Minda6, Reference Keen and Lakic30 ] for details).

2·3. Quasihyperbolic metric

For higher dimensions, however, we cannot define hyperbolic metric for general domains. Quasihyperbolic metrics were introduced by F.W. Gehring and B. Palka [ Reference Gehring and Palka21 ] as a substitute for it. For a domain $G\subsetneq {\mathbb{R}^n}$ , the quasihyperbolic metric ${k}_G$ is defined by

\[{k}_{G}(x,y)=\inf_{\gamma \in \Gamma} \int_{\gamma}\frac{|dt|}{d_G(t)}, \quad x, y \in G,\]

where $\Gamma$ is the family of all rectifiable curves in G joining x and y. Note here that the inequality

\[j_G(x,y)\leqslant k_G(x,y)\]

holds for an arbitrary $G\subsetneq {\mathbb{R}^n}$ and all $x,y \in G$ [ Reference Gehring and Palka21 , lemma 2·1].

2·4. Uniform domains

A proper subdomain G of ${{\mathbb{R}}^n}$ is called uniform if there exist positive constants a and b with the following property [ Reference Gehring and Osgood20, Reference Martio and Sarvas37 ]: for every pair of points $x_1, x_2\in G,$ there is a rectifiable curve $\gamma$ joining $x_1$ and $x_2$ in G in such a way that $\ell(\gamma)\leqslant a|x_1-x_2|$ and that $\min\{\ell(\gamma_1), \ell(\gamma_2)\}\leqslant b\, d_G(x)$ for each $x\in\gamma,$ where $\gamma_j$ is the part of $\gamma$ between $x_j$ and x for each $j=1,2,$ $\ell(\gamma)$ denotes the length of the curve $\gamma$ and $d_G(x)$ is the Euclidean distance to the boundary of G from x. The class of uniform domains can also be defined in terms of a comparison inequality between two metrics [ Reference Gehring and Osgood20, Reference Vuorinen55 ]Footnote ² a subdomain G of ${{\mathbb{R}}^n}$ with non-empty boundary is uniform if and only if there exists a constant $c\geqslant 1$ such that

(2·5) \begin{equation} k_G(x,y) \leqslant c\, j_G(x,y)\end{equation}

for all $x,y \in G$ , where $k_G$ and $j_G$ are the quasihyperbolic and distance-ratio metrics, respectively. Note that $j_G(x,y)\leqslant k_G(x,y)$ holds for every domain G and all $x,y \in G$ by [ Reference Gehring and Palka21 , lemma 2·1].

2·6. Ferrand’s metric

Since the definition of the quasihyperbolic metric relies on the Euclidean metric, it is not defined for all subdomains of the Möbius space and therefore it is not Möbius invariant. To overcome this shortcoming, Ferrand [ Reference Ferrand12 ] modified the definition as follows. For a subdomain G of ${\overline{\mathbb R}^n}$ with $\mathrm{card\,} (\partial G) \geqslant 2,$ define a density function

\[w_G (x)=\sup_{a,b \in \partial G} \frac{|a-b|}{|x-a|\,|x-b|},\quad x \in G \setminus \{\infty\}\,,\]

and the metric $\sigma_G$ in G,

(2·7) \begin{equation}\sigma_{G}(x,y)=\inf_{\gamma \in \Gamma}\int_{\gamma} w_G (t) |dt|,\end{equation}

where $\Gamma$ is the family of all rectifiable curves in G joining x and y. The following result is due to Ferrand [ Reference Ferrand12 , p.122] and $\sigma_G(x,y)$ is now called the Ferrand metric [ Reference Hariri, Klén and Vuorinen23 , Chapter 5].

1. (i) $\sigma_{G}$ is a Möbius invariant metric;

2. (ii) When G is either ${{\mathbb{B}}^n}$ or ${{\mathbb{H}}^n},$ $\sigma_{G}$ coincides with the hyperbolic metric $h_{G}$ ;

3. (iii) ${k}_G \leqslant \sigma_G \leqslant 2 {k}_G$ for every domain $G \subsetneq {{\mathbb{R}}^n}$ .

We remark that the metric $\sigma_G$ was recently studied by D. A. Herron and P. K. Julian [ Reference Herron and Julian26 ].

2·9. Möbius metric

Let $G \subset {\overline{\mathbb R}^n}$ be an open set with $\mathrm{card\,} (\partial G) \geqslant 2$ . The Möbius metric on G is defined as follows ([ Reference Vuorinen55 , pp.115–116], Seittenranta [ Reference Seittenranta47 ]):

(2·10) \begin{equation} \delta_G(x,y)\;:\!=\;\log (1+ m_G (x,y))\,, \quad m_G (x,y)\;:\!=\;\sup_{a,b \in \partial G} |a,x,b,y|\,.\end{equation}

Note that the Möbius metric $\delta_{G}$ coincides with the hyperbolic metric $h_{G}$ when G is either ${{\mathbb{B}}^n}$ or ${{\mathbb{H}}^n}$ [ Reference Vuorinen55 , lemma 8·39]. A metric very similar to the Möbius metric is the Apollonian metric of Beardon [ Reference Beardon5 ].

2·11. Chordal distance-ratio metric

For a proper subdomain G of ${\overline{\mathbb R}^n}$ we define the chordal (spherical) distance-ratio metric by

\[{\,\hat{j}}_G(x, y) = \log\left(1+\frac{q(x,y)}{\min\{{\hat{d}}_G(x), {\hat{d}}_G(y)\}}\right),\]

where

\[{\hat{d}}_G(x)=\inf_{a\in\partial G}q(x,a).\]

The triangle inequality for this metric follows from [ Reference Seittenranta47 , lemma 2·2].

The following results are due to Seittenranta [ Reference Seittenranta47 ].

Proof. The fact that $\delta_G$ satisfies the triangle inequality, assertions (i) and (iii) follow from theorems 3·3, 3·4 and 3·12 in [ Reference Seittenranta47 ], respectively. In order to show assertion (ii), we introduce the auxiliary metric

\[ j^*_G(x,y)= \log\left(1+\frac{q(x,y)}{{\hat{d}}_G(x)}\right) +\log\left(1+\frac{q(x,y)}{{\hat{d}}_G(y)}\right).\]

Theorem 3·6 in [ Reference Seittenranta47 ] means the inequality $\delta_G(x,y)\leqslant j^*_G(x,y)$ for $x,y\in G.$ It is easy to verify the inequalities ${\,\hat{j}}_G(x,y)\leqslant j^*_G(x,y)\leqslant 2{\,\hat{j}}_G(x,y).$ Thus assertion (ii) now follows.

As a consequence of the previous lemma, we have the following inequality, which will be used in the proof of Theorem 1·2 later:

(2·13) \begin{equation}j_G(x,y)\leqslant 2{\,\hat{j}}_G(x,y),\quad x,y\in G\subsetneq{\mathbb{R}}^n.\end{equation}

We note that there is no constant $c=c(n)>0$ depending only on n such that $j_G(x,y)\geqslant c{\,\hat{j}}_G(x,y),\; x,y\in G,$ holds for all proper subdomains G of ${\mathbb{R}}^n.$ The following result follows also from the previous lemma.

Proof. By the Möbius invariance, we may assume that $G\subset{\mathbb{R}}^n.$ Then $j_G\leqslant \delta_G$ by Lemma 2·12 (iii). Therefore, it is enough to show that $(G,j_G)$ is proper in this case. For $a\in G$ and $0<r,$ we have to show that the set $B=\{x\in G\;:\; j_G(x,a)<r\}$ is relatively compact. It is enough to show that B is bounded and $\mathrm{dist}(B, \partial G)>0.$ The inequality $\log(1+|x-a|/d_G(a))\leqslant r$ holds for $x\in B$ and thus $|x-a|\leqslant d_G(a)(e^r-1),$ which proves that B is bounded. On the other hand, the inequality $\log(1+|x-a|/d_G(x))\leqslant r$ holds for $x\in B.$ Note that $d_G(x)\geqslant d_G(a)/2$ if $|x-a|\leqslant d_G(a)/2.$ For $x\in B$ with $|x-a|\geqslant d_G(a)/2,$ we thus have $d_G(x)\geqslant |x-a|/(e^r-1)\geqslant d_G(a)/(e^r-1).$ Therefore, we have shown $\mathrm{dist}(B,\partial G)\geqslant \min\{d_G(a)/2, d_G(a)/(e^r-1)\}>0$ as required.

2·15. Möbius uniform domains

We now consider a Möbius invariant characterisation of uniform domains. As we saw above, uniform domains in ${\mathbb{R}}^n$ are characterised by the condition (2·5) in terms of quasihyperbolic and distance-ratio metrics. These two metrics are invariant under similarity transformations but unfortunately not under Möbius transformations. To overcome this lack of invariance we apply Ferrand’s Möbius invariant metric $\sigma_G$ and the Möbius metric $\delta_G\,.$

Definition 2·16 ([ Reference Seittenranta47 ]). We say that a domain $G \subset {\overline{\mathbb R}^n}\,$ with $\mathrm{card\,}({\overline{\mathbb R}^n}\setminus G)\geqslant 2$ is Möbius uniform, if there exists a constant $c\geqslant1$ such that for all $x,y \in G$

\[\sigma_G(x,y)\leqslant c \,\delta_G(x,y)\,.\]

Note that Definition 2·5 only applies to subdomains of ${{\mathbb{R}}^n}$ whereas Definition 2.16 applies to subdomains of $ {\overline{\mathbb R}^n} .$ Indeed, we have the following result.

Proof. From Lemmas 2·8 and 2·12 it follows that if G is Möbius uniform with a constant $c_1$ , then it is uniform in the sense of (2·5) with the constant $2 c_1\,.$ Conversely, from Lemmas 2·8 and 2·12 it follows that if G is uniform in the sense of (2·5) with a constant $c_2$ , then it is Möbius uniform with the the constant $2 c_2\,.$

Therefore, we will use the shorter term “uniform” below for both uniform domains and Möbius uniform domains unless we want to emphasise which definition is intended.

We end this section with a proof of Lemma 1·10.

Proof of Lemma 1·10. By assumption, there is a number $r>0$ such that the punctured disk $0<|z-a|<r$ is contained in G. It is enough to prove the assertions for $a=0$ and $r=1.$ By assumption, we can find a finite boundary point b of G so that

\[m_G(z,z_0)\geqslant |0,z,b,z_0|=\frac{|b||z-z_0|}{|z||b-z_0|}\geqslant\frac{|b||z_0|}{2|z||b-z_0|}\;=\!:\;\frac{C}{|z|}\]

for $z\in G$ with $0<|z|<|z_0|/2.$ Hence,

\[\delta_G(z,z_0)=\log(1+m_G(z,z_0))\geqslant\log(1+C/|z|)=\log\frac1{|z|}+O(1)\]

as $z\to 0.$ Next we estimate $w_G(z)$ from above for $0<|z|\leqslant 1/4.$ For $b\in\partial G\setminus\{0\},$ we have $|z-b|/|b|\leqslant 1+|z|/|b|\leqslant 1+|z|$ and $|z-b|/|b|\geqslant 1-|z|/|b|\geqslant 1-|z|$ and thus

\[\frac{16}5\leqslant\frac1{|z|(1+|z|)}\leqslant\frac{|b|}{|z||z-b|}\leqslant \frac1{|z|(1-|z|)}=\frac1{|z|}+\frac1{1-|z|}\leqslant\frac1{|z|}+\frac43\]

for $0<|z|\leqslant 1/2.$ For $b_1,b_2\in\partial G\setminus\{0\},$ we have $|z-b_j|\geqslant |b_j|-|z|\geqslant 3|b_j|/4\geqslant 3/4$ and

\[\frac{|b_1-b_2|}{|z-b_1||z-b_2|}\leqslant \frac{|z-b_2|+|z-b_1|}{|z-b_1||z-b_2|}=\frac{1}{|z-b_1|}+\frac{1}{|z-b_2|}\leqslant \frac{8}3\]

as $z\to0.$ Hence, we obtain $w_G(z)\leqslant 1/|z|+4/3$ for $0<|z|\leqslant 1/4.$ For a given $z_0,$ we take a point $z_1\in G$ so that $|z_1|\leqslant \min\{|z_0|,1/4\}.$ Then, for $0<|z|<|z_1|,$ we have

\[\sigma_G(z,z_0)\leqslant\sigma_G(z,z_1)+\sigma_G(z_1,z_0)\leqslant \int_\gamma \frac{|dt|}{|t|}+O(1)=\log\frac1{|z|}+O(1),\]

where $\gamma$ is the curve going from $z_1$ to the point $(|z_1|/|z|)z$ along the circle $|t|=|z_1|$ and then going to z radially. Since $\delta_G(z,z_0)\leqslant\sigma_G(z,z_0),$ (1·11) follows.

Secondly, we prove (1·12). For simplicity, we further assume that $1,\infty\in\partial G.$ (For the general case, we may use a suitable Möbius transformation to reduce to this case.) Then

\[{\mathbb{D}}^*=\{z\in{\mathbb{C}}\;:\; 0<|z|<1\}\subset G\subset {\mathbb{C}}\setminus\{0,1\}\]

and therefore

\[\rho_{{\mathbb{D}}^*}(z)\geqslant\rho_G(z)\geqslant \rho_{{\mathbb{C}}\setminus\{0,1\}}(z)\]

for $0<|z|<1.$ Since

\[\rho_{{\mathbb{D}}^*}(z)=\frac1{|z|\log(1/|z|)}\quad\textrm{and}\quad\rho_{{\mathbb{C}}\setminus\{0,1\}}(z)=\frac1{|z|(C_0+\log(1/|z|))},\]

where $C_0=1/\rho_{{\mathbb{C}}\setminus\{0,1\}}(-1)$ (see [ Reference Keen and Lakic30 ] for instance), we have

\[\rho_G(z)=\frac1{|z|\log(1/|z|)}+O\left(\frac1{|z|\log^2(1/|z|)}\right)\]

as $z\to0.$ Noting the fact that the real function $1/[t\log^2t]$ is integrable over $(0,1/2],$ we obtain the required asymptotics (1·12) as required.

3·6. Definition

We say that a closed set $C \subset {{\mathbb{R}}^n}$ satisfies the continuum criterion at $x \in C$ if there exists a continuum $K \subset \{x\} \cup \left({\overline{\mathbb R}^n} \setminus C\right)$ such that

\[{\textsf{M}}(\Delta(K, C;\;{\overline{\mathbb R}^n} \setminus C))< \infty.\]

We write ${\textsf{M}}(x,\, C) < \infty$ if this holds, and otherwise we write ${\textsf{M}}(x,\, C)=\infty.$

We now recall that a continuum is a compact connected set in ${\overline{\mathbb R}^n}$ containing at least two points. We note that ${\textsf{M}}(x_0,\,C)=\infty$ if a continuum $C_0 \subset C$ contains $x_0.$ In fact, the sphere $|x-x_0|=r$ meets both K and C for all small enough $r>0$ in this case. A simple application of the following lemma implies that

\[{\textsf{M}}(\Delta(K,C;\;{\overline{\mathbb R}^n}\setminus C))\geqslant{\textsf{M}}(\Delta(K,C;\;{\overline{\mathbb R}^n}))=\infty\]

for every continuum K with $x_0\in K\subset({\overline{\mathbb R}^n}\setminus C)\cup\{x_0\}.$ Here we have used the relation $\Delta(K,C;\;{\overline{\mathbb R}^n}\setminus C)<\Delta(K,C;\;{\overline{\mathbb R}^n})$ and Lemma 3·1.

We now define the notion of M-domains.

By the above observation, a point $x\in \partial G$ satisfies the condition ${\textsf{M}}(x,{\overline{\mathbb R}^n}\setminus G)<\infty$ only if the singleton $\{x\}$ is a connected component of $\partial G.$ On the other hand, any isolated point x of $\partial G$ satisfies $M(x,{\overline{\mathbb R}^n}\setminus G)<\infty.$

We need the following result in the proof of Theorem 1·1. Our proof is similar to that of [ Reference Martio35 , lemma 3·5].

Lemma 3·9. Let G be a domain in ${\overline{\mathbb R}^n}\,.$ Suppose that a point $x_0\in\partial G\setminus\{\infty\}$ and a continuum K in $G\cup\{x_0\}$ with $x_0\in K$ satisfy the condition ${\textsf{M}}(\Delta(K,\partial G;\;G))<\infty.$ Then

\[\lim_{r\to0} {\textsf{M}}(\Delta(K\cap \overline{B}^n(x_0,r),\partial G;\; G))=0.\]

Proof. If $\partial G=\{x_0\},$ the assertion trivially holds. Thus we may assume that $\partial G$ contains at least two points. By the conformal invariance of the capacity, we may assume that $\infty\in\partial G.$ For brevity, we write ${\overline{B}}(r)=\overline{B}^n(x_0,r)$ and $S(r)=\partial B(r)$ throughout the proof. Let $M_0={\textsf{M}}(\Delta(K,\partial G;\;G))<\infty$ and choose $r_0>0$ large enough so that $K\subset B(r_0).$ For a decreasing sequence $r_j\;(j=0,1,2,\dots\!)$ with $r_j\to0\;(j\to\infty),$ consider the ring $R_j=\{x\in{\mathbb{R}}^n\;:\; r_{j+1}<|x-x_0|<r_j\}.$ We can choose such a sequence so that

\[c_j\;:\!=\;\mathrm{cap}\, R_j=\left(\frac{\omega_{n-1}}{\log(r_j/r_{j+1})}\right)^{1/(n-1)}\quad {\textrm{ satisfies} } \qquad \sum_{j=0}^\infty c_j<\infty\,.\]

For instance, for $c_j=2^{-j},$ we define $r_j$ recursively by the formula

\[r_{j+1}=r_j \exp\left(-\omega_{n-1} \,{c_j^{1-n}}\right)=r_j\exp\left(-\omega_{n-1}2^{(n-1)j}\right)\]

for $j=0,1,2,\dots.$ It is obvious that $r_j\to0$ as $j\to\infty$ for this choice. Let $K_j=K\cap\overline{R_j}$ and denote by $\Delta_j$ the family of curves joining $K_j$ and $\partial G$ in the set $\{x\in G\;:\; r_{j+2}<|x-x_0|<r_{j-1}\}$ for $j=1,2,\dots.$ Then the families $\Delta_{N+3j}\;(j=0,1,2,\dots\!)$ are disjointly supported and contained in the family $\Delta(K,\partial G;\; G)$ for $N=1,2,3,\dots.$ By Lemma 3·1 (ii) we obtain

\[\sum_{j=0}^\infty {\textsf{M}}(\Delta_{N+3j})\leqslant{\textsf{M}}(\Delta(K,\partial G;\; G))=M_0\quad (N=1,2,3,\dots\!)\]

and hence

\[\sum_{j=1}^\infty {\textsf{M}}(\Delta_{j})\leqslant 3M_0.\]

For a given number $\eta>0,$ take a large enough integer $N>0$ so that

\[\sum_{j=N}^\infty {\textsf{M}}(\Delta_{j})<\eta\quad\textrm{and}\quad\sum_{j=N-1}^\infty c_j<\eta.\]

By construction, we easily see that the curve family $\Delta(K_j,\partial G;\; G)\setminus\Delta_j$ is minorised by the family

\[\Delta(S(r_j),S(r_{j-1});\; R_{j-1})\cup \Delta(S(r_{j+2}),S(r_{j+1});\; R_{j+1}).\]

Thus, by Lemma 3·1 (i), we obtain

\begin{align*}{\textsf{M}}(\Delta(K_j, & \partial G;\; G)) \\& \leqslant {\textsf{M}}(\Delta_j)+{\textsf{M}}(\Delta(K_j,\partial G;\; G)\setminus\Delta_j)) \\& \leqslant {\textsf{M}}(\Delta_j)+{\textsf{M}}(\Delta(S(r_j),S(r_{j-1});\; R_{j-1}))+{\textsf{M}}(\Delta(S(r_{j+2}),S(r_{j+1});\; R_{j+1})) \\&={\textsf{M}}(\Delta_j)+\mathrm{cap}\, R_{j-1}+\mathrm{cap}\, R_{j+1}.\end{align*}

Therefore, we finally have

\begin{eqnarray*}{\textsf{M}}(\Delta(K\cap{\overline{B}}(r_N),\partial G;\; G))&\leqslant& {\textsf{M}}(\Delta(\{x_0\},\partial G;\; G))+\sum_{j=N}^\infty \Big[{\textsf{M}}(\Delta_j)+c_{j-1}+c_{j+1}\Big] \\& <& 0+\eta+\eta+\eta=3\eta.\end{eqnarray*}

Hence we obtain ${\textsf{M}}(\Delta(K\cap \overline{B}^n(x_0,r),\partial G;\; G))<3\eta$ for $0<r\leqslant r_N.$

The next theorem due to Martio [ Reference Martio35 , theorem 3·4] will also be used in Section 4.

Lemma 3·10. Let G be a proper subdomain of ${\overline{\mathbb R}^n}$ and fix a point $a\in G.$ For a boundary point $x_0$ of G with $x_0\ne\infty,$ set

\[L(\varepsilon)=\inf_K {\textsf{M}}(\Delta(K,\partial G;\; G)),\]

where the infimum is taken over all continua K joining a and the sphere $S^{n-1}(x_0,\varepsilon)$ in G. Then ${\textsf{M}}(x_0,{\overline{\mathbb R}^n}\setminus G)=\infty$ if and only if $L(\varepsilon)\to \infty$ as $\varepsilon\to 0^+.$

It is clear that M-domains are invariant under Möbius transformations and conformal mappings. We next give an example of an M-domain which does not have uniformly perfect boundary.

3·11. Example

Let $\{s_k\}$ and $\{r_k\}\;(k=1,2,3,\dots\!)$ be two sequences of positive numbers converging to 0 monotonically with the following property:

\[(\!*\!) \quad \alpha_k\;:\!=\;s_k-r_k-(s_{k+1}+r_{k+1})>0.\]

Then the closed balls ${\overline{B}}_k={\overline{B}}^n {(s_ke_1,r_k)}$ , $k=1,2,\ldots$ , are disjoint because $\mathrm{dist}({\overline{B}}_k,{\overline{B}}_{k+1})=\alpha_k>0,$ where $e_1=(1,0,\dots,0)\in{\mathbb{R}}^n.$ Let $C=\{0\}\cup\bigcup_{k=1}^\infty {\overline{B}}_k$ and $K_0=\{x=(x_1,\dots,x_n)\in{\mathbb{R}}^n\;:\; x_1\leqslant 0\}\cup\{\infty\}.$ Note that the ring $R_k=\{x\;:\; r_k<|x-s_ke_1|<r_k'\}$ separates C, where $r_k'=r_k+\min\{\alpha_{k-1},\alpha_k\}.$ Observe that $\alpha_{k-1}\geqslant\alpha_{k}$ if and only if $2s_k-s_{k-1}-s_{k+1}\leqslant r_{k+1}-r_{k-1}.$ This condition is fulfilled when $\{s_k\}$ is convex.

1. (1) The domain $G={\overline{\mathbb R}^n}\setminus(K_0\cup C)$ is an M-domain because every connected component of $K_0\cup C$ is a continuum. However, $\partial G$ is not uniformly perfect when $\limsup_{k\to\infty}(r_k'/r_k)=\infty.$ For instance, we can choose a convex sequence $\{s_k\}$ with $2s_{k+1}\leqslant s_k$ (such as $s_k=2^{-k}$ ) and let $r_k=2^{-k}s_k$ for $k\geqslant 1.$ Then

   \[r_{k+1}/r_k=s_{k+1}/(2s_k)\leqslant 1/4, \quad r_k'=2^{k}r_k-(2^{k+1}+1)r_{k+1}\]
   and thus
   \[\frac{r_k'}{r_k}\geqslant 2^k-\frac14(2^{k+1}+1)=2^{k-1}-2^{-2}\to+\infty\]
   as $k\to\infty.$

2. (2) Let $G={\overline{\mathbb R}^n}\setminus C.$ Suppose that the sequence of rings $A_k=\{x\;:\; s_k-r_{k}<|x|<s_k+r_k\}$ satisfies the condition $\limsup_{k\to\infty}{{\textrm{mod}}\,} A_k=\infty.$ For instance, we can take $s_k=2^{-k^2}, r_k=s_k-2s_{k+1}.$ Then ${\textsf{M}}(0,C)=\infty.$ Indeed, for each k and $t\in (s_k-r_{k}, s_k+r_k),$ the sphere $|x|=t$ intersects C by definition. Hence, for any continuum K with $0\in K\subset G\cup\{0\},$ Lemma 3.7 now yields

   \[{\textsf{M}}(\Delta(K, \partial G;\; G))\geqslant {\textsf{M}}(\Delta(K,C;\;{\overline{\mathbb R}^n}))\geqslant c_n\,\log\frac{s_k+r_k}{s_k-r_k}\]
   for sufficiently large k. By the assumption, we have ${\textsf{M}}(\Delta(K,\partial G;\;G))=\infty.$ In this case, the singleton $\{0\}$ is a connected component of $\partial G$ but the condition ${\textsf{M}}(0, {\overline{\mathbb R}^n}\setminus G)=\infty$ is satisfied.

3. (3) Let $G={\overline{\mathbb R}^n}\setminus C$ again. Then

   \[\Delta(K_0,C;\; G)\subset\bigcup_{k=0}^\infty \Delta_k,\]
   where $\Delta_k=\Delta(K_0,{\overline{B}}_k;\;{\overline{\mathbb R}^n})$ for $k\geqslant 1$ and $\Delta_0=\Delta(K_0,\{0\};\;{\overline{\mathbb R}^n}).$ Note that $\beta_0\;:\!=\;{\textsf{M}}(\Delta_0)=0.$ Since the ring $R(K_0,{\overline{B}}_k)$ contains $R_k$ as a subring, we have
   \[{\textsf{M}}(\Delta_k)=\mathrm{cap}\, R(K_0,{\overline{B}}_k)\leqslant\mathrm{cap}\, R_k=\omega_{n-1}({{\textrm{mod}}\,} R_k)^{1-n}=\omega_{n-1}\left(\log\frac{r_k'}{r_k}\right)^{1-n}.\]

Let $D_k=\{x\;:\; |x-s_k|<s_k\}$ for $k\geqslant 1$ and $H=\{x\;:\; x_1>0\}={\overline{\mathbb R}^n}\setminus K_0.$ Then

\[{\textsf{M}}(\Delta_k)=\mathrm{cap}\, (H, B_k)\leqslant \mathrm{cap}(D_k,B_k)=\omega_{n-1}\left(\log\frac{s_k}{r_k}\right)^{1-n}\;=\!:\;\beta_k\]

for $k\geqslant 1.$ If $\sum_k \beta_k<+\infty,$ we have

\[{\textsf{M}}(\Delta(K_0,C;\; G))\leqslant\sum_{k=0}^\infty {\textsf{M}}(\Delta_k)\leqslant\sum_{k=0}^\infty\beta_k<+\infty.\]

Hence ${\textsf{M}}(0,\partial G)<\infty$ in this case. For instance, if we choose $s_k$ and $r_k$ so that $r_k=s_k e^{-k^2}$ then $\beta_k=\omega_{n-1}k^{2-2n}$ satisfies the above condition. Hence, ${\textsf{M}}(0, {\overline{\mathbb R}^n}\setminus G)<\infty.$ This gives an example of a non-isolated boundary point of a domain which does not satisfy the M-condition.

3·12. Open problem

It is well known that the Hausdorff dimension of the boundary of a domain with uniformly perfect boundary is positive [ Reference Järvi and Vuorinen29 ]. We do not know whether the boundary of an M-domain has positive Hausdorff dimension.

4·8. Proof of Theorem 1·1

The part (i) $\Rightarrow$ (ii) is obvious. We show now that (ii) implies (iii) by contradiction. Suppose that G is not an M-domain, namely, ${\textsf{M}}(x_0,{\overline{\mathbb R}^n}\setminus G)<\infty$ for some $x_0\in\partial G.$ By the conformal invariance, we may assume that $x_0\ne\infty.$ We write $B(r)=B^n(x_0,r)$ and ${\overline{B}}(r)={\overline{B}}^n(x_0,r)$ for brevity. By definition, there is a continuum K with $x_0\in K\subset G\cup\{x_0\}$ such that $M_0\;:\!=\;{\textsf{M}}(\Delta(K,\partial G;\; G))<\infty.$ Take a point $x_1$ from $K\cap G$ and fix it. Let $r_1=|x_1-x_0|$ and $K_1=K.$ For each $x\in K\cap B(r_1)$ and $r\in(0,|x-x_0|),$ let $K_1(x,r)$ be the connected component of $K_1\setminus B(r)$ containing x. Note that $K_1(x,r)$ is a continuum. By construction, $K_1(x,r)\subset K_1(x,r')$ for $0<r'<r<|x-x_0|.$ We set

\[C_1=C(x_1,K_1)\;:\!=\;\bigcup_{0<r<r_1} K_1(x_1,r).\]

Then, $C_1$ is connected and, for $x,y\in C_1,$ we have $x,y\in K_1(x_1,r)$ for some $0<r<r_0.$ In particular, for such a pair of points x, y and r,

\[\mu_G(x,y)\leqslant {\textsf{M}}(\Delta(K_1(x_1,r),\partial G;\; G))\leqslant {\textsf{M}}(\Delta(K_1,\partial G;\; G)).\]

We also see that $x_0\in\overline{C_1}.$ Indeed, otherwise $\overline{C_1}$ would be a continuum in $K\setminus{\overline{B}}(\varepsilon)$ for small enough $\varepsilon>0$ and thus $K_1(x_1,\varepsilon)\supset\overline{C_1}\supset C_1.$ Since $K_1(x_1,\varepsilon)\subset C_1,$ the set $C_1$ would be closed and have a positive distance to $K\setminus C_1,$ which would violate connectedness of K.

Let $K_2$ be the connected component of the compact set $K_1\cap{\overline{B}}(r_1/2)$ containing $x_0.$ Since $x_0\in\overline{C_1},$ we have $C_1\cap K_2\ne\emptyset.$ Take a point $x_2$ from $C_1\cap K_2$ and fix it. As before, set $C_2=C(x_2,K_2).$ Then $C_2\subset C_1\cap K_2.$ Repeating this procedure, we define sequences of points $x_j,$ continua $K_j$ and connected sets $C_j$ inductively with the following properties:

1. (i) $K_j\subset {\overline{B}}(r_1 2^{1-j});$

2. (ii) $x_j\in C_j\subset C_{j-1}\cap K_j$

3. (iii) $x_0\in \overline{C_j}\subset K_j$ and

4. (iv) $\mu_G(x,y)\leqslant {\textsf{M}}(\Delta(K_j,\partial G;\; G))$ for all $x,y\in C_j.$

In particular, we observe that

\[\mu_G(x_j,x_k)\leqslant {\textsf{M}}(\Delta(K_j,\partial G;\; G)),\quad j\leqslant k.\]

By Lemma 3·9, we have

\[{\textsf{M}}(\Delta(K_j,\partial G;\; G))\leqslant {\textsf{M}}(\Delta(K\cap {\overline{B}}(r_12^{1-j}),\partial G;\; G))\to 0 \quad (j\to\infty).\]

Hence, we conclude that $\{x_j\}$ is a Cauchy sequence in $(G,\mu_G).$ Suppose that this sequence is convergent; that is, $\mu_G(x_j, x_\infty)\to 0$ as $j\to\infty$ for some $x_\infty\in G.$ On the other hand, since $|x_j-x_0|\leqslant r_12^{1-j},$ we have $x_j\to x_0$ in ${\overline{\mathbb R}^n}.$ Lemma 4·6 now implies that $x_\infty=x_0\in\partial G,$ which is a contradiction. Therefore, $(G,\mu_G)$ is not complete.

Finally, we prove that (iii) implies (i). If $\mathrm{cap}\,\partial G=0,$ then

\[{\textsf{M}}(\Delta(K,{\overline{\mathbb R}^n}\setminus G;\; G))={\textsf{M}}(\Delta(K,\partial G;\; G))=0,\]

which is not allowed by condition (iii). Therefore, $(G,\mu_G)$ is a metric space under the assumption (iii). Suppose next that the set $X=\{x\in G\;:\; \mu_G(x,a)\leqslant r_0\}$ is not compact for some $a\in G$ and $r_0>0.$ Then there is a point $x_0\in \partial X\cap(\partial G).$ We may assume that $x_0\ne\infty.$ For every $\varepsilon>0,$ there exists a point $x\in X\cap B^n(x_0,\varepsilon).$ By definition of X, ${\textsf{M}}(\Delta(K,\partial G;\; G))\leqslant r_0$ for a continuum K in $G\cup\{x_0\}$ with $a,x\in K.$ Therefore, under the notation in Lemma 3·10, we obtain $L(\varepsilon)\leqslant r_0.$ However, the lemma implies that ${\textsf{M}}(x_0,{\overline{\mathbb R}^n}\setminus G)<\infty.$ By contradiction, we have shown that (iii) implies (i).

Next we prove our second result.

4·9. Proof of Theorem 1·2

Since the uniform perfectness is Möbius invariant (Lemma 3·4), we may assume that $\infty\in\partial G$ and thus $G\subset{{\mathbb{R}}^n}$ and $\mathrm{diam}(\partial G)=+\infty.$

First suppose that the boundary $\partial G$ of G is uniformly perfect. Lemma 3·5 implies that the complement $E={\overline{\mathbb R}^n}\setminus G$ is also uniformly perfect. By a theorem of Järvi and Vuorinen [ Reference Järvi and Vuorinen29 ], E satisfies the metric thickness condition. Vuorinen [ Reference Vuorinen54 ] proved that for such a domain G there exists a constant $b_1>0$ such that for all $x,y\in G$

\[\mu_G(x,y)\geqslant b_1 {\,\hat{j}}_G(x,y).\]

Applying (2·13), we obtain (1·3) with $b=b_1/4.$

We next suppose (1·3). Then by Lemma 2·12 (iii), we have $\mu_G(x,y)\geqslant b\, j_G(x,y).$ Let $E={\overline{\mathbb R}^n}\setminus G$ and

\[0<c< c_0\;:\!=\; \exp\left[-2\left(\frac{2\omega_{n-1}}{b\log 3}\right)^{1/(n-1)}\right].\]

We prove now that $\{x\;:\; cr\leqslant |x-a|\leqslant r\}\cap E\ne \emptyset$ for every $a\in E\setminus\{\infty\}$ and $r>0.$ Suppose, to the contrary, that $\{x\;:\; cr\leqslant |x-a|\leqslant r\}\cap E= \emptyset$ for some $a \in E,\;a\ne\infty,$ and $r>0.$ Set $C_1=\{x\in{{\mathbb{R}}^n}\;:\; |x-a|\leqslant cr\}$ and $C_2=\{x\in{\overline{\mathbb R}^n}\;:\; |x-a|\geqslant r\}.$ Then the assumption implies that the set E decomposes into the two non-empty sets $E_1=E\cap C_1$ and $E_2=E\cap C_2.$ Pick two points x, y from the sphere $S=S^{n-1}(a,\rho)$ so that $|x-y|=2\rho,$ where $\rho=\sqrt c \, r.$ We take a curve $C_{xy}^0$ joining x and y in S. Then, by the subadditivity and monotonicity of the modulus (Lemma 3·1), we obtain

\begin{eqnarray*}\mu_G(x,y) &\leqslant & {{\textsf{M}}}(\Delta(C_{xy}^0, E)) \\ &\leqslant& {{\textsf{M}}}(\Delta(C_{xy}^0, E_1))+{{\textsf{M}}}(\Delta(C_{xy}^0, E_2)) \\&\leqslant& {{\textsf{M}}}(\Delta(S, C_1;\; G_1))+{{\textsf{M}}}(\Delta(S, C_2;\;G_2)),\end{eqnarray*}

where $G_1=\{x\;:\; |x-a|<\rho\}$ and $G_2=\{x\;:\; |x-a|>\rho\}.$ As is well known [ Reference Vuorinen55 , (5·10), (5·14)],

\begin{eqnarray*}{{\textsf{M}}}(\Delta(S, C_1;\; G_1))={{\textsf{M}}}(\Delta(S, C_2;\;G_2))=\frac{\omega_{n-1}}{(\log 1/\sqrt c)^{n-1}},\end{eqnarray*}

we have

\[\mu_G(x,y)\leqslant \frac{2\omega_{n-1}}{(-\log\sqrt c)^{n-1}},\]

where $\omega_{n-1}$ is the $(n -1)$ -dimensional area of $\mathbb{S}^{n-1}$ . On the other hand, since $d_G(x)\leqslant |x-a|=\rho$ and $d_G(y)\leqslant |y-a|=\rho,$ we obtain

\[j_G(x,y)=\log\left(1+\frac{|x-y|}{\min\{d_G(x),d_G(y)\}}\right)\geqslant\log\left(1+\frac{2\rho}{\rho}\right)=\log 3.\]

Thus we have $b\log 3\leqslant 2\omega_{n-1}/(\!-\log\sqrt c)^{n-1},$ that is,

$$ \left[c\geqslant \exp\left[\!-2(2\omega_{n-1}/b\log 3)^{1/(n-1)}\right]=c_0,\right]$$

a contradiction.

In the case when G is either ${{\mathbb{B}}^n}$ of ${{\mathbb{H}}^n},$ the metric $\mu_G(x,y)$ has the explicit expression in terms of the hyperbolic metric $h_G$ [ Reference Vuorinen55 , theorem 8·6]

(4·10) \begin{equation} \mu_G(x,y)= 2^{n-1}\,\tau_n \left(\frac1{\sinh^2\left(\frac12 h_{G}(x,y)\right)}\right)= \gamma_n \left(\coth^2\Big(\frac{h_{G}(x,y)}2\Big)\right).\end{equation}

The decreasing homeomorphism $\mu\;:\; (0, 1] \Longrightarrow [0,\infty)$ is defined by

\begin{eqnarray*}\mu(r)=\frac{\pi}{2}\frac{{\mathcal{K}}\left(\sqrt{1-r^2}\right)}{{\mathcal{K}}(r)}\,, \quad {\mathcal{K}}(r)=\int_{0}^{\pi/2} \frac{dt}{\sqrt{1-r^2\sin^2 t}} \,,\end{eqnarray*}

for $ r \in (0,1)\,, \, \mu(1)=0\,$ . Now the Grötzsch capacity for $n=2$ can be expressed as follows

(4·11) \begin{equation} \gamma_2(s)= \frac{2 \pi}{ \mu(1/s)}, \quad s >1\,.\end{equation}

In conjunction with the above relations (4·10), (4·11), when G is the unit disk $\mathbb{B}^2={\mathbb{D}}$ in ${\mathbb{C}},$ we obtain the expression

\begin{equation} \mu_{{\mathbb{D}}}(z,w)=\gamma_2 \left(\frac{1}{\tanh \frac{1}{2}h_{{\mathbb{D}}}(z,w)}\right)=\frac{2\pi}{\mu\left({\tanh \frac{1}{2}h_{{\mathbb{D}}}(z,w)}\right)},\quad z,w \in {\mathbb{D}}.\nonumber\end{equation}

The following estimate will be used later.

Lemma 4·13.

\[\mu\!\left(\tanh x\right)< \frac{\pi^2}{4x},\quad x>0.\]

Proof. From [ Reference Anderson, Vamanamurthy and Vuorinen2 , (5·29)], we note the inequality

\[\mu (r)<\frac{\pi^2}{4\,\mathrm{artanh}\, \sqrt[4]{r}}\]

for $0<r<1.$ Let $v=\left(\tanh x\right)^{1/4}\in(0,1)$ for $x>0.$ Since $0 <\tanh x=v^4<v<1,$ we obtain $x<\,\mathrm{artanh}\, v.$ Hence,

\[\mu(\tanh x) =\mu\!\left(v^4\right)< \frac{\pi^2}{4\,\mathrm{artanh}\, v}<\frac{\pi^2}{4x}.\]

We are now ready to show our third result.

4·14. Proof of Theorem 1·6

Assume that G is a Möbius uniform domain in ${\overline{\mathbb R}^n}.$ By Möbius invariance of Definition 2·16, we may assume that $G\subset{\mathbb{R}}^n.$ By virtue of Lemmas 2·8 and 2·12, the uniformity assumption reads

\[k_G(x,y)\leqslant c \, j_G(x,y),\quad x,y\in G\]

for a positive constant c. By [ Reference Vuorinen55 , lemma 8·32 (ii)] (see also [ Reference Hariri, Klén and Vuorinen23 , lemma 10·7]) there are positive constants $b_1, b_2$ depending only on n such that

\[\mu_G(x,y)\leqslant b_1k_G(x,y)+b_2\]

for all $x,y\in G.$ In view of Lemma 2·12, we have the required inequality with $d_j=cb_j$ (j = 1, 2).

Next we assume that the inequality (1·7) holds for a simply connected domain G in ${\overline{\mathbb{C}}}$ with non-degenerate boundary. We can also assume that $G\subset{\mathbb{C}}.$ Then, as is well known, the Koebe one-quarter theorem leads to the inequality $k_G(x,y)\leqslant 2h_G(x,y).$ By the Riemann mapping theorem, there is a conformal homeomorphism $f\;:\;G\to\mathbb{B}^2={\mathbb{D}}.$ Since $\mu_G$ and $h_G$ are conformally invariant, we obtain the formula

\[\mu_G(x,y)=\mu_{\mathbb{D}}(f(x),f(y))=\frac{2\pi}{\mu\left({\tanh \frac{1}{2}h_{{\mathbb{D}}}(f(x),f(y))}\right)}=\frac{2\pi}{\mu\left({\tanh \frac{1}{2}h_{G}(x,y)}\right)}.\]

We now apply Lemma 4·13 to get

\[\mu_G(x,y)\geqslant \frac 4\pi \, h_G(x,y)\geqslant \frac 2\pi k_G(x,y).\]

Combining this with (1·7) and Lemma 2·12, we have

\[k_G(x,y)\leqslant \frac\pi 2 \mu_G(x,y)\leqslant \frac \pi 2(2d_1 j_G(x,y)+d_2).\]

Now a result of Gehring and Osgood [ Reference Gehring and Osgood20 ] implies that G is uniform.

4·15. Open problem

As pointed out above, in the case of planar simply connected domains the modulus metric can be expressed as a function of the hyperbolic metric. We do not know, whether for a general hyperbolic planar domain, the hyperbolic metric has a minorant in terms of the modulus metric.

5·4. Proof of Theorem 5·3

Choose Möbius transformations $f_1, f_2$ such that $0, \infty \in \partial f_1(G_1)$ and $0, \infty \in \partial f_2(G_2)\,.$ Then

\[g= f_2 \circ f \circ f_1^{-1}:\; f_1(G_1) \longrightarrow f_2(G_2) \]

is K-quasiregular and by Theorem 5·2 we have

\begin{eqnarray*}k_{f_2(G_2)}(g(x), g(y)) \leqslant d_3 \, \max \{ k_{f_1(G_1)}(x,y)^{\alpha} , k_{G_1}(x,y)\}\,, \quad \alpha=K^{1/(1-n)}\,. \nonumber\end{eqnarray*}

Because $f_1(G_1), f_2(G_2) \subset \mathbb{R}^n\,,$ we obtain by Lemma 2·8 (iii) a similar inequality for the $\sigma$ metric, with a bit different constants.

5·5. Example

To show that the condition $\partial G_2$ be uniformly perfect cannot be dropped from Theorem 5·3, we consider the analytic function $g(z)= \exp\left(({z+1})/({z-1})\right)$ which maps the unit disk $\mathbb{B}^2$ onto $\mathbb{B}^2\setminus \{0\}\,.$ Let $G_1=\mathbb{B}^2$ and $G_2=\mathbb{B}^2\setminus \{0\}$ , and let $x_j=(e^j-1)/(e^j+1)$ for $j=1,2,\ldots$ . Then $u_j=g(x_j)=\exp(-e^j).$ The standard formula for the hyperbolic distance [ Reference Beardon4 , pp.38–40], [ Reference Vuorinen55 , (2·17)] shows that

\[h_{G_1}(x_j,x_{j+1})=\int_{x_j}^{x_{j+1}}\frac{2dx}{1-x^2}=2\,\mathrm{artanh}\, (x_{j+1})-2\,\mathrm{artanh}\, (x_j)=1\]

where as

\[k_{G_2}(g(x_j),g(x_{j+1}))=\int_{u_{j+1}}^{u_j}\frac{du}{u}=e^{j+1}-e^j=(e-1)e^j\rightarrow +\infty\]

as $j \rightarrow \infty$ . Thus by (i) and (ii) of Lemma 2·8, when $j \rightarrow \infty$ , $\sigma_{G_2}(g(x_j),g(x_{j+1}))\rightarrow +\infty$ while $\sigma_{G_1}(x_j,x_{j+1})=h_{G_1}(x_j,x_{j+1})=1$ . This demonstrates that uniform perfectness is needed in Theorem 5·3.