Geometric Properties of the Triangular Ratio Metric and Related Metrics

Abstract

We study the inclusion relation of the triangular ratio metric balls and the Cassinian metric balls in subdomains of $\mathbb {R}^n$. Moreover, we study distortion properties of Möbius transformations with respect to the triangular ratio metric in the punctured unit ball.

1 Introduction

In geometric function theory, various metrics relative to the boundary of domains in which families of functions are defined have been introduced and played important roles in the studies of geometric and analytic properties of these functions. In the planar case, the hyperbolic metric serves as an important example of such metrics [3, 15]. The so-called hyperbolic-type metrics, defined as generalizations of the hyperbolic metric of the planar domains to subdomains of higher-dimensional Euclidean space, share some but not all properties of the hyperbolic metric [5, 8]. Examples of well-known hyperbolic-type metrics include the quasihyperbolic metric, distance ratio metric, and Apollonian metric.

Most of the hyperbolic-type metrics belong to the family of relative metrics. A relative metric is a metric that is evaluated in a domain $D\subsetneq {\mathbb {R}^n}$ relative to its boundary. In 2002, Hästö [7] introduced the generalized relative metric named as the $M-$relative metric which is defined on a domain $D\subsetneq {\mathbb {R}^n}$ by the quantity

$$\begin{aligned} \rho _{M,D}(x,y)=\sup \limits _{a\in \partial {D}}\frac{|x-y|}{M(|x-a|,|y-a|)}, \end{aligned}$$

where M is continuous in $(0,\infty )\times (0,\infty )$ and $\partial D$ is the boundary of D. For $M(\alpha ,\beta )=\alpha +\beta$, the corresponding relative metric is the so-called triangular ratio metric

$$\begin{aligned} s_D(x,y)=\sup \limits _{a\in \partial {D}}\frac{|x-y|}{|x-a|+|y-a|}. \end{aligned}$$

The triangular ratio metric has been recently investigated in [4, 9,10,11, 16]. Another example of generalized relative metric is the Cassinian metric defined by the choice $M(\alpha ,\beta )=\alpha \beta$, i.e.,

$$\begin{aligned} c_D(x,y)=\sup \limits _{a\in \partial {D}}\frac{|x-y|}{|x-a||y-a|}. \end{aligned}$$

The geometric properties of the Cassinian metric have been studied in [13, 14, 17].

In this paper, we continue to study the geometric properties of the triangular ratio metric and Cassinian metric. In particular, we investigate the inclusion relation of the triangular ratio metric balls and the Cassinian metric balls in subdomains of $\mathbb {R}^n$. Also, we study distortion properties of Möbius transformations with respect to the triangular ratio metric in the punctured unit ball. By using the comparison between the triangular ratio metric and Ibragimov’s metric, we show the quasiconformality of bilipschitz mappings in Ibragimov’s metric.

2 Hyperbolic-Type Metrics

In this section, we collect the definitions and some basic properties of various hyperbolic-type metrics. We always denote by D the proper subdomain of the Euclidean space ${{\mathbb {R}}}^n$ and write $d(x)= d(x,\partial D)$ for the distance from x to the boundary of the domain D, and let $d_{xy}=\min \{d(x),d(y)\}$.

2.1 Hyperbolic Metric

The hyperbolic metrics $\rho _{\mathbb {H}^n}$ and $\rho _{\mathbb {B}^n}$ of the upper half space ${\mathbb {H}^n} = \{ (x_1,\ldots ,x_n)\in {\mathbb {R}^n}: x_n>0 \}$ and of the unit ball ${\mathbb {B}^n}= \{ z\in {\mathbb {R}^n}: |z|<1 \}$ are, respectively, defined as follows [2]: for $x,y\in \mathbb {H}^n$

$$\begin{aligned} \,\text {ch}{\rho _{\mathbb {H}^n}(x,y)}=1+\frac{|x-y|^2}{2x_ny_n}, \end{aligned}$$

(2.1)

and for $x,y\in \mathbb {B}^n$

$$\begin{aligned} \,\text {sh}{\frac{\rho _{\mathbb {B}^n}(x,y)}{2}}=\frac{|x-y|}{\sqrt{1-|x|^2}{\sqrt{1-|y|^2}}}. \end{aligned}$$

(2.2)

2.2 Distance Ratio Metric

For all $x,y\in D$, the distance ratio metric $j_G$ is defined as

$$\begin{aligned} j_D(x,y)=\log \left( 1+\frac{|x-y|}{d_{xy}} \right) . \end{aligned}$$

This metric was introduced by Gehring and Palka [6] in a slightly different form and in the above form in [20]. It follows from [21, Lemma 2.41(2)] and [1, Lemma 7.56] that

$$\begin{aligned} j_D(x,y)\le \rho _D(x,y)\le 2 j_D(x,y) \end{aligned}$$

for $D\in \{\mathbb {B}^n,\mathbb {H}^n\}$ and all $x,y\in D$.

2.3 Quasihyperbolic Metric

For all $x,\,y\in D$, the quasihyperbolic metric $k_D$ is defined as

$$\begin{aligned} k_D(x,y)=\inf _{\gamma }\int _{\gamma }\frac{1}{d(z,\partial D)}|dz|, \end{aligned}$$

where the infimum is taken over all rectifiable arcs $\gamma$ joining x to y in D [6]. It is well known that

$$\begin{aligned} j_D(x,y)\le k_D(x,y) \end{aligned}$$

for all $x,y\in D.$

2.4 Point Pair Function

We define for $x,y\in D\subsetneq \mathbb {R}^n$ the point pair function

$$\begin{aligned} p_D(x,y)=\frac{|x-y|}{\sqrt{|x-y|^2+4\,d(x)\,d(y)}}. \end{aligned}$$

This point pair function was introduced in [4] where it turned out to be a very useful function in the study of the triangular ratio metric. However, the function $p_G$ is generally not a metric.

2.5 Ibragimov’s Metric

For a domain $D \subsetneq \mathbb {R}^{n}$, Ibragimov’s metric is defined as

$$\begin{aligned} u_{D}(x,y) = 2\, \log \frac{|x-y|+\max \{d(x),d(y)\}}{\sqrt{d(x)\,d(y)}}, \quad \quad x,y \in D. \end{aligned}$$

Several authors have studied comparison inequalities between Ibragimov’s metric and the hyperbolic metric as well as some hyperbolic-type metrics [12, 19, 22, 23].

3 Inclusion Properties

In this section, we study inclusion relation between triangular ratio metric balls and other hyperbolic-type metric balls. Let (D, d) be a metric space. A metric ball $B_{d}(x,r)$ is a set

$$\begin{aligned} B_{d}(x,r)=\{y\in D:d(x,y)<r\}. \end{aligned}$$

Our first theorem shows the inclusion relation between the triangular ratio metric balls $B_s$ and the Cassinian metric balls $B_c$.

Theorem 3.1

For arbitrary $x\in D\subsetneq \mathbb {R}^n$ and $t\in (0,1)$,

$$\begin{aligned} B_{c}(x,r)\subset B_{s}(x,t)\subset B_{c}(x,R), \end{aligned}$$

where $r=\frac{2t}{(1+2t)d(x)}$ and $R=\frac{2t}{(1-t)d(x)}$. Moreover, $R/r\rightarrow 1$ as $t\rightarrow 0$.

Proof

For all $x,y\in D$, it is easy to see that

$$\begin{aligned} \inf _{z\in \partial D}|x-z||y-z|\le d_{xy}(d_{xy}+|x-y|). \end{aligned}$$

By the definition of the Cassinian metric, we obtain

$$\begin{aligned} c_{D}(x,y)=\frac{|x-y|}{\inf _{z\in \partial D}|x-z||y-z|}\ge \frac{|x-y|}{d_{xy}(d_{xy}+|x-y|)}, \end{aligned}$$

and hence,

$$\begin{aligned} |x-y|\le \frac{c_{D}(x,y)d_{xy}^2}{1-c_{D}(x,y)d_{xy}}<\frac{rd_{xy}^2}{1-rd(x)\wedge d(y)}<2td_{xy}. \end{aligned}$$

By the definition of the triangular ratio metric, we have

$$\begin{aligned} s_{D}(x,y)=\frac{|x-y|}{\inf _{z\in \partial D}|x-z|+|y-z|}\le \frac{|x-y|}{d(x)+d(y)}\le \frac{|x-y|}{2d_{xy}}<t. \end{aligned}$$

Hence, we obtain $B_{c}(x,r)\subset B_{s}(x,t)$. As for the inclusion $B_{s}(x,t)\subset B_{c}(x,R)$, let $y\in B_{s}(x,t)$, then

$$\begin{aligned} \frac{|x-y|}{2d(y)+|x-y|}\le s_{D}(x,y)< t, \end{aligned}$$

which implies that $|x-y|<\frac{2td(y)}{1-t}$ and

$$\begin{aligned} c_{D}(x,y)\le \frac{|x-y|}{d(x)d(y)}<\frac{2t}{(1-t)d(x)}. \end{aligned}$$

Clearly,

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{R}{r}=1. \end{aligned}$$

$\square$

Theorem 3.2 shows the inclusion between the triangular ratio metric balls and distance ratio metric balls, which was conjectured in [11, Conjecture 7.7].

Theorem 3.2

For arbitrary $x\in D\subsetneq \mathbb {R}^n$ and $t\in (0,1)$,

$$\begin{aligned} B_{j}(x,r)\subset B_{s}(x,t)\subset B_{j}(x,R), \end{aligned}$$

where $r=\log (1+2t)$ and $R=\log (1+\frac{2t}{1-t})$. Moreover, $R/r \rightarrow 1$ as $t \rightarrow 0$.

Proof

Suppose that $y\in B_{j}(x,r)$. Then,

$$\begin{aligned} \log (1+\frac{|x-y|}{d_{xy}})=j_{D}(x,y)<r= \log (1+2t), \end{aligned}$$

which implies that

$$\begin{aligned} |x-y|< 2td_{xy}. \end{aligned}$$

Since

$$\begin{aligned} \inf _{z\in \partial D}|x-z|+|y-z|\ge 2 d_{xy}, \end{aligned}$$

by the definition of the triangular ratio metric we have

$$\begin{aligned} s_{D}(x,y)=\frac{|x-y|}{\inf _{z\in \partial D}|x-z|+|y-z|}\le \frac{|x-y|}{2d_{xy}}< t. \end{aligned}$$

Hence, $y\in B_{s}(x,t)$. Now we prove the second inclusion. It follows from triangle inequality that $\inf \limits _{z\in \partial D}|x-z|+|y-z|\le 2d_{xy}+|x-y|$, and

$$\begin{aligned} t>s_{D}(x,y)\ge \frac{|x-y|}{2d_{xy}+|x-y|}, \end{aligned}$$

which implies

$$\begin{aligned} \frac{|x-y|}{d_{xy}}<\frac{2t}{1-t}. \end{aligned}$$

Hence, the second inclusion holds. It is easy to check that

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{R}{r}=1. \end{aligned}$$

$\square$

From the well-known inequalities [1, Theorem 7.56]

$$\begin{aligned} j_{ {\mathbb {B}^n} }(x,y)\le \rho _{ {\mathbb {B}^n} }(x,y)\le 2j_{ {\mathbb {B}^n} }(x,y), \end{aligned}$$

it follows that

$$\begin{aligned} B_{\rho }(x,r)\subset B_{j}(x,r)\subset B_{\rho }(x,2r). \end{aligned}$$

(3.1)

Theorem 3.3

Let $x\in \mathbb {B}^n$ and $t\in (0,1)$. Then,

$$\begin{aligned} B_{\rho }(x,r)\subset B_{s}(x,t)\subset B_{\rho }(x,R), \end{aligned}$$

where $r=\log (1+2t)$ and $R=2\log (1+\frac{2t}{1-t})$. Moreover, $R/r \rightarrow 2$ as $t \rightarrow 0$ .

Proof

By Theorem 3.2, we have $B_{s}(x,t)\subset B_{j}(x,\log (1+\frac{2t}{1-t}))$, which together with the right-hand side of (3.1) implies the second inclusion with $R=2\log (1+\frac{2t}{1-t})$. Similarly, Theorem 3.2 together with the left-hand side of (3.1) implies

$$\begin{aligned} B_{\rho }(x,r)\subset B_{j}(x,r)\subset B_{s}(x,\frac{\mathrm{e}^r-1}{2}). \end{aligned}$$

That is, $B_{\rho }(x,r)\subset B_{s}(x,t)$ with $r=\log (1+2t)$. By l’Hôpital rule, it is easy to see that

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{R}{r}=2. \end{aligned}$$

$\square$

In a convex domain $D\subset \mathbb {R}^n$, we recall the following inequality [4, Lemma 3.14]

$$\begin{aligned} s_{D}(x,y)\le p_{D}(x,y)\le \sqrt{2}s_{D}(x,y),\quad \text{ for }\quad x,y\in D. \end{aligned}$$

It follows immediately that

$$\begin{aligned} B_{p}(x,r)\subset B_{s}(x,r)\subset B_{p}(x,\sqrt{2} r). \end{aligned}$$

Similarly, in a convex domain $D\subset \mathbb {R}^n$, the inequality [9, Theorem 2.17]

$$\begin{aligned} s_{D}(x,y)\le v_{D}(x,y)\le \pi s_{D}(x,y) \end{aligned}$$

implies the inclusion

$$\begin{aligned} B_{v}(x,r)\subset B_{s}(x,r)\subset B_{v}(x,\pi r). \end{aligned}$$

Lemma 3.4

[18, Corollary 3.4] For $x\in \mathbb {B}^n$ and $r>0$,

$$\begin{aligned} B_{j}(x,r)\subset B_{k}(x,t)\subset B_{j}(x,R), \end{aligned}$$

where

$$\begin{aligned} r=\max \{\log (1+(1+|x|)\sinh \frac{t}{4}),\log (1+(1-|x|)\frac{\mathrm{e}^{t/2}-1}{2})\} \end{aligned}$$

and

$$\begin{aligned} R=\log (1+(1+|x|)\frac{\mathrm{e}^t-1}{2}). \end{aligned}$$

Theorem 3.5

Let $x\in \mathbb {B}^n$ and $t\in (0,1)$. Then, the following inclusion relation holds:

$$\begin{aligned} B_{k}(x,r)\subset B_{s}(x,t)\subset B_{k}(x,R), \end{aligned}$$

where $r=\log (1+\frac{4t}{1+|x|})$ and $R=\max \{R_{1},R_{2}\}$ with

$$\begin{aligned} R_{1}=4 \,\text {arsh}(\frac{2t}{(1+|x|)(1-t)}), \quad R_{2}=2\log (1+\frac{4t}{(1-|x|)(1-t)}). \end{aligned}$$

Proof

By Lemma 3.4, it is easy to see that

$$\begin{aligned} B_{k}(x,r)\subset B_{j}(x,\log (1+(1+|x|)\frac{\mathrm{e}^{r}-1}{2})), \end{aligned}$$

and by Theorem 3.2, $B_{j}(x,r)\subset B_{s}(x,\frac{\mathrm{e}^r-1}{2})$, then we have

$$\begin{aligned} B_{k}(x,r)\subset B_{s}(x,(1+|x|)\frac{\mathrm{e}^r-1}{4}). \end{aligned}$$

Similarly, $B_{k}(x,r)\subset B_{s}(x,t)$ with $r=\log (1+\frac{4t}{1+|x|})$.

Again from Theorem 3.2 and Lemma 3.4, it follows that

$$\begin{aligned} B_{s}(x,t)\subset B_{j}(x,\log (1+\frac{2t}{1-t})) \end{aligned}$$

and

$$\begin{aligned} B_{j}(x,t)\subset B_{k}(x,\max \{4\,\text {arsh}\frac{\mathrm{e}^t-1}{1+|x|},2\log (1+\frac{2(\mathrm{e}^t-1)}{1-|x|})\}). \end{aligned}$$

Hence, the second inclusion holds with $R=\max \{R_{1},R_{2}\}$, where

$$\begin{aligned} R_{1}=4 \,\text {arsh}(\frac{2t}{(1+|x|)(1-t)})\quad \text{ and }\quad R_{2}=2\log (1+\frac{4t}{(1-|x|)(1-t)}). \end{aligned}$$

$\square$

Lemma 3.6

[17, Theorem 5.4] For given $a\in \mathbb {R}^n$, let domain $D=\mathbb {R}^n{\setminus }\{a\}$, $x\in D$ and $0<t<1/(2|x-a|)$. Then, we have the following inclusion relation

$$\begin{aligned} B_{j}(x,r)\subset B_{c}(x,t)\subset B_{j}(x,R), \end{aligned}$$

where $r=\log (1+t|x-a|)$ and $R=\log (\frac{1-t|x-a|}{1-2t|x-a|})$. Moreover, $R/r \rightarrow 1$ as $t \rightarrow 0$.

The following improved inclusion relation between the Cassinian metric balls and the distance ratio metric balls was conjectured in [17, Conjecture 5.5].

Theorem 3.7

Let $D\subsetneq \mathbb {R}^n$ be a domain and $x\in D$. For $0<t<\frac{1}{d(x)}$, the following inclusion holds:

$$\begin{aligned} B_{j}(x,r)\subset B_{c}(x,t)\subset B_{j}(x,R), \end{aligned}$$

where $r=\log (1+td(x))$ and $R=\log \left( \frac{1}{1-td(x)}\right)$. Moreover, $R/r \rightarrow 1$ as $t\rightarrow 0$.

Proof

Suppose that $y\in B_j(x,r)$. Then, $j_{D}(x,y)<r$, and

$$\begin{aligned} \log \left( 1+\frac{|x-y|}{\min \{d(x),d(y)\}}\right) <\log \left( 1+td(x)\right) . \end{aligned}$$

On simplification, we get

$$\begin{aligned} |x-y|<td(x)\min \{d(x),d(y)\}<td(x)d(y), \end{aligned}$$

which together with the inequality

$$\begin{aligned} \inf _{p\in \partial D}|x-p||y-p|\ge d(x)d(y) \end{aligned}$$

implies

$$\begin{aligned} c_D(x,y)=\frac{|x-y|}{\inf \limits _{p\in \partial D}|x-p||y-p|}<t. \end{aligned}$$

Hence, $y\in B_c(x,t)$ and $B_j(x,r)\subset B_{c}(x,t)$.

Now we prove the second inclusion. Let $p_0\in \partial D$ with $|x-p_0|=d(x)$. The triangle inequality yields $|y-p_0|\le d(x)+|x-y|$, and then,

$$\begin{aligned} \inf _{p\in \partial D}|x-p||y-p|\le |x-p_0||y-p_0|\le d(x)(d(x)+|x-y|). \end{aligned}$$

Similarly,

$$\begin{aligned} \inf _{p\in \partial D}|x-p||y-p|\le d(y)(d(y)+|x-y|). \end{aligned}$$

Combining the above two inequalities, we have

$$\begin{aligned} \inf _{p\in \partial D}|x-p||y-p|\le d_{xy}(d_{xy}+|x-y|), \end{aligned}$$

and then, for $y\in B_c(x,t)$,

$$\begin{aligned} \frac{|x-y|}{d_{xy}(d_{xy}+|x-y|)}\le \frac{|x-y|}{\inf _{p\in \partial D}|x-p||y-p|}<t, \end{aligned}$$

which implies

$$\begin{aligned} \frac{|x-y|}{d_{xy}}<\frac{td_{xy}}{1-t d_{xy}}. \end{aligned}$$

Therefore,

$$\begin{aligned} j_D(x,y)&=\log \left( 1+\frac{|x-y|}{d_{xy}}\right) \\&\le \log \frac{1}{1-td_{xy}}<R \end{aligned}$$

and $y\in B_j(x,R)$. Hence, the second inclusion holds. Clearly, one can see that $R/r\rightarrow 1$ as $t\rightarrow 0$. $\square$

Before proving Theorem 3.8, we recall the following inequality [6, Lemma 2.1]:

$$\begin{aligned} j_{D}(x,y)\le k_{D}(x,y)\quad \text{ for } \text{ all }\quad x,y\in D. \end{aligned}$$

(3.2)

Theorem 3.8

Let $D\subsetneq \mathbb {R}^n$ be a domain and $x\in D$. For $t<\frac{1}{2d(x)}$, we have

$$\begin{aligned} B_{k}(x,r)\subset B_{c}(x,t)\subset B_{k}(x,R), \end{aligned}$$

where $r=\log (1+td(x))$ and $R=\log (\frac{1-td(x)}{1-2td(x)})$. Moreover, $R/r\rightarrow 1$ as $t\rightarrow 0$.

Proof

For arbitrary $y\in B_{k}(x,r)$, we have $k_{D}(x,y)<r$. Inequality (3.2) implies $j_{D}(x,y)<r$, and then $B_{k}(x,r)\subset B_{j}(x,r)$. Since $B_{j}(x,r)\subset B_{c}(x,t)$ by Theorem 3.7, the first inclusion follows.

Let $z\in \partial D$ such that $c_{D}(x,y)=c_{\mathbb {R}^n\backslash \{z\}}(x,y)$. Since $D\subset \mathbb {R}^n\backslash \{z\}$, it follows from the domain monotonicity of the distance ratio metric that

$$\begin{aligned} j_{\mathbb {R}^n\backslash \{z\}}(x,y)\le j_{D}(x,y). \end{aligned}$$

Hence, we have $j_{\mathbb {R}^n\backslash \{z\}}(x,y)<r$. By Lemma 3.6, we obtain

$$\begin{aligned} c_{\mathbb {R}^n\backslash \{z\}}(x,y)<t \end{aligned}$$

and $c_{D}(x,y)<t$, which implies $B_{j}(x,r)\subset B_{c}(x,t)$.

For the second inclusion relation, let $y\in B_c(x,t)$. It follows from Theorem 3.7 that $y\in B_{j}(x,\log (1/(1-td(x))))$ and then

$$\begin{aligned} |x-y|<\frac{td^2(x)}{1-td(x)}. \end{aligned}$$

Since $t<{1}/{(2d(x))}$, we have $|x-y|<d(x)$. By [21, Lemma 3.7],

$$\begin{aligned} k_{D}(x,y)&\le \log (1+\frac{|x-y|}{d(x)-|x-y|})\\&<\log (1+\frac{td(x)}{1-2td(x)})\\&=\log (\frac{1-td(x)}{1-2td(x)}). \end{aligned}$$

It is easy to see that

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{R}{r}=1. \end{aligned}$$

$\square$

4 Distortion Property of Möbius Transformations

The distortion property of the triangular ratio metric under Möbius transformations of the unit ball has been studied in [4, 10]. In this section, we study the similar property but under Möbius transformations of a punctured unit ball.

For $a\in \mathbb {R}^n\backslash \{0\}$, let $a^*=\frac{a}{|a|^2}$, $0^*=\infty$, and $\infty ^*=0$. Let

$$\begin{aligned} \sigma _{a}(x)=a^*+s^2(x-a^*)^*, \quad s^2=|a^*|^2-1 \end{aligned}$$

be the inversion in the sphere $S^{n-1}(a*,s)$.

Let f be a Möbius transformation of the unit ball. Since the triangular ratio metric $s_{D}$ is invariant under orthogonal transformations, it follows from [2, Theorem 3.5.1] that

$$\begin{aligned} s_{\mathbb {B}^n}(f(x),f(y))=s_{\mathbb {B}^n}(\sigma _{a}(x),\sigma _{a}(y))\quad \text{ for } \quad x,y,a\in \mathbb {B}^n. \end{aligned}$$

Theorem 4.1

Let $a\in \mathbb {B}^n$ and $f:\mathbb {B}^n\backslash \{0\}\rightarrow \mathbb {B}^n\backslash \{a\}$ be a Möbius transformation with $f(0)=a$. Then, for $x,y\in \mathbb {B}^n\backslash \{0\}$, it holds

$$\begin{aligned} \frac{1-|a|}{1+|a|}s_{\mathbb {B}^n\backslash \{0\}}(x,y)\le s_{\mathbb {B}^n\backslash \{a\}}(f(x),f(y))\le \frac{1+|a|}{1-|a|}s_{\mathbb {B}^n\backslash \{0\}}(x,y). \end{aligned}$$

Proof

If $a=0$, i.e., $f(0)=0$, then f is a rotation and preserves the triangular ratio metric. Now we suppose that $a\ne 0$ and then $f(a)=0$.

$$\begin{aligned} s_{\mathbb {B}^n\backslash \{0\}}(x,y)=\frac{|x-y|}{P}, \end{aligned}$$

where

$$\begin{aligned} P=\min \{|x|+|y|,\inf _{w\in \partial \mathbb {B}^n}|x-w|+|y-w|\} \end{aligned}$$

and

$$\begin{aligned} s_{\mathbb {B}^n\backslash \{a\}}(\sigma _{a}(x),\sigma _{a}(y))= \frac{|\sigma _{a}(x)-\sigma _{a}(y)|}{T}, \end{aligned}$$

with

$$\begin{aligned} T=\min \{|\sigma _{a}(x)-a|+|\sigma _{a}(y)-a|,\inf _{z\in \partial \mathbb {B}^n}|\sigma _{a}(x)-\sigma _{a}(z)|+|\sigma _{a}(z)-\sigma _{a}(y)|\}. \end{aligned}$$

We first prove the right-hand side inequality.

If $T=\inf \limits _{z\in \partial \mathbb {B}^n}|\sigma _{a}(x)-\sigma _{a}(z)|+ |\sigma _{a}(z)-\sigma _{a}(y)|$, then the distortion of the triangular ratio metric under Möbius transformations of the punctured unit ball is the same as the case of the unit ball [4, Theorem 3.31].

Now we suppose that $T=|\sigma _{a}(x)-a|+|\sigma _{a}(y)-a|$. Then,

$$\begin{aligned} s_{\mathbb {B}^n\backslash \{a\}}(\sigma _{a}(x),\sigma _{a}(y))&=\frac{|\sigma _{a}(x)-\sigma _{a}(y)|}{|\sigma _{a}(x)-a|+|\sigma _{a}(y)-a|}\\&=\frac{\frac{s^2|x-y|}{|x-a^*||y-a^*|}}{\frac{s^2|x|}{|x-a^*||a^*|}+ \frac{s^2|y|}{|y-a^*||a^*|}}\\&=\frac{|x-y|}{\frac{|y-a^*|}{|a^*|}|x|+\frac{|x-a^*|}{|a^*|}|y|}\\&=\frac{|x-y|}{\beta |x|+\gamma |y|}, \end{aligned}$$

where $\beta =\frac{|y-a^*|}{|a^*|}$ and $\gamma =\frac{|x-a^*|}{|a^*|}$. Clearly,

$$\begin{aligned} |a^*|-1\le |x-a^*|,|y-a^*|\le |a^*|+1 \end{aligned}$$

which together with $\beta ,\gamma \ge 1-|a|$ implies

$$\begin{aligned} s_{\mathbb {B}^n\backslash \{a\}}(\sigma _{a}(x),\sigma _{a}(y))\le \frac{1}{1-|a|}\frac{|x-y|}{|x|+|y|}\le \frac{1+|a|}{1-|a|}s_{\mathbb {B}^n \backslash \{0\}}(x,y). \end{aligned}$$

Next we prove the left-hand side of the inequality. If $P=\inf \limits _{w\in \partial \mathbb {B}^n}|x-w|+|y-w|$, the distortion of the triangular ratio metric under Möbius transformations of the punctured unit ball is the same as the case of the unit ball [4, Theorem 3.31]. Now we assume $P=|x|+|y|$. Then,

$$\begin{aligned} s_{\mathbb {B}^n\backslash \{0\}}(x,y)&=\frac{|x-y|}{|x|+|y|}\\&=\frac{\frac{|\sigma _{a}(x)-\sigma _{a}(y)||x-a^*||y-a^*|}{s^2}}{\frac{|\sigma _{a}(x)-a||x-a^*||a^*|}{s^2}+\frac{|\sigma _{a}(y)-a||y-a^*||a^*|}{s^2}}\\&=\frac{|\sigma _{a}(x)-\sigma _{a}(y)|}{\frac{1}{\beta }|\sigma _{a}(x)-a|+ \frac{1}{\gamma }|\sigma _{a}(y)-a|}, \end{aligned}$$

where $\beta ,\gamma \le 1+|a|$. Therefore,

$$\begin{aligned} s_{\mathbb {B}^n\backslash \{0\}}(x,y)&\le (1+|a|)\frac{|\sigma _{a}(x)-\sigma _{a}(y)|}{|\sigma _{a}(x)-a|+|\sigma _{a}(y)-a|}\\&\le (1+|a|)s_{\mathbb {B}^n\backslash \{a\}}(\sigma _{a}(x),\sigma _{a}(y))\\&\le \frac{1+|a|}{1-|a|}s_{\mathbb {B}^n\backslash \{a\}}(\sigma _{a}(x),\sigma _{a}(y)). \end{aligned}$$

$\square$

5 Quasiconformality of a Bilipschitz Mapping in Ibragimov’s Metric

Bilipschitz mappings with respect to the triangular ratio metric have been studied in [10]. In this section, we use the comparison inequality between the triangular ratio metric and Ibragimov’s metric to investigate the quasiconformality of bilipschitz mappings in Ibragimov’s metric.

Theorem 5.1

[10, Theorem 4.4] Let $G\subsetneq {\mathbb {R}^n}$ be a domain and let $f:G\rightarrow fG\subset {\mathbb {R}^n}$ be a sense-preserving homeomorphism, satisfying L-bilipschitz condition with respect to the triangular ratio metric, i.e.,

$$\begin{aligned} s_G(x,y)/L\le s_{fG}(f(x),f(y))\le L s_G(x,y), \end{aligned}$$

holds for all $x, y\in G.$ Then, f is quasiconformal with the linear dilatation $H(f)\le L^2.$

Lemma 5.2

[22, Theorem 3.10] Let $D\subsetneq \mathbb {R}^n$. For $x,y\in D$,

$$\begin{aligned} (2\log 3)s_{D}(x,y)\le u_{D}(x,y)\le 3\log \frac{1+s_{D}(x,y)}{1-s_{D}(x,y)}, \end{aligned}$$

and the inequalities are sharp.

Theorem 5.3

Let $D\subsetneq \mathbb {R}^n$ be a domain and $f:D\rightarrow fD\subset \mathbb {R}^n$ is a sense-preserving homeomorphism satisfying the L-bilipschitz condition in Ibragimov’s metric

$$\begin{aligned} \frac{1}{L}u_D(x,y)\le u_{f(D)}(f(x),f(y))\le L u_{D}(x,y). \end{aligned}$$

(5.1)

Then, f is a quasiconformal mapping with the linear dilatation $H(f)\le \frac{9L^2}{\log ^2 3}$.

Proof

Since, by Lemma 5.2,

$$\begin{aligned} u_{D}(x,y)\le 3\log \frac{1+s_{D}(x,y)}{1-s_{D}(x,y)}=3\log \left( 1+\frac{2s_{D}(x,y)}{1-s_{D}(x,y)}\right) , \end{aligned}$$

for arbitrary $\varepsilon >0$, there exists $\delta >0$ such that for $x,y\in D$ satisfying $s_{D}(x,y)<\delta$, we have

$$\begin{aligned} (2\log 3)s_{D}(x,y)\le u_{D}(x,y)\le 6(1+\varepsilon )s_{D}(x,y). \end{aligned}$$

By Lemma 5.2, we have

$$\begin{aligned} s_{fD}(f(x),f(y))&\le \frac{1}{2\log 3}u_{fD}(f(x),f(y))\\&\le \frac{L}{2\log 3}u_{D}(x,y)\\&\le \frac{3(1+\epsilon )L}{\log 3}s_{D}(x,y). \end{aligned}$$

Similarly,

$$\begin{aligned} s_{fD}((f(x),f(y))&\ge \frac{1}{6(1+\epsilon )}u_{fD}(f(x),f(y))\\&\ge \frac{1}{6L(1+\epsilon )}u_{D}(x,y)\\&\ge \frac{\log 3}{3(1+\epsilon )L}s_{D}(x,y). \end{aligned}$$

Therefore, an L-bilipschitz mapping under Ibragimov’s metric is a $\frac{3L(1+\varepsilon )}{\log 3}$-bilipschitz mapping under the triangle ratio metric. It follows from Theorem 5.1 that f is a quasiconformal mapping with the linear dilatation $H(f)\le (\frac{3L(1+\varepsilon )}{\log 3})^2$. Let $\varepsilon \rightarrow 0$, we obtain the desired linear dilatation. $\square$