Functional Inequalities for Metric-Preserving Functions with Respect to Intrinsic Metrics of Hyperbolic Type

Abstract

We obtain functional inequalities for functions which are metric-preserving with respect to one of the following intrinsic metrics in a canonical plane domain: hyperbolic metric or some restrictions of the triangular ratio metric, respectively, of a Barrlund metric. The subadditivity turns out to be an essential property, being possessed by every function that is metric-preserving with respect to the hyperbolic metric and also by the composition with some specific function of every function that is metric-preserving with respect to some restriction of the triangular ratio metric or of a Barrlund metric. We partially answer an open question, proving that the hyperbolic arctangent is metric-preserving with respect to the restrictions of the triangular ratio metric on the unit disk to radial segments and to circles centered at origin.

1. Introduction

Metric-preserving functions have been studied in general topology from a theoretical point of view and have applications in fixed point theory [1,2], as well as in metric geometry to construct new metrics from known metrics, as the metrics $\frac{d}{1+d}$, $log\left(1+d\right)$ and the $\alpha -$ snowlake $d^{\alpha }$, $\alpha \in \left(0,1\right)$ associated to each metric d [3,4,5]. The theory of metric-preserving functions, that can be traced back to Wilson and Blumenthal, has been developed by Borsík, Doboš, Piotrowski, Vallin [6,7,8,9] and others, being recently generalized to semimetric spaces and quasimetric spaces [10] (see also [11,12,13]). As we will show below, there is a strong connection between metric-preserving functions and subadditive functions. The theory of subadditive function is well-developed [14,15], the functional inequality corresponding to subadditivity being viewed as a natural counterpart of Cauchy functional equation [16,17].

Given a function $f: \left[0,\infty \right) \to \left[0,\infty \right)$, it is said that f is metric-preserving if for every metric space $\left(X,d\right)$ the function $f\circ d$ is also a metric on X, i.e., f transfers every metric to a metric The function $f: \left[0,\infty \right) \to \left[0,\infty \right)$ is called amenable if $f^{-1}\left(\left\{0\right\}\right)=\left\{0\right\}$. If there exists some metric space $\left(X,d\right)$ such that the function $f\circ d$ is also a metric on X, then $f: \left[0,\infty \right) \to \left[0,\infty \right)$ is amenable. The symmetry axiom of a metric is obviously satisfied by $f\circ d$ whenever d is a metric. Given f amenable, f is metric-preserving if and only if $f\circ d$ satisfies triangle inequality whenever d is a metric.

Each of the following properties is known to be a sufficient condition for an amenable function to be metric-preserving [10,11]:

• f is concave;
• f is nondecreasing and subadditive;
• f is tightly bounded (that is, there exists $a>0$ such that $f\left(x\right)\in \left[a,2a\right]$ for every $x>0$).

For instance, every function $f: \left[0,\infty \right) \to \left[0,\infty \right)$ with $f\left(0\right)=0$ for which $\frac{f\left(t\right)}{t}$ is nonincreasing on $\left(0,\infty \right)$ is subadditive. In particular, if $f: \left[0,\infty \right) \to \left[0,\infty \right)$ with $f\left(0\right)=0$ is concave, then f is nondecreasing [18] and Jensen inequality shows that $\frac{f\left(t\right)}{t}$ is nonincreasing on $\left(0,\infty \right)$; hence f is nondecreasing and subadditive.

One proves that every metric-preserving function $f:\left[0,\infty \right)\to \left[0,\infty \right)$ is subadditive, using a particular choice of the metric d, e. g. the usual metric on $\mathbb{R}$. However, a subadditive amenable function $f:\left[0,\infty \right)\to \left[0,\infty \right)$ need not be metric-preserving, as in the case of $f\left(t\right)=\frac{t}{1+t^2}$ [11]. Recall that a function $f:\left[0,\infty \right)\to \left[0,\infty \right)$ which is convex and vanishes at the origin is subadditive if and only if f is linear ([11] Theorem 3.5).

We are interested in the following problem: given a particular metric d on a subset A of the complex plane, find necessary conditions satisfied by amenable functions $f:\left[0,\infty \right)\to \left[0,\infty \right)$ for which $f\circ d$ is a metric. In other terms, we look for solutions of the functional inequality

$$f\left(d(x,z)\right)\le f\left(d(x,y)\right)+f\left(d(y,z)\right)\mathrm{for}\mathrm{all}x,y,z\in A.$$

If we can find for every $a,b\in \left[0,\infty \right)$ some points $x,y,z\in A$ such that $d(x,y)=a$, $d(y,z)=b$ and $d(x,z)=a+b$, then f is subadditive on $\left[0,\infty \right)$. For some metrics d it could be difficult or impossible to find such points.

We will consider the cases where d is a hyperbolic metric, a triangular ratio metric or some other Barrlund metric. Recall that all these metrics belong to the class of intrinsic metrics, which is recurrent in the study of quasiconformal mappings [4].

The hyperbolic metric ${\rho }_{\mathbb{D}}$ on the unit disk $\mathbb{D}$ is given by

$$tanh\frac{{\rho }_{\mathbb{D}}\left(x,y\right)}{2}=\frac{\left|x-y\right|}{\left|1-\overline{x}y\right|},$$

that is, ${\rho }_{\mathbb{D}}\left(x,y\right)=2\mathrm{arctanh}{p}_{\mathbb{D}}\left(x,y\right)$, where $p_{\mathbb{D}}\left(x,y\right)=\frac{\left|x-y\right|}{\left|1-\overline{x}y\right|}$ is the pseudo-hyperbolic distance and we denoted by $\mathrm{arctanh}$ the inverse of the hyperbolic tangent tanh [19]. The hyperbolic metric ${\rho }_{\mathbb{H}}$ on the upper half plane $\mathbb{H}$ is given by

$$tanh\frac{{\rho }_{\mathbb{H}}\left(x,y\right)}{2}=\frac{\left|x-y\right|}{\left|x-\overline{y}\right|}.$$

For every simply-connected proper subdomain $\Omega$ of $\mathbb{C}$ one defines, via Riemann mapping theorem, the hyperbolic metric ${\rho }_{\Omega }$ on $\Omega$. We prove that, given $f:\left[0,\infty \right)\to \left[0,\infty \right)$, if $f\circ {\rho }_{\Omega }$ is a metric on $\Omega$, then f is subadditive. In the other direction, if $f:\left[0,\infty \right)\to \left[0,\infty \right)$ is amenable, nondecreasing and subadditive, then $f\circ {\rho }_{\Omega }$ is a metric on $\Omega$.

The triangular ratio metric $s_G$ of a given proper subdomain $G\subset \mathbb{C}$ is defined as follows for $x,y\in G$ [20]

$$s_G(x,y)=\underset{z\in \partial G}{sup}\frac{|x-y|}{|x-z|+|z-y|}.$$

(1)

For the triangular ratio metric $s_{\mathbb{H}}$ on the half-plane, it is known that $s_{\mathbb{H}}\left(x,y\right)=tanh\frac{{\rho }_{\mathbb{H}}\left(x,y\right)}{2}$ for all $x,y\in \mathbb{H}$. If $F:\left[0,1\right)\to \left[0,\infty \right)$ and $F\circ s_{\mathbb{H}}$ is a metric on the upper half-plane $\mathbb{H}$, we show that $F\circ tanh$ is subadditive on $\left[0,\infty \right)$.

The triangular ratio metric $s_{\mathbb{D}}\left(x,y\right)$ on the unit disk can be computed analytically as $s_{\mathbb{D}}\left(x,y\right)=\frac{|x-y|}{|x-z_0|+|z_0-y|}$, where $z_0\in \partial \mathbb{D}$ is the root of the algebraic equation

$$\overline{xy}{z}^4-\left(\overline{x}+\overline{y}\right)z^3+\left(x+y\right)z-xy=0$$

for which $|x-z|+|z-y|$ has the least value [21]. However, a simple explicit formula for $s_{\mathbb{D}}\left(x,y\right)$ is not available in general.

As $\mathrm{arctanh}{s}_{\mathbb{H}}$ is a metric on the upper half-plane $\mathbb{H}$, it is natural to ask if $\mathrm{arctanh}{s}_{\mathbb{D}}$ is a metric on the unit disk $\mathbb{D}$. The answer is unknown, but we prove that some restrictions of $\mathrm{arctanh}{s}_{\mathbb{D}}$ are metrics, namely the restriction to each radial segment of the unit disk and the restriction to each circle $\left|z\right|=\rho <1$. Given $f:\left[0,1\right)\to \left[0,\infty \right)$ such that the restriction of $f\circ s_{\mathbb{D}}$ to some radial segment of the unit disk $\mathbb{D}$ is a metric, we prove that $f\circ tanh$ is subadditive on $\left[0,\infty \right)$. If a continuous amenable function $F:\left[0,1\right)\to \left[0,\infty \right)$ is metric-preserving with respect to the restriction of the triangular ratio metric $s_{\mathbb{D}}$ to every circle $\left|z\right|=r<1$, $r\in \left(0,1\right)$, we prove that for every $a,b\in \left[0,\infty \right)$ we have

$$F\left(\frac{sinh\left(a\right)+sinh\left(b\right)}{\sqrt{1+{\left(sinh\left(a\right)+sinh\left(b\right)\right)}^2}}\right)\le F\left(tanh\left(a\right)\right)+F\left(tanh\left(b\right)\right).$$

For a proper subdomain $G\subset {\mathbb{R}}^n,$ for a number $p\ge 1,$ and for points $x,y\in G,$ let

$$b_{G,p}(x,y)=\underset{z\in \partial G}{sup}\frac{|x-y|}{\sqrt[p]{{|x-z|}^p+{|z-y|}^p}}.$$

(2)

The above formula defines a metric, as shown by A. Barrlund [22] for $G={\mathbb{R}}^n\backslash \left\{0\right\}$ and by P. Hästö [5] in the general case. This metric is called a Barrlund metric and is studied in [23]. In addition, the limit case $p=\infty$ is considered and it is shown that the formula

$$b_{G,\infty }(x,y)=\underset{z\in \partial G}{sup}\frac{|x-y|}{max\left(|x-z|,|z-y|\right)},x,y\in G$$

defines a metric. Note that $b_{G,p}$ is invariant to similarities for every $p\in \left[1,\infty \right]$ and that for $p=1$ the Barrlund metric coincides with the triangular ratio metric. We will consider Barrlund metrics with $p=2$ on canonical domains in plane, the upper half plane and the unit disk, that have explicit formulas [23]. For $G\in \left\{\mathbb{H},\mathbb{D}\right\}$, assuming that $F\circ b_{G,2}$ is a metric on some subset $A\subset G$ which is a ray, a line or a circle, we obtain a functional inequality satisfied by F, under the form of the subadditivity of a composition $F\circ \phi$, where the function $\phi$ depends only on A.

2. The Case of Hyperbolic Metrics

Let $\mathbb{D}$ be the unit disk with the hyperbolic metric ${\rho }_{\mathbb{D}}$.

Let $f:\left[0,\infty \right)\to \left[0,\infty \right)$ be amenable. If f is subadditive and nondecreasing, then $f\circ {\rho }_{\mathbb{D}}$ is a metric on the unit disk $\mathbb{D}$.

Proposition 1.

If $f:\left[0,\infty \right)\to \left[0,\infty \right)$ and $f\circ {\rho }_{\mathbb{D}}$ is a metric on the unit disk $\mathbb{D}$, then f is subadditive.

Proof. 

Let $f:\left[0,\infty \right)\to \left[0,\infty \right)$ such that $f\circ {\rho }_{\mathbb{D}}$ is a metric on the unit disk $\mathbb{D}$.

Denote $g\left(t\right)=f\left(2\mathrm{arctanh}\left(t\right)\right)$, $t\in \left[0,\infty \right)$. Then $g:\left[0,\infty \right)\to \left[0,\infty \right)$ satisfies

$$g\left(p_{\mathbb{D}}\left(x,y\right)\right)\le g\left(p_{\mathbb{D}}\left(x,z\right)\right)+g\left(p_{\mathbb{D}}\left(z,y\right)\right)\mathrm{for}\mathrm{all}x,y,z\in \mathbb{D}.$$

Since the pseudo-hyperbolic metric is invariant to the Möbius automomorphisms of the unit disk, it suffices to consider the case $z=0$. In conclusion, $g\circ p_{\mathbb{D}}$ satisfies the triangle inequality if and only if

$$g\left(p_{\mathbb{D}}\left(x,y\right)\right)\le g\left(p_{\mathbb{D}}\left(x,0\right)\right)+g\left(p_{\mathbb{D}}\left(0,y\right)\right)\mathrm{for}\mathrm{all}x,y\in \mathbb{D}.$$

(3)

However, $p_{\mathbb{D}}\left(x,0\right)=\left|x\right|$, $p_{\mathbb{D}}\left(0,y\right)=\left|y\right|$; hence, $g\left(p_{\mathbb{D}}\left(x,0\right)\right)+g\left(p_{\mathbb{D}}\left(0,y\right)\right)=g\left(\left|x\right|\right)+g\left(\left|y\right|\right)$.

On the other hand, $p_{\mathbb{D}}{\left(x,y\right)}^2=\frac{{\left|x-y\right|}^2}{{\left|x-y\right|}^2+\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)}=\frac{{\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }{1+{\left|x\right|}^2·{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }$, where $\alpha :=\measuredangle \left(x0y\right)$.

Let $x,y\in \mathbb{D}$. Since

$$\frac{d}{d\alpha }\left(\frac{{\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }{1+{\left|x\right|}^2·{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }\right)=\frac{2\left|x\right|\left|y\right|\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)}{{\left(1+{\left|x\right|}^2·{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha \right)}^2}sin\alpha ,$$

the function $\alpha \mapsto \frac{{\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }{1+{\left|x\right|}^2·{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }$ is nonincreasing on $[-\pi ,0]$ and nondecreasing on $[0,\pi ]$.

Then $\alpha \mapsto \frac{{\left|x\right|}^2+{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }{1+{\left|x\right|}^2·{\left|y\right|}^2-2\left|x\right|·\left|y\right|cos\alpha }$ attains its maximum if and only if $cos\alpha =-1$, in which case $p_{\mathbb{D}}\left(x,y\right)=\frac{\left|x\right|+\left|y\right|}{1+\left|x\right|\left|y\right|}$.

If (3) holds, then

$$g\left(\frac{r+s}{1+rs}\right)\le g\left(r\right)+g\left(s\right)\mathrm{for}\mathrm{all}r,s\in \left[0,1\right).$$

(4)

Conversely, if (4) holds and g is nondecreasing, then (3) is satisfied.

Let $\phi :\mathbb{C}\backslash \left\{-1\right\}\to \mathbb{C}\backslash \left\{-1\right\}$, $\phi \left(w\right)=\frac{1-w}{1+w}$. Then ${\phi }^{-1}=\phi$ and $\phi \left(\frac{z_1+z_2}{1+z_1{z}_2}\right)=\phi \left(z_1\right)\phi \left(z_2\right)$ for every $z_1,z_2\in \mathbb{C}\backslash \left\{-1\right\}$ with $z_1{z}_2\in \mathbb{C}\backslash \left\{-1\right\}$. Note that $\phi \left(\left[0,1\right)\right)=(0,1]$.

Denoting $\rho =\phi \left(r\right)$ and $\sigma =\phi \left(s\right)$, (4) is equivalent to

$$\left(g\circ \phi \right)\left(\rho ·\sigma \right)\le \left(g\circ \phi \right)\left(\rho \right)+\left(g\circ \phi \right)\left(\sigma \right)\mathrm{for}\mathrm{all}\rho ,\sigma \in (0,1].$$

Now, denoting $ln\rho =-u$ and $ln\sigma =-v$, we see that the above condition reduces to

$$\left(g\circ \phi \circ exp\right)\left(-u-v\right)\le \left(g\circ \phi \circ exp\right)\left(-u\right)+\left(g\circ \phi \circ exp\right)\left(-v\right)\mathrm{for}\mathrm{all}u,v\in \left[0,\infty \right).$$

Denote $h\left(t\right):=\left(g\circ \phi \circ exp\right)\left(-t\right)$, $t\in \left[0,\infty \right)$. Condition (4) holds if and only if h is subadditive on $\left[0,\infty \right)$.

We note that $\left(\phi \circ exp\right)\left(-t\right)=tanh\frac{t}{2}$ for $t\in \left[0,\infty \right)$. Then for all $t\in \left[0,\infty \right)$ we have $h\left(t\right)=g\left(tanh\frac{t}{2}\right)=f\left(t\right)$.

We proved that f is subadditive. □

Remark 1.

The functional equation associated to the inequality (4) $g\left(\frac{r+s}{1+rs}\right)=g\left(r\right)+g\left(s\right)$, $r,s\in \left[0,1\right)$ reduces via the substitution $h\left(t\right)=g\left(tanh\frac{t}{2}\right))$ to the Cauchy equation $h\left(u+v\right)=h\left(u\right)+h\left(v\right)$, $u,v\in \left[0,\infty \right)$. Extending h to an odd function, we may assume that h is additive on $\mathbb{R}$. If g is bounded on one side on a set of positive Lebesgue measure, then h is linear [16]; hence, there exists some positive constant c such that $g\left(t\right)=c\mathrm{arctanh}\left(t\right)$, $t\in \left[0,\infty \right)$.

Let $\mathbb{H}$ be the upper half-plane with the hyperbolic metric ${\rho }_{\mathbb{H}}$. We are interested in the amenable functions $f:\left[0,\infty \right)\to \left[0,\infty \right)$ for which $f\circ {\rho }_{\mathbb{H}}$ is a metric on $\mathbb{H}$. Consider the Cayley transform $T:\mathbb{H}\to \mathbb{D}$, $T\left(z\right)=\frac{z-i}{z+i}$, which is a bijective conformal map. Noting that for all $x,y\in \mathbb{H}$ we have ${\rho }_{\mathbb{H}}\left(x,y\right)={\rho }_{\mathbb{D}}\left(T\left(x\right),T\left(y\right)\right)$, it follows that $f\circ {\rho }_{\mathbb{H}}$ is a metric on $\mathbb{H}$ if and only if $f\circ {\rho }_{\mathbb{D}}$ is a metric on $\mathbb{D}$. From Proposition 1 we get the following

Corollary 1.

If $f:\left[0,\infty \right)\to \left[0,\infty \right)$ is amenable and $f\circ {\rho }_{\mathbb{H}}$ is a metric on upper half-plane $\mathbb{H}$, then f is subadditive.

More generally, for every proper simply-connected subdomain $\Omega$ of $\mathbb{C}$ there exists, by Riemann mapping theorem, a conformal mapping $T_{\Omega }:\Omega \to \mathbb{D}$. The hyperbolic metric ${\rho }_{\Omega }$ on $\Omega$ is defined by ${\rho }_{\Omega }\left(x,y\right)={\rho }_{\mathbb{D}}\left(T_{\Omega }\left(x\right),T_{\Omega }\left(y\right)\right)$.

Clearly, $f\circ {\rho }_{\Omega }$ is a metric on $\Omega$ if and only if $f\circ {\rho }_{\mathbb{D}}$ is a metric on $\mathbb{D}$. Now Proposition 1 leads to following generalization of itself.

Theorem 1.

Let Ω be a proper simply-connected subdomain of $\mathbb{C}$ and ${\rho }_{\Omega }$ be the hyperbolic metric on Ω. If $f:\left[0,\infty \right)\to \left[0,\infty \right)$ and $f\circ {\rho }_{\Omega }$ is a metric on Ω, then f is subadditive.

Corollary 2.

Let Ω be a proper simply-connected subdomain of $\mathbb{C}$ and ${\rho }_{\Omega }$ be the hyperbolic metric on Ω. Let $f:\left[0,\infty \right)\to \left[0,\infty \right)$ amenable and nondecreasing. Then $f\circ {\rho }_{\Omega }$ is a metric on Ω if and only if f is subadditive on $\left[0,\infty \right)$.

3. The Case of Unbounded Geodesic Metric Spaces

We can give another proof of Theorem 1, based on geometric arguments in geodesic metric spaces. The main idea is that in a geodesic metric space the distance is additive along geodesics.

A topological curve $\gamma :I\to X$ in a metric space $\left(X,d\right)$, where $I\subset \mathbb{R}$ is an interval, is called a geodesic if $L\left({\left.\gamma \right|}_{\left[t_1,t_2\right]}\right)=d\left(\gamma \left(t_1\right),\gamma \left(t_2\right)\right)$ for every subinterval $\left[t_1,t_2\right]\subset I$, i.e., the length of every arc of the geodesic is equal to the distance between the endpoints of the arc. A metric space is called a geodesic metric space if every pair of its points can be joined by a geodesic path.

Lemma 1.

In a geodesic metric space $\left(X,d\right)$ that is unbounded, for every positive numbers a and b there exists some points $x,y,z\in X$ such that $d\left(x,y\right)=a$, $d\left(y,z\right)=b$ and $d\left(x,z\right)=a+b$.

Proof. 

Let $a,b$ be positive numbers. Fix an arbitrary point $x\in X$. As $\left(X,d\right)$ is unbounded, there exists a point $w\in X$ such that $d(x,w)>a+b$. As $\left(X,d\right)$ is a geodesic metric space, there exists a geodesic path joining x and w in X. We may assume that this path is parameterized by arc-length, let us denote it by $\gamma :\left[0,L\right]\to X$, where $L=L\left(\gamma \right)=d\left(x,w\right)$. Then the length of the restriction of $\gamma$ to $\left[0,t\right]$ is $L\left({\left.\gamma \right|}_{\left[0,t\right]}\right)=t$, where $t\in \left[0,L\right]$. Since $L=d\left(x,w\right)>a+b$, we may consider $\gamma \left(a\right)=:y$ and $\gamma \left(a+b\right)=:z$. By the definition of a geodesic curve, $d\left(x,y\right)=L\left({\left.\gamma \right|}_{\left[0,a\right]}\right)=a$, $d\left(x,z\right)=L\left({\left.\gamma \right|}_{\left[0,a+b\right]}\right)=a+b$ and $d\left(x,y\right)=L\left({\left.\gamma \right|}_{\left[a,a+b\right]}\right)=b$. □

Proposition 2.

If the geodesic metric space $\left(X,d\right)$ is unbounded, then every function $f:\left[0,\infty \right)\to \left[0,\infty \right)$ which is metric-preserving with respect to d must be subadditive on $\left[0,\infty \right)$.

Proof. 

Let $\left(X,d\right)$ be a geodesic metric space that is unbounded. Assume that $f:\left[0,\infty \right)\to \left[0,\infty \right)$ is metric-preserving with respect to d.

We have to prove that $f\left(a+b\right)-\left[f\left(a\right)+f\left(b\right)\right]\le 0$ for all nonnegative numbers a and b. If $a=0$ and $b=0$ this inequality is obvious. Let a and b be positive numbers. By Lemma 1, there exists some points $x,y,z\in X$ such that $d\left(x,y\right)=a$, $d\left(y,z\right)=b$ and $d\left(x,z\right)=a+b$. Then $f\left(a+b\right)-\left[f\left(a\right)+f\left(b\right)\right]=\left(f\circ d\right)(x,z)-\left[\left(f\circ d\right)(x,y)+\left(f\circ d\right)(y,z)\right]$$\le 0$, since $f\circ d$ satisfies triangle inequality. □

Remark 2.

The metric space $\left(\Omega ,{\rho }_{\Omega }\right)$, where Ω is a proper simply-connected subdomain of $\mathbb{C}$ and ${\rho }_{\Omega }$ is the hyperbolic metric on Ω, is a geodesic metric space and is unbounded. By Proposition 2 we get another proof for Theorem 1. Note that the pseudo-hyperbolic distance on Ω is not additive along geodesics of $\left(\Omega ,{\rho }_{\Omega }\right)$ [19].

4. The Case of Triangular Ratio Metric on a Canonical Plane Domain

The triangular ratio metric $s_G$ of a given proper subdomain $G\subset {\mathbb{R}}^n$ is defined as follows for $z_1,z_2\in G$ [20]

$$s_G(z_1,z_2)=\underset{z\in \partial G}{sup}\frac{|z_1-z_2|}{|z_1-z|+|z-z_2|}.$$

(5)

Note that $s_G(z_1,z_2)\le 1$ for all $z_1,z_2\in G$. If no segment joining two points in G intersects the boundary $\partial G$, as it is the case if G is convex, then $s_G(z_1,z_2)<1$ for all $z_1,z_2\in G$.

We have $s_G(z_1,z_2)=\frac{|z_1-z_2|}{\underset{z\in \partial G}{inf}\left(|z_1-z|+|z-z_2|\right)}$ and the infimum $\underset{z\in \partial G}{inf}\left(|z_1-z|+|z-z_2|\right)$ is always attained. A point $u\in \partial G$ is called a Ptolemy–Alhazen point for $z_1,z_2\in G$ if a light ray from $z_1$ is reflected at u on the circle, such that the reflected ray goes through the point $z_2$. Every point at which $\underset{z\in \partial G}{inf}\left(|z_1-z|+|z-z_2|\right)$ is attained is a Ptolemy–Alhazen point for $z_1,z_2\in G$.

For a subset A of a convex domain G we may study the functional inequality

$$F\left(s_G\left(x,y\right)\right)\le F\left(s_G\left(x,z\right)\right)+F\left(s_{{}_G}\left(z,y\right)\right)\mathrm{for}\mathrm{all}x,y,z\in A,$$

(6)

where $F:\left[0,1\right)\to \left[0,\infty \right)$.

4.1. The Triangular Ratio Metric on the Upper Half-Plane

Recall that $s_{\mathbb{H}}\left(x,y\right)=tanh\frac{{\rho }_{\mathbb{H}}\left(x,y\right)}{2}=\frac{\left|x-y\right|}{\left|1-\overline{x}y\right|}$ for all $x,y\in \mathbb{H}$, that is $s_{\mathbb{H}}$ coincides with the pseudo-hyperbolic distance $p_{\mathbb{H}}$.

Using the subadditivity of functions preserving the hyperbolic metric of the upper half-plane, we get the following

Proposition 3.

If $F:\left[0,1\right)\to \left[0,\infty \right)$ and F is metric-preserving with respect to $s_{\mathbb{H}}$ on the upper half-plane $\mathbb{H}$, then $F\circ tanh$ is subadditive on $\left[0,\infty \right)$.

Proof. 

$F\left(s_{\mathbb{H}}\left(x,y\right)\right)=F\left(tanh\frac{{\rho }_{\mathbb{H}}\left(x,y\right)}{2}\right)=G\left({\rho }_{\mathbb{H}}\left(x,y\right)\right)$ for all $x,y\in \mathbb{H}$, where we denoted $G\left(t\right):=F\left(tanh\frac{t}{2}\right)$, $t\in \left[0,\infty \right)$.

Note that $F\circ s_{\mathbb{H}}$ is a metric on $\mathbb{H}$ if and only if $G\circ {\rho }_{\mathbb{H}}$ is a metric on $\mathbb{H}$.

If $F\circ s_{\mathbb{H}}$ is a metric on $\mathbb{H}$, by Corollary 1, it follows that G is subadditive on $\left[0,\infty \right)$. Then $H=F\circ tanh$ is also subadditive on $\left[0,\infty \right)$, as $H\left(t\right)=G\left(2t\right)$ for all $t\in \left[0,\infty \right)$. □

Remark 3.

Note that tanh maps $\left[0,\infty \right)$ onto $\left[0,1\right)$ and is subadditive on $\left[0,\infty \right)$. If $F:\left[0,1\right)\to \left[0,\infty \right)$ is subadditive on $\left[0,1\right)$, then $F\circ tanh$ is subadditive on $\left[0,\infty \right)$. The converse is not true, as it is shown in the case $F\left(t\right)=\mathrm{arctanh}\left(t\right)$, $t\in \left[0,1\right)$. Note that

$$\mathrm{arctanh}\left(a\right)+\mathrm{arctanh}\left(b\right)=\mathrm{arctanh}\left(\frac{a+b}{1+ab}\right)\le \mathrm{arctanh}\left(a+b\right)\mathrm{for}\mathrm{all}a,b\in \left[0,1\right).$$

4.2. The Triangular Ratio Metric on the Unit Disk

There is an open conjecture stating that $\mathrm{arctanh}{s}_{\mathbb{D}}$ is a metric on the unit disk [23]. Note that $\mathrm{arctanh}{s}_{\mathbb{D}}$ is a metric on $A\subset \mathbb{D}$ if and only if

$$s_{\mathbb{D}}\left(x,z\right)\le \frac{s_{\mathbb{D}}\left(x,y\right)+s_{\mathbb{D}}\left(y,z\right)}{1+s_{\mathbb{D}}\left(x,y\right)s_{\mathbb{D}}\left(y,z\right)}\mathrm{for}\mathrm{all}x,y,z\in A.$$

(7)

We will prove that $\mathrm{arctanh}$ is metric-preserving with respect to the restrictions of the triangular ratio metric on the unit disk to radial segments, respectively, to circles centered at the origin.

Lemma 2

([23] Theorem 2.2). For $x,y\in \mathbb{D}$, $s_{\mathbb{D}}\left(x,y\right)\le \frac{\left|x-y\right|}{2-\left|x+y\right|}$. Equality holds if and only if $0,x$ and y are collinear.

Lemma 3.

The following addition formula holds: if $-1<r\le s\le t<1$, then

$$\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,t\right)=\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,s\right)+\mathrm{arctanh}{s}_{\mathbb{D}}\left(s,t\right).$$

The restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to each diameter of the unit disk is a metric.

Proof. 

Recall that $\mathrm{arctanh}\left(u\right)=\frac{1}{2}log\frac{1+u}{1-u}$ for all $u\in \left(-1,1\right)$. In particular, if $-1<u\le v<1$, then

$$\mathrm{arctanh}{s}_{\mathbb{D}}\left(u,v\right)=\frac{1}{2}log\frac{1+\frac{(v-u)}{2-\left(v+u\right)}}{1-\frac{(v-u)}{2-\left(v+u\right)}}=\frac{1}{2}log\frac{1-u}{1-v}$$

Let $-1<r\le s\le t<1$. Then

$$\begin{array}{ccc}\hfill \mathrm{arctanh}{s}_{\mathbb{D}}\left(r,s\right)+\mathrm{arctanh}{s}_{\mathbb{D}}\left(s,t\right)& =& \frac{1}{2}log\frac{1-r}{1-s}+\frac{1}{2}log\frac{1-s}{1-t}\hfill \\ & =& \frac{1}{2}log\frac{1-r}{1-t}=\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,t\right).\hfill \end{array}$$

Since $s_{\mathbb{D}}$ is invariant to rotations around the origin (more generally, $s_{\mathbb{D}}$ is invariant to similarities), it suffices to prove that the restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to the intersection of the unit disk with the real axis is a metric. This follows from the above addition formula, observing that $0\le max\left(\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,s\right),\mathrm{arctanh}{s}_{\mathbb{D}}\left(s,t\right)\right)\le \mathrm{arctanh}{s}_{\mathbb{D}}\left(r,t\right).$ □

We prove that the restriction of the triangular ratio metric of the unit disk $s_{\mathbb{D}}$ to each radial segment of the unit disk takes all values between 0 and 1.

Lemma 4.

For every $\lambda \in \left(0,1\right)$ and $r\in \left[0,\frac{1}{2}\right)$, there exists $s\in \left(r,\frac{1}{2}\right)$ such that $s_{\mathbb{D}}\left(r,s\right)=\lambda$.

Proof. 

Let $\lambda \in \left(0,1\right)$ and $r\in \left[0,\frac{1}{2}\right)$.

Assume that $s\in \left(r,1\right)$. Then $s_{\mathbb{D}}\left(r,s\right)=\lambda$ if and only if $\frac{s-r}{1-\left(s+r\right)}=\lambda$, i.e., $s=\frac{\left(1-\lambda \right)r+\lambda }{1+\lambda }$.

Note that $s=\frac{\left(1-\lambda \right)r+\lambda }{1+\lambda }$ implies $s-r=\frac{\lambda \left(1-2r\right)}{1+\lambda }>0$ and $s-\frac{1}{2}=\frac{(1-\lambda )\left(2r-1\right)}{2\left(1+\lambda \right)}<0$. □

Proposition 4.

Let $f:\left[0,1\right)\to \left[0,\infty \right)$.

(1) If the restriction of $f\circ s_{\mathbb{D}}$ to some radial segment of the unit disk $\mathbb{D}$ is a metric, then $f\circ tanh$ is subadditive on $\left[0,\infty \right)$.

(2) If f is amenable and $f\circ tanh$ is subadditive and nondecreasing on $\left[0,\infty \right)$, then the restriction of $f\circ s_{\mathbb{D}}$ to every diameter of the unit disk $\mathbb{D}$ is a metric.

Proof. 

(1) Let $F:=f\circ tanh:\left[0,\infty \right)\to \left[0,\infty \right)$.

As $s_{\mathbb{D}}$ is invariant to rotations around the origin, we may assume that the restriction of $f\circ s_{\mathbb{D}}=F\circ \mathrm{arctanh}{s}_{\mathbb{D}}$ to $\left\{x+iy:x\in \left[0,1\right),y=0\right\}$ is a metric.

Since $F\left(0\right)=0$, it suffices to prove that F is subadditive on $\left(0,\infty \right)$.

Let $a,b\in \left(0,\infty \right)$. We prove that $F\left(a+b\right)\le F\left(a\right)+F\left(b\right)$.

Denote $tanha=\lambda$ and $tanhb=\mu$.

Fix $r\in \left[0,\frac{1}{2}\right)$. By Lemma 4, there exists $s\in \left(r,\frac{1}{2}\right)$such that$s_{\mathbb{D}}\left(r,s\right)=\lambda .$ Applying again Lemma 4, we get $t\in \left(s,\frac{1}{2}\right)$such that$s_{\mathbb{D}}\left(s,t\right)=\mu$.

The addition formula $\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,t\right)=\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,s\right)+\mathrm{arctanh}{s}_{\mathbb{D}}\left(s,t\right)$ shows that $\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,t\right)=a+b$.

Since the restriction of $F\circ \mathrm{arctanh}{s}_{\mathbb{D}}$ to $\left\{x+iy:x\in \left[0,1\right),y=0\right\}$ is a metric,

$$F\left(\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,t\right)\right)\le F\left(\mathrm{arctanh}{s}_{\mathbb{D}}\left(r,s\right)\right)+F\left(\mathrm{arctanh}{s}_{\mathbb{D}}\left(s,t\right)\right),$$

i.e., $F\left(a+b\right)\le F\left(a\right)+F\left(b\right)$, q.e.d.

(2) The function $F:=f\circ tanh:\left[0,\infty \right)\to \left[0,\infty \right)$ is amenable, subadditive and nondecreasing self-mapping of $\left[0,\infty \right)$; therefore, F is metric-preserving. By Lemma 3, the restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to every diameter of the unit disk $\mathbb{D}$ is a metric. Then the restriction of $f\circ s_{\mathbb{D}}=F\circ \mathrm{arctanh}{s}_{\mathbb{D}}$ to every diameter of the unit disk $\mathbb{D}$ is a metric. □

Remark 4.

Let $f:\left[0,1\right)\to \left[0,\infty \right)$. If f is amenable, subadditive and nondecreasing on $\left[0,1\right)$, then $f\circ s_{\mathbb{D}}$ is a metric on the entire unit disk $\mathbb{D}$ and obviously $f\circ tanh$ is amenable, subadditive and nondecreasing on $\left[0,\infty \right)$.

We prove that each restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to a circle centered at origin and contained in the unit disk is a metric.

If $x,y\in \mathbb{D}$ with $\left|x\right|=\left|y\right|=\rho <1$ it is known from ([24] Remark 3.14) that, denoting ${\omega }_0:=2arccos\rho$ and $2\alpha :=\measuredangle (x,0,y)$, we have

$$s_{\mathbb{D}}(x,y)=\left\{\begin{array}{c}\rho ,\mathrm{if}2\alpha >{\omega }_0\\ \frac{\rho sin\alpha }{\sqrt{1+{\rho }^2-2\rho cos\alpha }},\mathrm{if}2\alpha \le {\omega }_0\end{array}\right..$$

(8)

If $2\alpha ={\omega }_0$, then $\frac{\rho sin\alpha }{\sqrt{1+{\rho }^2-2\rho cos\alpha }}=\frac{\rho \sqrt{1-{\rho }^2}}{\sqrt{1+{\rho }^2-2{\rho }^2}}=\rho$.

We will see that the restriction of the triangular ratio metric of the unit disk $s_{\mathbb{D}}$ to any circle $\left|z\right|=\rho <1$ takes all values from 0 and $\rho$.

Lemma 5.

Let $\rho \in \left(0,1\right)$. Denote ${\omega }_0:=2arccos\rho$ and let $f:\left[0,\frac{{\omega }_0}{2}\right]\to \mathbb{R}$ be defined by $f\left(\theta \right)=\frac{\rho sin\theta }{\sqrt{1+{\rho }^2-2\rho cos\theta }}$. Then:

(1) f is increasing on $\left[0,\frac{{\omega }_0}{2}\right]$ and $f^{\prime \prime }\left(\theta \right)<0$ for all $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$. In particular,

$f\left(\theta \right)\le f\left(\frac{{\omega }_0}{2}\right)=\rho$ for all $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$ and for every $\lambda \in \left[0,\rho \right]$ there exists an unique $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$ such that $f\left(\theta \right)=\lambda$.

(2) The function $h:\left[0,\frac{{\omega }_0}{2}\right]\to \mathbb{R}$ defined by $h\left(\theta \right):=\frac{f^{\prime }\left(\theta \right)}{1-f^2\left(\theta \right)}$ is decreasing on $\left[0,\frac{{\omega }_0}{2}\right]$.

(3) For every $k\in \left(0,{\omega }_0\right)$ the function $R_k\left(\theta \right):=\frac{f\left(\theta \right)+f\left(k-\theta \right)}{1+f\left(\theta \right)f\left(k-\theta \right)}$, with

$max\left(0,k-\frac{{\omega }_0}{2}\right)\le \theta \le min\left(k,\frac{{\omega }_0}{2}\right)$, is increasing on $\left[max\left(0,k-\frac{{\omega }_0}{2}\right),\frac{k}{2}\right]$ and decreasing on $\left[\frac{k}{2},min\left(k,\frac{{\omega }_0}{2}\right)\right].$

Proof. 

(1) We have

$$f^{\prime }\left(\theta \right)=\frac{\rho \left(cos\theta -\rho \right)\left(1-\rho cos\theta \right)}{{\left(1+{\rho }^2-2\rho cos\theta \right)}^{\frac{3}{2}}},\theta \in \left[0,\frac{{\omega }_0}{2}\right].$$

We note that $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$ implies $cos\theta -\rho \ge 0$; hence, $f^{\prime }\left(\theta \right)\ge 0$, with equality if and only if $\theta =\frac{{\omega }_0}{2}$. Then f is increasing on $\left[0,\frac{{\omega }_0}{2}\right]$; therefore, $f\left(\theta \right)\le f\left(\frac{{\omega }_0}{2}\right)$ for all $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$. Note that $f\left(\frac{{\omega }_0}{2}\right)=\rho$.

By the intermediate value property and the monotonicity of f, for every $\lambda \in \left[0,\rho \right]=\left[f\left(0\right),f\left(\frac{{\omega }_0}{2}\right)\right]$ there exists an unique $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$ such that $f\left(\theta \right)=\lambda$. Moreover, in this case $cos\theta =t\in \left[\rho ,1\right]$ is the unique root from $\left[\rho ,1\right]$ of the quadratic function $Q\left(t\right)={\rho }^2{t}^2-2{\lambda }^2\rho t+{\lambda }^2-{\rho }^2+{\lambda }^2{\rho }^2$.

Furthermore,

$$f^{\prime \prime }\left(\theta \right)=\frac{-\rho sin\theta }{{\left({\rho }^2-2\left(cos\theta \right)\rho +1\right)}^{\frac{5}{2}}}\left({\rho }^2{\left(\rho -cos\theta \right)}^2+\left(1-{\rho }^2\right)\left(1-\rho cos\theta \right)\right)\le 0$$

for all $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$, with equality if and only if $\theta =0.$

(2) By computation we get

$$h\left(\theta \right)=\frac{\rho \left(cos\theta -\rho \right)}{\left(1-\rho cos\theta \right)\sqrt{1+{\rho }^2-2\rho cos\theta }}\ge 0,$$

for all $\theta \in \left[0,\frac{{\omega }_0}{2}\right]$. Then ${\rho }^{-2}{\left(h\left(\theta \right)\right)}^2=H\left(cos\theta \right)$, where

$$H\left(t\right)=\frac{{\left(t-\rho \right)}^2}{{\left(1-\rho t\right)}^2\left(1+{\rho }^2-2\rho t\right)},t\in [\rho ,1].$$

We have

$$H^{\prime }\left(t\right)=\frac{2\left(t-\rho \right)}{{\left(1-\rho t\right)}^3{\left(1+{\rho }^2-2\rho t\right)}^2}\left[\left(1-{\rho }^2\right)\left(1+{\rho }^2-2\rho t\right)+\rho \left(\rho -t\right)\left(\rho t-1\right)\right].$$

Then, $H^{\prime }\left(t\right)\ge 0$ for all $t\in [\rho ,1)$, with $H^{\prime }\left(t\right)=0$ only for $t=\rho$, therefore, H is increasing on $[\rho ,1]$. This shows that h is decreasing on $\left[0,\frac{{\omega }_0}{2}\right]$.

(3) Let $k\in \left(0,{\omega }_0\right)$. Denote $a=max\left(0,k-\frac{{\omega }_0}{2}\right)$ and $b=min\left(k,\frac{{\omega }_0}{2}\right)$. Note that $a<b$ and $R_k\left(\theta \right)$ is well-defined on $\left[a,b\right]$. Moreover, $\frac{k}{2}\in \left(a,b\right)$. If $k\in \left(0,\frac{{\omega }_0}{2}\right]$, then $a=0$ and $b=k$. If $k\in \left(\frac{{\omega }_0}{2},{\omega }_0\right)$, then $a=k-\frac{{\omega }_0}{2}$ and $b=\frac{{\omega }_0}{2}$.

The derivative

$$\begin{array}{ccc}\hfill R_k^{\prime }\left(\theta \right)& =& \frac{\left(1-{\left(f\left(\theta \right)\right)}^2\right)\left(1-{\left(f\left(k-\theta \right)\right)}^2\right)}{{\left(1+f\left(\theta \right)f\left(k-\theta \right)\right)}^2}\left(\frac{f^{\prime }\left(\theta \right)}{1-{\left(f\left(\theta \right)\right)}^2}-\frac{f^{\prime }(k-\theta )}{1-{\left(f\left(k-\theta \right)\right)}^2}\right)\hfill \\ & =& \frac{\left(1-{\left(f\left(\theta \right)\right)}^2\right)\left(1-{\left(f\left(k-\theta \right)\right)}^2\right)}{{\left(1+f\left(\theta \right)f\left(k-\theta \right)\right)}^2}\left(h\left(\theta \right)-h(k-\theta )\right)\hfill \end{array}$$

vanishes if and only if $\theta =k-\theta$, i.e., $\theta =\frac{k}{2}$. Since h is decreasing on $\left[0,\frac{{\omega }_0}{2}\right]\supset \left[a,b\right]$, we have $R_k^{\prime }\left(\theta \right)>0$ for $\theta \in \left[a,\frac{k}{2}\right)$ and $R_k^{\prime }\left(\theta \right)<0$ for $\theta \in \left(\frac{k}{2},b\right]$; hence, $R_k$ is increasing on $\left[a,\frac{k}{2}\right]$ and decreasing on $\left[\frac{k}{2},b\right]$. □

Remark 5.

The maximum of $R_k$ on $\left[a,b\right]$ is $\frac{2f\left(\frac{k}{2}\right)}{1+f^2\left(\frac{k}{2}\right)}$. The minimum of $R_k$ on $\left[a,b\right]$ is

$$min\left(R_k\left(a\right),R_k\left(b\right)\right)=\left\{\begin{array}{c}f\left(k\right),\mathrm{if}k\in \left(0,\frac{{\omega }_0}{2}\right]\\ \frac{f\left(\frac{{\omega }_0}{2}\right)+f\left(k-\frac{{\omega }_0}{2}\right)}{1+f\left(\frac{{\omega }_0}{2}\right)f\left(k-\frac{{\omega }_0}{2}\right)},\mathrm{if}k\in \left(\frac{{\omega }_0}{2},{\omega }_0\right)\end{array}\right..$$

Theorem 2.

The restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to each circle $\left|z\right|=\rho <1$ is a metric.

Proof. 

Let $x,y,z\in \mathbb{D}$ with $\left|x\right|=\left|y\right|=\left|z\right|=\rho <1$.

Denote $2\alpha :=\measuredangle (x,0,y)$, $2\beta :=\measuredangle (y,0,z)$ and $2\gamma :=\measuredangle (z,0,x)$, where

$\alpha ,\beta ,\gamma \in (0,\pi /2]$. Set ${\tilde{\rho }}_{\alpha }:=s_{\mathbb{D}}(x,y)$, ${\tilde{\rho }}_{\beta }:=s_{\mathbb{D}}(y,z)$ and ${\tilde{\rho }}_{\gamma }:=s_{\mathbb{D}}(z,x)$. Without loss of generality, we may assume that the triangle $\Delta xyz$ is positively oriented.

Note that $max\left({\tilde{\rho }}_{\alpha },{\tilde{\rho }}_{\beta },{\tilde{\rho }}_{\gamma }\right)\le \rho <1$.

The expression

$$\begin{array}{c}\mathrm{arctanh}{s}_{\mathbb{D}}(x,y)+\mathrm{arctanh}{s}_{\mathbb{D}}(y,z)-\mathrm{arctanh}{s}_{\mathbb{D}}(x,z)=\\ {\displaystyle \frac{1}{2}}ln\left(1+{\displaystyle \frac{2\left(1+{\tilde{\rho }}_{\alpha }{\tilde{\rho }}_{\beta }\right)}{\left(1-{\tilde{\rho }}_{\alpha }\right)\left(1-{\tilde{\rho }}_{\beta }\right)\left(1+{\tilde{\rho }}_{\gamma }\right)}}\left({\displaystyle \frac{{\tilde{\rho }}_{\alpha }+{\tilde{\rho }}_{\beta }}{1+{\tilde{\rho }}_{\alpha }{\tilde{\rho }}_{\beta }}}-{\tilde{\rho }}_{\gamma }\right)\right)\end{array}$$

has the same sign as $E\left(\alpha ,\beta ,\gamma \right):={\tilde{\rho }}_{\alpha }+{\tilde{\rho }}_{\beta }-{\tilde{\rho }}_{\gamma }-{\tilde{\rho }}_{\alpha }{\tilde{\rho }}_{\beta }{\tilde{\rho }}_{\gamma }$.

By symmetry, $E\left(\alpha ,\beta ,\gamma \right)=E\left(\beta ,\alpha ,\gamma \right)$, $E\left(\alpha ,\gamma ,\beta \right)=E\left(\gamma ,\alpha ,\beta \right)$ and $E\left(\beta ,\gamma ,\alpha \right)=E\left(\gamma ,\beta ,\alpha \right)$.

We have to prove that $E\left(\alpha ,\beta ,\gamma \right)$, $E\left(\beta ,\gamma ,\alpha \right)$ and $E\left(\alpha ,\gamma ,\beta \right)$ are always nonnegative.

We will consider $f:\left[0,\frac{{\omega }_0}{2}\right]\to \mathbb{R}$ be defined by $f\left(\theta \right)=\frac{\rho sin\theta }{\sqrt{1+{\rho }^2-2\rho cos\theta }}$ as in the previous lemma.

Case 1. Assume that $2\left(\alpha +\beta \right)\le {\omega }_0$.

Obviously, $2\alpha \le {\omega }_0$, $2\beta \le {\omega }_0$ and $2\gamma =2\left(\alpha +\beta \right)\le {\omega }_0$; hence, ${\tilde{\rho }}_{\phi }=f\left(\phi \right)$ for $\phi \in \left\{\alpha ,\beta ,\gamma \right\}$.

We have $0<\alpha <\gamma \le \frac{{\omega }_0}{2}$ and $0<\beta <\gamma \le \frac{{\omega }_0}{2}$; therefore, ${\tilde{\rho }}_{\alpha }<{\tilde{\rho }}_{\gamma }$ and ${\tilde{\rho }}_{\beta }<{\tilde{\rho }}_{\gamma }$, since f is increasing on $\left[0,\frac{{\omega }_0}{2}\right]$.

Then $E\left(\alpha ,\gamma ,\beta \right)={\tilde{\rho }}_{\alpha }\left(1-{\tilde{\rho }}_{\beta }{\tilde{\rho }}_{\gamma }\right)+{\tilde{\rho }}_{\gamma }-{\tilde{\rho }}_{\beta }>0$ and $E\left(\beta ,\gamma ,\alpha \right)={\tilde{\rho }}_{\beta }\left(1-{\tilde{\rho }}_{\alpha }{\tilde{\rho }}_{\gamma }\right)+{\tilde{\rho }}_{\gamma }-{\tilde{\rho }}_{\alpha }>0.$

We may write

$$E\left(\alpha ,\beta ,\gamma \right)=\left(1+f\left(\alpha \right)f\left(\gamma -\alpha \right)\right)\left(\frac{f\left(\alpha \right)+f\left(\gamma -\alpha \right)}{1+f\left(\alpha \right)f\left(\gamma -\alpha \right)}-f\left(\gamma \right)\right),$$

where $0<\alpha <\gamma$.

Fix $\gamma \in \left[0,\frac{{\omega }_0}{2}\right]$. With the notation from Lemma 5 (3), we have

$$E\left(\alpha ,\beta ,\gamma \right)=\left(1+f\left(\alpha \right)f\left(\gamma -\alpha \right)\right)\left(R_{\gamma }\left(\alpha \right)-f\left(\gamma \right)\right).$$

Since $R_{\gamma }$ is increasing on $\left[0,\frac{\gamma }{2}\right]$ and decreasing on $\left[\frac{\gamma }{2},\gamma \right]$, we have $R_{\gamma }\left(\theta \right)>R_k\left(0\right)=R_k\left(\gamma \right)=f\left(\gamma \right)$ for all $\theta \in \left(0,\gamma \right)$.

Then $E\left(\alpha ,\beta ,\gamma \right)>0$.

Case 2. Assume that $2\alpha \le {\omega }_0$ and $2\beta \le {\omega }_0$ and $2\left(\alpha +\beta \right)>{\omega }_0$.

Subcase 2.1. Assume that $2\left(\alpha +\beta \right)\le \pi$.

Then $2\gamma =2\left(\alpha +\beta \right)>{\omega }_0$; hence, ${\tilde{\rho }}_{\gamma }=\rho \ge max\left\{{\tilde{\rho }}_{\alpha },{\tilde{\rho }}_{\beta }\right\}$.

We see that $E\left(\alpha ,\gamma ,\beta \right)={\tilde{\rho }}_{\alpha }\left(1-{\tilde{\rho }}_{\beta }{\tilde{\rho }}_{\gamma }\right)+{\tilde{\rho }}_{\gamma }-{\tilde{\rho }}_{\beta }>0$ and

$E\left(\beta ,\gamma ,\alpha \right)={\tilde{\rho }}_{\beta }\left(1-{\tilde{\rho }}_{\alpha }{\tilde{\rho }}_{\gamma }\right)+{\tilde{\rho }}_{\gamma }-{\tilde{\rho }}_{\alpha }>0$.   Now

$$\begin{array}{ccc}\hfill E\left(\alpha ,\beta ,\gamma \right)& =& E\left(\alpha ,\beta ,\frac{{\omega }_0}{2}\right)=f\left(\alpha \right)+f\left(\beta \right)-\rho -\rho f\left(\alpha \right)f\left(\beta \right)\hfill \\ & =& f\left(\beta \right)\left(1-\rho f\left(\alpha \right)\right)+f\left(\alpha \right)-\rho .\hfill \end{array}$$

Since f is increasing on $\left[0,\frac{{\omega }_0}{2}\right]$ and $\frac{{\omega }_0}{2}-\alpha <\beta \le \frac{{\omega }_0}{2}$, we have $E\left(\alpha ,\beta ,\frac{{\omega }_0}{2}\right)$$>E\left(\alpha ,\frac{{\omega }_0}{2}-\alpha ,\frac{{\omega }_0}{2}\right).$

Let ${\alpha }^{\prime }=\alpha$, ${\beta }^{\prime }=\frac{{\omega }_0}{2}-\alpha$ and ${\gamma }^{\prime }=\frac{{\omega }_0}{2}$. As $2\left({\alpha }^{\prime }+{\beta }^{\prime }\right)={\omega }_0$, according to Case 1 we have $E\left({\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime }\right)>0$, i.e., $E\left(\alpha ,\frac{{\omega }_0}{2}-\alpha ,\frac{{\omega }_0}{2}\right)>0.$

The latter two inequalities show that $E\left(\alpha ,\beta ,\gamma \right)>0$.

Subcase 2.2. Assume that $2\left(\alpha +\beta \right)>\pi$.

Then $2\gamma =2\pi -2\left(\alpha +\beta \right)$. We compare $2\gamma$ to ${\omega }_0$.

Subcase 2.2.1. Assume that $2\gamma \le {\omega }_0$.

Note that ${\omega }_0\ge \frac{2\pi }{3}$, that is $0<\rho \le \frac{1}{2}$.

For each $\phi \in \left\{\alpha ,\beta ,\gamma \right\}$ we have $2\phi \le {\omega }_0$; hence, ${\tilde{\rho }}_{\phi }=f\left(\phi \right)$.

Since $\beta =\pi -\gamma -\alpha$, we have $E\left(\alpha ,\beta ,\gamma \right)=\left(1+f\left(\alpha \right)f\left(\pi -\gamma -\alpha \right)\right)\left(R_{\pi -\gamma }\left(\alpha \right)-f\left(\gamma \right)\right)$, where $0<\pi -\gamma -\frac{{\omega }_0}{2}\le \alpha \le \frac{{\omega }_0}{2}\le \pi -\gamma$. We have $\frac{\pi -\gamma }{2}=\frac{\alpha +\beta }{2}\le \frac{{\omega }_0}{2}$ with equality only if $2\alpha =2\beta ={\omega }_0$. In addition, $\frac{\pi -\gamma }{2}\ge \pi -\gamma -\frac{{\omega }_0}{2}$, with equality only if $2\alpha =2\beta ={\omega }_0$.

If $2\alpha =2\beta ={\omega }_0$, then ${\tilde{\rho }}_{\alpha }={\tilde{\rho }}_{\beta }=\rho$; hence, $E\left(\alpha ,\beta ,\gamma \right)=2\rho -{\tilde{\rho }}_{\gamma }-{\rho }^2{\tilde{\rho }}_{\gamma }=\rho \left(1-\rho {\tilde{\rho }}_{\gamma }\right)+\rho -{\tilde{\rho }}_{\gamma }>0$

and $E\left(\alpha ,\gamma ,\beta \right)=E\left(\beta ,\gamma ,\alpha \right)=\rho +{\tilde{\rho }}_{\gamma }-\rho -{\rho }^2{\tilde{\rho }}_{\gamma }={\tilde{\rho }}_{\gamma }\left(1-{\rho }^2\right)>0.$

Letting aside the case $2\alpha =2\beta ={\omega }_0$, we get

$$0<\pi -\gamma -\frac{{\omega }_0}{2}<\frac{\pi -\gamma }{2}<\frac{{\omega }_0}{2}.$$

Fix $\gamma \in \left[0,\frac{{\omega }_0}{2}\right]$. From Lemma 5 (3), $R_{\pi -\gamma }$ is increasing on $\left[\pi -\gamma -\frac{{\omega }_0}{2},\frac{\pi -\gamma }{2}\right]$ and decreasing on $\left[\frac{\pi -\gamma }{2},\frac{{\omega }_0}{2}\right]$.

The minimum of $R_{\pi -\gamma }$ on $\left[\pi -\gamma -\frac{{\omega }_0}{2},\frac{{\omega }_0}{2}\right]$ is $\frac{f\left(\frac{{\omega }_0}{2}\right)+f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}{1+f\left(\frac{{\omega }_0}{2}\right)f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}$, attained at both endpoints of the interval.

It follows that $R_{\pi -\gamma }\left(\theta \right)-f\left(\gamma \right)\ge \frac{f\left(\frac{{\omega }_0}{2}\right)+f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}{1+f\left(\frac{{\omega }_0}{2}\right)f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}-f\left(\gamma \right)>f\left(\frac{{\omega }_0}{2}\right)-f\left(\gamma \right)>0$

for every $\theta \in \left[\pi -\gamma -\frac{{\omega }_0}{2},\frac{{\omega }_0}{2}\right]$; hence, $E\left(\alpha ,\beta ,\gamma \right)>0$.

Due to the symmetry of the assumptions $2\phi \le {\omega }_0$ for $\phi \in \left\{\alpha ,\beta ,\gamma \right\}$ and $\alpha +\beta +\gamma =\pi$, it follows similarly that $E\left(\alpha ,\gamma ,\beta \right)>0$ and $E\left(\beta ,\gamma ,\alpha \right)>0$.

Subcase 2.2.2. Assume that $2\gamma >{\omega }_0$. Then ${\tilde{\rho }}_{\gamma }=\rho$.

We have $E\left(\alpha ,\gamma ,\beta \right)={\tilde{\rho }}_{\alpha }+{\tilde{\rho }}_{\gamma }-{\tilde{\rho }}_{\beta }-{\tilde{\rho }}_{\alpha }{\tilde{\rho }}_{\beta }{\tilde{\rho }}_{\gamma }={\tilde{\rho }}_{\alpha }\left(1-\rho {\tilde{\rho }}_{\beta }\right)+\rho -{\tilde{\rho }}_{\beta }>0$ and $E\left(\beta ,\gamma ,\alpha \right)={\tilde{\rho }}_{\beta }\left(1-\rho {\tilde{\rho }}_{\alpha }\right)+\rho -{\tilde{\rho }}_{\alpha }>0$.

Note that $2\alpha \le {\omega }_0$, $2\beta \le {\omega }_0$ and $2\left(\alpha +\beta \right)>\pi$ imply ${\omega }_0>\frac{\pi }{2}$.

We have $2\alpha \le {\omega }_0$, $2\beta \le {\omega }_0$ and $\frac{\pi }{2}<\alpha +\beta =\pi -\gamma \le min\left\{{\omega }_0,\pi -\frac{{\omega }_0}{2}\right\}$. Here $min\left({\omega }_0,\pi -\frac{{\omega }_0}{2}\right)={\omega }_0$ if $\frac{\pi }{2}<{\omega }_0\le \frac{2\pi }{3}$ and $min\left({\omega }_0,\pi -\frac{{\omega }_0}{2}\right)=\pi -\frac{{\omega }_0}{2}$ if $\frac{2\pi }{3}\le {\omega }_0<\pi$.

We write

$$E\left(\alpha ,\beta ,\gamma \right)=\left(1+f\left(\alpha \right)f\left(\pi -\gamma -\alpha \right)\right)\left(R_{\pi -\gamma }\left(\alpha \right)-\rho \right),$$

where $\pi -\gamma -\frac{{\omega }_0}{2}\le \alpha \le \frac{{\omega }_0}{2}<\pi -\gamma$. In addition, $\pi -\gamma -\frac{{\omega }_0}{2}\le \frac{\pi -\gamma }{2}\le \frac{{\omega }_0}{2}$.

Note that $\pi -\gamma -\frac{{\omega }_0}{2}=\frac{\pi -\gamma }{2}$ if and only if $2\alpha =2\beta ={\omega }_0$, in which case ${\tilde{\rho }}_{\alpha }={\tilde{\rho }}_{\beta }=\rho$; hence,

$$E\left(\alpha ,\beta ,\gamma \right)=2\rho -\rho -{\rho }^3=\rho \left(1-{\rho }^2\right)>0.$$

Similarly, $\frac{\pi -\gamma }{2}=\frac{{\omega }_0}{2}$ if and only if $2\alpha =2\beta ={\omega }_0$.

We may assume that $\pi -\gamma -\frac{{\omega }_0}{2}<\frac{\pi -\gamma }{2}<\frac{{\omega }_0}{2}$. Fix $\gamma \in \left[\frac{{\omega }_0}{2},\frac{\pi }{2}\right]$.

As in Subcase 2.2.1. $R_{\pi -\gamma }$ is increasing on $\left[\pi -\gamma -\frac{{\omega }_0}{2},\frac{\pi -\gamma }{2}\right]$ and decreasing on $\left[\frac{\pi -\gamma }{2},\frac{{\omega }_0}{2}\right]$.

The minimum of $R_{\pi -\gamma }$ on $\left[\pi -\gamma -\frac{{\omega }_0}{2},\frac{{\omega }_0}{2}\right]$ is $\frac{f\left(\frac{{\omega }_0}{2}\right)+f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}{1+f\left(\frac{{\omega }_0}{2}\right)f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}.$

It follows that $R_{\pi -\gamma }\left(\theta \right)-\rho \ge \frac{f\left(\frac{{\omega }_0}{2}\right)+f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}{1+f\left(\frac{{\omega }_0}{2}\right)f\left(\pi -\gamma -\frac{{\omega }_0}{2}\right)}-\rho >f\left(\frac{{\omega }_0}{2}\right)-\rho =0$

for every $\theta \in \left[\pi -\gamma -\frac{{\omega }_0}{2},\frac{{\omega }_0}{2}\right]$, in particular $E\left(\alpha ,\beta ,\gamma \right)>0$.

Case 3. Assume that $2\alpha \ge {\omega }_0$ or $2\beta \ge {\omega }_0$.

We may consider that $2\alpha \ge {\omega }_0$, the other case being analogous. Then ${\tilde{\rho }}_{\alpha }=\rho \ge max\left\{{\tilde{\rho }}_{\beta },{\tilde{\rho }}_{\gamma }\right\}$.

We have $E\left(\alpha ,\beta ,\gamma \right)={\tilde{\rho }}_{\beta }\left(1-\rho {\tilde{\rho }}_{\gamma }\right)+\rho -{\tilde{\rho }}_{\gamma }>0$ and $E\left(\alpha ,\gamma ,\beta \right)={\tilde{\rho }}_{\gamma }\left(1-\rho {\tilde{\rho }}_{\beta }\right)+\rho -{\tilde{\rho }}_{\beta }>0.$

It remains to analyze the sign of $E\left(\beta ,\gamma ,\alpha \right)={\tilde{\rho }}_{\beta }+{\tilde{\rho }}_{\gamma }-\rho -\rho {\tilde{\rho }}_{\beta }{\tilde{\rho }}_{\gamma }$.

Case 3.1. Assume that $2\beta \ge {\omega }_0$ or $2\gamma \ge {\omega }_0$.

If $2\beta \ge {\omega }_0$, then ${\tilde{\rho }}_{\beta }=\rho$ and $E\left(\beta ,\gamma ,\alpha \right)=\rho +{\tilde{\rho }}_{\gamma }-\rho -{\rho }^2{\tilde{\rho }}_{\gamma }={\tilde{\rho }}_{\gamma }\left(1-{\rho }^2\right)>0$.

Similarly, if $2\gamma \ge {\omega }_0$, then ${\tilde{\rho }}_{\gamma }=\rho$ and $E\left(\beta ,\gamma ,\alpha \right)={\tilde{\rho }}_{\beta }+\rho -\rho -{\rho }^2{\tilde{\rho }}_{\beta }={\tilde{\rho }}_{\beta }\left(1-{\rho }^2\right)>0$.

Case 3.2. Assume that $2\beta \le {\omega }_0$ and $2\gamma \le {\omega }_0$.

We cannot have $2\left(\alpha +\beta \right)\le \pi$, since this implies $2\gamma =2\left(\alpha +\beta \right)>2\alpha \ge {\omega }_0$, a contradiction. Then $2\left(\alpha +\beta \right)>\pi$ and $2\gamma =2\pi -2\left(\alpha +\beta \right)$.

In Subcase 2.2.2. we proved that $E\left(\alpha ,\beta ,\gamma \right)>0$ under the following assumptions: $2\left(\alpha +\beta +\gamma \right)=2\pi$, $2\alpha \le {\omega }_0$, $2\beta \le {\omega }_0$ and $2\gamma \ge {\omega }_0$.

In the present case, $2\left(\alpha +\beta +\gamma \right)=2\pi$, $2\beta \le {\omega }_0$, $2\gamma \le {\omega }_0$ and $2\alpha \ge {\omega }_0$; hence $E\left(\beta ,\gamma ,\alpha \right)>0$. □

Corollary 3.

Let $F:\left[0,1\right)\to \left[0,\infty \right)$ with $F^{-1}\left(\left\{0\right\}\right)=\left\{0\right\}$. If $F\circ tanh$ is subadditive and nondecreasing on $\left[0,\infty \right)$, then the restriction of $F\circ s_{\mathbb{D}}$ to every circle $\left|z\right|=r<1$ is a metric. Moreover, if F is subadditive and nondecreasing on $\left[0,1\right)$, then the restriction of $F\circ s_{\mathbb{D}}$ to the entire unit disk is a metric.

Proof. 

Let $r\in \left(0,1\right)$. Theorem 2 shows that the restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to the circle $\left|z\right|=r$ is a metric.

The function $G:=F\circ tanh$ is metric-preserving. Therefore, $F\circ s_{\mathbb{D}}=G\circ \mathrm{arctanh}{s}_{\mathbb{D}}$ is a metric on the circle $\left|z\right|=r$. □

Remark 6.

Triangle inequality for the restriction of $\mathrm{arctanh}{s}_{\mathbb{D}}$ to any circle $\left|z\right|=r$ with $r\in \left(0,1\right)$ is always strict, as we see from the proof of Theorem 2. Therefore, given $a,b\in \left(0,\infty \right)$ we cannot find $x,y,z$ on a circle $\left|z\right|=r$, $r\in \left(0,1\right)$ such that $\mathrm{arctanh}{s}_{\mathbb{D}}\left(x,y\right)=a$, $\mathrm{arctanh}{s}_{\mathbb{D}}\left(y,z\right)=b$ and $\mathrm{arctanh}{s}_{\mathbb{D}}\left(x,z\right)=a+b$. This prevents us from obtaining the subadditivity of $F\circ tanh$ under the assumption that $F:\left[0,1\right)\to \left[0,\infty \right)$ with $F^{-1}\left(\left\{0\right\}\right)=\left\{0\right\}$ is metric-preserving with respect to the restriction of the triangular ratio metric $s_{\mathbb{D}}$ to every circle $\left|z\right|=r<1$.

We prove a functional inequality similar to (4) satisfied by continuous functions $F:\left[0,1\right)\to \left[0,\infty \right)$ with $F^{-1}\left(\left\{0\right\}\right)=\left\{0\right\}$ which are metric-preserving with respect to the restriction of the triangular ratio metric $s_{\mathbb{D}}$ to every circle $\left|z\right|=r<1$.

Theorem 3.

Assume that the continuous amenable function $F:\left[0,1\right)\to \left[0,\infty \right)$ is metric-preserving with respect to the restriction of the triangular ratio metric $s_{\mathbb{D}}$ to every circle $\left|z\right|=r<1$, $r\in \left(0,1\right)$.Then, for every $\lambda ,\mu \in \left[0,1\right)$, the following inequality holds:

$$F\left(\frac{\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}}{\sqrt{{\left(\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}\right)}^2+\left(1-{\lambda }^2\right)\left(1-{\mu }^2\right)}}\right)\le F\left(\lambda \right)+F\left(\mu \right).$$

(9)

Equivalently, for every $a,b\in \left[0,\infty \right)$ we have

$$F\left(\frac{sinh\left(a\right)+sinh\left(b\right)}{\sqrt{1+{\left(sinh\left(a\right)+sinh\left(b\right)\right)}^2}}\right)\le F\left(tanh\left(a\right)\right)+F\left(tanh\left(b\right)\right).$$

(10)

Proof. 

For $\lambda =0$ or $\mu =0$ the inequality is trivial. Fix $\lambda ,\mu \in \left(0,1\right)$.

Denote by $C_r$ the circle $\left|z\right|=r<1$.

For every $r$ with $max\left(\lambda ,\mu \right)<r<1$ there exist $x_r$, $y_r$, $z_r\in C_r$ such that, denoting $2\alpha :=\measuredangle (x_r,0,y_r)$ and $2\beta :=\measuredangle (y_r,0,z_r)$, the following conditions are satisfied:

(i) $\alpha ,\beta \in \left(0,\frac{1}{2}arccosr\right)$;

(ii) $d\left(x_r,y_r\right)=\lambda$ and $d\left(y_r,z_r\right)=\mu$.

Using the Formula (8) we look for $\alpha \in \left(0,\frac{1}{2}arccosr\right)$, i.e., with $cos\alpha \in \left(\sqrt{\frac{1+r}{2}},1\right)$, such that

$$\frac{rsin\alpha }{\sqrt{1+r^2-2rcos\alpha }}=\lambda .$$

The above requirements are satisfied if and only if $cos\alpha =\frac{1}{r}\left({\lambda }^2+\sqrt{\left(1-{\lambda }^2\right)\left(r^2-{\lambda }^2\right)}\right)$. Then we compute $sin\alpha =\frac{\lambda }{r}\frac{1-r^2}{\sqrt{1-{\lambda }^2}+\sqrt{r^2-{\lambda }^2}}$. Taking $x_r\in C_r$ arbitrary and $y_r=x_r{e}^{i\alpha }$, it follows that $d\left(x_r,y_r\right)=\lambda$.

Similarly, we find an unique $\beta \in \left(0,\frac{1}{2}arccosr\right)$ such that $\frac{rsin\beta }{\sqrt{1+r^2-2rcos\beta }}=\mu$ and obtain

$cos\beta =\frac{1}{r}\left({\mu }^2+\sqrt{\left(1-{\mu }^2\right)\left(r^2-{\mu }^2\right)}\right)$ and $sin\beta =\frac{\mu }{r}\frac{1-r^2}{\sqrt{1-{\mu }^2}+\sqrt{r^2-{\mu }^2}}$. Now, taking $z_r=y_r{e}^{i\beta }$, it follows that $d\left(y_r,z_r\right)=\mu$.

Denote $2\gamma :=\measuredangle (z_r,0,x_r)$. Since $\alpha ,\beta \in \left(0,\frac{1}{2}arccosr\right)$, it follows that $2\left(\alpha +\beta \right)\in \left(0,2arccosr\right)\subset \left(0,\pi \right)$; therefore, $2\gamma =2\left(\alpha +\beta \right)<2arccosr$. Using (8), it follows that

$$d\left(x_r,z_r\right)=\frac{rsin\left(\alpha +\beta \right)}{\sqrt{1+r^2-2rcos\left(\alpha +\beta \right)}}.$$

Computing $sin\left(\alpha +\beta \right)$ and $cos\left(\alpha +\beta \right)$, and using the notations

$H\left(r,\tau \right)=\sqrt{1-{\tau }^2}+\sqrt{r^2-{\tau }^2}$ and $K\left(r,\tau \right)={\tau }^2+\sqrt{\left(1-{\tau }^2\right)\left(r^2-{\tau }^2\right)}$, we obtain

$$d(x_r,z_r)=\frac{1-r^2}{r}\frac{A(r,\lambda ,\mu )}{B(r,\lambda ,\mu )}$$

where

$$A(r,\lambda ,\mu )=\lambda \frac{K(r,\mu )}{H\left(r,\lambda \right)}+\mu \frac{K(r,\lambda )}{H\left(r,\mu \right)}$$

and

$$B(r,\lambda ,\mu )={\left(1+r^2-\frac{2}{r}K\left(r,\lambda \right)K(r,\mu )-\lambda \mu {(1-r^2)}^2\frac{1}{H\left(r,\lambda \right)H\left(r,\mu \right)}\right)}^{1/2}.$$

Let the function $F:\left[0,1\right)\to \left[0,\infty \right)$ be amenable and metric-preserving with respect to the restriction of the triangular ratio metric $s_{\mathbb{D}}$ to every circle $\left|z\right|=r<1$.

Let $r$ satisfying $max\left(\lambda ,\mu \right)<r<1$. The triangle inequality $F\left(d\left(x_r,z_r\right)\right)\le F\left(d\left(x_r,y_r\right)\right)+F\left(d\left(y_r,z_r\right)\right)$ may be written as

$$F\left(\frac{1-r^2}{r}\frac{A(r,\lambda ,\mu )}{B(r,\lambda ,\mu )}\right)\le F\left(\lambda \right)+F\left(\mu \right).$$

(11)

We compute

$$\underset{r\nearrow 1}{lim}\frac{1-r^2}{r}\frac{A(r,\lambda ,\mu )}{B(r,\lambda ,\mu )}=\frac{\lambda \sqrt{\left(1-{\mu }^2\right)}+\mu \sqrt{\left(1-{\lambda }^2\right)}}{\sqrt{2\lambda \mu \sqrt{\left(1-{\lambda }^2\right)\left(1-{\mu }^2\right)}+1-{\lambda }^2{\mu }^2}}.$$

We have $\frac{\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}}{\sqrt{2\lambda \mu \sqrt{\left(1-{\lambda }^2\right)\left(1-{\mu }^2\right)}+1-{\lambda }^2{\mu }^2}}=\frac{\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}}{\sqrt{{\left(\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}\right)}^2+\left(1-{\lambda }^2\right)\left(1-{\mu }^2\right)}}\in \left[0,1\right)$ whenever $\lambda ,\mu \in \left[0,1\right)$.

If F is continuous on $\left[0,1\right)$, then letting r tend to 1 from below in inequality (11) we get (9).

Let $a,b\in \left[0,\infty \right)$. Denote $\lambda =tanh\left(a\right)$ and $\mu =tanh\left(b\right)$. Then

$\frac{\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}}{\sqrt{2\lambda \mu \sqrt{\left(1-{\lambda }^2\right)\left(1-{\mu }^2\right)}+1-{\lambda }^2{\mu }^2}}=\frac{sinh\left(a\right)+sinh\left(b\right)}{\sqrt{1+{\left(sinh\left(a\right)+sinh\left(b\right)\right)}^2}}$. Since the function $tanh:\left[0,\infty \right)\to \left[0,1\right)$ is surjective, the inequalities (9) and (10) are equivalent. □

Remark 7.

Since sinh is supradditive and the function $x\mapsto \frac{x}{\sqrt{1+x^2}}$ is increasing on $\mathbb{R}$, we have

$$\frac{sinh\left(a\right)+sinh\left(b\right)}{\sqrt{1+{\left(sinh\left(a\right)+sinh\left(b\right)\right)}^2}}\le \frac{sinh\left(a+b\right)}{\sqrt{1+{\left(sinh\left(a+b\right)\right)}^2}}=tanh\left(a+b\right).$$

If F is nonincreasing, then inequality (10) implies the subadditivity of the function $F\circ tanh$ on $\left[0,\infty \right)$. If F is nondecreasing, then inequality (10) is implied by the subadditivity of the function $F\circ tanh$ on $\left[0,\infty \right)$.

Numerical experiments show that $\frac{sinh\left(a\right)+sinh\left(b\right)}{\sqrt{1+{\left(sinh\left(a\right)+sinh\left(b\right)\right)}^2}}$ and $tanh\left(a+b\right)$ are close to each other for all $a,b\in \left[0,\infty \right)$.

Lemma 6.

For every $a,b\in \left[0,\infty \right)$ we have

$$0\le tanh\left(a+b\right)-\frac{sinh\left(a\right)+sinh\left(b\right)}{\sqrt{1+{\left(sinh\left(a\right)+sinh\left(b\right)\right)}^2}}\le \frac{2\sqrt{t_0\left(t_0-1\right)}}{2t_0-1}-\frac{2\sqrt{t_0-1}}{\sqrt{4t_0-3}},$$

where $t_0$ is the unique real positive root of the polynomial $P\left(t\right)=16t^4-16t^3-56t^2+80t-27$.

Equivalently, for all $\lambda ,\mu \in \left[0,1\right)$

$$0\le \frac{\lambda +\mu }{1+\lambda \mu }-\frac{\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}}{\sqrt{{\left(\lambda \sqrt{1-{\mu }^2}+\mu \sqrt{1-{\lambda }^2}\right)}^2+\left(1-{\lambda }^2\right)\left(1-{\mu }^2\right)}}\le \frac{2\sqrt{t_0\left(t_0-1\right)}}{2t_0-1}-\frac{2\sqrt{t_0-1}}{\sqrt{4t_0-3}}$$

Proof. 

Let $E(x,y)=tanh\left(x+y\right)-\frac{sinh\left(x\right)+sinh\left(y\right)}{\sqrt{1+{\left(sinh\left(x\right)+sinh\left(y\right)\right)}^2}}$, where $x,y\in \left[0,\infty \right)$. We observe that $E(x,y)$ tends to zero as $x\to 0$ or $y\to 0$, respectively, as $x\to \infty$ or $y\to \infty$. Then there exists the maximum of E on $\left[0,\infty \right)\times \left[0,\infty \right)$, attained at some point $\left(x_0,y_0\right)\in \left[0,\infty \right)\times \left[0,\infty \right)$.

The partial derivatives $\frac{\partial E}{\partial x}\left(x,y\right)=\frac{1}{\left(cosh(x+y\right){)}^2}-\frac{cosh\left(x\right)}{{\left(1+{\left(sinh\left(x\right)+sinh\left(y\right)\right)}^2\right)}^{3/2}}$ and $\frac{\partial E}{\partial y}\left(x,y\right)=\frac{1}{\left(cosh(x+y\right){)}^2}-\frac{cosh\left(y\right)}{{\left(1+{\left(sinh\left(x\right)+sinh\left(y\right)\right)}^2\right)}^{3/2}}$ vanish at $\left(x_0,y_0\right)$; hence, $x_0=y_0$ and $x_0>0$ is a solution of the equation

$$\frac{1}{\left(cosh(2x\right){)}^2}=\frac{cosh\left(x\right)}{{\left(1+4{\left(sinh\left(x\right)\right)}^2\right)}^{3/2}}.$$

Using the change of variable $\left(cosh(x\right){)}^2=t$, the above equation transforms into $t{\left(2t-1\right)}^4={(4t-3)}^3$. However, $t{\left(2t-1\right)}^4-{(4t-3)}^3=\left(t-1\right)P\left(t\right)$ and $\left(cosh(x_0\right){)}^2>1$ is a root of P. It turns out that P has one positive root $t_0$, one negative root and two complex nonreal roots. Then $cosh\left(x_0\right)=\sqrt{t_0}$ and

$$\begin{array}{ccc}\hfill max\left\{E(x,y):(x,y)\in \left[0,\infty \right)\times \left[0,\infty \right)\right\}& =& E\left(x_0,x_0\right)=tanh\left(2x_0\right)-\frac{2sinh\left(x_0\right)}{\sqrt{1+4{\left(sinh\left(x_0\right)\right)}^2}}\hfill \\ & =& \frac{2\sqrt{t_0\left(t_0-1\right)}}{2t_0-1}-\frac{2\sqrt{t_0-1}}{\sqrt{4t_0-3}}.\hfill \end{array}$$

□

Remark 8.

Using the approximate value $t_0\cong 1.6638$ we get $\frac{2\sqrt{t_0\left(t_0-1\right)}}{2t_0-1}-\frac{2\sqrt{t_0-1}}{\sqrt{4t_0-3}}\cong 0.050705$.

5. The Case of Barrlund Metric with p = 2 on a Canonical Plane Domain

We will consider Barrlund metrics on canonical domains in plane: the upper half plane and the unit disk. For $p=2$ and $G\in \left\{\mathbb{H},\mathbb{D}\right\}$ explicit formulas for $b_{G,p}$ have been proved in [23], as follows:

$$b_{\mathbb{H},2}(z_1,z_2)=\frac{\sqrt{2}|z_1-z_2|}{\sqrt{|z_1-z_2{|}^2+{\left(\mathrm{Im}(z_1+z_2)\right)}^2}}\mathrm{for}\mathrm{all}{z}_1,z_2\in \mathbb{H}$$

and

$$b_{\mathbb{D},2}(w_1,w_2)=\frac{|w_1-w_2|}{\sqrt{2+|w_1{|}^2+|w_2{|}^2-2|w_1+w_2|}}\mathrm{for}\mathrm{all}{w}_1,w_2\in \mathbb{D}.$$

Using parallelogram’s rule, we can write

$$b_{\mathbb{D},2}(w_1,w_2)=\frac{\sqrt{2}|w_1-w_2|}{\sqrt{|w_1-w_2{|}^2+{\left(2-|w_1+w_2|\right)}^2}}.$$

We can see that $b_{\mathbb{H},2}\left(\mathbb{H}\times \mathbb{H}\right)=\left[0,\sqrt{2}\right)$ and $b_{\mathbb{D},2}\left(\mathbb{D}\times \mathbb{D}\right)=\left[0,\sqrt{2}\right)$.

Next, we study the restrictions of $b_{\mathbb{H},2}$ to vertical rays $V_{x_0}:$($Re\left(z\right)=x_0$ and $Im\left(z\right)>0$), $x_0\in \mathbb{R}$, to rays through origin $O_m:\left(Im\left(z\right)=mRe(z\right)$ and $Im\left(z\right)>0$), $m\in {\mathbb{R}}^{\ast }$ and to horizontal lines $L_c:(Im\left(z\right)=c)$, $c\in \left(0,\infty \right)$.

Proposition 5.

Let $F:\left[0,1\right)\to \left[0,\infty \right)$ be an amenable function and $\psi :\mathbb{R}\to \left(-1,1\right)$, $\psi \left(t\right)=\frac{e^t-1}{\sqrt{e^{2t}+1}}$. The following are equivalent:

(1) F is metric-preserving with respect to the restriction of $b_{\mathbb{H},2}$ to every ray $V_{x_0}$, $x_0\in \mathbb{R}$;

(2) F is metric-preserving with respect to the restriction of $b_{\mathbb{H},2}$ to some ray $V_{x_0}$, $x_0\in \mathbb{R}$;

(3) $F\circ \left|\psi \right|$ is subadditive on $\mathbb{R}$.

Proof. 

Consider the ray $V_{x_0}:$ ($Re\left(z\right)=x_0$ and $Im\left(z\right)>0$), $x_0\in \mathbb{R}$. For $z_1=x_0+i{y}_1,z_2=x_0+i{y}_2\in R_{x_0}$, denoting $\frac{y_2}{y_1}=e^u$, $u\in \mathbb{R}$ we have

$$b_{\mathbb{H},2}\left(z_1,z_2\right)=\frac{\left|y_1-y_2\right|}{\sqrt{y_1^2+y_2^2}}=\frac{\left|1-\frac{y_2}{y_1}\right|}{\sqrt{1+{\left(\frac{y_2}{y_1}\right)}^2}}=\frac{\left|e^u-1\right|}{\sqrt{e^{2u}+1}}=\psi \left(u\right).$$

The functions $F\circ \left({\left.b_{\mathbb{H},2}\right|}_{R_{x_0}}\right)$ and $F\circ \psi$ are well-defined, since $\psi \left(u\right)\in \left[0,1\right)$ for every $u\in \mathbb{R}$.

For $z_k=x_0+i{y}_k\in V_{x_0}$, $k=1,2,3$ denote $\frac{y_2}{y_1}=e^u$ and $\frac{y_3}{y_2}=e^v$, where $u,v\in \mathbb{R}$.

With these notations, the triangle inequality

$$\left(F\circ b_{\mathbb{H},2}\right)\left(z_1,z_3\right)\le \left(F\circ b_{\mathbb{H},2}\right)\left(z_1,z_2\right)+\left(F\circ b_{\mathbb{H},2}\right)\left(z_2,z_3\right)$$

(12)

is equivalent to

$$\left(F\circ \psi \right)\left(u+v\right)\le \left(F\circ \psi \right)\left(u\right)+\left(F\circ \psi \right)\left(v\right).$$

(13)

$\left(1\right)\Rightarrow \left(2\right)$ is obvious.

$\left(2\right)\Rightarrow \left(3\right)$ Assume that F is metric-preserving with respect to the restriction of $b_{\mathbb{H},2}$ to some fixed ray $V_{x_0}$.

For every $u,v\in \mathbb{R}$ there exist $z_k=x_0+i{y}_k\in R_{x_0}$, $k=1,2,3$ such that $\frac{y_2}{y_1}=e^u$ and $\frac{y_3}{y_2}=e^v$. Since $z_1,z_2$, $z_3$ satisfy (12), it follows that (13) holds. Therefore, $F\circ \psi$ is subadditive on $\mathbb{R}$.

$\left(3\right)\Rightarrow \left(1\right)$ Assume that $F\circ \psi$ is subadditive on $\mathbb{R}$. Fix $x_0\in \mathbb{R}$. For every $z_k=x_0+i{y}_k\in V_{x_0}$, $k=1,2,3$ there exist $u,v\in \mathbb{R}$ such that $\frac{y_2}{y_1}=e^u$ and $\frac{y_3}{y_2}=e^v$. Since u and v satisfy (13), we obtain the triangle inequality (12). It follows that the restriction of $F\circ b_{\mathbb{H},2}$ to $V_{x_0}$ is a metric, q.e.d. □

Proposition 6.

Let $O_m$: $\left(Im\left(z\right)=mRe(z\right)$ and $Im\left(z\right)>0$), $m\in \mathbb{R}\backslash \left\{0\right\}$. Let $F:\left[0,\sqrt{2}\right)\to \left[0,\infty \right)$ be an amenable function and ${\varkappa }_m:\mathbb{R}\to (-\sqrt{2},\sqrt{2})$, ${\varkappa }_m\left(u\right)=\frac{tanh\left(\frac{u}{2}\right)\sqrt{2}}{\sqrt{{\left(tanh\left(\frac{u}{2}\right)\right)}^2+\frac{m^2}{m^2+1}}}$. Then F is metric-preserving with respect to the restriction of $b_{\mathbb{H},2}$ to the ray $S_m$ if and only if $F\circ \left|{\varkappa }_m\right|$ is subadditive on $\mathbb{R}$.

Proof. 

For $z_1=x_1+im{x}_1,z_2=x_2+im{x}_2\in O_m$, denoting $\frac{x_2}{x_1}=e^u$, $u\in \mathbb{R}$ we have

$$\begin{array}{ccc}\hfill b_{\mathbb{H},2}\left(z_1,z_2\right)& =& \frac{\left|x_1-x_2\right|\sqrt{2(m^2+1)}}{\sqrt{\left(1+m^2\right){\left(x_1-x_2\right)}^2+m^2{\left(x_1+x_2\right)}^2}}\hfill \\ & =& \frac{\left|1-\frac{x_2}{x_1}\right|\sqrt{2(m^2+1)}}{\sqrt{\left(1+m^2\right){\left(1-\frac{x_2}{x_1}\right)}^2+m^2{\left(1+\frac{x_2}{x_1}\right)}^2}}\hfill \\ & =& \frac{\left|e^u-1\right|\sqrt{2(m^2+1)}}{\sqrt{\left(1+m^2\right){\left(1-e^u\right)}^2+m^2{\left(1+e^u\right)}^2}}\hfill \\ & =& \frac{\left|tanh\left(\frac{u}{2}\right)\right|\sqrt{2}}{\sqrt{{\left(tanh\left(\frac{u}{2}\right)\right)}^2+\frac{m^2}{m^2+1}}}={\varkappa }_m\left(u\right).\hfill \end{array}$$

The functions $F\circ \left({\left.b_{\mathbb{H},2}\right|}_{S_m}\right)$ and $F\circ \left|{\varkappa }_m\right|$ are well-defined, since $\left|{\varkappa }_m\left(u\right)\right|\in \left[0,\sqrt{2}\right)$ for every $u\in \mathbb{R}$.

For $z_k=\left(1+im\right)x_k\in O_m$, $k=1,2,3$ denote $\frac{x_2}{x_1}=e^u$ and $\frac{x_3}{x_2}=e^v$, where $u,v\in \mathbb{R}$.

With these notations, the triangle inequality

$$\left(F\circ b_{\mathbb{H},2}\right)\left(z_1,z_3\right)\le \left(F\circ b_{\mathbb{H},2}\right)\left(z_1,z_2\right)+\left(F\circ b_{\mathbb{H},2}\right)\left(z_2,z_3\right)$$

(14)

is equivalent to

$$\left(F\circ \left|{\varkappa }_m\right|\right)\left(u+v\right)\le \left(F\circ \left|{\varkappa }_m\right|\right)\left(u\right)+\left(F\circ \left|{\varkappa }_m\right|\right)\left(v\right).$$

(15)

If F is metric-preserving with respect to the restriction of $b_{\mathbb{H},2}$ to the ray $O_m$, then for every $u,$$v\in \mathbb{R}$ we find $z_k=\left(1+im\right)x_k\in S_m$, $k=1,2,3$ such that $\frac{x_2}{x_1}=e^u$ and $\frac{x_3}{x_2}=e^v$ and applying (14) we get (15). Conversely, if $F\circ {\varkappa }_m$ is subadditive on $\mathbb{R}$, then for every $z_k=\left(1+im\right)x_k\in O_m$, $k=1,2,3$ we find $u,$$v\in \mathbb{R}$ such that $\frac{x_2}{x_1}=e^u$ and $\frac{x_3}{x_2}=e^v$ and applying (14) we get (15). □

Proposition 7.

Let $F:\left[0,\sqrt{2}\right)\to \left[0,\infty \right)$ and $c>0$. Denote ${\phi }_c\left(t\right)=\frac{\sqrt{2}t}{\sqrt{t^2+4c^2}}$, $t\in \mathbb{R}$. The restriction of $F\circ b_{\mathbb{H},2}$ to the line $Im\left(z\right)=c$ is a metric if and only if $F\circ \left|{\phi }_c\right|$ is subadditive on $\mathbb{R}$.

Proof. 

By our assumption, for all $x_1,x_2,x_3\in \mathbb{R}$ we have $F\left(b_{\mathbb{H},2}(x_1+ic,x_3+ic)\right)$$\le F\left(b_{\mathbb{H},2}(x_1+ic,x_2+ic)\right)+F\left(b_{\mathbb{H},2}(x_2+ic,x_3+ic)\right)$, i.e.,

$$F\left(\frac{\sqrt{2}|x_1-x_3|}{\sqrt{|x_1-x_3{|}^2+4c^2}}\right)\le F\left(\frac{\sqrt{2}|x_1-x_2|}{\sqrt{|x_1-x_2{|}^2+4c^2}}\right)+F\left(\frac{\sqrt{2}|x_2-x_3|}{\sqrt{|x_2-x_3{|}^2+4c^2}}\right).$$

Let ${\phi }_c\left(t\right)=\frac{\sqrt{2}t}{\sqrt{t^2+4c^2}}$, $t\in \left[0,\infty \right)$. The above inequality is equivalent to

$$\left(F\circ \left|{\phi }_c\right|\right)\left(x_1-x_3\right)\le \left(F\circ \left|{\phi }_c\right|\right)\left(x_1-x_2\right)+\left(F\circ \left|{\phi }_c\right|\right)\left(x_2-x_3\right)\mathrm{for}\mathrm{all}{x}_1,x_2,x_3\in \mathbb{R}.$$

(16)

Note that $\left|{\phi }_c\left(t\right)\right|\in \left[0,\sqrt{2}\right)$ for every $t\in \left[0,\infty \right)$; therefore, the functions $F\circ \left({\left.b_{\mathbb{H},2}\right|}_{Im\left(z\right)=c}\right)$ and $F\circ \left|{\phi }_c\right|$ are well-defined.

We see that $F\circ \left|{\phi }_c\right|$ satisfies (16) if and only if $F\circ \left|{\phi }_c\right|$ is subadditive on $\mathbb{R}$. □

Next, we study metric-preserving functions with respect to the restriction of the Barrlund distance on the unit disk $b_{\mathbb{D},2}$ to some one-dimensional manifolds, such as radial segments, diameters or circles centered at origin.

Proposition 8.

Let $F:\left[0,1\right)\to \left[0,\infty \right)$ be an amenable function and $\psi :\mathbb{R}\to \left(-1,1\right)$, $\psi \left(t\right)=\frac{e^t-1}{\sqrt{e^{2t}+1}}$. The following are equivalent:

(1) F is metric-preserving with respect to the restriction of $b_{\mathbb{D},2}$ to every radial segment in the unit disk;

(2) F is metric-preserving with respect to the restriction of $b_{\mathbb{D},2}$ to some radial segment in the unit disk;

(3) $F\circ \left|\psi \right|$ is subadditive on $\mathbb{R}$.

Proof. 

Obviously, $\left(1\right)\Rightarrow \left(2\right)$. Using the invariance of the Barrlund distance on the unit disk with respect to rotations around the origin, it follows that $\left(2\right)\Rightarrow \left(1\right)$, since $\left(1\right)$ holds if and only if F is metric-preserving with respect to the restriction of $b_{\mathbb{D},2}$ to the intersection $I=\left[0,1\right)$ between the unit disk and the non-negative semiaxis.

In order to prove that (2) and (3) are equivalent, we may assume without loss of generality that the radial segment in (2) is $I=\left[0,1\right)$. For $z_1=x_1,z_2=x_2\in I$, denoting $\frac{1-x_2}{1-x_1}=e^u$, $u\in \mathbb{R}$ we have

$$\begin{array}{ccc}\hfill b_{\mathbb{D},2}\left(z_1,z_2\right)& =& \frac{\left|x_1-x_2\right|}{\sqrt{2+x_1^2+x_2^2-2\left(x_1+x_2\right)}}=\frac{\left|\left(1-x_2\right)-\left(1-x_1\right)\right|}{\sqrt{{\left(1-x_2\right)}^2+{\left(1-x_1\right)}^2}}\hfill \\ & =& \frac{\left|e^u-1\right|}{\sqrt{e^{2u}+1}}=\psi \left(u\right).\hfill \end{array}$$

For $z_k=x_k\in I$, $k=1,2,3$ denote $\frac{1-x_2}{1-x_1}=e^u$ and $\frac{1-x_3}{1-x_2}=e^v$, where $u,v\in \mathbb{R}$.

With these notations, the triangle inequality

$$\left(F\circ b_{\mathbb{D},2}\right)\left(z_1,z_3\right)\le \left(F\circ b_{\mathbb{D},2}\right)\left(z_1,z_2\right)+\left(F\circ b_{\mathbb{D},2}\right)\left(z_2,z_3\right)$$

(17)

is equivalent to

$$\left(F\circ \left|\psi \right|\right)\left(u+v\right)\le \left(F\circ \left|\psi \right|\right)\left(u\right)+\left(F\circ \left|\psi \right|\right)\left(v\right).$$

(18)

Assume that F is metric-preserving with respect to the restriction of $b_{\mathbb{D},2}$ to the radial segment I. For every $u,$$v\in \mathbb{R}$ we find $z_k=x_k\in I$, $k=1,2,3$ such that $\frac{1-x_2}{1-x_1}=e^u$ and $\frac{1-x_3}{1-x_2}=e^v$. Indeed, we may choose any $x_1$ between $max\left\{0,1-e^{-u},1-e^{-u-v}\right\}$ and 1. Then $\frac{1-x_2}{1-x_1}=e^u$ if and only if $x_2=1-e^u\left(1-x_1\right)$, but $0<1-x_1<e^{-u}$; therefore, $0<x_2<1$. Moreover, $0<1-x_1<e^{-u-v}$ implies $x_2>1-e^{-v}$. Since $\frac{1-x_3}{1-x_2}=e^v$ if and only if $x_3=1-e^v\left(1-x_2\right)$, where $0<1-x_2<e^{-v}$, it follows that $0<x_3<1$. Now applying (17) we get (18).

Conversely, if $F\circ \left|\psi \right|$ is subadditive on $\mathbb{R}$, then for every $z_k=x_k\in I$, $k=1,2,3$ we find $u,$$v\in \mathbb{R}$ such that $\frac{1-x_2}{1-x_1}=e^u$ and $\frac{1-x_3}{1-x_2}=e^v$ and applying (18) we get (17). □

We give a sufficient condition for a function to be metric-preserving with respect to the restriction of the Barrlund metric $b_{\mathbb{D},2}$ to some diameter of the unit disk, under the form of a functional inequality.

Proposition 9.

Let $F:\left[0,\sqrt{2}\right)\to \left[0,\infty \right)$. Assume that the restriction of $F\circ b_{\mathbb{D},2}$ to some diameter of the unit disk is a metric. Then

$$F\left(\frac{r\sqrt{2}}{\sqrt{r^2+1}}\right)\le 2F\left(\frac{r}{\sqrt{r^2-2r+2}}\right)\mathrm{for}\mathrm{all}r\in \left[0,1\right).$$

(19)

Proof. 

Since $b_{\mathbb{D},2}$ is invariant to rotations around the origin, if a function is metric-preserving with respect to the restriction of the Barrlund metric $b_{\mathbb{D},2}$ to some diameter of the unit disk, then that function is metric-preserving with respect to the restriction of the Barrlund metric $b_{\mathbb{D},2}$ to every diameter of the unit disk. We may assume that the given diameter is on the real axis.

Note that $b_{\mathbb{D},2}\left(0,w\right)=b_{\mathbb{D},2}\left(0,-w\right)=\frac{\left|w\right|}{\sqrt{2+{\left|w\right|}^2-2\left|w\right|}}$ and $b_{\mathbb{D},2}\left(w,-w\right)=\frac{2\left|w\right|}{\sqrt{2+{\left|2w\right|}^2}}=\frac{\left|w\right|\sqrt{2}}{\sqrt{1+{\left|w\right|}^2}}$.

The above inequality writes as

$$\left(F\circ b_{\mathbb{D},2}\right)\left(r,-r\right)\le \left(F\circ b_{\mathbb{D},2}\right)\left(0,r\right)+\left(F\circ b_{\mathbb{D},2}\right)\left(0,-r\right)\mathrm{for}\mathrm{all}r\in \left[0,1\right),$$

which is true, due to the assumption that the restriction of $F\circ b_{\mathbb{D},2}$ to the diameter $\left\{x+i0:-1<x<1\right\}$ of the unit circle is a metric. □

Remark 9.

Let $F:\left[0,\sqrt{2}\right)\to \left[0,\infty \right)$. Assume that the restriction of $F\circ b_{\mathbb{D},2}$ to some radial segment of the unit disk is a metric. By Proposition 8, $F\circ \left|\psi \right|$ is subadditive on $\mathbb{R}$, where $\psi :\mathbb{R}\to \left[0,1\right)$, $\psi \left(t\right)=\frac{e^t-1}{\sqrt{e^{2t}+1}}$. In particular, $F\left(\frac{\left|e^{2a}-1\right|}{\sqrt{e^{4a}+1}}\right)\le 2F\left(\frac{\left|e^a-1\right|}{\sqrt{e^{2a}+1}}\right)$ for every $a\in \mathbb{R}$. If $r\in \left[0,1\right)$, denoting $e^a=1-r$, the previous inequality becomes

$$F\left(\frac{r\left(2-r)\right)}{\sqrt{{\left(r-1\right)}^4+1}}\right)\le 2F\left(\frac{r}{\sqrt{r^2-2r+2}}\right)$$

(20)

Note that $\frac{r\sqrt{2}}{\sqrt{r^2+1}}\le \frac{r\left(2-r)\right)}{\sqrt{{\left(r-1\right)}^4+1}}$ for every $r\in \left[0,1\right)$. If F is nondecreasing on $\left[0,\sqrt{2}\right)$, then the above inequality (20) is stronger than (19).

Propositions 5 and 8 are very similar and, together with Propositions 6 and 7, have a common pattern.

Lemma 7.

Each of the functions $\psi :\mathbb{R}\to \left(-1,1\right)$ with $\psi \left(t\right)=\frac{e^t-1}{\sqrt{e^{2t}+1}}$, ${\varkappa }_m:\mathbb{R}\to \left(-\sqrt{2},\sqrt{2}\right)$ with ${\varkappa }_m\left(u\right)=\frac{tanh\left(\frac{u}{2}\right)\sqrt{2}}{\sqrt{{\left(tanh\left(\frac{u}{2}\right)\right)}^2+\frac{m^2}{m^2+1}}}$ and ${\phi }_c:\mathbb{R}\to \left(-\sqrt{2},\sqrt{2}\right)$ with ${\phi }_c\left(t\right)=\frac{\sqrt{2}t}{\sqrt{t^2+4c^2}}$, generically denoted by φ, has the following properties: it is odd on $\mathbb{R}$, nonnegative, increasing and concave on $\left[0,\infty \right)$; hence, it is subadditive on $\left[0,\infty \right)$.

Lemma 8.

If the restriction to $\left[0,\infty \right)$ of a function $\phi :\mathbb{R}\to \mathbb{R}$ is nonnegative, nondecreasing and subadditive and if $\left|\phi \right|$ is even on $\mathbb{R}$, then $\left|\phi \right|$ is subadditive on $\mathbb{R}$.

Proof. 

We prove that $\left|\phi \left(a+b\right)\right|\le \left|\phi \left(a\right)\right|+\left|\phi \left(b\right)\right|$ for every $a,b\in \mathbb{R}$. This is clear for $a,b\in \left[0,\infty \right)$, taking into account that the restriction to $\left[0,\infty \right)$ of $\phi$ is nonnegative and subadditive. If $a,b\in (-\infty ,0]$, using the fact that $\left|\phi \right|$ is even on $\mathbb{R}$ and the previous case we get $\left|\phi \left(a+b\right)\right|=\left|\phi \left(\left(-a\right)+\left(-b\right)\right)\right|\le \left|\phi \left(-a\right)\right|+\left|\phi \left(-b\right)\right|=\left|\phi \left(a\right)\right|+\left|\phi \left(b\right)\right|$. It remains to study the case where $a·b<0$. By symmetry, it suffice to assume that $a\le 0\le b$ and $a+b\ge 0$. Then $\left|\phi \left(a+b\right)\right|=\left|\phi \left(a-\left|b\right|\right)\right|=\phi \left(a-\left|b\right|\right)\le \phi \left(a\right)\le \left|\phi \left(a\right)\right|+\left|\phi \left(b\right)\right|$, since $\phi$ is nondecreasing and nonnegative on $\left[0,\infty \right)$. □

Corollary 4.

(a) If $F:\left[0,1\right)\to \left[0,\infty \right)$ is subadditive, then F is metric-preserving with respect to the restriction of $b_{\mathbb{H},2}$ to each vertical ray $V_{x_0}$ and with respect to the restriction of $b_{\mathbb{D},2}$ to each radial segment of the unit disk.

(b) If $F:\left[0,\sqrt{2}\right)\to \left[0,\infty \right)$ is subadditive, then F is metric-preserving with respect to the restrictions of $b_{\mathbb{H},2}$ to each oblique ray $S_m$ and to each horizontal line $L_c$.

Proof. 

Using Lemmas 7 and 8, we see that the modulus of each of the functions $\psi$, ${\varkappa }_m$ and ${\phi }_c$ is a subadditive function on $\mathbb{R}$. The composition of two subadditive functions is subadditive. For (a) Then we apply Propositions 5 and 8 for (a), respectively, Propositions 6 and 7 for (b). □

Finally, we obtain a characterization of functions F which are metric-preserving with respect to the restriction of $b_{\mathbb{D},2}$ to each circle centered at origin.

Proposition 10.

Let $F:\left[0,\sqrt{2}\right)\to \left[0,\infty \right)$ be an amenable function, $r\in \left[0,1\right)$ and ${\theta }_r:\mathbb{R}\to \mathbb{R}$, ${\theta }_r\left(t\right)=\frac{r\left|sint\right|\sqrt{2}}{\sqrt{r^2-2r\left|cost\right|+1}}$. Then the restriction of $F\circ b_{\mathbb{D},2}$ to the circle $\left|z\right|=r$ is a metric if and only if $F\circ {\theta }_r$ is subadditive on $\mathbb{R}$.

Proof. 

Denote by $C_r$ the circle $\left|z\right|=r$. For $z_1=r{e}^{i{\alpha }_1},$$z_2=r{e}^{i{\alpha }_2}\in C_r$, where ${\alpha }_1$, ${\alpha }_2\in \mathbb{R}$ we compute $|z_1-z_2|=2r\left|sin\frac{{\alpha }_1-{\alpha }_2}{2}\right|$ and $|z_1+z_2|=2r\left|cos\frac{{\alpha }_1-{\alpha }_2}{2}\right|$, therefore denoting $\frac{{\alpha }_1-{\alpha }_2}{2}=u$ we have

$$\begin{array}{ccc}\hfill b_{\mathbb{D},2}\left(z_1,z_2\right)& =& \frac{|z_1-z_2|}{\sqrt{2+|z_1{|}^2+|z_2{|}^2-2|z_1+z_2|}}=\frac{2r\left|sin\frac{{\alpha }_1-{\alpha }_2}{2}\right|}{\sqrt{2+2r^2-4r\left|cos\frac{{\alpha }_1-{\alpha }_2}{2}\right|}}\hfill \\ & =& \frac{r\left|sin\frac{{\alpha }_1-{\alpha }_2}{2}\right|\sqrt{2}}{\sqrt{r^2-2r\left|cos\frac{{\alpha }_1-{\alpha }_2}{2}\right|+1}}=\frac{r\left|sinu\right|\sqrt{2}}{\sqrt{r^2-2r\left|cosu\right|+1}}={\theta }_r\left(u\right).\hfill \end{array}$$

For $z_k=r{e}^{i{\alpha }_k}\in C_r$ with ${\alpha }_k\in \mathbb{R}$ for $k=1,2,3$, we denote $\frac{{\alpha }_1-{\alpha }_2}{2}=u$ and $\frac{{\alpha }_2-{\alpha }_3}{2}=v$. Then $u+v=\frac{{\alpha }_1-{\alpha }_3}{2}$.

With the above notations, the triangle inequality

$$\left(F\circ b_{\mathbb{D},2}\right)\left(z_1,z_3\right)\le \left(F\circ b_{\mathbb{D},2}\right)\left(z_1,z_2\right)+\left(F\circ b_{\mathbb{D},2}\right)\left(z_2,z_3\right)$$

(21)

is equivalent to

$$\left(F\circ {\theta }_r\right)\left(u+v\right)\le \left(F\circ {\theta }_r\right)\left(u\right)+\left(F\circ {\theta }_r\right)\left(v\right).$$

(22)

First we assume that $F\circ {\theta }_r$ is subadditive on $\mathbb{R}$. As above, for every $z_k=r{e}^{i{\alpha }_k}\in C_r$ with ${\alpha }_k\in \mathbb{R}$ for $k=1,2,3$, we denote $\frac{{\alpha }_1-{\alpha }_2}{2}=u$ and $\frac{{\alpha }_2-{\alpha }_3}{2}=v$ and applying (22) we get (21).

Now assume that F is metric-preserving with respect to the restriction of $b_{\mathbb{D},2}$ to the circle $C_r$. For arbitrary $u,$$v\in \mathbb{R}$ we look for ${\alpha }_k\in \mathbb{R}$, $k=1,2,3$ such that $\frac{{\alpha }_1-{\alpha }_2}{2}=u$ and $\frac{{\alpha }_2-{\alpha }_3}{2}=v$. It suffices to take ${\alpha }_3=0$, ${\alpha }_2=2v$ and ${\alpha }_1=2\left(u+v\right)$. Finally, applying (21) with $z_k=r{e}^{i{\alpha }_k}\in C_r$ for $k=1,2,3$ we get (22). □

6. Conclusions

In this paper, we investigated properties related to subadditivity of the functions transferring some special metrics to metrics, establishing connections between metric geometry and functional inequalities. For a metric space, $\left(X,d\right)$ such that $d\left(x,y\right)\in [0,T)$ for all $x,y\in X$, where $0<T\le \infty$, let us denote by $MP\left(X,d\right)$ the class of functions $f:[0,T)\to {\mathbb{R}}_{+}=[0,\infty )$ with the property that $f\circ d$ is a metric on d. It is known from the theory of metric-preserving functions that the intersection of all classes $MP\left(X,d\right)$ includes the class of all nondecreasing subadditive self-maps on ${\mathbb{R}}_{+}$ and is included in the class of all subadditive self-maps on ${\mathbb{R}}_{+}$. We obtained functional inequalities satisfied by functions in $MP\left(X,d\right)$ in several cases, where X is some subset of $G\in$$\left\{\mathbb{H},\mathbb{D}\right\}$ and d is the restriction to X of an intrinsic metric on G, namely the hyperbolic metric, the triangular ratio metric $s_G$ or the Barrlund metric $b_{G,2}$. We will denote by $Sa\left([0,T)\right)$ the class of functions $f:[0,T)\to {\mathbb{R}}_{+}$ that are subadditive. In addition, denote by $X_r$ the circle of radius $r\in \left(0,1\right)$ centered at origin.

We summarize in

Table 1. Synthesis of the main results.

The Set X	The Metric d	Result
proper simply-connected plane domain G	hyperbolic metric ${\rho }_G$	$f\in \mathcal{F}\Rightarrow f$$\in Sa\left({\mathbb{R}}_{+}\right)$
$\mathbb{H}$	$s_{\mathbb{H}}$	$f\in \mathcal{F}\Rightarrow f\circ tanh$$\in Sa\left({\mathbb{R}}_{+}\right)$
radial segment in $\mathbb{D}$	${\left.s_{\mathbb{D}}\right|}_{X\times X}$	$f\in \mathcal{F}\Rightarrow f\circ tanh$$\in Sa\left({\mathbb{R}}_{+}\right)$
vertical ray in $\mathbb{H}$: $x=x_0$, $y>0$	${\left.b_{\mathbb{H},2}\right|}_{X\times X}$	$f\in \mathcal{F}\iff f\circ \left|\psi \right|\in Sa\left(\mathbb{R}\right)$
ray through origin in $\mathbb{H}$, with slope $m>0$	${\left.b_{\mathbb{H},2}\right|}_{{}_{X\times X}}$	$f\in \mathcal{F}\iff f\circ \left|{\varkappa }_m\right|\in Sa\left(\mathbb{R}\right)$
horizontal line in $\mathbb{H}$: $y=c>0$	${\left.b_{\mathbb{H},2}\right|}_{{}_{X\times X}}$	$f\in \mathcal{F}\iff f\circ \left|{\phi }_c\right|\in Sa\left(\mathbb{R}\right)$
radial segment in $\mathbb{D}$	${\left.b_{\mathbb{D},2}\right|}_{{}_{X\times X}}$	$f\in \mathcal{F}\iff f\circ \left|\psi \right|\in Sa\left(\mathbb{R}\right)$
$X_r$	${\left.b_{\mathbb{D},2}\right|}_{{}_{X\times X}}$	$f\in \mathcal{F}\iff f\circ \left|{\theta }_r\right|\in Sa\left(\mathbb{R}\right)$

most of our results, excepting Proposition 2 and Theorem 3, namely Theorem 1 and Propositions 3, 4, 5, 6, 7, 8 and 10. Within the table we use the abbreviation $\mathcal{F}=MP\left(X,d\right)$:

We determined the functions $\psi$ (that is fixed), ${\varkappa }_m$, ${\phi }_c$ and ${\theta }_r$ (each depending only on the respective parameter).

Moreover, denoting $d_r=$${\left.s_{\mathbb{D}}\right|}_{X_r}$, we proved that every $f{\in }_{r\in \left(0,1\right)}MP\left(X_r,d_r\right)$ satisfies the functional inequality (10) related to the subadditivity of $f\circ tanh$, as follows. If f is nonincreasing on $\left[0,1\right)$ and satisfies (10), then $f\circ tanh$ is subadditive on ${\mathbb{R}}_{+}$. If f is nondecreasing on $\left[0,1\right)$ and $f\circ tanh$ is subadditive on ${\mathbb{R}}_{+}$, then f satisfies (10).

Since $\mathrm{arctanh}{s}_{\mathbb{H}}=\frac{1}{2}{\rho }_{\mathbb{H}}$ is a metric on $\mathbb{H}$, it would be interesting to know if $\mathrm{arctanh}{s}_{\mathbb{D}}$ is a metric on $\mathbb{D}$ ([23] Conjecture 2.1). We proved that $\mathrm{arctanh}{s}_{\mathbb{D}}$ induces a metric on each diameter of $\mathbb{D}$ and on each circle of radius $r\in \left(0,1\right)$ centered at origin. The above conjecture remains open.