IINTRODUCING A NEW INTRINSIC METRIC
OONA RAINIO AND MATTI VUORINEN
Abstract.
A new intrinsic metric called t -metric is introduced. Several sharp inequal-ities between this metric and the most common hyperbolic type metrics are proven forvarious domains G (cid:40) R n . The behaviour of the new metric is also studied under afew examples of conformal and quasiconformal mappings, and the differences betweenthe balls drawn with all the metrics considered are compared by both graphical andanalytical means. Introduction
In geometric function theory, one of the topics studied deals with the variation ofgeometric entities such as distances, ratios of distances, local geometry and measures ofsets under different mappings. For such studies, we need an appropriate notion of distancethat is compatible with the class of mappings studied. In classical function theory of thecomplex plane, one of the key concepts is the hyperbolic distance, which measures notonly how close the points are to each other but also how they are located inside thedomain with respect to its boundary.The hyperbolic distance also serves as a model when we need generalisations to sub-domains G of arbitrary metric spaces X . These generalized distances behave like thehyperbolic metric in the aspect that they define the Euclidean topology and, in partic-ular, we can cover compact subsets of G using balls of the generalized metrics. Thus,the boundary of the domain has a strong influence on the inner geometry of the domaindefined by some chosen metric.Since the classical hyperbolic geometry acts as a model, some of its key features areinherited by the generalizations but not all. For instance, it is desirable to study localbehaviour of functions and we need to have a metric that is locally comparable with theEuclidean geometry. Such a metric is here called an intrinsic metric . Note that there isno established definition for this concept and it is sometimes required, for instance, thatthe closures of the balls defined with an intrinsic metric never intersect the boundary ofthe domain. An example of an intrinsic metric is the following new metric, on which thiswork focuses. File: main.tex, printed: 2020-10-6, 1.12
Mathematics Subject Classification.
Primary 51M10; Secondary 30C65.
Key words and phrases.
Hyperbolic geometry, intrinsic geometry, intrinsic metrics, quasiconformalmappings, triangular ratio metric. a r X i v : . [ m a t h . M G ] O c t O. RAINIO AND M. VUORINEN
Definition 1.1.
Let G be some non-empty, open, proper and connected subset of ametric space X . Choose some metric η G defined in the closure of G and denote η G ( x ) = η G ( x, ∂G ) = inf { η G ( x, z ) | z ∈ ∂G } for all x ∈ G . The t -metric for a metric η G in adomain G is a function t G : G × G → [0 , ,t G ( x, y ) = η G ( x, y ) η G ( x, y ) + η G ( x ) + η G ( y ) , for all x, y ∈ G . Here, we mostly focus on the special case where G (cid:40) R n and η G is theEuclidean distance.Our work in this paper is motivated by the research of several other mathematicians.During the past thirty years, many intrinsic metrics have been introduced and studied[3, 4, 7, 10, 11]. It is noteworthy that each metric might be used to discover some intricatefeatures of mappings not detected by other metrics. Since our new metric differs slightlyfrom other intrinsic metrics and has a relatively simple definition, it could potentiallybe a great help for new discoveries about intrinsic geometry of domains. For instance,there is one inequality that is an open question for the triangular ratio metric, but couldpotentially be proved for the t -metric, see Conjecture 4.13 and Remark 4.14.Unlike several other hyperbolic type metrics, such as the triangular ratio metric or thehyperbolic metric itself, the t -metric does not have the property about the closed ballsnever intersecting with the boundary, see Theorem 5.5. This is an interesting aspect forthis metric clearly fulfills most of the others, if not all, properties of a hyperbolic typemetrics listed in [7, p. 79]. Consequently, we have found an intrinsic metric that does nothave one of the common properties of hyperbolic type metrics.In this paper, we will study this new metric and its connection to other metrics. InSection 3, we prove that the function of Definition 1.1 is really a metric and find the sharpinequalities between this metric and several hyperbolic type metrics, including also thehyperbolic metric, in different domains. In Section 4, we show how the t -metric behavesunder certain quasiconformal mappings and find the Lipschitz constants for M¨obius mapsbetween balls and half-spaces. Finally, in Section 5, we draw t -metric disks and comparetheir certain properties to those of other metric disks. Acknowledgements.
The research of the first author was supported by Finnish Con-cordia Fund. 2.
Preliminaries
In this section, we will introduce the definitions of a few different metrics and metricballs that will be necessary later on but, first, let us recall the definition of a metric.
Definition 2.1.
For any non-empty space G , a metric is a function η G : G × G → [0 , ∞ )that fulfills the following three conditions for all x, y, z ∈ G : (1) Positivity: η G ( x, y ) ≥
0, and η G ( x, y ) = 0 if and only if x = y , (2) Symmetry: η G ( x, y ) = η G ( y, x ), (3) Triangle inequality: η G ( x, y ) ≤ η G ( x, z ) + η G ( z, y ) . NTRODUCING A NEW INTRINSIC METRIC 3
Let η G be now some arbitrary metric. An open ball defined with it is B η ( x, r ) = { y ∈ G | η G ( x, y ) < r } and the corresponding closed ball is B η ( x, r ) = { y ∈ G | η G ( x, y ) ≤ r } .Denote the sphere of these balls by S η ( x, r ). For Euclidean metric, these notations are B n ( x, r ), B n ( x, r ) and S n − ( x, r ), respectively, where n is the dimension. In this paper,the unit ball B n = B n (0 , H n = { ( x , ..., x n ) ∈ R n | x n > } andthe open sector S θ = { x ∈ C | < arg( x ) < θ } with an angle θ ∈ (0 , π ) will be commonlyused as domains G . Note also that the unit vectors will be denoted by { e , ..., e n } .Let us now define the metrics needed for a domain G (cid:40) R n . Denote the Euclideandistance between the points x, y by | x − y | and let d G ( x ) = inf {| x − z | | z ∈ ∂G } . Supposethat the t -metric is defined with the Euclidean distance so that t G ( x, y ) = | x − y || x − y | + d G ( x ) + d G ( y )for all x, y ∈ G , if not otherwise specified.The following hyperbolic type metrics will be considered:The triangular ratio metric: s G : G × G → [0 , s G ( x, y ) = | x − y | inf z ∈ ∂G ( | x − z | + | z − y | ) , the j ∗ G -metric: j ∗ G : G × G → [0 , ,j ∗ G ( x, y ) = | x − y || x − y | + 2 min { d G ( x ) , d G ( y ) } , and the point pair function: p G : G × G → [0 , ,p G ( x, y ) = | x − y | (cid:112) | x − y | + 4 d G ( x ) d G ( y ) . Out of these hyperbolic type metrics, the triangular ratio metric was studied by P. H¨ast¨oin 2002 [9], and the two other metrics are more recent. As pointed out in [8], the j ∗ G -metricis derived from the distance ratio metric found by F.W. Gehring and B.G. Osgood in [6].Note that there are proper domains G in which the point pair function is not a metric [3,Rmk 3.1 p. 689].Define also the hyperbolic metric asch ρ H n ( x, y ) = 1 + | x − y | d H n ( x ) d H n ( y ) , x, y ∈ H n , sh ρ B n ( x, y )2 = | x − y | (1 − | x | )(1 − | y | ) , x, y ∈ B n O. RAINIO AND M. VUORINEN in the upper half-plane H n and in the Poincar´e unit disk B n [7, (4.8), p. 52 & (4.14), p.55]. In the two-dimensional space,th ρ H ( x, y )2 = th (cid:18)
12 log (cid:18) | x − y | + | x − y || x − y | − | x − y | (cid:19)(cid:19) = (cid:12)(cid:12)(cid:12)(cid:12) x − yx − y (cid:12)(cid:12)(cid:12)(cid:12) , th ρ B ( x, y )2 = th (cid:18)
12 log (cid:18) | − xy | + | x − y || − xy | − | x − y | (cid:19)(cid:19) = (cid:12)(cid:12)(cid:12)(cid:12) x − y − xy (cid:12)(cid:12)(cid:12)(cid:12) = | x − y | A [ x, y ] , where y is the complex conjugate of y and A [ x, y ] = (cid:112) | x − y | + (1 − | x | )(1 − | y | ) isthe Ahlfors bracket [7, (3.17) p. 39].Note that the following inequalities hold for the hyperbolic type metrics. Lemma 2.2. [8, Lemma 2.3, p. 1125]
For a proper subdomain G of R n , the inequality j ∗ G ( x, y ) ≤ p G ( x, y ) ≤ √ j ∗ G ( x, y ) holds for all x, y ∈ G . Lemma 2.3. [8, Lemma 2.1, p. 1124 & Lemma 2.2, p. 1125]
For a proper subdomain G of R n , the inequality j ∗ G ( x, y ) ≤ s G ( x, y ) ≤ j ∗ G ( x, y ) holds for all x, y ∈ G . Lemma 2.4. [7, p. 460]
For all x, y ∈ G ∈ { H n , B n } , (1) th ρ H n ( x, y )4 ≤ j ∗ H n ( x, y ) ≤ s H n ( x, y ) = p H n ( x, y ) = th ρ H n ( x, y )2 ≤ ρ H n ( x, y )4 , (2) th ρ B n ( x, y )4 ≤ j ∗ B n ( x, y ) ≤ s B n ( x, y ) ≤ p B n ( x, y ) ≤ th ρ B n ( x, y )2 ≤ ρ B n ( x, y )4 . t-Metric and Its Bounds Now, we will prove that our new metric is truly a metric in the general case.
Theorem 3.1.
For any metric space X , a domain G (cid:40) X and a metric η G defined in G ,the function t G is a metric.Proof. The function t G is a metric if it fulfills all the three conditions of Definition 2.1.Trivially, the first two conditions hold. Consider now a function f : [0 , ∞ ) → [0 , ∞ ), f ( x ) = x/ ( x + k ) where k > f is increasing in its whole domain[0 , ∞ ), x ≤ y ⇔ xx + k ≤ yy + k . NTRODUCING A NEW INTRINSIC METRIC 5
Because η G is a metric, η G ( x, y ) ≤ η G ( x, z ) + η G ( z, y ) for all x, y, z ∈ G . Furthermore, η G ( z ) ≤ min { η G ( x, z ) + η G ( x ) , η G ( z, y ) + η G ( y ) } . From these results, it follows that t G ( x, y ) = η G ( x, y ) η G ( x, y ) + η G ( x ) + η G ( y ) ≤ η G ( x, z ) + η G ( z, y ) η G ( x, z ) + η G ( z, y ) + η G ( x ) + η G ( y )= η G ( x, z ) η G ( x, z ) + η G ( z, y ) + η G ( x ) + η G ( y ) + η G ( z, y ) η G ( x, z ) + η G ( z, y ) + η G ( x ) + η G ( y ) ≤ η G ( x, z ) η G ( x, z ) + η G ( x ) + η G ( z ) + η G ( z, y ) η G ( z, y ) + η G ( y ) + η G ( z ) = t G ( x, z ) + t G ( z, y )for all x, y, z ∈ G . Thus, t G fulfills the triangle inequality. (cid:3) We now show that the method of proof of Theorem 3.1 can be used to prove that severalother functions are metrics, too.
Theorem 3.2. If G is a proper subset of a metric space X , η G some metric defined inthe closure of G and c G : G × G → [0 , ∞ ) some symmetric function such that, for all x, y, z ∈ G , c G ( x, z ) ≤ η G ( z, y ) + c G ( x, y ) , (3.3) then any function φ G : G × G → [0 , , defined as φ G ( x, x ) = 0 , φ G ( x, y ) = η G ( x, y ) η G ( x, y ) + c G ( x, y ) if x (cid:54) = y for all x, y ∈ G , is a metric in the domain G .Proof. Since η G is a metric and c G is both symmetric and non-negative, the function φ G trivially fulfills the first two conditions of Definition 2.1. Note that, by the triangleinequality of the metric η G and the inequality (3.3), the inequalities η G ( x, y ) ≤ η G ( x, z ) + η G ( z, y ) ,c G ( x, z ) ≤ η G ( z, y ) + c G ( x, y ) ,c G ( z, y ) ≤ η G ( x, z ) + c G ( x, y ) , hold for all x, y, z ∈ G . Now, φ G ( x, y ) = η G ( x, y ) η G ( x, y ) + c G ( x, y ) ≤ η G ( x, z ) + η G ( z, y ) η G ( x, z ) + η G ( z, y ) + c G ( x, y )= η G ( x, z ) η G ( x, z ) + η G ( z, y ) + c G ( x, y ) + η G ( z, y ) η G ( x, z ) + η G ( z, y ) + c G ( x, y ) ≤ η G ( x, z ) η G ( x, z ) + c G ( x, z ) + η G ( z, y ) η G ( z, y ) + c G ( z, y ) = φ G ( x, z ) + φ G ( z, y ) , so the function φ G fulfills the triangle inequality and it must be a metric. (cid:3) O. RAINIO AND M. VUORINEN
Remark 3.4. (1) If the function c G of Theorem 3.2 is strictly positive, the condition φ G ( x, x ) = 0 does not need to be separately specified. Namely, this condition followsdirectly from the fact that η G ( x, x ) = 0 for a metric η G . Note also that if c G is a nullfunction, the function φ G becomes the discrete metric.(2) If η G is a metric, then η αG is a metric, too, for 0 < α ≤
1, but this is not true for α >
Corollary 3.5.
The function ψ : B n × B n → [0 , , defined as ψ ( x, x ) = 0 , ψ ( x, y ) = | x − y || x − y | + c | x || y | if x (cid:54) = y, for all x, y ∈ B n with a constant < c ≤ , is a metric on the unit ball.Proof. Since now c | x | ( | z | − | y | ) ≤ | z | − | y | ≤ | z − y | ⇒ c | x || z | ≤ | z − y | + c | x || y | , for all x, y, z ∈ B n , the result follows from Theorem 3.2. (cid:3) Corollary 3.6. If G is a proper subset of a metric space X and η G is some metric definedin the closure of G such that η G ( x ) = inf { η G ( x, u ) | u ∈ ∂G } ≤ for all x ∈ G , then afunction υ G : G × G → [0 , , defined as υ G ( x, y ) = η G ( x, y ) η G ( x, y ) + c (cid:112) (1 + η G ( x ))(1 + η G ( y )) with a constant < c ≤ √ is a metric in the domain G .Proof. Fix c G ( x, y ) = c (cid:112) (1 + η G ( x ))(1 + η G ( y )). Now, c G ( x, z ) − c G ( x, y ) = c (cid:112) (1 + η G ( x ))(1 + η G ( z )) − c (cid:112) (1 + η G ( x ))(1 + η G ( y ))= c · (1 + η G ( x ))(1 + η G ( z )) − (1 + η G ( x ))(1 + η G ( y )) (cid:112) (1 + η G ( x ))(1 + η G ( z )) + (cid:112) (1 + η G ( x ))(1 + η G ( y ))= c (cid:112) η G ( x )( η G ( z ) − η G ( y )) (cid:112) η G ( z ) + (cid:112) η G ( y ) ≤ √ c ( η G ( z ) − η G ( y ))1 + 1 ≤ η G ( z ) − η G ( y ) ≤ η G ( z, y ) , so the inequality (3.3) holds for all x, y, z ∈ G and the result follows from Theorem 3.2. (cid:3) Corollary 3.7.
The function χ : B n × B n → [0 , , defined as χ ( x, y ) = | x − y || x − y | + c (cid:112) (2 − | x | )(2 − | y | ) for all x, y ∈ B n with a constant < c ≤ √ , is a metric on the unit ball.Proof. Follows from Corollary 3.6. (cid:3)
NTRODUCING A NEW INTRINSIC METRIC 7
Let us focus again on the t -metric. Since the result of Theorem 3.1 holds for any metric η G , the t -metric is trivially a metric also when defined for the Euclidean metric. Below,we will consider the t -metric in this special case only. Let us next prove the inequalitiesbetween the t -metric and the three hyperbolic type metrics defined earlier. Theorem 3.8.
For all domains G (cid:40) R n and all points x, y ∈ G , the following inequalitieshold:(1) j ∗ G ( x, y ) / ≤ t G ( x, y ) ≤ j ∗ G ( x, y ) ,(2) p G ( x, y ) / ≤ t G ( x, y ) ≤ p G ( x, y ) ,(3) s G ( x, y ) / ≤ t G ( x, y ) ≤ s G ( x, y ) .Furthermore, in each case the constants are sharp for some domain G .Proof. (1) t G ( x, y ) ≤ j ∗ G ( x, y ) follows trivially from the definitions of these metrics. If d G ( x ) ≤ d G ( y ), then d G ( y ) ≤ | x − y | + d G ( x ) and d G ( x ) + d G ( y ) ≤ | x − y | + 2 d G ( x ). Inthe same way, if d G ( y ) ≤ d G ( x ), then d G ( x ) + d G ( y ) ≤ | x − y | + 2 d G ( y ). It follows fromthis that d G ( x ) + d G ( y ) ≤ | x − y | + 2 min { d G ( x ) , d G ( y ) } . With this information, we can write j ∗ G ( x, y ) = | x − y || x − y | + 2 min { d G ( x ) , d G ( y ) } = 2 | x − y || x − y | + 2 min { d G ( x ) , d G ( y ) } + | x − y | + 2 min { d G ( x ) , d G ( y ) }≤ | x − y || x − y | + 2 min { d G ( x ) , d G ( y ) } + d G ( x ) + d G ( y ) ≤ | x − y || x − y | + d G ( x ) + d G ( y ) = 2 t G ( x, y ) . The equality t G ( x, y ) = j ∗ G ( x, y ) holds always when d G ( x ) = d G ( y ). For x = i and y = ki , lim k → + ( t H ( x, y ) /j ∗ H ( x, y )) = 1 /
2. Thus, the first inequality of the theorem andits sharpness follow.(2) From Lemma 2.2 and Theorem 3.8(1), it follows that t G ( x, y ) ≤ p G ( x, y ).Let us now prove that p G ( x, y ) / ≤ t G ( x, y ). This is clearly equivalent to | x − y | + d G ( x ) + d G ( y ) ≤ (cid:112) | x − y | + 4 d G ( x ) d G ( y ) . (3.9)Fix u = | x − y | , v = min { d G ( x ) , d G ( y ) } and k = | d G ( x ) − d G ( y ) | . The inequality (3.9) isnow u + 2 v + k ≤ (cid:112) u + 4 v ( v + k ) ⇔ k + 4 uv + 2 uk − u − v − kv ≤ . Define a function f ( k ) = k + 4 uv + 2 uk − u − v − kv . Since the inequality aboveis equivalent to f ( k ) ≤
0, we need to find out the greatest value of this function. There
O. RAINIO AND M. VUORINEN is no upper limit for u ≥ v ≥ ≤ k ≤ u . We can solve that f (cid:48) ( k ) = 2 k + 2 u − v = 0 ⇔ k = 6 v − u. Since f (0) = − u + 4 uv − v ≤ − u − v ≤ ,f (6 v − u ) = − u + 16 uv − v ≤ − v ≤ ,f ( u ) = − uv − v ≤ ,f ( k ) is always non-positive on the closed interval k ∈ [0 , u ] and, consequently, the in-equality p G ( x, y ) / ≤ t G ( x, y ) follows.For x = ki and y = i , lim k → + ( t H ( x, y ) /p H ( x, y )) = 1 / x = ki and y = 1 + ki ,lim k → + ( t H ( x, y ) /p H ( x, y )) = 1.(3) By the triangle inequality and Lemma 2.3 and Theorem 3.8(1), s G ( x, y )2 = | x − y | inf z ∈ ∂G ( | x − z | + | z − y | ) + inf z ∈ ∂G ( | x − z | + | z − y | ) ≤ | x − y | inf z ∈ ∂G ( | x − z | + | z − y | ) + d G ( x ) + d G ( y ) ≤ t G ( x, y ) ≤ j ∗ G ( x, y ) ≤ s G ( x, y ) . For x = ki and y = i , lim k → + ( t H ( x, y ) /s H ( x, y )) = 1 / x = ki and y = 1 + ki ,lim k → + ( t H ( x, y ) /s H ( x, y )) = 1. (cid:3) Proposition 3.10.
For any fixed domain G (cid:40) R n , the inequalities of Theorem 3.8 aresharp.Proof. If G is a proper subdomain, there must exist some ball B n ( x, r ) ⊂ G with S n − ( x, r ) ∩ ∂G (cid:54) = ∅ where r >
0. Fix z ∈ S n − ( x, r ) ∩ ∂G and y ∈ [ x, z ] so that | y − z | = kr with k ∈ (0 , d G ( x ) = r , d G ( y ) = kr , | x − y | = (1 − k ) r and inf z ∈ ∂G ( | x − z | + | z − y | ) =1 + k . Consequently, t G ( x, y ) = (1 − k ) r (1 − k ) r + r + kr = 1 − k ,j ∗ G ( x, y ) = p G ( x, y ) = s G ( x, y ) = 1 − k k . It follows thatlim k → + t G ( x, y ) j ∗ G ( x, y ) = lim k → + t G ( x, y ) j ∗ G ( x, y ) = lim k → + t G ( x, y ) j ∗ G ( x, y ) = lim k → + (cid:18) k (cid:19) = 12 , lim k → − t G ( x, y ) j ∗ G ( x, y ) = lim k → − t G ( x, y ) j ∗ G ( x, y ) = lim k → − t G ( x, y ) j ∗ G ( x, y ) = lim k → − (cid:18) k (cid:19) = 1 . Thus, regardless of how G is chosen, the inequalities of Theorem 3.8 are sharp. (cid:3) NTRODUCING A NEW INTRINSIC METRIC 9
Next, we will study the connection between the t -metric and the hyperbolic metric. Theorem 3.11.
For all x, y ∈ G ∈ { H n , B n } , the inequality
12 th ρ G ( x, y )2 ≤ t G ( x, y ) ≤ th ρ G ( x, y )2 holds and the constants here are sharp.Proof. In the case G = H n , the result follows directly from Lemma 2.4(1) and Theorem3.8(3).Suppose now that G = B n . From Lemma 2.4(2) and Theorem 3.8(1), it follows that t B n ( x, y ) ≤ j ∗ B n ( x, y ) ≤ th ρ B n ( x, y )2 . Thus, the inequality t B n ( x, y ) ≤ th( ρ B n ( x, y ) /
2) holds for all x, y ∈ B n and the sharpnessfollows because x = ke and y = − ke fulfill lim k → − ( t B n ( x, y ) / th( ρ B n ( x, y ) / t -metrics and the hyperbolic metric in the domain B n onlydepend on how the points x, y are located on the two-dimensional plane fixed by thesetwo points and the origin, we can assume without loss of generality that n = 2. Considernow the quotient t B ( x, y )th( ρ B ( x, y ) /
2) = A [ x, y ] | x − y | + 1 − | x | + 1 − | y | . By [1, Lemma 7.57.(1) p. 152], A [ x, y ] ≥ | x − y | + (1 − | x | )(1 − | y | ) for all x, y ∈ B . Itfollows that t B ( x, y )th( ρ B ( x, y ) / ≥ | x − y | + 1 − | x | − | y | + | x || y || x − y | + 1 − | x | + 1 − | y | ≥ | x − y | − | x | − | y | + 1 | x − y | − | x | − | y | + 2 ≥ , which proves that th( ρ B ( x, y ) / / ≤ t B ( x, y ). By the observation above, this also holdsin the more general case where n is not fixed. The inequality is sharp, too: For x = ke and y = − ke , lim k → + ( t B n ( x, y ) / th( ρ B n ( x, y ) / / (cid:3) Theorem 3.12.
For a fixed angle θ ∈ (0 , π ) and for all x, y ∈ S θ , the following resultshold:(1) t S θ ( x, y ) ≤ th( ρ S θ ( x, y ) / ≤ π/θ ) sin( θ/ t S θ ( x, y ) if θ ∈ (0 , π ) ,(2) t S θ ( x, y ) ≤ th( ρ S θ ( x, y ) / ≤ t S θ ( x, y ) if θ = π ,(3) ( π/θ ) t S θ ( x, y ) ≤ th( ρ S θ ( x, y ) / ≤ t S θ ( x, y ) if θ ∈ ( π, π ) .Proof. Follows from Theorems 3.8(3) and 3.11, and [12, Cor. 4.9, p. 9]. (cid:3) Quasiconformal Mappings and Lipschitz Constants
In this section, we will study the behaviour of the t -metric under different conformaland quasiconformal mappings in order to demonstrate how this metric works. Remark 4.1.
The t -metric is invariant under all similarity maps. In particular, the t -metric defined in a sector S θ is invariant under a reflection over the bisector of the sectorand a stretching x (cid:55)→ r · x with any r >
0. Consequently, this allows us to make certainassumptions when choosing the points x, y ∈ S θ .First, let us study how the t -metric behaves under a certain conformal mapping betweentwo sectors with angles at most π . Lemma 4.2. If α, β ∈ (0 , π ] and f : S α → S β , f ( z ) = z ( β/α ) , then for all x, y ∈ S α t S α ( x, y )2 ≤ t S β ( f ( x ) , f ( y )) ≤ β sin( α/ α sin( β/ t S α ( x, y ) if α ≤ β,β sin( α/ α sin( β/ t S α ( x, y ) ≤ t S β ( f ( x ) , f ( y )) ≤ t S α ( x, y ) otherwise.Proof. If α ≤ β , by Theorem 3.8(3) and [12, Lemma 5.11, p. 13], t S β ( f ( x ) , f ( y )) ≥ s S β ( f ( x ) , f ( y ))2 ≥ s S α ( x, y )2 ≥ t S α ( x, y )2 . (4.3)Suppose that α ≤ β still. Fix x = e hi and y = re ki , where 0 < h ≤ k < α and r ≥ t S β ( f ( x ) , f ( y )) t S α ( x, y ) = | − ( re ( k − h ) i ) βα | ( | − re ( k − h ) i | + sin( γ ) + r sin( µ )) | − re ( k − h ) i | ( | − ( re ( k − h ) i ) βα | + sin( βα γ ) + r βα sin( βα µ )) , (4.4)where µ = min { h, α − h } and γ = min { k, α − k } . This quotient is strictly decreasing withrespect to r and, since r ≥
1, it attains its maximum value when r = 1. Consequently,the quotient (4.4) has an upper limit ofsin( β α ( k − h ))(2 sin( k − h ) + sin( γ ) + sin( µ ))sin( k − h )(2 sin( β α ( k − h )) + sin( βα γ ) + sin( βα µ )) . (4.5)The value of the quotient above is at greatest, when k − h is at minimum and both γ and µ are at maximum. This happens when h < α/ k = α − h . Now, γ = µ = h andthe quotient (4.5) is sin( β α ( α − h ))(sin( α − h ) + sin( h ))sin( α − h )(sin( β α ( α − h )) + sin( βα h )) . Since the expression above is strictly increasing with respect to h and h < α/
2, themaximum value of the quotient (4.4) islim h → α − (cid:32) sin( β α ( α − h ))(sin( α − h ) + sin( h ))sin( α − h )(sin( β α ( α − h )) + sin( βα h )) (cid:33) = β sin( α/ α sin( β/ , (4.6)which, together with the inequality (4.3), proves the first part of our theorem. Supposenext that α > β instead. It can be now proved that the minimum value of the quotient NTRODUCING A NEW INTRINSIC METRIC 11 (4.4) is the same limit value (4.6) and, by Theorem 3.8(3) and [12, Lemma 5.11, p. 13], t S β ( f ( x ) , f ( y )) ≤ t S α ( x, y ). Thus, the theorem follows. (cid:3) Let us now consider a more general result than the one above. Namely, instead ofstudying a conformal power mapping, we can assume that, for domains G , G ⊂ R , themapping f : G → G = f ( G ) is a K -quasiconformal homeomorphism, see [13, Ch. 2].Let c ( K ) be as in [7, Thm 16.39, p. 313]. Now, c ( K ) ≥ K and c ( K ) → K →
1. See also the book [5] by F.W. Gehring and K. Hag.
Theorem 4.7. If α, β ∈ (0 , π ) and f : S α → S β = f ( S α ) is a K -quasiconformalhomeomorphism, the following inequalities hold for all x, y ∈ S α . (1) β c ( K ) K π sin( β/ t S α ( x, y ) K ≤ t S β ( f ( x ) , f ( y )) ≤ c ( K )( πα sin( α /K t S α ( x, y ) /K if α, β ∈ (0 , π ] , (2) 12 c ( K ) K t S α ( x, y ) K ≤ t S β ( f ( x ) , f ( y )) ≤ c ( K ) πβ ( πα sin( α /K t S α ( x, y ) /K if α ∈ (0 , π ) and β ∈ ( π, π ) , (3) 12 ( αc ( K ) π ) K t S α ( x, y ) K ≤ t S β ( f ( x ) , f ( y )) ≤ c ( K ) πβ t S α ( x, y ) /K if α, β ∈ [ π, π ) . Proof.
Follows from Theorem 3.8(3) and [12, Cor. 5.7, p. 12]. (cid:3)
Next, we will focus on the radial mapping, which is another example of a quasiconformalmapping, see [13, 16.2, p. 49].
Theorem 4.8. If f : G → G with G = B \{ } is the radial mapping defined as f ( z ) = | z | a − z for some < a < , then for all x, y ∈ G such that | x | = | y | , the sharp inequality t B \{ } ( x, y ) ≤ t B \{ } ( f ( x ) , f ( y )) ≤ a − t B \{ } ( x, y ) holds.Proof. Fix x = re ki and y = re − ki with 0 < r < < k < π/
2. Now, f ( x ) = r a e ki and f ( y ) = r a e − ki . Consider the quotient t G ( f ( x ) , f ( y )) t G ( x, y ) = | f ( x ) − f ( y ) | ( | x − y | + d G ( x ) + d G ( y )) | x − y | ( | f ( x ) − f ( y ) | + d G ( f ( x )) + d G ( f ( y ))) , (4.9)where | x − y | = 2 r sin( k ) , | f ( x ) − f ( y ) | = 2 r a sin( k ) ,d G ( x ) = d G ( y ) = min { r, − r } ,d G ( f ( x )) = d G ( f ( y )) = min { r a , − r a } . If 0 < r < r a < /
2, the quotient (4.9) is t G ( f ( x ) , f ( y )) t G ( x, y ) = r a ( r sin( k ) + r ) r ( r a sin( k ) + r a ) = 1 . If 0 < r ≤ / < r a <
1, the quotient (4.9) is t G ( f ( x ) , f ( y )) t G ( x, y ) = r a ( r sin( k ) + r ) r ( r a sin( k ) + 1 − r a ) , (4.10)which is decreasing with respect to k . Sincelim k → + (cid:18) r a ( r sin( k ) + r ) r ( r a sin( k ) + 1 − r a ) (cid:19) = r a r (1 − r a ) = r a − r a is increasing with respect to r and r ≤ /
2, the maximum value of the quotient (4.10) is1 / (2 a − k → − (cid:18) r a ( r sin( k ) + r ) r ( r a sin( k ) + 1 − r a ) (cid:19) = r a ( r + r ) r ( r a + 1 − r a ) = 2 r a is increasing with respect to r a and r a > /
2, so the quotient (4.10) is always more than1. If 1 / < r < r a <
1, the quotient (4.9) is t B \{ } ( f ( x ) , f ( y )) t B \{ } ( x, y ) = r a ( r sin( k ) + 1 − r ) r ( r a sin( k ) + 1 − r a ) , (4.11)which is decreasing with respect to k . Since r > / k → + (cid:18) r a ( r sin( k ) + 1 − r ) r ( r a sin( k ) + 1 − r a ) (cid:19) = r a (1 − r ) r (1 − r a ) , is decreasing with respect to r , the quotient (4.10) is less than 1 / (2 a − k → − (cid:18) r a ( r sin( k ) + 1 − r ) r ( r a sin( k ) + 1 − r a ) (cid:19) = r a − , which is clearly more than 1.Thus, the minimum value of the quotient (4.9) is 1 and the maximum value 1 / (2 a − (cid:3) Let us now find Lipschitz constants of a few different mappings for the t -metric. Theorem 4.12.
For all conformal mappings f : G → G = f ( G ) with G , G ∈{ H n , B n } , the inequality t G ( x, y ) ≤ t G ( f ( x ) , f ( y )) ≤ t G ( x, y ) holds for all x, y ∈ G . NTRODUCING A NEW INTRINSIC METRIC 13
Proof.
By Theorem 3.11 and the conformal invariance of the hyperbolic metric,12 t G ( x, y ) ≤
12 th ρ G ( x, y )2 = 12 th ρ G ( f ( x ) , f ( y ))2 ≤ t G ( f ( x ) , f ( y )) ≤ th ρ G ( f ( x ) , f ( y ))2 = th ρ G ( x, y )2 ≤ t G ( x, y ) . (cid:3) It follows from Theorem 4.12 that the Lipschitz constant Lip( f | G ) for the t -metric inany conformal mapping f : G → G = f ( G ), G , G ∈ { H n , B n } , is at most 2. Supposenow that h is the M¨obius transformation h : B → H , h ( z ) = (1 − z ) i/ (1 + z ). Since, for x = 0 and y = − kk +1 with 0 < k < k → − (cid:18) t H n ( h ( x ) , h ( y )) t B n ( x, y ) (cid:19) = lim k → − ( k + 1) = 2 , the Lipschitz constant Lip( h | B ) is equal to 2. However, for certain M¨obius transforma-tions, there might be a better constant than 2. For instance, the following conjecture issupported by several numerical tests. Conjecture 4.13.
For all a, x, y ∈ B , the M¨obius transformation T a : B → B , T a ( z ) =( z − a ) / (1 − az ) fulfills the inequality t B ( T a ( x ) , T a ( y )) ≤ (1 + | a | ) t B ( x, y ) . Remark 4.14.
It is also an open question whether the inequality of Conjecture 4.13 holdsfor the triangular ratio metric or so called Barrlund metric, but numerical tests suggestso, see [3, Conj. 1.6, p. 684] and [4, Conj. 4.3, p. 25].In the next few results, we will study a mapping f ∗ : S θ → S θ , f ∗ ( x ) = x/ | x | definedin some open sector S θ , and find its Lipschitz constants for the t -metric. Theorem 4.15. If θ ∈ (0 , π ] and f ∗ is the mapping f ∗ : S θ → S θ , f ∗ ( x ) = x/ | x | , theLipschitz constant Lip ( f ∗ | S θ ) for the t -metric is θ/ .Proof. Without loss of generality, we can fix x = e hi and y = re ki with 0 < h ≤ π/ h ≤ k < θ and r >
0. Since x ∗ = e hi and y ∗ = (1 /r ) e ki , it follows that t S θ ( x ∗ , y ∗ ) t S θ ( x, y ) = (cid:112) r − r cos( k − h ) + r sin( h ) + sin(min { k, θ − k } ) (cid:112) r − r cos( k − h ) + sin( h ) + r sin(min { k, θ − k } ) . To maximize this, we clearly need to choose k = θ/ r and h as small aspossible. If k = θ/
2, lim h → + , r → + t S θ ( x ∗ , y ∗ ) t S θ ( x, y ) = 1 + sin( θ/ , so the theorem follows. (cid:3) Theorem 4.16. If x ∗ = x/ | x | and y ∗ = y/ | y | , the equality s S θ ( x, y ) = s S θ ( x ∗ , y ∗ ) holdsin an open sector S θ with θ ∈ (0 , π ) .Proof. Fix x = e hi and y = re ki where r > < h ≤ k < θ . Clearly, x ∗ = x = e hi and y ∗ = (1 /r ) e ki . Suppose first that θ ≤ π . By the known solution to Heron’s problem,the infimum inf z ∈ ∂S θ ( | x − z | + | z − y | ) is min {| x − y | , | x − y (cid:48) |} , where y (cid:48) is the point y reflected over the left side of the sector θ . Clearly, | x − y | ≤ | x − y (cid:48) | ⇔ | e − hi − re ki | ≤ | e hi − re (2 θ − k ) i |⇔ | − re ( h + k ) i | ≤ | − re (2 θ − h − k ) i | ⇔ h + k ≤ θ − h − k ⇔ ( h + k ) / ≤ θ/ . By symmetry, we can suppose that ( h + k ) / ≤ θ/ z ∈ ∂S θ ( | x − z | + | z − y | ) = | x − y | but alsoinf z ∈ ∂S θ ( | x ∗ − z | + | z − y ∗ | ) = | x ∗ − y ∗ | . Now, s S θ ( x, y ) = | x − y || x − y | = | − re ( k − h ) i || − re ( k + h ) i | = | − re ( h − k ) i || − re − ( k + h ) i | = | r − e ( k − h ) i || r − e ( k + h ) i | = | − (1 /r ) e ( k − h ) i || − (1 /r ) e ( k + h ) i | = | x ∗ − y ∗ || x ∗ − y ∗ | = s S θ ( x ∗ , y ∗ ) . Consider now the case where θ > π . If k − h ≥ π , then s S θ ( x, y ) = 1 = s S θ ( x ∗ , y ∗ )always so suppose that k − h < π instead. This leaves us three possible options. If( h + k ) / ≤ π/
2, then ( h + k ) / < θ/ s S θ ( x, y ) = | x − y || x − y | = | x ∗ − y ∗ || x ∗ − y ∗ | = s S θ ( x ∗ , y ∗ ) , just like above. By symmetry, s S θ ( x, y ) = s S θ ( x ∗ , y ∗ ) also if ( k + h ) / ≥ θ − π/
2. If π/ < ( k + h ) / < θ − π/ s S θ ( x, y ) = | x − y || x | + | y | = | − re ( k − h ) i | r = | r − e ( k − h ) i | r = | − (1 /r ) e ( k − h ) i | /r = | x ∗ − y ∗ || x ∗ | + | y ∗ | = s S θ ( x ∗ , y ∗ ) . (cid:3) Theorem 4.17. If θ ∈ [ π, π ) and f ∗ is the mapping f ∗ : S θ → S θ , f ∗ ( x ) = x/ | x | , theLipschitz constant Lip ( f ∗ | S θ ) for the t -metric is 2.Proof. It follows from Theorems 4.16 and 3.8(3) that t S θ ( x, y )2 ≤ t S θ ( x ∗ , y ∗ ) ≤ t S θ ( x, y ) NTRODUCING A NEW INTRINSIC METRIC 15 for all x, y ∈ S θ . Since for x = e hi and y = re πi/ with h < π/ r > h,r → + t S θ ( x ∗ , y ∗ ) t S θ ( x, y ) = lim h,r → + (cid:32) (cid:112) r − r cos( π/ − h ) + r sin( h ) + 1 (cid:112) r − r cos( π/ − h ) + sin( h ) + r (cid:33) = 2 , and it follows that sup { t S θ ( x ∗ , y ∗ ) t S θ ( x, y ) | x, y ∈ S θ , x (cid:54) = y, θ ∈ [ π, π ) } = 2 . (cid:3) Comparison of Metric Balls
Next, we will graphically demonstrate the differences and similarities between the var-ious metrics considered in this paper by drawing for each metric several circles centeredat the same point but with different radii. In all of the figures of this section, the domain G ⊂ R is a regular five-pointed star and the circles have a radius of r = 1 / , ..., / G in the first figures, and then off the centerin the rest of the figures. All the figures in this section were drawn by using the contourplot function contour in R-Studio and choosing a grid of the size 1,000 × n -dimensional metric balls.For several hyperbolic type metrics, the metric balls of small radii resemble Euclideanballs, but the geometric structure of the boundary of the domain begins to affect theshape of these balls when their radii grow large enough, see [7, Ch. 13, pp. 239-259]. Byanalysing this phenomenon more carefully, we can observe, for instance, that the ballsare convex with radii less than some fixed r > Theorem 5.1.
If the domain G is a polygon, then the corner points of the circles S p ( x, r ) and S t ( x, r ) are located on the the angle bisectors of G .Proof. Suppose G has sides l and l that have a common endpoint k . Fix x ∈ G andchoose some point y ∈ G so that k is the vertex of G that is closest to y and there isno other side closer to y than l and l . Thus, d G ( y ) = min { d ( y, l ) , d ( y, l ) } and, fora fixed distance | x − y | , d G ( y ) is at maximum when d ( y, l ) = d ( y, l ). The condition d ( y, l ) = d ( y, l ) is clearly fulfilled when y is on the bisector of ∠ ( l , l ) and, the greaterthe d G ( y ), the smaller the distances p G ( x, y ) and t G ( x, y ) are now. Consequently, if the (a) s G -metric circles. (b) j ∗ G -metric circles. (c) p G -metric circles. (d) t G -metric circles. Figure 1.
Circles in a five-pointed star domain with different metrics.circle S p ( x, r ) or S t ( x, r ) has a corner point, it must be located on an angle bisector of G . (cid:3) However, it can been seen from Figures 2a and 2b that the circles with s G - and j ∗ G -metrics can have corner points also elsewhere than on the angle bisectors of the domain NTRODUCING A NEW INTRINSIC METRIC 17 (a) s G -metric circles. (b) j ∗ G -metric circles. (c) p G -metric circles. (d) t G -metric circles. Figure 2.
Circles in a five-pointed star domain with different metrics. G . We also notice that the circles in Figure 2b clearly differ those in Figures 2a and 2c.This can be described with the concept of starlikeness, which is a looser form of convexity.Namely, a set K is starlike with respect to a point x ∈ K if and only if the segment [ x, y ]belongs to K fully for every y ∈ K . In particular, the five-pointed star domain is starlike with respect to its center. The disks by the j ∗ G - and t G -metrics (Figures 2a and 2c) areclearly not starlike and, even if it cannot be clearly seen from Figure 2d, there are disksdrawn with the point pair function p G that are not starlike. Lemma 5.2.
There exist disks B j ∗ ( x, r ) , B p ( x, r ) , and B t ( x, r ) that are not starlike withrespect to their center.Proof. Consider a domain G = H ∪ { z ∈ C | − < Re( z ) < , − < Im( z ) ≤ } . Fix x = − i and y = 3 + i . Clearly, d G ( x ) = d G ( y ) = 1 and | x − y | = 3 √
2. Consequently, j ∗ G ( x, y ) = t G ( x, y ) = 33 + √ < . , p G ( x, y ) = 3 √ < . . The segment [ x, y ] does not clearly belong to G fully and no disk in G can contain thissegment. However, its end point y is clearly included in the disks B j ∗ ( x, . B t ( x, . B p ( x, . (cid:3) There are no disks or balls like this for the triangular ratio metric.
Lemma 5.3. [7, p. 206]
The balls B s ( x, r ) in any domain G (cid:40) R n are always starlikewith respect to their center x . For several common hyperbolic type metrics η G , the closed ball B η ( x, M ) with M = η G ( x, y ) and x, y ∈ G is always a compact subset of the domain G , see [7, p. 79]. Forinstance, the hyperbolic metric ρ G has this property [7, p. 192]. As can be seen from thefigures, the j ∗ -metric, the triangular ratio metric and the point pair function share thisproperty, too. Lemma 5.4.
The balls B j ∗ ( x, r ) , B p ( x, r ) and B s ( x, r ) touch the boundary of the domain G (cid:40) R n if and only if r = 1 .Proof. If the ball B η ( x, r ), η G ∈ { j ∗ G , p G , s G } , touches the boundary of G , then there issome point y ∈ S η ( x, r ) with d G ( y ) = 0 and j ∗ G ( x, y ) = p G ( x, y ) = s G ( x, y ) = 1. Thus, weneed to just prove that the balls with radius 1 always touch the boundary. Consider firstthe balls B j ∗ ( x, r ) and B p ( x, r ), with a radius r = 1. Since, for all the points y on theirboundary, j ∗ G ( x, y ) = 1 ⇔ { d G ( x ) , d G ( y ) } = 0 ⇔ d G ( x ) = 0 or d G ( y ) = 0 ,p G ( x, y ) = 1 ⇔ d G ( x ) d G ( y ) = 0 ⇔ d G ( x ) = 0 or d G ( y ) = 0 , the balls B j ∗ ( x,
1) and B p ( x,
1) touch the boundary of G .Consider yet the triangular ratio metric. Because only balls with radius r = 1 cantouch the boundary, B s ( x, ∩ ∂G = ∅ . However, if s G ( x, y ) = 1, there is some point z ∈ ∂G such that | x − y | = | x − z | + | z − y | . This means that z is on a line segment [ x, y ]and, since z / ∈ B s ( x, z must be arbitrarily close to the point y . Thus, d G ( y ) = 0 andthe ball B s ( x,
1) touches the boundary. (cid:3)
However, the t -metric differs from the hyperbolic type metrics in this aspect: the closureof a t -metric ball is a compact set, if and only if the radius of the ball is less than 1 / NTRODUCING A NEW INTRINSIC METRIC 19
Theorem 5.5.
The balls B t ( x, r ) touch the boundary of the domain G (cid:40) R n if and onlyif r ≥ .Proof. If B t ( x, r ) touches the boundary, there must be some y ∈ S t ( x, r ) such that d G ( y ) =0. Since d G ( x ) ≤ | x − y | + d G ( y ), it follows that r = t G ( x, y ) = | x − y || x − y | + d G ( x ) + d G ( y ) ≥ | x − y || x − y | + | x − y | + 0 + 0 = 12 . Thus, only balls B t ( x, r ) with a radius r ≥ can touch the boundary of G .Let us yet prove that the balls B t ( x, ) always touch the boundary of G . For any point y ∈ S t ( x, ), it holds that | x − y | = 1 / t G ( x, y ) = | x − y || x − y | + d G ( x ) + d G ( y ) = 12 ⇔ d G ( y ) = | x − y | − d G ( x ) . Since only balls B t ( x, r ) with r ≥ can touch the boundary of G , B t ( x, ) ∩ ∂G = ∅ and d G ( x ) ≥ /
2. Thus, d G ( y ) = 1 / − d G ( x ) ≤ d G ( y ) = 0 and the ball B t ( x, ) truly touches the boundary of G . (cid:3) The result above is visualized in Figures 1d and 2d.
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