aa r X i v : . [ m a t h . M G ] S e p TRIANGULAR RATIO METRIC IN THE UNIT DISK
OONA RAINIO AND MATTI VUORINEN
Abstract.
The triangular ratio metric is studied in a domain G ( R n , n ≥
2. Severalsharp bounds are proven for this metric, especially, in the case where the domain is theunit disk of the complex plane. The results are applied to study the H¨older continuityof quasiconformal mappings. Introduction
In geometric function theory, metrics are often used to define new types of geometriesof subdomains of the Euclidean, Hilbert, Banach and other metric spaces [6, 7, 13, 15].One can introduce a metric topology and build new types of geometries of a domain G ⊂ R n , n ≥
2, based on metrics. Since the local behavior of functions defined on G is an important area of study, it is natural to require that, given a point in G , a metricrecognizes points close to it from the boundary ∂G .Thus, certain constraints on metrics are necessary. A natural requirement is that thedistance defined by a metric for given two points x, y ∈ G takes into account both howfar the points are from each other and also their location with respect to the boundary.Indeed, we require that the closures of the balls defined by the metrics do not intersectthe boundary ∂G of the domain. We call these type of metrics intrinsic metrics . Ageneric example of an intrinsic metric is the hyperbolic metric [2] of a planar domainor its generalization, the quasihyperbolic metric [8] defined in all proper subdomains of G ( R n , n ≥ G isinvariant under conformal automorphisms of G . File: main.tex, printed: 2020-9-3, 1.25
Mathematics Subject Classification.
Primary 51M10; Secondary 30C65.
Key words and phrases.
H¨older continuity, hyperbolic geometry, midpoint rotation, quasiconformalmappings, triangular ratio metric.
In 2002, P. H¨ast¨o [12] introduced the triangular ratio metric , defined in a domain G ( R n as the function s G : G × G → [0 , s G ( x, y ) = | x − y | inf z ∈ ∂G ( | x − z | + | z − y | ) . This metric was studied recently in [3, 4, 11], and our goal here is to continue this inves-tigation. We introduce new methods for estimating the triangular ratio metric in termsof several other metrics and establish several results with sharp constants.In order to compute the value of the triangular ratio metric between points x and y ina domain G , we must find a point z on the boundary of G that gives the infimum for thesum | x − z | + | z − y | . This is a very trivial task if the domain is, for instance, a half-planeor a polygon, but solving the triangular ratio distance in the unit disk is a complicatedproblem with a very long history, see [4]. However, there are two special cases where thisproblem becomes trivial: If the points x and y in the unit disk are collinear with the originor at the same distance from the origin, there are explicit formulas for the triangular ratiometric.Since the points x and y can be always rotated around their midpoint to end up intoone of these two special cases, we can estimate the value of the triangular ratio metric,regardless of how the original points are located in the unit disk. This rotation can bedone either by using Euclidean or hyperbolic geometry, and the main result of this articleis to prove that both these ways give lower and upper limits for the value of the triangularratio metric. Note that while we study the midpoint rotation only in the two-dimensionaldisk, our results can be directly extended into the case B n , n ≥
3, for the point z givingthe infimum is always on the same two-dimensional disk as x , y , and the origin.The structure of this article is as follows. First, we show a few simple ways to findbounds for the triangular ratio metric in Section 3. We define the Euclidean midpointrotation and prove the inequalities related to it in Section 4 and then do the same for thehyperbolic midpoint rotation in Section 5. Finally, in Section 6, we explain how findingbetter bounds for the triangular ratio metric can be useful for studying K -quasiconformalmappings in the unit disk. Acknowledgements.
The authors are indebted to Professor Masayo Fujimura andProfessor Marcelina Mocanu for their kind help in connection with the proof of Theorem4.4. The research of the first author was supported by Finnish Concordia Fund.2.
Preliminaries
Let G be some non-empty, open, proper and connected subset of R n . For all x ∈ G , d G ( x ) is the Euclidean distance d ( x, ∂G ) = inf {| x − z | | z ∈ ∂G } . Other than thetriangular ratio metric defined earlier, we will need the following hyperbolic type metrics: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 ) } , RIANGULAR RATIO METRIC IN THE UNIT DISK 3 the point pair function p G : G × G → [0 , ,p G ( x, y ) = | x − y | p | x − y | + 4 d G ( x ) d G ( y )and the Barrlund metric b G,p : G × G → [0 , ∞ ), b G,p ( x, y ) = sup z ∈ ∂G | x − y | ( | x − z | p + | z − y | p ) /p . Note that the function p G is not metric in all domains [3, Rmk 3.1 p. 689].The hyperbolic metric is defined 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 in the upper half-plane H n and in the Poincar unit disk B n , respectively [9, (4.8), p. 52;(4.14), p. 55]. In the two-dimensional unit disk,th ρ B ( x, y )2 = th( 12 log( | − xy | + | x − y || − xy | − | x − y | )) = | x − y − xy | = | x − y | A [ x, y ] , where y is the complex conjugate of y and A [ x, y ] = p | x − y | + (1 − | x | )(1 − | y | ) isthe Ahlfors bracket [9, (3.17) p. 39]. The hyperbolic segment between points x and y is denoted by J [ x, y ], while Euclidean lines, line segments, balls and spheres are writtenin forms L ( x, y ), [ x, y ], B n ( x, r ) and S n − ( x, r ), respectively, just like in [9, pp. vii-xi].Note that if the center x or the radius r is not specified in the notations B n ( x, r ) and S n − ( x, r ), it means that x = 0 and r = 1. The hyperbolic ball is denoted by B nρ ( q, R ),as in the following lemma. Lemma 2.1. [9, (4.20) p. 56]
The equality B nρ ( q, R ) = B n ( j, h ) holds, if j = q (1 − t )1 − | q | t , h = (1 − | q | ) t − | q | t and t = th (cid:18) R (cid:19) . In order to use the results of Section 5, the formula for the hyperbolic midpoint isneeded.
Theorem 2.2. [17, Thm 1.4, p.3]
For all x, y ∈ B , the hyperbolic midpoint q of J [ x, y ] with ρ B ( x, q ) = ρ B ( q, y ) = ρ B ( x, y ) / is given by q = y (1 − | x | ) + x (1 − | y | )1 − | x | | y | + A [ x, y ] p (1 − | x | )(1 − | y | ) . Furthermore, the next results will be useful when studying the triangular ratio metricin the unit disk.
O. RAINIO AND M. VUORINEN
Theorem 2.3. [9, p. 460]
For all x, y ∈ B n , 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 . Theorem 2.4. [4, p. 138]
For all x, y ∈ B n , the radius drawn to the point z giving theinfimum inf z ∈ S n − ( | x − z | + | z − y | ) bisects the angle ∡ XZY . Lemma 2.5. [9, 11.2.1(1) p. 205]
For all x, y ∈ B n , s B n ( x, y ) ≤ | x − y | − | x + y | , where the equality holds if the points x, y are collinear with the origin. Theorem 2.6. [10, Thm 3.1, p. 276] If x = h + ki ∈ B with h, k > , then s B ( x, x ) = | x | if | x − | > ,s B ( x, x ) = k p (1 − h ) + k ≤ | x | otherwise. Bounds for triangular ratio metric
In this section, we will introduce a few different upper and lower bounds for the tri-angular ratio metric in the unit disk B , using Barrlund metric and a special lower limitfunction. There are numerous similar results already in literature, but we complementthem and prove that our inequalities are sharp by showing that they have the best possibleconstant. First, we introduce the following inequality: Lemma 3.1.
For all y ∈ G , the inequality s G ( x, y ) ≤ | x − y | d G ( x ) + q | x − y | + d G ( x ) − d G ( x ) p | x − y | − d G ( y ) , holds, if the domain G is starlike with respect to x ∈ G and d G ( x ) + d G ( y ) ≤ | x − y | .Proof. Let G be starlike with respect to x ∈ G and consider an arbitrary y ∈ G . Clearly, B n ( x, d G ( x )) , B n ( y, d G ( y )) ⊂ G . It also follows from the starlikeness of G that the con-vex hull ∪ u ∈ B n ( y,d G ( y )) [ x, u ] must belong to G . Fix u, v ∈ S n − ( y, d G ( y )), u = v on thesame plane with the points x, y so that the lines L ( x, u ) and L ( y, v ) are tangents of S n − ( y, d G ( y )), and fix z ∈ S n − ( x, d G ( x )) ∩ [ x, u ].By the starlikeness of G , ∪ s ∈ B n ( y,d G ( y )) [ x, s ] ⊂ G , so it follows that z fulfills | x − z | + | z − y | ≤ inf z ∈ ∂G ( | x − z | + | z − y | ) ⇔ s G ( x, y ) ≤ | x − y || x − z | + | z − y | . RIANGULAR RATIO METRIC IN THE UNIT DISK 5
Here, | x − z | = d G ( x ) and, with the information that | u − y | = | y − v | = d G ( y ) and ∡ XU Y = ∡ Y V X = π/
2, we can conclude that | z − y | = q | x − y | + d G ( x ) − d G ( x ) p | x − y | − d G ( y ) . Thus, the lemma follows. (cid:3)
Remark 3.2.
The same method as the proof of Lemma 3.1 can be also applied intothe case where G is convex. In that case, J = ∪ s ∈ B n ( x,d G ( x )) , t ∈ B n ( y,d G ( y )) [ s, t ] ⊂ G for all x, y ∈ G , so s G ( x, y ) ≤ | x − y || x − z | + | z − y | where z is chosen from ∂J so that | x − z | + | z − y | is at minimum. By finding the valueof this sum, we end up with the result s G ( x, y ) ≤ p G ( x, y ), which holds by [9, Lemma11.6(1), p. 197].Let us now focus on the Barrlund metric. Lemma 3.3. [5, Thm 3.6 p. 7]
For all x, y ∈ G ( R n , s G ( x, y ) ≤ b G, ( x, y ) ≤ − /p s G ( x, y ) . Theorem 3.4. [5, Thm 3.15 p. 11]
For all x, y ∈ B , b B , ( x, y ) = | x − y | p | x | + | y | − | x + y | . Lemma 3.5.
For all x, y ∈ B , √ b B , ( x, y ) ≤ s B ( x, y ) ≤ b B , ( x, y ) . Furthermore, this inequality is sharp.Proof.
The inequality follows from Lemma 3.3. Let x = 0 and y = k with 0 < k <
1. ByLemma 2.5 and Theorem 3.4, s B ( x, y ) = k − k and b B , ( x, y ) = k √ k − k , so we will have limit valueslim k → + s B ( x, y ) b B , ( x, y ) = lim k → + (cid:18) √ k − k − k (cid:19) = 1 √ k → − s B ( x, y ) b B , ( x, y ) = 1 . Thus, the sharpness follows. (cid:3)
Let us next study the connection between Barrlund metric and two other hyperbolictype metrics that can used to bound the value of the triangular ratio metric in the unitdisk, see Theorem 2.3.
O. RAINIO AND M. VUORINEN
Theorem 3.6.
For all x, y ∈ B , the sharp inequality b B , ( x, y ) ≤ j ∗ B ( x, y ) ≤ b B , ( x, y ) holds.Proof. The inequality follows from Lemma 3.5, Theorem 2.3 and [11, Thm 2.9(1), p.1129]. By Theorem 3.4, j ∗ B ( x, y ) b B , ( x, y ) = p | x | + | y | − | x + y || x − y | + 2 − | x | − | y | . For x = 0 and y = k with 0 < k < k → − j ∗ B ( x, y ) b B , ( x, y ) = lim k → − (cid:18) √ k − k (cid:19) = 12 . and, for x = − k and y = k with 0 < k < k → − j ∗ B ( x, y ) b B , ( x, y ) = lim k → − r k ! = 1 . Thus, the sharpness follows. (cid:3)
Theorem 3.7.
For all x, y ∈ B , the sharp inequality √ b B , ( x, y ) ≤ p B ( x, y ) ≤ √
10 + √ b B , ( x, y ) holds.Proof. Consider now the quotient p B ( x, y ) b B , ( x, y ) = s | x | + | y | − | x + y || x − y | + 4(1 − | x | )(1 − | y | ) . (3.8)By Lemma 3.5 and [9, 11.16(1), p. 203], b B , ( x, y ) / √ ≤ s B ( x, y ) ≤ p B ( x, y ) holds forall x, y ∈ B . This inequality is sharp, because, for x = 0 and y = k ,lim k → + p B ( x, y ) b B , ( x, y ) = lim k → + √ k + 2 k + 22 − k ! = 1 √ . Without loss of generality, fix x = h and y = je µi with 0 ≤ h ≤ j < < µ < π .The quotient (3.8) is now p B ( x, y ) b B , ( x, y ) = s h + j − p h + j + 2 hj cos( µ ) h + j − hj cos( µ ) + 4(1 − h )(1 − j ) . RIANGULAR RATIO METRIC IN THE UNIT DISK 7
This is decreasing with respect to cos( µ ), so we can assume that µ = π and cos( µ ) = − p B ( x, y ) b B , ( x, y ) = s (1 + h ) + (1 − j ) ( h + j ) + 4(1 − h )(1 − j ) = s (1 + h ) + (1 − h − q ) (2 h + q ) + 4(1 − h )(1 − h − q ) , where q = j − h ≥
0. The quotient above is clearly decreasing with respect to q . Thus,let us fix j = h . It follows that p B ( x, y ) b B , ( x, y ) = r h h − h + 4 = r h h − h + 2 ≡ p f ( h ) , where f : [0 , → R , f ( h ) = (1 + h ) / (4 h − h + 2). By differentiation, for 0 ≤ h < f ′ ( h ) = ∂∂h (cid:18) h h − h + 2 (cid:19) = − ( h + h − h − h + 1) = 0 ⇔ h = √ − . Since f (0 . > f (0 . <
0, the quotient (3.8) has a maximum q f (( √ − /
2) =( √
10 + √ / (cid:3) Finally, we will introduce one special function defined in the punctured unit disk.
Definition 3.9.
For x, y ∈ B \{ } , definelow( x, y ) = | x − y | min {| x − y ∗ | , | x ∗ − y |} , where x ∗ = x/ | x | and y ∗ = y/ | y | . Remark 3.10.
The low-function is not a metric on the punctured unit disk: By choosingpoints x = 0 . y = − . z = 0 .
1, we will have0 . ≈ low( x, y ) > low( x, z ) + low( z, y ) ≈ . , so the triangle inequality does not hold.Furthermore, because A [ x, y ] = | x || y − x ∗ | for x, y ∈ B n \ { } , it follows that(3.11) th ρ B ( x, y )2 = | x − y || x || y − x ∗ | ≥ low( x, y ) , see [1, 7.44(20)]. Note also that, by [1, 7.42(1)], the left hand side of (3.11) defines ametric.This low-function is a suitable lower bound for the triangular ratio metric, as the nexttheorem states. Lemma 3.12.
For all x, y ∈ B \{ } , the inequality s B ( x, y ) ≥ low( x, y ) holds. O. RAINIO AND M. VUORINEN
Proof.
Suppose that | x − y ∗ | ≤ | x ∗ − y | and fix z ∈ [ x, y ∗ ] ∩ S . Clearly, d ( y, S ) < d ( y ∗ , S ) ⇔ − | y | < | y ∗ | − | y | − ⇔ | y | − | y | = 1 | y | ( | y | − > . It follows from this that s B ( x, y ) ≥ | x − y || x − z | + | z − y | > | x − y || x − z | + | z − y ∗ | = | x − y || x − y ∗ | = low( x, y ) . (cid:3) As a lower bound for s B ( x, y ), the low-function is essentially sharp, when max {| x | , | y |} →
1. However, the low-function does not give any useful upper limits for the triangular ratiometric, unless we limit the absolute value of the points inspected from below. This canbe seen our next theorem.
Theorem 3.13.
For all x, y ∈ B \{ } , the triangular ratio metric and its lower boundfulfill sup { s B ( x, y )low( x, y ) | max {| x | , | y |} ≥ r } ≤ r r , where the equality holds if max {| x | , | y |} = r .Proof. Consider the quotient s B ( x, y )low( x, y ) = min {| x − y ∗ | , | x ∗ − y |} inf z ∈ S ( | x − z | + | z − y | ) . (3.14)Fix x, y ∈ B such that 0 < | x | ≤ | y | and choose z ∈ S so that it gives the infimum inthe denominator of the quotient (3.14). Let k = ∡ ZOX and k = ∡ Y ZO , where thepoint o is the origin. Note that, by Theorem 2.4, ∡ XZO = ∡ OZY , so it follows that0 ≤ k ≤ k ≤ π/
2. We can write thatinf z ∈ S ( | x − z | + | z − y | ) = p | x | + 1 − | x | cos( k ) + p | y | + 1 − | y | cos( k ) . Furthermore, | x − y ∗ | = s | x | + 1 | y | − | x || y | cos( k + k ) , | x ∗ − y | = s | y | + 1 | x | − | y || x | cos( k + k ) . RIANGULAR RATIO METRIC IN THE UNIT DISK 9
Now, we can find an upper bound for the quotient (3.14): s B ( x, y )low( x, y ) ≤ | x − y ∗ | inf z ∈ S ( | x − z | + | z − y | ) ≤ sup ≤ k ≤ k ≤ π/ p | x | + 1 / | y | − | x | / | y | ) cos( k + k ) p | x | + 1 − | x | cos( k ) + p | y | + 1 − | y | cos( k )= inf ≤ k ≤ k ≤ π/ p | x | + 1 − | x | cos( k ) + p | y | + 1 − | y | cos( k ) p | x | + 1 / | y | − | x | / | y | ) cos( k + k ) ! − ≤ inf ≤ k ≤ k ≤ π/ s | x | + 1 − | x | cos( k ) | x | + 1 / | y | − | x | / | y | ) cos( k + k )+ inf ≤ k ≤ k ≤ π/ s | y | + 1 − | y | cos( k ) | x | + 1 / | y | − | x | / | y | ) cos( k + k ) ! − = s | x | + 1 − | x || x | + 1 / | y | − | x | / | y | ) + s | y | + 1 − | y || x | + 1 / | y | − | x | / | y | ) ! − = (cid:18) − | x | / | y | − | x | + 1 − | y | / | y | − | x | (cid:19) − = 1 / | y | − | x | − | x | − | y | . (3.15)Let us yet find another upper bound for the quotient (3.15). It can be shown bydifferentiation that the function f : (0 , → R , f ( | x | ) = 1 / | y | − | x | − | x | − | y | is increasing. It follows from this that | x | ≤ | y | ⇔ f ( | x | ) ≤ f ( | y | ) ⇔ / | y | − | x | − | x | − | y | ≤ / | y | − | y | − | y | − | y | = 1 + | y | | y | . Thus, for all x, y ∈ B such that 0 < | x | ≤ | y | , the quotient (3.14) fulfills the inequality s B ( x, y )low( x, y ) ≤ / | y | − | x | − | x | − | y | ≤ | y | | y | . (3.16)Fix now x = 1 / y = 1 / j with 0 < j < /
2. The quotient (3.14) is now s B ( x, y )low( x, y ) = 3 + 2 j (2 + 4 j )(1 − j ) = 1 + | y | | y | (1 − j ) = 11 − j · | y | | y | , and it has a limit value lim j → + s B ( x, y )low( x, y ) = 1 + | y | | y | . Thus, the inequality (3.16) is sharp and this result proves thatsup s B ( x, y )low( x, y ) = 1 + max {| x | , | y |} {| x | , | y |} . Since the quotient (1 + k ) / (2 k ) is decreasing for k ∈ (0 , (cid:3) The low-function yields a lower limit for also other hyperbolic type metrics.
Lemma 3.17.
For all x, y ∈ B \{ } , the following inequalities hold and are sharp:1. low( x, y ) ≤ √ j ∗ B ( x, y ) ,2. low( x, y ) ≤ p B ( x, y ) ,3. low( x, y ) ≤ b B , ( x, y ) .Furthermore, there is no c > such that low( x, y ) ≥ c · d ( x, y ) for all x, y ∈ B \{ } , where d ∈ { j ∗ B , p B , b B , } .Proof. The inequalities follow from Theorem 2.3, Lemmas 3.5 and 3.12, and [11, Thm2.9(1), p. 1129]. Fix x = k and y = − k with 0 < k <
1. Sincelim k → − low( x, y ) j ∗ B ( x, y ) = lim k → − (cid:18) kk + 1 (cid:19) = 1 , lim k → − low( x, y ) p B ( x, y ) = lim k → − (cid:18) k √ k − k + 1 k + 1 (cid:19) = 1 , lim k → − low( x, y ) b B , ( x, y ) = lim k → − √ kk + 1 ! = 1 , the inequalities are sharp. The latter part of the lemma follows the fact that the limitvalues above are all 0 if k → − instead. (cid:3) Euclidean Midpoint Rotation
In this section, we introduce the Euclidean midpoint rotation. Finding the value of thetriangular ratio distance for two points in the unit disk is a trivial problem, if the pointsare collinear with the origin or at same distance from it, see Lemma 2.5 and Theorem 2.6.Since any two points can always be rotated around their midpoint into one of these twopositions, this transformation gives us a simple way to estimate the value of the triangularratio metric of the original points.
Definition 4.1.
Euclidean midpoint rotation.
Choose distinct points x, y ∈ B . Let k = ( x + y ) /
2, and r = | x − k | = | y − k | . Let x , y ∈ S ( k, r ), x = y , so that | x | = | y | and the points x , k, y are collinear. Fix then x , y ∈ S ( k, r ) so that x , k, y are collinear, | x | = | k | + r and | y | = | k | − r . Note that x , y , y ∈ B always but x isnot necessarily in B . See Figure 1.For all x, y ∈ B , x = y , such that x ∈ B , the inequality s B ( x , y ) ≤ s B ( x, y ) ≤ s B ( x , y ) RIANGULAR RATIO METRIC IN THE UNIT DISK 11 holds, as we will prove in Theorems 4.15 and 4.16. If x / ∈ B , s B ( x , y ) is not defined butthe first part of this inequality holds. In order to prove this result, let us next introducea few results needed to find the value of s G -diameter of a closed disk in some domain G . ko x y x y xy Figure 1.
Euclidean midpoint rotation
Proposition 4.2.
For a fixed point x ∈ G and a fixed direction of −→ xy , the value of s G ( x, y ) is increasing with respect to | x − y | .Proof. Let x, y ∈ G and t ∈ [ x, y ] ∩ G . Choose z ∈ ∂G so that s G ( x, t ) = | x − t || x − z | + | z − t | . Because the function f : (0 , ∞ ) → R , f ( µ ) = ( u + µ ) / ( v + µ ) with constants 0 < u ≤ v isincreasing, s G ( x, t ) ≤ | x − t | + | t − y || x − z | + | z − t | + | t − y | = | x − y || x − z | + | z − t | + | t − y | ≤ s G ( x, y ) . Thus, the result follows. (cid:3)
Proposition 4.3.
A function f : [0 , π/ → R , f ( µ ) = p u − v cos( µ ) + p u + v cos( µ ) , where u, v > are constants, is increasing on the interval µ ∈ [0 , π/ . Proof.
Let s = cos( µ ), so that the function f can be written as g : [0 , → R , g ( s ) = √ u − vs + √ u + vs . By differentiation, g ′ ( s ) = v (cid:18) √ u + vs − √ u − vs (cid:19) ≤ , and it follows that the function g is decreasing on the interval s ∈ [0 , s =cos( µ ) is decreasing, too, with respect to µ , the function f is increasing. (cid:3) Theorem 4.4.
Suppose that j ≤ k < j + r < z . Choose x, y ∈ S ( j, r ) so that ∡ ZKX = µ with ≤ µ ≤ π/ and k ∈ [ x, y ] . Then the quotient | x − y || x − z | + | z − y | (4.5) is decreasing with respect to µ .Proof. Suppose without loss of generality that j = 0 and r = 1. First, we will considerthe special case where k = j = 0. From the condition k ∈ [ x, y ], it follows that x, y are theendpoints of a diameter of S and therefore | x − y | = 2 for all angles µ . Since | x | = | y | = 1and z = 1 + d , we obtain by the law of cosines | x − z | = p d ) − d ) cos( µ ) , | z − y | = p d ) + 2(1 + d ) cos( µ ) . The sum | x − z | + | z − y | can be described with the function f of Proposition 4.3 ifthe constants u, v are replaced with 1 + (1 + d ) > d ) >
0, respectively.By Proposition 4.3, this function f is increasing with respect to µ ∈ [0 , π/ /f ( µ ), it follows that it must be decreasing withrespect to µ .Suppose now that S ( j, r ) is still the unit circle S , but let 0 < k <
1. The equation ofthe line L ( x, y ) can be written as t + xyt = x + y (4.6)with t ∈ C as variable. Here, x can be written as e θi with 0 ≤ θ < π/
2. Furthermore, theline L ( x, y ) must contain k and, by substituting t = k in (4.6), we will have y = x − kkx − e θi − kke θi − . Consider now a function h : [0 , π ) → R , h ( θ ) = | e θi − ( e θi − k ) / ( ke θi − || e θi − z | + | z − ( e θi − k ) / ( ke θi − | , which clearly depicts the values of the quotient (4.5). For all θ ∈ [0 , π/ y = e ϕi = e θi − kke θi − ⇒ h ( θ ) = h ( − ϕ )(4.7) RIANGULAR RATIO METRIC IN THE UNIT DISK 13
The function h fulfills h (0) = h ( π ) = 1 /z , which is clearly its maximum value. If θ = 0,then so is µ , so the maximum of the quotient (4.5) is at µ = 0. By Rolle’s theorem, thereis a critical point ˜ θ such that f ′ (˜ θ ) = 0. By the property (4.7), ˜ θ is the solution of e θi = e − θi − kke − θi − . Thus, e θi + e − θi k ⇒ Re( e θ ) = k ⇒ µ = π µ = π/
2. Because thereare no other points where the derivative h ′ is 0 at the open interval 0 < θ < π/ µ ∈ [0 , π/ µ = 0 and minimumat µ = π/ µ , regardlessof if k = j or k > j . (cid:3) Theorem 4.8.
Fix S n − ( j, r ) ⊂ R n and z ∈ R n so that d = | z − j | − r > . Then sup x,y ∈ S n − ( j,r ) | x − y || x − z | + | z − y | = rr + d . Proof.
Suppose without loss of generality that n = 2, j = 0, r = 1 and z = d + 1 ∈ (1 , ∞ ).By symmetry, we can assume that the points x, y ∈ S fulfill 0 ≤ arg( x ) ≤ π/ x ) < arg( y ) < π . We will next prove the theorem by inspecting the quotient (4.5) afew different cases separately.Consider first the case where arg( x ) = 0. Now, x = 1 and y = e ϕi for some 0 < ϕ < π .It follows that | x − y || x − z | + | z − y | = | − e ϕi | d + | d − e ϕi | = (cid:18) d | − e ϕi | + | d − e ϕi || − e ϕi | (cid:19) − . Since both of the quotients d/ | − e ϕi | and | d − e ϕi | / | − e ϕi | obtain clearly theirminimum with ϕ = π , so the quotient (4.5) is at maximum within limitation x = 1 when y = − x ) = θ = 0 and arg( y ) ≤ π . Now, we can rotate the points x, y by the angle θ clockwise about the origin. This transformation does not affect thedistance | x − y | but decreases distances | x − z | and | z − y | , so it increases the value of thethe quotient (4.5). Since x maps into 1 in the rotation, this transformation lead to thefirst case studied above.Finally, consider the case where arg( x ) = 0 and π < arg( y ) < π . Now, ( x, y ) ∩ ( − , = ∅ , so we can choose a point k ∈ ( x, y ) ∩ ( − , − < k <
0, we can always reflectthe points x, y over the imaginary axis so that the quotient (4.5) increases. Thus, we can suppose that 0 ≤ k <
1. By Theorem 4.4, the quotient is decreasing with respect to ∡ ZKX = µ ∈ [0 , π/ µ = 0. It follows that x = 1 and y = − x = 1 and y = −
1. In thegeneral case x, y ∈ S ( j, r ), this means that x = j + r and y = j − r . Since the value ofthe quotient (4.5) is now r/ ( r + d ), the result follows. (cid:3) Corollary 4.9.
The s G -diameter of a closed ball J = B n ( k, r ) in a domain G ( R n is s G ( J ) = r/ ( r + d ) , where d = d ( J, ∂G ) .Proof. Clearly, s G ( J ) = sup x,y ∈ J s G ( x, y ) = sup x,y ∈ J (cid:18) sup z ∈ ∂G | x − y || x − z | + | z − y | (cid:19) = sup z ∈ ∂G (cid:18) sup x,y ∈ J | x − y || x − z | + | z − y | (cid:19) = sup z ∈ ∂G (cid:18) sup x,y ∈ J s R n \{ z } ( x, y ) (cid:19) = sup z ∈ ∂G s R n \{ z } ( J ) . Trivially, s R n \{ z } ( J ) is at maximum when the distance d ( z, J ) is at minimum. Thus, s G ( J ) = sup x,y ∈ J | x − y || x − z | + | z − y | , (4.10)where z ∈ ∂G such that d = d ( z, J ) = d ( J, ∂G ). It follows from Proposition 4.2 that, forall distinct x, y ∈ J , we can choose s, t ∈ ∂J , s = t , such that [ s, t ] = L ( x, y ) ∩ J and s G ( s, t ) ≥ s G ( x, y ). Thus, the points x, y giving the supremum in (4.10) must belong to S n − ( k, r ). By Theorem 4.8, it follows from this that s G ( J ) = sup x,y ∈ S n − ( k,r ) | x − y || x − z | + | z − y | = rr + d . (cid:3) Corollary 4.11.
The s B n -diameter of a set J = B n ( k, r ) ⊂ B n is s B n ( J ) = r/ (1 − | k | ) .Proof. Follows directly from Corollary 4.9. (cid:3)
Corollary 4.12.
For all x, y ∈ B n such that | y | ≤ | x | , the inequality s B n ( x, y ) ≤ | x | holds.Proof. Since y ∈ J = B n ( | x | ), s B n ( x, y ) ≤ s B n ( J ) and, by Corollary 4.11, s B n ( J ) = | x | . (cid:3) Consider yet the following situation.
Lemma 4.13.
For all points x ∈ B \{ } and y ∈ B ( | x | ) non-collinear with the origin, s B ( x, y ) < s B ( x, y ′ ) , where y ′ = xe ψi and ψ = arcsin (cid:18) | x − y | | x | (cid:19) . RIANGULAR RATIO METRIC IN THE UNIT DISK 15
Proof.
Since | y ′ | = | xe ψi | = | x | and | x − y ′ | = | x || − e ψi | = 2 | x | sin( ψ ) = | x − y | , thepoint y ′ is chosen from S ( | x | ) ∩ S ( x, | x − y | ). By symmetry, we can assume that y ′ is theintersection point closer to y . Fix z so that it gives the infimum inf z ∈ S ( | x − z | + | z − y | ).If µ ′ = ∡ ZXY ′ , then µ = ∡ ZXY = µ ′ + ∡ Y ′ XY > µ ′ . Clearly, by the law of cosines, s B ( x, y ) = | x − y || x − z | + | z − y | = | x − y || x − z | + p | x − y | + | x − z | − | x − y || x − z | cos( µ ) < | x − y || x − z | + p | x − y | + | x − z | − | x − y || x − z | cos( µ ′ ) = | x − y ′ || x − z | + | z − y ′ |≤ s B ( x, y ′ ) , so the lemma follows. (cid:3) Let us now focus on the results related to the Euclidean midpoint rotation.
Proposition 4.14.
Consider two triangles △ Y XZ and △ Y X Z with obtuse angles ∡ Y XZ and ∡ Y X Z . Let k and k be the midpoints of sides XY and X Y , respectively.Suppose that | x − y | = | x − y | , | k − z | ≤ | k − z | and ∡ ZKX ≤ ∡ Z K X . Then, | x − z | + | z − y | ≤ | x − z | + | z − y | . Proof.
Let r = | x − k | = | x − k | , m = | k − z | , m = | k − z | , µ = ∡ ZKX and µ = ∡ Z K X , see Figure 2. By the law of cosines, | x − z | + | z − y | = p r + m − rm cos( µ ) + p r + m + 2 rm cos( µ ) , | x − z | + | z − y | = q r + m − rm cos( µ ) + q r + m + 2 rm cos( µ ) . Furthermore, by Proposition 4.3, a function f : [0 , π/ → R , f ( µ ) = p u − v cos( µ ) + p u + v cos( µ ) , where u, v >
0, is increasing with respect to µ ∈ [0 , π/ µ, µ ∈ [0 , π/ ∡ Y XZ and ∡ Y X Z . Thus, it followsfrom µ ≤ µ and m ≤ m that | x − z | + | z − y | = p r + m − rm cos( µ ) + p r + m + 2 rm cos( µ ) ≤ p r + m − rm cos( µ ) + p r + m + 2 rm cos( µ ) ≤ q r + m − rm cos( µ ) + q r + m + 2 rm cos( µ )= | x − z | + | z − y | . (cid:3) Theorem 4.15.
For all x, y ∈ B , s B ( x, y ) ≥ s B ( x , y ) ≥ | x − y | p | x − y | + (2 − | x + y | ) . yx zk µ y x z k µ Figure 2.
The triangles △ Y XZ and △ Y X Z of Proposition 4.14 Proof.
Fix k = ( x + y ) / r = | x − k | . Suppose that k = 0, for otherwise s B ( x, y ) = s B ( x , y ) holds trivially. Without loss of generality, let 0 < k < ∡ XKZ = ν ∈ [0 , π/ ∡ Y KZ = π + ν , x = k + ri and y = k − ri . There are two possible cases;either the infimum inf z ∈ S ( | x − z | + | z − y | ) is given by one point z ∈ S or there aretwo possible points z ∈ S .Suppose first that the infimum inf z ∈ S ( | x − z | + | z − y | ) is given by only one point.By symmetry, this point must be z = 1. Fix u = r + (1 − k ) and v = 2 r (1 − k ) andconsider the function f of Proposition 4.3 for a variable ν . Now, we will haveinf z ∈ S ( | x − z | + | z − y | ) ≤ | x − | + | − y | = f ( ν ) ≤ f ( π/
2) = | x − | + | − y | = inf z ∈ S ( | x − z | + | z − y | ) , from which the inequality s B ( x, y ) ≥ s B ( x , y ) follows.Consider yet the case where there are two points giving the infimum inf z ∈ S ( | x − z | + | z − y | ). By symmetry, we can fix z so that 0 < arg( z ) ≤ π/
2. Now, the infimuminf z ∈ S ( | x − z | + | z − y | ) is given by some point z such that 0 ≤ arg( z ) ≤ arg( z ). If x, y are collinear with the origin, by Lemma 2.5 and Corollary 4.11, s B ( x, y ) = | x − y | − | x + y | = r − k = s B ( B n ( k, r )) ≥ s B ( x , y ) . RIANGULAR RATIO METRIC IN THE UNIT DISK 17 If x, y, △ Y XZ and △ Y X Z exists. The sides XY and X Y are both the length of 2 r and have a common midpoint k . It follows fromTheorem 2.4 and the inequality 0 < arg( z ) ≤ arg( z ) ≤ π/ ∡ Y XZ and ∡ Y X Z are obtuse, | k − z | ≤ | k − z | and ∡ ZKX ≤ ∡ Z KX . By Proposition 4.14, | x − z | + | z − y | ≤ | x − z | + | z − y | , so the inequality s B ( x, y ) ≥ s B ( x , y ) follows.Thus, s B ( x, y ) ≥ s B ( x , y ) holds in every cases and, by Theorem 2.6, s B ( x , y ) ≥ r p r + (1 − k ) = | x − y | p | x − y | + (2 − | x + y | ) , which proves the latter part of the theorem. (cid:3) Theorem 4.16.
Let x, y ∈ B with k = ( x + y ) / and r = | x − k | . If r + k < , s B ( x, y ) ≤ s B ( x , y ) = | x − y | − | x + y | < . Proof. If r + k <
1, then x , y ∈ B and, by Lemma 2.5, s B n ( x, y ) ≤ | x − y | − | x + y | = | x − y | − | x + y | = s B ( x , y ) . (cid:3) Hyperbolic Midpoint Rotation
In this section, we consider the hyperbolic midpoint rotation. The idea behind it is thesame as the one of the Euclidean midpoint rotation, for our aim is still to rotate the pointsaround their midpoint in order to estimate their triangular ratio distance. However, nowthe rotation is done by using the hyperbolic geometry of the unit circle instead of thesimpler Euclidean method.
Definition 5.1.
Hyperbolic midpoint rotation.
Choose distinct points x, y ∈ B . Let q be their hyperbolic midpoint and R = ρ B ( x, q ) = ρ B ( y, q ). Let x , y ∈ S ρ ( q, R ) so that | x | = | y | but x = y . Fix then x , y ∈ S ρ ( q, R ) so that x , y are collinear with theorigin and | y | < | q | < | x | . See Figure 3.The main result of this section is the inequality s B ( x , y ) ≤ s B ( x, y ) ≤ s B ( x , y ) . This holds for all distinct x, y ∈ B for the values of s B ( x , y ) and s B ( x , y ) are alwaysdefined. The first part of this inequality is proved in Theorem 5.11 and the latter part inTheorem 5.12, and the formula of for the value of s B ( x , y ) is in Theorem 5.3. Note that,according to numerical tests, the hyperbolic midpoint rotation gives better estimates for s B ( x, y ) than the Euclidean midpoint rotation or the point pair function, see Conjecture5.13. o q x y x y x y Figure 3.
Hyperbolic midpoint rotation
Lemma 5.2.
Choose x, y ∈ B so that their hyperbolic midpoint is < q < . Let t = th( R/
2) = th( ρ B ( x, y ) / . Then x = q (1 + t )1 + q t + t (1 − q )1 + q t i and y = x . Proof.
By Lemma 2.1, S ρ ( q, R ) = S ( j, h ) with j = q (1 − t )1 − q t and h = (1 − q ) t − q t . To find x and y , we need to find the intersection points of S ( j, h ) and S ( c, d ), where S ( c, d ) ⊥ S and c >
1. Now, c = ( q + d ) = 1 + d , from which it follows that d = 1 − q q and c = 1 + q q . Clearly, x = y since both j, c ∈ R . Let x = u + ri and y = u − ri . Now, h = r +( u − j ) and d = r + ( c − u ) . Thus, h − ( u − j ) = d − ( c − u ) ⇔ h − u + 2 ju − j = d − c + 2 cu − u ⇔ u = h − j − d + c c − j ) . RIANGULAR RATIO METRIC IN THE UNIT DISK 19
Since h − j = (1 − q ) t (1 − q t ) − q (1 − t ) (1 − q t ) = ( t − q )(1 − q t )(1 − q t ) = t − q − q t , − d + c = − (1 − q ) q + (1 + q ) q = 4 q q = 1 ,h − j − d + c = t − q − q t + 1 = (1 + t )(1 − q )1 − q t , c − j ) = 2( 1 + q q − q (1 − t )1 − q t ) = (1 + q )(1 − q t ) − q (1 − t ) q (1 − q t )= (1 − q )(1 + q t ) q (1 − q t ) , we will have u = h − j − d + c c − j ) = q (1 + t )(1 − q )(1 − q t )(1 − q )(1 − q t )(1 + q t ) = q (1 + t )1 + q t . From the equality h = r + ( u − j ) , it follows that r = p h − ( u − j ) = s (1 − q ) t (1 − q t ) − (cid:18) q (1 + t )1 + q t − q (1 − t )1 − q t (cid:19) = s (1 − q ) t (1 − q t ) − (cid:18) q (1 + t )(1 − q t ) − q (1 − t )(1 + q t )1 − q t (cid:19) = s (1 − q ) t (1 − q t ) − (cid:18) qt (1 − q )1 − q t (cid:19) = s (1 − q ) t (1 + q t ) (1 − q t ) − q t (1 − q ) (1 − q t ) = s t (1 − q ) (1 − q t ) (1 − q t ) = s t (1 − q ) (1 + q t ) = t (1 − q )1 + q t . (cid:3) Theorem 5.3.
For all x, y ∈ B with a hyperbolic midpoint q ∈ B \{ } and t =th( ρ B ( x, y ) / , s B ( x , y ) = s | q | + t | q | t if | q | < t ,s B ( x , y ) = t (1 + | q | ) p (1 + t )(1 + | q | t ) ≤ s | q | + t | q | t otherwise . Proof.
Suppose without loss of generality that 0 < q < x = u + ri and y = x . FromLemma 5.2, it follows that | x | = √ u + r = s q (1 + t ) (1 + q t ) + t (1 − q ) (1 + q t ) = s ( q + t )(1 + q t )(1 + q t ) = s q + t q t , | x − | = | u + ri − | > ⇔ ( u −
12 ) + r > ⇔ u + r > u ⇔ q (1 + t ) (1 + q t ) + t (1 − q ) (1 + q t ) > q (1 + t )1 + q t ⇔ q (1 + t ) + t (1 − q ) = ( t + q )(1 + q t ) > q (1 + t )(1 + q t ) ⇔ t + q > q (1 + t ) ⇔ q (1 − q ) < t (1 − q ) ⇔ q < t , and (1 − u ) + r = (cid:18) − q (1 + t )1 + q t (cid:19) + t (1 − q ) (1 + q t ) = (1 − q ) (1 + t )(1 + q t )(1 + q t ) = (1 − q ) (1 + t )1 + q t ⇒ r p (1 − u ) + r = t (1 − q ) p q t (1 − q )(1 + q t ) √ t = t (1 + q ) p (1 + t )(1 + q t ) . The result follows now from Theorem 2.6. (cid:3)
Theorem 5.4. [16, Prop. 3.1, p. 447]
The hyperbolic midpoint of J [0 , b ] is [0 , b ] ∩ J [ c, d ] for all c, d ∈ S such that b ∈ L ( c, d ) and c, d are non-collinear with the origin. Theorem 5.5.
If hyperbolic segments J [ u i , v i ] ⊂ B , i = 1 , ..., n , are of the same hyper-bolic length and have a common hyperbolic midpoint q , all their Euclidean counterparts [ u i , v i ] intersect at the same point.Proof. Choose distinct points u , v ∈ B that are non-collinear with the origin. Let q be their hyperbolic midpoint, R = ρ B ( u , v ) and k = L (0 , q ) ∩ L ( u , v ). Fix j, h as inLemma 2.1. Now, u , v ∈ S ( j, h ), J [ u , v ] ⊥ S ( j, h ) and u , v , j are non-collinear. Itfollows from Theorem 5.4 that the hyperbolic midpoint of J [ j, k ] is [ j, k ] ∩ J [ u , v ]. Since0 , j, q are collinear and k ∈ L (0 , q ), [ j, k ] ∩ J [ u , v ] = L (0 , q ) ∩ J [ u , v ] = q . Thus, q is the hyperbolic midpoint of J [ j, k ]. It follows that k only depends on q and j , so theintersection point L (0 , q ) ∩ L ( u i , v i ) must be same for all indexes i , as can be seen inFigure 4. If u i , v i are collinear with the origin for some index i , then k ∈ [ u i , v i ] trivially.Thus, the theorem follows. (cid:3) RIANGULAR RATIO METRIC IN THE UNIT DISK 21 o j q k Figure 4.
Hyperbolic circle S ρ ( q, R ) with the points j, q, k of Theorem 5.5 Corollary 5.6.
For all x, y ∈ B , there is a point k = L ( x, y ) ∩ L ( x , y ) ∩ L ( x , y ) . Proof.
Follows from Theorem 5.5. (cid:3)
Theorem 5.7.
For all x, y ∈ B that are non-collinear with the origin and have a hyper-bolic midpoint q , the distance | x − y | is decreasing with respect to the angle ∡ ([0 , q ] , [ x, y ]) .Proof. Consider a hyperbolic circle S ρ ( q, R ), where R = ρ B ( x, q ), and let S ( j, h ) bethe corresponding Euclidean circle. By Lemma 2.1, we see that the points 0 , j, q arecollinear. Fix k as in Corollary 5.6 and let u = | j − k | . Denote θ = ∡ ([ x, y ] , [0 , q ]) = ∡ ([ x, y ] , [ j, k ]) ∈ [0 , π/ u does not depend on the angle θ . Itfollows that | x − y | = 2 p h − u sin ( θ ) is decreasing with respect to θ . (cid:3) Corollary 5.8.
For all x, y ∈ B , | x − y | ≤ | x − y | ≤ | x − y | .Proof. Follows from Theorem 5.7. (cid:3)
Corollary 5.9.
For all x, y ∈ B , | x − y | ≤ − | q | ) t − | q | t ≤ ρ B ( x, y ) / . where q is the hyperbolic midpoint of J [ x, y ] , and t = th( ρ B ( x, y ) / . Proof.
By fixing h as in Lemma 2.1, we will have | x − y | = 2 h = 2(1 − | q | ) t − | q | t ≤ t = 2th( ρ B ( x, y ) / , so the result follows from Corollary 5.8. (cid:3) Remark 5.10.
The inequality | x − y | ≤ ρ B ( x, y ) /
4) can be also found in [9, (4.25),p. 57].
Theorem 5.11.
For all x, y ∈ B , s B ( x, y ) ≥ s B ( x , y ) .Proof. Let q be the hyperbolic midpoint of J [ x, y ] and R = ρ B ( x, q ). If q = 0, s B ( x, y ) = s B ( x , y ) holds trivially. Thus, choose x, y ∈ B so that 0 < q <
1. Now, either theinfimum inf z ∈ S ( | x − z | + | z − y | ) is given by one point z ∈ S or two points on S .If there is only one point giving the infimum inf z ∈ S ( | x − z | + | z − y | ), it must clearlybe z = 1. Let B ( j, h ) = B ρ ( q, R ) like in Lemma 2.1, and fix k as in Corollary 5.6. Bysymmetry, we can assume that ∡ KX = µ ∈ [0 , π/ µ = π/
2, then x = x and y = y . Now, it follows from Theorem 4.4 that s B ( x, y ) ≥ | x − y || x − | + | − y | ≥ | x − y || x − | + | − y | = s B ( x , y ) . Suppose now that there are two possible points on z ∈ S for inf z ∈ S ( | x − z | + | z − y | ).By symmetry, let Im( x ) > ≤ arg( x ) ≤ arg( x ). Fix z so that Im( z ) > ∡ OZ X = ∡ Y Z O , where o is the origin. By Theorem 2.4, this point z gives theinfimum inf z ∈ S ( | x − z | + | z − y | ). Denote yet ψ = ∡ Y X Z , which is clearly anobtuse angle.By Corollary 5.8, we can fix y ′ ∈ [ x, y ] so that | x − y ′ | = | x − y | . Let z ∈ S withIm( z ) < Im( x ) so that ∡ Y ′ XZ = ψ . Clearly, | x − z | ≤ | x − y | . By Proposition 4.2, itfollows that s B ( x, y ) ≥ s B ( x, y ′ ) ≥ | x − y ′ || x − z | + | z − y ′ | = | x − y ′ || x − z | + p | x − y ′ | + | x − z | − | x − y ′ || x − z | cos( ψ ) ≥ | x − y || x − z | + p | x − y | + | x − z | − | x − y || x − z | cos( ψ )= | x − y || x − z | + | z − y | = s B ( x , y ) (cid:3) Theorem 5.12.
For all x, y ∈ B , s B ( x, y ) ≤ s B ( x , y ) = (1 + | q | ) t | q | t , RIANGULAR RATIO METRIC IN THE UNIT DISK 23 where q is the hyperbolic midpoint of J [ x, y ] , and t = th( ρ B ( x, y ) / .Proof. Let q be the hyperbolic midpoint of J [ x, y ]. Fix then R = ρ B ( x, q ) and j, h, t asin Lemma 2.1. Now, B ( j, h ) = B ρ ( q, R ) and t = th( ρ B ( x, y ) / s B ( x, y ) ≤ s B ( B ρ ( q, R )) = s B ( B ( j, h )) = h − | j | = (1 − | q | ) t − | q | t − | q | (1 − t )= (1 − | q | ) t − | q | + | q | t − | q | t = (1 − | q | )(1 + | q | ) t (1 − | q | )(1 + | q | t ) = (1 + | q | ) t | q | t . Since | j | = | x + y | / h = | x − y | /
2, by Lemma 2.5, s B ( x , y ) = | x − y | − | x + y | = h − | j | , so the theorem follows. (cid:3) According to numerous computer tests, the following result holds.
Conjecture 5.13.
For all x, y ∈ B ,1. s B ( x , y ) ≥ s B ( x , y ) ,2. s B ( x , y ) ≤ s B ( x , y ) ,3. s B ( x , y ) ≤ p B ( x, y ) ,where the points x i , y i , i = 0 , ..., , are as in Definitions 4.1 and 5.1. Thus, by this conjecture, the hyperbolic midpoint rotation gives sharper estimationsfor s B ( x, y ) than the Euclidean midpoint rotation or the point pair function.6. H¨older continuity
In this section, we show how finding better upper bounds for the triangular ratio metricin the unit disk is useful when studying quasiconformal mappings. The behaviour ofthe distance between two points x, y ∈ B n under a K -quasiconformal homeomorphism f : B n → B n = f ( B n ) has been studied earlier in numerous works, for instance, see [9,Thm 16.14, p. 304]. Our next theorem illustrates how finding a good upper limit for thevalue of the triangular ratio metric can give new information regarding this question. Theorem 6.1. If f : B → B = f ( B ) is a K -quasiconformal map, the inequality | f ( x ) − f ( y ) | ≤ − /K (cid:18) s B ( x, y )1 + s B ( x, y ) (cid:19) /K , holds for all x, y ∈ B .Proof. Define a homeomorphism ϕ K : [0 , → [0 ,
1] as in [9, (9.13), p. 167] for
K > ϕ K ( r ) ≤ − /K r /K = 4 − / (2 K ) (cid:16) r (cid:17) /K , (6.2) where 0 ≤ r ≤ K ≥
1. Let f be as above, x, y ∈ B and t = th( ρ B ( x, y ) / s B ( f ( x ) , f ( y )) ≤ th ρ B ( f ( x ) , f ( y ))2 ≤ ϕ K (cid:18) th ρ B ( x, y )2 (cid:19) = ϕ K (cid:18) t t (cid:19) ≤ − / (2 K ) (cid:18) t t (cid:19) /K ≤ − / (2 K ) (cid:18) s B ( x, y )1 + s B ( x, y ) (cid:19) /K . By [9, Lemma 11.12, p. 201; Prop. 11.15, p. 202], it follows from the inequality abovethat | f ( x ) − f ( y ) | ≤ s B ( f ( x ) , f ( y )) ≤ − /K (cid:18) s B ( x, y )1 + s B ( x, y ) (cid:19) /K , which proves the theorem. (cid:3) Thus, as we see from Theorem 6.1, finding a suitable upper bound for the value of s B ( x, y ) can help us estimating the distance of the points x, y under the K -quasiconformalmapping f . Corollary 6.3. If f is as in Theorem 6.1, the inequality | f ( x ) − f ( y ) | ≤ − /K p | x − y | + 4(1 − | x | )(1 − | y | ) | x − y || x − y | + 2(1 − | x | )(1 − | y | ) ! /K . holds for all x, y ∈ B .Proof. It follows from Theorems 6.1 and 2.3 that | f ( x ) − f ( y ) | ≤ − /K (cid:18) s B ( x, y )1 + s B ( x, y ) (cid:19) /K ≤ − /K (cid:18) p B ( x, y )1 + p B ( x, y ) (cid:19) /K = 2 − /K p | x − y | + 4(1 − | x | )(1 − | y | ) | x − y | | x − y | + 4(1 − | x | )(1 − | y | ) ! /K = 2 − /K p | x − y | + 4(1 − | x | )(1 − | y | ) | x − y || x − y | + 2(1 − | x | )(1 − | y | ) ! /K . (cid:3) Corollary 6.4. If f is as in Theorem 6.1, the inequality | f ( x ) − f ( y ) | ≤ − /K (cid:18) (2 − | x + y | ) | x − y | − | x + y | + | x | + | y | (cid:19) /K . holds for all x, y ∈ B .Proof. Follows from Theorem 6.1 and Lemma 2.5, and the fact that | x + y | + | x − y | =2 | x | + 2 | y | . (cid:3) RIANGULAR RATIO METRIC IN THE UNIT DISK 25
Corollary 6.5. If f is as in Theorem 6.1, all x, y ∈ B fulfill | f ( x ) − f ( y ) | ≤ − /K (cid:18) (1 + | q | )(1 + | q | t ) t (1 + | q | t ) + (1 + | q | ) t (cid:19) /K , where q is the hyperbolic midpoint of J [ x, y ] , and t = th( ρ B ( x, y ) / .Proof. Follows from Theorems 6.1 and 5.12. (cid:3)
Remark 6.6.
Neither of Corollaries 6.4 and 6.3 is better than the other for all points x, y ∈ B . For x = 0 . y = 0 . i , the limit in Corollary 6.4 is sharper than the onein Corollary 6.3 and, for x = 0 . y = 0 . i , the opposite holds. However, accordingto numerical tests related to Conjecture 5.13, the result in Corollary 6.5 is always betterthan the ones in Corollaries 6.4 and 6.3.By restricting how the point pair x, y is chosen from B , we can find yet better estimates. Corollary 6.7. If f is as in Theorem 6.1, the inequality | f ( x ) − f ( y ) | ≤ − /K (cid:18) | x − y | − r (cid:19) /K . holds for all x, y ∈ B such that | x + y | / ≤ r .Proof. Now, 2 − | x + y | − | x + y | + | x | + | y | ≤ − | x + y | − | x + y | + | x + y | / − | x + y | / ≤ − r so the result follows from Corollary 6.4. (cid:3) Corollary 6.8.
Let A = { x, y ∈ B | | x + y | ≤ } . For all x, y ∈ A , | f ( x ) − f ( y ) | ≤ − /K | x − y | /K , where f is as in Theorem 6.1.Proof. Follows from Corollary 6.7. (cid:3)
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