Intrinsic metrics in polygonal domains

We study inequalities between the hyperbolic metric and intrinsic metrics in convex polygonal domains in the complex plane. A special attention is paid to the triangular ratio metric in rectangles. A local study leads to investigation of the relationship between the conformal radius at an arbitrary point of a planar domain and the distance of the point to the boundary.


Introduction
During the past few decades there has been considerable interest in the study of metrics defined in subdomains G of R n , n ≥ 2. In geometric function theory the most useful metrics are intrinsic metrics.Distances in these metrics between two points measure not only how far the points are from each other, but also how close they are to the boundary of the domain.The hyperbolic distance ρ G (x, y), x, y ∈ G, of a planar domain G, is ideal for this purpose, in particular, because of its conformal invariance.There are many reason to study metrics: A. Papadopoulos [P, pp.42-48] lists twelve different metrics commonly used in geometric function theory.We mention two factors motivating this research: (a) In the higher dimensional case n ≥ 3 there is no metric, equally flexible as the hyperbolic metric is in the planar case.Thus many authors have introduced metrics of hyperbolic-type which share some but not all properties of the hyperbolic metric [HKV].(b) In the planar case, estimating the hyperbolic distances of a given planar domain is a difficult problem [BM].Therefore, finding concrete estimates for the hyperbolic metric in terms of simpler metric is needed; these estimates should take into account both the geometric structure and the metric properties of the boundary.
We study here the topic (b), continuing the earlier work [DNRV, R, RV, HKV].In particular, we investigate the case when the domain is a polygonal domain in the plane and compare the value of the hyperbolic metric to the triangular ratio metric s G [HKV], see Section 2 for the definition.In the case when the polygonal domain G is a rectangle, our main result is Theorem 4.7.We also formulate a conjecture about the sharp constant in this theorem (Conjecture 4.10).
It also turns out that the methodology of our proofs leads to the study of the conformal radius of a domain and its connection with the distance to the boundary.In Section 5 we study some concrete domains and find the maximum of their conformal radii.In Section 6 we investigate the ratio 2d G (u)/r G (u) for u ∈ G, where G is a polygonal domain, d G (u) = dist(u, ∂G) and r G (u) is the conformal radius of G at u. Also we prove that the maximum of the ratio is attained on some graph which consists of those points u ∈ G for which d G (u) = |u − z j |, j = 1, 2 for at least two distinct points z 1 = z 2 , z j ∈ ∂G (Theorem 6.1).Then we give some examples of polygonal domains and calculate the maximal value of 2d G (u)/r G (u); the obtained values give the lower estimates for the best constant C = C(G) in the inequality

Preliminaries
Let C = C ∪ {∞} be the extended complex number.Also let B 2 be the unit disk and H 2 be the upper half plane in the complex plane C. The special Möbius transformation [A] (2.2) Suppose that G is a proper subdomain in R 2 .Denote by d G (x) the Euclidean distance dist(x, ∂G) = inf{|x − z| : z ∈ ∂G} between the point x and the boundary of G.The triangular ratio metric s G : G × G → [0, 1] is defined by [HKV, RV] (2.3) Using the above notation, we can also record the formulas for the hyperbolic metric ρ G in these domains G ∈ {H 2 , B 2 } as [B, BM], [HKV,(4.8),p. 52 & (4.14) Note that the hyperbolic metric enjoys the following conformal invariance property: If G is a domain and f : Thus, the hyperbolic metric ρ G can be defined in any planar simply connected domain in terms of a conformal mapping of the domain onto the unit disk [BM,Thm 6.3 p. 26].In particular, the hyperbolic metric is invariant under Möbius transformations.In terms of complex numbers, the formulas of the hyperbolic metric for the two cases H 2 and B 2 can be simplified to where y is the complex conjugate of y.
2.6.Conformal radius.The conformal radius r D (z 0 ) of a simply connected domain D with a nondegenerate boundary at a point z 0 = ∞ is a well-known characteristic playing an important role in geometric function theory and applications.By definition, r D (z 0 ) is the radius of a disk centered at the origin which can be mapped onto D conformally by a function F with the normalization F (0) = z 0 and F (0) > 0 (see, e.g.[GOL, PS, BF, BKN, So]).Let g : D → B 2 be a conformal mapping of D onto B 2 and f be its inverse.If g(z 0 ) = ζ, then [PS,p. 162] (2.7) It is well known that the conformal radius r D (z 0 ) and the distance to the boundary d D (z 0 ) are equivalent in the sense that for every simply connected domain D (2.9) The left estimation follows from the Schwarz lemma and the right one is a corollary of the Koebe quarter theorem.It should be noted that both inequalities (2.9) above are sharp.The first inequality is sharp when f (z) = z and D = B

Nearest point for a rectangle
Consider the rectangle R with vertices ±k ± i, where k ≥ 1.From the definition (2.3) of s-metric we see that if a point x ∈ R is fixed then the infimum inf z∈∂G (|x − z| + |z − y|) is attained at some points z lying on a side of ∂R.This side depends on the location of y.
In Fig. 1 we show, for a fixed x ∈ R, the subdomains of R such that if y belongs to one of these subdomains, then z belongs to the corresponding side of R which is a common side of R and of the subdomain.Among the four subdomains, two are trapezoids and other two are triangles.The lower and the upper subdomains are trapezoids if and only if Re x 2 − Im x 2 < k 2 − 1; in the case Re x 2 − Im x 2 > k 2 − 1 trapezoids are the left and right subdomains.The case Re x 2 − Im x 2 = k 2 − 1 corresponds to a subdivision of R into four triangles.In Fig. 1 we show a subdivision for k = 1.4,x = 0.7 − 0.4i.
We can also describe the geometry of disks and circles in the s-metric.Every circle centered at x of radius r is either a Euclidean circle or a piecewise smooth Jordan curve.In Fig. 2 we show such curves for the case k = 1.4,x = 0.7 − 0.4i, r = 0.1j, 1 ≤ j ≤ 9. We note that the corner points of these curves are on the segments separating R into four parts described above (see Fig. 1).Besides, the disks are convex sets in the Euclidean metric.

Hyperbolic metric versus triangle ratio metric in rectangle
Our goal is to find for the rectangle R the best constants C 1 , C 2 > 0 such that the inequality (4.1) -1.0 -0.5 0.5 1.0 holds.Now we will find the limit ) and give its geometric interpretation.
Consider the segments Σ 1 and Σ 2 which are parts of the bisectors of the angles at the vertices −k − i and −k + i of R, respectively, connecting these vertices with the point −(k − 1) on the real axis; similarly, let also Σ 3 and Σ 4 join the vertices k − i and k + i with the point k − 1 on the real axis.Consider also the segment Σ 0 with endpoints ±(k − 1).
These five segments subdivide R into four subdomains, two trapezoids and two triangles (see Fig. 3); in the case k = 1 the trapezoids degenerate into triangles and Σ 0 degenerates into a point.
We see that if a point v is in one of these subdomains, then the point ζ 0 , where the minimal value of |ζ − v|, ζ ∈ ∂R, is attained, lies on the side of R which is common with this subdomain.If v is on one of the segments described above and v = ±(k − 1), then the shortest distance is attained at two sides of R. At last, if v = ±(k − 1) then there are three sides where the minimal value of |ζ − v|, ζ ∈ ∂R, is attained; in the case k = 1, we have the square, the points ±(k − 1) coincide with its center and the minimal distance is attained at all four the sides.
We can describe this situation as follows.If v is an interior point of one subregion, then the largest disk contained in R and centered at v touches ∂R at one point lying on the of R which is common with this subdomain.If v ∈ ∪ 4 j=0 Σ j and v = ±(k − 1), then the largest disk touches two of the sides, and if v = ±(k − 1), then it touches ∂R at three points.These three case are shown on Fig. 3. (If k = 1, then the largest disk touches all the four sides of R.) Lemma 4.2.We have Proof.Let f : R → H 2 be a conformal mapping of R onto the upper half plane H 2 .Then, by the definition, Let v be a point from the lower trapezoidal part of R (other three cases can be considered similarly).Then Therefore, . Since we obtain (4.3).
We see that if (4.1) holds, then by Lemma 4.2 we get Therefore, we obtain an estimation of the conformal radius r R (v) through d R (v).On the other hand, if for some point v ∈ R we have and for convex domains, with the help of (2.10), we deduce that Given k ≥ 1, let λ ∈ (0, 1) be the unique root of the equation (4.4) where K(λ) is as defined in (2.5).Then for every z 1 which is an interior point of the segment with endpoints z 0 , ζ 0 we have Moreover, if either (i) holds and D is distinct from a half plane, or (ii) is valid and ∂D is distinct from the union of two ray coming from the point ζ 0 and passing through the points a and b, then the inequality in (4.6) is strict.
Proof.Since d D and r D are invariant under the shifts of the plane, without loss of generality we can assume that where the mapping ψ : z → αz.We note that ψ(z 1 ) = z 0 and r , we obtain (4.6).If either (i) holds and D is distinct from a half plane, or (ii) is valid and ∂D is distinct from the union of two ray coming from the point ζ 0 and passing through the points a and b, then D = G and, therefore, r G (z 0 ) < r D (z 0 ), so the inequality in (4.6) is strict. where Both inequalities of (4.8) are sharp.Moreover, equality in the second inequality holds if and only if v = ±(k − 1).
Proof.The lower estimation in (4.8) is evident.We will prove the upper one.First we will show that the maximal value of 2d R (v)/r R (v), v ∈ R, is attained only on the segment with endpoints ±(v −1).Using the statement (i) of Lemma 4.5, we see that the maximum of the value 2d By the statement (ii) of Lemma 4.5 we conclude that v is on the segment Σ 0 with endpoints ±(k − 1).
For v ∈ Σ 0 we have d R (v) = 1.Therefore, to find the maximal value of 2d R (v)/r R (v) we only need to find the minimal value of the conformal radius r is attained at the endpoints of the segment, i.e. at the points ±(k − 1).Now we will find the value r R (k − 1).To calculate it we replace, for convenience, R with its image R 2 = [−α, α] × [0, 2αk] under the mapping z → iα(k − z) where the constant α > 0 will be fixed below; we note that, under the mapping, the the point k − 1 goes to iα.This mapping does not change the value of the conformal radius.It is known that R 2 can be mapped onto the upper half-plane H 2 such that its vertices α(−1 + 2ik), −α, α, and α(1 + 2ik) will go to −1/λ, −1, 1, 1/λ; here λ ∈ (0, 1) is the unique root of the equation (4.4).If we put α = K(λ), then we can use as such a mapping the Jacobi elliptic function f (z) = sn (z, λ) which maps R 2 onto H 2 with the correspondence of boundary points indicated above [AF,p.358].By the definition of the conformal radius, we have Since f (z) = cn (z, λ)dn (z, λ) and f (i K(λ)) is a pure imaginary number, we find We also have d R 2 (i K(λ)) = K(λ).Since the ratio of the distance to the boundary and the conformal radius is invariant under linear conformal mappings, we have As we showed above, this is the biggest value of 2d R (v)/r R (v) over R.
Corollary 4.10.We have and the inequalities are sharp.
Conjecture 4.11.Let C(λ) be defined as in (4.9).Then for all points u, v lying in the These inequalities are sharp.
This corresponds to the values λ ≥ λ 0 := 3 − 2 √ 2 = 0.171572875 . .., where λ 0 is the unique root of the equation (4.4) for k = 1 [AVV, p.81, Table 5.1].We can also consider a rectangle with k < 1.It is evident that this case is easily reduced to the case k ≥ 1 because we can apply the mapping z Therefore, let us define C(λ) as follows: The graph of C(λ) for 0 < λ < 1 is shown in Fig. 4. We have The first limiting case corresponds to a square and the second one matches to the case of a half-strip.

Conformal radius of specific domains and its maximum
In this section, we determine the maximal value of the conformal radius for some domains which are conformal images of the unit disk B 2 .These domains can be found in [KT] and its references.We note that exact values of the maximal value for conformal radius of some special domains can be found in [PS, Tables of some functionals].
Example 5.1.Consider the univalent function w , where the branch of the square root is chosen such that √ 1 = 1.The function φ 1 maps the open unit disk B 2 onto the domain D 1 which is the right-half of the lemniscate of Bernoulli.Hence, the inverse mapping z = ϕ 1 (w) = w(w + 2) maps D 1 onto B 2 .Since ϕ 1 (0) = 0 and ϕ 1 (0) = 2 > 0, we have r φ 1 (B 2 ) (0) = 0.5.Now we will find the maximum of the conformal radius of D 1 .If w = φ 1 (z), where z = te iθ , t ∈ [0, 1) and θ ∈ [0, 2π), then by (2.7) the conformal radius of D 1 at the point w is equal to and the equality holds if and only if θ = π.The maximal value of the function (1 − t) 1/2 (1 + t)/2 is attained at t 0 = 1/3 and is equal to 2 √ 6/9.Thus we see that the maximum of r D 1 (w) in B 2 is equal to 2 √ 6/9 = 0.544331 . . .which is attained at the point w , where the branch of the square root is chosen such that √ 1 = 1.We note that this function is closely connected with the function inverse to the Joukowsky transform.It maps the unit disk B 2 onto the crescent-shaped region D 2 , see Fig. 5

(a). The inverse function
Now we will find the maximal value of the conformal radius r D 2 (w) for points w lying on the real axis.Let w = φ 2 (t), t ∈ (−1, 1), then, with the help of (2.7), we obtain Simple analysis shows that the maximal value of the expression in the right-hand side equals M = 1.17117 . .., it is attained at the point t 0 = 0.319968 . . .which is the minimal positive root of the equation 4t 8 + 12t 6 + t 4 − 10t 2 + 1 = 0.The corresponding point of maximum of the conformal radius is equal to w = φ 2 (t 0 ) = 0.369911 . ... Therefore, we found that max This function φ 5 maps the unit disk B 2 onto the domain D 5 , bounded by a Booth lemniscate.We have Now we will investigate the behavior of ψ(τ ), τ ∈ [0, 1].Simple analysis shows that for 0 ≤ α ≤ 1/3 the function ψ decreases on [0,1].Therefore, ψ(t 2 ) ≤ ψ(0) = 1 and the maximal value of the conformal radius r D 5 (w) equals 1; it is attained at the origin.
It is attained at the points .

Comparison of the conformal radius and the distance to the boundary for some domains
It is easy to see that we can apply the above method, based on Lemma 4.5, to obtain sharp two-sided estimations of 2d D (•)/r D (•) for other convex planar domains D.
First consider a bounded convex n-gon P and its decomposition into parts: P = ∪ n j=1 P j such that z ∈ P j if and only if z is closer to the j-th side of P , than to others.Then we investigate the set ∪ n j=1 ∂P j .It is a graph and after removing from it the boundary points of P and open, i.e. without endpoints, edges which have nonempty intersection with ∂P we obtain its subgraph; denote it by Gr(P ).For example, if we take as then Gr(P ) consists of the segment Σ 0 (see Section 4).If P admits an inscribed circle (a circle which is tangent to all sides of P ), then Gr(P ) consists of the point which is the center of the circle.
Lemma 4.5 immediately implies the following result.
Theorem 6.1.The maximum of 2d P (u)/r P (u), u ∈ P , is attained at some point of the graph Gr(P ).
Analyzing some concrete domains and classes of domains (see the examples below), we can suggest the following conjecture.
Conjecture 6.2.The maximum of 2d P (u)/r P (u), u ∈ P , is attained at a vertex of the graph Gr(P ).Remark 6.3.We note that the statement of Theorem 6.1 is also valid for most unbounded convex polygonal domains P , excluding those with degenerate (empty) Gr(P ) such as strips and infinite sectors.Actually, if P is unbounded, then ∂P contains two rays, L 1 and L 2 .If the rays are not parallel, for simplicity, we can assume that their continuations intersect at the origin.Consider the set A of points in P such that maximal disks centered at these points and contained in P touch both the rays; it is also a ray coming from a point z 0 .Then applying the mappings z → αz, α > 0, we show as in the proof of Lemma 4.5 that when approaching the origin by the set A the value of 2d P (u)/r P (u) increases.Therefore the maximal value of 2d P (u)/r P (u), u ∈ A, is attained at the vertex z 0 of Gr(P ).If L 1 and L 2 are parallel, then the ray A is parallel to them and we can apply shifts z → z + h instead of the mappings z → αz, α > 0.
2) Isosceles triangles.For an arbitrary triangle ∆, the upper bound of 2d ∆ (•)/r ∆ (•) is attained at the center of its inscribed circle.Here we describe the situation for the case of an isosceles triangle ∆ α with vertices 1, ±i cot(απ) and angles (1 − 2α)π, απ and απ, where α ∈ (0, 1/2).The conformal mapping of the right half plane onto ∆ α is given by Here 2 F 1 is the Gaussian hypergeometric function [AVV,Ch 1].The center of the inscribed circle for ∆ α is at the point (1 − tan 2 (απ/2))/2 and its preimage x = x(α) under the mapping G α can be found from the equation Then the maximal value of 2d ∆α (•)/r ∆α (•) is equal to From the graph of M 1 (α) (see Fig. 6 and Remark 6.5) we can see that the maximal value of M 1 (α) is attained at the point α 0 = 1/3 corresponding to the case of regular triangle; possibly this can be proved strictly analytically.The corresponding value is In Table 1, we give some values of M 1 (α), where α ∈ (0, 1/2).
3) Half-strip.Ω a := {z : | Re z| < a and Im z > 0, 0 < a ∈ R}; The upper estimate 2d Ωa (•)/r Ωa (•) is attained at the intersection point of the bisectors of the two angles at its two finite vertices.The function mapping the half-strip Ω a onto the upper half-plane is g(z) = sin(πz/2a).The intersection point of the bisectors is ia.It follows from (2.8) that 4) Arbitrary convex triangle with a vertex at infinity.Let Λ α be a symmetric triangle bounded with the segment [−i, i] and two rays going from the points ±i and forming the angles απ with the segment, where α ∈ (1/2, 1).The upper estimate for 2d Λα (•)/r Λα (•) is attained at the intersection point of the bisectors of the two angles at its two finite vertices.The function P α mapping the right half plane onto Λ α has the form The intersection point of the bisectors of the angles at the finite vertices is at the point tan(απ/2), therefore we can find x = x(α) which is the preimage of tan(απ/2) under the mapping P α from the equation Then the maximal value of 2d Λα (•)/r Λα (•) is equal to Some values of M 2 (α) also are given in Table 2: Remark 6.5.We can combine Examples 2) and 4) above by defining M 3 (α) as follows for all α ∈ (0, 1): 6 shows the graph of M 3 (α) for α ∈ (0, 1).

5)
Rhomb.Let Π be a rhomb with acute angle δ and length of side 1.The upper estimate of 2d Π (•)/r Π (•) is attained at its center.From [PS,Ch.1,§ 1.22] we have that the conformal radius at its center is equal to .
The graph of φ(δ) is given on Fig. 7.

6)
Trapezoid.Let us denote an arbitrary trapezoid by T. a) For long trapezoids, the maximum of 2d T (•)/r T ( • ) is attained at its midsegment, more precisely, between the intersection points of bisectors of the angles at two adjacent vertices, lying on different parallel sides.Most likely, the maximum is attained at one of the intersection points of its bisectors.This case requires additional investigation.b) For short trapezoids, the maximum of 2d T (•)/r T (•) is attained at the segment joining the intersection points of bisectors of the angles at two adjacent vertices, lying on different parallel sides.This segment is on the bisector of the angle formed by the straight lines, containing the nonparallel sides of the trapezoid.Most likely, the maximum is attained at one of the intersection points of the bisectors described above.This case also requires additional investigation.
c) The transitional case is when the considered trapezoid has an inscribed circle.Then the upper estimate is attained at its center.7) Arbitrary convex quadrilateral.The situations is close to the case of trapezoids.The only difference that, in the non-trapezoidal case, instead of a part of midsegment we consider the segment of bisector, which form the straight lines, containing the corresponding pair of nonparallel sides.
8) Regular n-gon.Let N be a regular n-gon with the side length a.The upper estimate of 2d N (•)/r N (•) is attained at its center.Again, from [PS,Ch.1,§ 1.22] we have that the conformal radius at its center is equal to The distance from the center to the boundary is equal to the radius of the inscribed circle: This value grows as n increases and tends to 2, as n tends to infinity (the case of a circle).
For n equal to 3, 4 and 6 this formula gives values for equilateral triangle, square and regular hexagon, respectively.9) Polygons having inscribed circles.The upper estimate is attained at the center of the inscribed circle.Now we will estimate the ratio 2d D 1 (•)/r D 1 (•) and 2d D 2 (•)/r D 2 (•) for non-polygonal domains described in Examples 5.1 and 5.2, respectively.
10) The domain D 1 which is the image of the unit disk under the mapping w = φ 1 (z) = √ 1 + z − 1 is convex.From Lemma 4.5 it follows that the maximum of the ratio 2d D 1 (w)/r D 1 (w), w ∈ D 1 is attained at some point w which is real, i.e. −1 < w < √ 2 − 1.Now we will find d D 1 (w) for such w.
11) The domain D 2 which is the image of the unit disk under the mapping w = φ 2 (z) = z + √ 1 + z 2 − 1.We will also find the maximum of the ratio 2d D 2 (w)/r D 2 (w) but, because of the fact that this domain is not convex, we will confine ourselves to considering the case of real w.

Figure 1 .
Figure 1.For a fixed point x in a rectangle, there are four subdomains with the following property.If y is a point of a subdomain, then the point z providing the minimum of |x − z| + |z − y| is on the common side of the subdomain and of the rectangle.
Lemma 4.5.Let D be a convex planar domain with nonempty boundary ∂D and suppose that one of the following conditions is valid: (i) The circle centered at a point z 0 ∈ D of radius d D (z 0 ) touches ∂D at some point ζ 0 ; (ii) The boundary ∂D contains two linear segments [a, ζ 0 ] and [ζ 0 , b] and the circle centered at a point z 0 ∈ D of radius d D (z 0 ) touches ∂D at two points ζ 1 ∈ [a, ζ 0 ] and ζ 2 ∈ [ζ 0 , b].

Figure 4 .
Figure 4.The graph of the function C(λ).