MATEMATIQKI VESNIK

60 (2008), 9–22

UDK 515.124

originalni nauqni rad research paper

LOCAL LIPSCHITZ PROPERTY FOR THE CHEBYSHEV CENTER MAPPING OVER N-NETS

Pyotr N. Ivanshin and Evgenii N. Sosov

Abstract. We prove local Lipschitz property of the map which puts in correspondence to each exact N-net its Chebyshev center. If dimension of Euclidean or Lobachevsky space is greater than 1 and the net consists of more than 2 points we show that this map is not Lipschitz in a neighbourhood of the space of all 2-nets embedded into the space of N-nets endowed with Hausdorff metric.

1. Definitions and notation

We assume the following notation.

ℝ_{+} – the set of all non-negative real numbers.

(X, ρ) – metric space with metric ρ.

xy = ρ(x, y), xZ = inf{xu : u ∈ Z} for x, y ∈ X, Z ⊂ X.

ω(x, y) – the set such that if non-empty (for example in case (X, ρ) is a convex metric space) then for any z ∈ ω(x, y), 2xz = 2yz = xy for x, y ∈ X (in the Euclidean space it is the middle-point of the interval [x, y]).

B[x, r] (B(x, r), S(x, r)) – a closed ball (open ball, sphere) centered at x ∈ (X, ρ) with radius r ≥ 0.

For each positive integer N, denote by Σ_{N}(X) the class of all nonempty subsets Y of X with at most N points; each element of it will be called an N-net [1]. Let also Σ_{N}^{∗}(X) stand for the class of all Y ∈ Σ_{N}(X) with exactly N points. Clearly, Σ_{N}^{∗}(X) = Σ_{N}(X) \ Σ_{N−1}(X); each element of it will be referred to as an exact N-net.

Let B(X) stand for the set of all non-empty bounded closed subsets of the space X. α: B(X)×B(X) → ℝ_{+}, α(M, T) = max{max{xT : x ∈ M}, max{tM : t ∈ T }} is the Hausdorff metric on the set B(X) (cf. [2], p. 223). It is also a metric on the set Σ_{N}(X) ⊂ B(X). Let us denote for brevity by Σ_{N}(X) the metric space (Σ_{N}(X), α).

AMS Subject Classification: 54E40, 52C35.

Keywords and phrases: Chebyshev center, local Lipschitz property, N-net, Hausdorff metric.

B_{α}(M, r) – open ball centered at the point M ∈ (Σ_{N}(X), α), with radius r > 0.

D[M] – diameter of the set M ∈ Σ_{N}(X).

Consider M ∈ Σ_{N}(X). Assume that R(M) = inf{sup[yx : y ∈ M] : x ∈ X} is Chebyshev radius of the N-net M. The point cheb(M) ∈ X is called a Chebyshev center if sup[x cheb(M) : x ∈ M] = R(M) [1].

Let (X, ρ) and (X_{1}, ρ_{1}) be metric spaces. The mapping f : X → X_{1} will be called (ρ, ρ_{1})-Lipschitz (or, simply, Lipschitz when ρ and ρ_{1} are understood) if there is a constant L ≥ 0 (called the Lipschitz constant of f) such that ρ_{1}(f(x), f(y)) ≤ Lρ(x, y), for all x, y ∈ X. When L = 1, we say that f is (ρ, ρ_{1})-nonexpansive (or, simply, nonexpansive when ρ and ρ_{1} are understood)(cf. [3], p. 10).

The notation given below will be used in Euclidean space as well as in Lobachevsky one.

co(M) – convex hull of the set M.

[x, y] ((x, y), (x, y]) – closed (open, half-closed) interval of the end-points x, y. Π(x_{1}, …, x_{m+1}) – m-plane defined by points x_{1}, …, x_{m+1}. Λ(x, y) – ray with the vertex in the point x comprising y ≠ x.

2. Statement of the problem

The notion of Chebyshev center is of a fundamental importance in many areas of Analysis and Geometry (fixed point and approximation theory, metrical geometry, etc.). A natural problem to be addressed here is the study of local and global Lipschitz properties of the mapping cheb: B(X) → X over different classes K of B(X). Despite its obvious importance, this problem was not completely solved until now even in the easiest case of Euclidean space.

It is well known that the Chebyshev center mapping cheb: B(X) → X is continuous, whenever X is Euclidean or Lobachevsky; cf. [4], [5]. Moreover, restriction of the map cheb to the set of all balls of the space with inner metric is a nonexpansive map [7]. Nevertheless the restriction of the map cheb to the set (B(ℝ^{2}), α) is not Lipschitz even in a neighbourhood of a closed disk [8].

Thus we arrive to the following (rather complicated in general case) task: Find classes of subsets (different from one consisting of the points or balls) of the given metric space such that for any class K the map cheb: (K, α) → X, M ↦ cheb(M) is locally Lipschitz.

It seems natural to solve this problem first for Euclidean space (X, ρ) (X may be infinite-dimensional and ρ is the standard metric) and K = Σ_{N}(X) or K = Σ_{N}^{∗}(X). The brief formulation of the result achieved by the authors is as follows:

i) if X is Euclidean the Chebyshev center mapping cheb: Σ_{N}^{∗}(X) → X is locally Lipschitz;

ii) if the dimension of Euclidean or Lobachevsky space is greater than 1 and N > 2 the mapping in question is not Lipschitz in any neighbourhood of the class Σ_{2}(X) ⊂ Σ_{N}(X).

3. Statements of the main results

Here we present the detailed formulation of our results. For N = 2 lemma 1 gives the answer to the problem stated in the previous paragraph. This lemma is a part of lemma 2 from [8].

Lemma 1. Let the metric space (X, ρ) be such that a) for each x, y ∈ X, ω(x, y) is a singleton, and b) 2ω(p, x)ω(p, y) ≤ xy, for all x, y, p ∈ X. Then, the inequalities

cheb(M) cheb(Z) ≤ α(M, Z) ≤ cheb(M) cheb(Z) + (D[M] + D[Z])/2

hold true for any M, Z ∈ Σ_{2}(X).

We give the answer to our question for Euclidean line ℝ in lemma 2.

Lemma 2. Let X be the Euclidean line. Then the map cheb: (Σ_{N}(X), α) → X is nonexpansive.

The global Lipschitz property of the map cheb vanishes if the dimension of the space is greater than 1 and N > 2.

Lemma 3. The following properties hold:

i) if the dimension of the Euclidean or Lobachevsky space X is greater than 1 and N > 2, the Chebyshev center mapping cheb: U → X is not Lipschitz for any neighborhood U of the class Σ_{2}(X) ⊂ Σ_{N}(X);

ii) if X is Euclidean, the Chebyshev center mapping cheb: Σ_{N}(X) → X is not uniformly continuous.

The following lemmas help us to prove local Lipschitz property of the map cheb in case N = 3.

Lemma 4. Let X be an Euclidean space. Then for any pair M = {u, v, w} and Z = {u, v, z} ∈ Σ_{3}^{∗} such that w ∈ (u, z) we get inequality

cheb(M) cheb(Z) ≤ Lα(M, Z).

Here L = 1/(2 sin(φ)) if φ = ∠vuw ≠ 0 and ∠uvw are acute angles; and L = 1/2 in other cases.

Lemma 5. Let X be an Euclidean space, M = {u, v, w} ∈ Σ_{3}^{∗} and ε = min[ab : a ≠ b, a b ∈ M]/8. Then there exists a constant L > 0 such that for all z_{1}, z_{2} ∈ B(w, ε) the following inequality holds true

cheb(W_{1}) cheb(W_{2}) ≤ Lα(W_{1}, W_{2}).

Here W_{1} = {u, v, z_{1}}, W_{2} = {u, v, z_{2}}.

Now we are ready to prove a local Lipschitz property of the map cheb for exact 3-nets.

Corollary 1. Let X be an Euclidean space, M ∈ Σ_{3}^{∗} and ε = min[ab : a ≠ b, a b ∈ M]/8. Then cheb: (B_{α}(M, ε), α) → X is a Lipschitz map.

Theorem 1 is an extension of this result to the case N > 3 in Euclidean plane.

Theorem 1. Let X be the Euclidean plane, N > 3, M ∈ Σ_{N}^{∗} and ε = min[ab : a ≠ b, a b ∈ M]/8. Then cheb: (B_{α}(M, ε), α) → X is a Lipschitz map.

Then we turn to the case of Euclidean space of dimension greater than 2.

Lemma 6. Assume that X is an Euclidean space of dimension greater than 1 and a N-net M = {x_{1}, …, x_{N}} ∈ Σ_{N}^{∗} defines an (N − 1)-dimensional simplex. Then there exists a constant L > 0 such that for each N-net Z = {x_{1}, …, x_{N−1}, y_{N}} meeting the condition x_{N} ∈ (x_{1}, y_{N}) the inequality

cheb(M) cheb(Z) ≤ Lα(M, W)

holds true.

Lemma 7. Let X be an Euclidean space of dimension greater than 1, M = {x_{1}, …, x_{N}} ∈ Σ_{N}^{∗}(X) be an (N −1)-dimensional simplex and W = M ⋃{x_{N+1}} ∈ Σ_{N+1}^{∗}(X) be such that x_{N+1} ∈ Π(x_{1}, …, x_{N})\M and

ε = min[min[ab : a ≠ b, a, b ∈ {x_{1}, …, x_{N−1}}], x_{N}Π(x_{1}, …, x_{N−1})]/8.

Then

(i) there exists a constant L > 0 such that for all z_{1}, z_{2} ∈ B(x_{N}, ε) the inequality

cheb(Z_{1}) cheb(Z_{2}) ≤ Lα(Z_{1}, Z_{2})

holds true, where Z_{1} = {x_{1}, …, x_{N−1}, z_{1}}, Z_{2} = {x_{1}, …, x_{N−1}, z_{2}}.

(ii) there exist constants δ > 0 and L > 0 such that for all y_{1}, y_{2} ∈ B(x_{N+1}, δ) the inequality

cheb(Y_{1}) cheb(Y_{2}) ≤ Lα(Y_{1}, Y_{2}),

holds, where Y_{1} = {x_{1}, …, x_{N}, y_{1}}, Y_{2} = {x_{1}, …, x_{N}, y_{2}}.

Theorem 2. Let (X, ρ) be an Euclidean space and N ≥ 2 be arbitrary fixed. Then, for each M ∈ Σ_{N}^{∗}(X) there exists ε = ε(M) > 0 such that cheb: B_{α}(M, ε) → X is Lipschitz.

It turns out that Lipschitz property of the map cheb holds for two N-nets which lying sufficiently far from each other.

Proposition 1. Let X be the Euclidean plane and N > 2. Then for any M, Z ∈ Σ_{N}(X) such that B(cheb(M), R(M)) ∩ B(cheb(Z), R(Z)) = ∅ the inequality

cheb(M) cheb(Z) ≤ Lα(M, Z)

holds true, where L = (1 + \sqrt{5})/2 for N > 3 and L = \sqrt{2} for N = 3.

Proposition 2. (i) Consider an Euclidean space X and a 3-net M = {u, v, w}. Let the 3-net Z = {u, v, z} ∈ Σ_{3}^{∗}(X) be such that:

1. co({u, v, w}) ∩ co({u, v, z}) = [u, v];

2. if angles ∠(uwv), ∠(uzv) are acute then α(M, Z) < wz.

Then the inequality

cheb(M) cheb(Z) ≤ α(M, Z)

holds true.

(ii) Let X be the Euclidean plane. Then the inequality

cheb(M) cheb(Z) ≤ 2α(M, Z)

holds for all M = {u, v, w} and Z = {u, q, z} ∈ Σ_{3}(X)\Σ_{2}(X) such that

co({u, v, w}) ∩ co({u, v, z}) = {u}.

4. Proofs of the results

Proof of Lemma 2. Consider two arbitrary N-nets M = {x_{1}, …, x_{N}}, Z = {y_{1}, …, y_{N}}. Let us assume that x_{1} ≤ ⋯ ≤ x_{N}, y_{1} ≤ ⋯ ≤ y_{N} and x_{1} ≤ y_{1}. Then Lemma 1 together with the definitions of the Hausdorff metric and of the Chebyshev center gives us the inequality

cheb(M) cheb(Z) = cheb({x_{1}, x_{N}}) cheb({y_{1}, y_{N}})
≤ α({x_{1}, x_{N}}, {y_{1}, y_{N}}) ≤ α(M, Z).

Proof of Lemma 3. (i) It suffices to verify the statement in case N = 3. Let x, y be two different points in Euclidean plane (Lobachevsky plane) and S_{+} be the semicircle constructed on the interval [x, y] as diameter of the circle. Consider a non-fixed point z ∈ S_{+} such that yz ∈ (0, xy/2). Now find a point u meeting the following conditions: a) z ∈ [x, u]; b) the point y lies on a circle based on [x, u] as diameter. Consider 3-nets M = {x, y, z}, W = {x, y, u} ∈ Σ_{3}^{∗} in Euclidean space (or in Lobachevsky space). Then by construction v = cheb(M) = ω(x, y), w = cheb(W) = ω(x, u) and uz = α(M, W). In Euclidean plane for any constant L > 0 we can choose a point z ∈ S_{+} so that 2Lyz < xy. Then cheb(M) cheb(W) = vw = (xy)(uz)/(2zy) > Luz = Lα(M, W). This proves lemma 3 for Euclidean space.

Now let p be the base of a perpendicular to the interval [x, w] passing through point v and ψ be the angle with vertex w of the triangle △xvw in Lobachevsky plane. The triangles △xvw, △wpv are right-angled by construction. Hence, we get formulae (cf. (4a), (4b) from [8], p. 58) tanh(pv) = sinh(pw) tan(ψ), tanh(vx) = sinh(vw) tan(ψ). Then the fraction sinh(vw)/ sinh(pw) = tanh(vx)/ tanh(pv) tends to ∞ as (pv → 0). But then cheb(M) cheb(W)/α(M, W) = (vw)/(2pw) also infinitely increases as (pv → 0). This completes the proof for Lobachevsky space.

(ii) Let us use the notation of the previous part of the proof. At the same time let us introduce new points x_{n} = y+n(x−y), z_{n} = Λ(u, x_{n})∩S(ω(x_{n}, y), x_{n}y/2) and triples M_{n} = {x_{n}, y, z_{n}}, Z_{n} = {x_{n}, y, u} where we identify points with their radiusvectors and n = 1, 2, . . .. Then α(M_{n}, Z_{n}) → 0 but cheb(M_{n}) cheb(Z_{n}) → yu as (n → ∞). This completes the proof of lemma 3.

Proof of Lemma 4. Let φ be a nonzero acute angle. Then

cheb(M) cheb(W) ≤ wz/2 = α(M, W)/2.

Let φ ≠ 0 be acute angle, p – base of the perpendicular to the ray Λ(u, w) passing through v, q – point on the ray Λ(u, w) such that the interval [q, v] is perpendicular to the interval [u, v]. Then the set of points of the ray Λ(u, w) can be represented as follows: Λ(u, w) = (u, p] ∪ (p, q) ∪ (Λ(u, w)\(u, q)). Let us consider position of cheb(M) depending on w. If w ∈ (u, p] then c = cheb(M) = ω(u, v). If w ∈ (p, q) then cheb(M) ∈ (c, b), where the point b ∈ (u, q] is such that the interval [b, c] is perpendicular to the ray Λ(u, v) and cheb(M) = pw/(2 sin(φ)). If w ∈ Λ(u, w)\(u, q) then cheb(M) = ω(u, w). Let us investigate all possibilities for location of the points w, z on the ray Λ(u, w).

If either w, z ∈ (u, p] or w, z ∈ Λ(u, w)\(u, q) then cheb(M) cheb(W) = 0.

In case w ∈ (u, p), z ∈ (p, q) we have cheb(M) cheb(W) = pz/(2 sin(φ)) = α(M, W)/(2 sin(φ)).

If w ∈ (u, p), z ∈ Λ(u, w)\(u, q) then cheb(M) cheb(W) = c cheb(W) ≤ cb + b cheb(W) = pq/(2 sin(φ)) + qz/2 ≤ pz/(2 sin(φ)) ≤ α(M, W)/(2 sin(φ)).

If w, z ∈ (p, q) then cheb(M) cheb(W) = (pz − pw)/(2 sin(φ)) ≤

≤ α(M, W)/(2 sin(φ)).

If w ∈ (p, q), z ∈ Λ(u, w)\(u, q) then cheb(M) cheb(W) ≤ cheb(M)b + b cheb(W) = (pq − pw)/(2 sin(φ)) + qz/2 = α(M, W)/(2 sin(φ)).

Thus in all considered cases we can fix 1/(2 sin(φ)) as Lipschitz constant L. This completes the proof of the lemma.

Proof of Lemma 5. Let triangles defined by 3-nets W_{1}, W_{2} be not acute. Then with the help of lemma 1 we get the desired inequality with constant L = 1/2. Now let X be a Euclidean plane and 3-net W_{1} define an acute triangle. If the angle ∠uz_{2}v is blunt then there exists a point z_{3} ∈ (z_{1}, z_{2}) such that the angle ∠uz_{3}v is right and cheb({u, v, z}) = cheb(W_{2}). Thus without loss of generality one can assume that the triangle given by the 3-net W_{2} is not blunt. Let us consider all possible cases of the location of the point z_{2} ∈ B(w, ε).

1. Let a point z_{2} ≠ z_{1} belong to the set co{u, v, z_{1}} or to the closed angle with vertex z_{1} and sides on the rays Λ(v, z_{1}), Λ(u, z_{1}). If z_{2} ∈ Λ(u, z_{1}) ∪ Λ(v, z_{1}) then we get the desired inequality from Lemma 4. In the other case let us denote by p a point of intersection of the rays Λ(v, z_{2}), Λ(u, z_{1}) and W_{3} = {u, v, p}. Note that the angle ∠z_{1}pz_{2} is not acute. Now we use Lemma 4 and triangle inequality. Thus we get constants L_{1} = sup{1/(2 sin(∠z_{1}uv)) : z_{1} ∈ B(w, ε)} < ∞ and L_{2} = sup{1/(2 sin(∠z_{2}vu)) : z_{2} ∈ B(w, ε)} < ∞ such that

cheb(W_{1}) cheb(W_{2}) ≤ cheb(W_{1}) cheb(W_{3}) + cheb(W_{3}) cheb(W_{2})

≤ L_{1}α(W_{1}, W_{3}) + L_{2}α(W_{3}, W_{2}) ≤ (L_{1} + L_{2})α(W_{1}, W_{2}).

So, in this case L = L_{1} + L_{2}.

2. Now let a point z_{2} ≠ z_{1} belong to the open angle with vertex z_{1} and sides on the rays Λ(z_{1}, u) and Λ(v, z_{1}) (the case in which z_{2} ≠ z_{1} belongs to the open acute angle with vertex z_{1} and sides on the rays Λ(z_{1}, v) and Λ(u, z_{1}) is similar to this one). Let us denote by p the intersection point of the rays Λ(v, z_{2}), Λ(u, z_{1}) and W_{3} = {u, v, p}. If the angle ∠z_{1}pz_{2} is not acute we can use the same considerations as in the first case. If the angle ∠z_{1}pz_{2} is acute then we have inequalities α(W_{1}, W_{3}) = pz_{1} = pv sin(∠pvz_{1})/ sin(∠uz_{1}v) ≤ vz_{1} sin(∠pvz_{1})/ sin(∠uz_{1}v) ≤ z_{1}z_{2}/ sin(∠uz_{1}v) ≤ L_{3}α(W_{1}, W_{2}), here L_{3} = sup{1/(sin(∠uz_{1}v)) : z_{1} ∈ B(w, ε)} < ∞ by Lemma 5. Now we again use this inequality, triangle inequality and Lemma 4. Thus we get

cheb(W_{1}) cheb(W_{2}) ≤ cheb(W_{1}) cheb(W_{3}) + cheb(W_{3}) cheb(W_{2})
≤ L_{1}α(W_{1}, W_{3}) + L_{2}α(W_{3}, W_{2})
≤ L_{1}α(W_{1}, W_{3}) + L_{2}(α(W_{1}, W_{3}) + α(W_{1}, W_{2})) ≤ Lα(W_{1}, W_{2}).

Here L = L_{3}(L_{1} + L_{2}) + L_{2}, L_{1} = sup{1/(2 sin(∠z_{1}uv)) : z_{1} ∈ B(w, ε)} < ∞, L_{2} = sup{1/(2 sin(∠z_{2}vu)) : z_{2} ∈ B(w, ε)} < ∞.

Now let X be Euclidean space of dimension greater than 2 and W_{4} be an orthogonal projection of W_{2} on the plane containing W_{1}; note that if W_{1} lies on the line then W_{4} = W_{2}. Then the previous considerations provide us with the constant L_{4} > 0 such that for all z_{1}, z_{2} ∈ B(w, ε) the inequality

cheb(W_{1}) cheb(W_{4}) ≤ L_{4}α(W_{1}, W_{4}) ≤ L_{4}α(W_{1}, W_{2})

holds true. Note that the inequality

cheb(W_{4}) cheb(W_{2}) ≤ α(W_{1}, W_{2})

follows from purely geometric considerations. It suffices now to apply triangle inequality to complete the proof.

Proof of Corollary 1. Fix an arbitrary pair of nets W_{1} = {x, y, z}, W_{2} = {u, v, w} ∈ B_{α}(M, ε) and put W_{3} = {x, y, w}, W_{4} = {x, v, w}. Then Lemma 5, definition of the Hausdorff metric and triangle inequality provide us with the constants L_{1}, L_{2}, L_{3} > 0 such that

cheb(W_{1}) cheb(W_{2}) ≤ cheb(W_{1}) cheb(W_{3}) + cheb(W_{3}) cheb(W_{4})+

+ cheb(W_{4}) cheb(W_{2}) ≤ L_{1}α(W_{1}, W_{3}) + L_{2}α(W_{3}, W_{4}) + L_{3}α(W_{4}, W_{2})

≤ (L_{1} + L_{2} + L_{3})α(W_{1}, W_{2}).

Thus cheb: (B_{α}(M, ε), α) → X is a Lipschitz map. This completes the proof.

Proof of Theorem 1. (i) Let us consider two arbitrary N-nets of the special kind Z_{1} = {x_{1}, x_{2}, …, x_{N}}, Z_{2} = {y_{1}, x_{2}, …, x_{N}} ∈ B_{α}(M, ε) and introduce the parametrisation x = x(s) by the length of the interval [x_{1}, y_{1}] so that x_{1} = x(0), y_{1} = x(x_{1}y_{1}). If z ∈ (x_{1}, y_{1}] ∩ B[cheb(Z_{1}), R(Z_{1})] then cheb(Z_{1}) = cheb({z, x_{2}, …, x_{N}}. Hence we may assume that x_{1} ∈ S(cheb(Z_{1}), R(Z_{1})). Let us put in correspondence to any s ∈ [0, x_{1}y_{1}] an N-net Z(s) = {x(s), x_{2}, …, x_{N}} and a convex polygon Q(s), given by vertices Z(s) ∩ S(cheb(M(s)), R(M(s))). For any s ∈ [x_{1}y_{1}] divide the polygon Q(s) into triangles {△_{1}(s), …, △_{k(s)}(s)} with common vertex x(s) by connecting this vertex by intervals with all other vertices of Q(s). Now for each s ∈ [x_{1}y_{1}] cheb(Q(s)) = cheb(M(s)) belongs either to the interior of one of the acute triangles or to the side common to two adjacent triangles (it is possible that these triangles are one and the same or even they are just points on an interval) from the division {△_{1}(s), …, △_{k(s)}(s)}. Consider two cases.

1. Let cheb(Z_{1}) belong to the interior of the triangle △x_{1}ab ∈ {△_{1}(0), …, △_{k(0)}(0)}. Then by continuity of the mapping cheb one can find a minimal number s_{1} ∈ (0, x_{1}y_{1}] such that either cheb(Z_{2}) = cheb({y_{1}, a, b}) in case s_{1} = x_{1}y_{1}, or cheb(Z(s_{1})) = cheb({x(s_{1}), a, b}) lies on the side of one or two of triangles of the division {△_{1}(s_{1}), …, △_{k(s_{1})}(s_{1})} in case s_{1} ∈ (0, x_{1}y_{1}).

2. Let cheb(Z_{1}) belong to the common side of the triangles △x_{1}ab, △x_{1}ac ∈ {△_{1}(0), …, △_{k(0)}(0)}. Then continuity of the map cheb implies the existence of a minimal s_{1} ∈ (0, x_{1}y_{1}] such that either cheb(Z_{2}) = cheb({y_{1}, a, b}) or cheb(Z_{2}) = cheb({y_{1}, a, c}) if s_{1} = x_{1}y_{1}, or cheb(Z(s_{1})) = cheb({x(s_{1}), a, b}) or cheb(Z(s_{1})) = cheb({x(s_{1}), a, c}) lies on the side of one or two of triangles of the division {△_{1}(s_{1}), …, △_{k(s_{1})}(s_{1})} in case s_{1} ∈ (0, x_{1}y_{1}).

In either of the considered cases Lemma 5 and compactness of the interval [x_{1}, y_{1}] imply the existence of the constant L_{1} > 0 such that cheb(Z_{1}) cheb(Z(s_{1})) ≤ L_{1}α(Z_{1}, Z(s_{1})). If s_{1} ≠ x_{1}y_{1} then for cheb(Z(s_{1})) we can proceed in similar way and get s_{2}. Continuing this process we get from Lemma 5 constants 0 < s_{1} < ⋯ < s_{m} = x_{1}y_{1}, L_{1}, …, L_{m} > 0 such that

cheb(Z_{1}) cheb(Z_{2}) ≤ cheb(Z_{1}) cheb(Z(s_{1})) + cheb(Z(s_{1})) cheb(Z(s_{2})) + ⋯ +
+ cheb(Z(s_{m−1})) cheb(Z_{2}) ≤ L_{1}α(Z_{1}, Z(s_{1})) + ⋯ +
+ L_{m}α(Z(s_{m−1}), Z_{2}) ≤ (L_{1} + ⋯ + L_{m})α(Z_{1}, Z_{2}).

This completes the proof of the Theorem 1 in this special case.

(ii) Choose arbitrary nets Z_{1} = {x_{1}, x_{2}, …, x_{N}}, Z_{2} = {y_{1}, y_{2}, …, y_{N}} ∈ B_{α}(M, ε) and fix Z_{3} = {y_{1}, x_{2}, …, x_{N}}, …, Z_{N+1} = {y_{1}, y_{2}, …, y_{N−1}, x_{N}}. Then triangle inequality, Lemma 5 and the definition of the Hausdorff metric give us constants L_{1} …, L_{N−1} > 0 such that

cheb(Z_{1}) cheb(Z_{2}) ≤ cheb(Z_{1}) cheb(Z_{3}) + cheb(Z_{3}) cheb(Z_{4}) + ⋯ +
+ cheb(Z_{N+1}) cheb(Z_{2}) ≤ L_{1}α(Z_{1}, Z_{3}) + L_{2}α(Z_{3}, Z_{4}) + ⋯ +
+ L_{N−1}α(Z_{N+1}, Z_{2}) ≤ (L_{1} + ⋯ + L_{N−1})α(Z_{1}, Z_{2}).

Thus cheb: (B_{α}(M, ε), α) → X is a Lipschitz map.

Proof of Lemma 6. The proof is by induction. Lemma 4 implies consistence of the statement in the case N = 3. Assume now that Lemma 6 holds true for all numbers less or equal than N − 1. Let us show that the statement holds true for N. Let p ≠ x_{1} be an intersection point of the ray Λ(x_{1}, x_{N}) with the sphere S(cheb({x_{1}, …, x_{N−1}}), R({x_{1}, …, x_{N−1}})). and q be a point on Λ(x_{1}, x_{N}) such that there exists k ∈ {2, …, N} such that

x_{k} ∈ S(cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, q}), R({x_{1}, …, x̂_{k}, …, x_{N−1}, q})),

(here cap on the symbol means that this symbol must be excluded) closest to x_{1}. Then the set of points of the ray Λ(x_{1}, x_{N}) can be represented as follows: Λ(x_{1}, x_{N}) = (x_{1}, p] ∪ (p, q) ∪ (Λ(x_{1}, x_{N})\(x_{1}, q)). Consider all possible locations of the points x_{N}, y_{N} on Λ(x_{1}, x_{N}).

If x_{N}, y_{N} ∈ (x_{1}, p] then cheb(M) cheb(Z) = 0. If x_{N}, y_{N} ∈ (Λ(x_{1}, x_{N})\(x_{1}, q)) then there exist k, j ∈ {2, …, N} such that cheb(M) = cheb({x_{1}, …, x̂_{k}, …, x_{N}}), cheb(Z) = cheb({x_{1}, …, x̂_{j}, …, x_{N−1}, y_{N}}). Moreover by induction assumption there exists a constant L > 0 such that

cheb(M) cheb(Z) ≤ Lα({x_{1}, …, x̂_{k}, …, x_{N}}, {x_{1}, …, x̂_{j}, …, x_{N−1}, y_{N}})
≤ Lα(M, Z).

If x_{N} ∈ (x_{1}, q), y_{N} ∈ (p, q) then cheb(M) cheb(Z) ≤ α(M, Z)/(2 cos(ψ)), here ψ is an angle between the ray Λ(x_{1}, x_{N}) and the normal to the scale (x_{1}, …, x_{N−1}) of the simplex.

Let b be an intersection-point of the boundary of the simplex (x_{1}, …, x_{N}) with normal to the scale (x_{1}, …, x_{N−1}) passing through the point cheb({x_{1}, …, x_{N−1}}) which does not belong to this face. If x_{N} ∈ (x_{1}, q), y_{N} ∈ (Λ(x_{1}, x_{N})\(x_{1}, q)) then there exist k, j ∈ {2, …, N} such that b = cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, q}), cheb(Z) = cheb({x_{1}, …, x̂_{j}, …, x_{N−1}, y_{N}}). Moreover by assumption there exist a constant L_{1} > 0 such that

cheb(M) cheb(Z) ≤ cheb(M)b + b cheb(Z) ≤ q{p, x_{N}}/(2 cos ψ)+

+ cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, q}) cheb({x_{1}, …, x̂_{j}, …, x_{N−1}, y_{N}})

≤ q{p, x_{N}}/(2 cos ψ) + L_{1}α({x_{1}, …, x̂_{k}, …, x_{N−1}, q},
{x_{1}, …, x̂_{j}, …, x_{N−1}, y_{N}}) ≤ Lα(M, Z),

where L = (1 + L_{1})/(2 cos ψ). This completes the proof.

Proof of Lemma 7. The proof is by induction.

(i) Lemma 5 implies that statement (i) of Lemma 7 holds true for N = 3. Assume now that it holds for all numbers less or equal than N − 1 and prove it for N. Consider two cases.

1. Let there exists k ∈ {1, …, N −1} such that pz_{1} ≤ z_{1}z_{2}, where p is the intersection point of the ray Λ(x_{k}, z_{2}) with the (N−2)-plane Π({x_{1}, …, x̂_{k}, …, x_{N−1}, z_{1}}).

Let us denote by Z_{3} the set {x_{1}, …, x_{N−1}, p}. Now Lemma 6 implies the existence of a constant L_{2} > 0 such that cheb(Z_{3}) cheb(Z_{2}) ≤ L_{2}α(Z_{3}, Z_{2}).

By induction assumption there exists a constant L_{1} > 0 such that

cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, z_{1}}) cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, p}) L_{1}α({x_{1}, …, x̂_{k}, …, x_{N−1}, z_{1}}, {x_{1}, …, x̂_{k}, …, x_{N−1}, p}) ≤ L_{1}α(Z_{1}, Z_{3}).

Let ψ stand for the angle between normal vector to the scale (x_{1}, …, x_{N−1}) directed into the simplex and normal vector to the scale (x_{1}, …, x̂_{k}, …, x_{N−1}, z_{1}) directed outside of the simplex. Now purely geometrical considerations imply the inequality

cheb(Z_{1}) cheb(Z_{3}) ≤ ≤ max[1/ sin ψ, 1] cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, z_{1}}) cheb({x_{1}, …, x̂_{k}, …, x_{N−1}, p}).

Thus

cheb(Z_{1}) cheb(Z_{2}) ≤ cheb(Z_{1}) cheb(Z_{3}) + cheb(Z_{3}) cheb(Z_{2}) ≤ ≤ L_{1} max[1/ sin ψ, 1]α(Z_{1}, Z_{3})+L_{2}α(Z_{3}, Z_{2}) ≤ (L_{1} max[1/ sin ψ, 1]+2L_{2})α(Z_{1}, Z_{2}).

2. Let the assumption of case 1 be wrong, j ∈ {1, …, N − 1} and denote by p_{j} the intersection point of the ray Λ(x_{j}, z_{2}) and the (N − 2)-plane Π(x_{1}, …, x̂_{j}, …, x_{N−1}, z_{1}). Now choose k ∈ {1, …, N − 1} such that p_{k}z_{1} = min[p_{j}z_{1} : j ∈ {1, …, N − 1}] and consider Z_{3} = {x_{1}, …, x_{N−1}, p_{k}}.

Note that the angle ∠z_{1}p_{k}z_{2} is acute and introduce the inequality

α(Z_{1}, Z_{3}) = p_{k}z_{1} = p_{k}x_{k} sin ∠p_{k}x_{k}z_{1}/ sin ∠p_{k}z_{1}x_{k} ≤ ≤ x_{k}z_{1} sin ∠p_{k}x_{k}z_{1}/ sin ∠p_{k}z_{1}x_{k} ≤ z_{1}z_{2}/ sin ∠p_{k}z_{1}x_{k} ≤ L_{3}α(Z_{1}, Z_{2}),

where general assumptions imply that L_{3} = sup{1/ sin ∠p_{k}z_{1}x_{k} : z_{1} ∈ B(x_{N}, ε)} < ∞. This inequality, the triangle one, Lemma 6 and similar estimate for cheb(Z_{1}) cheb(Z_{3}) of the first case put together give us constants L_{1}, L_{2} > 0 such that

cheb(Z_{1}) cheb(Z_{2}) ≤ cheb(Z_{1}) cheb(Z_{3}) + cheb(Z_{3}) cheb(Z_{2}) ≤
≤ L_{1}α(Z_{1}, Z_{3}) + L_{2}α(Z_{3}, Z_{2})
≤ L_{1}α(Z_{1}, Z_{3}) + L_{2}(α(Z_{1}, Z_{3}) + α(Z_{1}, Z_{2})) ≤ Lα(Z_{1}, Z_{2}),

where L = L_{3}(L_{1} + L_{2}) + L_{2}. Thus statement (i) of Lemma 7 holds true.

(ii) Lemma 5 provides us with the proof of statement (ii) of Lemma 7 for N = 2. Let us assume that statement (ii) holds true also for all natural numbers less or equal than N −1 and prove it for N. Let 0 < δ < min[ab : a ≠ b, a, b ∈ W]/8. Consider three cases.

1. Let [y_{1}, y_{2}] ∩ Π(x_{1}, …, x_{N}) = ⊘. Then our statement holds by compactness of the interval [y_{1}, y_{2}] and previously proved statement (i) of Lemma 7.

2. Let [y_{1}, y_{2}] ⊂ Π(x_{1}, …, x_{N}). Introduce the natural parametrisation y = y(s) of the interval [y_{1}, y_{2}] by length so that y_{1} = y(0), y_{2} = y(y_{1}y_{2}). If y ∈ (y_{1}, y_{2}] ∩ B[cheb(Y_{1}), R(Y_{1})] then cheb(Y_{1}) = cheb({x_{1}, x_{2}, …, x_{N}, y}).

Hence we can assume that y_{1} ∈ S(cheb(Y_{1}), R(Y_{1})). Consider for any s ∈ [0, y_{1}y_{2}] (N + 1)-net Y (s) = {x_{1}, x_{2}, …, x_{N}, y(s)} and convex polygon Q(s) with vertices Y (s) ∩ S(cheb(Y (s)), R(Y (s))). Now we divide polygon Q(s) for each s ∈ [x_{1}, y_{1}] into simplices {△_{1}(s), △_{k(s)}(s)} with the common vertex y(s), here 1 ≤ k(s) ≤ 2. For any s ∈ [y_{1}, y_{2}] cheb(Q(s)) = cheb(Y (s)) belongs either to the interior of one simplex or to the common face of two simplices of the division {△_{1}(s), △_{k(s)}(s)} (the degenerate cases are also possible). Consider two cases.

A. Let cheb(Y_{1}) belong to the interior of the simplex (a_{1}, …, a_{N−1}, y_{1}) ∈ {△_{1}(0), △_{k(0)}(0)}.

Then—since the mapping cheb is continuous—one can find a minimal s_{1} ∈ (0, y_{1}y_{2}] such that either cheb(Y_{2}) = cheb({a_{1}, a_{2}, …, a_{N−1}, y_{2}}) for s_{1} = y_{1}y_{2} or cheb(Y (s_{1})) = cheb({y(s_{1}), x_{2}, …, x_{N}}) belongs to the scale of one or two simplices of the division {△_{1}(s_{1}), △_{k(s_{1})}(s_{1})} for s_{1} ∈ (0, y_{1}y_{2}).

B. Let cheb(Y_{1}) belong to the common face of the simplices (a_{1}, …, a_{N−2}, b, y_{1}), (a_{1}, …, a_{N−2}, c, y_{1}) ∈ {△_{1}(0), △_{k(0)}(0)}.

Then again continuity of the mapping cheb implies the existence of the minimal s_{1} ∈ (0, y_{1}y_{2}], such that either cheb(Y_{2}) = cheb({a_{1}, …, a_{N−2}, b, y_{2}}) or cheb(Y_{2}) = cheb({a_{1}, …, a_{N−2}, c, y_{2}}) for s_{1} = y_{1}y_{2} or cheb(Y (s_{1})) = = cheb({a_{1}, …, a_{N−2}, b, y(s_{1})}) or cheb(Y (s_{1})) = cheb({a_{1}, …, a_{N−2}, c, y(s_{1})}) belongs to the scale of one or two simplices of the division {△_{1}(s_{1}), △_{k(s_{1})}(s_{1})} for s_{1} ∈ (0, y_{1}y_{2}).

In either case compactness of the interval [y_{1}, y_{2}] and statement (i) of Lemma 7 or induction hypothesis provide us with the constant L_{1} > 0 such that cheb(Y_{1}) cheb(Y (s_{1})) ≤ L_{1}α(Y_{1}, Y (s_{1})). If s_{1} ≠ y_{1}y_{2} then similar considerations applied to cheb(Y (s_{1})) give us s_{2}. Continuing this process we get the set of constants 0 < s_{1} < ⋯ < s_{i} = y_{1}y_{2}, L_{1}, …, L_{i} > 0 such that

cheb(Y_{1}) cheb(Y_{2}) ≤ cheb(Y_{1}) cheb(Y (s_{1})) + cheb(Y (s_{1})) cheb(Y (s_{2})) + ⋯ +

+ cheb(Y (s_{i−1})) cheb(Y_{2}) ≤ L_{1}α(Y_{1}, Y (s_{1})) + ⋯ + L_{i}α(Y (s_{i−1}), Y_{2})

≤ (L_{1} + ⋯ + L_{i})α(Y_{1}, Y_{2}).

Thus our statement holds true in this special case.

3. There exists a unique point u such that u = [y_{1}, y_{2}] ∩ Π(x_{1}, …, x_{N}).

Since triangle inequality holds true it suffices to prove the statement for the interval [y_{1}, u]. Let ỹ_{1} denote orthogonal projection of y_{1} onto Π(x_{1}, …, x_{N}). Then geometrical considerations give us the inequality

cheb(Y_{1}) cheb({x_{1}, x_{2}, …, x_{N}, ỹ_{1}}) ≤ α(Y_{1}, {x_{1}, x_{2}, …, x_{N}, ỹ_{1}}). Now case 2 of the proof implies existence of the constant L > 0 such that

cheb(Y_{1}) cheb(Y_{2}) ≤ cheb(Y_{1}) cheb({x_{1}, x_{2}, …, x_{N}, ỹ_{1}})+
+ cheb({x_{1}, x_{2}, …, x_{N}, ỹ_{1}}) cheb({x_{1}, x_{2}, …, x_{N}, u})

≤ α(Y_{1}, {x_{1}, x_{2}, …, x_{N}, ỹ_{1}}) + Lα({x_{1}, x_{2}, …, x_{N}, ỹ_{1}}, {x_{1}, x_{2}, …, x_{N}, u})

≤ (1 + L)α(Y_{1}, {x_{1}, x_{2}, …, x_{N}, u}).

This completes the proof of the Lemma.

Proof of Theorem 2. (i) Let us consider two arbitrary N-nets of the special kind: Z_{1} = {x_{1}, x_{2}, …, x_{N}}, Z_{2} = {y_{1}, x_{2}, …, x_{N}} ∈ B_{α}(M, ε) and introduce a parametrisation x = x(s) by the length of the interval [x_{1}, y_{1}] such that x_{1} = x(0), y_{1} = x(x_{1}y_{1}). If z ∈ (x_{1}, y_{1}] ∩ B[cheb(Z_{1}), R(Z_{1})] then cheb(Z_{1}) = cheb({z, x_{2}, …, x_{N}}. Hence, we can assume that x_{1} ∈ S(cheb(Z_{1}), R(Z_{1})). Let us put in correspondence the N-net Z(s) = {x(s), x_{2}, …, x_{N}} and the convex polygon Q(s) defined by its vertices Z(s) ∩ S(cheb(Z(s)), R(Z(s))) to any s ∈ [0, x_{1}y_{1}]. Now for any s ∈ [x_{1}y_{1}] divide the polygon Q(s) into simplices {△_{1}(s), …, △_{k(s)}(s)} with the common vertex x(s). Now for any s ∈ [x_{1}y_{1}] cheb(Q(s)) = cheb(Z(s)) either belongs to the interior of one simplex or to the common plane of the two of that of the division {△_{1}(s), …, △_{k(s)}(s)} (the degenerate cases are also possible). Consider two cases.

1. Let cheb(Z_{1}) belong to the interior of the simplex

(x_{1}, a_{1}, …, a_{N−1}) ∈ {△_{1}(0), …, △_{k(0)}(0)}.

Then—since the Chebyshev center depends on the set continuously—one can find a minimal s_{1} ∈ (0, x_{1}y_{1}] such that either cheb(Z_{2}) = cheb({y_{1}, a_{1}, …, a_{N−1}}) for s_{1} = x_{1}y_{1}, or cheb(Z(s_{1})) = cheb({x(s_{1}), a_{1}, …, a_{N−1}}) belongs to the common face of one or two simplices of the division {△_{1}(s_{1}), …, △_{k(s_{1})}(s_{1})} if s_{1} ∈ (0, x_{1}y_{1}).

2. Let cheb(Z_{1}) belong to the face of the simplex (x_{1}, a_{1}, …, a_{N−2}, b),

(x_{1}, a_{1}, …, a_{N−2}, c) ∈ {△_{1}(0), …, △_{k(0)}(0)}.

Then again as in the first case there exists a minimal s_{1} ∈ (0, x_{1}y_{1}] such that either cheb(Z_{2}) = cheb({y_{1}, a_{1}, …, a_{N−2}, b}) or cheb(Z_{2}) = cheb({y_{1}, a_{1}, …, a_{N−2}, c}) if s_{1} = x_{1}y_{1}, or cheb(Z(s_{1})) = cheb({x(s_{1}), a_{1}, …, a_{N−2}, b}) or cheb(Z(s_{1})) = cheb({x(s_{1}), a_{1}, …, a_{N−2}, c}) lies on the face of one or two simplices of the division {△_{1}(s_{1}), …, △_{k(s_{1})}(s_{1})} if s_{1} ∈ (0, x_{1}y_{1}).

In either case Lemma 7 and compactness of the interval [x_{1}, y_{1}] imply existence of the constant L_{1} > 0 such that cheb(Z_{1}) cheb(Z(s_{1})) ≤ L_{1}α(Z_{1}, Z(s_{1})). If s_{1} ≠ x_{1}y_{1} then we consider a similar construction for cheb(Z(s_{1})) and get the number s_{2}. Continuing the process we get the constants 0 < s_{1} < ⋯ < s_{i} = x_{1}y_{1}, L_{1}, …, L_{i} > 0 such that

cheb(Z_{1}) cheb(Z_{2}) ≤ cheb(Z_{1}) cheb(Z(s_{1})) + cheb(Z(s_{1})) cheb(Z(s_{2})) + ⋯ +

+ cheb(Z(s_{i−1})) cheb(Z_{2}) ≤ L_{1}α(Z_{1}, Z(s_{1})) + ⋯ + L_{i}α(Z(s_{i−1}), Z_{2}) ≤ ≤ (L_{1} + ⋯ + L_{i})α(Z_{1}, Z_{2}).

This completes the proof of the statement in this special case.

(ii) Consider arbitrary Z_{1} = {x_{1}, x_{2}, …, x_{N}}, Z_{2} = {y_{1}, y_{2}, …, y_{N}} ∈ B_{α}(M, ε) and put Z_{3} = {y_{1}, x_{2}, …, x_{N}}, …, Z_{N+1} = {y_{1}, y_{2}, …, y_{N−1}, x_{N}}. Again by Lemma 6, definition of the Hausdorff metric and triangle inequality together imply existence of the constants L_{1} …, L_{N−1} > 0 such that

cheb(Z_{1}) cheb(Z_{2}) ≤ cheb(Z_{1}) cheb(Z_{3}) + cheb(Z_{3}) cheb(Z_{4}) + ⋯ +

+ cheb(Z_{N+1}) cheb(Z_{2}) ≤ L_{1}α(Z_{1}, Z_{3}) + L_{2}α(Z_{3}, Z_{4}) + ⋯ + + L_{N−1}α(Z_{N+1}, Z_{2}) ≤ (L_{1} + ⋯ + L_{N−1})α(Z_{1}, Z_{2}).

Thus cheb: (B_{α}(M, ε), α) → X is a Lipschitz map. This completes the proof.

Proof of Proposition 1. Let us denote by r = R(M), R = R(Z), t = cheb(M) cheb(Z) − R − r and assume that r ≤ R. Now the definition of the Hausdorff metric and the inclusion cheb(Z) ∈ co(Z) imply that \sqrt{(R + r + t)^{2} + R^{2}}−r ≤ α(M, Z) if N > 3 and \sqrt{(R + t)^{2} + R^{2}} ≤ α(M, Z) if N = 3. On the other hand cheb(M) cheb(Z) = R + r + t ≤ (1 + \sqrt{5})(\sqrt{(R + r + t)^{2} + R^{2})} − r)/2. These inequalities put together complete the proof.

Proof of Proposition 2. (i) In the case both angles ∠(uwv), ∠(uzv) are not acute cheb(M) cheb(Z) = 0. Assume then that the angle ∠(uwv) is acute one. Consider a 3-net Z_{1} = {u, v, z_{1}}, where the point z_{1} is constructed rotating the point z over the line Π(u, v) by angle equal to one adjacent to that between the halfplanes Π_{+}(u, v, w) and Π_{+}(u, v, z). Then the conditions of the propostion and geometrical considerations imply that α(M, Z) = α(M, Z_{1}) and cheb(M) cheb(Z) ≤ cheb(M) cheb(Z_{1}). Consider two angles ∠(vuw) and ∠(uvw) for the 3-net M. Assume now that the angle ∠(vuw) is acute or non-zero and ∠(vuw) ≤ ∠(uvw). Then cheb(M)ω(u, v) ≤ vw/2 ≤ α(M, Z_{1})/2. Another case (the acute angle ∠(uzv)) is considered similarly to the stated one. In the rest of the cases ω(u, v) cheb(Z_{1}) = 0. So using all the inequalities found in this proof we get

cheb(M) cheb(Z) ≤ cheb(M) cheb(Z_{1})
≤ cheb(M)ω(u, v) + ω(u, v) cheb(Z_{1}) ≤ α(M, Z).

Thus statement (i) of the proposition holds true.

(ii) Assume without loss of generality that [v, q] is the interval of the minimal length of the intervals v, q, [v, z], [w, q], [w, z] such that its intersection with the interior of the set co(M) ∪ co(Z) is empty. Then using the definition of Hausdorff metric and the first part of the statement we find the following inequality:

cheb(M) cheb(Z) ≤ cheb(M) cheb({u, v, q}) + cheb({u, v, q}) cheb(Z)
≤ α(M, {u, v, q}) + α({u, v, q}, Z) ≤ 2α(M, Z).

This completes the proof of Proposition 2.

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(received 12.02.2007, in revised form 01.11.2007)

Pyotr N. Ivanshin, N.G. Chebotarev’s RIMM, Kazan State University, 420008, Kazan, Universitetskaya, 17, Russia

E-mail: Pyotr.Ivanshin@ksu.ru

Evgenii N. Sosov, N.G. Chebotarev’s RIMM, Kazan State University, 420008, Kazan, Universitetskaya, 17, Russia

E-mail: Evgenii.Sosov@ksu.ru