By Alexander Brudnyi, Yuri Brudnyi

**Read or Download [Article] Metric Spaces with Linear Extensions Preserving Lipschitz Condition PDF**

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**Extra info for [Article] Metric Spaces with Linear Extensions Preserving Lipschitz Condition**

**Example text**

1 below. 3) λ( ⊕p {T k }1≤i≤n ) ≤ 256 λ( ⊕p {H2 }1≤i≤n ). 29 for the case Mi = H2 for all 1 ≤ i ≤ n. 6 for trees with edges of length one. We begin with establishing the desired quasi-isometric embedding of T k into 2 H . In the formulation of this result, d and ρ are, respectively, the path metrics on T k and H2 = {x ∈ R2 : x2 > 0}. 1. 5) BA−1 ≤ 256. METRIC SPACES WITH LINEAR EXTENSIONS 247 Proof. 6) ρ0 (x, y) := max log 1 + i=1,2 |xi − yi | . min (x2 , y2 ) The following result establishes an equivalence of this to the hyperbolic metric ρ for pairs of points far enough from each other.

1) K := sup ργ Lip(M) γ <∞ where K depends only on cΓ , N = NM and R = RM . Proof. Set Bγ := BR (γ) and c Bγ := M \ Bγ and define dγ (m) := dist(m, c Bγ ), m ∈ M. 2) supp dγ ⊂ Bγ and dγ Lip(M) ≤ 1. Let now φ: R+ → [0, 1] be continuous, equal to one on [0, R/2], zero on [R, ∞) and be linear on [R/2, R]. 3) φ ◦ dγ . 1 only at most µ terms here are nonzero at every point. 4) Lip(M) ≤ 4µ/R. On the other hand, every m ∈ M is contained in some ball BR/2 (γ) of the cover 1 2 B. For this γ (φ ◦ dγ )(m) ≥ φ(R/2) = 1 and therefore s ≥ 1.

6) we have d(m , m) ≤ ri < dist(Fi \ {m}, Fj \ {m}) ≤ d(m , m ). Combining these we prove that Ni ∈ Ext(Fi , F∞ ) and Ni ≤ 2. 5). 7) holds and the proof is complete. 2. In this proof the properness of M is not used. 3. Let U be a finite cover of a compact set C ⊂ M by open sets. Then there is a partition of unity {ρU }U∈U on C subordinate to U such that every ρU is Lipschitz with a constant depending only on the cover. Let us recall its proof: Define dU : M → R+ by dU (m) := dist(m, M \ U ), m ∈ M .