By Alexander Brudnyi, Yuri Brudnyi
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Additional info for [Article] Metric Spaces with Linear Extensions Preserving Lipschitz Condition
10) we get (Ef )(m) = (Ei fi )(m) = f (m). We now show that E ∈ Ext(S, C) and E is bounded by a constant depending only on C. , [Ke]. So every subset of ∪ni=1 Ui of diameter at most δ lies in one of the Ui . Using this we first establish the corresponding Lipschitz estimate for m , m ∈ ( ∪ni=1 Ui ) ∩ C with d(m , m ) ≤ δ. 18) In this case both m , m ∈ Ui0 for some i0 . 19) (Ef )(m) = (ρi Ei fi )(m). Ui ∩Ui0 =∅ In this sum each ρi is Lipschitz with a constant L(C) depending only on C and 0 ≤ ρi ≤ 1.
8 251 METRIC SPACES WITH LINEAR EXTENSIONS We now introduce the required embedding I : T k → H2 beginning with its definition on the subset V k ⊂ T k of vertices; namely, we let v ∈ V k. 5). 22) it suffices to work with the metric space (R2+ , ρ0 ). So we have to compare ρ0 (c(v ), c(w)) with the distance d(v , w) in the tree T k . 3. The ρ0 -distance between c(v ) and c(v + ) equals log n. Proof. 19) log 1 + |c2 (v ) − c2 (v + )| min (c2 (v ), c2 (v + )) = log (1 + n − 1) = log n. 18). Since 0 ≤ δ1 (v ) ≤ k, this is at most log n.
Proof. 5 with M := l1n , S := Zn1 and the dilation φ: x → 1 1 n n −1 equal 1 and 2 x. Then φ(S) = 2 Z ⊃ Z and the Lipschitz constants of φ and φ 2 2, respectively. 3) λ(l1n ) = λ(Zn1 ). 19 with p = 1. 4. 4) λ(lpn ) = λ(Znp ) ≥ c0 n p−2 1 1 . , Zn1 (l) := Zn1 ∩ [ − l, l]n . 5. 5) λ(Sn , Zn1 (l)) ≥ c1 n with c1 > 0 independent of n. Proof. 3 λ(Zn1 ) = sup λ(F ) F where F runs through all finite subsets F ⊂ Zn . On the other hand λ(Zn1 ) ≥ sup λ(Zn1 (l)). 6) λ(Zn1 (l)) = sup λ(F ). F⊂Zn1 (l) These three relations imply that λ(Zn1 ) = sup λ(Zn1 (l)).