# Download e-book for iPad: [Article] Metric Spaces with Linear Extensions Preserving by Alexander Brudnyi, Yuri Brudnyi By Alexander Brudnyi, Yuri Brudnyi

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Additional resources for [Article] Metric Spaces with Linear Extensions Preserving Lipschitz Condition

Example text

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)). 2) this gives for some l = l(n) λ(Zn1 (l(n))) > c0 √ n. 6) with l := l(n) we then find Sn ⊂ Zn1 (l(n)) such that for l = l(n) c0 √ n. λ(Sn , Zn1 (l)) := inf{ E : E ∈ Ext(Sn , Zn1 (l))} ≥ 2 The result has been established.

2. 1) Ext(S, V) = ∅. Here and above, S and V are regarded as metric subspaces of MΓ . 1) implies statement (b). Proof. Our argument is based on the following result. Let Znp denote Zn regarded as a metric subspace of lpn . 3. 2) √ λ(Zn1 ) ≥ c0 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 .

7) ri+1 < min{ri , dist(Fi+1 \ {m}, ∪s