[Article] Metric Spaces with Linear Extensions Preserving by Alexander Brudnyi, Yuri Brudnyi

By Alexander Brudnyi, Yuri Brudnyi

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

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Let C ⊂ (l2n , 0) be a closed convex set containing 0, and pC (x) be the (unique) closest to x point from C. , [BL, Sect. 3) pC (x) − pC ( y) 2 ≤ x − y 2. 4) (Ef )(x) := ( f ◦ pC )(x), x ∈ l2n . Since pC is the identity on C and pC (0) = 0 as 0 ∈ C, this operator belongs to METRIC SPACES WITH LINEAR EXTENSIONS 265 Ext(C, l2n ). 2) is established. 7) 1 − p1 2 . In order to prove the lower estimate we need the following result where Banach spaces are regarded as pointed metric spaces with m∗ = 0.

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. Let now Gn := (Zn , E n ) be the graph whose set of edges is given by E n := {(x, y): x, y ∈ Zn , x − y l1n = 1}. 5. 5) is contained in Vn . The metric graph MΓn is then a (metric) subspace of the space l1n , but it also can and will be regarded below as a subspace of l2n with the path metric induced by the Euclidean metric. 6. 8) √ λ(Sn , Vn ) ≥ c n; here Sn and Vn are regarded as subspaces of MΓ .

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. Let now Gn := (Zn , E n ) be the graph whose set of edges is given by E n := {(x, y): x, y ∈ Zn , x − y l1n = 1}.

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