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**

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Then Li ∈ Ext(Fi , Gi ) and Li ≤ 2, since |(Li f )(m)−(Li f )(m )| = | f (mi )−f (m )| ≤ f Lip(Fi ) d (mi , m )≤2 f Lip(Fi ) d (m, m ). 5) does not hold for {Gi } substituted for {Fi }, then there is a sequence Ei ∈ Ext(Gi , Bi ) such that supi Ei < ∞. 5). The proof will be now finished by the following argument. Choose a subsequence Fik ⊂ Bik := Bri (m), k ∈ N, such that k rik+1 < min{rik , dist(Fik+1 \ {m}, ∪s

2 imply that λ(S) = lim λ(Mi ). 3 give λ(M ) ≤ λ(S) = sup λ(F ) F where F runs over all finite subspaces of S. 2 has been completed. 5. From condition (b) of the corollary it follows that the sequence {(φj (S), d)}j∈N γ-converges to (M , d). Since φ is a dilation of M , every φj , j > 1, is a dilation of M , as well. Thus the sequence {(φj (S), d)}j∈N δ-converges to (S, d). 2(b). 1. (a) Let us recall that the relative extension constant λ(S, M ) where S ⊂ M is determined by the formula λ(S, M ) := inf{ T : T ∈ Ext(S, M )}.

8) is done. 9) Lip(M) = E γ fγ Lip(Bγ ) . 10) Fγ ργ . 11) (Fγ − Fγˆ )ργ := Gγˆ := γ Fγ γˆ ργ . γ Then we can write for every γˆ Ef = Fγˆ + Gγˆ . 13) Fγˆ Lip(M) ≤ λR f Lip(Γ) . 14) m ∈ Bγ ∩ Bγˆ . Lip(Γ) , In fact, we have for these m |Fγ γˆ (m)| = |(Eγ fγ − Eγˆ fγˆ )(m)| ≤ | f (γ) − f (γˆ )| + |(Eγ fγ )(m) − (Eγ fγ )(γ)| + |(Eγˆ fγˆ )(m) − (Eγˆ fγˆ )(γˆ )|. 6) to estimate the right-hand side we get |Fγ γˆ (m)| ≤ λR f ˆ) Lip(Γ) (d (γ, γ + d(m, γ) + d(m, γˆ )) ≤ 4RλR f Lip(Γ) . We apply this to estimate ∆Gγˆ := Gγˆ (m) − Gγˆ (m ) provided that m, m ∈ Bγˆ .