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

By Alexander Brudnyi, Yuri Brudnyi

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8) and estimate Ef Lip(M) . Given γˆ ∈ Γ we can write (Eγ fγ )(γˆ )ργ (γˆ ). (Ef )(γˆ ) = Bγ γˆ Since Eγ is an extension from Bγ ∩ Γ we get (Eγ fγ )(γˆ ) = fγ (γˆ ) = f (γˆ ). 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 )|.

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 )}. 3) λ(S, M ) = sup λ(F , M ) F where F runs over all finite subspaces of S. 4) Emin = λ(S, M ). (c) The same argument allows to establish the following fact. The set function S → λ(S) defined on closed subspaces of M is lower semicontinuous in the Hausdorff metric. 7. 6. Trees with all edges of length one. 1) E ≤ cn with c independent of S and S ; recall that p = 1 or ∞.

Ui ∩Ui0 =∅ In this sum each ρi is Lipschitz with a constant L(C) depending only on C and 0 ≤ ρi ≤ 1. 9) (recall that Ei fi = 0 if S ∩ Ui = ∅). If now m ∈ Ui with S ∩ Ui = ∅, then for arbitrary mi ∈ S ∩ Ui |(Ei fi )(m)| ≤ |(Ei fi )(m) − (Ei fi )(mi )| + |(Ei fi )(mi )| + | f (mi ) − f (m∗ )| ≤ A f Lip0 (S) d (m, mi ) ≤ A f Lip0 (S) (d (m, mi ) + d(mi , m∗ )). 20) |(Ei fi )(m)| ≤ 2A diam C f Lip0 (S) . 21) |(Ef )(m ) − (Ef )(m )| ≤ An(2L(C) diam C + 1) f Lip0 (S) d (m , m ), provided m , m ∈ Ui0 ∩ C.

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