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The classical Busemann-Feller lemma in Euclidean space says the following.

Let $K\subset \mathbb{R}^n$ be a closed convex set.
Then

  1. for any point $x\in \mathbb{R}^n$ there exists unique nearest point in $K$; let us denote it by $p(x)$.

  2. the map $p\colon \mathbb{R}^n \to K$ does not increase distances, i.e. is 1-Lipschitz.

Question. Does this result hold in an $n$-dimensional hyperbolic space? A reference would be helpful.

Remark. I think part (1) is true for hyperbolic and spherical spaces; in the spherical space one should require in addition that $K$ is contained in an open half-sphere. I think part (2) is not true in the spherical space.

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    $\begingroup$ It is true in every CAT(0) space. Must be in Bridson-Haefliger. $\endgroup$
    – markvs
    Commented Jul 7, 2021 at 7:12
  • $\begingroup$ Are you sure about (1) holding in spherical space? If $\pm p \in \mathbf{S}^n$ are antipodal and $D_r(p)$ is a closed disc around $p$, then the nearest points to $-p$ form $\partial D_r(p)$, no? $\endgroup$
    – Leo Moos
    Commented Jul 7, 2021 at 14:13
  • $\begingroup$ @LeoMoos: to be more precise one should consider everything in open half-sphere rather than sphere. $\endgroup$
    – asv
    Commented Jul 7, 2021 at 14:31
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    $\begingroup$ Note that in $\mathbb{R}^n$, if (1) is true for a compact set $K$ then $K$ must be convex. This is interstingly not true in $\mathbb{H}^n$, where horospheres are not flat. You can see that as several variants of convexity that no longer match (using either geodesic or busemann functions). $\endgroup$
    – Benoît Kloeckner
    Commented Jul 8, 2021 at 9:57

1 Answer 1

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In any Hadamard space, projection into convex sets is non-expansive; see Proposition 2.4(4) in Metric spaces of non-positive curvature by Bridson and Haefliger.

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    $\begingroup$ Also found in Ballmann's lectures on Spaces of non-positive curvature, Corollary 5.6. $\endgroup$
    – Benoît Kloeckner
    Commented Jul 8, 2021 at 9:53

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