2007
DOI: 10.1017/s0305004107000199
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A Denjoy–Wolff theorem for Hilbert metric nonexpansive maps on polyhedral domains

Abstract: For a polyhedral domain ⊂ R n , and a Hilbert metric nonexpansive map T : → which does not have a fixed point in , we prove that the omega limit set ω(x; T ) of any point x ∈ is contained in a convex subset of the boundary ∂ . We also identify a class of order-preserving homogeneous of degree one maps on the interior of the standard cone R n + which demonstrate that there are Hilbert metric nonexpansive maps on an open simplex with omega limit sets that can contain any convex subset of the boundary.

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Cited by 14 publications
(18 citation statements)
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References 14 publications
(23 reference statements)
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“…Note that if Σ is finite dimensional and its closure (in the usual topology) is strictly convex, then each convex subset of ∂Σ reduces to a single point. Conjecture 1.1 was shown to hold in case Σ has a strictly convex closure by Beardon [7], and for polytopes by Lins [34]. Further supporting evidence was obtained in [2,28,35,41].…”
Section: Introductionmentioning
confidence: 78%
“…Note that if Σ is finite dimensional and its closure (in the usual topology) is strictly convex, then each convex subset of ∂Σ reduces to a single point. Conjecture 1.1 was shown to hold in case Σ has a strictly convex closure by Beardon [7], and for polytopes by Lins [34]. Further supporting evidence was obtained in [2,28,35,41].…”
Section: Introductionmentioning
confidence: 78%
“…By using Theorem 2.2 in [11] directly or by using Theorem 2.1 with K 1 = K 2 = K p , we see that co(ω(E; T )) ⊂ ∂C E . It follows that there exists a nonempty, proper subset J ⊂ {1, 2, .…”
Section: T (X) =mentioning
confidence: 97%
“…In almost all applications in analysis, Σ is not strictly convex, Theorem 1.1 is not applicable and Theorem 1.2 is somewhat weak. Motivated by these deficiencies, Lins [11] has proved the following. [11].)…”
Section: Introductionmentioning
confidence: 99%
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“…Now, reparametrize the earthquake ray {E t µ X} t≥0 as {E t µ X} t≥0 by setting E t µ X = E (d/d) exp(2t) µ X. It follows from (14) and (15) that for any > 0 there is a T > 0 depending on , Γ, X and µ such that…”
Section: Now (7) Can Be Rewritten Asmentioning
confidence: 99%