2016
DOI: 10.1137/15m1051257
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Proximal Point Algorithms on Hadamard Manifolds: Linear Convergence and Finite Termination

Abstract: Abstract. In the present paper, we consider inexact proximal point algorithms for finding singular points of multivalued vector fields on Hadamard manifolds. The rate of convergence is shown to be linear under the mild assumption of metric subregularity. Furthermore, if the sequence of parameters associated with the iterative scheme converges to 0, then the convergence rate is superlinear. At the same time, the finite termination of the inexact proximal point algorithm is also provided under a weak sharp minim… Show more

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Cited by 34 publications
(13 citation statements)
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“…Note that for θ k = 0 in (7), Algorithm 1 merges into algorithm introduced in [5]. The error criterion (7) was introduced in the celebrated paper [38] to analyze an inexact version of the proximal point method to find zeroes of maximal monotone operators in linear context; see [46,47] for a generalization to Riemannian setting.…”
Section: Inexact Proximal Point Methods For Variational Inequalitiesmentioning
confidence: 99%
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“…Note that for θ k = 0 in (7), Algorithm 1 merges into algorithm introduced in [5]. The error criterion (7) was introduced in the celebrated paper [38] to analyze an inexact version of the proximal point method to find zeroes of maximal monotone operators in linear context; see [46,47] for a generalization to Riemannian setting.…”
Section: Inexact Proximal Point Methods For Variational Inequalitiesmentioning
confidence: 99%
“…Remark. In [46,47] is presented an inexact version of the proximal point method for to find singularity of a vector field on Hadamard manifolds. These papers differs from the present paper in two ways, namely, [46,47] use only absolute summable error criteria and the enlargement X ǫ of X was not considered.…”
Section: Convergence Analysismentioning
confidence: 99%
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“…Thus, S m ++ is a Hadamard manifold manifold of nonpositive curvature everywhere [22,36]. The dimension of S m ++ is equal to m(m + 1)/2 [21, Proposition 2.1].…”
Section: Example 41mentioning
confidence: 99%