2012
DOI: 10.1016/j.aim.2011.09.005
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Weighted extremal domains and best rational approximation

Abstract: Let f be holomorphically continuable over the complex plane except for finitely many branch points contained in the unit disk. We prove that best rational approximants to f of degree n, in the L 2 -sense on the unit circle, have poles that asymptotically distribute according to the equilibrium measure on the compact set outside of which f is single-valued and which has minimal Green capacity in the disk among all such sets. This provides us with n-th root asymptotics of the approximation error. By conformal ma… Show more

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Cited by 30 publications
(39 citation statements)
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References 43 publications
(98 reference statements)
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“…That the infimum is indeed attained in the right hand side of (10) follows from (8) and the fact that RAB(n) has a solution. Define A f , the Hankel operator with symbol f , by…”
Section: Duality In Meromorphic Approximationmentioning
confidence: 75%
See 3 more Smart Citations
“…That the infimum is indeed attained in the right hand side of (10) follows from (8) and the fact that RAB(n) has a solution. Define A f , the Hankel operator with symbol f , by…”
Section: Duality In Meromorphic Approximationmentioning
confidence: 75%
“…It may seem artificial to favor the hyperbolic geodesic segment linking the branchpoints among all possible cuts. However, this one turns out to attract almost all poles of best rational approximants (see [8] for this and generalizations to finitely many branchpoints) and also of best meromorphic approximants (see [6,Theorem 10.1] and Corollary 3 below), which makes it in some sense the natural singular set of the function. We need additional facts from AAK theory that shed light on singular vectors of Hankel operators with continuous symbol.…”
Section: Lemmamentioning
confidence: 97%
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“…We note that in the case treated in [1] an additional assumption about the mutual arrangement of E and A f was imposed. In the general case, when no additional assumptions are made, the answer in the problem can be negative even for m = 1: in place of (1) we may have the inequality…”
mentioning
confidence: 99%