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arXiv · 1711.03715

Bounded Point Evaluations For Certain Polynomial And Rational Modules

Abstract

Let $K$ be a compact subset of the complex plane $\mathbb C.$ Let $P(K)$ and $R(K)$ be the closures in $C(K)$ of analytic polynomials and rational functions with poles off $K,$ respectively. Let $A(K) \subset C(K)$ be the algebra of functions that are analytic in the interior of $K$. For $1\le t <\infty,$ let $P^t(1, ϕ_1,...,ϕ_N,K)$ be the closure of $P(K)+P(K)ϕ_1+...+P(K)ϕ_N$ in $L^t(dA|_K),$ where $dA|_K$ is the area measure restricted to $K$ and $ϕ_1,...,ϕ_N\in L^t(dA|_K).$ Let $HP(ϕ_1,...,ϕ_N,K)$ be the closure of $P(K)ϕ_1+...+P(K)ϕ_N +R(K)$ in $C(K),$ where $ϕ_1,...,ϕ_N\in C(K).$ In this paper, we prove if $R(K)\ne C(K),$ then there exists an analytic bounded point evaluation for both $P^t(1, ϕ_1,...,ϕ_N,K)$ and $HP(ϕ_1,...,ϕ_N,K)$ for certain smooth functions $ϕ_1,...,ϕ_N,$ in particular, for $\bar z,\bar z^2,...,\bar z^N.$ We show that $A(K)\subset HP(\bar z,\bar z^2,...,\bar z^N,K)$ if and only if $R(K) = A(K).$ In particular, $C(K) \ne HP(\bar z,\bar z^2,...,\bar z^N,K)$ unless $R(K) = C(K).$ We also give an example of $K$ showing the results are not valid if we replace $\bar z^n$ by certain $ϕ_n,$ that is, there exist $K$ and a function $ϕ\in A(K)$ such that $R(K) \ne A(K),$ but $A(K) = HP (ϕ,K).$

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BibTeXRIS

Liming Yang. 2017-11-10. Bounded Point Evaluations For Certain Polynomial And Rational Modules. https://arxiv.org/abs/1711.03715

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