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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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The resolvent star algebra is dense in C_0(R)

Statement

Fix z∉R and let r(x)=(x−z)−1. Let A be the ∗-algebra generated by r in C0(R), that is, the uniform closure of the linear span of the products rkr‾ j with k+j≥1. Then A=C0(R). Under Countable Choice (The Axiom of Countable Choice (ACω)), consequently, if self-adjoint An,A satisfy RAn(z)→RA(z) strongly (respectively in norm) at one nonreal z, then the same convergence holds at every nonreal z′, in particular at z‾.

Facts & Assumptions

[A2]

Let X be a compact Hausdorff space and let B⊆C(X,C) be a self-adjoint complex function algebra containing the constants, separating points, with no common zero. Then B is uniformly dense in C(X,C) (Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense).

[A3]

Under Countable Choice, with Dz:=RA(z)−RA′(z) for self-adjoint A,A′ and nonreal z, the resolvent identity gives Dz′=[I+(z′−z)RA(z)]−1Dz[I−(z′−z)RA′(z′)] for nonreal z′, the first factor being I+(z−z′)RA(z′) and the second I+(z−z′)RA′(z′); these affine transforms of resolvents are bounded with norms at most 1+∣z′−z∣/∣Im⁡z′∣ (Resolvent and spectrum of an unbounded operator, Resolvent of a self-adjoint operator: nonreal resolvents and the estimate, Norm and strong resolvent convergence, The Axiom of Countable Choice (ACω)).

Proof

technique · direct

Given: z∉R, r(x)=(x−z)−1 and the ∗-algebra A generated by r.

1.1A1

Extend r to R∗ by r(∞)=0; then r∈C0(R)⊆C(R∗) and r separates the points of R∗: it is injective on R, and r(x)≠0=r(∞).

1.2A3

Parameter independence: for nonreal z′ the difference Dz′=RA(z′)−RA′(z′) factors as in [A3] through Dz, with both outer factors of norm at most 1+∣z′−z∣/∣Im⁡z′∣, since they are I+(z−z′)RA(z′) and I+(z−z′)RA′(z′) and ∥RA(z′)∥,∥RA′(z′)∥≤1/∣Im⁡z′∣; hence Dz′→0 in norm whenever Dz→0 in norm, and Dz′x→0 for every x whenever Dzy→0 for every y, because bounded operators preserve both modes of convergence.

2.1A1A2step 1.1

The algebra B:={f+c1:f∈A,c∈C}⊆C(R∗) contains the constants and r,r‾, is self-adjoint, separates points by step 1.1, and has no common zero because of the constant function 1; hence B is uniformly dense in C(R∗) by [A2].

3.1A1step 2.1

Therefore A=C0(R): given f∈C0(R) and ε>0, density of B gives f=c1+g with g∈A and ∥f−c1−g∥∞<ε; evaluating at ∞, where f(∞)=0, gives ∣c∣<ε, so ∥f−g∥∞<2ε and g∈A.

4.1

The density claim is step 3.1 and the consequence is step 1.2, which also covers z′=z‾. ∎

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