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Born truncation error for a small-norm volume integral operator

Statement

Assume Countable Choice and the data of The three-dimensional volume integral scattering model with $0 < ∣ D ∣ < \infty$ . If $ρ : = ∥ K_{k} ∥_{L^{2} \to L^{2}} < 1$ , then for every $f \in L^{2} (D; C)$ the equation $u = f + K_{k} u$ has a unique solution. For each integer $m \geq 0$ , define the Born partial sum $u_{m} : = \sum_{j = 0}^{m} K_{k}^{j} f$ . Then $∥ u - u_{m} ∥_{2} \leq \frac{ρ^{m + 1}}{1 - ρ} ∥ f ∥_{2} .$ No compactness, analytic continuation, or physical uniqueness hypothesis is used in this small-norm branch.

Facts & Assumptions

Given: Countable Choice, $f \in L^{2} (D; C)$ , and $K_{k}$ with operator norm $ρ < 1$ .

[F1]

$K_{k}$ is a bounded linear operator on complex $L^{2} (D)$ (The volume integral operator is well defined and bounded on L2).

[F2]

Complex $L^{2} (D)$ is complete under Countable Choice (Complex Lp completeness and almost-everywhere subsequences, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

For every integer $j \geq 0$ , $∥ K_{k}^{j} f ∥_{2} \leq ρ^{j} ∥ f ∥_{2}$ , by repeated application of the operator norm inequality in [F1].

Proof

Given: The data of the Statement, with $ρ = ∥ K_{k} ∥ < 1$ .

1.1givenF2F3algebra

Define $u_{N} : = \sum_{j = 0}^{N} K_{k}^{j} f$ . For $M > N$ , the triangle inequality and [F3] give $∥ u_{M} - u_{N} ∥_{2} \leq ∥ f ∥_{2} \sum_{j = N + 1}^{M} ρ^{j} \leq ∥ f ∥_{2} ρ^{N + 1} / (1 - ρ)$ . This tends to zero as $N \to \infty$ , so $(u_{N})$ is Cauchy and converges to some $u \in L^{2} (D; C)$ by [F2].

2.1F1F3step 1.1algebra

Finite telescoping gives $(I - K_{k}) u_{N} = f - K_{k}^{N + 1} f$ . By [F3], $∥ K_{k}^{N + 1} f ∥_{2} \leq ρ^{N + 1} ∥ f ∥_{2} \to 0$ ; boundedness in [F1] lets $K_{k} u_{N} \to K_{k} u$ . Taking limits yields $(I - K_{k}) u = f$ . If $(I - K_{k}) v = 0$ , then $∥ v ∥_{2} = ∥ K_{k} v ∥_{2} \leq ρ ∥ v ∥_{2}$ , and $ρ < 1$ forces $v = 0$ , so this solution is unique.

3.1F2F3step 1.1algebra∎

With $m$ fixed, [F3] and the geometric-series identity give $∥ u - u_{m} ∥_{2} \leq \sum_{j = m + 1}^{\infty} ∥ K_{k}^{j} f ∥_{2} \leq ∥ f ∥_{2} \sum_{j = m + 1}^{\infty} ρ^{j} = ρ^{m + 1} ∥ f ∥_{2} / (1 - ρ)$ . The argument includes $ρ = 0$ and $f = 0$ .

Depends on

Used by

A small-norm ball scatterer has a certified Born tail Example

Dependency tree · two levels

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Sources

Otmar Scherzer and Leonidas Mindrinos, Inverse Problems: Lecture Notes, Sommersemester 2015, §5.2.6 (standard reference, not scraped)
Habib Ammari, Brian Fitzpatrick, and Sanghyeon Yu, Mathematical and Computational Methods in Photonics: Tutorial Notes, §1.3 (standard reference, not scraped)