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Collective compactness gives eventual Fredholm invertibility and solution error bounds

Statement

Assume the Axiom of Choice. Let $X \neq {0}$ be a Banach space (Banach space), let $K : X \to X$ be compact, and let $(K_{n})_{n \geq 1}$ be bounded linear operators such that $\overline{⋃_{n \geq 1} K_{n} (B_{X})}$ is compact for the closed unit ball $B_{X} : = {x \in X : ∥ x ∥ \leq 1}$ and $K_{n} x \to K x$ in norm for every $x \in X$ . Suppose $I - K$ is injective. Then there are $n_{0} \geq 1$ and $C < \infty$ such that $I - K_{n}$ is boundedly invertible for every $n \geq n_{0}$ and $\sup_{n \geq n_{0}} ∥ (I - K_{n})^{- 1} ∥ \leq C .$ For any $f, f_{n} \in X$ , if $u$ and $u_{n}$ solve $(I - K) u = f$ and $(I - K_{n}) u_{n} = f_{n}$ , then for $n \geq n_{0}$ , $∥ u_{n} - u ∥ \leq C (∥ (K_{n} - K) u ∥ + ∥ f_{n} - f ∥) .$ The theorem does not assert that $∥ K_{n} - K ∥ \to 0$ .

Facts & Assumptions

Given: AC, the Banach space $X \neq {0}$ , compact $K$ , bounded linear $K_{n}$ , compact closure of the union of their unit-ball images, pointwise norm convergence $K_{n} x \to K x$ , and injectivity of $I - K$ .

[A1]

The Axiom of Choice and AC supplies the countable and dependent choices used in Banach integration: AC supplies Countable Choice and Dependent Choice, including the prescribed countable selections used below.

[F1]

Fredholm alternative for identity minus compact: under AC, for a compact operator on a Banach space, injectivity of $I - T$ implies surjectivity and a bounded inverse.

[F2]

For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice: under Countable Choice and Dependent Choice, a compact metric space is sequentially compact.

[F3]

A compact subset of a metric space is closed and bounded: a compact subset of a normed space is bounded and closed.

[F4]

Compact linear operator: compactness means that every bounded set has image with compact closure; in particular each $K_{n}$ is compact when its unit-ball image has compact closure.

[F5]

The operator norm as the least bound and as the unit-sphere or unit-ball supremum: for a bounded linear operator, its norm is the supremum of the norms of its image on the closed unit ball.

Proof

Given: The data in the Statement.

1.1givenF1

Since $K$ is compact and $I - K$ is injective, [F1] gives a bounded inverse $(I - K)^{- 1}$ .

1.2givenA1F2choose

We claim there are $n_{0}$ and $c > 0$ such that $∥ (I - K_{n}) x ∥ \geq c ∥ x ∥$ for all $x \in X$ and $n \geq n_{0}$ . Otherwise, for each integer $m \geq 1$ , failure of the bound $c = 1 / m$ for all $n \geq m$ gives an index $n_{m} \geq m$ and a unit vector $x_{m}$ with $∥ (I - K_{n_{m}}) x_{m} ∥ < 1 / m$ ; [A1] permits choosing such a sequence. Then $x_{m} = (I - K_{n_{m}}) x_{m} + K_{n_{m}} x_{m} .$ The sequence $(K_{n_{m}} x_{m})$ lies in the compact closure of the union in the Statement, so [F2] gives a convergent subsequence, say $K_{n_{m_{j}}} x_{m_{j}} \to y$ . The displayed identity and $∥ (I - K_{n_{m_{j}}}) x_{m_{j}} ∥ \to 0$ imply $x_{m_{j}} \to y$ , hence $∥ y ∥ = 1$ .

2.1givenF3F5step 1.2algebra

By [F3], choose $M < \infty$ bounding the norms of all vectors in the compact union closure. Every $K_{n}$ maps the unit ball into this closure, so [F5] gives $∥ K_{n} ∥ \leq M$ for all $n$ . Along the subsequence in step 1.2, $∥ K_{n_{m_{j}}} x_{m_{j}} - K_{n_{m_{j}}} y ∥ \leq M ∥ x_{m_{j}} - y ∥ \to 0,$ while pointwise convergence for the fixed vector $y$ gives $K_{n_{m_{j}}} y \to K y$ . Therefore the same sequence converges to both $y$ and $K y$ , so $K y = y$ . This contradicts injectivity of $I - K$ because $∥ y ∥ = 1$ .

3.1givenF1F4step 1.2step 2.1

The contradiction in step 2.1 establishes the eventual lower bound claimed in step 1.2, so each $I - K_{n}$ is injective for $n \geq n_{0}$ . Its unit-ball image is contained in the compact closure from the Statement, so $K_{n}$ is compact by [F4]; applying [F1] to each $K_{n}$ makes $I - K_{n}$ surjective with bounded inverse. For $z = (I - K_{n}) x$ , the same lower bound gives $∥ (I - K_{n})^{- 1} z ∥ = ∥ x ∥ \leq c^{- 1} ∥ z ∥$ , hence the inverse norms are uniformly at most $C : = c^{- 1}$ .

4.1givenstep 3.1algebra∎

Subtracting the equations gives $(I - K_{n}) (u_{n} - u) = f_{n} - f + (K_{n} - K) u .$ Applying the inverse from step 3.1 and the triangle inequality yields the stated error estimate.

Depends on

Used by

Dependency tree · two levels

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Sources

Andreas Neubauer and Ronny Ramlau, Lecture Notes: Integral Equations and Boundary Value Problems, §3.3 (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)