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✓ 11 results · all verified · 10 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Nystrom product integration and fredholm error

1 · Prerequisites

2 · Summary

This page sets up finite-rank Nyström operators and proves their equivalence with the sampled grid equations. It treats two stability routes separately: an operator-norm perturbation estimate and an AC-dependent collectively compact Fredholm argument. For composite midpoint sampling, it proves pointwise consistency, the solution error bound under smoothness of the actual integrand, and the limitation of operator-norm convergence on the full space of continuous functions. A second section develops piecewise-linear product integration for the kernel $∣ x - y ∣^{- 1 / 2}$ , including the improper weights, their exact closed forms, the kernel's improper L1 translation modulus, collective compactness, consistency, and a Fredholm error bound. The page keeps the choice assumptions explicit and makes no operator-norm convergence claim for point sampling on $C ([0, 1])$ .

Affine normalization to a general interval

The item formulas below use $[0, 1]$ as a normalized coordinate. They give the approved results on every nondegenerate compact interval $I = [α, β]$ as follows. Put $L : = β - α > 0$ and define the isometry $U : C (I) \to C ([0, 1]), (U u) (t) : = u (α + L t) .$ Its inverse is $(U^{- 1} v) (x) = v ((x - α) / L)$ . Here $C (I)$ and $C ([0, 1])$ denote real continuous functions. The intervals $I$ and $[0, 1]$ are nonempty compact metric spaces by Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Pointwise addition and real scalar multiplication preserve continuity, and the absolute-value triangle inequality, homogeneity and separation give the supremum norm on each space (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Their induced supremum metrics are complete by $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ , so both spaces are Banach (Banach space). The affine coordinate map is a bijection, and pointwise evaluation makes $U$ linear; for every $u \in C (I)$ , $∥ U u ∥_{\infty} = \sup_{t \in [0, 1]} ∣ u (α + L t) ∣ = \sup_{x \in I} ∣ u (x) ∣ = ∥ u ∥_{\infty},$ so $U$ is an isometry for these norms. For a continuous kernel $k_{I} : I^{2} \to R$ , set $\tilde{k} (t, s) : = L k_{I} (α + L t, α + L s) .$ For the affine integral substitution used here, a tagged partition of $[0, 1]$ maps to the partition of $I$ obtained by $s \mapsto α + L s$ ; its mesh is multiplied by $L$ , and each corresponding Riemann-sum term is multiplied by $L$ . Taking the defining limits therefore gives $\int_{α}^{β} g (y) d y = L \int_{0}^{1} g (α + L s) d s$ for continuous $g$ . For the half-order improper integrals the same identity holds first on each truncated interval, and then in the limit as the cutoff tends to zero. Thus, for every $u \in C (I)$ , $U K_{I} U^{- 1} = \tilde{K}, (K_{I} u) (x) : = \int_{α}^{β} k_{I} (x, y) u (y) d y,$ where $\tilde{K}$ is the unit-interval integral operator with kernel $\tilde{k}$ . More generally, if a finite quadrature on $I$ has nodes $z_{j} \in I$ and weights $ω_{j}$ , write $s_{j} = (z_{j} - α) / L$ and ${\tilde{ω}}_{j} = ω_{j} / L$ . Then the conjugated Nyström operator is exactly $(U K_{Q, I} U^{- 1} v) (t) = \sum_{j} {\tilde{ω}}_{j} \tilde{k} (t, s_{j}) v (s_{j}), (K_{Q, I} u) (x) = \sum_{j} ω_{j} k_{I} (x, z_{j}) u (z_{j}) .$ Thus the transformed finite grid matrix is the same matrix, the right-hand sides and solution errors have the same supremum norms, and operator norms, inverse norms, compactness, collective compactness, injectivity, and invertibility are preserved by $U$ : the operator-norm equality follows since $U$ maps the unit ball onto the unit ball isometrically, and compact closures are preserved by this isometric bijection. The operator-norm perturbation route and the collectively compact route therefore apply on $C (I)$ with exactly their stated hypotheses after this normalization. For the composite midpoint rule, the physical panel width is $Δ = L / N$ and its normalized width is $1 / N$ ; if $g_{x} (y) = k_{I} (x, y) u (y)$ has the stated uniform second-derivative bound $M_{2} : = \sup_{x, y \in I} ∣ \partial_{y}^{2} g_{x} (y) ∣$ , the midpoint residual from Composite midpoint error bound for a twice continuously differentiable function becomes $∥ K_{N, I} u - K_{I} u ∥_{\infty} \leq \frac{L Δ^{2}}{24} M_{2} .$ Equivalently, the transformed integrand $L g_{x} (α + L \cdot)$ has second derivative bounded by $L^{3} M_{2}$ , and $(1 / N)^{2} = Δ^{2} / L^{2}$ .

For the half-order kernel $k_{I} (x, y) = ∣ x - y ∣^{- 1 / 2} a_{I} (x, y),$ the transformed regular factor is $\tilde{a} (t, s) = L^{1 / 2} a_{I} (α + L t, α + L s)$ , because $d y = L d s$ and $∣ x - y ∣^{- 1 / 2} = L^{- 1 / 2} ∣ t - s ∣^{- 1 / 2}$ . The physical product weights use the hats $ϕ_{N, j}^{I} (y) : = ϕ_{N, j} ((y - α) / L)$ on nodes $x_{j} = α + j Δ$ and satisfy $W_{N, j} (x) : = \int_{α}^{β} ∣ x - y ∣^{- 1 / 2} ϕ_{N, j}^{I} (y) d y = L^{1 / 2} w_{N, j} ⁣ (\frac{x - α}{L}) .$ The improper singular functional has $∥ S_{x}^{I} ∥ = S_{x}^{I} (1) = 2 \sqrt{x - α} + 2 \sqrt{β - x} \leq 2 \sqrt{2 L} .$ Consequently the product weights are nonnegative and their exact sum and stability bound are $\sum_{j} W_{N, j} (x) = 2 \sqrt{x - α} + 2 \sqrt{β - x} \leq 2 \sqrt{2 L};$ the closed forms in Closed forms for the half-order piecewise-linear product weights apply in normalized coordinates and are multiplied by $L^{1 / 2}$ . The half-order kernel translation modulus scales as $τ_{I} (d) = L^{1 / 2} τ (d / L)$ , by the same change of variables, so it still tends to zero as $d ↓ 0$ . Suppose each $g_{x} (y) : = a_{I} (x, y) u (y)$ extends to a $C^{2}$ function on a neighborhood of $I$ and put $M_{2, I} : = \sup_{x, y \in I} ∣ \partial_{y}^{2} g_{x} (y) ∣ < \infty .$ In normalized coordinates the regular factor is ${\tilde{g}}_{t} (s) = L^{1 / 2} g_{α + L t} (α + L s)$ , so $\sup_{t, s} ∣ \partial_{s}^{2} {\tilde{g}}_{t} (s) ∣ = L^{5 / 2} M_{2, I}$ and the normalized panel width is $1 / N = Δ / L$ . Applying the unit-interval interpolation bound from Pointwise linear interpolation error from a second-derivative bound and the collective compact solution estimate gives the physical product-interpolation residual and Fredholm solution bound $∥ K_{N, I} u - K_{I} u ∥_{\infty} \leq \frac{\sqrt{2}}{4} L^{1 / 2} M_{2, I} Δ^{2},$ $∥ u_{N, I} - u ∥_{\infty} \leq C (∥ f_{N, I} - f ∥_{\infty} + \frac{\sqrt{2}}{4} L^{1 / 2} M_{2, I} Δ^{2}),$ with the same uniform inverse bound $C$ as for the conjugated unit-interval problem. Setting $L = 1$ recovers every formula in the assigned items; the unit-interval examples are the canonical normalized cases.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Nyström quadrature operators and their reconstructed equations

Definition

Let $X : = C ([0, 1], R)$ be the real continuous functions on $[0, 1]$ (The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space) with the supremum norm $∥ u ∥_{\infty} : = \sup_{0 \leq x \leq 1} ∣ u (x) ∣$ . The interval $[0, 1]$ is a nonempty compact metric space by Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Pointwise addition and real scalar multiplication preserve continuity. The supremum is finite by $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ ; the pointwise absolute-value inequalities give $∥ u + v ∥_{\infty} \leq ∥ u ∥_{\infty} + ∥ v ∥_{\infty}$ and $∥ c u ∥_{\infty} = ∣ c ∣ ∥ u ∥_{\infty}$ , and $∥ u ∥_{\infty} = 0$ holds exactly when $u = 0$ . Thus this is a real norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Its induced metric is the complete supremum metric of $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ , so $X$ is Banach (Banach space). Fix a continuous real kernel $k : [0, 1]^{2} \to R$ . The integral $\int_{0}^{1}$ below is the ordinary Riemann integral, equivalently the Riemann–Stieltjes integral with integrator $y \mapsto y$ (Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral). For $u \in X$ , define $(K u) (x) : = \int_{0}^{1} k (x, y) u (y) d y, 0 \leq x \leq 1.$ The integrand is continuous in $y$ , so this integral exists by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion.

A finite quadrature rule on $[0, 1]$ consists of an integer $m \geq 1$ , nodes $y_{1}, \dots, y_{m} \in [0, 1]$ , and real weights $w_{1}, \dots, w_{m}$ . It acts on a function $g$ defined at the nodes by $Q (g) : = \sum_{j = 1}^{m} w_{j} g (y_{j}) .$ Its Nyström operator is the finite-sum rule $(K_{Q} u) (x) : = Q (k (x, \cdot) u (\cdot)) = \sum_{j = 1}^{m} w_{j} k (x, y_{j}) u (y_{j}) .$ In particular, for the composite midpoint rule with $N \geq 1$ panels, $h : = 1 / N$ , $y_{j} : = (j + \frac{1}{2}) h$ for $0 \leq j < N$ , and $w_{j} = h$ , the operator is denoted $K_{N}$ . The nodes and weights are those of Composite midpoint rule on a uniform grid.

For a right-hand side $f \in X$ , the Nyström equation is $u_{Q} - K_{Q} u_{Q} = f .$ This definition asserts neither consistency nor convergence of $K_{Q}$ to $K$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Finite-node Nyström equations are equivalent to their reconstructed equations

Statement

Let $X : = C ([0, 1], R)$ (The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space), let $m \geq 1$ , let $y_{1}, \dots, y_{m} \in [0, 1]$ , and let $ψ_{1}, \dots, ψ_{m} \in X$ . Define the finite-rank linear map $(F u) (x) : = \sum_{j = 1}^{m} ψ_{j} (x) u (y_{j})$ and, for $f \in X$ , define the square matrix and vector $M_{i j} : = δ_{i j} - ψ_{j} (y_{i}), b_{i} : = f (y_{i}), 1 \leq i, j \leq m .$ Then the solutions of $(I - F) u = f$ are in bijection with the solutions $c \in R^{m}$ of $M c = b$ , via $u (x) = f (x) + \sum_{j = 1}^{m} ψ_{j} (x) c_{j}, c_{i} = u (y_{i}) .$ The nullspaces of $I - F$ and $M$ are linearly isomorphic by $c \mapsto \sum_{j} ψ_{j} c_{j}$ . In particular, $I - F$ is bijective if and only if $M$ is nonsingular.

Facts & Assumptions

Given: The space $X$ , nodes $y_{i}$ , continuous coefficient functions $ψ_{i}$ , and $f \in X$ as in the Statement.

[F1]

The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space: members of $X$ are continuous real-valued functions on $[0, 1]$ , so finite linear combinations of the $ψ_{j}$ and $f$ lie in $X$ .

[F2]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : a linear map on the finite-dimensional space $R^{m}$ is injective if and only if it is surjective, hence a square matrix is nonsingular exactly when its kernel is zero.

Proof

Given: The data in the Statement, the operator $F$ , the matrix $M$ , and the vector $b$ .

1.1givenF1algebra

If $u$ solves $(I - F) u = f$ , put $c_{i} : = u (y_{i})$ . Evaluating $u (x) - \sum_{j} ψ_{j} (x) u (y_{j}) = f (x)$ at $x = y_{i}$ gives $c_{i} - \sum_{j} ψ_{j} (y_{i}) c_{j} = f (y_{i})$ for every $i$ , which is exactly $M c = b$ .

2.1givenF1step 1.1algebraconstruct

Conversely, if $M c = b$ , define $u (x) : = f (x) + \sum_{j} ψ_{j} (x) c_{j}$ . By [F1] this lies in $X$ . The matrix equation gives $u (y_{i}) = f (y_{i}) + \sum_{j} ψ_{j} (y_{i}) c_{j} = c_{i}$ for each $i$ , so $u = f + F u$ and it solves $(I - F) u = f$ . These constructions are inverse because each reconstructed function has the prescribed grid values.

3.1givenF2step 2.1algebra∎

If $M c = 0$ and $c \neq 0$ , set $v (x) : = \sum_{j} ψ_{j} (x) c_{j}$ . The homogeneous matrix equations imply $v (y_{i}) = c_{i}$ ; hence $v \neq 0$ and $F v = v$ . Thus a nonzero matrix kernel gives a nonzero kernel of $I - F$ . Conversely, if $(I - F) v = 0$ and $v \neq 0$ , then $c_{i} : = v (y_{i})$ satisfies $M c = 0$ ; if $c = 0$ , the formula $v = F v = \sum_{j} ψ_{j} v (y_{j})$ would give $v = 0$ , a contradiction. Therefore the two kernels are simultaneously trivial. By [F2] this is equivalent to $M$ being nonsingular; [step 2.1] then gives bijective solvability of the reconstructed equation exactly when the matrix is nonsingular.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A continuous kernel defines a compact integral operator on C of an interval

Statement

Assume Countable Choice and Dependent Choice. Let $X : = C ([0, 1], R)$ with $∥ u ∥_{\infty} : = \sup_{x \in [0, 1]} ∣ u (x) ∣$ (The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space, $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ ) and let $k : [0, 1]^{2} \to R$ be continuous. The map $(K u) (x) : = \int_{0}^{1} k (x, y) u (y) d y$ uses the ordinary Riemann integral of Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral and defines a bounded linear operator $K : X \to X$ , and it is compact (Compact linear operator).

Facts & Assumptions

Given: Countable Choice, Dependent Choice, $X = C ([0, 1], R)$ , and a continuous kernel $k$ on the compact square $[0, 1]^{2}$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) and The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain: these are the two explicit choice hypotheses in the real Arzelà–Ascoli theorem used below.

[F1]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: every continuous real-valued function on a compact interval is Riemann integrable.

[F2]

Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ : finite sums and scalar multiples of Riemann-integrable functions are integrable and their integrals are the corresponding finite sums and scalar multiples.

[F3]

Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error: if two integrable functions differ everywhere by at most $η$ , then their integrals on an interval of length $L$ differ by at most $L η$ .

[F4]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous function on a compact metric space is uniformly continuous.

[F5]

Arzelà--Ascoli for real $C (K)$ under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded: under Countable Choice and Dependent Choice, an equicontinuous pointwise bounded family in $C (K, R)$ on a compact metric space has compact closure in the supremum metric.

[F6]

$C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ : every continuous real-valued function on a compact metric space is bounded, so $M$ is finite.

[F7]

Compact linear operator: a bounded linear operator is compact when the image of its closed unit ball has compact closure.

[F8]

The interval $[0, 1]$ and square $[0, 1]^{2}$ with their Euclidean metrics are nonempty compact metric spaces (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F9]

On the nonempty compact interval $[0, 1]$ (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), let $X = C ([0, 1], R)$ with its supremum norm. Pointwise addition and real scalar multiplication preserve continuity. The supremum is finite by $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ ; the pointwise absolute-value inequalities give $∥ u + v ∥_{\infty} \leq ∥ u ∥_{\infty} + ∥ v ∥_{\infty}$ and $∥ c u ∥_{\infty} = ∣ c ∣ ∥ u ∥_{\infty}$ , and $∥ u ∥_{\infty} = 0$ holds exactly when $u = 0$ . Thus this is a real norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms). Its induced metric is the complete supremum metric of $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ , so $X$ is Banach (Banach space).

[F10]

A linear map between normed spaces is bounded when its values satisfy a uniform norm bound (A bounded linear operator between normed spaces).

Proof

Given: The data in the Statement, $M : = \sup_{(x, y) \in [0, 1]^{2}} ∣ k (x, y) ∣$ , and the modulus $ω (δ) : = \sup {∣ k (x, y) - k (x^{'}, y) ∣ : ∣ x - x^{'} ∣ \leq δ, y \in [0, 1]}$ .

1.1givenF1F2F3F6F8F10

By [F8], the domain of $k$ is a compact metric space, so [F6] makes $M$ finite. For each $u \in X$ and $x \in [0, 1]$ , the function $y \mapsto k (x, y) u (y)$ is continuous, so its Riemann integral exists by [F1]. Linearity [F2] makes $K$ linear. Since $∣ k (x, y) u (y) ∣ \leq M ∥ u ∥_{\infty}$ , [F3] applied against the zero function gives $∣ K u (x) ∣ \leq M ∥ u ∥_{\infty}$ , so [F10] makes $K$ bounded.

2.1givenF3F4F8step 1.1

For $x, x^{'} \in [0, 1]$ , [F3] applied to the two continuous integrands gives $∣ K u (x) - K u (x^{'}) ∣ \leq \sup_{y \in [0, 1]} ∣ k (x, y) - k (x^{'}, y) ∣ ∥ u ∥_{\infty} \leq ω (∣ x - x^{'} ∣) ∥ u ∥_{\infty} .$ By [F4] on the compact square from [F8], $ω (δ) \to 0$ as $δ ↓ 0$ , so $K u$ is continuous and $K$ maps $X$ into $X$ .

3.1step 1.1step 2.1

For every $u$ in the closed unit ball, step 1.1 gives $∥ K u ∥_{\infty} \leq M$ , while step 2.1 gives the common equicontinuity modulus $ω$ . Thus the image of the unit ball is pointwise bounded and equicontinuous.

4.1A1F5F7F8F9step 3.1∎

By [A1] and [F5], applied on the compact interval from [F8], the image of the unit ball has compact closure in the supremum metric. [F9] supplies the Banach norm structure on $X$ with this same metric, so [F7] makes $K$ compact.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

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.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The half-order diagonal singularity defines a bounded improper integral

Statement

For $x \in [0, 1]$ , $g \in C ([0, 1], R)$ , and $0 < ε < 1$ , let $S_{x, ε} (g) : = \int_{[0, 1] ∖ (x - ε, x + ε)} ∣ x - y ∣^{- 1 / 2} g (y) d y,$ where the integral is the sum of the ordinary Riemann integrals on the at most two remaining closed intervals, and an empty interval contributes zero. The limit $S_{x} (g) : = \lim_{ε ↓ 0} S_{x, ε} (g)$ exists. The map $S_{x} : C ([0, 1], R) \to R$ is linear and positive, and its operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum) is $∥ S_{x} ∥ = S_{x} (1) = 2 \sqrt{x} + 2 \sqrt{1 - x} \leq 2 \sqrt{2} .$

Facts & Assumptions

Given: $x \in [0, 1]$ , $g \in C ([0, 1], R)$ , and the truncated Riemann integrals $S_{x, ε} (g)$ from the Statement.

[F1]

Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral: $\int_{a}^{b}$ denotes the ordinary Riemann integral, realized as the Riemann–Stieltjes integral with the identity integrator.

[F2]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous real-valued function on each remaining closed interval has an ordinary Riemann integral.

[F3]

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ : pointwise inequalities between integrable functions on an interval pass to their integrals.

[F4]

The two FTC forms for a Riemann–Stieltjes integral with a $C^{1}$ integrator: for the identity integrator, an antiderivative evaluates the integral; since $(2 \sqrt{r})' = r^{- 1 / 2}$ on $r > 0$ , $\int_{δ}^{ε} r^{- 1 / 2} d r = 2 (\sqrt{ε} - \sqrt{δ})$ for $0 < δ < ε$ .

[F5]

The reals are complete: every Cauchy sequence of real numbers converges.

[F6]

Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ : finite Riemann integrals are linear in their integrands.

[F7]

$C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ : every continuous real-valued function on $[0, 1]$ is bounded, so $M : = ∥ g ∥_{\infty}$ is finite.

[F8]

[F9]

The interval $[0, 1]$ with its usual metric is a nonempty compact metric space (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F10]

[F11]

The operator norm of a bounded linear map is the supremum of its values on the closed unit ball (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

Given: The data in the Statement, $M : = ∥ g ∥_{\infty}$ , and $0 < δ < ε < 1$ .

1.1givenF1F2F7F9

By [F9], [F7] applies and $M = ∥ g ∥_{\infty}$ is finite. The integrand $y \mapsto ∣ x - y ∣^{- 1 / 2} g (y)$ is continuous on each closed interval in $[0, 1] ∖ (x - ε, x + ε)$ , because that set stays away from $x$ . Thus the truncated integrals are finite and defined by [F1].

1.2givenF2F3F4F5F8algebra

For $m \geq 2$ , put $a_{m} : = S_{x, 1 / m} (g)$ . By [F8], the difference $a_{m} - a_{ℓ}$ integrates only over the two annuli between the cutoffs $1 / m$ and $1 / ℓ$ . On each annulus, $∣ g (y) ∣ \leq M$ ; by [F2] and [F3], its absolute integral is bounded by $M \int_{\min (1 / m, 1 / ℓ)}^{\max (1 / m, 1 / ℓ)} r^{- 1 / 2} d r$ . Adding at most two sides gives $∣ a_{m} - a_{ℓ} ∣ \leq 4 M ∣ m^{- 1 / 2} - ℓ^{- 1 / 2} ∣$ , so $(a_{m})$ is Cauchy and has a limit $L$ by [F5]. For any $ε > 0$ , choose $m$ with $1 / m < ε$ ; comparing the two cutoffs gives $∣ S_{x, ε} (g) - a_{m} ∣ \leq 4 M \sqrt{ε}$ . Letting $m \to \infty$ along such choices yields $∣ S_{x, ε} (g) - L ∣ \leq 4 M \sqrt{ε}$ , so $S_{x, ε} (g) \to L$ as $ε ↓ 0$ .

2.1givenF3F4F8step 1.2algebra

For every $ε$ , comparison on the left and right of $x$ with the positive function $r^{- 1 / 2}$ gives $∣ S_{x, ε} (g) ∣ \leq M (2 \sqrt{x} + 2 \sqrt{1 - x}) .$ Taking the limit from step 1.2 proves the same bound for $S_{x} (g)$ . For $g = 1$ , the truncated integrals increase to the sum of the two endpoint integrals by [F8], so $S_{x} (1) = 2 \sqrt{x} + 2 \sqrt{1 - x}$ . Since $(2 x - 1)^{2} \geq 0$ , we have $2 \sqrt{x (1 - x)} \leq 1$ , and therefore $S_{x} (1)^{2} = 4 (1 + 2 \sqrt{x (1 - x)}) \leq 8$ . Thus $S_{x} (1) \leq 2 \sqrt{2}$ .

3.1F3F6F9F10F11step 2.1algebra∎

Each truncated functional is linear and nonnegative on nonnegative $g$ ; limits preserve those equalities and inequalities, so $S_{x}$ is linear and positive. By [F10] it is a linear functional on a normed space. Step 2.1 gives $∣ S_{x} (g) ∣ \leq S_{x} (1) ∥ g ∥_{\infty}$ , and the input $g = 1$ attains equality; [F11] therefore gives $∥ S_{x} ∥ = S_{x} (1)$ .

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Translations of the half-order kernel are continuous in the improper L1 norm

Statement

Put $s (t) : = ∣ t ∣^{- 1 / 2}$ for $t \in [- 1, 1] ∖ {0}$ . For $x, x^{'} \in [0, 1]$ , define $T (x, x^{'}) : = \int_{0}^{1} ∣ s (x - y) - s (x^{'} - y) ∣ d y$ as an improper Riemann integral split at the distinct points among $x$ and $x^{'}$ ; an endpoint value at a singular point is omitted. If $x = x^{'}$ , set $T (x, x) = 0$ . These improper integrals exist and are finite. The modulus $τ (d) : = \sup {T (x, x^{'}) : x, x^{'} \in [0, 1], ∣ x - x^{'} ∣ \leq d}$ is finite and satisfies $τ (d) \to 0$ as $d ↓ 0$ .

Facts & Assumptions

Given: $x, x^{'} \in [0, 1]$ , the singular function $s (t) = ∣ t ∣^{- 1 / 2}$ for $t \neq 0$ , and the improper Riemann integral convention in the Statement.

[F1]

Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral defines the ordinary integrals on nonsingular closed intervals, with the identity as integrator. Here an improper integral means the limit of the truncated integrals, by the convention in the Statement; existence of that limit is proved below.

[F2]

The two FTC forms for a Riemann–Stieltjes integral with a $C^{1}$ integrator: for $0 < δ < η$ , $\int_{δ}^{η} r^{- 1 / 2} d r = 2 (\sqrt{η} - \sqrt{δ})$ .

[F3]

The reals are complete: every Cauchy sequence of real numbers has a real limit.

[F4]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous function on a closed interval has an ordinary Riemann integral.

[F5]

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ and If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ : truncated interval integrals add, and pointwise bounds between integrable functions give the corresponding integral bounds.

[F6]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous function on the compact interval $[- 1, 1]$ is uniformly continuous.

[F7]

The Cauchy-sequence reals have the least-upper-bound property: every nonempty bounded set of reals has a finite supremum.

[F8]

The interval $[- 1, 1]$ with its usual metric is a nonempty compact metric space (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

Given: The data in the Statement, $d : = ∣ x - x^{'} ∣$ , and $0 < η < 1$ .

1.1givenF2F3F4F5algebra

If $x = x^{'}$ , then the integrand is zero away from the omitted singular point and $T (x, x^{'}) = 0$ . Suppose $x \neq x^{'}$ . Split $[0, 1]$ at $x$ and $x^{'}$ . On the interior of each resulting interval the integrand is continuous by [F4]. Near a singular endpoint $c \in {x, x^{'}}$ , take a cutoff $ρ < ∣ x - x^{'} ∣ / 3$ . The other singular term is bounded there by $(2 ∣ x - x^{'} ∣ / 3)^{- 1 / 2}$ , while the term singular at $c$ has tail integral at most $2 \sqrt{ρ}$ by [F2] after the lower cutoff tends to zero. Thus each omitted endpoint tail is at most $2 \sqrt{ρ} + 2 ρ (2 ∣ x - x^{'} ∣ / 3)^{- 1 / 2}$ , which tends to zero. Taking cutoffs $ρ_{m} = ∣ x - x^{'} ∣ / (4 (m + 1))$ for integers $m \geq 0$ gives a Cauchy sequence of truncated sums by additivity and comparison [F5]; [F3] gives a finite limit. The same tail bound compares any sufficiently small cutoff with this cofinal sequence, so the improper limit exists. The argument also applies when a singular point is 0 or 1, with only one adjacent interval.

1.2givenF1F2F4algebra

For $t \in [- 1, 1]$ , define $s_{η} (t) : = \min {η^{- 1 / 2}, ∣ t ∣^{- 1 / 2}}$ when $t \neq 0$ and $s_{η} (0) : = η^{- 1 / 2}$ . This is continuous. Direct integration using [F2] gives $\int_{- 1}^{1} ∣ s (t) - s_{η} (t) ∣ d t = 2 \int_{0}^{η} (t^{- 1 / 2} - η^{- 1 / 2}) d t = 2 \sqrt{η} .$ The two translated tails on $y \in [0, 1]$ are each bounded by this full tail.

2.1givenF1F5F6F7F8step 1.1step 1.2algebra∎

For every $y$ outside the finitely many singular points, the triangle inequality gives $∣ s (x - y) - s (x^{'} - y) ∣ \leq ∣ s (x - y) - s_{η} (x - y) ∣ + ∣ s_{η} (x - y) - s_{η} (x^{'} - y) ∣ + ∣ s_{η} (x^{'} - y) - s (x^{'} - y) ∣ .$ Integrating truncated intervals and taking the limits from step 1.1, the first and third terms each contribute at most $2 \sqrt{η}$ by step 1.2. The middle term is at most $ω_{η} (d) : = \sup_{∣ t - t^{'} ∣ \leq d} ∣ s_{η} (t) - s_{η} (t^{'}) ∣$ , since the interval has length one. Thus $T (x, x^{'}) \leq 4 \sqrt{η} + ω_{η} (d) .$ By [F8], the domain $[- 1, 1]$ is compact, so [F6] gives $ω_{η} (d) \to 0$ as $d ↓ 0$ for fixed $η$ ; also $ω_{η} (d) \leq 2 η^{- 1 / 2}$ , so the preceding bound bounds every $T (x, x^{'})$ . The set in the definition of $τ (d)$ is nonempty because it contains $T (x, x) = 0$ , so [F7] gives its finite supremum. Given $ϵ > 0$ , first choose $η$ with $4 \sqrt{η} < ϵ / 2$ , then choose $d_{0} > 0$ with $ω_{η} (d_{0}) < ϵ / 2$ . For every $d \leq d_{0}$ , the same bound gives $τ (d) \leq ϵ$ , proving $τ (d) \to 0$ .

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The operator-norm Nyström perturbation bound

Statement

Use $X$ , $K$ , and the finite quadrature Nyström operator $K_{Q}$ from Nyström quadrature operators and their reconstructed equations. Suppose $A : = I - K$ has a bounded inverse, put $R : = ∥ A^{- 1} ∥$ , and let $ε : = ∥ K_{Q} - K ∥$ as an operator on $X$ . If $q : = R ε < 1$ , then $I - K_{Q}$ has a bounded inverse and $∥ (I - K_{Q})^{- 1} ∥ \leq \frac{R}{1 - q} .$ For $f, f_{Q} \in X$ , let $u$ and $u_{Q}$ solve $(I - K) u = f$ and $(I - K_{Q}) u_{Q} = f_{Q}$ . Then $∥ u_{Q} - u ∥_{\infty} \leq \frac{R}{1 - q} (∥ (K_{Q} - K) u ∥_{\infty} + ∥ f_{Q} - f ∥_{\infty}) .$ When $Q$ is point sampling on all of $C ([0, 1], R)$ , this theorem is conditional: its hypothesis is an actual operator-norm bound, not merely convergence of the quadrature on each fixed continuous integrand.

Facts & Assumptions

Given: The Nyström operators in the Statement, a bounded inverse $A^{- 1}$ , and $q = ∥ A^{- 1} ∥ ∥ K_{Q} - K ∥ < 1$ .

[F1]

Nyström quadrature operators and their reconstructed equations: $K_{Q}$ is the finite-sum map $u \mapsto \sum_{j} w_{j} k (\cdot, y_{j}) u (y_{j})$ , with continuous coefficient functions $x \mapsto w_{j} k (x, y_{j})$ .

[F2]

Finite-node Nyström equations are equivalent to their reconstructed equations: the finite system is nonsingular exactly when the reconstructed operator $I - K_{Q}$ is bijective.

[F3]

Neumann series and small perturbations of bounded inverses: if $∥ R ∥ < 1$ , then $I - R$ is invertible with inverse norm at most $(1 - ∥ R ∥)^{- 1}$ ; if $A$ and $E$ are bounded and $A^{- 1}$ is bounded with $∥ A^{- 1} E ∥ < 1$ , then $A + E = A (I + A^{- 1} E)$ is invertible.

[F4]

The operator norm as the least bound and as the unit-sphere or unit-ball supremum: $∥ T ∥$ is defined for a bounded linear operator by its supremum on the unit ball.

[F5]

Composition satisfies |ST|\le|S|,|T|: for bounded linear operators $S, T$ , $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ .

[F6]

A bounded linear operator between normed spaces: a linear map is bounded when one constant bounds its output norm by that constant times the input norm.

[F7]

The square $[0, 1]^{2}$ is compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), so $k$ is bounded by $C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ . If $M : = \sup_{(x, y) \in [0, 1]^{2}} ∣ k (x, y) ∣$ , the integral formula and Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error give $∥ K u ∥_{\infty} \leq M ∥ u ∥_{\infty}$ , and Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ makes $K$ linear. The finite-sum formula and Triangle inequality for finite sums, Finite sums and finite products, by recursion, Laws of finite sums and finite products give $∥ K_{Q} u ∥_{\infty} \leq M ∥ u ∥_{\infty} \sum_{j} ∣ w_{j} ∣$ and make $K_{Q}$ linear. Hence $∥ (K_{Q} - K) u ∥_{\infty} \leq M (1 + \sum_{j} ∣ w_{j} ∣) ∥ u ∥_{\infty}$ and $∥ (I - K) u ∥_{\infty} \leq (1 + M) ∥ u ∥_{\infty}$ ; thus $K$ , $K_{Q}$ , $K_{Q} - K$ , and $A = I - K$ are bounded linear maps in the sense of [F6].

Proof

Given: The operators and right-hand sides in the Statement, $E : = K - K_{Q}$ , and $A : = I - K$ .

1.1givenF1F2F3F4F5F6F7algebra

By [F7] and [F6], $K_{Q} - K$ is bounded, so its norm $ε$ is defined by [F4]. Put $E : = K - K_{Q}$ ; then $I - K_{Q} = A + E = A (I + A^{- 1} E)$ and $∥ A^{- 1} E ∥ \leq ∥ A^{- 1} ∥ ∥ E ∥ = R ε = q < 1$ by [F5]. Applying both parts of [F3], with $R_{0} : = - A^{- 1} E$ , gives invertibility and $∥ (I - K_{Q})^{- 1} ∥ = ∥ (I + A^{- 1} E)^{- 1} A^{- 1} ∥ \leq \frac{R}{1 - ∥ A^{- 1} E ∥} \leq \frac{R}{1 - q} .$ The coefficient functions $x \mapsto w_{j} k (x, y_{j})$ are continuous because $k$ is continuous, so [F2] applies and the associated finite system has the same unique solvability.

2.1givenF1F4step 1.1algebra

Subtracting $(I - K) u = f$ from $(I - K_{Q}) u_{Q} = f_{Q}$ in the form $(I - K_{Q}) (u_{Q} - u) = f_{Q} - f + (K_{Q} - K) u$ and applying the inverse from step 1.1 yields the stated error estimate by the triangle inequality.

3.1step 1.1step 2.1∎

The two displayed conclusions are exactly the inverse and solution bounds obtained in steps 1.1 and 2.1. This proof uses the norm hypothesis $ε$ itself and asserts no norm convergence of the point-sampling operators.

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Composite midpoint Nyström operators are collectively compact and consistent

Statement

Assume AC. Let $k : [0, 1]^{2} \to R$ be continuous and let $K$ be its integral operator from Nyström quadrature operators and their reconstructed equations. For each $N \geq 1$ , let $K_{N}$ use the composite midpoint rule with $h = 1 / N$ . Then each $K_{N}$ is a bounded linear map $X \to X$ , $K$ is compact, the family $(K_{N})_{N \geq 1}$ is collectively compact (the union of the images of the unit ball has compact closure in $X$ ), and $∥ K_{N} u - K u ∥_{\infty} ⟶ 0 for every u \in X .$

Facts & Assumptions

Given: AC, the continuous kernel $k$ , its integral operator $K$ , and the composite midpoint operators $K_{N}$ on $X = C ([0, 1], R)$ .

[A1]

The Axiom of Choice and AC supplies the countable and dependent choices used in Banach integration: AC supplies both Countable Choice and Dependent Choice, as required by the local compactness and Arzelà–Ascoli suppliers.

[F1]

Nyström quadrature operators and their reconstructed equations: $K_{N} u (x) = h \sum_{j = 0}^{N - 1} k (x, (j + \frac{1}{2}) h) u ((j + \frac{1}{2}) h)$ .

[F2]

A continuous kernel defines a compact integral operator on C of an interval: under Countable Choice and Dependent Choice, the continuous-kernel integral operator $K : X \to X$ is bounded and compact.

[F3]

Composite midpoint rule on a uniform grid: the midpoint rule on $N$ uniform panels uses $h = 1 / N$ , one midpoint per panel, and weight $h$ .

[F4]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: the kernel $k$ and, for each fixed $u \in X$ , the function $(x, y) \mapsto k (x, y) u (y)$ are uniformly continuous on the compact square.

[F5]

Arzelà--Ascoli for real $C (K)$ under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded: under Countable Choice and Dependent Choice, a pointwise bounded equicontinuous family in $C ([0, 1], R)$ has compact closure.

[F6]

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , and Uniformly close integrable functions have integrals differing by at most the interval length times their uniform error: midpoint panel integrals add, and a uniformly bounded continuous panel error has integral bounded by panel length times its supremum.

[F7]

$C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ : the continuous kernel is bounded on the compact square, so $M : = ∥ k ∥_{\infty} < \infty$ .

[F8]

[F9]

[F10]

A linear map between normed spaces is bounded when one constant bounds its output norm by that constant times the input norm (A bounded linear operator between normed spaces).

[F11]

Additivity and scalar homogeneity of finite sums make the midpoint finite-sum formula linear in $u$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

Proof

Given: AC, $N \geq 1$ , $h = 1 / N$ , $M : = ∥ k ∥_{\infty}$ , and the operators in the Statement.

1.1A1F2

Compactness of $K$ is [F2], whose Countable Choice and Dependent Choice hypotheses follow from [A1].

1.2givenF1F3F4F7F8F9F10F11algebra

By [F8], the square is compact, so [F7] makes $M = ∥ k ∥_{\infty}$ finite. Each finite sum in [F1] is continuous in $x$ , so $K_{N} u \in X$ . The formula is linear in $u$ by [F11], and for arbitrary $u \in X$ it gives $∣ K_{N} u (x) ∣ \leq h N M ∥ u ∥_{\infty} = M ∥ u ∥_{\infty}$ , so $K_{N}$ is bounded in the Banach norm of [F9] by [F10]. For $∥ u ∥_{\infty} \leq 1$ , [F1] and [F3] give $∣ K_{N} u (x) ∣ \leq M$ for every $x$ . For $x, x^{'} \in [0, 1]$ they also give $∣ K_{N} u (x) - K_{N} u (x^{'}) ∣ \leq \sup_{y \in [0, 1]} ∣ k (x, y) - k (x^{'}, y) ∣ .$ Uniform continuity of $k$ on the square makes the right side a common modulus tending to zero as $∣ x - x^{'} ∣ \to 0$ . Thus the union of all unit-ball images is pointwise bounded and equicontinuous.

2.1A1F5F8F9step 1.2

By [F5] on the compact interval supplied by [F8] and [A1], the closure of that union is compact, so $(K_{N})$ is collectively compact. The Banach space $X$ is normed by [F9], and its metric is the supremum metric used here.

3.1givenF1F3F4F6F8algebra∎

Fix $u \in X$ and put $G_{u} (x, y) : = k (x, y) u (y)$ . By [F4] on the compact square from [F8], its uniform modulus in the $y$ variable, $ω_{u} (ρ) : = \sup {∣ G_{u} (x, y) - G_{u} (x, y^{'}) ∣ : x \in [0, 1], ∣ y - y^{'} ∣ \leq ρ}$ , tends to zero as $ρ ↓ 0$ . On each midpoint panel $I_{j} = [j h, (j + 1) h]$ with midpoint $m_{j}$ , [F6] gives $∣ h G_{u} (x, m_{j}) - \int_{I_{j}} G_{u} (x, y) d y ∣ \leq h ω_{u} (h / 2) .$ Adding the $N$ panel bounds and using $N h = 1$ yields $∥ K_{N} u - K u ∥_{\infty} \leq ω_{u} (h / 2) \to 0$ .

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Piecewise-linear product integration for a half-order diagonal singularity

Definition

Let $X : = C ([0, 1], R)$ and let $a : [0, 1]^{2} \to R$ be continuous. Use $S_{x}$ from The half-order diagonal singularity defines a bounded improper integral to define $(K u) (x) : = S_{x} (y \mapsto a (x, y) u (y)) .$ For an integer $n \geq 1$ , set $h : = 1 / n$ and $y_{j} : = j h$ for $j = 0, \dots, n$ . Define the nonnegative piecewise-linear hat functions by $ϕ_{n, 0} (y) : = \max (1 - y / h, 0), ϕ_{n, n} (y) : = \max (1 - (1 - y) / h, 0),$ and, for $1 \leq j < n$ , $ϕ_{n, j} (y) : = \max (1 - ∣ y - y_{j} ∣ / h, 0) .$ Their piecewise-linear interpolant is $I_{n} g : = \sum_{j = 0}^{n} g (y_{j}) ϕ_{n, j} .$ The product weights and product-integration operator are $w_{n, j} (x) : = S_{x} (ϕ_{n, j}), (K_{n} u) (x) : = \sum_{j = 0}^{n} w_{n, j} (x) a (x, y_{j}) u (y_{j}) = S_{x} (I_{n} (y \mapsto a (x, y) u (y))) .$ The kernel factor $s (x, y) = ∣ x - y ∣^{- 1 / 2}$ is integrated in the weights; the smooth factor $a (x, y) u (y)$ is interpolated. This item defines the pointwise formulas; the subsequent results on this page establish their mapping properties, stability, and error estimates.

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Closed forms for the half-order piecewise-linear product weights

Statement

Use the nodes, hats, and weights $w_{n, j} (x) = S_{x} (ϕ_{n, j})$ of Piecewise-linear product integration for a half-order diagonal singularity. For $x \in [0, 1]$ , $0 \leq α \leq β \leq 1$ , and $u, v \in R$ , put $C : = u + v x$ and, for $r \geq 0$ , define $P_{x, u, v} (r) : = 2 C \sqrt{r} + \frac{2}{3} v r \sqrt{r}, Q_{x, u, v} (r) : = 2 C \sqrt{r} - \frac{2}{3} v r \sqrt{r} .$ Write $r_{+} : = \max (r, 0)$ and set $J_{x} [α, β; u, v] : = Q_{x, u, v} ((x - α)_{+}) - Q_{x, u, v} ((x - β)_{+}) + P_{x, u, v} ((β - x)_{+}) - P_{x, u, v} ((α - x)_{+}) .$ Then this is the exact improper Riemann integral $J_{x} [α, β; u, v] = \int_{α}^{β} ∣ x - y ∣^{- 1 / 2} (u + v y) d y,$ with the integral split at $x$ if $x \in [α, β]$ . In particular, for $h = 1 / n$ and $y_{j} = j h$ , $w_{n, 0} (x) = J_{x} [0, h; 1, - h^{- 1}],$ $w_{n, j} (x) = J_{x} [y_{j - 1}, y_{j}; - y_{j - 1} / h, h^{- 1}] + J_{x} [y_{j}, y_{j + 1}; y_{j + 1} / h, - h^{- 1}], 1 \leq j < n,$ and $w_{n, n} (x) = J_{x} [y_{n - 1}, 1; - y_{n - 1} / h, h^{- 1}] .$ The formulas include $x$ at a panel endpoint and the one-panel case $n = 1$ .

Facts & Assumptions

Given: $x \in [0, 1]$ , $0 \leq α \leq β \leq 1$ , $u, v \in R$ , and the hat functions and weights from the product-integration definition.

[F1]

Piecewise-linear product integration for a half-order diagonal singularity: each hat is affine on each adjacent panel, the product weight is $w_{n, j} (x) = S_{x} (ϕ_{n, j})$ , and the hat formulas are specified at both endpoints.

[F2]

The half-order diagonal singularity defines a bounded improper integral: for every continuous $g$ on $[0, 1]$ , $S_{x} (g)$ exists as the improper Riemann integral of $∣ x - y ∣^{- 1 / 2} g (y)$ .

[F3]

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ : for $r \geq 0$ , $\sqrt{r}$ is the unique nonnegative real whose square is $r$ (the case $n = 2$ ).

[F4]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : sums, scalar multiples, and products of differentiable functions obey their usual derivative formulas.

[F5]

The two FTC forms for a Riemann–Stieltjes integral with a $C^{1}$ integrator: a continuous integrand with an antiderivative on a closed interval has integral equal to the endpoint difference; the identity integrator gives the ordinary Riemann integral.

[F6]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: a continuous function on a closed interval has an ordinary Riemann integral.

[F7]

[F8]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : differentiating a composition differentiable on the relevant open interval multiplies the outer derivative by the inner derivative.

Proof

Given: The data in the Statement, $C = u + v x$ , and the functions $P_{x, u, v}$ and $Q_{x, u, v}$ .

1.1givenF3F4F8algebra

Fix $r > 0$ and put $q : = \sqrt{r}$ , which is positive by [F3] because $q^{2} = r > 0$ . For $s > 0$ , rationalizing gives $∣ \sqrt{s} - q ∣ = ∣ s - r ∣ / (\sqrt{s} + q) \leq ∣ s - r ∣ / q$ , so $\sqrt{s} \to q$ as $s \to r$ . For $s \neq r$ the difference quotient is $(\sqrt{s} - q) / (s - r) = 1 / (\sqrt{s} + q)$ , and $∣ 1 / (\sqrt{s} + q) - 1 / (2 q) ∣ = ∣ q - \sqrt{s} ∣ / (2 q (\sqrt{s} + q)) \leq ∣ s - r ∣ / (2 q^{3})$ , so $(\sqrt{r})^{'} = 1 / (2 \sqrt{r})$ . By the product rule [F4], $P_{x, u, v}^{'} (r) = C r^{- 1 / 2} + v \sqrt{r} = r^{- 1 / 2} (C + v r)$ and $Q_{x, u, v}^{'} (r) = C r^{- 1 / 2} - v \sqrt{r} = r^{- 1 / 2} (C - v r)$ . Since $u + v y = C + v (y - x)$ , [F8] makes $P_{x, u, v} (y - x)$ an antiderivative of $(y - x)^{- 1 / 2} (u + v y)$ for $y > x$ . Since $u + v y = C - v (x - y)$ , [F8] makes $- Q_{x, u, v} (x - y)$ an antiderivative of $(x - y)^{- 1 / 2} (u + v y)$ for $y < x$ .

2.1givenF2F3F5F6F7step 1.1algebra

By [F3], for every $ϵ > 0$ , $0 \leq δ < ϵ^{2}$ implies $0 \leq \sqrt{δ} < ϵ$ ; hence $\sqrt{δ} \to 0$ and both $P_{x, u, v} (δ)$ and $Q_{x, u, v} (δ)$ tend to zero as $δ ↓ 0$ . If $x > α$ , evaluate the left integral on $[α, \min (β, x - δ)]$ for sufficiently small $δ > 0$ using [F6], [F5], and step 1.1. As $δ ↓ 0$ , its value tends to $Q_{x, u, v} ((x - α)_{+}) - Q_{x, u, v} ((x - β)_{+})$ . If $x \leq α$ , the left contribution and this expression are zero. Likewise, if $x < β$ , evaluate the right integral on $[\max (α, x + δ), β]$ for sufficiently small $δ > 0$ ; its limit is $P_{x, u, v} ((β - x)_{+}) - P_{x, u, v} ((α - x)_{+})$ . If $x \geq β$ , the right contribution and this expression are zero. Thus both one-sided integrals exist and have the displayed values. Splitting truncated intervals by [F7] and passing to their finite limits gives exactly $J_{x} [α, β; u, v]$ .

3.1givenF1step 2.1algebra∎

On $[0, h]$ , $ϕ_{n, 0} (y) = 1 - y / h$ ; for an interior index $j$ , its left and right affine pieces are $(y - y_{j - 1}) / h$ and $(y_{j + 1} - y) / h$ ; on $[y_{n - 1}, 1]$ , $ϕ_{n, n} (y) = (y - y_{n - 1}) / h$ . Substituting these coefficients into the integral formula of step 2.1 gives the three displayed formulas for $w_{n, j}$ . The endpoint cases use positive parts in $J$ , so a singularity at a split point contributes a zero primitive value, and when $n = 1$ the interior-index range is empty while both endpoint formulas remain valid.

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Smooth-solution error bound for composite midpoint Nyström discretization

Statement

Assume AC. Let $k : [0, 1]^{2} \to R$ be continuous, $K$ its integral operator, and $K_{N}$ its composite midpoint Nyström operator with panel width $h = 1 / N$ . Suppose $I - K$ is injective and $u$ solves $(I - K) u = f$ . Assume that $g_{x} (y) : = k (x, y) u (y)$ is twice continuously differentiable in $y$ for every $x$ , with the uniform bound $M_{2} : = \sup_{x, y \in [0, 1]} ∣ \partial_{y}^{2} g_{x} (y) ∣ < \infty .$ Then there are $N_{0}$ and $C < \infty$ such that for every $N \geq N_{0}$ the Nyström matrix is nonsingular, its reconstruction $u_{N}$ solving $(I - K_{N}) u_{N} = f_{N}$ exists for each $f_{N} \in X$ , and $∥ u_{N} - u ∥_{\infty} \leq C (∥ f_{N} - f ∥_{\infty} + \frac{h^{2}}{24} M_{2}) .$ In the standard Nyström method $f_{N} = f$ . The regularity assumption is on the actual function $k (x, \cdot) u (\cdot)$ , not on every function in the unit ball of $C ([0, 1], R)$ .

Facts & Assumptions

Given: AC, $k$ , $K$ , $K_{N}$ , $u$ , $f$ , $f_{N}$ , and $M_{2}$ as in the Statement.

[F1]

Nyström quadrature operators and their reconstructed equations: the Nyström equation is $(I - K_{N}) u_{N} = f_{N}$ .

[F2]

Finite-node Nyström equations are equivalent to their reconstructed equations: nonsingularity of the finite grid matrix is equivalent to bijective solvability of the reconstructed equation.

[F3]

A continuous kernel defines a compact integral operator on C of an interval: the continuous-kernel operator $K$ is compact; AC supplies its Countable Choice and Dependent Choice hypotheses by AC supplies the countable and dependent choices used in Banach integration.

[F4]

Composite midpoint Nyström operators are collectively compact and consistent: each composite midpoint operator is bounded linear, the family is collectively compact, and $K_{N} x \to K x$ in norm for every $x \in X$ .

[F5]

Collective compactness gives eventual Fredholm invertibility and solution error bounds: under AC, compactness of $K$ , collective compactness and pointwise convergence of $K_{N}$ , and injectivity of $I - K$ give eventual bounded inverses with a uniform bound and the residual solution estimate.

[F6]

Composite midpoint error bound for a twice continuously differentiable function: for a $C^{2}$ real integrand $g$ on $[0, 1]$ , the composite midpoint rule satisfies $∣ \int_{0}^{1} g - Q_{N} g ∣ \leq h^{2} ∥ g^{''} ∥_{\infty} / 24$ .

[F7]

[F8]

The constant function $1$ belongs to $X$ and is nonzero by the definition of continuous real-valued functions (The space $C (K, R)$ of continuous real-valued functions on a nonempty compact metric space), so $X \neq {0}$ .

Proof

Given: AC, the continuous kernel and exact solution in the Statement, and $h = 1 / N$ .

1.1givenF3F4F5F7F8

By [F7] and [F8], the space $X$ satisfies the Banach and nonzero hypotheses of [F5]. By [F3], $K$ is compact; [F4] gives bounded linear $K_{N}$ with the required collective compactness and pointwise convergence; the Statement assumes $I - K$ injective. Thus [F5] supplies $N_{0}$ and a common bound $C$ for $(I - K_{N})^{- 1}$ for all $N \geq N_{0}$ .

1.2givenF6

For each fixed $x$ , apply [F6] to the $C^{2}$ function $g_{x} (y) = k (x, y) u (y)$ . The common derivative bound gives $∣ K_{N} u (x) - K u (x) ∣ \leq h^{2} M_{2} / 24$ for every $x$ , hence $∥ K_{N} u - K u ∥_{\infty} \leq h^{2} M_{2} / 24$ .

2.1F1F2step 1.1step 1.2algebra∎

The function-equation inverse from step 1.1 and [F2] make the finite matrix nonsingular and define its unique reconstruction $u_{N}$ . Subtracting the exact and discrete equations gives $(I - K_{N}) (u_{N} - u) = f_{N} - f + (K_{N} - K) u .$ Applying the inverse bound $C$ from step 1.1, then the triangle inequality and step 1.2, proves the stated estimate.

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Stability, collective compactness, and consistency of half-order product integration

Statement

Assume AC and let $a : [0, 1]^{2} \to R$ be continuous. For $K$ and $K_{n}$ from Piecewise-linear product integration for a half-order diagonal singularity, put $A : = ∥ a ∥_{\infty}$ and $h : = 1 / n$ . Then both operators are bounded linear maps from $X = C ([0, 1], R)$ to itself, $K$ is compact, and $(K_{n})$ is collectively compact and converges pointwise in norm to $K$ . Their product weights satisfy $w_{n, j} (x) \geq 0, \sum_{j = 0}^{n} ∣ w_{n, j} (x) ∣ = 2 \sqrt{x} + 2 \sqrt{1 - x} \leq 2 \sqrt{2}$ for every $x \in [0, 1]$ . If each $g_{x} (y) : = a (x, y) u (y)$ extends to a $C^{2}$ function on a neighborhood of $[0, 1]$ and $M_{2} : = \sup_{x, y} ∣ \partial_{y}^{2} g_{x} (y) ∣ < \infty$ , then $∥ K_{n} u - K u ∥_{\infty} \leq \frac{\sqrt{2}}{4} M_{2} h^{2} .$

Facts & Assumptions

Given: AC, continuous $a$ , the scalar functionals $S_{x}$ , hat functions $ϕ_{n, j}$ , and operators $K, K_{n}$ defined in the preceding item.

[A1]

The Axiom of Choice and AC supplies the countable and dependent choices used in Banach integration: AC supplies the two choice hypotheses of real Arzelà–Ascoli.

[F1]

Piecewise-linear product integration for a half-order diagonal singularity: $K_{n} u (x) = S_{x} (I_{n} g_{x})$ for $g_{x} (y) = a (x, y) u (y)$ , and $w_{n, j} (x) = S_{x} (ϕ_{n, j})$ .

[F2]

The half-order diagonal singularity defines a bounded improper integral: $S_{x}$ is positive, linear, and $∣ S_{x} (g) ∣ \leq 2 \sqrt{2} ∥ g ∥_{\infty}$ ; it is integration against $∣ x - y ∣^{- 1 / 2}$ in the improper-Riemann sense.

[F3]

Translations of the half-order kernel are continuous in the improper L1 norm: the finite modulus $τ (d) : = \sup_{∣ x - x^{'} ∣ \leq d} \int_{0}^{1} ∣ s (x - y) - s (x^{'} - y) ∣ d y$ tends to zero as $d ↓ 0$ .

[F4]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: continuous $a$ and $g_{u} (x, y) = a (x, y) u (y)$ on the compact square are uniformly continuous.

[F5]

[F6]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ and Pointwise linear interpolation error from a second-derivative bound: for $g \in C^{2}$ on a neighborhood of a panel of width $h$ , the linear interpolant error is at most $\frac{1}{2} ∥ g^{''} ∥_{\infty} (y - y_{j}) (y_{j + 1} - y) \leq h^{2} ∥ g^{''} ∥_{\infty} / 8$ .

[F7]

$C (K, R)$ is complete in the supremum metric for every nonempty compact metric space $K$ : every continuous $a$ on the compact square is bounded, so $A : = ∥ a ∥_{\infty} < \infty$ .

[F8]

[F9]

[F10]

A linear map between normed spaces is bounded when a constant bounds its output norm by that constant times the input norm (A bounded linear operator between normed spaces).

[F11]

Additivity and scalar homogeneity of finite sums make the finite-sum interpolation formula linear in its input (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[F12]

A bounded linear operator is compact if the image of its closed unit ball has compact closure (Compact linear operator).

Proof

Given: AC, the definitions in [F1], $A = ∥ a ∥_{\infty}$ , $h = 1 / n$ , and $s (t) : = ∣ t ∣^{- 1 / 2}$ for $0 < ∣ t ∣ \leq 1$ .

1.1givenF1F2F7F8F9algebra

By [F8], the square is compact, so [F7] makes $A = ∥ a ∥_{\infty}$ finite. In the Banach normed space [F9], the hat functions in [F1] are nonnegative and on each panel the two adjacent hats add to $1$ , while all other hats vanish. Hence $\sum_{j = 0}^{n} ϕ_{n, j} = 1$ pointwise. Positivity and linearity in [F2] give $w_{n, j} (x) \geq 0$ and $\sum_{j} ∣ w_{n, j} (x) ∣ = \sum_{j} w_{n, j} (x) = S_{x} (1) = 2 \sqrt{x} + 2 \sqrt{1 - x} \leq 2 \sqrt{2} .$ For every $u \in X$ , the same bound shows $∥ K_{n} u ∥_{\infty} \leq 2 \sqrt{2} A ∥ u ∥_{\infty}$ .

1.2F3

For $d : = ∣ x - x^{'} ∣$ , [F3] gives an upper bound $τ (d)$ for the improper kernel translation integral, with $τ (d) \to 0$ as $d ↓ 0$ .

2.1F1F2F4step 1.1

For fixed $u$ , [F4] makes $(x, y) \mapsto g_{x} (y)$ uniformly continuous. By step 1.1 the hats are nonnegative and sum to $1$ , so on a panel $[y_{j}, y_{j + 1}]$ , $I_{n} g_{x} (y)$ is a convex combination of the two endpoint values. Hence $∥ I_{n} g_{x} - g_{x} ∥_{\infty} \to 0$ uniformly in $x$ , and the operator bound in [F2] yields $∥ K_{n} u - K u ∥_{\infty} \leq 2 \sqrt{2} ∥ I_{n} g_{x} - g_{x} ∥_{\infty} \to 0$ .

2.2F1F2F3F4F5F8F9F10F11F12A1step 1.1step 1.2

Write $g_{x} (y) = a (x, y) u (y)$ . The identities in [F1], the translation bound in [F3], and the operator bound in [F2] show $∣ K u (x) - K u (x^{'}) ∣ \leq A ∥ u ∥_{\infty} τ (d) + 2 \sqrt{2} ∥ u ∥_{\infty} ω_{a} (d),$ where $τ (d)$ is the translation bound from step 1.2 and $ω_{a} (d) : = \sup_{∣ x - x^{'} ∣ \leq d, y} ∣ a (x, y) - a (x^{'}, y) ∣ \to 0$ by [F4] on the compact square from [F8]. Replacing $g_{x}$ by $I_{n} g_{x}$ gives the same bound for $K_{n} u$ : by step 1.1 the interpolant is a convex combination of nodal values, so $∥ I_{n} g_{x} ∥_{\infty} \leq ∥ g_{x} ∥_{\infty}$ and $∥ I_{n} (g_{x} - g_{x^{'}}) ∥_{\infty} \leq ∥ g_{x} - g_{x^{'}} ∥_{\infty}$ . Thus $K$ and all $K_{n}$ map into $X$ and have a common equicontinuity modulus. The formula [F1] and linearity of $S_{x}$ in [F2] make $K$ linear; its operator bound follows from [F2]. Each $K_{n}$ is linear by [F11], and step 1.1 bounds it. By [F9] and [F10], both are bounded linear maps. Their images of the unit ball are uniformly bounded and equicontinuous. By [F5] and [A1], $K$ is compact by [F12], and $(K_{n})$ is collectively compact.

3.1F1step 2.2

For each fixed $n, j$ , take $a \equiv 1$ and $u = ϕ_{n, j}$ in the preceding continuity argument; then $K_{n} u (x) = w_{n, j} (x)$ , so each weight function is continuous. Consequently the coefficient functions $x \mapsto w_{n, j} (x) a (x, y_{j})$ in the finite-node representation are continuous.

4.1F2F6∎

Under the $C^{2}$ extension hypothesis, [F6] bounds $∥ I_{n} g_{x} - g_{x} ∥_{\infty} \leq M_{2} h^{2} / 8$ uniformly in $x$ . Applying [F2] gives $∥ K_{n} u - K u ∥_{\infty} \leq 2 \sqrt{2} M_{2} h^{2} / 8 = \sqrt{2} M_{2} h^{2} / 4$ .

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Fredholm error bound for half-order product integration

Statement

Assume AC. Let $a : [0, 1]^{2} \to R$ be continuous, let $K, K_{n}$ be the half-order product-integration operators of Piecewise-linear product integration for a half-order diagonal singularity, and suppose $I - K$ is injective. Let $u$ solve $(I - K) u = f$ , and assume each $g_{x} (y) : = a (x, y) u (y)$ extends to a $C^{2}$ function on a neighborhood of $[0, 1]$ , with $M_{2} : = \sup_{x, y \in [0, 1]} ∣ \partial_{y}^{2} (a (x, y) u (y)) ∣ < \infty .$ Then there are $n_{0}$ and $C < \infty$ such that for every $n \geq n_{0}$ , the finite Nyström grid matrix is nonsingular, its reconstructed solution $u_{n}$ exists, and $∥ u_{n} - u ∥_{\infty} \leq C (∥ f_{n} - f ∥_{\infty} + \frac{\sqrt{2}}{4} M_{2} h^{2}), h = 1 / n,$ whenever $(I - K_{n}) u_{n} = f_{n}$ . In particular, for the exact right-hand side $f_{n} = f$ , the solution error is $O (h^{2})$ under the stated smoothness of the regular factor times the exact solution.

Facts & Assumptions

Given: AC, the continuous $a$ , the operators $K, K_{n}$ , an exact solution $u$ , and the uniform second-derivative bound in the Statement.

[A1]

The Axiom of Choice and AC supplies the countable and dependent choices used in Banach integration: AC supplies the Countable Choice and Dependent Choice needed by the local Arzelà–Ascoli result.

[F1]

Piecewise-linear product integration for a half-order diagonal singularity: $K$ and $K_{n}$ are the improper-integral and weighted-hat interpolation operators, and $K_{n}$ has the finite-node form $\sum_{j} ψ_{j} (x) u (y_{j})$ with coefficient functions $ψ_{j} (x) = w_{n, j} (x) a (x, y_{j})$ .

[F2]

Stability, collective compactness, and consistency of half-order product integration: $K$ and each $K_{n}$ are bounded linear maps, $K$ is compact, $K_{n}$ is collectively compact and pointwise norm convergent to $K$ , each coefficient function $ψ_{j}$ is continuous, and $∥ K_{n} u - K u ∥_{\infty} \leq \sqrt{2} M_{2} h^{2} / 4$ .

[F3]

Collective compactness gives eventual Fredholm invertibility and solution error bounds: under AC and injectivity of $I - K$ , the operators $I - K_{n}$ are boundedly invertible for all sufficiently large $n$ with one uniform inverse bound, and their solution errors satisfy the residual estimate.

[F4]

Finite-node Nyström equations are equivalent to their reconstructed equations: bijective solvability of the reconstructed equation is equivalent to nonsingularity of its finite grid matrix.

[F5]

Proof

Given: The hypotheses in the Statement and $h = 1 / n$ .

1.1givenA1F2F3F5

By [F5], $X$ is a nonzero Banach space. By [F2], $K$ is compact and $(K_{n})$ consists of bounded linear maps that are collectively compact and pointwise norm convergent to $K$ . With AC and the assumed injectivity of $I - K$ , [F3] gives $n_{0}$ and $C$ such that $I - K_{n}$ is invertible and $∥ (I - K_{n})^{- 1} ∥ \leq C$ for all $n \geq n_{0}$ .

2.1step 1.1F1F2F4

The inverse in step 1.1 gives a unique solution to $(I - K_{n}) u_{n} = f_{n}$ . By [F2] each coefficient function in [F1] is continuous, so [F4] applies: the finite grid matrix is nonsingular and its solution reconstructs exactly this $u_{n}$ .

3.1givenF2F3step 1.1algebra∎

Subtracting $(I - K) u = f$ from $(I - K_{n}) u_{n} = f_{n}$ gives $(I - K_{n}) (u_{n} - u) = f_{n} - f + (K_{n} - K) u .$ Apply the inverse bound in [F3] and then the consistency bound in [F2]: $∥ u_{n} - u ∥_{\infty} \leq C (∥ f_{n} - f ∥_{\infty} + ∥ K_{n} u - K u ∥_{\infty}) \leq C (∥ f_{n} - f ∥_{\infty} + \frac{\sqrt{2}}{4} M_{2} h^{2}) .$ This is the claimed error estimate.

5 · Examples, counterexamples and false statements

None yet.

Nystrom product integration and fredholm error

1 · Prerequisites

2 · Summary

Affine normalization to a general interval

3 · Logical flowchart

4 · Definitions, theorems and proofs

Nyström quadrature operators and their reconstructed equations

Definition

Finite-node Nyström equations are equivalent to their reconstructed equations

Statement

Facts & Assumptions

Proof

A continuous kernel defines a compact integral operator on C of an interval

Statement

Facts & Assumptions

Proof

Collective compactness gives eventual Fredholm invertibility and solution error bounds

Statement

Facts & Assumptions

Proof

The half-order diagonal singularity defines a bounded improper integral

Statement

Facts & Assumptions

Proof

Translations of the half-order kernel are continuous in the improper L1 norm

Statement

Facts & Assumptions

Proof

The operator-norm Nyström perturbation bound

Statement

Facts & Assumptions

Proof

Composite midpoint Nyström operators are collectively compact and consistent

Statement

Facts & Assumptions

Proof

Piecewise-linear product integration for a half-order diagonal singularity

Definition

Closed forms for the half-order piecewise-linear product weights

Statement

Facts & Assumptions

Proof

Smooth-solution error bound for composite midpoint Nyström discretization

Statement

Facts & Assumptions

Proof

Stability, collective compactness, and consistency of half-order product integration

Statement

Facts & Assumptions

Proof

Fredholm error bound for half-order product integration

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources