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

Statement

Assume AC and let a:[0,1]2→R be continuous. For K and Kn from Piecewise-linear product integration for a half-order diagonal singularity, put A:=∥a∥∞ and h:=1/n. Then both operators are bounded linear maps from X=C([0,1],R) to itself, K is compact, and (Kn) is collectively compact and converges pointwise in norm to K. Their product weights satisfy wn,j(x)≥0,∑j=0n∣wn,j(x)∣=2x+21−x≤22 for every x∈[0,1]. If each gx(y):=a(x,y)u(y) extends to a C2 function on a neighborhood of [0,1] and M2:=sup⁡x,y∣∂y2gx(y)∣<∞, then ∥Knu−Ku∥∞≤24M2h2.

Facts & Assumptions

Given: AC, continuous a, the scalar functionals Sx, hat functions ϕn,j, and operators K,Kn 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: Knu(x)=Sx(Ingx) for gx(y)=a(x,y)u(y), and wn,j(x)=Sx(ϕn,j).

[F2]

The half-order diagonal singularity defines a bounded improper integral: Sx is positive, linear, and ∣Sx(g)∣≤22∥g∥∞; 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′∣≤d∫01∣s(x−y)−s(x′−y)∣dy 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 gu(x,y)=a(x,y)u(y) on the compact square are 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 on [0,1] has compact closure in X.

[F6]

Higher derivatives and the classes Ck and C∞ and Pointwise linear interpolation error from a second-derivative bound: for g∈C2 on a neighborhood of a panel of width h, the linear interpolant error is at most 12∥g′′∥∞(y−yj)(yj+1−y)≤h2∥g′′∥∞/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∥∞<∞.

[F9]

On the nonempty compact interval [0,1] (Heine-Borel in Rn: with the Euclidean metric a subset of Rn 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∥∞≤∥u∥∞+∥v∥∞ and ∥cu∥∞=∣c∣∥u∥∞, and ∥u∥∞=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 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∥∞, h=1/n, and s(t):=∣t∣−1/2 for 0<∣t∣≤1.

1.1givenF1F2F7F8F9algebra

By [F8], the square is compact, so [F7] makes A=∥a∥∞ 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 ∑j=0nϕn,j=1 pointwise. Positivity and linearity in [F2] give wn,j(x)≥0 and ∑j∣wn,j(x)∣=∑jwn,j(x)=Sx(1)=2x+21−x≤22. For every u∈X, the same bound shows ∥Knu∥∞≤22A∥u∥∞.

1.2F3

For d:=∣x−x′∣, [F3] gives an upper bound τ(d) for the improper kernel translation integral, with τ(d)→0 as d↓0.

2.1F1F2F4step 1.1

For fixed u, [F4] makes (x,y)↦gx(y) uniformly continuous. By step 1.1 the hats are nonnegative and sum to 1, so on a panel [yj,yj+1], Ingx(y) is a convex combination of the two endpoint values. Hence ∥Ingx−gx∥∞→0 uniformly in x, and the operator bound in [F2] yields ∥Knu−Ku∥∞≤22∥Ingx−gx∥∞→0.

2.2F1F2F3F4F5F8F9F10F11F12A1step 1.1step 1.2

Write gx(y)=a(x,y)u(y). The identities in [F1], the translation bound in [F3], and the operator bound in [F2] show ∣Ku(x)−Ku(x′)∣≤A∥u∥∞τ(d)+22∥u∥∞ωa(d), where τ(d) is the translation bound from step 1.2 and ωa(d):=sup⁡∣x−x′∣≤d,y∣a(x,y)−a(x′,y)∣→0 by [F4] on the compact square from [F8]. Replacing gx by Ingx gives the same bound for Knu: by step 1.1 the interpolant is a convex combination of nodal values, so ∥Ingx∥∞≤∥gx∥∞ and ∥In(gx−gx′)∥∞≤∥gx−gx′∥∞. Thus K and all Kn map into X and have a common equicontinuity modulus. The formula [F1] and linearity of Sx in [F2] make K linear; its operator bound follows from [F2]. Each Kn 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 (Kn) is collectively compact.

3.1F1step 2.2

For each fixed n,j, take a≡1 and u=ϕn,j in the preceding continuity argument; then Knu(x)=wn,j(x), so each weight function is continuous. Consequently the coefficient functions x↦wn,j(x)a(x,yj) in the finite-node representation are continuous.

4.1F2F6∎

Under the C2 extension hypothesis, [F6] bounds ∥Ingx−gx∥∞≤M2h2/8 uniformly in x. Applying [F2] gives ∥Knu−Ku∥∞≤22M2h2/8=2M2h2/4.

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