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A tensor product of two positive Gaussian rules has a derivative error bound

Statement

Let a<b and c<d and let [a,b], [c,d] be compact intervals. Let μ,ν be finite positive measures on them, respectively. Assume each has valid Gaussian moment data of orders 2n−1 and 2n for the same n and measure: the order-2n−1 datum defines its n-node Gaussian rule Qμ,Qν, while the stronger datum supplies the Hermite remainder. (It suffices that the measure has support with at least n+1 points and finite absolute moments through degree 2n.) Write Mμ=μ([a,b]), Mν=ν([c,d]), and Kμ=1(2n)!R∫pμ2 dμ,Kν=1(2n)!R∫pν2 dν, where pμ,pν are their monic node polynomials. Suppose G:[a,b]×[c,d]→R and its pure partial derivatives through order 2n in each coordinate are continuous on the rectangle. Then the tensor rule QμQνG satisfies ∣∫[a,b]∫[c,d]G(x,y) dν(y)dμ(x)−QμQνG∣≤KμMν∥∂x2nG∥∞+MμKν∥∂y2nG∥∞.

Facts & Assumptions

Given: Two finite positive order-2n Gaussian moment data, their n-node rules, and a function with the stated continuous pure partial derivatives.

[F1]

Positive-measure Gaussian quadrature moment data and The m-node Gaussian quadrature rule for a positive measure: order-2n data require finite absolute moments through 2n; discarding the last moment condition gives the same measure's order-2n−1 data used to define the Gaussian rules.

[F2]

Gaussian quadrature is exact through degree 2m−1 and has positive weights: each rule has positive weights summing to its measure's total mass.

[F3]

Gaussian quadrature has the Hermite mean-value remainder: for u∈C2n, each one-coordinate error has absolute value at most Kμ∥u(2n)∥∞ or Kν∥u(2n)∥∞, respectively.

[F5]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: each continuous partial derivative on the compact rectangle is uniformly continuous.

[F6]

The Borel sigma-algebra of a topological space and Integrable real and complex functions, and their integrals: continuous bounded functions on the compact intervals are Borel measurable and integrable for the finite measures.

[F7]

Monotonicity and nonnegative homogeneity of the nonnegative integral and The Lebesgue integral is linear on L1(μ): ∣∫v dν∣≤Mν∥v∥∞, bounded integrands are integrable, and finite integral differences distribute linearly.

[F9]

For a nonempty subset of Rn with n≥1, compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent: the rectangle is closed and bounded in R2, hence compact, and its continuous real-valued partial derivatives attain finite extrema, so their supremum norms are finite.

Proof

1.1F1F4F5F6F7F9givenconstruct

By [F9], the continuous partial derivatives are bounded on the compact rectangle; they are Borel and their integrals exist by [F1, F6]–[F7]. Define H(x)=∫[c,d]G(x,y)dν(y). For 0≤j<2n, the difference quotient of ∂xjG in x differs from ∂xj+1G(x,y) by a quantity tending uniformly in y to zero: by [F4], the quotient is the average of ∂xj+1G(u,y) between x and x+h, and [F5] makes that derivative uniformly continuous. Integrating the uniform error changes it by at most Mν times its supremum by [F7], so induction gives H∈C2n and H(2n)(x)=∫∂x2nG(x,y)dν(y).

1.2F2F3F8givenalgebra

Expand Qμ(H)−QμQνG=∑iwi(∫G(xi,y)dν(y)−Qν(G(xi,⋅))). The positive weights in [F2] sum to Mμ, and [F3] bounds each parenthesized error by Kν∥∂y2nG∥∞. Therefore this term has absolute value at most MμKν∥∂y2nG∥∞.

2.1F3step 1.1algebra

Apply [F3] to H under μ. The derivative formula in step 1.1 gives ∣∫H dμ−Qμ(H)∣≤KμMν∥∂x2nG∥∞.

3.1step 2.1step 1.2algebra∎

The iterated integral minus the tensor rule is the sum of the errors in steps 2.1 and 1.2. The triangle inequality gives the stated bound.

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