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A smooth finite-variation integrand and its deterministic error bound

Example

On [0,1]2, take f(x,y)=xy and the one-point set P=((1/4,1/2)). Then VHK,1(f)=3,D1∗(P)=78,∣QP(f)−I2(f)∣=18, so the Koksma–Hlawka bound is valid for this smooth integrand.

Facts & Assumptions

Given: The function f(x,y)=xy on the unit square and the point set containing (1/4,1/2).

[F1]

For Cd functions, upper-anchor Hardy–Krause variation is the sum of the integrals of the absolute mixed derivatives on all nonempty upper faces. (Mixed-derivative formula for smooth Hardy–Krause variation)

[F2]

The unit square has volume 1; grid cells' volumes sum to that volume, and for the constant-one function every lower and upper Darboux sum equals the sum of the cell volumes, so its integral is 1. (Axis-parallel rectangles in Rm and their volume, Grid partitions of a rectangle in Rm, their cells, refinements and mesh, Lower and upper Darboux sums over a grid partition in Rm, The lower and upper Darboux integrals over a nondegenerate rectangle in Rm)

[F3]

A continuous function on the square is Riemann integrable, and its integrable sections have iterated integrals equal to the multiple integral. (Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable, Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections)

[F4]

Star discrepancy is the supremum over u∈[0,1]d of the absolute difference between the fraction of nodes in [0,u) and its volume. (Star discrepancy, digital nets, and digital sequences)

[F5]

If G is continuous on [a,b], differentiable on (a,b), and its interior derivative has a Riemann-integrable extension, then its integral is G(b)−G(a). (Newton–Leibniz needs only continuity on [a,b], differentiability on (a,b), and a Riemann-integrable extension of the interior derivative)

[F6]

For a finite nonempty point multiset and an integrand of finite upper-anchor Hardy–Krause variation, ∣QP(f)−Id(f)∣≤DN∗(P)VHK,1(f). (Koksma–Hlawka integration bound with upper-anchor variation)

Verification

technique · direct
1.1F1F2givenalgebra

The upper faces are f(x,1)=x and f(1,y)=y, and the full mixed derivative is ∂xyf=1. By [F1], VHK,1(f)=∫011 dx+∫011 dy+∫[0,1]21 dx dy=1+1+1=3; the one-dimensional integrals and the square integral of 1 equal the respective volumes by [F2].

1.2F3F4F5givenalgebra

At u=(1/4,1/2) the point is excluded from the half-open anchored box, so the discrepancy there is 1/8. If u1≤1/4 or u2≤1/2, the box is empty and its volume is at most 1/2. If both coordinates exceed those thresholds, the box contains the point and its error is 1−u1u2≤1−(1/4)(1/2)=7/8; values approach 7/8 as u↓(1/4,1/2). Thus D1∗(P)=7/8. Also QP(f)=1/8. By [F3], the double integral equals the iterated Riemann integral, and [F5] evaluates it as I2(f)=∫01y(∫01x dx)dy=1/4.

2.1F6step 1.1step 1.2∎

The error is ∣1/8−1/4∣=1/8≤(7/8)⋅3=D1∗(P)VHK,1(f), consistent with [F6].

Depends on

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