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Mixed-derivative formula for smooth Hardy–Krause variation

Statement

Let d≥1 and let f:[0,1]d→R be the continuous extension to the closed cube of a function on (0,1)d whose mixed partial derivatives of every order at most d exist and extend continuously to [0,1]d. This is the Cd hypothesis here, including agreement of the boundary values of f with its order-zero extension. For u={j1<⋯<jq}, write ∂uf for the iterated derivative ∂jq⋯∂j1f, interpreted on the closed cube by its continuous extension. For each nonempty u, let fu(xu)=f(xu;1−u) be the restriction to the face adjacent to the upper corner. Then VHK,1(f)=∑∅≠u⊆{1,…,d}∫[0,1]∣u∣∣∂uf(xu;1−u)∣ dxu, where VHK,1 is the upper-anchor variation of Hardy–Krause variation anchored at the upper corner, and ∂u differentiates once in each coordinate in u.

Facts & Assumptions

Given: The cube dimension and the stated continuous mixed-partial hypothesis on f.

[F1]

The upper-anchor Hardy–Krause variation is the sum of the Vitali variations of the face restrictions fu(xu)=f(xu;1−u) over all nonempty coordinate subsets u. (Hardy–Krause variation anchored at the upper corner)

[F2]

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)

[F4]

A continuous real function on a closed nondegenerate rectangle is Riemann integrable. (Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable)

[F5]

Iterated Riemann integrals of a Riemann-integrable function on a product rectangle equal its multiple integral whenever the sections in the iterated order are integrable; continuous functions have such sections. (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections)

[F6]

Riemann integrals are linear and monotone, satisfy the absolute-value estimate, and are additive under each interior coordinate cut; a constant c integrates over R to cvol⁡(R) by the Darboux definition and rectangle-volume convention. (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in Rm, The lower and upper Darboux integrals over a nondegenerate rectangle in Rm, Axis-parallel rectangles in Rm and their volume)

[F7]

For a bounded Riemann-integrable function, all tagged grid sums converge with mesh to its integral. (The multidimensional Darboux and tagged-mesh definitions of the Riemann integral agree)

[F8]

For every η>0 there is a positive integer n with 1/n<η. (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε)

[F9]

The real numbers form an ordered field, and for real a,b, ∣a+b∣≥∣a∣−∣b∣, obtained from the reverse triangle inequality by applying it to a+b and a. (The reals form a totally ordered field, The reverse triangle inequality)

Proof

1.1F2F3F4F5F6given

Fix a nonempty u={j1<⋯<jq} and a product cell R=∏j∈u[aj,bj] in a finite coordinate grid of its face. Put g=fu and h=∂jq⋯∂j1g, using the continuous derivative extension. Apply [F2] successively in coordinates j1,…,jq. If a line lies on the boundary because some other coordinates are fixed at 0 or 1, approximate those fixed coordinates by interior values and pass to the limit: every intermediate mixed partial and its derivative have continuous extensions, hence are uniformly continuous by [F3]. The integrands therefore converge uniformly along the line, so the integral difference is bounded by the interval length times a uniform error tending to zero; endpoint values converge as well. At each application the finite alternating differences in the remaining coordinates commute with the integral, yielding exactly the alternating sum over the 2q vertices. By [F4] and [F5] this iterated integral equals the product-rectangle Riemann integral, so Δ(g;R)=∫Rh. Consequently ∣Δ(g;R)∣≤∫R∣h∣. Summing over cells and using the absolute-value estimate and finite grid-cell additivity from [F6] gives ∑R∣Δ(g;R)∣≤∫[0,1]q∣h∣. Taking the supremum over grids proves VV(g)≤∫∣h∣.

2.1F1F3F4F6F7F8F9step 1.1algebra

Conversely, fix ε>0. By [F3], choose δ1>0 so that ∣h(x)−h(y)∣<ε/2 whenever d2(x,y)<δ1. By [F7] applied to the continuous, hence integrable, function ∣h∣ ([F4]), choose δ2>0 so every tagged grid of mesh below δ2 has sum within ε/2 of ∫∣h∣. By [F8], choose n with η=1/n<min⁡{δ1/(q+1),δ2} and take the uniform product grid of side width η. For any two points in one cell, each coordinate difference is at most η, so [F3] gives d2(x,y)2≤qη2<δ12 because q<(q+1)2; hence d2(x,y)<δ1. Tag each cell at its lower corner ξR. By [F6], linearity and the constant-integral formula give ∫Rh=h(ξR)vol⁡(R)+∫R(h−h(ξR)). Since ∣h(x)−h(ξR)∣<ε/2 throughout R, monotonicity and the absolute-value estimate in [F6] give ∣∫R(h−h(ξR))∣≤∫R∣h−h(ξR)∣≤(ε/2)vol⁡(R). Applying [F9] to a=h(ξR)vol⁡(R) and b=∫R(h−h(ξR)) yields ∣∫Rh∣≥∣h(ξR)∣vol⁡(R)−(ε/2)vol⁡(R). Summing, using that cell volumes total 1, and applying [F7] to the chosen tags gives ∑R∣∫Rh∣≥∑R∣h(ξR)∣vol⁡(R)−ε/2≥∫∣h∣−ε. The variation supremum is at least this sum, and arbitrary ε gives VV(g)≥∫∣h∣.

3.1F1step 1.1step 2.1∎

Thus VV(fu)=∫∣∂uf(xu;1−u)∣ for each nonempty u. Summing these finitely many equalities and using [F1] gives the displayed formula.

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