Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Finite samples outside a uniform BV bound

Example

Let Ω=B((1/2,1/2),1/4)⊂(0,1)2, let Ω′=B((1/2,1/2),1/8), and fix m=1/2, M=3/2, and contrast a=1/4. For every integer N≥1, put q=⌈N⌉, choose N distinct cell centers from the q×q grid in [7/16,9/16]2, set rN=1/(32q), and define γN=1+a1EN, where EN is the union of the radius-rN disks about those centers. Then every γN is a real, uniformly elliptic, individually finite-BV conductivity equal to 1 outside Ω′, but TV(γN;Ω)≥(πa/32)N. Hence the family has no common finite BV bound R and is not contained in any one class X2(R) as N varies. This is a non-applicability example for a fixed uniform BV theorem, not a counterexample to its conclusion for members of that class.

Facts & Assumptions

Given: the disk domain and parameters in the Example.

[F1]

A conductivity in X2(R) must be uniformly elliptic, equal to the fixed background outside Ω′, and satisfy the displayed BV norm bound. (Calderón admissible conductivity classes)

[F2]

For u∈BV(Ω), TV(u;Ω)=sup⁡{∫Ωu div⁡φ dx:φ∈Cc∞(Ω;R2), ∣φ(x)∣≤1}; symmetry under φ↦−φ makes this the supremum of absolute pairings. (Test-field formula and L1 lower semicontinuity of spatial TV)

[F3]

A function is in BV(Ω) when its distributional derivative is a finite vector Radon measure, and its total variation is the mass of that measure; the distributional derivative is linear and annihilates constants. (Spatial BV and distributional total variation)

Proof

Given: N≥1 and the grid in the Example.

1.1F1givenalgebra

The square [7/16,9/16]2 has side 1/8 and lies in Ω′ because its corner radius from (1/2,1/2) is 2/16<1/8. The grid spacing is sN=1/(8q) and rN=sN/4. Distinct cell centers are separated by at least sN>2rN, and each disk lies inside its cell; hence the disks are disjoint and contained in Ω′. Thus 1≤γN≤1+a=5/4<3/2 and γN=1 on Ω∖Ω′.

1.2F2F3givenalgebra

For a disk B(c,r) and a smooth compactly supported vector field φ, polar coordinates give ∫B(c,r)div⁡φ dx=r∫02πφ(c+r(cos⁡θ,sin⁡θ))⋅(cos⁡θ,sin⁡θ) dθ. Indeed, in polar variables div⁡φ=ρ−1∂ρ(ρφr)+ρ−1∂θφθ; integrating against ρ dρ dθ makes the angular derivative integrate to zero and the inner radial term vanish. The distributional definition therefore gives D1B(c,r)=−ν H1∣∂B(c,r), whose variation has mass 2πr. Summing these measures over the disjoint disks shows D1EN is finite with variation mass 2πrNN, so [F3] gives 1EN∈BV(Ω) and [F2] gives the stated upper bound.

2.1F2F3step 1.1step 1.2construct

The annuli of radii (rN/2,3rN/2) about the chosen centers are pairwise disjoint: their outer radii sum to 3rN=3sN/4<sN, while distinct centers are at least sN apart. They lie in Ω, since each center is within 2/16 of (1/2,1/2) and 2/16+3rN/2≤2/16+3/64<1/4. On each annulus choose a smooth radial cutoff equal to one at radius rN, vanishing outside the annulus and between zero and one. The sum of the corresponding radial unit-vector fields is smooth, compactly supported in Ω, has norm at most one, and equals the outward normal on every circle. Its flux is 2πrNN, so [F2] gives the matching lower bound. Since DγN=aD1EN by [F3] and a>0, step 1.2 gives TV(1EN;Ω)=2πrNN and TV(γN;Ω)=2πarNN=πaN/(16q).

3.1F1step 1.1step 2.1algebra∎

Since q=⌈N⌉≤2N, step 2.1 gives TV(γN;Ω)≥(πa/32)N. The value and collar bounds were checked in step 1.1, and every finite union of disks has the finite perimeter computed in step 2.1. Thus each conductivity is individually finite-BV while the total variations diverge, so no fixed R contains the whole family in X2(R).

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