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Weak Elliptic Maximum Principles and Holder Regularity — Examples

1 · Prerequisites

2 · Summary

These companions compute, test and stress the maximum-principle and De Giorgi--Nash--Moser statements of the main page. A smooth radial eigenfunction of −Δ−1 on the ball of radius π shows that the weak maximum principle fails without the zero-order sign condition, while the quadratic subsolution on the disc exhibits the agreement of the classical and the weak principles in the smooth case, and a dyadic oscillation example reads off the Holder exponent from the oscillation ratio. The measurable-coefficient annulus example realizes a Holder-regular weak solution with a discontinuous radial derivative, showing that the De Giorgi--Nash conclusion cannot be improved to C1, and the zero class of H1 illustrates why the estimates control essential extrema of a class rather than the pointwise extrema of an arbitrary representative. Two Harnack counterexamples exhibit the necessity of nonnegativity and of the additive forcing term, a disconnected domain shows that the global comparison needs connectedness, and a degenerate positive-semidefinite coefficient matrix produces a nonconstant weak solution with an interior zero set, isolating uniform ellipticity as a hypothesis rather than a convenience.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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The weak and the classical maximum principles agree on a smooth subsolution

Example

Example. On the unit disc Ω=B1(0)⊂R2 consider L0=−Δ (so aij=δij and b=c=0 in Uniformly elliptic divergence-form operators and their sesquilinear forms) and u(x)=∣x∣2−1. Then Δu=4≥0, so u is subharmonic in the classical sense (Subharmonic and superharmonic functions in rn), and u=0 on ∂Ω while u≤0 in Ω. The classical weak maximum principle for the Laplacian (Weak maximum principle for the laplacian) gives sup⁡Ωu=sup⁡∂Ωu=0, and the weak maximum principle Weak maximum principle for coercive divergence-form equations gives the same conclusion, because u is also a weak subsolution of L0u=0 in the sense of Weak subsolutions and supersolutions of a divergence-form equation with sup⁡∂Ωu+=0.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the unit disc Ω=B1(0)⊂R2; the coefficients aij=δij, b=c=0; and the function u(x)=∣x∣2−1.

[F1]

Classical differentiation gives Diu=2xi and Δu=4 on Ω, so Δu≥0: u is subharmonic in the sense of Subharmonic and superharmonic functions in rn, and u∈C2(Ω‾)∩H01(Ω) because u≤0 on Ω with equality exactly on ∂Ω and u is a polynomial (Bounded C^k domains and boundary charts, The kernel of the trace is the closure of the test functions for the zero-trace identification).

[F2]

Classical weak maximum principle for the Laplacian: for a bounded nonempty open Ω and u∈C2(Ω)∩C(Ω‾) with Δu≥0, max⁡Ω‾u=max⁡∂Ωu (Weak maximum principle for the laplacian).

[F3]

Classical-to-weak consistency: if u∈C2(Ω‾)∩H01(Ω) and Lu=f with f∈C(Ω‾), then a(u,v)=∫Ωfvˉ for every v∈H01(Ω), for the sesquilinear form a of Uniformly elliptic divergence-form operators and their sesquilinear forms (Classical solutions satisfy the weak formulation).

[F4]

Alternative direct integration by parts: for v∈H01(Ω) and u∈C2(Ω‾), the Sobolev Gauss-Green formula gives ∫Ω∇u⋅∇v dx=−∫Ωv Δu dx, the boundary term vanishing because Tv=0 (The Gauss-Green integration-by-parts formula with Sobolev traces, Weak subsolutions and supersolutions of a divergence-form equation).

[F5]

Weak maximum principle for coercive divergence-form equations: under its hypotheses, a weak subsolution u of L0u=0 on a bounded C1 domain satisfies ess sup⁡Ωu≤sup⁡∂Ωu+ (Weak maximum principle for coercive divergence-form equations).

Verification

1.1F1F2

The classical side. By [F1], u∈C2(Ω)∩C(Ω‾) with Δu=4≥0 and u=0 on ∂Ω, u≤0 in Ω; the classical weak maximum principle [F2] therefore gives max⁡Ω‾u=max⁡∂Ωu=0, and the values u(re1)=r2−1↑0 as r↑1 give sup⁡Ωu=ess sup⁡Ωu=0 by continuity.

2.1step 1.1F1F3F4

u is a weak subsolution. Since L0u=−Δu=−4, [F3] (or, equivalently, the direct integration by parts of [F4]) gives a0(u,v)=∫Ω(−4)v dx=−4∫Ωv dx≤0 for every nonnegative v∈H01(Ω), where a0(w,v)=∫Ω∇w⋅∇v dx; moreover u∈H01(Ω) with u≤0, so (u−0)+=0∈H01(Ω) and sup⁡∂Ωu+=0 in the boundary-order convention of Weak subsolutions and supersolutions of a divergence-form equation. Thus u is a weak subsolution of L0u=0 with zero positive boundary supremum.

3.1step 1.1step 2.1F2F5∎

Agreement of the two principles. Applying [F5] to the weak subsolution of step 2.1 gives ess sup⁡Ωu≤sup⁡∂Ωu+=0, which agrees with the value 0 computed in step 1.1; the approaching boundary values and continuity in that step supply the reverse inequality, and the example uses only the explicit polynomial, the two maximum principles and the classical-to-weak consistency.

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Measurable coefficients with a Holder-regular weak solution

Example

Example. On the annulus Ω={x∈R2:14<∣x∣<1} let a(x)={1,∣x∣≤12,4,∣x∣>12,u(x)={log⁡∣x∣,14<∣x∣≤12,14log⁡∣x∣−34log⁡2,12<∣x∣<1. Then a is measurable, bounded and uniformly elliptic on Ω with θ=1 and Ma=4, and u∈H1(Ω)∩C0,1(Ω) is continuous across ∣x∣=12 but has a discontinuous radial derivative there (it drops from 2 to 12), so u∉C1(Ω). The a.e. flux a∇u has the smooth representative x/∣x∣2 on Ω, which is divergence-free, and it realizes u as a weak solution of −div⁡(a∇u)=0 in the local sense of Local weak solutions of a divergence-form operator. The coefficient is not continuous, yet u is Holder continuous of every exponent α<1, in accordance with De Giorgi-Nash interior Holder regularity for divergence-form equations; the example also shows that this conclusion cannot be improved to C1.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the annulus Ω={x∈R2:14<∣x∣<1}; the radial coefficient a equal to 1 for ∣x∣<12 and 4 for ∣x∣>12; and the radial function u defined by the two displayed formulas.

[F1]

Uniform ellipticity and boundedness: a is measurable, 1≤a≤4 on Ω, and the matrix A=a Id satisfies ∣ξ∣2≤⟨A(x)ξ,ξ⟩≤16∣ξ∣2 for all ξ∈R2, so the ellipticity constant is θ=1 and the coefficient bound is Ma=4 in the convention of Uniformly elliptic divergence-form operators and their sesquilinear forms (The notation Hk and the reserved zero-boundary symbol).

[F2]

Regularity of the pieces: on each of the open annuli U1={14<∣x∣<12} and U2={12<∣x∣<1} the function u is smooth and radial, with ∇u=1r∂u∂rx and ∂ru=1/r on U1, ∂ru=1/(4r) on U2; the glued function lies in H1(Ω) with these a.e. gradients. Indeed u=G(∣x∣), where G(t)=−log⁡4 for t≤1/4, G(t)=log⁡t for 1/4<t≤1/2, and G(t)=14log⁡t−34log⁡2 for t>1/2. This is a globally 4-Lipschitz scalar function. Since x↦∣x∣ is smooth on Ω‾ with gradient x/∣x∣ and belongs to H1(Ω), the Sobolev chain rule establishes the asserted membership and gradient (Chain rule for globally Lipschitz scalar maps of Sobolev functions) (Weak derivative of a locally integrable function, Holder's inequality for integrals, including the endpoint cases).

[F3]

Continuity and differentiability across the interface: at ∣x∣=12 the first formula gives log⁡12=−log⁡2 and the second gives 14log⁡12−34log⁡2=−log⁡2, so u is continuous there; the radial derivative is 2 from the inner side and 12 from the outer side, so the derivative is discontinuous and u∉C1(Ω), while Ω is bounded away from the origin in polar coordinates, so u is Lipschitz and hence Holder of every exponent α<1 (Local Hölder and scaled C-two-alpha norms on balls).

[F4]

The flux: a∇u=x/∣x∣2 a.e. on Ω, because a(r)∂ru(r)=1/r for both branches of a and of u; the field x/∣x∣2 is smooth on Ω, div⁡(x/∣x∣2)=0 there, and x/∣x∣2=∇log⁡∣x∣ (Local weak solutions of a divergence-form operator).

[F5]

Local weak solutions: u∈H1(Ω) is a local weak solution of −div⁡(a∇u)=0 when ∫Ωa(x)∇u⋅∇v dx=0 for every v∈H01(Ω); the De Giorgi-Nash theorem gives, for such a solution with measurable uniformly elliptic coefficients, a Holder representative with exponent depending only on n,θ,Ma (Local weak solutions of a divergence-form operator, De Giorgi-Nash interior Holder regularity for divergence-form equations).

Verification

1.1givenF1

The coefficient satisfies the structural hypotheses. By [F1] the coefficient is measurable, bounded by 4 and bounded below by 1, so the associated divergence-form operator with A=a Id is uniformly elliptic with θ=1 and Ma=4; in particular the hypotheses of the De Giorgi-Nash theorem are satisfied although a is not continuous.

2.1step 1.1F2F3F4F5

The flux is divergence-free and realizes the weak equation. By [F2]-[F4], a∇u=x/∣x∣2 a.e. on Ω. This smooth field has divergence 2/∣x∣2−2∣x∣2/∣x∣4=0. For v∈Cc∞(Ω), integration by parts therefore gives ∫Ωa∇u⋅∇v=0. Approximate an arbitrary v∈H01(Ω) by these compact smooth tests; Cauchy--Schwarz passes the integral because x/∣x∣2∈L2(Ω). Thus the identity holds for every H01 test; hence u is a local weak solution of −div⁡(a∇u)=0 in the sense of [F5].

3.1step 2.1F3F5∎

The conclusions about regularity. Since u is Lipschitz on Ω by [F3], it is Holder continuous of every exponent α<1, consistently with the De Giorgi-Nash conclusion but with no C1 regularity: the radial derivative jumps from 2 to 12 at ∣x∣=12, so u∉C1(Ω); the example therefore exhibits a weak solution whose regularity comes from the structure constants alone, while the measurable coefficient fails to be continuous. All verifications use the explicit formulas and the cited interface items, with no choice principle beyond the declared Axiom of Choice and Countable Choice.

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The weak maximum principle needs the zero-order sign condition

Statement refuted

Statement refuted. Let L be a uniformly elliptic divergence-form operator on a bounded smooth domain with bounded coefficients. Without a sign condition on its zero-order coefficient, every weak subsolution of Lu=0 satisfies ess sup⁡Ωu≤sup⁡∂Ωu+, where boundary order means (u−k)+∈H01(Ω).

Counterexample. Assume the Axiom of Choice and Countable Choice. Let Ω=Bπ(0)⊂R3, aij=δij, b=0, c=−1, and L=−Δ−1 in the convention of Uniformly elliptic divergence-form operators and their sesquilinear forms. Set u(x)=sin⁡∣x∣/∣x∣ for x≠0 and u(0)=1. Then u∈C∞(Ω‾)∩H01(Ω), Lu=0, u>0 in Ω and u=0 on ∂Ω. Thus u is a weak solution and a weak subsolution, but ess sup⁡Ωu=1>0=sup⁡∂Ωu+. The sign condition in Weak maximum principle for coercive divergence-form equations fails: for every nonzero nonnegative ζ∈Cc∞(Ω), ∫Ωcζ dx<0.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the ball Ω=Bπ(0)⊂R3; the coefficients aij=δij, bi=0, c=−1; and the function u(x)=sin⁡∣x∣/∣x∣ for x≠0, u(0)=1.

[F1]

The operator L=−Δ−1 is the divergence-form operator with aij=δij, b=0, c=−1 in the convention of Uniformly elliptic divergence-form operators and their sesquilinear forms; the ball Bπ(0) is a bounded C∞ domain with trace T (Bounded C^k domains and boundary charts, Weak subsolutions and supersolutions of a divergence-form equation).

[F2]

Assume the Axiom of Choice and Countable Choice. Classical-to-weak consistency: if Ω is a bounded C1 domain, u∈C2(Ω‾)∩H01(Ω) and Lu=f with f∈C(Ω‾), then u is a weak solution of Lu=f in the sense of Weak Dirichlet solutions for a divergence-form operator, i.e. a(u,v)=∫Ωfv‾ dx for every v∈H01(Ω) (Classical solutions satisfy the weak formulation).

[F3]

Assume the Axiom of Choice. {w∈H1(Ω):Tw=0}=H01(Ω) (The kernel of the trace is the closure of the test functions); consequently a class in C(Ω‾)∩H1(Ω) vanishing on ∂Ω has zero trace and belongs to H01(Ω).

[F4]

The smooth radial function r↦sin⁡r/r on (0,∞) has the convergent power series ∑j≥0(−1)jr2j/(2j+1)!, so u extends to a C∞ function on R3 with u(0)=1 (The notation Hk and the reserved zero-boundary symbol for the Sobolev class notation).

Counterexample

1.1givenF4algebra

The function u is smooth and solves the equation classically. Because the power series ∑j≥0(−1)jr2j/(2j+1)! converges everywhere, u is the C∞ function x↦∑j≥0(−1)j∣x∣2j/(2j+1)! on R3, with u(x)>0 for ∣x∣<π and u=0 on the sphere ∣x∣=π. On r>0 one computes u′(r)=cos⁡r/r−sin⁡r/r2 and u′′(r)=−sin⁡r/r−2cos⁡r/r2+2sin⁡r/r3, hence the radial Laplacian satisfies u′′(r)+2u′(r)/r=−sin⁡r/r=−u(r); by continuity this identity Δu=−u holds on all of Ω, and Lu=−Δu−u=0 there.

2.1step 1.1F1F2F3

The boundary conditions and the weak equation. Since u is smooth on the closed ball and u=0 on ∂Ω, its trace vanishes; by [F3] u∈H01(Ω), and with u∈C2(Ω‾) and f=0∈C(Ω‾), [F2] exhibits u as a weak solution of Lu=0 in the sense of Weak subsolutions and supersolutions of a divergence-form equation.

3.1step 2.1F1algebra

The supremum is larger than the boundary supremum. As a weak solution u is also a weak subsolution; since u(x)>0 for ∣x∣<π and u(0)=1, while u(x)<1 for every x≠0 (because sin⁡r<r on (0,π)), the essential supremum is ess sup⁡Ωu=1. On the boundary u=0≥0 is nonnegative, so u+=u and (u−0)+=u∈H01(Ω); thus sup⁡∂Ωu+=0 in the boundary-order convention of [F1]. Hence ess sup⁡Ωu=1>0=sup⁡∂Ωu+, and the refuted statement fails for this weak subsolution.

4.1step 1.1step 3.1F1∎

The sign condition fails. For every nonnegative ζ∈Cc∞(Ω) with ζ≢0, the sign functional is ∫Ω(cζ+biDiζ)dx=−∫Ωζ dx<0, so the weak sign condition required by Weak maximum principle for coercive divergence-form equations does not hold. This is a direct adaptation of the adverse-zero-order obstruction in [S] Section II.2; the radial eigenfunction and its weak verification are computed here, and the example uses only the explicit function, the trace theorem and the classical-to-weak consistency, so no choice principle beyond the declared Axiom of Choice and Countable Choice is used.

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The Harnack inequality requires nonnegativity

Statement refuted

Statement refuted. There is a positive constant C such that every weak solution u∈H1(B1(0);R) of −Δu=0 on the unit ball B1(0)⊂R2 satisfies sup⁡B1/2(0)u≤Cinf⁡B1/2(0)u.

Counterexample. Take u(x)=x1, the first coordinate. Then u is harmonic, hence a weak solution of −Δu=0, but sup⁡B1/2(0)u=12,inf⁡B1/2(0)u=−12, so sup⁡≤Cinf⁡ fails for every positive constant C: the right-hand side is negative while the left-hand side is 12. The nonnegativity hypothesis in Harnack inequality for nonnegative weak solutions cannot be omitted; the theorem assumes a nonnegative class on its domain.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the unit ball B1(0)⊂R2; the linear function u(x)=x1.

[F1]

Harmonic linear functions are weak solutions: Δx1=0 classically, so ∫B1∇u⋅∇v dx=0 for every v∈H01(B1) by the divergence theorem, and u is a local weak solution of −Δu=0 in the sense of Local weak solutions of a divergence-form operator with the coefficients of Uniformly elliptic divergence-form operators and their sesquilinear forms (aij=δij, b=c=0).

[F2]

On the open half-ball B1/2(0), the values u(x)=x1 approach 1/2 along xj=(1/2−1/j,0) and −1/2 along yj=(−1/2+1/j,0) for j>2. Thus sup⁡B1/2u=1/2 and inf⁡B1/2u=−1/2, although neither boundary value is attained; by continuity these also equal the essential extrema (The average of a locally integrable function over a Euclidean ball, The essential supremum of a measurable function with respect to a measure).

[F3]

The Harnack statement: for a nonnegative weak solution of L0u=−F one has ess sup⁡BR/2u≤C(ess inf⁡BR/2u+R2−n/q∥F∥Lq(B2R)); the sign hypothesis is used in the proof through the test functions with uβ for negative exponents and through the weak Harnack inequality (Harnack inequality for nonnegative weak solutions).

Counterexample

1.1givenF1

The linear function is a weak solution. By [F1] u(x)=x1 is harmonic on B1(0) and hence a weak solution of −Δu=0 in the local sense; in particular it belongs to H1(B1(0)) and is smooth.

2.1step 1.1F2

The extrema have opposite signs. By [F2], sup⁡B1/2(0)u=12 and inf⁡B1/2(0)u=−12; therefore for every positive constant C one has sup⁡B1/2(0)u=12>−12C=Cinf⁡B1/2(0)u, so no positive constant satisfies the claimed comparison.

3.1step 2.1F2F3algebra∎

The nonnegativity hypothesis is essential. The function takes both positive and negative values on the half-ball: by [F2], its supremum is 1/2 and its infimum is −1/2. For every positive Harnack constant C, Cinf⁡B1/2u=−C/2<1/2=sup⁡B1/2u, so the displayed comparison fails. The theorem uses nonnegativity in the weak-Harnack argument [F3]. All verifications use the explicit linear function, with no choice principle beyond the declared Axiom of Choice and Countable Choice.

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Worked oscillation decay and its Hölder modulus

Example

Example. Let n≥1, let Ω⊆Rn be open and let u:Ω→R have finite oscillation on every compactly contained ball and satisfy osc⁡Br(x)u≤θosc⁡B2r(x)u for all B2r(x)⋐Ω, in the setting of Geometric oscillation decay implies a Hölder modulus.

  1. If θ=12, then α0=1 and the lemma uses the capped exponent α=12 to give ∣u(x)−u(y)∣≤2 (∣x−y∣/R)1/2osc⁡BR(x0)u for x,y∈BR/2(x0).
  2. If θ=2−1/3, then α0=α=13 and ∣u(x)−u(y)∣≤41/3(∣x−y∣/R)1/3osc⁡BR(x0)u; smaller exponents have the corresponding constant 4α′.

Facts & Assumptions

Given: An open set Ω⊆Rn, a function u:Ω→R with finite oscillation on every compactly contained ball satisfying osc⁡Br(x)u≤θosc⁡B2r(x)u whenever B2r(x)⋐Ω, and the two values θ=12 and θ=2−1/3.

[F1]

Oscillation-to-modulus conversion: under the stated finite-oscillation hypothesis, put α0=log⁡(1/θ)/log⁡2 and α=min⁡{α0,1/2}; then ∣u(x)−u(y)∣≤4α(∣x−y∣/R)αosc⁡BR(x0)u for all x,y∈BR/2(x0), with 4α′ for every 0<α′<α (Geometric oscillation decay implies a Hölder modulus).

[F2]

Real powers with positive base satisfy 2−a=1/2a, log⁡(2a)=alog⁡2 and log⁡(1/θ)=−log⁡θ for θ∈(0,1) (Real powers for positive bases, with the zero-base positive-exponent convention); the Hölder seminorm on balls is defined as in Local Hölder and scaled C-two-alpha norms on balls.

[F3]

The De Giorgi oscillation reduction produces a ratio θ∈(0,1) on balls satisfying its doubled-ball condition. Reserving this interior margin permits dyadic iteration; the resulting power-law conversion is the one illustrated in [F1] (De Giorgi oscillation reduction: one half-level set is small).

Verification

1.1givenF1F2algebra

The case θ=1/2. Since 1/θ=2, α0=log⁡2/log⁡2=1 and the capped exponent is α=1/2; [F1] gives ∣u(x)−u(y)∣≤2 (∣x−y∣/R)1/2osc⁡BR(x0)u for x,y∈BR/2(x0).

2.1step 1.1givenF1F2F3algebra∎

The case θ=2−1/3. Here 1/θ=21/3, so α0=log⁡(21/3)/log⁡2=13=α; [F1] gives ∣u(x)−u(y)∣≤41/3(∣x−y∣/R)1/3osc⁡BR(x0)u on BR/2(x0), so u is 13-Hölder there with the displayed constant. For the De Giorgi reduction, [F3] records the extra interior margin before the analogous dyadic conversion is used.

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Degenerate ellipticity allows nonconstant solutions with interior zero sets

Statement refuted

Statement refuted. The strong maximum principle of Strong maximum principle for weak elliptic solutions holds for every divergence-form equation with bounded measurable symmetric coefficient matrix that is positive semidefinite, ∑i,j=1naij(x)ξiξj≥0 for a.e. x∈Ω and all ξ∈Rn: degeneracy of the coefficients does not affect the conclusion.

Counterexample. On Ω=B1(0)⊂R3 take A=diag⁡(1,0,0) (so the matrix is positive semidefinite but degenerate: the uniform ellipticity inequality fails for ξ=e2) and u(x)=x2+:=max⁡{x2,0}. Then u is Lipschitz, nonnegative and nonconstant on Ω, and its zero set contains the lower half-ball {x2<0} of positive measure. Its weak gradient is (0,1{x2>0},0), so A∇u=0 a.e. and the weak identity ∫B1A∇u⋅∇v dx=0 holds for every v∈H01(B1) — the integral identity of Local weak solutions of a divergence-form operator applied to the degenerate matrix. The analogue of Strong maximum principle for weak elliptic solutions fails: uniform ellipticity is a hypothesis of the theorem.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the unit ball B1(0)⊂R3; the coefficient matrix A=diag⁡(1,0,0); and the function u(x)=x2+.

[F1]

The coefficient matrix is bounded, measurable, symmetric and positive semidefinite, but not uniformly elliptic: ξTAξ=ξ12≥0 for all ξ, while for ξ=e2 one has ξTAξ=0<θ∣ξ∣2 for every θ>0, so the ellipticity hypothesis of Uniformly elliptic divergence-form operators and their sesquilinear forms fails (The notation Hk and the reserved zero-boundary symbol).

[F2]

The function u(x)=x2+ is Lipschitz and nonnegative on B1(0)⊂R3, nonconstant, and its zero set contains the lower half-ball {x∈B1(0):x2<0}, an open interior set of positive measure; moreover u(x)>0 on the upper half-ball (Weak derivative of a locally integrable function, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[F3]

The weak gradient of u is Du=(0,1{x2>0},0) a.e. on B1(0)⊂R3, so u∈H1(B1(0)). For A=diag⁡(1,0,0) one has ADu=0 a.e., hence the degenerate form a0(u,v)=∫B1ADu⋅Dv dx is zero for every v∈H01(B1) (Weak derivative of a locally integrable function, Local weak solutions of a divergence-form operator).

[F4]

The conclusion that fails: for uniformly elliptic coefficients, a nonnegative weak solution of L0u=0 on a connected open set is either zero a.e. or strictly positive a.e., and its continuous representative has no interior zero unless it vanishes identically (Strong maximum principle for weak elliptic solutions, Zero-set propagation for a nonnegative Holder weak solution).

Counterexample

1.1givenF1F2F3

The degenerate matrix and the function. By [F1] the matrix satisfies the stated boundedness and positive-semidefiniteness but violates uniform ellipticity, and by [F2] the function u(x)=x2+ is nonnegative, nonconstant and has a half-ball of interior zeros; by [F3] its weak gradient is (0,1{x2>0},0) and ADu=0, so u∈H1(B1(0)).

2.1step 1.1F3

The weak identity holds. For every v∈H01(B1(0)), the matrix-vector product is ADu=0 a.e.; therefore ∫B1(0)ADu⋅Dv dx=0 directly. No distributional derivative of a sign function is involved. Hence u is a weak solution of the degenerate equation in the integral sense of Local weak solutions of a divergence-form operator.

3.1step 2.1F4∎

The strong maximum principle fails. The function of step 1.1 is a nonnegative nonconstant weak solution of the degenerate equation whose zero set contains a half-ball of positive measure, so its a.e.-class is neither zero nor strictly positive and its representative has interior zeros; this contradicts the conclusion of [F4] for uniformly elliptic coefficients. The example therefore shows that uniform ellipticity cannot be dropped from Strong maximum principle for weak elliptic solutions and from the De Giorgi-Nash machinery of the page. All verifications use the explicit function and the cited definitions, with no choice principle beyond the declared Axiom of Choice and Countable Choice.

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The Harnack estimate needs an additive forcing term

Statement refuted

Statement refuted. For every n≥1, every nonnegative weak solution u∈H1(B1(0)⊂Rn) of −Δu=f with f∈L∞(B1(0)) satisfies the forcing-free comparison sup⁡B1/2(0)u≤Cinf⁡B1/2(0)u with a constant C independent of f and u.

Counterexample. For any n≥1, define u(x)=∣x∣2/(2n) and f≡−1 on Ω=B3(0)⊂Rn, and restrict them to B1(0) for the refuted estimate. Then −Δu=f classically, u≥0, and on B1/2(0) the infimum is 0 while the supremum 1/(8n)>0 is approached as ∣x∣↑1/2. Hence the forcing-free comparison fails for every constant C; the additive term in Harnack inequality for nonnegative weak solutions and Weak Harnack inequality for nonnegative supersolutions cannot be omitted.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an integer n≥1; Ω=B3(0)⊂Rn; the function u(x)=∣x∣2/(2n); and the constant source f≡−1.

[F1]

Elementary differentiation: Diu=xi/n and Δu=2n/(2n)=1, so −Δu=−1=f classically on B3(0); the Laplacian is the operator −Δ with aij=δij, b=c=0 in the convention of Uniformly elliptic divergence-form operators and their sesquilinear forms (Local weak solutions of a divergence-form operator).

[F2]

On the open half-ball B1/2(0), u has infimum 0 attained at the origin, while its supremum 1/(8n) is approached along xj=(1/2−1/j)e0 for j>2 and is not attained; continuity makes this equal to the essential supremum (The average of a locally integrable function over a Euclidean ball, The essential supremum of a measurable function with respect to a measure).

[F3]

For n≥2, the displayed Harnack statements carry the forcing additively: ess sup⁡BR/2u≤C(ess inf⁡BR/2u+R2−n/q∥F∥Lq(B2R)) for weak solutions of L0u=−F, and the analogous bound with the same additive structure holds for nonnegative supersolutions (Harnack inequality for nonnegative weak solutions, Weak Harnack inequality for nonnegative supersolutions, Weak subsolutions and supersolutions of a divergence-form equation). The polynomial counterexample itself is valid in every n≥1 by [F1]-[F2].

Counterexample

1.1givenF1

The function solves the equation and is nonnegative. By [F1], u is smooth on B3(0) with −Δu=−1=f, so its restriction to B1(0) is a weak solution in the local sense; u≥0 and f≡−1 is bounded.

2.1step 1.1F2

The extrema on the half ball. By [F2], inf⁡B1/2(0)u=0 and sup⁡B1/2(0)u=1/(8n)>0; hence for every real constant C one has sup⁡B1/2(0)u=1/(8n)>0=C⋅0=Cinf⁡B1/2(0)u, so the forcing-free comparison sup⁡≤Cinf⁡ fails for every C.

3.1step 2.1F3∎

For n≥2, the additive term repairs the estimate and cannot be dropped. With F=−f=1, the solution is defined on Ω=B3(0), so the doubled ball B2(0)⋐Ω is admissible in [F3]. Its source norm is ∥F∥Lq(B2)=∣B2∣1/q, and the estimate with R=1 has the nonzero additive term ∣B2∣1/q; the polynomial still has zero infimum and positive supremum on B1/2. Thus no finite constant can replace the additive term by Cinf⁡. Steps 1.1–2.1 already verify the forcing-free failure for every n≥1, independently of invoking [F3]. All verifications use the explicit polynomial and the cited statements, with no choice principle beyond the declared Axiom of Choice and Countable Choice.

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The essential supremum precedes the Holder representative in De Giorgi theory

Example

Example. On Ω=B1(0)⊂R2 let u be the zero class of H1(Ω) (the class of the function that vanishes a.e.), and let u^=1{0} be the representative that equals 1 at the origin and 0 elsewhere. Then:

  1. u is a weak solution of −Δu=0 on Ω (Local weak solutions of a divergence-form operator);
  2. u^ differs from the zero function on the Lebesgue-null set {0}, so u and u^ determine the same class and the same weak derivatives (Weak differentiation ignores null-set changes);
  3. sup⁡Ωu^=1 while ess sup⁡Ωu=0 (The essential supremum of a measurable function with respect to a measure), so the pointwise supremum of an arbitrary representative is not the quantity controlled by the local boundedness estimate De Giorgi local boundedness of homogeneous subsolutions or by the Harnack bound Harnack inequality for nonnegative weak solutions. To read these class estimates as pointwise bounds, use the continuous representative produced by De Giorgi-Nash interior Holder regularity for divergence-form equations; its pointwise and essential extrema agree.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the unit disc Ω=B1(0)⊂R2; the zero class u∈H1(Ω) and the representative u^=1{0}.

[F1]

The local weak formulation: u is a local weak solution of −Δu=0 on Ω if ∫Ω∇u⋅∇v dx=0 for every v∈H01(Ω); the zero class satisfies this identically (Local weak solutions of a divergence-form operator).

[F2]

Weak derivatives depend only on the class: two Lloc1 representatives of the same class have the same weak derivatives, and the set {0} is Lebesgue-null, so u^ and the zero function determine the same class (Weak differentiation ignores null-set changes, The space Lp(μ) as the quotient by null functions).

[F3]

Essential versus pointwise suprema: the essential supremum of a class is the infimum of the essential bounds, hence ess sup⁡Ωu=0 for the zero class, whereas the pointwise supremum of the particular function u^ is sup⁡Ωu^=1 (The essential supremum of a measurable function with respect to a measure, Local Hölder and scaled C-two-alpha norms on balls).

[F4]

The estimates of the page are stated for essential extrema of classes: the local boundedness theorem bounds ess sup⁡BρRu by an Lp mean of the class, and the Harnack inequality bounds ess sup⁡BR/2u by ess inf⁡BR/2u (De Giorgi local boundedness of homogeneous subsolutions, Harnack inequality for nonnegative weak solutions, De Giorgi-Nash interior Holder regularity for divergence-form equations).

Verification

1.1givenF1

The zero class is a weak solution. For every v∈H01(Ω) one has ∫Ω∇u⋅∇v dx=0 because ∇u=0 a.e. for the zero class, so [F1] exhibits u as a local weak solution of −Δu=0 on Ω; equivalently, the classical zero solution restricted to Ω.

2.1step 1.1F2

The two representatives differ on a null set. The set {0} has Lebesgue measure zero, so u^=0 a.e. and u^ represents the class u; by [F2] u^ and the zero function have the same weak derivatives, so every weak formulation tested against u^ gives the same value as against the zero function.

3.1step 2.1F3F4∎

The suprema differ, so only the essential supremum is controlled. By [F3], ess sup⁡Ωu=0 while sup⁡Ωu^=1: the pointwise supremum of the particular representative u^ exceeds the essential supremum of the class. The local boundedness and Harnack estimates of [F4] control only essential extrema of the class, so they cannot be applied to an arbitrary pointwise representative; the class estimates give pointwise bounds for the Holder representative produced by De Giorgi-Nash interior Holder regularity for divergence-form equations, which for the zero class is the zero function and for which pointwise and essential extrema agree. All verifications use the explicit functions and the cited interface items, with no choice principle beyond the declared Axiom of Choice and Countable Choice.

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The global Harnack comparison needs connectedness

Statement refuted

Statement refuted. For every open set Ω⊆Rn (connected or not) and every nonnegative weak solution u of −Δu=0 on Ω, one has sup⁡Ωu≤Cinf⁡Ωu with a constant C depending only on Ω.

Counterexample. Let Ω=B1(0)∪B1(3e1)⊂R2, a disconnected open set, and define u=0 on B1(0) and u=1 on B1(3e1). Then u is constant on each connected component, hence a weak solution of −Δu=0 on Ω (Local weak solutions of a divergence-form operator), and it is nonnegative. But sup⁡Ωu=1 and inf⁡Ωu=0, so no finite constant C satisfies sup⁡≤Cinf⁡. The local estimate Harnack inequality for nonnegative weak solutions applies on each ball without a connectedness assumption; the two independent component values show why connectedness is necessary for comparisons across components. The cited A finite interior ball chain propagates weak Harnack bounds gives comparisons on compact connected positive-measure subsets when n≥3; it is not invoked for this two-dimensional witness and does not assert a whole-domain bound from connectedness alone.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the open set Ω=B1(0)∪B1(3e1)⊂R2 with e1=(1,0); and the function u equal to 0 on B1(0) and to 1 on B1(3e1).

[F1]

The set Ω is open as a union of open balls, and its two connected components B1(0) and B1(3e1) are disjoint because ∣3e1∣=3>2, so Ω is disconnected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[F2]

Locally constant H1 classes are weak solutions: if u∈H1(Ω) is constant on each connected component of an open set Ω, then u is locally constant on Ω, so ∇u=0 a.e. and ∫Ω∇u⋅∇v dx=0 for every v∈H01(Ω), which is the local weak formulation of −Δu=0 with aij=δij, b=c=0 (Local weak solutions of a divergence-form operator, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F3]

The extrema of u on Ω: since u takes only the values 0 and 1, sup⁡Ωu=1 and inf⁡Ωu=0 (The essential supremum of a measurable function with respect to a measure).

[F4]

Local Harnack applies in dimensions n≥2 on balls whose doubled balls are compactly contained in the domain (Harnack inequality for nonnegative weak solutions). The cited finite-chain lemma assumes n≥3 and compares extrema on compact connected positive-measure subsets of a connected open domain (A finite interior ball chain propagates weak Harnack bounds); that lemma is not applied to the present n=2 domain.

Counterexample

1.1givenF1F2

The function is a nonnegative weak solution. By [F1] the components of Ω are the two disjoint balls; u is constant on each of them, hence locally constant. It is in H1(Ω) because it is bounded on the finite-measure set Ω and its distributional gradient is zero; by [F2] it is a nonnegative weak solution of −Δu=0 on Ω.

2.1step 1.1F3

The comparison fails. By [F3], sup⁡Ωu=1 and inf⁡Ωu=0; hence for every finite constant C one has sup⁡Ωu=1>0=C⋅0=Cinf⁡Ωu, so no finite constant satisfies the claimed comparison, however the two components are normalized.

3.1step 1.1step 2.1F1F4∎

The geometric obstruction to a chain is direct. Every ball contained in Ω lies in one component: if it met both balls, the segment between such points would lie in that ball but would cross the gap outside Ω. Overlapping balls must therefore lie in the same component, and induction along any finite overlap chain prevents it from joining the two components. Local Harnack in [F4] remains valid on interior balls in either component. The finite-chain lemma is not used in dimension two, and connectedness alone is not asserted to yield a comparison over all of Ω. The contradiction in step 2.1 is already complete.

Sources