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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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