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Harnack inequality for nonnegative weak solutions

Statement

Assume Countable Choice and the Axiom of Choice. Let n≥2, let Ω⊆Rn be open, let A and L0 be as in De Giorgi local boundedness of homogeneous subsolutions, and let F∈Llocq(Ω) with q>n/2. Let u∈H1(Ω;R) satisfy u≥0 a.e. and be a weak solution of L0u=−F, i.e. a0(u,φ)=−∫ΩF φ dxfor every real φ∈Cc∞(Ω). Then for every ball BR(x0) with B2R(x0)⋐Ω, ess sup⁡BR/2(x0)u≤C(ess inf⁡BR/2(x0)u+R 2−n/q∥F∥Lq(B2R(x0))),C=C(n,q,θ,Ma), independent of R and x0; in the homogeneous case F=0 this is ess sup⁡BR/2u≤Cess inf⁡BR/2u, the Harnack inequality. For n=2 every finite q>1 is allowed. The two essential extrema are taken over the same ball, so no regularity of u is needed for the statement; the additive forcing term is essential and the estimate is not claimed without it.

Facts & Assumptions

Given: Countable Choice and the Axiom of Choice; an open set Ω⊆Rn, n≥2; uniformly elliptic measurable symmetric coefficients A with constants θ,Ma; the principal operator L0u=−Di(aijDju) with form a0; a source F∈Llocq(Ω), q>n/2; a nonnegative u∈H1(Ω;R) with a0(u,φ)=−∫ΩFφ dx for every real φ∈Cc∞(Ω); a ball BR(x0) with B2R(x0)⋐Ω.

[F1]

Both roles of a local solution: the identity a0(u,φ)=−∫ΩFφ against real compactly supported smooth tests gives both the subsolution inequality and the supersolution inequality for the equation L0u=−F, with the appropriate inequality directions (Weak subsolutions and supersolutions of a divergence-form equation, Uniformly elliptic divergence-form operators and their sesquilinear forms).

[F2]

Assume the Axiom of Choice. Local boundedness with a scale-correct source: for every ball BS(y)⋐Ω, every 0<ρ<1 and every p>0, ess sup⁡BρS(y)w≤C1[(1∣BS(y)∣∫BS(y)wp)1/p+S2−n/q∥f+∥Lq(BS(y))] for every nonnegative weak subsolution w of a0(w,⋅)≤∫f ⋅ dx with f∈Llocq(Ω), where C1=C1(n,q,θ,Ma,ρ,p) (De Giorgi local boundedness with a scale-correct forcing term).

[F3]

Assume the Axiom of Choice. Weak Harnack inequality: for every ball BS(y) with B2S(y)⋐Ω and every 0<p<n/(n−2) when n≥3, or every finite p>0 when n=2, S−n/p∥w∥Lp(BS(y))≤C2(ess inf⁡BS/2(y)w+S2−n/q∥G∥Lq(B2S(y))) for every nonnegative weak supersolution w of L0w=−G with G∈Llocq(Ω), where C2=C2(n,q,θ,Ma,p); the range contains s0:=min⁡{p0,1/2}, where p0=p0(n,θ,Ma)>0 is produced by the Moser iteration (Weak Harnack inequality for nonnegative supersolutions, Moser iteration for positive supersolutions: negative-power and logarithmic comparison).

[F4]

Assume the Axiom of Choice. Averaging and the elementary comparison of the negative part of the source: 1∣BS(y)∣∫BS(y)wpdx=S−n∥w∥Lp(BS(y))p/∣B1∣, and ∥(−F)+∥Lq=∥F−∥Lq≤∥F∥Lq (The average of a locally integrable function over a Euclidean ball, The space Lp(μ) as the quotient by null functions, The essential supremum of a measurable function with respect to a measure).

[F5]

Assume the Axiom of Choice. The n=2 substitute: the critical embedding W01,2↪Lκ for every finite κ replaces the 2∗ embedding in both quoted theorems (The Sobolev inequality for zero-boundary Sobolev closures on open sets, The critical Sobolev embedding into every finite Lq).

Proof

technique · direct; combine the forcing local-boundedness estimate for the subsolution role of $u$ with the weak Harnack inequality for its supersolution role, both on the same ball $B_R$, and compare the two source terms
1.1givenF1F4

The forcing source in the subsolution role. By [F1] the solution u is a nonnegative weak subsolution with source −F, whose positive part is (−F)+=F−; by [F4], ∥F−∥Lq(BR(x0))≤∥F∥Lq(B2R(x0)).

2.1step 1.1F2F3F4

Chaining local boundedness with the weak Harnack inequality. Fix s0:=min⁡{p0,1/2} from [F3], which is an admissible weak-Harnack exponent in both the n≥3 and n=2 ranges, and apply local boundedness [F2] to u on BR(x0) with ρ=1/2. Its source term satisfies R2−n/q∥F−∥Lq(BR)≤R2−n/q∥F∥Lq(B2R) by [F4]. Apply weak Harnack [F3] to the supersolution u on the same ball with p=s0 and G=F; after converting the normalized mean to the stated norm, (1∣BR∣∫BRus0)1/s0≤C2′(ess inf⁡BR/2u+R2−n/q∥F∥Lq(B2R)), where C2′=C2/∣B1∣1/s0. Substituting this bound into the local estimate gives coefficient C1C2′ on the infimum and C1(C2′+1) on the source term; thus C:=C1(C2′+1) works for both and depends only on n,q,θ,Ma.

3.1step 2.1F5algebra∎

The homogeneous case and the n=2 clause. If F=0 the same two steps give ess sup⁡BR/2u≤C1C2′(ess inf⁡BR/2u), the Harnack inequality; the extremal balls agree, so no regularity is used. For n=2 the same proof applies with [F5] in place of the 2∗ embedding in both quoted theorems and with every finite p, so the range 0<p<n/(n−2) becomes unbounded. All arguments use Countable Choice and the Axiom of Choice only through the suppliers named above.

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