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Zero-set propagation for a nonnegative Holder weak solution

Statement

Assume Countable Choice and the Axiom of Choice. Let n≥3, let Ω⊆Rn be connected and open, let A,L0 be as in De Giorgi local boundedness of homogeneous subsolutions, and let u∈H1(Ω;R) with u≥0 a.e. be a weak solution of L0u=0. Let u∗ be a continuous representative of u on Ω (such a representative exists by De Giorgi-Nash interior Holder regularity for divergence-form equations) and let x0∈Ω with u∗(x0)=0. Then u∗≡0 on Ω; equivalently the zero set {u∗=0} is both relatively open and relatively closed in Ω. The same argument shows: if u∈H1(Ω) is merely a nonnegative weak supersolution of L0u≥0 and a representative of u is continuous at a point x0 with value 0, then u=0 a.e. on a neighbourhood of x0; the global conclusion then needs a continuous representative on all of Ω.

Facts & Assumptions

Given: Countable Choice and the Axiom of Choice; a connected open set Ω⊆Rn, n≥3; uniformly elliptic measurable symmetric coefficients A with constants θ,Ma; the principal operator L0u=−Di(aijDju) with form a0; a nonnegative weak solution u∈H1(Ω;R) of L0u=0; a continuous representative u∗ of u; a point x0∈Ω with u∗(x0)=0.

[F1]

Assume the Axiom of Choice. Weak Harnack inequality at p=1: for every ball BS(y) with B2S(y)⋐Ω one has S−n∥w∥L1(BS(y))≤C1(ess inf⁡BS/2(y)w+S2−n/q∥F∥Lq(B2S(y))) for every nonnegative supersolution w of L0w=−F with F∈Llocq, q>n/2, with C1=C1(n,q,θ,Ma) (Weak Harnack inequality for nonnegative supersolutions).

[F2]

Assume the Axiom of Choice. A class in H1 with ∫B∣w∣ dx=0 on an open ball B vanishes a.e. on B; and the essential infimum of a nonnegative class over a ball is the infimum of any continuous representative over that ball, so that ess inf⁡BS/2(z)u≤u∗(x0)=0 whenever x0∈BS/2(z) and u≥0 a.e. (The space Lp(μ) as the quotient by null functions, The average of a locally integrable function over a Euclidean ball, The essential supremum of a measurable function with respect to a measure, Local Hölder and scaled C-two-alpha norms on balls).

[F3]

Assume the Axiom of Choice. Continuity and connectedness: the zero set of a continuous function is relatively closed, and a nonempty subset of a connected topological space that is both relatively open and relatively closed is the whole space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, Local Hölder and scaled C-two-alpha norms on balls).

[F4]

Assume the Axiom of Choice. Supersolution and subsolution vocabulary: a weak solution of L0u=0 is in particular a nonnegative weak supersolution of L0u≥0, and L0u=0 weakly means a0(u,v)=0 for every v∈H01(Ω) (Weak subsolutions and supersolutions of a divergence-form equation, Uniformly elliptic divergence-form operators and their sesquilinear forms).

Proof

technique · direct; the weak Harnack inequality with $p=1$ turns the vanishing of $u^*$ at a point into the vanishing of $u$ on a whole ball, which makes the zero set relatively open, while continuity makes it relatively closed; connectedness then forces the zero set to be all of $\Omega$
1.1givenF1F2

The zero set is relatively open. Fix a radius S>0 with B2S(x0)⋐Ω; such an S exists because Ω is open and x0∈Ω. Apply the weak Harnack inequality [F1] to u on the ball BS(x0) with p=1 and F=0: S−n∫BS(x0)u dx≤C1ess inf⁡BS/2(x0)u. Since x0∈BS/2(x0) and u≥0 a.e. with continuous representative u∗ vanishing at x0, [F2] gives ess inf⁡BS/2(x0)u≤u∗(x0)=0; hence ∫BS(x0)u dx=0 and therefore u=0 a.e. on BS(x0) by [F2]. Since u∗ is continuous and agrees with u a.e. on the ball BS(x0), the set where u∗≠0 is open and of measure zero in BS(x0); it must be empty, so u∗=0 on all of BS(x0).

2.1step 1.1F1F2F3F4

The zero set is relatively closed and the second assertion. The set Z:={x∈Ω:u∗(x)=0} is the preimage of the closed set {0} under the continuous map u∗, hence relatively closed in Ω by [F3]. For the second assertion, suppose only that u∈H1(Ω) is a nonnegative weak supersolution of L0u≥0 and that a representative is continuous at x0 with value 0; then the same computation with F=0 and the continuity of the representative at the single point x0 gives ∫BS(x0)u dx=0 for some S>0, hence u=0 a.e. on BS(x0), which is the local conclusion; the global conclusion needs a representative continuous on all of Ω so that [F3] applies to the whole zero set.

3.1step 2.1F3∎

Conclusion by connectedness. The set Z is nonempty (it contains x0), relatively open by step 1.1 and relatively closed by step 2.1; since Ω is connected, [F3] gives Z=Ω, that is u∗≡0 on Ω, which is the assertion. All arguments use Countable Choice and the Axiom of Choice only through the suppliers named above.

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