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The global Harnack comparison needs connectedness

Statement refuted

Statement refuted. For every open set $Ω \subseteq R^{n}$ (connected or not) and every nonnegative weak solution $u$ of $- Δ u = 0$ on $Ω$ , one has $\sup_{Ω} u \leq C \inf_{Ω} u$ with a constant $C$ depending only on $Ω$ .

Counterexample. Let $Ω = B_{1} (0) \cup B_{1} (3 e_{1}) \subset R^{2}$ , a disconnected open set, and define $u = 0$ on $B_{1} (0)$ and $u = 1$ on $B_{1} (3 e_{1})$ . Then $u$ is constant on each connected component, hence a weak solution of $- Δ u = 0$ on $Ω$ (Local weak solutions of a divergence-form operator), and it is nonnegative. But $\sup_{Ω} u = 1$ and $\inf_{Ω} u = 0$ , so no finite constant $C$ satisfies $\sup \leq C \inf$ . 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 $n \geq 3$ ; 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 $Ω = B_{1} (0) \cup B_{1} (3 e_{1}) \subset R^{2}$ with $e_{1} = (1, 0)$ ; and the function $u$ equal to $0$ on $B_{1} (0)$ and to $1$ on $B_{1} (3 e_{1})$ .

[F1]

The set $Ω$ is open as a union of open balls, and its two connected components $B_{1} (0)$ and $B_{1} (3 e_{1})$ are disjoint because $∣ 3 e_{1} ∣ = 3 > 2$ , so $Ω$ is disconnected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[F2]

Locally constant $H^{1}$ classes are weak solutions: if $u \in H^{1} (Ω)$ is constant on each connected component of an open set $Ω$ , then $u$ is locally constant on $Ω$ , so $\nabla u = 0$ a.e. and $\int_{Ω} \nabla u \cdot \nabla v d x = 0$ for every $v \in H_{0}^{1} (Ω)$ , which is the local weak formulation of $- Δ u = 0$ with $a^{i j} = δ^{i j}$ , $b = c = 0$ (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 $1$ , $\sup_{Ω} u = 1$ and $\inf_{Ω} u = 0$ (The essential supremum of a measurable function with respect to a measure).

[F4]

Local Harnack applies in dimensions $n \geq 2$ on balls whose doubled balls are compactly contained in the domain (Harnack inequality for nonnegative weak solutions). The cited finite-chain lemma assumes $n \geq 3$ 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 $H^{1} (Ω)$ 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 $- Δ u = 0$ on $Ω$ .

2.1step 1.1F3

The comparison fails. By [F3], $\sup_{Ω} u = 1$ and $\inf_{Ω} u = 0$ ; hence for every finite constant $C$ one has $\sup_{Ω} u = 1 > 0 = C \cdot 0 = C \inf_{Ω} u$ , 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.

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Sources

Brian Krummel, Consequences of De Giorgi-Nash-Moser (4 March 2016; complete 7-page notes) (standard reference, not scraped)
Leon Simon, Lectures on Partial Differential Equations (Stanford University; complete author scan, 118 sheets reproducing the 223 printed pages of the manuscript, two logical pages per sheet) (standard reference, not scraped)