A finite interior ball chain propagates weak Harnack bounds

Given: Countable Choice and the Axiom of Choice; a connected open set $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$; uniformly elliptic measurable symmetric coefficients $A$ with constants $\theta ,M_a$; a source $F\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^q(\mathrm{\Omega })$, $q>n\mathrm{/}2$; a nonnegative weak solution $u\in H^1(\mathrm{\Omega };\mathbb{R})$ of $L_{0}u=-F$; a compact connected set $K⋐\mathrm{\Omega }$ with positive Lebesgue measure.

Assume the Axiom of Choice. Harnack inequality on doubled balls: for every ball $B_R(x)$ with $B_{2R}(x)⋐\mathrm{\Omega }$, ${\mathrm{ess\hspace{0.17em}sup}}_{B_{R\mathrm{/}2}(x)}u\le C_1({\mathrm{ess\hspace{0.17em}inf}}_{B_{R\mathrm{/}2}(x)}u+R^{2-n\mathrm{/}q}\mathrm{\parallel }F{\mathrm{\parallel }}_{L^q(B_{2R}(x))})$ with $C_1=C_1(n,q,\theta ,M_a)$ (Harnack inequality for nonnegative weak solutions, Weak subsolutions and supersolutions of a divergence-form equation).

Assume the Axiom of Choice. Compactness and containment: since $K$ is compact and $\mathrm{\Omega }$ is open, $\mathrm{dist}(K,{\mathbb{R}}^n\setminus \mathrm{\Omega })>0$, so a finite family of balls $B_{2R_i}(x_i)⋐\mathrm{\Omega }$, $x_i\in K$, can be chosen with the half-balls $B_{R_i\mathrm{/}2}(x_i)$ covering $K$ (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Since $K$ is connected, the finite cover by the relative open sets $K\cap B_{R_i\mathrm{/}2}(x_i)$ has a connected intersection graph: otherwise the unions corresponding to two components of the graph would separate $K$. If two such relative open sets intersect, the corresponding open balls intersect in a nonempty open set and hence in a set of positive Lebesgue measure (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

Assume Countable Choice. The zero extension of $u\in L^2(\mathrm{\Omega })$ lies in $L^2({\mathbb{R}}^n)\subset L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$; applying the cited Lebesgue-point theorem and restricting to $\mathrm{\Omega }$ gives a full-measure Lebesgue set (Lebesgue points and the Lebesgue set of an $L_{loc}^1$ class, Almost every point is a Lebesgue point of a locally integrable function). At a Lebesgue point $x$ in a ball $B$, ${\mathrm{ess\hspace{0.17em}inf}}_{B}u\le u(x)\le {\mathrm{ess\hspace{0.17em}sup}}_{B}u$: if either inequality failed, the averages of $\mathrm{\mid }u(y)-u(x)\mathrm{\mid }$ over sufficiently small balls centered at $x$ would stay bounded below by a positive constant.

For two measurable balls $B_i,B_j$ with $\mathrm{\mid }{B}_i\cap B_j\mathrm{\mid }>0$, ${\mathrm{ess\hspace{0.17em}inf}}_{B_i}u\le {\mathrm{ess\hspace{0.17em}sup}}_{B_j}u$; otherwise a real number strictly between them would be both an almost-everywhere lower bound on $B_i$ and an almost-everywhere upper bound on $B_j$, impossible on their positive-measure intersection (The essential supremum of a measurable function with respect to a measure).

If $M_j\le C_1(M_{j+1}+g_j)$ for $j=1,\dots ,N-1$ and $C_1\ge 1$, then $M_1\le C_1^N(M_N+{\sum }_{j=1}^{N-1}{g}_j)$ by expanding the finite recurrence. [algebra]