[proofplan] We derive Harnack's inequality from the Weak Harnack Inequality, which provides separate one-sided $L^p$-to-pointwise estimates for subsolutions and supersolutions. The argument has three stages: 1. **Local estimate on a single ball.** Since $u \ge 0$ solves $Lu = 0$, the function $u$ is simultaneously a subsolution and a supersolution. We apply the local boundedness estimate (subsolution bound) to control $\sup_{B_R} u$ from above, and the Weak Harnack Inequality (supersolution bound) to control $\inf_{B_R} u$ from below. Combining these two one-sided bounds on concentric balls yields a local Harnack inequality: $\sup_{B_R} u \le C \inf_{B_R} u$ whenever $B_{4R} \subset U$. 2. **Ball-chaining from $V$ to cover the compact set.** We construct a finite chain of overlapping balls connecting any two points of $V$ while staying inside $U$, and iterate the local Harnack estimate along the chain. 3. **Geometric control of the chain length.** The number of balls in the chain depends only on $n$, $\operatorname{dist}(V, \partial U)$, and $\operatorname{diam}(V)$, so the final constant is uniform over $V$. [/proofplan] [step:Local Boundedness Estimate for Subsolutions] We recall the local boundedness estimate for subsolutions (a consequence of Moser iteration applied to $u$ tested against powers of itself with cut-off functions). **Local Boundedness Estimate.** Let $B_{4R}(x_0) \subset U$ and suppose $u \in H^1(B_{4R}(x_0))$ satisfies $u \ge 0$ and $Lu \le 0$ weakly in $B_{4R}(x_0)$. Then for every $p > 0$, there exists a constant $C_1 = C_1(n, \theta, \Lambda, p) > 0$ such that \begin{align*} \sup_{B_R(x_0)} u \le C_1 \left(\frac{1}{\mathcal{L}^n(B_{2R}(x_0))} \int_{B_{2R}(x_0)} u^p \, d\mathcal{L}^n\right)^{1/p}. \end{align*} This estimate converts an $L^p$ average of $u$ over the larger ball $B_{2R}(x_0)$ into a pointwise supremum bound on the smaller ball $B_R(x_0)$. The key feature is that it holds for every $p > 0$, including small values of $p$. We apply this with $p$ chosen to match the exponent provided by the Weak Harnack Inequality in the next step. [/step] [step:Weak Harnack Inequality for Supersolutions] We recall the Weak Harnack Inequality for supersolutions (established via Moser iteration on negative powers of $u$, combined with the John-Nirenberg lemma). **Weak Harnack Inequality.** Let $B_{4R}(x_0) \subset U$ and suppose $u \in H^1(B_{4R}(x_0))$ satisfies $u \ge 0$ and $Lu \ge 0$ weakly in $B_{4R}(x_0)$. Then there exist $p_0 = p_0(n, \theta, \Lambda) > 0$ and $C_2 = C_2(n, \theta, \Lambda) > 0$ such that \begin{align*} \left(\frac{1}{\mathcal{L}^n(B_{2R}(x_0))} \int_{B_{2R}(x_0)} u^{p_0} \, d\mathcal{L}^n\right)^{1/p_0} \le C_2 \inf_{B_R(x_0)} u. \end{align*} This is the deeper of the two estimates: it asserts that the $L^{p_0}$ average of a non-negative supersolution controls the pointwise infimum from above, preventing the function from being large on average while vanishing at a point. [/step] [step:Local Harnack Inequality on a Single Ball] Since $u \ge 0$ satisfies $Lu = 0$ in $U$, the function $u$ is both a subsolution ($Lu \le 0$) and a supersolution ($Lu \ge 0$). Fix a ball $B_{4R}(x_0) \subset U$. We apply the Local Boundedness Estimate from Step 1 with $p = p_0$ (the exponent from the Weak Harnack Inequality), and the Weak Harnack Inequality from Step 2, both on the same concentric balls $B_R(x_0) \subset B_{2R}(x_0) \subset B_{4R}(x_0)$. From the Local Boundedness Estimate (Step 1) applied to the subsolution $u$ with $p = p_0$: \begin{align*} \sup_{B_R(x_0)} u \le C_1 \left(\frac{1}{\mathcal{L}^n(B_{2R}(x_0))} \int_{B_{2R}(x_0)} u^{p_0} \, d\mathcal{L}^n\right)^{1/p_0}. \end{align*} From the Weak Harnack Inequality (Step 2) applied to the supersolution $u$: \begin{align*} \left(\frac{1}{\mathcal{L}^n(B_{2R}(x_0))} \int_{B_{2R}(x_0)} u^{p_0} \, d\mathcal{L}^n\right)^{1/p_0} \le C_2 \inf_{B_R(x_0)} u. \end{align*} Substituting the second inequality into the first eliminates the $L^{p_0}$ integral entirely: \begin{align*} \sup_{B_R(x_0)} u \le C_1 \cdot C_2 \inf_{B_R(x_0)} u. \end{align*} Setting $C_0 := C_1 \cdot C_2$, we obtain the **local Harnack inequality**: for every ball $B_{4R}(x_0) \subset U$, \begin{align*} \sup_{B_R(x_0)} u \le C_0 \inf_{B_R(x_0)} u, \end{align*} where $C_0 = C_0(n, \theta, \Lambda)$ is independent of $u$, $x_0$, and $R$. [/step] [step:Ball-Chaining Construction] We now pass from the local estimate on a single ball to the global estimate on the compact set $V \subset\subset U$. The geometric construction proceeds as follows. **Setup.** Define \begin{align*} \delta := \operatorname{dist}(V, \partial U) > 0, \qquad D := \operatorname{diam}(V). \end{align*} Since $V \subset\subset U$, we have $\delta > 0$. Set the radius \begin{align*} R := \frac{\delta}{8}. \end{align*} For every $x_0 \in V$, the ball $B_{4R}(x_0)$ has radius $4R = \delta/2$. Every point $z \in B_{4R}(x_0)$ satisfies $\operatorname{dist}(z, \partial U) \ge \operatorname{dist}(x_0, \partial U) - |z - x_0| \ge \delta - \delta/2 = \delta/2 > 0$, so $B_{4R}(x_0) \subset U$. In particular, the local Harnack inequality from Step 3 applies at every center in $V$. **Constructing the finite cover.** Since $V$ is compact, finitely many balls $B_R(w_1), \ldots, B_R(w_m)$ with centers $w_j \in V$ cover $V$. Since $U$ is connected and open, for any two centers $w_i, w_j$, there is a continuous path $\gamma_{ij} : [0,1] \to U$ from $w_i$ to $w_j$. The image $\gamma_{ij}([0,1])$ is compact in $U$, so $\operatorname{dist}(\gamma_{ij}([0,1]), \partial U) > 0$. Since there are only finitely many pairs $(i,j)$ with $1 \le i, j \le m$, we define \begin{align*} \delta_0 := \min_{1 \le i < j \le m} \operatorname{dist}(\gamma_{ij}([0,1]),\, \partial U) > 0. \end{align*} Redefine $R := \min(\delta, \delta_0) / 8$. With this choice, $B_{4R}(z) \subset U$ for every $z$ lying on any of the connecting paths $\gamma_{ij}$. **Extracting a chain between two centers.** Fix any two cover centers $w_i, w_j$. Along the path $\gamma_{ij}$, we extract a chain of points. Set $z_1 := w_i$. Given $z_k \in \gamma_{ij}([0,1])$, define $t_k := \sup\{t : \gamma_{ij}(t) \in \overline{B_R(z_k)}\}$. If $\gamma_{ij}(t_k) = w_j$, the chain terminates with $z_k$ as the last center. Otherwise, set $z_{k+1} := \gamma_{ij}(t_k)$, so $|z_{k+1} - z_k| = R$. The chain terminates after at most $N_{ij} \le \ell(\gamma_{ij}) / R + 1$ steps, where $\ell(\gamma_{ij})$ denotes the arc length. Since each path $\gamma_{ij}$ is continuous on $[0,1]$ and hence rectifiable (after possible reparametrization to a Lipschitz path, which is always possible in an open connected subset of $\mathbb{R}^n$), the length $\ell(\gamma_{ij})$ is finite. Setting \begin{align*} N := \max_{1 \le i < j \le m} N_{ij}, \end{align*} we obtain a uniform bound on the chain length, depending only on the paths $\gamma_{ij}$ (which depend on $V$, $U$, and $n$), but not on $u$. **Overlap property.** By construction, consecutive chain centers $z_k, z_{k+1}$ satisfy $|z_{k+1} - z_k| = R$, so $z_{k+1} \in B_R(z_k)$ and $B_R(z_{k+1}) \cap B_R(z_k) \neq \varnothing$. Since $\operatorname{dist}(z_k, \partial U) \ge 4R$ (by the choice of $R$), the ball $B_{4R}(z_k) \subset U$, and the local Harnack inequality from Step 3 applies on each $B_R(z_k)$. [/step] [step:Iterating Along the Chain] Fix arbitrary $x, y \in V$. By the covering from Step 4, there exist indices $i, j$ with $x \in B_R(w_i)$ and $y \in B_R(w_j)$. Let $z_1 = w_i, z_2, \ldots, z_K = w_j$ be the chain of centers along $\gamma_{ij}$ extracted in Step 4, where $K \le N$. At each link of the chain, both $z_k$ and $z_{k+1}$ belong to $B_R(z_k)$ (since $|z_{k+1} - z_k| = R$). The local Harnack inequality $\sup_{B_R(z_k)} u \le C_0 \inf_{B_R(z_k)} u$ therefore gives \begin{align*} u(z_k) \le \sup_{B_R(z_k)} u \le C_0 \inf_{B_R(z_k)} u \le C_0\, u(z_{k+1}). \end{align*} Chaining these $K - 1$ inequalities: \begin{align*} u(w_i) = u(z_1) \le C_0\, u(z_2) \le C_0^2\, u(z_3) \le \cdots \le C_0^{K-1}\, u(z_K) = C_0^{K-1}\, u(w_j). \end{align*} Since $x \in B_R(w_i)$, one application of the local Harnack at $B_R(w_i)$ gives $u(x) \le \sup_{B_R(w_i)} u \le C_0 \inf_{B_R(w_i)} u \le C_0\, u(w_i)$. Similarly, $y \in B_R(w_j)$ gives $u(w_j) \le \sup_{B_R(w_j)} u \le C_0 \inf_{B_R(w_j)} u \le C_0\, u(y)$. Combining these three estimates: \begin{align*} u(x) \le C_0\, u(w_i) \le C_0^{K}\, u(w_j) \le C_0^{K+1}\, u(y) \le C_0^{N+1}\, u(y). \end{align*} Since $x$ and $y$ were arbitrary points in $V$, we take the supremum over $x \in V$ and the infimum over $y \in V$: \begin{align*} \sup_V u \le C_0^{N+1} \inf_V u. \end{align*} [/step] [step:Uniformity of the Constant] Set $C := C_0^{N+1}$. It remains to verify that $C$ depends only on the quantities listed in the theorem statement, and not on $u$ or the particular points $x, y$. The local Harnack constant $C_0 = C_0(n, \theta, \Lambda)$ depends only on the dimension, ellipticity ratio, and coefficient bounds (as established in Step 3). The chain length bound $N$ was constructed in Step 4 as the maximum over finitely many chains connecting pairs of cover centers $w_i, w_j \in V$. The cover $\{B_R(w_j)\}_{j=1}^m$ and the connecting paths $\{\gamma_{ij}\}$ depend only on the geometry of $V$ and $U$: specifically on $\operatorname{diam}(V)$, $\operatorname{dist}(V, \partial U)$, and $n$ (through the covering number $m$ and the arc lengths of the connecting paths). Neither $N$ nor $m$ depends on $u$, $x$, or $y$. In particular, $N$ can be bounded explicitly. The covering number satisfies $m \le (D/R + 1)^n$ where $D = \operatorname{diam}(V)$, and each chain has length at most $\ell(\gamma_{ij})/R + 1$. Since each connecting path can be chosen with length $\ell(\gamma_{ij}) \lesssim D + \delta$ (where $\delta = \operatorname{dist}(V, \partial U)$ and the implicit constant depends on $n$ and the topology of $U$), we obtain \begin{align*} C = C_0^{N+1}, \end{align*} which depends only on $n$, $\theta$, $\Lambda$, $\operatorname{diam}(V)$, and $\operatorname{dist}(V, \partial U)$, completing the proof. $\blacksquare$ [/step]