[proofplan]
We prove a local comparison first: using the mean-value property and nonnegativity, any two points within distance $r = \frac{1}{4}\operatorname{dist}(V, \partial U)$ satisfy $u(x) \le 2^n u(y)$. We then chain finitely many such local comparisons along a path in the [connected](/page/Connectedness), [compact](/page/Compact%20Space) set $\overline{V}$ to obtain the global two-sided bound with constant $C = 2^{nN(V)}$, where $N(V)$ is the number of chain links.
[/proofplan]
[step:Fix an interior radius $r > 0$ depending only on $V$ and $U$]
Define
\begin{align*}
r := \tfrac{1}{4}\operatorname{dist}(V, \partial U).
\end{align*}
Since $V \Subset U$, the closure $\overline{V}$ is compact and contained in $U$, so $\operatorname{dist}(V, \partial U) > 0$. Hence $r > 0$. For every $z \in V$, the ball $B(z, 2r)$ satisfies $B(z, 2r) \subset U$, because $\operatorname{dist}(z, \partial U) \ge \operatorname{dist}(V, \partial U) = 4r > 2r$.
[/step]
[step:Establish the local comparison $u(x) \le 2^n u(y)$ for $|x - y| \le r$]
Fix $x, y \in V$ with $|x - y| \le r$. Since $B(x, 2r) \subset U$ and $u$ is [harmonic](/page/Laplace%27s%20Equation) and nonnegative in $U$, the [mean-value property](/theorems/31) applies on $B(x, r)$:
\begin{align*}
u(x) = \frac{1}{\mathcal{L}^n(B(0, r))} \int_{B(x, r)} u(z) \, d\mathcal{L}^n(z).
\end{align*}
Since $u \ge 0$ and $\mathcal{L}^n(B(0, 2r)) = 2^n \mathcal{L}^n(B(0, r))$, replacing the normalising volume by the larger quantity $\mathcal{L}^n(B(0, 2r))$ only decreases the expression:
\begin{align*}
u(x) \ge \frac{1}{\mathcal{L}^n(B(0, 2r))} \int_{B(x, r)} u(z) \, d\mathcal{L}^n(z).
\end{align*}
Because $|x - y| \le r$, the triangle inequality gives $B(y, r) \subset B(x, 2r)$. Since $u \ge 0$, enlarging the domain of integration from $B(x, r)$ to $B(x, 2r)$ and then restricting to $B(y, r)$ yields:
\begin{align*}
u(x) &\ge \frac{1}{\mathcal{L}^n(B(0, 2r))} \int_{B(x, 2r)} u(z) \, d\mathcal{L}^n(z) \ge \frac{1}{\mathcal{L}^n(B(0, 2r))} \int_{B(y, r)} u(z) \, d\mathcal{L}^n(z) \\
&= \frac{\mathcal{L}^n(B(0, r))}{\mathcal{L}^n(B(0, 2r))} \cdot \frac{1}{\mathcal{L}^n(B(0, r))} \int_{B(y, r)} u(z) \, d\mathcal{L}^n(z) = 2^{-n}\, u(y),
\end{align*}
where the last equality uses $\mathcal{L}^n(B(0, 2r)) = 2^n \mathcal{L}^n(B(0, r))$ and the mean-value property at $y$. Exchanging the roles of $x$ and $y$ gives $u(y) \ge 2^{-n} u(x)$. Therefore
\begin{align*}
2^{-n}\, u(y) \le u(x) \le 2^n\, u(y) \qquad \text{whenever } |x - y| \le r.
\end{align*}
[guided]
The goal is to show that two nearby evaluations of $u$ cannot differ by more than a factor of $2^n$. The mechanism is purely volumetric: the mean-value property expresses $u(x)$ as an average, and by inflating the averaging ball we lose only a factor equal to the volume ratio.
Concretely, the [mean-value property](/theorems/31) gives
\begin{align*}
u(x) = \frac{1}{\mathcal{L}^n(B(0, r))} \int_{B(x, r)} u(z) \, d\mathcal{L}^n(z).
\end{align*}
Since $u \ge 0$, we can replace the denominator by the larger volume $\mathcal{L}^n(B(0, 2r)) = 2^n \mathcal{L}^n(B(0, r))$ without making the fraction negative --- it only shrinks:
\begin{align*}
u(x) \ge \frac{1}{\mathcal{L}^n(B(0, 2r))} \int_{B(x, r)} u(z) \, d\mathcal{L}^n(z).
\end{align*}
Why enlarge to $2r$? Because $|x - y| \le r$ implies $B(y, r) \subset B(x, 2r)$ by the triangle inequality: for any $w \in B(y, r)$, $|w - x| \le |w - y| + |y - x| < r + r = 2r$. Since $u \ge 0$, the integral over the larger ball $B(x, 2r)$ dominates that over $B(y, r)$:
\begin{align*}
u(x) &\ge \frac{1}{\mathcal{L}^n(B(0, 2r))} \int_{B(x, 2r)} u(z) \, d\mathcal{L}^n(z) \ge \frac{1}{\mathcal{L}^n(B(0, 2r))} \int_{B(y, r)} u(z) \, d\mathcal{L}^n(z).
\end{align*}
Factoring out the volume ratio and applying the mean-value property at $y$:
\begin{align*}
u(x) \ge \frac{\mathcal{L}^n(B(0, r))}{\mathcal{L}^n(B(0, 2r))} \cdot \frac{1}{\mathcal{L}^n(B(0, r))} \int_{B(y, r)} u(z) \, d\mathcal{L}^n(z) = 2^{-n}\, u(y).
\end{align*}
Exchanging $x$ and $y$ (the argument is symmetric) gives $u(y) \ge 2^{-n} u(x)$, so
\begin{align*}
2^{-n}\, u(y) \le u(x) \le 2^n\, u(y) \qquad \text{whenever } |x - y| \le r.
\end{align*}
The factor $2^n$ is sharp for this ball-doubling argument; it comes entirely from the volume scaling $\mathcal{L}^n(B(0, 2r))/\mathcal{L}^n(B(0, r)) = 2^n$.
[/guided]
[/step]
[step:Connect arbitrary points in $V$ by a finite chain of $r$-steps]
Since $V$ is connected and open in $\mathbb{R}^n$, it is [path-connected](/page/Path%20Connectedness). Since $\overline{V}$ is compact, for any $x, y \in V$ there exists a continuous path $\gamma: [0, 1] \to V$ from $x$ to $y$. The image $\gamma([0, 1])$ is compact, so it can be covered by finitely many balls of radius $r/2$. Extracting a chain along $\gamma$, we obtain points $x = x_0, x_1, \ldots, x_N = y$ in $V$ with $|x_{j+1} - x_j| \le r$ for each $j = 0, \ldots, N - 1$. The number of chain links $N = N(V)$ depends only on $V$ (through $r$ and the covering number of $\overline{V}$), not on the particular points $x, y$.
[/step]
[step:Iterate the local comparison along the chain to obtain the global bound]
Applying the local estimate from the second step along each link of the chain:
\begin{align*}
u(x) = u(x_0) \le 2^n\, u(x_1) \le 2^{2n}\, u(x_2) \le \cdots \le 2^{nN}\, u(x_N) = 2^{nN}\, u(y).
\end{align*}
Reversing the roles of $x$ and $y$ gives $u(y) \le 2^{nN}\, u(x)$. Setting $C := 2^{nN(V)}$, which depends only on $n$ and $V$, we obtain
\begin{align*}
\frac{1}{C}\, u(y) \le u(x) \le C\, u(y) \qquad \text{for all } x, y \in V.
\end{align*}
Since this holds for all $x, y \in V$, taking the supremum over $x$ and the infimum over $y$ yields $\sup_V u \le C \inf_V u$, completing the proof.
[/step]
