The strategy is to compare $u$ with the harmonic [function](/page/Function) having the same [boundary](/page/Boundary) values on a ball, and use the mean-value hypothesis to show they agree. **Step 1: Setup.** Fix $x_0 \in \Omega$ and $r > 0$ with $\overline{B(x_0, r)} \subseteq \Omega$. Let $v \in C^2(B(x_0, r)) \cap C(\overline{B(x_0, r)})$ be the unique solution of \begin{align*} \Delta v &= 0 \quad \text{in } B(x_0, r), \quad v = u \quad \text{on } \partial B(x_0, r), \end{align*} which exists by [Poisson's Formula For The Ball](/theorems/576). Define $w := u - v$ on $\overline{B(x_0, r)}$. **Step 2: $w$ satisfies the mean-value property.** [claim:Difference Satisfies Mean Value Property] $w$ satisfies the ball mean-value property on $B(x_0, r)$: for every $y \in B(x_0, r)$ and every $\rho > 0$ with $\overline{B(y, \rho)} \subseteq B(x_0, r)$, \begin{align*} w(y) &= \frac{1}{\mathcal{L}^n(B(y,\rho))}\int_{B(y,\rho)} w(z)\,d\mathcal{L}^n(z). \end{align*} [/claim] [proof] By hypothesis, $u$ satisfies the ball mean-value property. By the [Mean Value Property](/theorems/31), the harmonic function $v$ also satisfies the ball mean-value property. Since the mean-value property is linear ($w = u - v$ inherits it), $w$ satisfies it on $B(x_0, r)$. [/proof] **Step 3: $w$ vanishes.** [claim:w Is Identically Zero] $w \equiv 0$ on $\overline{B(x_0, r)}$. [/claim] [proof] Since $w = u - v = 0$ on $\partial B(x_0, r)$ and $w$ is continuous on $\overline{B(x_0, r)}$, the maximum $M := \max_{\overline{B(x_0,r)}} w$ is attained. If $M > 0$, it must be attained at some interior point $y_0 \in B(x_0, r)$ (since $w = 0$ on $\partial B$). But then for any ball $\overline{B(y_0, \rho)} \subseteq B(x_0, r)$: \begin{align*} M = w(y_0) = \frac{1}{\mathcal{L}^n(B(y_0,\rho))}\int_{B(y_0,\rho)} w(z)\,d\mathcal{L}^n(z) \le M, \end{align*} with equality only if $w \equiv M$ on $B(y_0, \rho)$. The same connectivity argument as in the [Strong Maximum Principle](/theorems/32) (the [set](/page/Set) $\{w = M\}$ is open by the mean-value property and closed by [continuity](/page/Continuity)) gives $w \equiv M$ on $B(x_0, r)$, contradicting $w = 0$ on $\partial B(x_0, r)$. Hence $M \le 0$. The same argument applied to $-w$ gives $\min w \ge 0$. Therefore $w \equiv 0$. [/proof] **Step 4: Conclusion.** Since $u = v$ on $B(x_0, r)$ and $v$ is harmonic (hence $C^\infty$), $u$ is $C^\infty$ and $\Delta u = 0$ on $B(x_0, r)$. Since $x_0$ and $r$ were arbitrary (subject to $\overline{B(x_0,r)} \subseteq \Omega$), $u$ is harmonic on $\Omega$.