[proofplan] We establish two formulas: the spherical mean-value formula and the ball mean-value formula. For the spherical formula, we define the average of $u$ over $\partial B(z,r)$ via the unit sphere parametrisation, differentiate under the integral to obtain $\phi'(r)$, convert the resulting integral back to $\partial B(z,r)$ using the tangential Jacobian, and apply [Green's first identity](/theorems/30) with $\Delta u = 0$ to conclude $\phi'(r) = 0$. The constant value of $\phi$ is identified as $u(z)$ by the [Dominated Convergence Theorem](/theorems/511) as $r \to 0^+$. The ball formula then follows by integrating the spherical formula in polar coordinates and dividing by $\mathcal{L}^n(B(z,r))$. [/proofplan] [step:Define the spherical average on the unit sphere] Let $r_0 > 0$ be such that $\overline{B(z, r_0)} \subseteq U$. For $r \in (0, r_0)$, define the spherical average of $u$ over $\partial B(z,r)$ via the unit sphere parametrisation $y = z + r\omega$: \begin{align*} \phi: (0, r_0) &\to \mathbb{R} \\ r &\mapsto \frac{1}{\mathcal{H}^{n-1}(\partial B(0,1))} \int_{\partial B(0,1)} u(z + r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} [guided] The goal is to show that $u(z)$ equals the average of $u$ over $\partial B(z,r)$. Rather than working directly on the sphere $\partial B(z,r)$, which changes with $r$, we pull back to the fixed domain $\partial B(0,1)$ via the substitution $y = z + r\omega$. This lets us study $\phi$ as a function of one real variable $r$ and use calculus to show it is constant. Define the spherical average of $u$ over $\partial B(z,r)$ via the unit sphere parametrisation $y = z + r\omega$: \begin{align*} \phi: (0, r_0) &\to \mathbb{R} \\ r &\mapsto \frac{1}{\mathcal{H}^{n-1}(\partial B(0,1))} \int_{\partial B(0,1)} u(z + r\omega)\,d\mathcal{H}^{n-1}(\omega), \end{align*} where $r_0 > 0$ satisfies $\overline{B(z, r_0)} \subseteq U$. The integrand $\omega \mapsto u(z + r\omega)$ is continuous on the compact set $\partial B(0,1)$, so $\phi$ is well-defined. [/guided] [/step] [step:Differentiate under the integral to compute $\phi'(r)$] [claim:Differentiation under the integral] For all $r \in (0, r_0)$, \begin{align*} \phi'(r) = \frac{1}{\mathcal{H}^{n-1}(\partial B(0,1))} \int_{\partial B(0,1)} \nabla u(z + r\omega) \cdot \omega\,d\mathcal{H}^{n-1}(\omega). \end{align*} [/claim] [proof] For each $\omega \in \partial B(0,1)$, the chain rule gives $\frac{d}{dr}u(z + r\omega) = \nabla u(z + r\omega) \cdot \omega$. Since $u \in C^2(U)$ and $\overline{B(z, r_0)}$ is compact, the derivative is bounded uniformly in $r$ and $\omega$: \begin{align*} |\nabla u(z + r\omega) \cdot \omega| \le \|\nabla u\|_{L^\infty(\overline{B(z,r_0)})} < \infty. \end{align*} This constant serves as an integrable dominator on $(\partial B(0,1), \mathcal{H}^{n-1})$, since $\mathcal{H}^{n-1}(\partial B(0,1)) < \infty$. The [Dominated Convergence Theorem](/theorems/511) applied to the difference quotients $\frac{u(z + (r+h)\omega) - u(z + r\omega)}{h}$ as $h \to 0$ justifies differentiating under the integral. [/proof] [/step] [step:Convert the integral to $\partial B(z,r)$ via the tangential Jacobian] [claim:Tangential Jacobian of the scaling map] For any integrable $f: \partial B(z,r) \to \mathbb{R}$, \begin{align*} \int_{\partial B(z,r)} f(y)\,d\mathcal{H}^{n-1}(y) = r^{n-1}\int_{\partial B(0,1)} f(z + r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} [/claim] [proof] Define $\Psi: \mathbb{R}^n \to \mathbb{R}^n$ by $\Psi(\omega) := z + r\omega$, so that $\Psi(\partial B(0,1)) = \partial B(z,r)$. The total derivative is $D\Psi_\omega = rI_n$ for every $\omega$. Let $\sigma: A \to \mathbb{R}^n$ be any injective $C^1$ immersion parametrising $\partial B(0,1)$. By the chain rule, $J(\Psi \circ \sigma)(y) = r\,J\sigma(y)$, so the Gram matrix satisfies \begin{align*} J(\Psi \circ \sigma)(y)^\top\,J(\Psi \circ \sigma)(y) = r^2\,J\sigma(y)^\top\,J\sigma(y). \end{align*} Taking the determinant of this $(n-1) \times (n-1)$ matrix yields $\det(r^2\,J\sigma^\top J\sigma) = r^{2(n-1)}\det(J\sigma^\top J\sigma)$. The tangential Jacobian from the [Substitution Formula for Hausdorff Integrals](/theorems/868) is therefore \begin{align*} J_M\Psi = \frac{\sqrt{r^{2(n-1)}\det(J\sigma^\top J\sigma)}}{\sqrt{\det(J\sigma^\top J\sigma)}} = r^{n-1}, \end{align*} which is constant on $\partial B(0,1)$. The result is independent of the choice of parametrisation by the [Substitution Formula for Hausdorff Integrals](/theorems/868). Substituting into the change-of-variables formula gives the claim. [/proof] Using the tangential Jacobian with $f(y) = \nabla u(y) \cdot \frac{y-z}{r}$ and noting that $\frac{y-z}{r} = \omega$ under $y = z + r\omega$, the expression from the differentiation step becomes \begin{align*} \phi'(r) = \frac{1}{\mathcal{H}^{n-1}(\partial B(0,1))\,r^{n-1}}\int_{\partial B(z,r)} \nabla u(y) \cdot \nu(y)\,d\mathcal{H}^{n-1}(y), \end{align*} where $\nu(y) := (y - z)/r$ is the outward unit normal on $\partial B(z,r)$. [guided] We want to relate the integral over $\partial B(0,1)$ back to $\partial B(z,r)$. Define $\Psi: \mathbb{R}^n \to \mathbb{R}^n$ by $\Psi(\omega) := z + r\omega$, so that $\Psi(\partial B(0,1)) = \partial B(z,r)$. The total derivative is $D\Psi_\omega = rI_n$ for every $\omega$. Let $\sigma: A \to \mathbb{R}^n$ be any injective $C^1$ immersion parametrising $\partial B(0,1)$. By the chain rule, $J(\Psi \circ \sigma)(y) = r\,J\sigma(y)$, so the Gram matrix satisfies \begin{align*} J(\Psi \circ \sigma)(y)^\top\,J(\Psi \circ \sigma)(y) = r^2\,J\sigma(y)^\top\,J\sigma(y). \end{align*} Taking the determinant of this $(n-1) \times (n-1)$ matrix yields $\det(r^2\,J\sigma^\top J\sigma) = r^{2(n-1)}\det(J\sigma^\top J\sigma)$. The tangential Jacobian from the [Substitution Formula for Hausdorff Integrals](/theorems/868) is therefore \begin{align*} J_M\Psi = \frac{\sqrt{r^{2(n-1)}\det(J\sigma^\top J\sigma)}}{\sqrt{\det(J\sigma^\top J\sigma)}} = r^{n-1}, \end{align*} which is constant on $\partial B(0,1)$. The result is independent of the choice of parametrisation by the [Substitution Formula for Hausdorff Integrals](/theorems/868). The change-of-variables formula then gives, for any integrable $f: \partial B(z,r) \to \mathbb{R}$: \begin{align*} \int_{\partial B(z,r)} f(y)\,d\mathcal{H}^{n-1}(y) = r^{n-1}\int_{\partial B(0,1)} f(z + r\omega)\,d\mathcal{H}^{n-1}(\omega). \end{align*} We apply this with $f(y) = \nabla u(y) \cdot \frac{y-z}{r}$, noting that $\frac{y-z}{r} = \omega$ under $y = z + r\omega$: \begin{align*} \int_{\partial B(0,1)} \nabla u(z + r\omega) \cdot \omega\,d\mathcal{H}^{n-1}(\omega) = \frac{1}{r^{n-1}}\int_{\partial B(z,r)} \nabla u(y) \cdot \frac{y-z}{r}\,d\mathcal{H}^{n-1}(y). \end{align*} Dividing by $\mathcal{H}^{n-1}(\partial B(0,1))$ and recognising $\nu(y) := (y - z)/r$ as the outward unit normal on $\partial B(z,r)$: \begin{align*} \phi'(r) = \frac{1}{\mathcal{H}^{n-1}(\partial B(0,1))\,r^{n-1}}\int_{\partial B(z,r)} \nabla u(y) \cdot \nu(y)\,d\mathcal{H}^{n-1}(y). \end{align*} The integrand $\nabla u(y) \cdot \nu(y)$ is the normal derivative $\frac{\partial u}{\partial \nu}$ on $\partial B(z,r)$, which is the quantity that appears in Green's identity. [/guided] [/step] [step:Apply Green's first identity to show $\phi'(r) = 0$] [claim:Constancy of the spherical average] $\phi'(r) = 0$ for all $r \in (0, r_0)$, so $\phi$ is constant. [/claim] [proof] The integral $\int_{\partial B(z,r)} \nabla u(y) \cdot \nu(y)\,d\mathcal{H}^{n-1}(y)$ equals $\int_{\partial B(z,r)} \frac{\partial u}{\partial \nu}\,d\mathcal{H}^{n-1}$. By the [first Green identity](/theorems/30) (part (i)) applied to the bounded open set $B(z,r)$ with $C^1$ boundary: \begin{align*} \int_{\partial B(z,r)} \frac{\partial u}{\partial \nu}\,d\mathcal{H}^{n-1} = \int_{B(z,r)} \Delta u(y)\,d\mathcal{L}^n(y) = 0, \end{align*} since $\Delta u = 0$ in $U \supseteq \overline{B(z,r)}$. Therefore $\phi'(r) = 0$. [/proof] [guided] We have expressed $\phi'(r)$ as a constant multiple of $\int_{\partial B(z,r)} \frac{\partial u}{\partial \nu}\,d\mathcal{H}^{n-1}$. The [first Green identity](/theorems/30) states that for a bounded open set $V$ with $C^1$ boundary and $v \in C^2(\overline{V})$: \begin{align*} \int_{\partial V} \frac{\partial v}{\partial \nu}\,d\mathcal{H}^{n-1} = \int_{V} \Delta v\,d\mathcal{L}^n. \end{align*} We apply this with $V = B(z,r)$ and $v = u$. We verify: $B(z,r)$ is a bounded open set with smooth ($C^\infty$) boundary $\partial B(z,r)$, and $u \in C^2(U)$ with $\overline{B(z,r)} \subseteq U$, so $u \in C^2(\overline{B(z,r)})$. The hypotheses are satisfied. Since $\Delta u = 0$ throughout $U$, the right-hand side vanishes: \begin{align*} \int_{\partial B(z,r)} \frac{\partial u}{\partial \nu}\,d\mathcal{H}^{n-1} = \int_{B(z,r)} \Delta u(y)\,d\mathcal{L}^n(y) = 0. \end{align*} Therefore $\phi'(r) = 0$ for all $r \in (0, r_0)$, so $\phi$ is constant on $(0, r_0)$. [/guided] [/step] [step:Evaluate the limit as $r \to 0^+$ to identify $\phi \equiv u(z)$] [claim:Limit identification] $\lim_{r \to 0^+} \phi(r) = u(z)$. [/claim] [proof] Let $(r_k)_{k \ge 1}$ be any sequence in $(0, r_0)$ with $r_k \to 0$. Define $f_k: \partial B(0,1) \to \mathbb{R}$ by $f_k(\omega) := u(z + r_k\omega)$. For each $\omega$, $z + r_k\omega \to z$ as $k \to \infty$, so $f_k(\omega) \to u(z)$ by continuity of $u$. Since $u \in C^2(U)$ and $\overline{B(z, r_0)}$ is compact, $|f_k(\omega)| \le \|u\|_{L^\infty(\overline{B(z,r_0)})} =: M < \infty$ for all $k$ and $\omega$. The constant $M$ is integrable on $(\partial B(0,1), \mathcal{H}^{n-1})$. By the [Dominated Convergence Theorem](/theorems/511): \begin{align*} \lim_{k \to \infty} \int_{\partial B(0,1)} u(z + r_k\omega)\,d\mathcal{H}^{n-1}(\omega) = \int_{\partial B(0,1)} u(z)\,d\mathcal{H}^{n-1}(\omega) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1)). \end{align*} Dividing by $\mathcal{H}^{n-1}(\partial B(0,1))$ gives $\lim_{k \to \infty}\phi(r_k) = u(z)$. Since $(r_k)$ was arbitrary, $\lim_{r \to 0^+}\phi(r) = u(z)$. [/proof] [guided] Since $\phi$ is constant on $(0, r_0)$, the value of $\phi(r)$ for any $r \in (0, r_0)$ equals $\lim_{r \to 0^+} \phi(r)$. We compute this limit using the [Dominated Convergence Theorem](/theorems/511). Let $(r_k)_{k \ge 1}$ be any sequence in $(0, r_0)$ with $r_k \to 0$, and define $f_k: \partial B(0,1) \to \mathbb{R}$ by $f_k(\omega) := u(z + r_k\omega)$. We verify the hypotheses of DCT: 1. **Pointwise convergence:** For each $\omega \in \partial B(0,1)$, $z + r_k\omega \to z$ as $k \to \infty$, so $f_k(\omega) \to u(z)$ by continuity of $u$. 2. **Integrable dominator:** Since $u \in C^2(U)$ and $\overline{B(z, r_0)}$ is compact, $|f_k(\omega)| \le \|u\|_{L^\infty(\overline{B(z,r_0)})} =: M < \infty$ for all $k$ and $\omega$. The constant function $M$ is integrable on $(\partial B(0,1), \mathcal{H}^{n-1})$ because $\mathcal{H}^{n-1}(\partial B(0,1)) < \infty$. By the [Dominated Convergence Theorem](/theorems/511): \begin{align*} \lim_{k \to \infty} \int_{\partial B(0,1)} u(z + r_k\omega)\,d\mathcal{H}^{n-1}(\omega) = \int_{\partial B(0,1)} u(z)\,d\mathcal{H}^{n-1}(\omega) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1)). \end{align*} Dividing by $\mathcal{H}^{n-1}(\partial B(0,1))$ gives $\lim_{k \to \infty}\phi(r_k) = u(z)$. Since the sequence $(r_k)$ was arbitrary, $\lim_{r \to 0^+}\phi(r) = u(z)$. Together with the constancy of $\phi$, this gives $\phi(r) = u(z)$ for all $r \in (0, r_0)$. [/guided] [/step] [step:Conclude the spherical mean-value formula] By the constancy of $\phi$ and the limit identification, $\phi(r) = u(z)$ for all $r \in (0, r_0)$. Expressing $\phi(r)$ back in terms of $\partial B(z,r)$ using the tangential Jacobian (with $f = u$) and the identity $\mathcal{H}^{n-1}(\partial B(z,r)) = r^{n-1}\,\mathcal{H}^{n-1}(\partial B(0,1))$: \begin{align*} u(z) = \phi(r) = \frac{1}{\mathcal{H}^{n-1}(\partial B(0,1))}\int_{\partial B(0,1)} u(z + r\omega)\,d\mathcal{H}^{n-1}(\omega) = \frac{1}{\mathcal{H}^{n-1}(\partial B(z,r))}\int_{\partial B(z,r)} u(y)\,d\mathcal{H}^{n-1}(y). \end{align*} This is the spherical mean-value formula. [/step] [step:Integrate the spherical formula in polar coordinates to obtain the ball formula] Decompose $\mathcal{L}^n$ on $B(z,r)$ in polar coordinates $y = z + s\omega$ with $s \in (0, r)$ and $\omega \in \partial B(0,1)$, using the Jacobian factor $s^{n-1}$ from the tangential Jacobian: \begin{align*} \int_{B(z,r)} u(y)\,d\mathcal{L}^n(y) = \int_0^r \int_{\partial B(0,1)} u(z + s\omega)\,s^{n-1}\,d\mathcal{H}^{n-1}(\omega)\,d\mathcal{L}^1(s). \end{align*} By the spherical formula, $\int_{\partial B(0,1)} u(z + s\omega)\,d\mathcal{H}^{n-1}(\omega) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1))$ for each $s \in (0, r)$. Substituting: \begin{align*} \int_{B(z,r)} u(y)\,d\mathcal{L}^n(y) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1))\int_0^r s^{n-1}\,d\mathcal{L}^1(s) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1))\,\frac{r^n}{n}. \end{align*} By the [Volume of the Ball](/theorems/871), $\mathcal{L}^n(B(z,r)) = \frac{r^n}{n}\,\mathcal{H}^{n-1}(\partial B(0,1))$. Dividing both sides by $\mathcal{L}^n(B(z,r))$: \begin{align*} u(z) = \frac{1}{\mathcal{L}^n(B(z,r))}\int_{B(z,r)} u(y)\,d\mathcal{L}^n(y). \end{align*} [guided] Having established the spherical formula, the ball formula follows by integration. The idea is to write the ball integral as an iterated integral: first integrate over the sphere of radius $s$, then integrate over $s \in (0,r)$. Decompose $\mathcal{L}^n$ on $B(z,r)$ using polar coordinates $y = z + s\omega$ with $s \in (0, r)$ and $\omega \in \partial B(0,1)$. The tangential Jacobian identity gives $d\mathcal{L}^n(y) = s^{n-1}\,d\mathcal{H}^{n-1}(\omega)\,d\mathcal{L}^1(s)$, so: \begin{align*} \int_{B(z,r)} u(y)\,d\mathcal{L}^n(y) = \int_0^r \int_{\partial B(0,1)} u(z + s\omega)\,s^{n-1}\,d\mathcal{H}^{n-1}(\omega)\,d\mathcal{L}^1(s). \end{align*} For each fixed $s \in (0,r)$, the inner integral is precisely the spherical average times $\mathcal{H}^{n-1}(\partial B(0,1))$. By the spherical mean-value formula established above: \begin{align*} \int_{\partial B(0,1)} u(z + s\omega)\,d\mathcal{H}^{n-1}(\omega) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1)). \end{align*} Substituting and evaluating the remaining radial integral: \begin{align*} \int_{B(z,r)} u(y)\,d\mathcal{L}^n(y) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1))\int_0^r s^{n-1}\,d\mathcal{L}^1(s) = u(z)\,\mathcal{H}^{n-1}(\partial B(0,1))\,\frac{r^n}{n}. \end{align*} By the [Volume of the Ball](/theorems/871), the volume of $B(z,r)$ is $\mathcal{L}^n(B(z,r)) = \frac{r^n}{n}\,\mathcal{H}^{n-1}(\partial B(0,1))$. Dividing both sides by this quantity: \begin{align*} u(z) = \frac{1}{\mathcal{L}^n(B(z,r))}\int_{B(z,r)} u(y)\,d\mathcal{L}^n(y). \end{align*} This completes the proof of both mean-value formulas. [/guided] [/step]