[proofplan] We show that $u$ coincides with its mollification $\eta_\varepsilon * u$ on every compactly contained subdomain $U_\varepsilon := \{x \in U : \operatorname{dist}(x, \partial U) > \varepsilon\}$. The key computation uses the coarea formula to decompose the volume convolution into iterated surface integrals, then applies the spherical mean-value property (MV) to collapse each surface integral to a point evaluation. Since $\eta_\varepsilon * u \in C^\infty(U_\varepsilon)$ and every interior point belongs to $U_\varepsilon$ for sufficiently small $\varepsilon$, the identity $u = \eta_\varepsilon * u$ forces $u \in C^\infty(U)$. [/proofplan] [step:Fix a radial mollifier and write out the convolution on $U_\varepsilon$] Let $\eta \in C_c^\infty(B(0,1))$ be a radial [mollifier](/page/Standard%20Mollifier) with $\int_{\mathbb{R}^n} \eta \, d\mathcal{L}^n = 1$. For $\varepsilon > 0$, define $\eta_\varepsilon(x) := \varepsilon^{-n} \eta(x/\varepsilon)$, and set $U_\varepsilon := \{x \in U : \operatorname{dist}(x, \partial U) > \varepsilon\}$. For $x \in U_\varepsilon$, the [convolution](/page/Convolution) $u^\varepsilon := \eta_\varepsilon * u$ is given by \begin{align*} u^\varepsilon(x) = \int_U \eta_\varepsilon(x - y)\, u(y)\, d\mathcal{L}^n(y). \end{align*} Since $\operatorname{supp} \eta_\varepsilon \subset \overline{B}(0,\varepsilon)$ and $x \in U_\varepsilon$, the ball $B(x,\varepsilon) \subset U$, so the integration restricts to $B(x,\varepsilon)$: \begin{align*} u^\varepsilon(x) = \frac{1}{\varepsilon^n} \int_{B(x,\varepsilon)} \eta\!\left(\frac{|x - y|}{\varepsilon}\right) u(y)\, d\mathcal{L}^n(y), \end{align*} where we used that $\eta$ is radial, so $\eta\!\left(\frac{x - y}{\varepsilon}\right) = \eta\!\left(\frac{|x - y|}{\varepsilon}\right)$. [guided] We want to show that a continuous [function](/page/Function) $u$ satisfying the mean-value property is in fact smooth. The strategy is to express $u$ as a convolution with a smooth kernel, which automatically inherits the smoothness of the kernel. Let $\eta \in C_c^\infty(B(0,1))$ be a radial [mollifier](/page/Standard%20Mollifier) with $\int_{\mathbb{R}^n} \eta \, d\mathcal{L}^n = 1$. For $\varepsilon > 0$, define \begin{align*} \eta_\varepsilon: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \varepsilon^{-n} \eta(x/\varepsilon), \end{align*} and set $U_\varepsilon := \{x \in U : \operatorname{dist}(x, \partial U) > \varepsilon\}$. For $x \in U_\varepsilon$, the [convolution](/page/Convolution) $u^\varepsilon := \eta_\varepsilon * u$ is given by \begin{align*} u^\varepsilon(x) = \int_U \eta_\varepsilon(x - y)\, u(y)\, d\mathcal{L}^n(y). \end{align*} Why does the integration restrict to $B(x,\varepsilon)$? Because $\operatorname{supp} \eta_\varepsilon \subset \overline{B}(0,\varepsilon)$, the integrand $\eta_\varepsilon(x - y)$ vanishes unless $|x - y| \le \varepsilon$. Since $x \in U_\varepsilon$, we have $\operatorname{dist}(x, \partial U) > \varepsilon$, so $B(x,\varepsilon) \subset U$ and the integral becomes \begin{align*} u^\varepsilon(x) = \frac{1}{\varepsilon^n} \int_{B(x,\varepsilon)} \eta\!\left(\frac{|x - y|}{\varepsilon}\right) u(y)\, d\mathcal{L}^n(y), \end{align*} where we used the radiality of $\eta$: since $\eta$ depends only on $|x|$, we have $\eta\!\left(\frac{x - y}{\varepsilon}\right) = \eta\!\left(\frac{|x - y|}{\varepsilon}\right)$. [/guided] [/step] [step:Decompose the volume integral into surface integrals via the coarea formula] We apply the [coarea formula](/page/Geometric%20Measure%20Theory%20II%3A%20Area%20and%20Coarea%20Formulas) to the Lipschitz function $y \mapsto r = |x - y|$ on $B(x,\varepsilon)$. Since $|\nabla_y |x - y|| = 1$ for $\mathcal{L}^n$-a.e. $y \neq x$, the coarea formula gives \begin{align*} u^\varepsilon(x) = \frac{1}{\varepsilon^n} \int_0^\varepsilon \left(\int_{\partial B(x,r)} \eta\!\left(\frac{|x - y|}{\varepsilon}\right) u(y)\, d\mathcal{H}^{n-1}(y)\right) d\mathcal{L}^1(r). \end{align*} On the sphere $\partial B(x,r)$, the quantity $|x - y| = r$ is constant, so the factor $\eta(r/\varepsilon)$ exits the inner integral: \begin{align*} u^\varepsilon(x) = \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \left(\int_{\partial B(x,r)} u(y)\, d\mathcal{H}^{n-1}(y)\right) d\mathcal{L}^1(r). \end{align*} [guided] The volume integral over the ball $B(x,\varepsilon)$ mixes contributions from all distances $r \in (0,\varepsilon)$. To exploit the mean-value property -- which is a statement about surface averages -- we need to decompose the volume integral into a family of surface integrals, one for each radius $r$. The [coarea formula](/page/Geometric%20Measure%20Theory%20II%3A%20Area%20and%20Coarea%20Formulas) provides this decomposition. For the Lipschitz function $y \mapsto r = |x - y|$ on $B(x,\varepsilon)$, the gradient satisfies $|\nabla_y |x - y|| = 1$ for $\mathcal{L}^n$-a.e. $y \neq x$, and the level sets $\{y : |x - y| = r\} = \partial B(x,r)$. The coarea formula states \begin{align*} \int_{B(x,\varepsilon)} g(y)\, d\mathcal{L}^n(y) = \int_0^\varepsilon \left(\int_{\partial B(x,r)} g(y)\, d\mathcal{H}^{n-1}(y)\right) d\mathcal{L}^1(r) \end{align*} for any integrable $g$. Applying this with $g(y) = \varepsilon^{-n} \eta(|x - y|/\varepsilon)\, u(y)$ gives \begin{align*} u^\varepsilon(x) = \frac{1}{\varepsilon^n} \int_0^\varepsilon \left(\int_{\partial B(x,r)} \eta\!\left(\frac{|x - y|}{\varepsilon}\right) u(y)\, d\mathcal{H}^{n-1}(y)\right) d\mathcal{L}^1(r). \end{align*} Now comes a key simplification: on the sphere $\partial B(x,r)$, every point $y$ satisfies $|x - y| = r$, so $\eta(|x - y|/\varepsilon) = \eta(r/\varepsilon)$ is constant with respect to $y$. We factor it out of the inner integral: \begin{align*} u^\varepsilon(x) = \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \left(\int_{\partial B(x,r)} u(y)\, d\mathcal{H}^{n-1}(y)\right) d\mathcal{L}^1(r). \end{align*} This is the form we need: the inner integral $\int_{\partial B(x,r)} u(y)\, d\mathcal{H}^{n-1}(y)$ is exactly the unnormalised surface integral that the mean-value property controls. [/guided] [/step] [step:Apply the mean-value property to collapse each surface integral to a point evaluation] For each $r \in (0,\varepsilon)$, the ball $B(x,r) \subset B(x,\varepsilon) \subset U$, so hypothesis (MV) applies on $\partial B(x,r)$: \begin{align*} \int_{\partial B(x,r)} u(y)\, d\mathcal{H}^{n-1}(y) = \mathcal{H}^{n-1}(\partial B(x,r))\, u(x). \end{align*} Substituting into the previous expression: \begin{align*} u^\varepsilon(x) &= \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \mathcal{H}^{n-1}(\partial B(x,r))\, u(x)\, d\mathcal{L}^1(r) \\ &= u(x) \cdot \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \mathcal{H}^{n-1}(\partial B(x,r))\, d\mathcal{L}^1(r). \end{align*} [guided] This is the step where the mean-value property does its work. For each radius $r \in (0,\varepsilon)$, the sphere $\partial B(x,r)$ lies inside $B(x,\varepsilon) \subset U$, so the ball $B(x,r) \subset U$ and the spherical mean-value property (MV) applies: \begin{align*} u(x) = \frac{1}{\mathcal{H}^{n-1}(\partial B(x,r))} \int_{\partial B(x,r)} u(y)\, d\mathcal{H}^{n-1}(y). \end{align*} Rearranging, $\int_{\partial B(x,r)} u(y)\, d\mathcal{H}^{n-1}(y) = \mathcal{H}^{n-1}(\partial B(x,r))\, u(x)$. Substituting: \begin{align*} u^\varepsilon(x) &= \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \mathcal{H}^{n-1}(\partial B(x,r))\, u(x)\, d\mathcal{L}^1(r) \\ &= u(x) \cdot \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \mathcal{H}^{n-1}(\partial B(x,r))\, d\mathcal{L}^1(r). \end{align*} The value $u(x)$ has been extracted as a multiplicative constant. The remaining integral involves only the mollifier $\eta$ and the surface area, not the function $u$. [/guided] [/step] [step:Evaluate the remaining integral to recover the normalisation constant $1$] We reverse the coarea decomposition. On the sphere $\partial B(x,r)$, the function $\eta_\varepsilon(x - y) = \varepsilon^{-n} \eta(r/\varepsilon)$ is constant in $y$, so \begin{align*} \frac{1}{\varepsilon^n} \eta\!\left(\frac{r}{\varepsilon}\right) \mathcal{H}^{n-1}(\partial B(x,r)) = \int_{\partial B(x,r)} \eta_\varepsilon(x - y)\, d\mathcal{H}^{n-1}(y). \end{align*} Integrating over $r \in (0,\varepsilon)$ and applying the coarea formula in reverse: \begin{align*} \frac{1}{\varepsilon^n} \int_0^\varepsilon \eta\!\left(\frac{r}{\varepsilon}\right) \mathcal{H}^{n-1}(\partial B(x,r))\, d\mathcal{L}^1(r) &= \int_{B(x,\varepsilon)} \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y) \\ &= \int_{\mathbb{R}^n} \eta_\varepsilon(z)\, d\mathcal{L}^n(z) = 1, \end{align*} where we substituted $z = x - y$ (translation invariance of $\mathcal{L}^n$) and used the normalisation $\int_{\mathbb{R}^n} \eta_\varepsilon\, d\mathcal{L}^n = 1$. Therefore $u^\varepsilon(x) = u(x)$. [/step] [step:Conclude smoothness of $u$ on $U$] We have shown $u = \eta_\varepsilon * u$ on $U_\varepsilon$ for every $\varepsilon > 0$. Since $\eta_\varepsilon \in C_c^\infty(\mathbb{R}^n)$, the standard result on [convolution](/page/Convolution) with smooth compactly supported kernels gives $\eta_\varepsilon * u \in C^\infty(U_\varepsilon)$. Hence $u \in C^\infty(U_\varepsilon)$ for every $\varepsilon > 0$. For any $x \in U$, choose $\varepsilon < \operatorname{dist}(x, \partial U)$ to place $x \in U_\varepsilon$. Since $u$ is $C^\infty$ on $U_\varepsilon$, $u$ is $C^\infty$ at $x$. As $x \in U$ was arbitrary, $u \in C^\infty(U)$. [/step]