I'm trying to work through the proof of this statement:
Suppose $\Omega \in \mathbb{R}^d$ is open with $B_R(x) \in \Omega$ and suppose $u \in C^2(\Omega)$. Define $$\varphi (r):= \int_{\partial B_r(x)} u(y) d\sigma(y), \hspace{2mm} r\in(0, R) $$ then $\varphi$ is differentiable with $$\varphi'(r) = \frac{r}{d}\int_{B_r(x)} \Delta u(y) dy$$ and $\lim_{r \rightarrow 0} \varphi(r) = u(x)$.
So I've finally managed to work through the first part of the proof. I'm stuck on showing the limit. The first line is $$|\varphi(r) - u(x)| = \Big|\int_{\partial B_r(x)} u(y) - u(x) d \sigma(y) \Big|$$
And I'm confused about how $u(x)$ has been taken into the integral. I realise the implication of taking $u(x)$ into the integral is that $\int_{\partial B_r(x)} d\sigma(y)$ = 1, but again not too sure why this is the case. I would think the integral over a surface would be the surface area... Any pointers here?
Note: all the integrals are average integrals, but I couldn't find the symbol online for an integral with a line through it!
$\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}$ The limit can be taken inside the integral without any kind of $\epsilon-\delta$ argument provided we write out the expression in a clever way. In any case, I will give two proofs of this limit (one using the dominated convergence theorem and another direct approach).
For each $r \in (0,R)$, note that \begin{align*} \varphi(r) = \avint_{\partial B(x,r)} u(y)\,\mathrm{d}\sigma(y) = \frac{1}{n \omega_n r^{n-1}} \int_{\partial B(x,r)} u(y)\,\mathrm{d}\sigma(y). \end{align*} Here, $\omega_n$ denotes the volume of the $n$-dimensional unit ball in $\mathbb{R}^n$. Then, after a change of variables, we see that \begin{align*} \varphi(r) = \frac{1}{n\omega_n} \int_{\partial B(0,1)} u(x+rz)\,\mathrm{d}\sigma(z). \end{align*} Now, by continuity, $u$ is bounded uniformly on the compact ball $$ \overline{B(x,R/2)} \subseteq \Omega. $$ In particular, $u(x+rz)$ is uniformly bounded (in $r$) for all $z \in \partial B(0,1)$. Again using the continuity assumption, $u(x+rz) \to u(x)$ as $r \to 0$ for every $z \in \partial B(0,1)$. Hence, by dominated convergence, we see that \begin{align*} \lim_{r \to 0} \varphi(r) = \lim_{r \to 0}\frac{1}{n\omega_n} \int_{\partial B(0,1)} u(x+rz)\,\mathrm{d}\sigma(z) = \frac{1}{n\omega_n} \int_{\partial B(0,1)} u(x)\,\mathrm{d}\sigma(z) = u(x). \end{align*}
Alternatively, we can proceed directly with an $\epsilon-\delta$ argument. Here, we use that the average integral satsifies \begin{align*} \avint_{\partial B(x,r)} u(x)\,\mathrm{d}\sigma(y) &= u(x)\avint_{\partial B(x,r)} \,\mathrm{d}\sigma(y)\\ &= u(x) \left[ \frac{1}{|\partial B(x,r)|} \int_{\partial B(x,r)} \,\mathrm{d}\sigma(y)\right]\\ &= u(x) \cdot \frac{|\partial B(x,r)|}{|\partial B(x,r)|}\\ &= u(x). \end{align*} Then, we write \begin{align*} \Big|\avint_{\partial B_r(x)} \left(u(y) - u(x)\right) \,\mathrm{d} \sigma(y) \Big| &\leq \avint_{\partial B(x,r)} |u(y)-u(x)|\,\mathrm{d}\sigma(y) \to 0\\ &= \frac{1}{n\omega_n r^{n-1}} \int_{\partial B_r(x)} |u(y)-u(x)|\,\mathrm{d}\sigma(y). \end{align*} Given $\epsilon > 0$, we can find $\delta > 0$ so small that $|u(x)-u(y)| < \epsilon$ whenever $|x-y| < \delta$. So, if $0 <r < \delta$, we see that \begin{align*} |\varphi(r) - u(x)| &\leq \frac{1}{n\omega_n r^{n-1}} \int_{\partial B_r(x)} |u(y)-u(x)|\,\mathrm{d}\sigma(y)\\ &\leq \frac{\epsilon}{n\omega_n r^{n-1}} \int_{\partial B_r(x)} \mathrm{d}\sigma(y)\\ &= \epsilon. \end{align*}