From PDE Evans, 2nd edition, pages 25-26.
THEOREM 2 (Mean Value Formulas for Laplace's equation). If $u \in C^2(U)$ is harmonic, then $$u(x)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B(x,r)}u \, dS=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B(x,r)} u \, dy$$ for each ball $B(x,r)\subset U$.
Proof. 1. Set $$\phi(r) := \avint_{\partial B(x,r)} u(y) \, dS(y) = \avint_{\partial B(0,1)} u(x+rz) \, dS(z).$$ Then $$\phi'(r)=\avint_{\partial B(0,1)} Du(x+rz) \cdot z \, dS(z)$$ and consequently, using Green's formulas from §C.2, we compute \begin{align}\phi'(r) &= \avint_{\partial B(x,r)} Du(y) \cdot \frac{u-x}r dS(y) \\ &= \avint_{\partial B(x,r)} \frac{\partial u}{\partial \nu} dS(y) \\ &= \frac rn \avint_{B(x,r)} \Delta u(y) \, dy = 0.\end{align}
Hence $\phi$ is constant, and so $$\phi(r)=\lim_{t \to 0} \phi(t) = \lim_{t \to 0} \avint_{\partial B(x,t)} u(y) \, dS(y) = u(x).$$
The proof continues on but I'm stopping here for this question I need to ask:
Can anyone explain the last line? I understand that $\phi$ is constant since we established that $\phi'(r)=0$ for all harmonic functions $u$. Hence, this is why we can say $\phi(r)=\lim_{t\to 0} \phi(t)$. But how can we establish the last equality, that is, $$\lim_{t \to 0} \avint_{\partial B(x,t)} u(y) \, dS(y)=u(x)?$$
Best Answer
$\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}$
$$ \avint_{\partial B(x,t)} \, dS(y) = 1 $$ and therefore
$$ \bigl| u(x) - \avint_{\partial B(x,t)} u(y) \, dS(y)\bigr| = \bigl| \avint_{\partial B(x,t)} (u(x) - u(y)) \, dS(y) \bigr| \le \avint_{\partial B(x,t)} |u(x) - u(y)| \, dS(y) \\ \le \max_{y \in \partial B(x,t)} \{ |u(x) - u(y)| \} \quad . $$
This converges to zero for $t \to 0$ because $u$ is continuous at $x$.