Mean Value Theorem for Laplace's equation

Question

For the integral $⨍_{\partial B(0,1)}u(x+rz)dS(z),$ shall we interpret it as $\frac{1}{m(\partial B(x,r))}\int_{\partial B(0,1)} u(x+rz)dS(z)$? I think there is some abuse of notation. Please correct me if I am wrong.

Answer 1

See page 702 in Evans for a definition. But you are right by definition

$$⨍_{\partial B(0,1)} u(x+rz) dS(z) = \frac{1}{m(\partial B(0,1))} \int_{\partial B(0,1)} u(x+rz)dS(z).$$

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To see that $$⨍_{\partial B(x,r)} u(y) dS(y) = ⨍_{\partial B(0,1)} u(x+rz) dS(z)$$ use the notation in Evans and the hint from Sven Pistre. Note that $n\alpha(n)$ denotes the surface area of the $(n-1)$-dimensional unit sphere and that the Jacobi Determinant of the change of variables is $r^{n-1}$.

$$\begin{align*}⨍_{\partial B(x,r)} u(y) dS(y) &= \frac{1}{n\alpha(n)r^{n-1}} \int_{\partial B(x,r)} u(y) dS(y) \\\\ &= \frac{r^{n-1}}{n\alpha(n)r^{n-1}} \int_{\partial B(0,1)} u(x+rz)dS(z) \\\\ &=\frac{1}{n\alpha(n)} \int_{\partial B(0,1)} u(x+rz)dS(z) \\\\ &=⨍_{\partial B(0,1)} u(x+rz) dS(z),\end{align*}$$

where we used a change of variables $y = x+rz$ in the second equality.