I just have a small question:
It is always stated that $$ \frac{1}{n \alpha (n)r^{n-1}}\int_{\partial B(0,r)} u(x) \text{d}x = \frac{1}{n \alpha (n)}\int_{\partial B(0,1)} u(rx) \text{d}x $$
Now I tried to justify this equation and set the map $\phi: \partial B(0,1) \rightarrow \partial B(0,r), x \mapsto rx.$ Then $\text{det}(D \phi (x)) = r^n$ so with the Rule $$ \int_{\partial B(0,r)} u(x) \text{d}x = \int_{\partial B(0,1)} u(\phi(x)) |\text{det}(D \phi (x))| \text{d}x $$ I get:
$$ \frac{1}{n \alpha (n)r^{n-1}}\int_{\partial B(0,r)} u(x) \text{d}x = \frac{r}{n \alpha (n)}\int_{\partial B(0,1)} u(rx) \text{d}x $$
Does someone see my error?
You calculated the Jacobian in the surrounding $n$-dimensional space, but you need the Jacobian on the $(n-1)$-dimensional surface you're integrating over. If that's not immediately clear, consider the case $n=1$ or $n=2$ to see what's going on.