Evans computes by change of variables $z = x + sw$: $$\int_{\partial B(0,1)} |u(x+sw)-u(x)|\,dS(w) = \frac{1}{s^{n-1}}\int_{\partial B(x,s)} |u(z)-u(x)|\,dS(z)$$
The power $n-1$ confuses me. Can someone work this computation in more detail? Thanks!
Evans computes by change of variables $z = x + sw$: $$\int_{\partial B(0,1)} |u(x+sw)-u(x)|\,dS(w) = \frac{1}{s^{n-1}}\int_{\partial B(x,s)} |u(z)-u(x)|\,dS(z)$$
The power $n-1$ confuses me. Can someone work this computation in more detail? Thanks!
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A simple example will show where $1/s^{n-1}$ comes form. Suppose $u(x)=0$ and $u=1$ on $\partial B(x,s).$ Then the the left side equals
$$\int_{\partial B(0,1)} 1 \,dS= S(\partial B(0,1))$$
while the right side equals
$$(1/s^{n-1})\int_{\partial B(x,s)} 1 \,dS= (1/s^{n-1})S(\partial B(x,s)).$$
Now a sphere of radius $s$ has surface area equal to $s^{n-1}$ times the surface area of the unit sphere. That's why the $1/s^{n-1}$ is there on the right side.