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Here I am attempting problem 3 in Chapter 2 of the book PDE by Evans. Thankfully I have found the solution on the internet

enter link description here

However, there is one line confuses me I was starring it for 3 long hours. (See page 3) enter image description here

I think this must got to do with the polar coordinate formula in appendix, i.e. enter image description here

I am really struggling seeing this, I just cannot see why everything including $\frac{1}{t^{n-2}}$ is in term of $x$. Could anyone help explaining the equality of the two integrals?

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  • $\begingroup$ Can you see it when $n=2$? $\endgroup$ – zhw. Jun 22 '15 at 17:08
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I think there is an error, as the LHS depends on $\epsilon$ and the RHS does not. On that same page, he calls

$$ I=\int_\epsilon^r\int_{\partial B(0,t)}\frac{1}{t^{n-2}}f\,dSdt\,\,\,\,\,\text{ and }\,\,\,\,\,J=\frac{1}{\epsilon^{n-2}}\int_{B(0,\epsilon)}f\,dy. $$

Then

$$ I=\int_\epsilon^r\int_{\partial B(0,t)}\frac{1}{|y|^{n-2}}f\,dS(y)dt=\int_{\epsilon\leq|y|\leq r}\frac{1}{|y|^{n-2}}f\,dy=\int_{B(0,r)}\frac{1}{|y|^{n-2}}f\,dy-\int_{B(0,\epsilon)}\frac{1}{|y|^{n-2}}f\,dy $$

since $t=|y|$ when $y\in\partial B(0,t)$ and $\bigcup_{t\in[\epsilon,r]}\partial B(0,t)=B(0,r)\backslash B(0,\epsilon)$. For $n\geq 2$ we have

$$ \left|\int_{B(0,\epsilon)}\frac{1}{|y|^{n-2}}f\,dy\right|\leq\max\,|f|\int_{0}^\epsilon\frac{1}{t^{n-2}}\int_{\partial B(0,t)}\,dS(y)dt=\max\,|f|\int_0^\epsilon n\alpha_nt\,dt\to 0 $$

as $\epsilon\to 0$. He earlier states $J\to 0$ as $\epsilon\to 0$. Hence,

$$ I+J\to \int_{B(0,r)}\frac{1}{|y|^{n-2}}f\,dy\,\,\text{ as }\,\,\epsilon\to 0 $$

and not as he states.

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  • $\begingroup$ Just the second last line. I think you meant $$ \left|\int_{B(0,\epsilon)}\frac{1}{|y|^{n-2}}f\,dy\right|\leq\max\,|f|\int_{0}^\epsilon\frac{1}{t^{n-2}}\int_{\partial B(0,t)}\,dS(y)dt=\max\,|f|\int_0^\epsilon n\alpha_nt\,dt\to 0 $$ Because we are integrating on $\partial B(0,t)$ so we are integrating w.r.t. $dS(y)$ instead of $y$. Right? $\endgroup$ – math101 Jun 23 '15 at 7:45
  • $\begingroup$ Silly me. It should be fixed now. $\endgroup$ – user31415926535 Jun 23 '15 at 12:56
  • $\begingroup$ Sure. And the max taken over all $y\in B(0,r)$ (which is finite given the continuity of $f$). $\endgroup$ – user31415926535 Jun 25 '15 at 16:09

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