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I am studying the paper "symmetry and non-uniformly elliptic operators - jean dolbeault, patricio felmer and regis monneau" and in the demonstration of the lemma 8 page: 5, we have the problem:

Suppose that $\Omega\subset\mathbb{R^n}$ is a bounded Lipschitz domain, $0\in\partial\Omega$. Consider $u\in C^1(\overline\Omega)$ such that $u(0)=0$, and let $\phi$ be a nonnegative nonzero radial test function, then $$f(0)A(0)\int_{\mathbb{R^n}}\phi(x)dx=\lim_{\varepsilon\rightarrow0}\int_{\varepsilon^{-1}\Omega}f(u(\varepsilon x))\phi(x)dx,$$ where $f\in C(\mathbb{R})$ and $$A(0)=\frac{|S^{n-1}|}{n},$$ if $0\in\Omega$ and $$A(0)=\lim_{\varepsilon\rightarrow0}\frac{|\Omega\cap B(0.\varepsilon)|}{\varepsilon^n},$$ if $0\not\in\Omega$. I don't know how to obtain the equality above.

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  • $\begingroup$ I think you mean $f(u(0))$ on the left hand side, or $f(\varepsilon x)$ on the right. Either that, or at least $u(0) = 0$ $\endgroup$ – Ray Yang Apr 17 '13 at 1:37
  • $\begingroup$ I'm sorry, I edited the question. You are right, we need the hypothesis that $u(0)=0$. $\endgroup$ – José Carlos Apr 17 '13 at 20:51
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Try substituting $y=\epsilon x$. Then the right integral becomes an integral over $\Omega$, namely $\mathop{\int}\limits_{\Omega} f(u(y)) \frac{1}{\epsilon^n}\phi(\frac{1}{\epsilon}y)dy$. As $\epsilon$ approaches 0, the modified $\phi$ function becomes like a delta function, and is close to zero everywhere except near the origin. So the integral approaches $\mathop{\int}\limits_{\Omega} f(u(0)) \frac{1}{\epsilon^n}\phi(\frac{1}{\epsilon}y)dy$. So we can pull $f$ out. Now, the remaining part would be the integral of a test function over all of $\mathbb{R}^n$, except we're integrating over $\Omega$. Now, the test function $\phi(\frac{1}{\epsilon}y)$ is nonzero only over $B(0,\epsilon)$, so we can replace $\Omega$ with $\Omega\cap B(0,\epsilon)$ in the integral.

As $\epsilon$ decreases, the set $\Omega\cap B(0,\epsilon)$ becomes more and more like a spherical sector (I.e. the cone over an open subset of the sphere). Since the test function is radially symmetric, the integral is equal to the integral over $\mathbb{R}^n$ times the proportion of the ball that $\Omega\cap B(0,\epsilon)$ takes up. That's where $A(0)$ comes from in the second case. I don't know why it appears the way it does in the first case.

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  • $\begingroup$ The first case is just the limiting value of $\frac{|B(0,\epsilon)|}{\epsilon^n}$ $\endgroup$ – Ray Yang Apr 18 '13 at 3:19
  • $\begingroup$ I wondered that, but it puts a hole in my argument, because I don't see why it should appear at all. $\endgroup$ – Brian Rushton Apr 18 '13 at 3:48
  • $\begingroup$ You're right. What you need is $A(0) = \lim_{\epsilon \rightarrow 0} \frac{ |\Omega \cap B(0,\epsilon)|}{|B(0,\epsilon)|}$ $\endgroup$ – Ray Yang Apr 18 '13 at 3:51
  • $\begingroup$ Thank you for the answer. The answer helped me a lot. $\endgroup$ – José Carlos Apr 20 '13 at 4:55

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