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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.