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Suppose $\Omega \subset {\mathbb R}^n$ be open and bounded with smooth boundary. Let $t_0 > 0$ be small enough so that for every $x \in \partial \Omega$, there exists a unique $y \in \Omega$ with $|x-y|=t_0=dist(y,\partial \Omega)$. Let $h \in L^1(\Omega)$ with $h \geq 0$. For $x \in \Omega$, let $n(x)$ be the inward normal vector to $\partial \Omega$. Let $\Omega_t = \{x\in \Omega \mid dist(x,\partial \Omega) >t\}$. I would like to show that there exists $C = C(\Omega)$ such that if $t \in (0,t_0)$, then $$ \int_{\partial \Omega} \int_0^t h(x+sn(x))\,ds \leq C\int_{\Omega \setminus \Omega_t} h\,dx.$$ This is not a homework problem.

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Is $\Omega$ compact? –  Paul Dec 30 '11 at 22:13
    
Yes, $\Omega$ is bounded. –  Stefan Smith Dec 30 '11 at 22:25
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If $\Omega$ is open as you say in the first line, it is not compact. –  Alex Becker Dec 31 '11 at 0:48
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$\Omega$ is open and bounded, and therefore has compact closure. –  Stefan Smith Dec 31 '11 at 17:15

2 Answers 2

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Fix $t<t_0$. Let $M=\partial \Omega\times (0,t)$ equipped with the structure of product manifold, and product measure. Also let $N=\Omega\setminus\overline{\Omega_t}$. We have a smooth bijective map $F:M\to N$ defined by $F(x,s)=x+sn(x)$. You would like to show that $\int_{M}h\circ F\le C\int_N h$ for any $h\in L_+^1(N)$. By the change of variables formula $\int_N h=\int_{M}(h\circ F)J_F$, this amounts to showing that the Jacobian determinant $J_F$ is bounded on $M$. The only questionable part of $F$ is $n(x)$, but $C^2$ smoothness of the boundary ensures that $n(x)$ is $C^1$ smooth, and therefore its contribution to the Jacobian is bounded.

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I'll define $C$ at the end. Define $\displaystyle f(t)=\int_{\partial \Omega}\int_{0}^{t}h(x+sn(x))ds-C\int_{\Omega- \Omega_t}hdx$. Then, $$f'(t)=\int_{\partial \Omega}h(x+tn(x))dx'-C\int_{\partial \Omega_t}hdx'.$$ Here, $dx'$ is surface area measure for the boundary. Since $$g(t)=\frac{\int_{\partial \Omega}h(x+tn(x))dx'}{\int_{\partial \Omega_t}hdx'}$$ is continues, as you can see $lim_{t\to 0}g(t)=1$, therefore $g$ achieves its maximum, $M$, on any compact interval, for example, $[0,t_0].$ Define $C:=M$. Therefore $f$ is non-increasing so we must have $f(t)\leq f(0)=0.$

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Tim: I am not sure this works, since I am assuming only $h \in L^1(\Omega)$, so $h|_{\partial \Omega_t}$ may not be well-defined. –  Stefan Smith Dec 31 '11 at 17:25

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