Prove that if $E$ has smooth bounday $\lim_{r \to 0}{\mathcal{H}^{n-1}(\partial (E+B(0,r)))} = \mathcal{H}^{n-1}{\partial E} $

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Prove that, if $E \subset \mathbb{R}^n$ is an open bounded set with $C^1$ boundary (if you want you can assume $C^\infty$ ), then the following holds

$$\lim_{r \to 0}{\mathcal{H}^{n-1}(\partial (E + B(0,r)) } = \mathcal{H}^{n-1}(\partial E)$$

Where

$$E + B(0,r) = \{ x + y \; : \; x \in E \; , ||y|| < r \} = \{ x \in \mathbb{R}^n \; : \; dist(x,E) < r \}$$

I don't know how to prove this since It's hard to find a parametrization for $\partial(E + B(0,r))$

The only idea that came to my mind was to use the function

$f(x) := dist(x,E)$ , or maybe $f(x) := (dist(x,E))^2$, hoping that it is at least $C^1$, in that case it should hold something like this

$\partial E =\partial f^{-1}(0)$ and $\partial (E + B(0,r)) = \partial(f^(-1)(r)$

In any case I don't know how to prove that $f$ is $C^1$.

Please don't use anything involving weak or distributional derivatives since I'm using this to prove a weaker version of Isoperimetric inequality avoiding to use any weak or distributional derivative.

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Assume that $\partial E$ is, say, $C^2$. Then $\partial E$ admits a $C^1$ outward pointing unit normal vector fields $\nu$. In particular, for small $r>0$ one has

$$\partial (E+ B(0, r)) = \{ x + r\nu (x) : x\in \partial E\}.$$

Let $F=(F_1, \cdots, F_n)$ be a local parametrization of $\partial E$, with $g_{ij} = \partial_{x^i}\cdot \partial_{x^j}$, where $F_i = F_i (x^1, \cdots, x^{n_1})$. Then

$$F^r = F + r\nu$$

is a local parametrization for $\partial (E+ B(0, r))$. Moreover, if $g_{ij} = \partial_i F \cdot \partial_j F$. Then

\begin{align} g^r_{ij}& = \partial_i F^r \cdot \partial _j F^r \\ &=(\partial_i F + r\partial_i \nu)\cdot (\partial_j F+ r\partial_j \nu) \\ &= g_{ij} + r^2 \partial_i \nu \partial_j \nu. \end{align}

(note $\partial_i F \cdot \nu = \partial_j \cdot \nu = 0$). Note also that $\partial_i\nu$ is uniformly bounded since its $C^0$. Hence the volume element $d\mu^r$ satisfies

$$ d\mu^r = \sqrt{\det g^r_{ij}} dx^1\cdots dx^{n-1} \to \sqrt{\det g_{ij}} dx^1\cdots dx^{n-1} = d\mu$$ as $r\to 0$. Hence $$\lim_{r \to 0}{\mathcal{H}^{n-1}(\partial (E + B(0,r)) } = \mathcal{H}^{n-1}(\partial E).$$