Proof of isoperimetric inequality in $\mathbb R^n$

117 Views Asked by At

I am trying to find a proof (using geometric measure theory) of the isoperimetric inequality in $\mathbb R^n,$ see here, but I discover that many proofs online either only tackles the 2-dimensional case, or only proves a much more difficult generalisation.

Where can I find a proof of the $n$-dimensional inequality itself?

1

There are 1 best solutions below

0
On

I know that is been a while, but I will post some info that can be useful for some users.

As is mentioned in wikipedia, one option is using the Sobolev inequality. Following Evans and Gariepy: take $S$ a bounded set of finite perimeter. Take $f_{k} \in C_{c}^{\infty}(\mathbb{R}^{n})$ so that $f_{k} \to \chi_{S}$ in $L^{1}$, $f_{k} \to \chi_{S}$ ($\mathcal{L}^{n}-$)a.e. and $\|Df_{k}\|(\mathbb{R}^{n}) \to \| D\chi_{S}\|(\mathbb{R}^{n})$. By the Sobolev inequality $$\|f_{k}\|_{L^{1^{*}}} \leq \frac{1}{n\omega_{n}^{\frac{1}{n}}} \|Df_{k}\|_{L^{1}}.$$ Thus, by Fatou's lemma and the properties of the $f_{k}$ $$\|\chi_{S}\|_{L^{1^{*}}}\leq \liminf_{k \to \infty}\|f_{k}\|_{L^{1^{*}}} \leq \lim_{k \to \infty} \frac{1}{n\omega_{n}^{\frac{1}{n}}} \|Df_{k}\|_{L^{1}}=\frac{1}{n\omega_{n}^{\frac{1}{n}}} \|D\chi_{S}\|_{L^{1}}.$$ I think that this is the usual proof in $\mathbb{R}^{n}$ (a similar one appears on Maggi's book).

The more geometric theoric proof that I know is given in Simon's notes on GMT (see Theorem 6.1 on Chapter 6 here). The "bad part" is that the results comes naturaly as a corollary of a big theorem, the Federer-Fleming deformation theorem. Assuming that Theorem, the proof is very short.