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I want to know if the following fact has a standard name and/or reference

Let $X$ be a subset of $\mathbb R^2$ and $B$ be a disc of the same area as $X$. Set $X_\epsilon$ to be the $\epsilon$-neigborhood of $X$. Then $$area\\,X_\epsilon\ge area\\,B_\epsilon.$$

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Leonid, you should post that as an answer. –  Qiaochu Yuan Apr 22 '10 at 3:08
1  
Hmm, I wonder what happened there...? –  Reid Barton Apr 27 '10 at 3:06

1 Answer 1

up vote 3 down vote accepted

This inequality is essentially equivalent to the Classical isoperimetric inequality. If you have a measurable body $X$ in $\mathbb{R}^n$ and a ball $B\subset \mathbb R^n$ of same volume then you have the following: $$Area(X)=\lim_{\epsilon \to 0} \frac{\operatorname{Vol}(X_{\epsilon})-\operatorname{Vol}(X)}{\epsilon}$$ $$Area(B)=\lim_{\epsilon \to 0} \frac{\operatorname{Vol}(B_{\epsilon})-\operatorname{Vol}(B)}{\epsilon}$$ Proving that $Area(X)\geq Area(B)$ follows from $\operatorname{Vol}(X_{\epsilon})\geq \operatorname{Vol}(B_{\epsilon})$, which is your inequality. ($n=2$) Now this follows from the Brunn Minkowski inequality because $$\operatorname{Vol}(X_{\epsilon})=\left(\operatorname{Vol}(X+\epsilon B)^{1/n}\right)^n \geq \left(\operatorname{Vol}(X)^{1/n}+\epsilon \operatorname{Vol}(B)^{1/n}\right)^n=\operatorname{Vol}(B_{\epsilon})$$

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