Brunn-Minkowski Inequality and the Basic Isomperimetric Inequality

\begin{theorem}[General Brunn-Minkowski inequality in $\mathbb{R}^n$] Let $0< \lambda <1$ and let X and Y be nonempty bounded measurable sets in $\mathbb{R}^n$ such that $(1- \lambda)X + \lambda Y$ is also measurable. Then, \begin{equation} \label{eq:8} \mu ((1- \lambda )X+ \lambda Y)^{\frac{1}{n}} \geq (1- \lambda) \mu(X)^{\frac{1}{n}} + \lambda \mu(Y)^{\frac{1}{n}} \end{equation} where $\mu$ is the Lebesgue measure. \end{theorem} \begin{remark} We note that in general the assumption that the set $(1-\lambda)X+ \lambda Y$ is measurable is necessary even given that both $X$ and $Y$ are measurable. However, for borel sets $X$ and $Y$, we get that $(1-\lambda)X+ \lambda Y$ is a continuous image of their product and so guaranteed being measurable. \end{remark} Now we show how the $Brunn-Minkowski$ theorem directly implies the very well-know isomperimetric inequlaity: \begin{corollary}[The Isoperimetric Inequality] Consider the convex body K with surface area $S(K)$. \\ Then, the quantity $C(X)= \frac{\mu{(K)}^{1/n}}{S(K)^{1/(n-1)}}$ is maximized on Euclidean balls. \begin{proof} By $Brunn-Minkowski$ inequality, we have \begin{equation} \begin{split} \mu{(K+ \epsilon B)} &\geq (\mu{(K)}^{1/n} + \epsilon \mu(B)^{1/n})^n \\&= \mu(K)(1+ \epsilon (\frac{\mu(B)}{\mu(K)})^{1/n})^n\\ & \geq \mu(K)(1+n \epsilon (\frac{\mu(B)}{\mu(K)})^{1/n}) \end{split} \end{equation} where we derived the last inequality by $(1+x)^n \geq 1+nx$ for $x \geq 0$. Now we lower bound the surface area \begin{equation} S(K) \geq n \mu(K)(\frac{\mu(B)}{\mu(K)})^{1/n} \end{equation} Now since $S(K) = n \mu(B)$ (by $Minkowski-Steiner$ formula), we get: \begin{equation} \begin{split} \frac{S(K)}{S(B)}&=\frac{S(K)}{n\mu(B)}\geq \frac{\mu(K)(\frac{\mu(B)}{\mu(K)})^{1/n}}{\mu(B)}\\ &= \mu(K)^{\frac{n-1}{n}} \mu(B)^{\frac{1-n}{n}}. \end{split} \end{equation} Finally, we get the isoperimetric inequality: \begin{equation} \frac{\mu(B)^{1/n}}{(S(K))^{1/{n-1}}} \geq \frac{\mu(K)^{1/n}}{S(K)^{1/{n-1}}} \end{equation} \end{proof} \end{corollary} \begin{corollary}[The Classical Isoperimetric Inequality in the Plane] \begin{equation} L^2 \geq 4 \pi A \end{equation} where $A$ is the area of a domain enclosed by a curve of length $L$. \end{corollary}


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