## The Brunn-Minkowski inequality and the isoperimetric inequality

In this short note, I will prove the Brunn-Minkowski inequality and use it to derive the isoperimetric inequailty. Of course, the theory is quite well-known and I am writing it for my own benefit.

Let ${A, B\subset \mathbb{R}^n}$, we define the sum

$\displaystyle A+B=\{a+b:a\in A, b\in B\}.$

The following properties are easy to prove and will be used in the proof of the Brunn-Minkowski inequality:

 Lemma 1 If ${A}$ is open, then for any ${B}$, ${A+B}$ is also open. If ${A, B}$ are bounded, then ${A+B}$ is bounded. If ${A=I_1\times \cdots \times I_n}$, ${B= J_1\times \cdots \times J_n}$, where ${I_i, J_i}$ are intervals, then ${A+B= (I_1+J_1)\times \cdots \times (I_n+J_n)}$.
 Theorem 2 (Brunn-Minkowski inequality) If ${A, B\subset \mathbb{R}^n}$ are bounded open sets, then $\displaystyle \mathrm{Vol}(A)^{\frac{1}{n}}+\mathrm{Vol}(B)^{\frac{1}{n}}\le \mathrm{Vol}(A+B)^{\frac{1 }{n}}.$

Proof: Suppose first ${A}$ and ${B}$ are of the form

$\displaystyle A= I_1\times \cdots \times I_n,\quad B= J_1\times \cdots \times J_n,$

where ${I_i, J_i}$ are open bounded intervals. Then

$\displaystyle \mathrm{Vol}(A)= \prod_{i=1}^na_i, \quad \mathrm{Vol}(B)= \prod_{i=1}^nb_i, \quad \textrm{and }\mathrm{Vol}(A+B)= \prod_{i=1}^n(a_i+b_i),$

where ${a_i}$ and ${b_i}$ are the lengths of ${I_i}$ and ${J_i}$ respectively. Therefore by the AM-GM inequality,

$\displaystyle \begin{array}{rl} \frac{\mathrm{Vol}(A)^{\frac{1}{n}} + \mathrm{Vol}(B)^{\frac{1}{n}}}{\mathrm{Vol}(A+B)^{\frac{1}{n}}} =&\frac{\left(\prod _{i=1}^n a_i\right) ^{\frac{1}{n}}+\left(\prod _{i=1}^n b_i\right)^{\frac{1}{n}}}{\left(\prod_{i=1}^n (a_i+b_i)\right)^{\frac{1}{n}}}\\ =& \left(\prod _{i=1}^n \frac{a_i}{a_i+b_i}\right) ^{\frac{1}{n}}+\left(\prod _{i=1}^n \frac{b_i}{a_i+b_i}\right)^{\frac{1}{n}}\\ \le& \frac{1}{n} \sum _{i=1}^n \frac{a_i}{a_i+b_i} +\frac{1}{n} \sum _{i=1}^n \frac{b_i}{a_i+b_i}\\ =&1. \end{array}$

Now suppose ${A=\bigcup _{i=1}^m A_i }$ and ${B= \bigcup _{i=1} ^p B_i}$ are disjoint unions of open (${n}$-dimensional) rectangular blocks. If ${m+p\le 2}$, the inequality has been proved. So, for induction purpose, assume ${m+p>2}$. We can wlog assume ${m\ge 2}$. Take a plane ${P}$ parallel to one of the coordinates planes, say $x^1=0$, which separates ${A_1}$ and ${A_2}$. Then ${P}$ separates ${A}$ into the disjoint union of ${A^+}$ and ${A^-}$:

$\displaystyle A^+= \bigcup _{i=1}^{m^+}A^+_i,\quad A^+= \bigcup _{i=1}^{m^-}A^-_i,$

with ${m^{\pm}. Suppose ${\frac{\mathrm{Vol}(A^+)}{\mathrm{Vol}(A)}=\alpha\le 1}$. Take a plane ${Q}$ parallel to the plane $x^1=0$ which separates ${B}$ into ${B^+}$ and ${B^-}$, with ${\frac{\mathrm{Vol}(B^+)}{\mathrm{Vol}(B)}=\alpha}$. So now

$\displaystyle B^+ = \bigcup _{i=1}^{p^+ }B_i^+, \quad B^- = \bigcup _{i=1}^{p^-}B_i^- ,$

with ${p^{\pm }\le p}$. Note that ${m^++p^+ and ${m^-+p^-, and so we can apply the induction hypothesis to conclude that

$\displaystyle \mathrm{Vol}(A^+)^{\frac{1}{n}}+\mathrm{Vol}(B^+)^{\frac{1}{n}}\le \mathrm{Vol}(A^++B^+)^{\frac{1}{n}}, \quad \textrm{and} \quad \mathrm{Vol}(A^-)^{\frac{1}{n}}+\mathrm{Vol}(B^-)^{\frac{1}{n}}\le \mathrm{Vol}(A^-+B^-)^{\frac{1}{n}}.$

On the other hand, it is easy to see that ${A^++B^+}$ and ${A^-+B^-}$ are disjoint (they are separated by ${P+Q}$), and hence

$\displaystyle \begin{array}{rl} \mathrm{Vol}(A+B) \ge &\mathrm{Vol}(A^++B^+)+\mathrm{Vol}(A^-+B^-)\\ \ge&\left( \mathrm{Vol}(A^+)^{\frac{1}{n}}+\mathrm{Vol}(B^+)^{\frac{1}{n}}\right)^n+\left( \mathrm{Vol}(A^-)^{\frac{1}{n}}+\mathrm{Vol}(B^-)^{\frac{1}{n}}\right)^n\\ =&\left( \left(\alpha\mathrm{Vol}(A)\right)^{\frac{1}{n}}+\left(\alpha\mathrm{Vol}(B)\right)^{\frac{1}{n}}\right)^n+\left( \left((1-\alpha)\mathrm{Vol}(A)\right)^{\frac{1}{n}}+ \left((1-\alpha)\mathrm{Vol}(B)\right)^{\frac{1}{n}}\right)^n\\ =&\left( \mathrm{Vol}(A)^{\frac{1}{n}}+ \mathrm{Vol}(B)^{\frac{1}{n}}\right)^n . \end{array}$

In the general case, take a sequence of (monotone increasing) open sets ${A_i, B_i}$ with ${A_i\uparrow A}$, ${B_i \uparrow B}$ (i.e. ${A_i\subset A_{i+1}}$ and ${\bigcup _{i=1}^\infty A_i=A}$), such that ${A_i}$, ${B_i}$ are of the form considered in the previous case. By applying the previous argument to ${A_i}$, ${B_i}$ and taking ${i\rightarrow \infty}$, we can get the result. $\Box$

 Remark 1 (Some remarks about the ${n}$-dimensional balls and spheres.) Let ${\Gamma}$ be the gamma function defined by $\displaystyle \Gamma(t)= \int_0^\infty x^{t-1} e^{-x}dx.$ It is known that ${\Gamma}$ has the following properties: $\displaystyle \Gamma(n)=(n-1)!\textrm{ for all } n\in \mathbb{N} , \quad \Gamma(n+1)= n \Gamma(n), \quad \textrm{and }\Gamma\left(\frac{1}{2}\right)=\sqrt \pi.$ It is also known that the ${n}$-dimensional area (see also here) of the (${n}$-dimensional) unit hypersphere ${\mathbb{S}^n\subset \mathbb{R}^{n+1}}$ is given by $\displaystyle s_n = \frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}$ and the ${n}$-dimensional volume of the ${n}$-dimensional unit ball ${\mathbb{B}^n\subset \mathbb{R}^n}$ is $\displaystyle b_n=\mathrm{Vol}(\mathbb{B}^n)= \frac{s_{n-1}}{n}= \frac{2\pi^{\frac{n}{2}}}{n\Gamma\left(\frac{n}{2}\right)}.$ For example, ${s_0=2, s_1=2\pi, s_2= 4\pi, s_3=2\pi^2}$, and ${ b_1=2, b_2=\pi, b_3=\frac{4\pi}{3}, b_4= \frac{\pi^2}{2}}$.
 Theorem 3 (Isoperimetric inequality) For any smooth closed hypersurface ${S}$ in ${\mathbb{R}^n}$, we have $\displaystyle \left(\frac{\mathrm{Vol}(\Omega)}{b_n}\right)^{\frac{1}{n}}\le \left(\frac{\mathrm{Area}(S)}{s_{n-1}}\right)^{\frac{1}{n-1}},$ where ${\Omega}$ is the region enclosed by ${S}$.

Proof: Let ${S_t= \{x+t\nu(x): x\in S \}}$ be the “parallel” hypersurface, where ${\nu(x)}$ is the inward unit normal of ${S}$ at ${x}$. For small ${t}$, ${S_t}$ is a smooth hypersurface which bounds a region ${\Omega_t}$. Let ${B_t}$ be the open ball with radius ${t}$ (centered at ${O}$). Then it can be seen that ${\Omega_t+B_t\subset \Omega}$. Therefore by the Brunn-Minkowski inequality (Theorem 2), we have (we’ll write ${|\Omega|=\mathrm{Vol}(\Omega)}$ and ${|S|=\mathrm{Area}(S)}$, which shouldn’t cause confusion.)

$\displaystyle \begin{array}{rl} |\Omega|\ge | \Omega_t+ B_t| \ge & \left( |\Omega_t|^{\frac{1}{n}}+ |B_t|^{\frac{1}{n}}\right)^n =\left( |\Omega_t|^{\frac{1}{n}}+b_n ^{\frac{1}{n}}t\right)^n. \end{array}$

Therefore

$\displaystyle \begin{array}{rl} |S|=& \displaystyle \lim_{t\rightarrow 0^+}\frac{|\Omega|-|\Omega_t|}{t}\\ \ge&\displaystyle \lim_{t\rightarrow 0^+}\frac{\left(|\Omega_t|^{\frac{1}{n}}+b_n^{\frac{1}{n}}t \right)^n- |\Omega_t|}{t}\\ =&n|\Omega|^{1-\frac{1}{n}}b_n^{\frac{1}{n}}. \end{array}$

This is equivalent to the isoperimetric inequality by Remark 1. $\Box$

 Remark 2 I believe a more refined version of the above argument can be used to prove some inequalities involving the so called curvature measures. I will try to work it out later.