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

  1. If {A} is open, then for any {B}, {A+B} is also open.
  2. If {A, B} are bounded, then {A+B} is bounded.
  3. 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}<m}. 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^+<m+p} and {m^-+p^-<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.
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