Isoperimetric inequality and OMT

The classic isoperimetric inequality

A very interesting application of optimal transport is a proof of the isoperimetric inequality. The first proof with an OMT argument is due to Gromov and the main tool is the Knothe's map. . This proof is based on an idea by Knothe . The classic isoperimetric inequality in $\mathbb {R} ^{n}$ states that the round ball has the minimal (n-1)-dimensional volume of the boundary among all the domains with a given fixed volume. This is equivalent to say that every set $E$ has a larger perimeter than the ball $B$ with the same volume. I will present this proof following the exposition given in chapter two in . The usually way to state this is the following:

$\operatorname {Per} (E)\geq n\omega _{n}^{1/n}|E|^{1-1/n}.$

Here $\omega _{n}$ is the volume of the unit ball in $\mathbb {R} ^{n}$. The idea of the proof is to construct a map T called Knothe transport and use it between the two densities: $\mu ={\mathcal {L}}_{B}$, the inequality will follow from some symmetries and consideration on the Jacobian determinant of this map.

The Knothe's transport

For this part I will follow the first chapter in . In some sense, it can be seen as a multidimension generalization of monotone rearrangement. Take any two measures $\mu ,\nu \in {\mathcal {P}}(\mathbb {R} )$ and define

$F(x)=\int _{-\infty }^{x}d\mu (t),{\text{ }}G(y)=\int _{-\infty }^{x}d\nu (t)$,

This maps may not be well defined, since at some points the measures may have a delta. For the purpose of this exposition we will assume that those functions are well defined, for the precise definition and convention to include the mass of the deltas in the integral we refer again to the first chapter of. It follow easily from the definition that those maps are not decreasing. We now define $G^{-1}(y)=\inf {\{t\in \mathbb {R} |G(t)>y\}}$. We are now ready to define the following monotone rearrangement map :

$T=G^{-1}\circ F:\mathbb {R} \longrightarrow \mathbb {R}$.

Note that this map is also not decreasing. In the case that our first density $\mu$ has no deltas then it can be shown that T is indeed a transport map (Theorem 1.4.7 ). We now move to the two dimensional case: the key ingredient to the Knothe transport map is what is known as the disintegration theorem (1.4.10 in.): Given $\mu \in {\mathcal {P}}(\mathbb {R} ^{2})$ and let $\mu _{1}=\pi _{1}\#\mu \in {\mathcal {P}}(\mathbb {R} )$ where $\pi _{1}$is the projection on the first component of $\mathbb {R} ^{2}$: $\pi _{1}(x_{1},x_{2})=x_{1}$, Then there exist an uncountable family of probability measures $(\mu _{x_{1}})_{x_{1}\in \mathbb {R} }\subset {\mathcal {P}}(\mathbb {R} )$ such that for any $\phi :\mathbb {R} ^{2}\longrightarrow \mathbb {R}$ continuous and bounded we have that:

$\int _{\mathbb {R} ^{2}}\phi (x_{1},x_{2})d\mu (x_{1},x_{2})=\int _{\mathbb {R} }{\Bigg (}\int _{\mathbb {R} }\phi (x_{1},x_{2})d\mu _{x_{1}}(x_{2}){\Bigg )}d\mu _{1}(x_{1})$. The disintegration of the measure $\mu$ is often also denoted as $\mu =\mu _{x_{1}}\otimes \mu _{1}=\mu _{x_{1}}(dx_{2})\otimes \mu _{1}(dx_{1})$. Now we are ready to construct the Knothe map.

Fix any two absolutely continuous measures in $\mathbb {R^{2}}$: $\mu _{(}x_{1},x_{2})=f(x_{1},x_{2})dx_{1}dx_{2}$ and $\mu _{(}x_{1},x_{2})=g(x_{1},x_{2})dx_{1}dx_{2}$, define $F_{1}(x_{1})=\int _{\mathbb {R} }f(x_{1},x_{2})dx_{2}$ and $G_{1}(y_{1})=\int _{\mathbb {R} }g(y_{1},y_{2})dy_{2}$. Note that using these notation we can write:

$\mu (x_{1},x_{2})={\frac {f(x_{1},x_{2})}{F_{1}(x_{1})}}dx_{2}\otimes F_{1}(x_{1})dx_{1}$ and $\nu (y_{1},y_{2})={\frac {g(y_{1},y_{2})}{G_{1}(y_{1})}}dy_{2}\otimes G_{1}(y_{1})dy_{1}$. Applying the monotone rearrangement we get a map $T_{1}:\mathbb {R} \longrightarrow \mathbb {R}$ that satisfies $T_{1}\#F_{1}dx_{1}=G_{1}dy_{1}.$, We want to send the disintegration of $\mu$ at $x_{1}$ to the disintegration of $\nu$ at the point $T(x_{1})$, in symbols let $T_{2}(x_{1},\cdot )\#{\Bigg (}{\frac {f(x_{1},\cdot )dx_{2}}{F_{1}(x_{1})}}{\Bigg )}={\frac {g(T_{1}(x_{1}),\cdot )dy_{2}}{G_{1}(T_{1}(x_{1}))}}$.

The Knothe map is now defined as:

$T(x_{1},x_{2})=(T_{1}(x_{1}),T_{2}(x_{1},x_{2}))$.

It is not hard to check that thismap indeed transports $\mu$ to $\nu$ (Theorem 1.4.13 of ). By monotonicity of the monotone rearrangement, assuming that the map T is differentiable, we can say that:

$\nabla T=\left[{\begin{array}{cc}\partial _{1}T_{1}\geq 0&\star \\0&\partial _{1}T_{1}\geq 0\\\end{array}}\right]$.

We can iterate the same construction and obtain a Knothe map on $\mathbb {R} ^{n}$, the recursive nature of this definition is well described in , the formal construction can be found here .

Proof of the classic isoperimetric inequality

We now present three key properties of the Knothe map from $\mu$ to $\nu$, this is Proposition 1.5.2 in . Let now $E\subset \mathbb {R} ^{n}$ be a bounded set with smooth boundary, $|E|$ its Lebesgue measure and the probability measures $\mu ={\frac {\chi _{E}}{|E|}}dx$ and $\nu ={\frac {\chi _{B_{1}}}{\omega _{n}}}$, where $\chi _{A}$ is the characteristic function of the set $A$ and $B_{1}$ is the unit ball in $\mathbb {R} ^{n}$. Denote with $T$ be a Knothe map from $\mu$ to $\nu$. First by just noticing that if $x\in E$ then $T(x)\in B_{1}$ we can conclude:

$\forall x\in E,|T(x)|\leq 1$

Thank to a change of variable and using the fact that the jacobian map $\nabla T$ is upper triangular and its diagonal entries are non negative (similar to the two dimensional case) it can be shown that:

$\det \nabla T={\frac {\omega _{n}}{|E|}}$ in $E$

Now since the matrix is upper triangular, it's very easy to compute the determinant as the product of the diagonal entries, we then get an estimate on the divergence of $T$:

$\operatorname {div} T\geq n(\det \nabla T)^{\frac {1}{n}}:$.

We are now ready to prove the classic isoperimetric inequality, this is Theorem 1.5.1 of . Denote by $n_{E}$ the outer unit normal of $\partial E$ and by $d\sigma$ the surface element of $\partial E$. We can now write thanks to the first property of the Knothe map:

$\operatorname {Per} (E)=\int _{\partial E}d\sigma \geq \int _{\partial E}|T|d\sigma$

As a straightforward application of Stokes theorem together with our lower bound for the divergence we get:

$\int _{\partial E}|T|d\sigma \geq \int _{\partial E}T\cdot n_{E}d\sigma =\int _{E}\operatorname {div} Tdx\geq n\int _{E}(\det \nabla T)^{\frac {1}{n}}dx$

We can now conclude with our explicit expression for the Jacobian of $T$ in $E$,

$n\int _{E}(\det \nabla T)^{\frac {1}{n}}dx=n\int _{E}{\Bigg (}{\frac {\omega _{n}}{|E|}}{\Bigg )}^{\frac {1}{n}}dx=n\omega _{n}^{1/n}|E|^{1-1/n}.$