Transportation

Matrix determinant lemma

Suppose we are given a square block matrix $A\in\Ma(n,\R)$: $$ A=\left(\begin{array}{cc} A_{11}&A_{12}\\ A_{21}&A_{22} \end{array}\right), $$ $A_{11}\in\Ma(l,\R)$, $A_{22}\in\Ma(m,\R)$, $l+m=n$. Assuming $A_{11}$ to be invertible, we get $$ \left(\begin{array}{cc} 1&0\\ -A_{21}A_{11}^{-1}&1 \end{array}\right) \left(\begin{array}{cc} A_{11}&A_{12}\\ A_{21}&A_{22} \end{array}\right) \left(\begin{array}{cc} 1&-A_{11}^{-1}A_{12}\\ 0&1 \end{array}\right) =\left(\begin{array}{cc} A_{11}&0\\ 0&A_{22}-A_{21}A_{11}^{-1}A_{12} \end{array}\right) $$

If $A$ is symmetric and $A_{11}$ invertible, then $A$ is positive definit, iff both $A_{11}$ and $A_{22}-A_{21}A_{11}^{-1}A_{21}^t$ are positive definit. In this case we have the matrix determinant lemma: $$ \det(A) =\det(A_{11})\det(A_{22}-A_{21}A_{11}^{-1}A_{21}^t) =\det(A_{22})\det(A_{11}-A_{21}^tA_{22}^{-1}A_{21}) $$
We will only make use of the particular case $l=1$: \begin{equation}\label{eq1}\tag{1} \det(A) =\det(A_{22})(a_{11}-A_{21}^tA_{22}^{-1}A_{21}) \leq a_{11}\det(A_{22}) \end{equation}

Transport inequality

$T_xT:T_xM\rar T_{T(x)}M$: $$ T_xT=Y_x(H_x+\Hess_x\theta), $$ where $Y_x\colon=T_{\nabla\theta}\exp_x$ and $H_x=\Hess_xd_{T(x)}^2/2$. Let $\a=\norm{\nabla\theta}$, $\g(t)\colon=\exp_x(t\nabla\theta)$, $E^1=\nabla\theta/\a$ and $E^1,E^2,\ldots,E^n$ an orthonormal basis of $T_xM$. The matrix of $T_xT$ with respect to this basis is given by $$ \left(\begin{array}{cc} 1&0\\ 0&Y \end{array}\right) \left( \left(\begin{array}{cc} 1&0\\ 0&K \end{array}\right) + \left(\begin{array}{cc} a&b^t\\ b&M \end{array}\right) \right) $$ where $H_x(E^1,E^1)=1$, $\Hess_x\theta(E^1,E^1)=a$, $b_j=\Hess_x\theta(E^j,E^1)$ for $j=2,\ldots,n$ and $$ K=(H_x(E^j,E^k))_{j,k=2}^n,\quad M=(\Hess_x\theta(E^j,E^k))_{j,k=2}^n~. $$ Since $T$ pushes $d\l=F\,d\s$ forward onto $d\nu=G\,d\s$ we get the following Monge-Ampere equation: $$ F(x)=G(T(x))\det T_xT~. $$ The $c$-concavity of $-\theta$ implies: $$ H_x+\Hess_x\theta\geq0~. $$ Finally by Bishop's comparism theorem (cf. I. Chavel p. 71) we have, presuming $\Ric\geq k(n-1)$: \begin{eqnarray} \det Y&\leq&v(\a)^n\colon=(\Sk(\a)/\a)^{n-1}\quad\mbox{and}\nonumber\\ \tr K&\leq&w(\a)\colon=(n-1)\a(\Ck/\Sk)(\a)\label{eq2}\tag{2}~. \end{eqnarray} where $\Ck\colon=\Sk^{\prime}$ and $$ \Sk(\a)\colon=\left\{\begin{array}{cl} \frac{\sin(\a\sqrt k)}{\sqrt k}&\mbox{if $k>0$}\\ \a&\mbox{if $k=0$}\\ \frac{\sinh(\a\sqrt{-k})}{\sqrt{-k}}&\mbox{if $k<0$} \end{array}\right. $$

If $\Ric\geq k(n-1)$ and $T(x)\colon=\exp_x(\nabla\theta)$ pushes forward the probability density $F$ onto $G$, which are both supported on a compact and geodesically convex domain $D$. Then: $$ n\int_D G^{1-1/n}\,d\s \leq\int_D\frac{(h\Sk^{n-1})^\prime(\a)}{\Sk^{n-1}(\a)}\,F^{1-1/n} -\int_D\frac{h(\a)}{\a}\la\nabla F^{1-1/n},\nabla\theta\ra~, $$ where $$ h(\a)\colon=\Big(n\int_0^\a\Sk(t)^{n-1}\,dt\Big)^{1/n}.~ $$
$\proof$ For any smooth function $\mu:(0,diam(D))\rar\R^+$ we get by \eqref{eq1} and \eqref{eq2}: \begin{eqnarray*} \frac1{G(T)} &=&\frac1{F}\det Y \det\left( \begin{array}{cc} 1+a& b^t\\ b&K+M \end{array} \right)\\ &\leq&\frac{v(\a)^n}{F}v(\a)^n \det\left( \begin{array}{cc} 1+a&0\\ 0&K+M \end{array} \right)\\ &=&\frac{v(\a)^n}{F} \det\left( \begin{array}{cc} (1+a)\mu(\a)^{1-n}&0\\ 0&(K+M)\mu(\a) \end{array} \right) \end{eqnarray*} By the AM-GM-inequality we have for all positive definit matrices $A$: \begin{equation}\label{eq3}\tag{3} (\det A)^{1/n}\leq n^{-1}\tr A \end{equation} and since $\D\theta=\tr\Hess\theta=a+\tr M$ we conclude by \eqref{eq2} that \begin{equation}\label{eq4}\tag{4} nG(T(x))^{-1/n} \leq F^{-1/n}v(\a)\Big( (1+a)\mu(\a)^{1-n}+\mu(\a)w(\a)+\mu(\a)(\D\theta-a) \Big)~. \end{equation} Let $\l$ and $\nu$ respectively be the measures with densits $F$ and $G$ respectively. Integration of the left hand side with respect to $\l$ yields $$ \int_D G(T)^{-1/n}\,d\l =\int_D G^{-1/n}\,d\l_T =\int_D G^{-1/n}\,d\nu =\int_D G^{1-1/n}~. $$ As for the right hand side we get by the divergence theorem \begin{eqnarray*} \int_D F^{-1/n}v(\a)\mu(\a)\D\theta\,d\l &=&\int_D F^{1-1/n}v(\a)\mu(\a)\D\theta\\ &=&-\int_D F^{1-1/n}(v\mu)^\prime(\a)\la\nabla\a,\nabla\theta\ra\\ &&-\int_D (v\mu)(\a)\la\nabla F^{1-1/n},\nabla\theta\ra\\ &&+\int_{\pa D}F^{1-1/n}v(\a)\mu(\a)\la\nabla\theta,N\ra \end{eqnarray*} 1. Since $F|\pa D=0$ the boundary term vanishes - if $D$ is geodesically convex, we have for all $x\in\pa D$: $\la\nabla\theta,N\ra\leq0$. Anyway, the integral over the boundary is non-positive. 2. For any vectorfield $X$ \begin{eqnarray*} \la\nabla\a,X\ra &=&X\norm{\nabla\theta} =X\la\nabla\theta,\nabla\theta\ra/2\a\\ &=&\la\nabla_X\nabla\theta,\nabla\theta\ra/\a =\Hess\theta(X,\nabla\theta)/\a \end{eqnarray*} and thus $\la\nabla\a,\nabla\theta\ra=\Hess\theta(\nabla\theta,\nabla\theta)/\a=a\a$. From these observations and \eqref{eq4} we now infere that \begin{eqnarray*} n\int_D G^{1-1/n} &\leq&\int_D F^{1-1/n}v(\a)\Big((1+a)\mu(\a)^{1-n}+\mu(\a)w(\a)-a\mu(\a)\Big)\\ &&-\int_D F^{1-1/n}(v\mu)^\prime(\a)a\a+(v\mu)(\a)\la\nabla F^{1-1/n},\nabla\theta\ra\\ &=&\int_D F^{1-1/n}\Big(v(\a)\mu(\a)^{1-n}-v(\a)\mu(\a)-(v\mu)^\prime(\a)\a\Big)a\\ &&+\int_D F^{1-1/n}\Big(\mu(\a)^{1-n}+\mu(\a)w(\a)\Big)v(\a)\\ &&-\int_D v(\a)\mu(\a)\la\nabla F^{1-1/n},\nabla\theta\ra\\ \end{eqnarray*} Eventually we choose $\mu$ to satisfy the differential equation $$ v(\a)\mu(\a)^{1-n}-v(\a)\mu(\a)-(v\mu)^\prime(\a)\a=0~. $$ For $h(\a)\colon=\a\mu(\a)v(\a)$ this amounts to \begin{eqnarray*} h^\prime(\a) &=&\mu(\a)v(\a)+(v\mu)^\prime(\a)\a =v(\a)\mu(\a)^{1-n}\\ &=&v(\a)(h(\a)/\a v(\a))^{1-n} =v(\a)^n\a^{n-1}h(\a)^{1-n} \end{eqnarray*} or equivalently: $$ n^{-1}(h^n)^\prime(\a)=v(\a)^n\a^{n-1}=\Sk(\a)^{n-1}~. $$ Thus putting $h(0)=0$, we obtain: \begin{equation}\label{eq5}\tag{5} h(\a)=\Big(n\int_0^\a\Sk(t)^{n-1}\,dt\Big)^{1/n}. \end{equation} which implies $v(\a)\mu(\a)=h(\a)/\a$ and \begin{eqnarray*} \Big(\mu(\a)^{1-n}+\mu(\a)w(\a)\Big)v(\a) &=&h^\prime(\a)+h(\a)w(\a)/\a\\ &=&h^\prime(\a)+(n-1)h(\a)(\Ck/\Sk)(\a)\\ &=&\frac{(h\Sk^{n-1})^\prime(\a)}{\Sk^{n-1}(\a)}~. \end{eqnarray*} $\eofproof$
Suppose $\Ric\geq k(n-1)$, $k\geq0$, and let $\l$ be the normalized Riemannian measure on a compact geodesically convex set $D$. Suppose $T(x)=\exp_x(\nabla\theta)$ pushes $\l$ forward onto $\nu(dx)=f(x)\,\l(dx)$, then $$ (n-1)\int_Du(\sqrt k\,\a)\,d\l \leq\int_D f\log f\,d\l, $$ where $u(t)\colon=1-\log(\sin(t)/t)-t\cot(t)$.
2. If $S(x)\colon=T^{-1}(x)=\exp_x(\nabla\phi)$, then: $$ \int f\log f\,d\mu +(n-1)\int u(\sqrt k\norm{\nabla\phi})f\,d\mu \leq-\int\la\nabla f,\nabla\phi\ra\,d\mu, $$
$\proof$ We have $F=I_D$ and $G=fI_D$ and thus by Monge-Ampere: for $x\in D$: $1=f(F(x))\det T_xT$ and thus: $\log f(T(x))=-\log\det T_xT$. Since for a positive definit matrix $A$: $\log\det A\leq n\log(\tr A/n)\leq\tr A-n$, we get with $\a=\norm{\nabla\theta}$ by \eqref{eq1} and \eqref{eq2}: \begin{eqnarray*} \log f(T(x)) &=&-\log\det T_xT \geq-\log v(\a)^n-\log\det\left(\begin{array}{cc} 1+a&0\\ 0&K+M \end{array}\right)\\ &\geq&-\log v(\a)^n-\log\det\left(\begin{array}{cc} (1+a)\mu(\a)^{1-n}&0\\ 0&(K+M)\mu(\a) \end{array}\right)\\ &\geq&n-\log v(\a)^n-\Big((1+a)\mu(\a)^{1-n}+\mu(\a)w(\a)+\mu(\a)(\D\theta-a)\Big) \end{eqnarray*} Integrating with respect to the normalized volume $\l$ we conclude by the divergence theorem: $$ \int_D\log f(T)\,d\l \geq n-\int_D\log v(\a)^n -\int_D\Big(\mu(\a)^{1-n}+\mu(\a)w(\a)\Big) -\int_D\Big(\mu(\a)^{1-n}-\mu(\a)-\mu^\prime(\a)\a\Big)a $$ This time we choose the auxilary function $\mu$ in such a way that $\mu(\a)^{1-n}-\mu(\a)-\mu^\prime(\a)\a=0$, i.e. $\mu(\a)=1$. It follows that \begin{eqnarray*} \int_D f\log f\,d\l &=&\int_D \log f\,d\nu =\int_D \log f\,d\l_F =\int_D \log f(F)\,d\l\\ &\geq&\int_D(n-1)-\log v(\a)^n-w(\a)\,d\l \end{eqnarray*} As for the other inequality we have by Monge-Ampere: $f(x)=\det T_xS$. By the same arguments as above (but now with $\a=\norm{\nabla\phi}$): $$ \log f=\log\det T_xS\leq-(n-1)u(\a)+\D\phi $$ Multiplying by $f$ and integrating with respect to $\l$, we get by the divergence theorem: \begin{eqnarray*} \int f\log f\,d\l &\leq&-(n-1)\int u(\a)f\,d\l +\int f\D\phi\,d\l\\ &\leq&-(n-1)\int u(\a)f\,d\l -\int\la\nabla f,\nabla\phi\ra\,d\l \end{eqnarray*}

Non-linear transport inequality

Instead of $\det^{1/n}A\leq\tr A/n$ we can use \begin{eqnarray*} \det\left( \begin{array}{cc} 1+a&0\\ 0&K+M \end{array} \right)^{1/n} &=&(1+a)^{1/n}\det^{1/(n-1)}(K+M)^{1-1/n}\\ &\leq&(1+a)^{1/n}\Big(\frac1{n-1}\tr(K+M)\Big)^{1-1/n}\\ &\leq&(n-1)^{1/n-1}(1+a)^{1/n}\Big(w+\D\theta-a\Big)^{1-1/n} \end{eqnarray*} and estimate $$ \int F^{1-1/n}v\Big(\frac{1+a}{\mu^{n-1}}\Big)^{1/n} \Big(\mu(w+\D\theta-a)\Big)^{1-1/n} $$ by Hölder's inequality (only the exponents $n$ and $n/(n-1)$ are conceivable): $$ \Big(\int F^{1-1/n}v\mu^{-n+1}(1+a)\Big)^{1/n} \Big(\int F^{1-1/n}v\mu(w+\D\theta-a)\Big)^{1-1/n} $$ and equality holds if and only if $(1+a)$ is a constant multiple of $\mu^n(w+\D\theta-a)$. By the divergence theorem we get for the second integral: $$ \int F^{1-1/n}v\mu w -\int F^{1-1/n}(v\mu+(v\mu)^\prime\a)a -v\mu\int\la\nabla F^{1-1/n},\nabla\theta\ra $$ Choosing again $\mu$ such that $v\mu+(v\mu)^\prime\a=v\mu^{-n+1}$ (i.e. $\mu=h/\a v$) and putting $x=\int F^{1-1/n}v\mu^{-n+1}a$: $$ \Big(\int F^{1-1/n}v\mu^{-n+1}+x\Big)^{1/n} \Big(\int F^{1-1/n}v\mu w-v\mu\la\nabla F^{1-1/n},\nabla\theta\ra-x\Big)^{1-1/n} $$ which we write as $(A+x)^{1/n}(B-x)^{1-1/n}$ and $$ A\colon=\int F^{1-1/n}v\mu^{-n+1},\quad B\colon=\int F^{1-1/n}v\mu w-v\mu\la\nabla F^{1-1/n},\nabla\theta\ra~. $$ The maximal value of this expression (as a function of $x$) is attained at $x_0\colon=(B-(n-1)A)$ and is given by: $(n-1)^{1-1/n}(A+B)/n$, which yields: $(n-1)^{1-1/n}\int G^{1-1/n}\leq(n-1)^{1-1/n}(A+B)/n$ and we are back to proposition. Thus we have a clumsy non-linear version of proposition -- the appearance of the function $a$ however is a pain!
If $\Ric\geq k(n-1)$ and $T(x)\colon=\exp_x(\nabla\theta)$ pushes forward the probability density $F$ onto $G$. Then \begin{eqnarray*} (n-1)^{1-1/n}\int G^{1-1/n}\,d\s &\leq& \Big(\int F^{1-1/n}v\mu^{1-n}(1+a)\Big)^{1/n}\\ &&\Big(\int F^{1-1/n}v(\mu w-\mu^{1-n}a)- v\mu\la\nabla F^{1-1/n},\nabla\theta\ra\Big)^{1-1/n}~. \end{eqnarray*}

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Last modified: Thu Apr 30 17:26:54 CEST 2020