$u(a^{\prime})\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{\leq}}u(a)+\frac{\epsilon}{2}\qquad\text{and}\qquad
z(a^{\prime})\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{\geq}}0\,.$
(17)

Let $x^{*}\in[a,b)$ be given by

$x^{*}\vcentcolon=\inf\\{x\in(a,b):\partial_{x}u{\downarrow}(x)=0\\}\,,$

which is well-defined by the observation above and the fact that
$\partial_{x}u{\downarrow}\leq 0$ in all of $I$ (since $u{\downarrow}$ is
nonincreasing and continuously differentiable). By the continuity of
$\partial_{x}u{\downarrow}$, we have that
$\partial_{x}u{\downarrow}(x^{*})=0$. Since $\partial_{x}u{\downarrow}(a)\leq
0$, it must be the case that either $a=x^{*}$ or

$\mathcal{L}\left\\{x\in(a,x^{*}):\partial_{x}\partial_{x}u{\downarrow}(x)\geq
0\right\\}>0\,.$ (18)

We consider each case separately. First, suppose $a=x^{*}$. We claim that for
all $\delta>0$, there exists $x\in(a,a+\delta)$ such that $z(x)\geq 0$.
Suppose for a contradiction that this is not the case, and fix $\delta>0$ such
that for all $x\in(a,a+\delta)$, $z(x)<0$. Then since
$z=\partial_{x}\partial_{x}u{\downarrow}$ almost everywhere, we conclude that
$\partial_{x}\partial_{x}u{\downarrow}<0$ a.e. in $(a,a+\delta)$. Hence
$\partial_{x}u{\downarrow}(x)<0$ for all $x\in(a,a+\delta)$, contradicting the
assumption that $a=x^{*}$ given the definition of $x^{*}$. Thus the claim
holds. Now, using the continuity of $u$, choose $\delta>0$ small enough and
choose $a^{\prime}\in(a,a+\delta)$ so that $z(a^{\prime})\geq 0$ and moreover
$u(a^{\prime})\leq u(a)+\epsilon/2$. This choice of $a^{\prime}$ satisfies
(17).

Second, suppose $a<x^{*}$ and (18) holds. Since
$z=\partial_{x}\partial_{x}u{\downarrow}$ almost everywhere, choose
$a^{\prime}\in(a,x^{*})$ such that $z(a^{\prime})\geq 0$, which is possible by
(18). Now, since $\partial_{x}u{\downarrow}<0$ for all
$x\in(a,a^{\prime})\subset(a,x^{*})$ by the choice of $x^{*}$, Lemma 5.22
implies that $u{\uparrow}$ is constant in $(a,a^{\prime})$. By the continuity
of $u{\uparrow}$, we conclude that $u{\uparrow}(a)=u{\uparrow}(a^{\prime})$
and hence, since $u{\downarrow}$ is nonincreasing, we have

$u(a^{\prime})=u{\uparrow}(a^{\prime})+u{\downarrow}(a^{\prime})\leq
u{\uparrow}(a)+u{\downarrow}(a)=u(a)\,,$

and again (17) is satisfied. This concludes the choice of $a^{\prime}$.

Now, we may assume without loss of generality that $\epsilon<\left\lvert
u(a)-u(b)\right\rvert/2$. Therefore our choice of $a^{\prime}$ yields an
interval $(a^{\prime},b)$ which, using (17) and recalling that $u(a)<u(b)$,
satisfies $u(a^{\prime})<u(b)$. Therefore, repeating a symmetric version of
the argument above yields a choice of $b^{\prime}\in(a^{\prime},b)$ satisfying

$u(b^{\prime})\geq u(b)-\frac{\epsilon}{2}\qquad\text{and}\qquad
z(b^{\prime})\leq 0\,.$ (19)

Combining (17) and (19), we conclude that

$v(b^{\prime})-v(a^{\prime})=\Big{[}u(b^{\prime})-u(a^{\prime})\Big{]}-\lambda\underbrace{\Big{[}z(b^{\prime})-z(a^{\prime})\Big{]}}_{\leq
0}\geq u(b)-u(a)-\epsilon=\left\lvert u(a)-u(b)\right\rvert-\epsilon\,.\qed$

######  Lemma 5.33.

Let $u\in D(\partial\mathcal{E}^{-})$, suppose $u$ is not continuous, and let
$z\in-\partial\mathcal{E}^{-}(u)$. Let $\lambda>0$. Then $u-\lambda z$ is not
continuous.

###### Proof.

Recall that, by Proposition 5.18, $u,z$ form a weak solution to the static
Neumann problem, and by Lemma 5.21 we have $u{\downarrow}\in H^{2}(I)$ with
$\partial_{x}u{\downarrow}=0$ on $\\{0,1\\}$ and
$z=\partial_{x}\partial_{x}u{\downarrow}$ in $L^{2}(I)$. In particular,
$u{\downarrow}$ is continuously differentiable by Fact 5.27.

Since $u{\downarrow}\in H^{2}(I)$ is continuous while $u$ is not continuous,
we conclude that $u{\uparrow}$ is not continuous and, since it is monotone, it
contains only jump discontinuities. Let $x_{0}\in I$ be a point of jump
discontinuity of $u{\uparrow}$, i.e. a point such that $L_{-}<L_{+}$ where

$L_{-}\vcentcolon=\lim_{x\to x_{0}^{-}}u{\uparrow}(x)\qquad\text{and}\qquad
L_{+}\vcentcolon=\lim_{x\to x_{0}^{+}}u{\uparrow}(x)\,.$

We first claim that $\partial_{x}u{\downarrow}(x_{0})=0$. Indeed, it is clear
that $\partial_{x}u{\downarrow}\leq 0$ since $u{\downarrow}$ is nonincreasing
and continuously differentiable. If we had
$\partial_{x}u{\downarrow}(x_{0})<0$, then by continuity
$\partial_{x}u{\downarrow}$ would be strictly negative in a neighbourhood of
$x_{0}$, in which case Lemma 5.22 would imply that $u{\uparrow}$ is constant
in a neighbourhood of $x_{0}$, contradicting the fact that $L_{-}<L_{+}$.
Hence the claim holds.

We now claim that, for all $\delta>0$, there exists $x\in(x_{0},x_{0}+\delta)$
such that $z(x)\leq 0$. Suppose for a contradiction that there exists
$\delta>0$ such that for all $x\in(x_{0},x_{0}+\delta)$, $z(x)>0$. Then, since
$z=\partial_{x}\partial_{x}u{\downarrow}$ a.e. and
$\partial_{x}u{\downarrow}(x_{0})=0$, we conclude that
$\partial_{x}u{\downarrow}>0$ in $(x_{0},x_{0}+\delta)$, contradicting the
fact that $\partial_{x}u{\downarrow}\leq 0$. Hence the claim holds. By the
same reasoning, we conclude that for all $\delta>0$ there exists
$x\in(x_{0}-\delta,x_{0})$ such that $z(x)\geq 0$.

Thus, for all $\delta>0$ there exist points
$x_{-}\in(x_{0}-\delta,x_{0}),x_{+}\in(x_{0},x_{0}+\delta)$ such that
$z(x_{+})-z(x_{-})\leq 0$ and hence

$\displaystyle(u-\lambda z)(x_{+})-(u-\lambda z)(x_{-})$
$\displaystyle=\Big{[}u{\uparrow}(x_{+})-u{\uparrow}(x_{-})\Big{]}+\Big{[}u{\downarrow}(x_{+})-u{\downarrow}(x_{-})\Big{]}-\lambda\Big{[}z(x_{+})-z(x_{-})\Big{]}$
$\displaystyle\geq\Big{[}u{\uparrow}(x_{+})-u{\uparrow}(x_{-})\Big{]}+\Big{[}u{\downarrow}(x_{+})-u{\downarrow}(x_{-})\Big{]}\,.$

Since $u{\uparrow}\to L_{+}$ as $x\downarrow x_{0}$ and $u{\uparrow}\to L_{-}$
as $x\uparrow x_{0}$ and $u{\uparrow}$ is nondecreasing, we have
$u{\uparrow}(x_{+})-u{\uparrow}(x_{-})\geq L_{+}-L_{-}$. By the continuity of
$u{\downarrow}$, we can let $\delta>0$ be small enough so that
$u{\downarrow}(x_{+})-u{\downarrow}(x_{-})\geq-\frac{L_{+}-L_{-}}{2}$. We
conclude that, for all sufficiently small $\delta>0$, there exist points
$x_{-}\in(x_{0}-\delta,x_{0}),x_{+}\in(x_{0},x_{0}+\delta)$ such that

$(u-\lambda z)(x_{+})-(u-\lambda z)(x_{-})\geq\frac{L_{+}-L_{-}}{2}\,,$

and thus $u-\lambda z$ is not continuous. ∎

######  Lemma 5.34 (Specialization of [Bre73, Theorem 4.4]).

Let $H$ be a Hilbert space and let $\tau:H\to[0,+\infty]$ be a convex, proper,
and lower semicontinuous functional such that
$\tau(\mathrm{Proj}_{\overline{D(A)}}x)\leq\tau(x)$ for all $x\in H$. Let
$A:H\to 2^{H}$ be a maximal monotone operator and let $S_{t}$ be the semigroup
generated by $-A$. Then the following are equivalent:

  1. 1.

$\tau((I+\lambda A)^{-1}x)\leq\tau(x)$ for all $x\in H$ and $\lambda>0$; and

  2. 2.

$\tau(S_{t}x)\leq\tau(x)$ for all $x\in\overline{D(A)}$ and $t\geq 0$.

In the statement above, $(I+\lambda A)^{-1}:H\to D(A)$ is the _resolvent_ of
$A$ (also denoted by $J_{\lambda}$, see e.g. [Eva10, p. 563]).

######  Proposition 5.35 ($\varphi$-monotonicity of solutions).

Let $u_{0}\in\mathcal{U}$ and let $\bm{u}\in C([0,+\infty);L^{2}(I))$ be the
solution to the gradient flow problem with initial data $u_{0}$. Then for all
$0\leq t_{1}\leq t_{2}<+\infty$, we have
$\varphi(\bm{u}(t_{1}))\geq\varphi(\bm{u}(t_{2}))$.

###### Proof.

This is a direct consequence of Lemma 5.34. Indeed, letting
$H\vcentcolon=L^{2}(I)$ and $A\vcentcolon=\partial\mathcal{E}^{-}$ (which is
maximal monotone as remarked earlier), we first observe that indeed
$\varphi(\mathrm{Proj}_{\overline{D(A)}}f)\leq\varphi(f)$ for all $f\in
L^{2}(I)$, since by 5.29 we have $\mathrm{Proj}_{\overline{D(A)}}f=f$. Thus it
suffices to show that for all $f\in L^{2}(I)$ and all $\lambda>0$,

$\varphi((I+\lambda
A)^{-1}f)\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{\leq}}\varphi(f)\,.$

But this is equivalent to showing that, for all $u\in D(A)$, $-z\in A(u)$ and
$\lambda>0$,

$\varphi(u)\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{\leq}}\varphi(u-\lambda
z)\,,$

which is precisely Lemma 5.31. ∎

######  Corollary 5.36 (Preservation of $H^{1}$ regularity).

Suppose $u_{0}\in H^{1}(I)$, and let $\bm{u}\in C([0,+\infty);L^{2}(I))$ be
the solution to the gradient flow problem with initial data $u_{0}$. Then
$\bm{u}(t)\in H^{1}(I)$ for all $t>0$.

###### Proof.

This is an immediate consequence of Proposition 5.35 and the definition of
$\varphi$. ∎

###  5.5 Preservation of Lipschitz regularity

It will also be useful to control the Lipschitz regularity of solutions. At a
high level, we follow a similar strategy as in Section 5.4. Recall that
$W^{1,\infty}(I)$ is equivalent to the space of Lipschitz real-valued
functions on $I$, up to identification of almost everywhere equal functions.
Define $\psi:L^{2}(I)\to[0,+\infty]$ by

$\psi(u)\vcentcolon=\begin{cases}\|\partial_{x}u\|_{L^{\infty}(I)}&\text{if
}u\in W^{1,\infty}(I)\\\ +\infty&\text{otherwise.}\end{cases}$

######  Fact 5.37.

Let $u\in L^{2}(I)$ and let $M\in\mathbb{R}_{\geq 0}$. Then $\psi(u)\leq M$ if
and only if $u=f$ a.e. for some $M$-Lipschitz function $f:I\to\mathbb{R}$.
Moreover, in this case $f$ is the (unique) continuous representative of $u$
and $\psi(u)$ is the Lipschitz constant of $f$.

######  Claim 5.38.

The functional $\psi$ is convex, proper and lower semicontinuous.

###### Proof.

Convexity and properness are straightforward; it remains to verify lower
semicontinuity. Since $L^{2}(I)$ is a metric space, it suffices to check
sequential lower semicontinuity. Let $(u_{n})_{n\in\mathbb{N}}$ be a sequence
in $L^{2}(I)$ such that $u_{n}\to u$ in $L^{2}(I)$. We need to show that

$\psi(u)\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{\leq}}\liminf_{n\to\infty}\psi(u_{n})\,.$

The only relevant case is when the RHS above is finite, so suppose there
exists a subsequence $(u_{n_{k}})_{k\in\mathbb{N}}$ such that
$\lim_{k\to\infty}\psi(u_{n_{k}})=M<+\infty$. By extracting a subsequence if
necessary, we may assume that $\psi(u_{n_{k}})<+\infty$, and thus
$u_{n_{k}}\in W^{1,\infty}(I)$, for every $k$. For simplicity and using Fact
5.37, fix for each $n_{k}$ the continuous representative of $u_{n_{k}}$, which
we denote by the same name. Then each $u_{n_{k}}$ is $M_{k}$-Lipschitz with
$M_{k}\to M$.

Since $u_{n_{k}}\to u$ in $L^{2}(I)$, it is standard that we may extract a
subsequence that converges to $u$ pointwise almost everywhere. Denote this
further subsequence again by $(u_{n_{k}})_{k}$, and let $N\subset I$ be a
measure zero set such that $u_{n_{k}}\to u$ pointwise in $I\setminus N$. Then
for each $x\neq y$ in $I\setminus N$, we have

$\left\lvert u(x)-u(y)\right\rvert=\lim_{k\to\infty}\left\lvert
u_{n_{k}}(x)-u_{n_{k}}(y)\right\rvert\leq\lim_{k\to\infty}M_{k}|x-y|=M|x-y|\,.$

By Lemma A.3, $u$ is a.e. equal to an $M$-Lipschitz function, so $\psi(u)\leq
M$ by Fact 5.37, as needed. ∎

######  Definition 5.39.

For any $w:I\to\mathbb{R}$ and distinct $x,y\in I$, let

$\operatorname{\mathsf{slope}}_{w}(x,y)\vcentcolon=\frac{w(y)-w(x)}{y-x}\,.$

######  Lemma 5.40.

Let $u\in D(\partial\mathcal{E}^{-})$ and let
$z\in-\partial\mathcal{E}^{-}(u)$. Then for all $\lambda>0$, $\psi(u-\lambda
z)\geq\psi(u)$.

###### Proof.

We follow the proof outline from Lemma 5.31, but some of the technical details
are different. As in that proof, we have $u{\downarrow}\in H^{2}(I)$ with
$\partial_{x}u{\downarrow}=0$ on $\\{0,1\\}$ and
$z=\partial_{x}\partial_{x}u{\downarrow}$ in $L^{2}(I)$.

We may assume that $u-\lambda z\in W^{1,\infty}(I)$, since otherwise
$\psi(u-\lambda z)=+\infty$ and there is nothing to prove. In particular, this
establishes that $u-\lambda z$ is continuous.

When $u$ is not continuous, Lemma 5.33 yields that $u-\lambda z$ is not
continuous, a contradiction. Therefore we may assume that $u$ is continuous.
Since both $u$ and $u-\lambda z$ are continuous, we conclude that $z$ is also
continuous. Let $v\vcentcolon=u-\lambda z$. By Fact 5.37, it suffices to show
that for every $(a,b)\subset I$ and all $\epsilon>0$, there exists
$(a^{\prime},b^{\prime})\subset I$ such that

$\left\lvert\operatorname{\mathsf{slope}}_{v}(a^{\prime},b^{\prime})\right\rvert\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{\geq}}\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert-\epsilon\,.$
(20)

If $u(a)<u(b)$, then the interval $(a^{\prime},b^{\prime})\subseteq(a,b)$
given by Lemma 5.32 with parameter $\epsilon(b-a)$ satisfies (20); and if
$u(a)=u(b)$, then any interval will do. Therefore suppose $u(a)>u(b)$, so in
particular $\operatorname{\mathsf{slope}}_{u}(a,b)<0$.

Note that a sufficient condition for (20) is that
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{\prime},b^{\prime})\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert-\epsilon$
with $u(a^{\prime})>u(b^{\prime})$, $z(a^{\prime})\leq 0$ and
$z(b^{\prime})\geq 0$, since in that case,

$\displaystyle\left\lvert\operatorname{\mathsf{slope}}_{v}(a^{\prime},b^{\prime})\right\rvert$
$\displaystyle=\frac{\left\lvert(u-\lambda z)(b^{\prime})-(u-\lambda
z)(a^{\prime})\right\rvert}{b^{\prime}-a^{\prime}}=\Big{|}\underbrace{u(b^{\prime})-u(a^{\prime})}_{<0}-\lambda\underbrace{(z(b^{\prime})-z(a^{\prime}))}_{\geq
0}\Big{|}\left(\frac{1}{b^{\prime}-a^{\prime}}\right)$
$\displaystyle\geq\frac{\left\lvert
u(b^{\prime})-u(a^{\prime})\right\rvert}{b^{\prime}-a^{\prime}}=\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{\prime},b^{\prime})\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert-\epsilon\,,$

which is (20). We now find $a^{\prime}$ and $b^{\prime}$ satisfying the
aforementioned conditions.

For each point $x\in I$, say $x$ is _left-favourable_ if every neighbourhood
$(x-\delta,x+\delta)$ contains a point $x^{\prime}$ such that
$z(x^{\prime})\leq 0$. Similarly, say $x$ is _right-favourable_ if every
neighbourhood $(x-\delta,x+\delta)$ contains a point $x^{\prime}$ such that
$z(x^{\prime})\geq 0$.

We claim that there exist points $a^{*}<b^{*}$ in $I$ such that $a^{*}$ is
left-favourable, $b^{*}$ is right-favourable, and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b^{*})\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert$
with $u(a^{*})>u(b^{*})$. Let us first show that this claim yields the desired
points $a^{\prime}$ and $b^{\prime}$, and then proceed to prove the claim.
Suppose we have $a^{*}$ and $b^{*}$ as claimed. Then by the continuity of $u$,
we may fix sufficiently small $\delta>0$ and find points
$a^{\prime}\in(a^{*}-\delta,a^{*}+\delta)$ and
$b^{\prime}\in(b^{*}-\delta,b^{*}+\delta)$ such that $a^{\prime}<b^{\prime}$,
$z(a^{\prime})\leq 0$, $z(b^{\prime})\geq 0$, $u(a^{\prime})>u(b^{\prime})$,
and, for
$\alpha\vcentcolon=\frac{\epsilon}{2}\cdot\frac{b^{*}-a^{*}}{u(a^{*})-u(b^{*})}$
and $\beta\vcentcolon=\frac{\epsilon(b^{*}-a^{*})}{2}$,

$\displaystyle\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{\prime},b^{\prime})\right\rvert$
$\displaystyle=\frac{u(a^{\prime})-u(b^{\prime})}{b^{\prime}-a^{\prime}}\geq\frac{u(a^{*})-u(b^{*})-\beta}{(b^{*}-a^{*})(1+\alpha)}\geq\frac{u(a^{*})-u(b^{*})}{b^{*}-a^{*}}(1-\alpha)-\frac{\epsilon/2}{1+\alpha}$
$\displaystyle\geq\frac{u(a^{*})-u(b^{*})}{b^{*}-a^{*}}-\epsilon=\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b^{*})\right\rvert-\epsilon\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert-\epsilon\,,$

as desired, where we used the inequality $\frac{1}{1+\alpha}\geq 1-\alpha$. We
now establish the existence of $a^{*}$ and $b^{*}$.

We first find $a^{*}$. If $a$ is left-favourable, then choose $a^{*}=a$; note
that, of course, we have $u(a^{*})>u(b)$ and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert$.
Otherwise, $a$ is not left-favourable, which gives some $\delta>0$ such that
$z(x)>0$ for every $x\in(a-\delta,a+\delta)$, and hence
$\partial_{x}\partial_{x}u{\downarrow}>0$ a.e. in $(a-\delta,a+\delta)$. We
now consider two cases. Recall that
$\operatorname{\mathsf{slope}}_{u}(a,b)<0$.

First, suppose
$\partial_{x}u{\downarrow}(a)\geq\operatorname{\mathsf{slope}}_{u}(a,b)$. Note
that it cannot be the case that
$\partial_{x}u{\downarrow}(x)>\operatorname{\mathsf{slope}}_{u}(a,b)$ for all
$x\in(a,b)$, since otherwise we would have

$u(b)=u{\uparrow}(b)+u{\downarrow}(b)\geq
u{\uparrow}(a)+u{\downarrow}(a)+\int_{(a,b)}\partial_{x}u{\downarrow}\odif
x>u(a)+(b-a)\operatorname{\mathsf{slope}}_{u}(a,b)=u(b)\,,$

a contradiction. Therefore we may let

$a^{*}\vcentcolon=\inf\left\\{x\in(a,b):\partial_{x}u{\downarrow}(x)\leq\operatorname{\mathsf{slope}}_{u}(a,b)\right\\}\,.$

By the continuity of $\partial_{x}u{\downarrow}$ (recall that
$u{\downarrow}\in H^{2}(I)$ is continuously differentiable), we conclude that
$\partial_{x}u{\downarrow}(a^{*})=\operatorname{\mathsf{slope}}_{u}(a,b)$. We
also observe that we must have $a^{*}>a$ since, as noted above, we have
$\partial_{x}\partial_{x}u{\downarrow}(x)>0$ a.e. in some neighbourhood
$(a-\delta,a+\delta)$, which implies that
$\partial_{x}u{\downarrow}(x)>\partial_{x}u{\downarrow}(a)\geq\operatorname{\mathsf{slope}}_{u}(a,b)$
for all $x\in(a,a+\delta)$. We claim that $a^{*}$ is left-favourable. Indeed,
otherwise there would be some $\delta>0$ such that $z(x)>0$ for all
$x\in(a^{*}-\delta,a^{*}+\delta)$, and since
$z=\partial_{x}\partial_{x}u{\downarrow}$ a.e. we would conclude that
$\partial_{x}u{\downarrow}(x)<\partial_{x}u{\downarrow}(a^{*})=\operatorname{\mathsf{slope}}_{u}(a,b)$
for $x\in(a^{*}-\delta,a^{*})$, contradicting the choice of $a^{*}$. Hence we
have found a left-favourable $a^{*}$ in this case; we claim that $a^{*}$ also
satisfies $u(a^{*})>u(b)$ and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert$.
The first inequality holds since

$\displaystyle u(a^{*})-u(b)$
$\displaystyle=u{\downarrow}(a^{*})+u{\uparrow}(a^{*})-\big{[}u(a)+(b-a)\operatorname{\mathsf{slope}}_{u}(a,b)\big{]}$
$\displaystyle\geq
u{\downarrow}(a)+\int_{(a,a^{*})}\partial_{x}u{\downarrow}\odif
x+u{\uparrow}(a)-u{\uparrow}(a)-u{\downarrow}(a)-(b-a)\operatorname{\mathsf{slope}}_{u}(a,b)$
$\displaystyle=\int_{(a,a^{*})}\left(\underbrace{\partial_{x}u{\downarrow}(x)}_{>\operatorname{\mathsf{slope}}_{u}(a,b)}-\operatorname{\mathsf{slope}}_{u}(a,b)\right)\odif
x-(b-a^{*})\underbrace{\operatorname{\mathsf{slope}}_{u}(a,b)}_{<0}$
$\displaystyle>0\,,$

and the second inequality holds since

$\displaystyle(b-a^{*})(b-a)\Big{(}\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert-\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert\Big{)}$
$\displaystyle\qquad=(b-a^{*})(b-a)\left(\frac{u(a^{*})-u(b)}{b-a^{*}}-\frac{u(a)-u(b)}{b-a}\right)$
$\displaystyle\qquad=\Big{(}u(a^{*})-u(b)\Big{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad=\Bigg{(}u{\uparrow}(a^{*})+u{\downarrow}(a)+\int_{(a,a^{*})}\underbrace{\partial_{x}u{\downarrow}(x)}_{>\operatorname{\mathsf{slope}}_{u}(a,b)}\odif
x-u(b)\Bigg{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad>\Big{(}u{\uparrow}(a)+u{\downarrow}(a)+(a^{*}-a)\operatorname{\mathsf{slope}}_{u}(a,b)-u(b)\Big{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad=\Big{(}u(a)+(a^{*}-a)\frac{u(b)-u(a)}{b-a}-u(b)\Big{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad=\Big{(}u(a)-u(b)\Big{)}\Big{(}(b-a)-(b-a^{*})-(a^{*}-a)\Big{)}$
$\displaystyle\qquad=0\,.$

Therefore in the first case we have found a left-favourable $a^{*}\in(a,b)$
such that $u(a^{*})>u(b)$ and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert$.

Second, suppose
$\partial_{x}u{\downarrow}(a)<\operatorname{\mathsf{slope}}_{u}(a,b)$. We
proceed similarly, but with points to the left of $a$ instead. Namely, it
cannot be the case that
$\partial_{x}u{\downarrow}(x)<\operatorname{\mathsf{slope}}_{u}(a,b)$ for all
$x\in(0,a)$, since by continuity this would imply that
$\partial_{x}u{\downarrow}(0)\leq\operatorname{\mathsf{slope}}_{u}(a,b)<0$, a
contradiction. Hence we may define

$a^{*}\vcentcolon=\sup\left\\{x\in(0,a):\partial_{x}u{\downarrow}(x)\geq\operatorname{\mathsf{slope}}_{u}(a,b)\right\\}\,.$

By continuity, we conclude that
$\partial_{x}u{\downarrow}(a^{*})=\operatorname{\mathsf{slope}}_{u}(a,b)$,
which also implies that $a^{*}<a$. We claim that $a^{*}$ is left-favourable.
Indeed, if it was not, then for some $\delta>0$ we would have $z(x)>0$ for all
$x\in(a^{*}-\delta,a^{*}+\delta)$, and since
$z=\partial_{x}\partial_{x}u{\downarrow}$ a.e. we would conclude that
$\partial_{x}u{\downarrow}(x)>\partial_{x}u{\downarrow}(a^{*})=\operatorname{\mathsf{slope}}_{u}(a,b)$
for all $x\in(a^{*},a^{*}+\delta)$, contradicting the choice of $a^{*}$. We
now claim that $a^{*}$ also satisfies $u(a^{*})>u(b)$ and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert$.

To prove the first inequality, we first observe that
$\partial_{x}u{\downarrow}(x)<\operatorname{\mathsf{slope}}_{u}(a,b)<0$ for
all $x\in(a^{*},a)$, which by Lemma 5.22 implies that $u{\uparrow}$ is
constant in $(a^{*},a)$. Since $u{\uparrow}$ is continuous (because $u$ and
$u{\downarrow}$ are), we conclude that $u{\uparrow}(a^{*})=u{\uparrow}(a)$.
Therefore we have

$u(a^{*})-u(a)=u{\downarrow}(a^{*})-u{\downarrow}(a)\geq 0$

since $u{\downarrow}$ is nonincreasing, and hence $u(a^{*})\geq u(a)>u(b)$. As
for the second inequality, we have

$\displaystyle(b-a^{*})(b-a)\Big{(}\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert-\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert\Big{)}$
$\displaystyle\qquad=(b-a^{*})(b-a)\left(\frac{u(a^{*})-u(b)}{b-a^{*}}-\frac{u(a)-u(b)}{b-a}\right)$
$\displaystyle\qquad=\Big{(}u(a^{*})-u(b)\Big{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad=\Bigg{(}u{\uparrow}(a)+u{\downarrow}(a)-\int_{(a^{*},a)}\underbrace{\partial_{x}u{\downarrow}(x)}_{<\operatorname{\mathsf{slope}}_{u}(a,b)}\odif
x-u(b)\Bigg{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad>\Bigg{(}u(a)-(a-a^{*})\frac{u(b)-u(a)}{b-a}-u(b)\Bigg{)}(b-a)-\Big{(}u(a)-u(b)\Big{)}(b-a^{*})$
$\displaystyle\qquad=\Big{(}u(a)-u(b)\Big{)}\Big{(}(b-a)-(b-a^{*})+(a-a^{*})\Big{)}$
$\displaystyle\qquad=0\,.$

Therefore in any case we have found a left-favourable $a^{*}\in(0,b)$ such
that $u(a^{*})>u(b)$ and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a,b)\right\rvert$.

Repeating an analogous argument for the right endpoint of the interval
$(a^{*},b)$, we find a right-favourable $b^{*}\in(a^{*},1)$ such that
$u(a^{*})>u(b^{*})$ and
$\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b^{*})\right\rvert\geq\left\lvert\operatorname{\mathsf{slope}}_{u}(a^{*},b)\right\rvert$,
which concludes the proof as explained above. ∎

######  Proposition 5.41 ($\psi$-monotonicity of solutions).

Let $u_{0}\in\mathcal{U}$ and let $\bm{u}\in C([0,+\infty);L^{2}(I))$ be the
solution to the gradient flow problem with initial data $u_{0}$. Then for all
$0\leq t_{1}\leq t_{2}<+\infty$, we have
$\psi(\bm{u}(t_{1}))\geq\psi(\bm{u}(t_{2}))$.

###### Proof.

As in the proof of Proposition 5.35, this follows from Lemmas 5.40 and 5.34. ∎

######  Corollary 5.42 (Preservation of Lipschitz regularity).

Suppose $u_{0}\in W^{1,\infty}(I)$, and let $\bm{u}\in
C([0,+\infty);L^{2}(I))$ be the solution to the gradient flow problem with
initial data $u_{0}$. Then $\bm{u}(t)\in W^{1,\infty}(I)$ for all $t>0$.

###### Proof.

This is an immediate consequence of Proposition 5.41 and the definition of
$\psi$. ∎

###  5.6 Exponential decay of directed Dirichlet energy

######  Proposition 5.43.

There exists a constant $K>0$ such that the following holds. Let
$u\in\mathcal{U}$. Then for all $t>0$,

$\mathcal{E}^{-}(P_{t}u)\leq e^{-Kt}\mathcal{E}^{-}(u)\,.$

###### Proof.

Let $u\in\mathcal{U}$ and let $\bm{u}(t)=P_{t}u$ be the corresponding solution
to the gradient flow problem. Recall that Proposition 5.17 gives, for all
$t>0$,

$\mathcal{E}^{-}(\bm{u}(t))=\mathcal{E}^{-}(u)+\int_{(0,t)}-\|\bm{u^{\prime}}(s)\|_{L^{2}(I)}^{2}\odif
s\,.$

It follows that $t\mapsto\mathcal{E}^{-}(\bm{u}(t))$ is absolutely continuous
on every interval $[0,T]$ with weak derivative
$\partial_{t}\mathcal{E}^{-}(\bm{u}(t))=-\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}^{2}$.
Moreover for a.e. $t>0$ we have that $\bm{u}(t),\bm{u^{\prime}}(t)$ form a
weak solution to the static Neumann problem, and thus
$\bm{u^{\prime}}(t)=\partial_{x}\partial_{x}\bm{u}(t){\downarrow}$ a.e. in $I$
and $\partial_{x}\bm{u}(t){\downarrow}=0$ on $\\{0,1\\}$ by Lemma 5.21.
Therefore the Poincaré inequality (for zero-on-the-boundary functions) yields

$\mathcal{E}^{-}(\bm{u}(t))=\frac{1}{2}\|\partial_{x}\bm{u}(t){\downarrow}\|_{L^{2}(I)}^{2}\leq\frac{1}{2}C\|\partial_{x}\partial_{x}\bm{u}(t){\downarrow}\|_{L^{2}(I)}^{2}=\frac{C}{2}\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}^{2}$

for some constant $C>0$. It follows that for a.e. $t>0$, we have

$\partial_{t}\mathcal{E}^{-}(\bm{u}(t))=-\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}^{2}\leq-\frac{2}{C}\mathcal{E}^{-}(\bm{u}(t))\,,$

which implies that for all $t>0$,

$\mathcal{E}^{-}(\bm{u}(t))\leq e^{-2t/C}\mathcal{E}^{-}(u)\,.\qed$

The exponential decay of $t\mapsto\mathcal{E}^{-}(P_{t}u)$ allows us to find a
Cauchy sequence in $(P_{t}u)_{t\geq 0}$, and thus establish its strong
convergence to some limit in $L^{2}(I)$. Later on, we will say more about this
limit by reasoning about the weak convergence of $P_{t}u$ (to the same limit).

######  Lemma 5.44 (Cauchy sequence).

Let $u\in\mathcal{U}$, and let $(u_{k})_{k\in\mathbb{N}}$ be given by
$u_{k}\vcentcolon=P_{k}u$. Then $(u_{k})_{k}$ is Cauchy as a sequence in
$L^{2}(I)$. As a consequence, $u_{k}\to u^{*}$ in $L^{2}(I)$ for some
$u^{*}\in L^{2}(I)$.

###### Proof.

The existence of a strong limit from the Cauchy property follows from the fact
that $L^{2}(I)$ is a complete normed space. Let us now establish the Cauchy
property. Fix any $n>m$ in $\mathbb{N}$. Then

$\displaystyle\|u_{n}-u_{m}\|_{L^{2}(I)}$
$\displaystyle=\left\|\int_{(m,n)}\bm{u^{\prime}}(t)\odif
t\right\|_{L^{2}(I)}$ (Absolute continuity)
$\displaystyle\leq\sum_{j=m}^{n-1}\int_{(j,j+1)}\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}\odif
t$ (Triangle inequality)
$\displaystyle\leq\sum_{j=m}^{n-1}\left(\int_{(j,j+1)}\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}^{2}\odif
t\right)^{1/2}$ (Jensen’s inequality)
$\displaystyle=\sum_{j=m}^{n-1}\left(\mathcal{E}^{-}(u_{j})-\mathcal{E}^{-}(u_{j+1})\right)^{1/2}$
(Proposition 5.17)
$\displaystyle\leq\sum_{j=m}^{\infty}\mathcal{E}^{-}(u_{j})^{1/2}$
($\mathcal{E}^{-}$ is nonnegative)
$\displaystyle\leq\sum_{j=m}^{\infty}\left(e^{-Kj}\mathcal{E}^{-}(u)\right)^{1/2}$
(Proposition 5.43)
$\displaystyle=\mathcal{E}^{-}(u)^{1/2}\frac{e^{-Km/2}}{1-e^{-K/2}}$
(Geometric series) $\displaystyle=A\mathcal{E}^{-}(u)^{1/2}e^{-Bm}$

for constants $A,B>0$ that only depend on the constant $K$ from Proposition
5.43. It follows that, for every $\epsilon>0$, there exists $N\in\mathbb{N}$
such that, for all $n>m>N$,

$\|u_{n}-u_{m}\|_{L^{2}(I)}\leq A\mathcal{E}^{-}(u)^{1/2}e^{-Bm}\leq
A\mathcal{E}^{-}(u)^{1/2}e^{-BN}<\epsilon\,,$

so $(u_{k})_{k}$ is Cauchy. ∎

######  Lemma 5.45 (Strong convergence).

Let $u\in\mathcal{U}$. Then $P_{t}u$ converges in $L^{2}(I)$ to some $u^{*}\in
L^{2}(I)$ as $t\to\infty$.

###### Proof.

Let $u^{*}$ be such that $(P_{k}u)_{k\in\mathbb{N}}$ converges to $u^{*}$ in
$L^{2}(I)$ as $k\to\infty$, as given by Lemma 5.44. We claim that $P_{t}u\to
u^{*}$. Let $\epsilon>0$, and let $N\in\mathbb{N}$ be such that
$\|P_{k}u-u^{*}\|_{L^{2}(I)}<\epsilon$ for all $k\geq N$, as given by the
convergence of the sequence $(P_{k}u)_{k}$. Then for any $t\geq N$, letting
$j\vcentcolon=\lfloor t\rfloor$, we have

$\|P_{t}u-u^{*}\|_{L^{2}(I)}\leq\|P_{j}u-u^{*}\|_{L^{2}(I)}+\|P_{t}u-P_{j}u\|_{L^{2}(I)}$

by the triangle inequality. We have $\|P_{j}u-u^{*}\|_{L^{2}(I)}<\epsilon$ by
the choice of $N$, and on the other hand,

$\displaystyle\|P_{t}u-P_{j}u\|_{L^{2}(I)}$
$\displaystyle=\left\|\int_{(j,t)}\bm{u^{\prime}}(s)\odif
s\right\|_{L^{2}(I)}$ (Absolute continuity)
$\displaystyle\leq\int_{(j,t)}\|\bm{u^{\prime}}(s)\|_{L^{2}(I)}\odif s$
(Triangle inequality)
$\displaystyle\leq(t-j)^{1/2}\left(\int_{(j,t)}\|\bm{u^{\prime}}(s)\|_{L^{2}(I)}^{2}\odif
s\right)^{1/2}$ (Jensen’s inequality)
$\displaystyle\leq\left(\mathcal{E}^{-}(P_{j}u)-\mathcal{E}^{-}(P_{t}u)\right)^{1/2}$
($t-j\in[0,1)$, Proposition 5.17) $\displaystyle\leq
e^{-Kj/2}\mathcal{E}^{-}(u)^{1/2}$ $\displaystyle\text{($\mathcal{E}^{-}\geq
0$, \lx@cref{creftype~refnum}{prop:exponential-decay})}\,.$

Thus by letting $N$ be large enough, we can ensure that

$\|P_{t}u-P_{j}u\|_{L^{2}(I)}\leq e^{-Kj/2}\mathcal{E}^{-}(u)^{1/2}\leq
e^{-KN/2}\mathcal{E}^{-}(u)^{1/2}<\epsilon\,,$

and hence $\|P_{t}u-u^{*}\|_{L^{2}(I)}<2\epsilon$. Thus $P_{t}u\to u^{*}$ in
$L^{2}(I)$. ∎

###  5.7 Nonexpansiveness and order preservation

Our goal in this section is to establish that the semigroup $P_{t}$ is
nonexpansive and order preserving. These properties will help us show that
applying $P_{t}$ to line restrictions in the multidimensional setting behaves
as expected by “making progress” toward monotonicity with each application.

In the context of PDEs, one desirable way to show that a property is preserved
through time is to differentiate in time, and then pass the derivative inside
the integral to exploit the definition of the PDE. However, we need to
establish some technical results before we can justify such calculations.

We start with the following lemma, slightly adapted from [CT80], which reveals
a close connection between order preservation and nonexpansiveness (here, in
the supremum norm) of operators. Hence our strategy will be to establish order
preservation, and conclude nonexpansiveness.

######  Lemma 5.46.

Let $(\Omega,\Sigma,\mu)$ be a finite measure space, and let $1\leq
p\leq+\infty$. Let $L^{\infty}(\Omega)\subset C\subset L^{p}(\Omega)$ have the
property that for all $f\in C$ and $r\in\mathbb{R}$, $f+r\in C$. Let $T:C\to
L^{p}(\Omega)$ be continuous as a map from $L^{p}(\Omega)$ to $L^{p}(\Omega)$
(the continuity requirement may be dropped if $p=\infty$) satisfying

$T(f+r)=T(f)+r\text{ a.\kern 1.00006pte.\ }$ (21)

for every $f\in C$ and $r\in\mathbb{R}$. Then the following are equivalent:

  1. (a)

For all $f,g\in C$, if $f\leq g$ a.e. then $Tf\leq Tg$ a.e. .

  2. (b)

For all $f,g\in C$, $(Tf-Tg)^{+}\leq\operatorname*{ess\,sup}(f-g)^{+}$ a.e. .

  3. (c)

For all $f,g\in C$, $\left\lvert Tf-
Tg\right\rvert\leq\operatorname*{ess\,sup}\left\lvert f-g\right\rvert$ a.e. .

###### Proof.

We show (a) $\implies$ (b) $\implies$ (c) $\implies$ (a). First, assume (a)
holds, and let $f,g\in C$. Suppose
$r\vcentcolon=\operatorname*{ess\,sup}(f-g)^{+}<+\infty$, since otherwise
there is nothing to prove. Note that $g+r\geq f$ and $g+r\geq g$ both a.e. .
Thus by (a) and (21), we have

$Tg+(Tf-Tg)^{+}=Tf\lor Tg\leq T(g+r)=Tg+r\text{ a.\kern 1.00006pte.\ }\,,$

so $(Tf-Tg)^{+}\leq r$ a.e. as needed. Next, suppose (b) holds, and let
$f,g\in C$. Let $r_{1}\vcentcolon=\operatorname*{ess\,sup}(f-g)^{+}$ and
$r_{2}\vcentcolon=\operatorname*{ess\,sup}(g-f)^{+}$; note that
$\operatorname*{ess\,sup}\left\lvert f-g\right\rvert=\max\\{r_{1},r_{2}\\}$,
so we may assume that $r_{1},r_{2}<+\infty$, since otherwise there is nothing
to show. Then indeed, using (b),

$\left\lvert Tf-Tg\right\rvert=\max\left\\{(Tf-Tg)^{+},(Tg-
Tf)^{+}\right\\}\leq\max\\{r_{1},r_{2}\\}=\operatorname*{ess\,sup}\left\lvert
f-g\right\rvert\text{ a.\kern 1.00006pte.\ ,}$

as needed. Finally, suppose (c) holds, and let $f,g\in C$ satisfy $f\leq g$
a.e. . First, suppose $f,g\in L^{\infty}(\Omega)$. Let
$r\vcentcolon=\operatorname*{ess\,sup}(g-f)$, so that $0\leq r<+\infty$ by
assumption. Then, by (c) and (21),

$\displaystyle\operatorname*{ess\,sup}\\{Tf-Tg+r\\}$
$\displaystyle=\operatorname*{ess\,sup}\\{T(f+r)-Tg\\}\leq\operatorname*{ess\,sup}\left\lvert
T(f+r)-Tg\right\rvert$ $\displaystyle\leq\operatorname*{ess\,sup}\left\lvert
f-g+r\right\rvert\leq r\,,$

so $Tf\leq Tg$ a.e. as needed.

Now, for general $f,g\in C$ with $f\leq g$ a.e. , we may approximate them by
sequences $(f_{n})_{n},(g_{n})_{n}$ in $L^{\infty}(\Omega)\subset C$ such that
$f_{n}\to f$ and $g_{n}\to g$ in $L^{p}(\Omega)$ and moreover $f_{n}\leq
g_{n}$ a.e. for each $n$; indeed, letting
$f_{n}\vcentcolon=\max\\{-n,\min\\{n,f\\}\\}$ and likewise for $g_{n}$, it is
immediate that $f_{n},g_{n}\in L^{\infty}(\Omega)$ with $f_{n}\leq g_{n}$ a.e.
, and the convergences $f_{n}\to f$ and $g_{n}\to g$ in $L^{p}(\Omega)$ follow
from the dominated convergence theorem: since $\left\lvert
f-f_{n}\right\rvert^{p}\to 0$ pointwise and $\left\lvert
f-f_{n}\right\rvert\leq\left\lvert f\right\rvert$ pointwise for each $n$, we
have $\lim_{n\to\infty}\int_{\Omega}\left\lvert
f_{n}-f\right\rvert^{p}\odif\mu=0$.

The previous case yields $Tf_{n}\leq Tg_{n}$ a.e. for each $n$, and we claim
that the continuity of $T$ implies that $Tf\leq Tg$ a.e. as well. Indeed, we
have $Tf_{n}\to Tf$ an $Tg_{n}\to Tg$ in $L^{p}(\Omega)$ and thus

$\displaystyle\left\|(Tf-Tg)^{+}-(Tf_{n}-Tg_{n})^{+}\right\|_{L^{p}(\Omega)}$
$\displaystyle\leq\left\|(Tf-Tg)-(Tf_{n}-Tg_{n})\right\|_{L^{p}(\Omega)}$
$\displaystyle\leq\|Tf-Tf_{n}\|_{L^{p}(\Omega)}+\|Tg-
Tg_{n}\|_{L^{p}(\Omega)}\to 0\,,$

which means that $(Tf_{n}-Tg_{n})^{+}\to(Tf-Tg)^{+}$ in $L^{p}(\Omega)$. But
since $(Tf_{n}-Tg_{n})^{+}=0$ a.e. and hence in $L^{p}(\Omega)$ for each $n$,
we conclude that $(Tf-Tg)^{+}=0$ a.e. , i.e. $Tf\leq Tg$ a.e. as needed. ∎

The following lemma is a specific formulation of the well-known Leibniz rule
for differentiating under the integral sign. It is a specialization of the
version stated in [Che], and can be proved by a standard argument using the
Fubini-Tonelli theorem.

######  Lemma 5.47 (Differentiating under the integral sign).

Let $(\Omega,\Sigma,\mu)$ be a measure space and let $[a,b]\subset\mathbb{R}$.
Let $f:[a,b]\times\Omega\to\mathbb{R}$ be a jointly measurable function
satisfying the following:

  1. 1.

$f(t,\cdot)\in L^{1}(\Omega)$ for a.e. $t\in(a,b)$.

  2. 2.

$f(\cdot,x)$ is AC for a.e. $x\in\Omega$. (Its a.e. defined partial derivative
is denoted by $\partial_{t}f$ as usual.)

  3. 3.

It holds that

$\int_{(a,b)}\int_{\Omega}\left\lvert\partial_{t}f(t,x)\right\rvert\odif\mu\odif
t<+\infty\,.$

Then the function $t\mapsto\int_{\Omega}f(t,x)\odif\mu$ is AC and

$\partial_{t}\int_{\Omega}f(t,x)\odif\mu=\int_{\Omega}\partial_{t}f(t,x)\odif\mu\qquad\text{for
a.\kern 1.00006pte.\ $t\in(a,b)$.}$

The following fact is a standard consequence of the definition of absolute
continuity:

######  Fact 5.48.

Let $J\subset\mathbb{R}$ be a compact interval. If $f,g:J\to\mathbb{R}$ are
AC, then $f\lor g$ is AC.

######  Fact 5.49 (See e.g. [EG15, Theorem 4.4]).

Let $J\subset\mathbb{R}$ be a compact interval. Let $1\leq p<\infty$ and let
$f\in W^{1,p}(J)$. Then $f^{+}\in W^{1,p}(J)$ and
$\partial_{x}(f^{+})=\chi_{\\{f>0\\}}\partial_{x}f$ a.e. in $J$.

We also use the following standard formulation of a chain rule for Sobolev
functions. This version follows e.g. from [EG15, Theorem 4.4], which is stated
for globally Lipschitz functions $F$, by using the fact that the image of $f$
on $J$ is bounded, so that the local Lipschitz condition suffices (e.g. extend
$F$ linearly outside the image of $f$ to obtain a $C^{1}(\mathbb{R})$,
globally Lipschitz function $\tilde{F}$).

######  Fact 5.50.

Let $J\subset\mathbb{R}$ be a compact interval. Let $1\leq p<\infty$, let
$f\in W^{1,p}(J)$, and let $F\in C^{1}(\mathbb{R})$ be locally Lipschitz. Then
$F\circ f\in W^{1,p}(J)$ and $\partial_{x}(F\circ
f)(x)=F^{\prime}(f(x))\partial_{x}f(x)$ for a.e. $x\in J$.

The following lemma essentially says that, given an element of a Bochner
space, we can get a handle on a concrete jointly measurable function
“representing” that element in a precise sense. This makes concrete the
intuitive expectation that a solution $\bm{u}$ to our PDEs, which maps each
point in time to an element of $L^{2}(I)$, should also give us a specific
value at each point in time and space “$\bm{u}(t,x)$”.

######  Lemma 5.51.

[DS58, Theorem 17, p. 198] Let $(S,\Sigma_{S},\mu)$ and
$(T,\Sigma_{T},\lambda)$ be measure spaces which are either both finite or
both positive and $\sigma$-finite, and let $(R,\Sigma_{R},\rho)$ be their
product. Let $1\leq p\leq\infty$ and let $F$ be a $\mu$-integrable function on
$S$ to $L^{p}(T,\Sigma_{T},\lambda,\mathcal{X})$ where $\mathcal{X}$ is a real
or complex Banach space. Then there is a $\rho$-measurable function $f$ on $R$
to $\mathcal{X}$, which is uniquely determined except for a set of
$\rho$-measure zero, such that $f(s,\cdot)=F(s)$ for $\mu$-almost all $s$ in
$S$. Moreover $f(\cdot,t)$ is $\mu$-integrable on $S$ for $\lambda$-almost all
$t$ and the integral $\int_{S}f(s,t)\mu(\odif s)$, as a function of $t$ is
equal to the element $\int_{S}F(s)\mu(\odif s)$ of
$L^{p}(T,\Sigma_{T},\lambda,\mathcal{X})$.

######  Corollary 5.52.

Let $[a,b]\subset\mathbb{R}$ be a compact interval endowed with the Lebesgue
measure, and let $F\in L^{1}(a,b;L^{2}(I))$. Then there exists a jointly
measurable function $f^{*}:[a,b]\times I\to\mathbb{R}$ satisfying

  1. 1.

$f^{*}(s,\cdot)=F(s)$ in $L^{2}(I)$ for a.e. $s\in(a,b)$.

  2. 2.

$f^{*}(\cdot,x)\in L^{1}(a,b)$ for all $x\in I$.

  3. 3.

For all $s\in[a,b]$, the functions $x\mapsto\int_{(a,s)}f^{*}(r,x)\odif r$ and
$\int_{(a,s)}F(r)\odif r$ are equal in $L^{2}(I)$.

###### Proof.

Let condition $\tilde{2}$ denote condition 2 with “all $x\in I$” replaced by
“a.e. $x\in I$”. We first apply Lemma 5.51 with $S=[a,b]$ and $T=I$, both
endowed with the Lebesgue measure, to obtain a function $f$ satisfying
conditions 1 and $\tilde{2}$, as well condition 3 for $s=b$.

We now show that condition 3 also holds for other values of $s\in[a,b)$ using
the a.e. uniqueness given by Lemma 5.51. For any $s\in[a,b)$, we may apply
that lemma with $\tilde{S}=[a,s]$ instead to obtain a function $\tilde{f}$
satisfying $x\mapsto\int_{(a,s)}{\tilde{f}}(r,x)\odif r=\int_{(a,s)}F(r)\odif
r$ in $L^{2}(I)$. But since both $g=\tilde{f}$ and $g=f$ satisfy that
$g(r,\cdot)=F(r)$ in $L^{2}(I)$ for a.e. $r\in(a,s)$, Lemma 5.51 implies that
$\tilde{f}=f$ except for a subset of $[a,s]\times I$ of joint measure zero. We
conclude that, for a.e. $x\in I$, we have $\tilde{f}(\cdot,x)=f(\cdot,x)$ a.e.
in $(a,s)$ and hence $\int_{(a,s)}f(r,x)\odif
r=\int_{(a,s)}{\tilde{f}}(r,x)\odif r$. It follows that
$x\mapsto\int_{(a,s)}f(r,x)\odif r=x\mapsto\int_{(a,s)}{\tilde{f}}(r,x)\odif
r=\int_{(a,s)}F(r)$ in $L^{2}(I)$, as claimed.

The final step is to construct a jointly measurable function
$f^{*}:[a,b]\times I\to\mathbb{R}$ satisfying condition 2 rather than just
$\tilde{2}$, while preserving conditions 1 and 3. Let $N\subset I$ be a
measure zero set such that $f(\cdot,x)\in L^{1}(a,b)$ for all $x\in I\setminus
N$. Define $f^{*}$ by

$f^{*}(s,x)\vcentcolon=\begin{cases}f(s,x)&\text{if }x\in I\setminus N\\\
0&\text{otherwise.}\end{cases}$

We note that $f^{*}$ is jointly measurable; indeed, this follows from the
facts that $f=f^{*}$ on $[a,b]\times(I\setminus N)$ and that the set
$[a,b]\times(I\setminus N)$ is jointly measurable. By construction, $f^{*}$
satisfies condition 2, and it is clear that $f^{*}$ also satisfies conditions
1 and 3 since $f$ does and $N$ is a null set. ∎

######  Lemma 5.53.

Let $0\leq a<b<+\infty$, and let $\bm{u}\in C([a,b];L^{2}(I))$ be the
restriction to domain $[a,b]$ of any solution to the gradient flow problem,
with $\bm{u^{\prime}}:[a,b]\to L^{2}(I)$ its weak derivative restricted in the
same way. Then there exists a jointly measurable function
$\bm{\tilde{u}^{\prime}}:[a,b]\times I\to\mathbb{R}$ satisfying

  1. 1.

$\bm{\tilde{u}^{\prime}}(t,\cdot)=\bm{u^{\prime}}(t)$ in $L^{2}(I)$ for a.e.
$t\in(a,b)$.

  2. 2.

$\bm{\tilde{u}^{\prime}}(\cdot,x)\in L^{1}(a,b)$ for all $x\in I$.

  3. 3.

For all $t\in[a,b]$, the functions
$x\mapsto\int_{(a,t)}\bm{\tilde{u}^{\prime}}(s,x)\odif s$ and
$\int_{(a,t)}\bm{u^{\prime}}(s)\odif s$ are equal in $L^{2}(I)$.

Moreover, fixing any representative of $\bm{u}(a)\in L^{2}(I)$, the function
$\bm{\tilde{u}}:[a,b]\times I\to\mathbb{R}$ given by

$\bm{\tilde{u}}(t,x)\vcentcolon=\bm{u}(a)(x)+\int_{(a,t)}\bm{\tilde{u}^{\prime}}(s,x)\odif
s$

is jointly measurable and satisfies

  1. 4.

For each $t\in[a,b]$, $\bm{\tilde{u}}(t,\cdot)=\bm{u}(t)$ in $L^{2}(I)$.

  2. 5.

For each $x\in I$, $\bm{\tilde{u}}(\cdot,x)$ is absolutely continuous with
weak derivative
$\partial_{t}\bm{\tilde{u}}(t,x)=\bm{\tilde{u}^{\prime}}(t,x)$.

###### Proof.

Apply Corollary 5.52 to $\bm{u^{\prime}}$, which is integrable since it is the
weak derivative of $\bm{u}$, to obtain jointly measurable
$\bm{\tilde{u}^{\prime}}:[a,b]\times I\to\mathbb{R}$ satisfying properties
1–3. We now verify that $\bm{\tilde{u}}$ satisfies properties 4 and 5\. Note
that these two properties then imply that $\bm{\tilde{u}}$ is a Carathéodory
function and hence jointly measurable (see e.g. [AB06, Lemma 4.51]).

For each $t\in[a,b]$, the definition of $\bm{u}$ (in particular its absolute
continuity) implies that

$\bm{u}(t)-\bm{u}(a)=\left[\bm{u}(0)+\int_{(0,t)}\bm{u^{\prime}}(s)\odif
s\right]-\left[\bm{u}(0)+\int_{(0,a)}\bm{u^{\prime}}(s)\odif
s\right]=\int_{(a,t)}\bm{u^{\prime}}(s)\odif s$

in $L^{2}(I)$. Property 3 implies that, for a.e. $x\in I$,

$\bm{u}(t)(x)=\bm{u}(a)(x)+\int_{(a,t)}\bm{\tilde{u}^{\prime}}(s,x)\odif
s=\bm{\tilde{u}}(t,x)\,,$

which is property 4. Finally, property 5 is an immediate consequence of the
definition of $\bm{\tilde{u}}$. ∎

We are now prepared to differentiate in time in order to establish that
$P_{t}$ is order preserving.

######  Proposition 5.54.

Let $u_{0},v_{0}\in H^{1}(I)$, and let $\bm{u},\bm{v}$ be the solutions to the
gradient flow problem with initial data $u_{0},v_{0}$ respectively. Then the
function $\Delta:[0,+\infty)\to[0,+\infty)$ given by

$\Delta(t)\vcentcolon=\frac{1}{2}\int_{I}\left[(\bm{u}(t)-\bm{v}(t))^{+}\right]^{2}\odif
x$

is nonincreasing.

###### Proof.

Let $0\leq a<b<+\infty$, so that it suffices to show that
$\Delta(a)\geq\Delta(b)$. Apply Lemma 5.53 to $\bm{u}$ and $\bm{v}$ to obtain
functions $\bm{\tilde{u}^{\prime}}$, $\bm{\tilde{u}}$,
$\bm{\tilde{v}^{\prime}}$, and $\bm{\tilde{v}}$ with the properties stated in
that lemma. Define the function $f:[a,b]\times I\to\mathbb{R}$ by

$f(t,x)\vcentcolon=\frac{1}{2}\left[(\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x))^{+}\right]^{2}\,.$

Note that $f$ is jointly measurable, since both $\bm{\tilde{u}}$ and
$\bm{\tilde{v}}$ are. We also obtain that, for each $t\in[a,b]$,

$\Delta(t)=\int_{I}f(t,x)\odif x\,.$

We now verify that $f$ satisfies the conditions of Lemma 5.47. Let us verify
the first condition. Since $\bm{\tilde{u}}(t,\cdot),\bm{\tilde{v}}(t,\cdot)\in
L^{2}(I)$, it follows that $f(t,\cdot)\in L^{1}(I)$, as desired.

We now verify the second condition. We already have that, for all $x\in I$,
$\bm{\tilde{u}}(\cdot,x)$ and $\bm{\tilde{v}}(\cdot,x)$ are AC. Fact 5.49
implies that, for all $x\in I$, the function
$t\mapsto(\bm{\tilde{u}}(\cdot,x)-\bm{\tilde{v}}(\cdot,x))^{+}$ is AC. But the
function $y\mapsto y^{2}$ is locally Lipschitz, so Fact 5.50 implies that
$f(\cdot,x)$ is AC for each $x$, so the second condition is satisfied.
Moreover, using Facts 5.50 and 5.49, its weak derivative is

$\displaystyle\partial_{t}f(t,x)$
$\displaystyle=\left[(\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x))^{+}\right]\partial_{t}\left[(\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x))^{+}\right]$
$\displaystyle=\left[(\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x))^{+}\right]\chi_{\\{\bm{\tilde{u}}(t,x)>\bm{\tilde{v}}(t,x)\\}}\left[\partial_{t}\bm{\tilde{u}}(t,x)-\partial_{t}\bm{\tilde{v}}(t,x)\right]$
$\displaystyle=\chi_{\\{\bm{\tilde{u}}(t,x)>\bm{\tilde{v}}(t,x)\\}}\left[\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x)\right]\left[\bm{\tilde{u}^{\prime}}(t,x)-\bm{\tilde{v}^{\prime}}(t,x)\right]\,.$

We claim that
$\int_{(a,b)}\int_{I}\left\lvert\partial_{t}f(t,x)\right\rvert\odif x\odif
t<+\infty$. Indeed, we have

$\displaystyle\int_{(a,b)}\int_{I}\left\lvert\partial_{t}f(t,x)\right\rvert\odif
x\odif t$
$\displaystyle\quad=\int_{(a,b)}\int_{I}\Big{|}\chi_{\\{\bm{\tilde{u}}(t,x)>\bm{\tilde{v}}(t,x)\\}}\left[\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x)\right]\left[\bm{\tilde{u}^{\prime}}(t,x)-\bm{\tilde{v}^{\prime}}(t,x)\right]\Big{|}\odif
x\odif t$
$\displaystyle\quad\leq\frac{1}{2}\int_{(a,b)}\int_{I}(\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x))^{2}\odif
x\odif
t+\frac{1}{2}\int_{(a,b)}\int_{I}(\bm{\tilde{u}^{\prime}}(t,x)-\bm{\tilde{v}^{\prime}}(t,x))^{2}\odif
x\odif t$
$\displaystyle\quad\leq\int_{(a,b)}\int_{I}\bm{\tilde{u}}(t,x)^{2}\odif x\odif
t+\int_{(a,b)}\int_{I}\bm{\tilde{v}}(t,x)^{2}\odif x\odif
t+\int_{(a,b)}\int_{I}\bm{\tilde{u}^{\prime}}(t,x)^{2}\odif x\odif
t+\int_{(a,b)}\int_{I}\bm{\tilde{v}^{\prime}}(t,x)^{2}\odif x\odif t\,.$

We claim that each of the four terms above is finite. First, using Tonelli’s
theorem, the definition of $\bm{\tilde{u}}$ and Jensen’s inequality, and
letting $L\vcentcolon=b-a$, we have

$\displaystyle\int_{(a,b)}\int_{I}\bm{\tilde{u}}(t,x)^{2}\odif x\odif t$
$\displaystyle=\int_{I}\int_{(a,b)}\left(\bm{u}(a)(x)+\int_{(a,t)}\bm{\tilde{u}^{\prime}}(s,x)\odif
s\right)^{2}\odif t\odif x$ $\displaystyle\leq
2\int_{I}\int_{(a,b)}\bm{u}(a)(x)^{2}\odif t\odif
x+2L\int_{I}\int_{(a,b)}\int_{(a,t)}\bm{\tilde{u}^{\prime}}(s,x)^{2}\odif
s\odif t\odif x$ $\displaystyle\leq 2\int_{(a,b)}\int_{I}\bm{u}(a)(x)^{2}\odif
x\odif
t+2L\int_{(a,b)}\int_{(a,b)}\int_{I}\bm{\tilde{u}^{\prime}}(s,x)^{2}\odif
x\odif s\odif t$
$\displaystyle=2L\|\bm{u}(a)\|_{L^{2}(I)}^{2}+2L^{2}\int_{(a,b)}\int_{I}\bm{\tilde{u}^{\prime}}(s,x)^{2}\odif
x\odif s\,,$

where in the last line we write $\|\bm{u}(a)\|_{L^{2}(I)}^{2}$, which is
finite, since $\bm{u}(a)\in L^{2}(I)$. Hence, since the argument for the
$\bm{v}$ terms proceeds identically, all we need to show is that

$\int_{(a,b)}\int_{I}\bm{\tilde{u}^{\prime}}(t,x)^{2}\odif x\odif
t\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{<}}+\infty\,.$

Since $\bm{\tilde{u}^{\prime}}(t,\cdot)=\bm{u^{\prime}}(t)$ in $L^{2}(I)$ for
a.e. $t\in(a,b)$, this is equivalent to showing that

$\int_{(a,b)}\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}^{2}\odif
t\stackrel{{\scriptstyle\mathclap{\mbox{\tiny?}}}}{{<}}+\infty\,.$ (22)

Recall that, by Definition 5.13, $\bm{u^{\prime}}\in L^{2}(0,b;L^{2}(I))$, so
in particular $\bm{u^{\prime}}\in L^{2}(a,b;L^{2}(I))$, which by definition
implies (22). Hence the claim holds and the third condition of Lemma 5.47 is
satisfied.

Therefore, Lemma 5.47 implies that $\Delta(\cdot)=\int_{I}f(\cdot,x)\odif x$
is AC and, for a.e. $t\in(a,b)$,

$\displaystyle\partial_{t}\Delta(t)$
$\displaystyle=\partial_{t}\int_{I}f(t,x)\odif
x=\int_{I}\partial_{t}f(t,x)\odif x$
$\displaystyle=\int_{I}\chi_{\\{\bm{\tilde{u}}(t,x)>\bm{\tilde{v}}(t,x)\\}}\left[\bm{\tilde{u}}(t,x)-\bm{\tilde{v}}(t,x)\right]\left[\bm{\tilde{u}^{\prime}}(t,x)-\bm{\tilde{v}^{\prime}}(t,x)\right]\odif
x$ (Shown above)
$\displaystyle=\int_{I}\chi_{\\{\bm{u}(t)>\bm{v}(t)\\}}\left[\bm{u}(t)-\bm{v}(t)\right]\left[\bm{u^{\prime}}(t)-\bm{v^{\prime}}(t)\right]\odif
x$ (By Lemma 5.53)
$\displaystyle=\int_{I}\left[\bm{u}(t)-\bm{v}(t)\right]^{+}\partial_{x}\partial_{x}\left[\bm{u}(t){\downarrow}-\bm{v}(t){\downarrow}\right]\odif
x$ (By Lemma 5.21).

By Corollary 5.36, $\bm{u}(t),\bm{v}(t)\in H^{1}(I)$ and hence are AC, and by
Fact 5.49, $\left[\bm{u}(t)-\bm{v}(t)\right]^{+}$ is AC. Therefore we can
integrate by parts and, using Fact 5.49 and the boundary condition
$\partial_{x}\bm{u}(t){\downarrow}=\partial_{x}\bm{v}(t){\downarrow}=0$ on
$\\{0,1\\}$ from Lemma 5.21, we obtain

$\displaystyle\partial_{t}\Delta(t)$
$\displaystyle=-\int_{I}\chi_{\\{\bm{u}(t)>\bm{v}(t)\\}}\left[\partial_{x}(\bm{u}(t)-\bm{v}(t))\right]\left[\partial_{x}(\bm{u}(t){\downarrow}-\bm{u}(t){\downarrow})\right]$
$\displaystyle=-\int_{I}\chi_{\\{\bm{u}(t)>\bm{v}(t)\\}}\left[\partial_{x}\bm{u}(t)-\partial_{x}\bm{v}(t)\right]\left[\partial_{x}^{-}\bm{u}(t)-\partial_{x}^{-}\bm{u}(t)\right]$
(By Proposition 5.25).

But for any numbers $\alpha,\beta\in\mathbb{R}$, it is the case that
$\alpha\geq\beta\implies(\alpha\land 0)\geq(\beta\land 0)$ and
$\alpha\leq\beta\implies(\alpha\land 0)\leq(\beta\land 0)$. Hence the
integrand above is pointwise nonnegative, and we conclude that
$\partial_{t}\Delta(t)\leq 0$ for a.e. $t\in(a,b)$. Hence
$\Delta(a)\geq\Delta(b)$, as needed. ∎

######  Corollary 5.55 (“Directed nonexpansiveness” of $P_{t}$).

Let $u,v\in L^{2}(I)$. Then for all $t>0$,

$\int_{I}\left[(P_{t}u-P_{t}v)^{+}\right]^{2}\odif
x\leq\int_{I}\left[(u-v)^{+}\right]^{2}\odif x\,.$

###### Proof.

This is an immediate consequence of Proposition 5.54 when $u,v\in H^{1}(I)$,
and the general case follows by approximating $u$ and $v$ by $H^{1}(I)$
functions and using the continuity of the map $w\mapsto P_{t}w$ from
$L^{2}(I)$ to $L^{2}(I)$. ∎

######  Corollary 5.56 ($P_{t}$ is order preserving).

Let $u,v\in L^{2}(I)$, and suppose $u\leq v$ a.e. . Then $P_{t}u\leq P_{t}v$
a.e. for all $t>0$.

###### Proof.

This is an immediate consequence of Corollary 5.55. ∎

######  Lemma 5.57.

Every nondecreasing $u\in L^{2}(I)$ is a stationary point of $P_{t}$, i.e.
$P_{t}u=u$ for all $t$.

###### Proof.

By definition, $u=u{\uparrow}+u{\downarrow}$ with $u{\uparrow}=u$ and
$u{\downarrow}=0$, so $\mathcal{E}^{-}(u)=0$. Thus for every $v\in L^{2}(I)$,

$\mathcal{E}^{-}(v)\geq 0=\mathcal{E}^{-}(u)+\left\langle
0,v-u\right\rangle\,,$

so $0\in\partial\mathcal{E}^{-}(u)$. Thus $\bm{u}(t)=u$ is the solution to the
gradient flow problem with initial state $u$. ∎

######  Observation 5.58.

Corollary 5.56 in particular implies that if $u\geq a$ a.e. for some
$a\in\mathbb{R}$, then $P_{t}u\geq P_{t}a=a$ a.e. for all $t>0$, the equality
by Lemma 5.57.

We also observe below that $P_{t}$ is degree one positively homogeneous and
additive when one argument is a constant function, as the following results
show.

######  Lemma 5.59.

Let $u\in\mathcal{U}$, $\beta\in\mathbb{R}$, and $v\vcentcolon=u+\beta$. Then
$v\in\mathcal{U}$ with $v{\uparrow}=u{\uparrow}+\beta$ and
$v{\downarrow}=u{\downarrow}$, and thus
$\mathcal{E}^{-}(v)=\mathcal{E}^{-}(u)$.

###### Proof.

This is a straightforward consequence of the definition of
$v{\uparrow},v{\downarrow}$. ∎

######  Lemma 5.60.

Let $u\in\mathcal{U}$, $\alpha>0$, and $v\vcentcolon=\alpha u$. Then
$v\in\mathcal{U}$ with $v{\uparrow}=\alpha u{\uparrow}$ and
$v{\downarrow}=\alpha u{\downarrow}$, and thus
$\mathcal{E}^{-}(v)=\alpha^{2}\mathcal{E}^{-}(u)$.

###### Proof.

This is a straightforward consequence of the definitions of
$v{\uparrow},v{\downarrow}$ and $\mathcal{E}^{-}$. ∎

######  Lemma 5.61.

Let $u\in D(\partial\mathcal{E}^{-})$, $\beta\in\mathbb{R}$, and
$v\vcentcolon=u+\beta$. Then
$\partial\mathcal{E}^{-}(u)=\partial\mathcal{E}^{-}(v)$.

###### Proof.

By symmetry, it suffices to prove that
$\partial\mathcal{E}^{-}(u)\subseteq\partial\mathcal{E}^{-}(v)$. Let
$z\in\partial\mathcal{E}^{-}(u)$. Then for all $w\in\mathcal{U}$, using Lemma
5.59 twice and the definition of subdifferential,

$\mathcal{E}^{-}(w)=\mathcal{E}^{-}(w-\beta)\geq\mathcal{E}^{-}(u)+\left\langle
z,w-\beta-u\right\rangle=\mathcal{E}^{-}(v)+\left\langle
z,w-v\right\rangle\,,$

and hence $z\in\partial\mathcal{E}^{-}(v)$ as desired. ∎

######  Lemma 5.62.

Let $u\in D(\partial\mathcal{E}^{-})$, $\alpha>0$, and $v\vcentcolon=\alpha
u$. Then $\partial\mathcal{E}^{-}(v)=\alpha\partial\mathcal{E}^{-}(u)$.

###### Proof.

By symmetry, it suffices to prove that
$\alpha\partial\mathcal{E}^{-}(u)\subseteq\partial\mathcal{E}^{-}(v)$. Let
$z\in\partial\mathcal{E}^{-}(u)$; we claim that $\alpha
z\in\partial\mathcal{E}^{-}(v)$. Indeed for all $w\in\mathcal{U}$, using Lemma
5.60 twice and the definition of subdifferential,

$\mathcal{E}^{-}(w)=\alpha^{2}\mathcal{E}^{-}\left(\tfrac{1}{\alpha}w\right)\geq\alpha^{2}\left(\mathcal{E}^{-}(u)+\left\langle
z,\tfrac{1}{\alpha}w-u\right\rangle\right)=\mathcal{E}^{-}(v)+\left\langle\alpha
z,w-v\right\rangle\,,$

and hence $\alpha z\in\partial\mathcal{E}^{-}(v)$ as desired. ∎

######  Proposition 5.63 (Effect of certain affine transformations on
$P_{t}$).

Let $u\in L^{2}(I)$, $\alpha>0$, and $\beta\in\mathbb{R}$. Then $P_{t}(\alpha
u+\beta)=\alpha P_{t}u+\beta$.

###### Proof.

We suffices to prove the statement for $u\in\mathcal{U}$; the general case
then follows by approximating $\mathcal{U}$ by $H^{1}(I)$ functions and the
continuity of $P_{t}$ from $L^{2}(I)$ to $L^{2}(I)$. For
$v\vcentcolon=u+\beta$, we observe that $\bm{v}(t)\vcentcolon=\bm{u}(t)+\beta$
with $\bm{v^{\prime}}(t)\vcentcolon=\bm{u^{\prime}}(t)$ is a solution to the
gradient flow problem with initial state $v$, since for each $t>0$ for which
$-\bm{u^{\prime}}(t)\in\partial\mathcal{E}^{-}(\bm{u}(t))$, Lemma 5.61 implies
that $-\bm{v^{\prime}}(t)\in\partial\mathcal{E}^{-}(\bm{v}(t))$.

Similarly, let $w\vcentcolon=\alpha u$. Then
$\bm{w}(t)\vcentcolon=\alpha\bm{u}(t)$ with
$\bm{w^{\prime}}(t)\vcentcolon=\alpha\bm{u^{\prime}}(t)$ is the solution to
the gradient flow problem with initial state $w$, since for each $t>0$ for
which $-\bm{u^{\prime}}(t)\in\partial\mathcal{E}^{-}(\bm{u}(t))$, Lemma 5.62
implies that $-\bm{w^{\prime}}(t)\in\partial\mathcal{E}^{-}(w)$. ∎

###  5.8 Convergence to monotone equilibrium

Since the directed Dirichlet energy $\mathcal{E}^{-}(P_{t}u)$ decays over time
and $P_{t}u$ converges to some limit as $t\to\infty$ (by Lemma 5.45), we
expect this limit to be a monotone function. Let us establish this fact and
other properties of that limit.

######  Lemma 5.64.

Let $u\in\mathcal{U}$ satisfy $\mathcal{E}^{-}(u)=0$. Then $u$ is
nondecreasing.

###### Proof.

Since $\mathcal{E}^{-}(u)=0$, we have $\partial_{x}u{\downarrow}=0$ a.e. in
$I$. Hence $u{\downarrow}$ is a constant function, while $u{\uparrow}$ is
nondecreasing by definition. ∎

######  Proposition 5.65.

Let $u\in L^{2}(I)$. Then there exists a nondecreasing $u^{*}\in\mathcal{U}$,
unique as an element of $L^{2}(I)$, such that $P_{t}u\to u^{*}$ in $L^{2}(I)$
as $t\to\infty$.

###### Proof.

Since $\mathcal{E}^{-}$ achieves its minimum (namely $0$, on e.g. constant
functions) and $P_{t}u\in D(\partial\mathcal{E}^{-})$ for all $t>0$, [AC84,
Theorem 2, p. 160] implies that there exists a minimizer $u^{*}$ of
$\mathcal{E}^{-}$ such that $P_{t}u\rightharpoonup u^{*}$ weakly in
$L^{2}(I)$. This means that $u^{*}\in\mathcal{U}$ with
$\mathcal{E}^{-}(u^{*})=0$, so by Lemma 5.64 $u^{*}$ is nondecreasing. By
Lemma 5.45 (which we may apply because, fixing any $t_{0}>0$, we have
$P_{t_{0}}u\in D(\partial\mathcal{E}^{-})\subseteq
D(\mathcal{E}^{-})=\mathcal{U}$), $P_{t}u$ also converges strongly in
$L^{2}(I)$, and it is standard that the weak and strong limits agree and that
this limit is unique. ∎

Therefore the following definition is justified:

######  Definition 5.66 (Monotone equilibrium).

Let $P_{\infty}:L^{2}(I)\to\mathcal{U}$ be the operator mapping each $u\in
L^{2}(I)$ to the unique (as an element of $L^{2}(I)$) nondecreasing
$u^{*}\in\mathcal{U}$ such that $P_{t}u\to u^{*}$ in $L^{2}(I)$ as
$t\to\infty$. We call $P_{\infty}u$ the _monotone equilibrium_ of $u$.

We now pass to the limit $P_{\infty}$ some useful properties of $P_{t}$.

######  Proposition 5.67 (“Directed nonexpansiveness” of $P_{\infty}$).

Let $u,v\in L^{2}(I)$. Then

$\int_{I}\left[(P_{\infty}u-P_{\infty}v)^{+}\right]^{2}\odif
x\leq\int_{I}\left[(u-v)^{+}\right]^{2}\odif x\,.$

###### Proof.

This follows by a standard limit argument as follows. Let $u,v\in L^{2}(I)$.
By Corollary 5.55,
$\|(P_{t}u-P_{t}v)^{+}\|_{L^{2}(I)}\leq\|(u-v)^{+}\|_{L^{2}(I)}$ for all
$t>0$. By definition of $P_{\infty}$, we have $P_{t}u\to P_{\infty}u$ and
$P_{t}v\to P_{\infty}v$ in $L^{2}(I)$ as $t\to\infty$. Using the triangle
inequality, we obtain

$\displaystyle\|(P_{t}u-P_{t}v)^{+}-(P_{\infty}u-P_{\infty}v)^{+}\|_{L^{2}(I)}$
$\displaystyle\leq\|(P_{t}u-P_{t}v)-(P_{\infty}u-P_{\infty}v)\|_{L^{2}(I)}$
$\displaystyle\leq\|P_{t}u-P_{\infty}u\|_{L^{(}I)}+\|P_{t}v-P_{\infty}v\|_{L^{2}(I)}\to
0\,,$

so $(P_{t}u-P_{t}v)^{+}\to(P_{\infty}u-P_{\infty}v)^{+}$ in $L^{2}(I)$ as
$t\to\infty$. Since
$\|(P_{t}u-P_{t}v)^{+}\|_{L^{2}(I)}\leq\|(u-v)^{+}\|_{L^{2}(I)}$ for all
$t>0$, we conclude that
$\|(P_{\infty}u-P_{\infty}v)^{+}\|_{L^{2}(I)}\leq\|(u-v)^{+}\|_{L^{2}(I)}$,
which gives the conclusion. ∎

######  Corollary 5.68 ($P_{\infty}$ is order preserving).

Let $u,v\in L^{2}(I)$, and suppose $u\leq v$ a.e. . Then $P_{\infty}u\leq
P_{\infty}v$ a.e. .

###### Proof.

This is a direct consequence of Proposition 5.67. ∎

######  Proposition 5.69.

$P_{\infty}$ is nonexpansive, and therefore continuous, as an $L^{2}(I)\to
L^{2}(I)$ map.

###### Proof.

Let $u,v\in L^{2}(I)$. In the proof of Proposition 5.67, we showed that
$\|(P_{t}u-P_{t}v)-(P_{\infty}u-P_{\infty}v)\|_{L^{2}(I)}\to 0$, which implies
that $\|P_{t}u-P_{t}v\|_{L^{2}(I)}\to\|P_{\infty}u-P_{\infty}v\|_{L^{2}(I)}$.
But $\|P_{t}u-P_{t}v\|_{L^{2}(I)}\leq\|u-v\|_{L^{2}(I)}$ for all $t>0$ since
$(P_{t})_{t\geq 0}$ is a nonexpansive semigroup, so the conclusion follows. ∎

######  Lemma 5.70.

For every nondecreasing $u\in L^{2}(I)$, we have $P_{\infty}u=u$.

###### Proof.

This follows from Lemma 5.57 along with the convergence $P_{t}u\to
P_{\infty}u$ in $L^{2}(I)$. ∎

Strong convergence to the monotone equilibrium, together with the preservation
of regularity results from Section 5.4, allows us to obtain regularity of the
monotone equilibrium as well:

######  Proposition 5.71 ($H^{1}$ regularity of the monotone equilibrium).

Let $u\in H^{1}(I)$. Then $P_{\infty}u\in H^{1}(I)$ with
$\varphi(P_{\infty}u)\leq\varphi(u)$.

###### Proof.

Recall the functional $\varphi$ from Section 5.4. Since $u\in H^{1}(I)$,
$\varphi(u)<+\infty$. By Proposition 5.35, we conclude that
$\varphi(P_{t}u)\leq\varphi(u)<+\infty$ for all $t>0$. Since $\varphi$ is
lower semicontinuous by Claim 5.30 and $P_{t}u\to P_{\infty}u$ in $L^{2}(I)$
by Lemma 5.45, we conclude that $\varphi(P_{\infty}u)\leq\varphi(u)<+\infty$.
Hence $P_{\infty}u\in H^{1}(I)$. ∎

######  Proposition 5.72 (Lipschitz regularity of the monotone equilibrium).

Let $u\in W^{1,\infty}(I)$. Then $P_{\infty}u\in W^{1,\infty}(I)$ with
$\psi(P_{\infty}u)\leq\psi(u)$.

###### Proof.

Recall the functional $\psi$ from Section 5.5. Since $u\in W^{1,\infty}(I)$,
$\psi(u)<+\infty$. By Proposition 5.41, we conclude that
$\psi(P_{t}u)\leq\psi(u)<+\infty$ for all $t>0$. Since $\psi$ is lower
semicontinuous by Claim 5.38 and $P_{t}u\to P_{\infty}u$ in $L^{2}(I)$ by
Lemma 5.45, we conclude that $\psi(P_{\infty}u)\leq\psi(u)<+\infty$. Hence
$P_{\infty}u\in W^{1,\infty}(I)$. ∎

We observe that $P_{\infty}$ also behaves nicely under the appropriate class
of affine transformations.

######  Proposition 5.73 (Effect of certain affine transformations on
$P_{\infty}$).

Let $u\in L^{2}(I)$, $\alpha>0$, and $\beta\in\mathbb{R}$. Then
$P_{\infty}(\alpha u+\beta)=\alpha P_{\infty}u+\beta$.

###### Proof.

Since $P_{t}v\to P_{\infty}v$ as $t\to\infty$ for each $v\in L^{2}(I)$,
applying Proposition 5.63 yields

$P_{\infty}(\alpha u+\beta)=\lim_{t\to\infty}P_{t}(\alpha
u+\beta)=\lim_{t\to\infty}\left[\alpha
P_{t}u+\beta\right]=\alpha\left[\lim_{t\to\infty}P_{t}u\right]+\beta=\alpha
P_{\infty}u+\beta\,.\qed$

We can finally conclude, via Lemma 5.46, that $P_{\infty}$ is nonexpansive in
the $L^{\infty}$ norm.

######  Proposition 5.74 ($P_{\infty}$ is nonexpansive in $L^{\infty}$ norm).

Let $u,v\in L^{2}(I)$. Then $\left\lvert
P_{\infty}u-P_{\infty}v\right\rvert\leq\operatorname*{ess\,sup}\left\lvert
u-v\right\rvert$ a.e. in $I$.

###### Proof.

We verify the conditions of Lemma 5.46 with $(\Omega,\Sigma,\mu)$ the set $I$
endowed with the Lebesgue measure, $p=2$, $C=L^{2}(I)$, and $T=P_{\infty}$. It
is clear that for all $f\in L^{2}(I)$ and $r\in\mathbb{R}$, $f+r\in L^{2}(I)$.
By Proposition 5.69, $P_{\infty}$ is a continuous $L^{2}(I)\to L^{2}(I)$ map.
Moreover, for all $f\in L^{2}(I)$ and $r\in\mathbb{R}$, the condition
$P_{\infty}(f+r)=P_{\infty}f+r$ holds by Proposition 5.73. Finally, for all
$f,g\in L^{2}(I)$ with $f\leq g$ a.e. , we have that $P_{\infty}f\leq
P_{\infty}g$ a.e. by Corollary 5.68. The conclusion follows. ∎

##  6 Directed transport-energy inequality

In this section, we establish a connection between the PDE studied above and
the Wasserstein distance between the initial state $u$ and its monotone
equilibrium $P_{\infty}u$, via the dynamical approach embodied by the
_Benamou-Brenier formula_. We follow the presentation and formalism of
[San15], and also refer to [AGS05, Vil09].

###  6.1 Preliminaries for optimal transport

We start by introducing notation and definitions relevant to the theory of
optimal transport.

##### Projections.

Say $X\times Y$ is a product space and $z=(x,y)$ is an element in this space.
Then the projection operator onto the first coordinate, denoted
interchangeably by $\pi_{1}$ or $\pi_{x}$ depending on the context, is given
by $\pi_{1}(z)=\pi_{x}(z)=x$. Similarly, $\pi_{2}(z)=\pi_{y}(z)=y$. We extend
this definition in the natural way projections from larger product spaces onto
smaller product spaces. For example, given the product space
$X_{1}\times\dotsm\times X_{d}$ and index set $I\subset[d]$,
$I=\\{i_{1},\dotsc,i_{n}\\}$, the operator $\pi_{I}$ projects any element in
this space down to an element of $X_{i_{1}}\times\dotsm\times X_{i_{n}}$. We
also use the shorthand $\pi_{-I}\vcentcolon=\pi_{[d]\setminus I}$.

Now, let $\Omega\subset\mathbb{R}^{d}$ be a Borel set and let $I\subseteq[d]$.
For a point $x\in\Omega$, we write the projections
$x_{I}\vcentcolon=\pi_{I}(x)$ and $x_{-I}\vcentcolon=\pi_{-I}(x)$, and we also
write $x=(x_{I},x_{-I})$. Then, for the set $\Omega$, we write the projections
$\Omega_{I}\vcentcolon=\\{x_{I}:x\in\Omega\\}$ and
$\Omega_{-I}\vcentcolon=\\{x_{-I}:x\in\Omega\\}$. For small sets $I$ or
$[d]\setminus I$, we also use shorthand notation such as
$x_{i}\vcentcolon=x_{\\{i\\}}$ and
$\Omega_{-i-j}\vcentcolon=\Omega_{-\\{i,j\\}}$.

##### Pushforward measure.

Given measurable spaces $(X,\Sigma_{X})$ and $(Y,\Sigma_{Y})$, measure $\mu$
on $X$, and measurable map $T:X\to Y$, the _pushforward measure_
${T}_{\\#}{\mu}$ on $Y$ is the measure satisfying

$\displaystyle({T}_{\\#}{\mu})(B)$ $\displaystyle=\mu(T^{-1}(B))$ for all
$B\in\Sigma_{Y}$, or equivalently,
$\displaystyle\int_{Y}\phi(y)\odif{({T}_{\\#}{\mu})(y)}$
$\displaystyle=\int_{X}\phi(T(x))\odif\mu(x)$ for all measurable
$\phi:Y\to\mathbb{R}$.

For a measure $\gamma$ on product space $X\times Y$, we say that
${(\pi_{1})}_{\\#}{\gamma}$ and ${(\pi_{2})}_{\\#}{\gamma}$, which are
measures on $X$ and $Y$ respectively, are the first and second _marginals_ of
$\gamma$, respectively.

If $(\Omega_{3},\Sigma_{3})$ is another measurable space and $S:Y\to Z$ a
measurable map, then it holds that ${(S\circ
T)}_{\\#}{\mu}={S}_{\\#}{({T}_{\\#}{\mu})}$.

##### Transport plans.

For two probability spaces $(\Omega_{1},\Sigma_{1},\mu_{1})$ and
$(\Omega_{2},\Sigma_{2},\mu_{2})$, we write $\Pi(\mu_{1},\mu_{2})$ for the set
of _couplings_ , or _transport plans_ , between $\mu_{1}$ and $\mu_{2}$,
namely probability measures $\gamma$ on the product space
$\Omega_{1}\times\Omega_{2}$ whose first and second marginals are $\mu_{1}$
and $\mu_{2}$, respectively.

##### Space of probability measures.

For any Borel set $\Omega\subset\mathbb{R}^{d}$, let $P(\Omega)$ denote the
space of all (Borel) probability measures on $\Omega$. We endow $P(\Omega)$
with the weak topology, which is the topology of weak convergence with respect
to bounded continuous functionals. Namely, we say $\mu_{n}$ converges weakly
to $\mu$ in $P(\Omega)$, and write $\mu_{n}\rightharpoonup\mu$, if
$\int_{\Omega}\phi\odif\mu_{n}\to\int_{\Omega}\phi\odif\mu$ for all bounded
continuous $\phi:\Omega\to\mathbb{R}$.

##### Wasserstein distances.

Let $\Omega\subset\mathbb{R}^{d}$ be a _bounded_ Borel set and let
$p\in[1,+\infty)$. Let $\mu,\nu\in P(\Omega)$. Given transport plan
$\gamma\in\Pi(\mu,\nu)$, we define the _cost_ $C_{p}(\gamma)$ by

$C_{p}(\gamma)\vcentcolon=\left(\int_{\Omega\times\Omega}|x-y|^{p}\odif\gamma(x,y)\right)^{1/p}\,.$

We then define the _$p$ -Wasserstein_ distance between $\mu$ and $\nu$ by

$W_{p}(\mu,\nu)\vcentcolon=\inf_{\gamma\in\Pi(\mu,\nu)}C_{p}(\gamma)\,,$

and we also often refer to the quantity
$W_{p}^{p}(\mu,\nu)\vcentcolon=W_{p}(\mu,\nu)^{p}$. It is standard that
$W_{p}(\cdot,\cdot)$ is a distance metric on $P(\Omega)$.

######  Fact 6.1 (Wasserstein distance metrizes weak convergence; see e.g.
[San15, Theorems 5.10 and 5.11]).

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded Borel set, and let
$p\in[1,\infty)$. Then for a sequence $(\mu_{n})_{n}$ in $P(\Omega)$ and
$\mu\in P(\Omega)$, we have $\mu_{n}\rightharpoonup\mu$ if and only if
$W_{p}(\mu_{n},\mu)\to 0$.

######  Corollary 6.2.

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded Borel set, and let
$p\in[1,\infty)$. Then for sequences $(\mu_{n})_{n}$ and $(\nu_{n})_{n}$ in
$P(\Omega)$ with $\mu_{n}\rightharpoonup\mu$ and $\nu_{n}\rightharpoonup\nu$
for $\mu,\nu\in P(\Omega)$, we have $W_{p}(\mu_{n},\nu_{n})\to
W_{p}(\mu,\nu)$. In other words, the $p$-Wasserstein distance is continuous in
(the weak topology on) $P(\Omega)$.

###### Proof.

By the triangle inequality, we have

$W_{p}(\mu,\nu)-W_{p}(\mu,\mu_{n})-W_{p}(\nu_{n},\nu)\leq
W_{p}(\mu_{n},\nu_{n})\leq
W_{p}(\mu_{n},\mu)+W_{p}(\mu,\nu)+W_{p}(\nu,\nu_{n})\,.$

By Fact 6.1, the LHS and RHS converge to $W_{p}(\mu,\nu)$ as $n\to\infty$, so
the conclusion follows. ∎

###  6.2 Optimal transport via Benamou-Brenier

Let $\\{\varrho_{t}:t\in[0,T]\\}$ be a family of measures on $\overline{I}$
and $\\{v_{t}:t\in[0,T]\\}$ be a family of _velocity fields_ such that
$v_{t}\in L^{1}(\varrho_{t})$ for each $t$. We start by defining what it means
for the family $(\varrho_{t},v_{t})$ to solve the _continuity equation_

$\partial_{t}\varrho_{t}+\partial_{x}(\varrho_{t}v_{t})=0\,.$

######  Definition 6.3 (Weak solution; see [San15, Section 4.1.2]).

Let $(\varrho_{t},v_{t})$ be a family of measure/velocity field pairs indexed
by $t\in[0,T]$ such that $v_{t}\in L^{1}(\varrho)$ for each $t$. We say that
$(\varrho_{t},v_{t})$ is a _weak solution to the continuity equation_ if, for
every test function $\psi\in C^{1}(\overline{I})$, the function
$t\mapsto\int_{\overline{I}}\psi\odif\varrho_{t}$ is absolutely continuous in
$t$ and, for a.e. $t$, we have

$\partial_{t}\int_{\overline{I}}\psi\odif\varrho_{t}=\int_{\overline{I}}(\partial_{x}\psi)v_{t}\odif\varrho_{t}\,.$
(23)

In this case, we call $\varrho_{0}$ and $\varrho_{T}$ the _initial_ and _final
states_ of the solution, respectively (this makes sense because the above
implies that $t\mapsto\varrho_{t}$ is continuous for the weak convergence of
measures).

######  Remark 6.4.

We will only work with absolutely continuous measures $\odif\varrho_{t}=u\odif
x$, $u\in L^{2}(I)$.

######  Remark 6.5.

Applying (23) with a constant test function $\psi$ shows that every solution
to the continuity equation is _mass conserving_. In particular, Proposition
6.8 below implies that this is true of the semigroup $P_{t}$; see Corollary
6.9.

######  Proposition 6.6 (Benamou-Brenier formula; see e.g. [San15, Theorem
5.28]).

Let $\mu,\nu$ be probability measures on $\overline{I}$. Then

$W_{2}^{2}(\mu,\nu)=\min\left\\{\int_{0}^{1}\|v_{t}\|_{L^{2}(\varrho_{t})}^{2}\odif
t:\partial_{t}\varrho_{t}+\partial_{x}(\varrho_{t}v_{t})=0,\varrho_{0}=\mu,\varrho_{1}=\nu\right\\}\,,$

where the constraint above means that $(\varrho_{t},v_{t})_{t\in[0,1]}$ is a
weak solution to the continuity equation with initial state $\mu$ and final
state $\nu$.

######  Condition 6.7.

$u\in\mathcal{U}$ is a.e. positive and bounded away from zero, and satisfies
$\int_{I}u\odif x=1$.

######  Proposition 6.8.

Let $u\in\mathcal{U}$ satisfy 6.7. Let $\bm{u}(t)$ be the solution to the
gradient flow problem with initial state $u$, and $\bm{u^{\prime}}(t)$ its
weak derivative for each $t\geq 0$. Let $T>0$ and define the measure/velocity
field family $(\varrho_{t},v_{t})_{t\in[0,T]}$ by

$\odif\varrho_{t}\vcentcolon=\bm{u}(t)\odif x\quad\text{and}\quad
v_{t}\vcentcolon=-\frac{\partial_{x}\bm{u}(t){\downarrow}}{\bm{u}(t)}$

for each $t\in[0,T]$. Then $(\varrho_{t},v_{t})$ is a weak solution to the
continuity equation with initial state $\bm{u}(0)\odif x$ and final state
$\bm{u}(T)\odif x$.

###### Proof.

We first note that, by 5.58, each $\bm{u}(t)$ is a.e. positive and bounded
away from zero, uniformly in $t$. In particular, this justifies the
denominator in the definition of $v_{t}$. Furthermore, recall that
$\bm{u}(t)\in\mathcal{U}$ for all $t\geq 0$ by Corollaries 5.20 and 5.15, so
the numerator in the definition of $v_{t}$ is defined up to sets of measure
zero for each $t$. (In fact, by Definitions 5.16, 5.18 and 5.21 we have
$\bm{u}(t){\downarrow}\in H^{2}(I)$ for a.e. $t>0$, in which case the
numerator is even pointwise well-defined.)

We now verify that $(\varrho_{t},v_{t})$ satisfies Definition 6.3. The initial
and final states are as claimed by construction. Let $\psi\in
C^{1}(\overline{I})$. Let $\bm{\tilde{u}^{\prime}},\bm{\tilde{u}}:[0,T]\times
I\to\mathbb{R}$ be obtained by applying Lemma 5.53 to $\bm{u}$ with time
domain $[0,T]$. Define the (jointly measurable) function $f:[0,T]\times
I\to\mathbb{R}$ by

$f(t,x)\vcentcolon=\psi(x)\bm{\tilde{u}}(t,x)\,.$

We claim that $f$ satisfies the properties of Lemma 5.47. It is clear that
$f(t,\cdot)\in L^{2}(I)\subset L^{1}(I)$ for each $t\in[0,T]$ since
$\bm{u}(t,\cdot)\in L^{2}(I)$ while $\psi$ is bounded, so the first property
is satisfied. Also, for each $x\in I$ we have that $\bm{\tilde{u}}(\cdot,x)$
is AC and hence so is $f(\cdot,x)$, so the second property is satisfied.
Finally, by properties 1 and 5 of Lemma 5.53 and the Cauchy-Schwarz
inequality,

$\int_{(0,T)}\int_{I}\left\lvert\partial_{t}f(t,x)\right\rvert\odif x\odif
t=\int_{(0,T)}\int_{I}\left\lvert\psi(x)\bm{\tilde{u}^{\prime}}(t,x)\right\rvert\odif
x\odif
t\leq\|\psi\|_{L^{2}(I)}\int_{(0,T)}\|\bm{u^{\prime}}(t)\|_{L^{2}(I)}\odif
t<+\infty\,,$

the last inequality since $\psi\in C^{1}(\overline{I})$ while
$\bm{u^{\prime}}\in L^{2}(0,T;L^{2}(I))$ by Definition 5.13. Hence the third
property is satisfied and Lemma 5.47 applies. Thus the function mapping each
$t\in[0,T]$ to

$\int_{\overline{I}}\psi\odif\varrho_{t}=\int_{I}\psi\bm{u}(t)\odif
x=\int_{I}\psi(x)\bm{\tilde{u}}(t,x)\odif x=\int_{I}f(t,x)\odif x$

is absolutely continuous and, for a.e. $t\in(0,T)$,

$\displaystyle\partial_{t}\int_{\overline{I}}\psi\odif\varrho_{t}$
$\displaystyle=\int_{I}\partial_{t}f(t,x)\odif
x=\int_{I}\psi(x)\bm{\tilde{u}^{\prime}}(t,x)\odif x$ (Property 5 of Lemma
5.53) $\displaystyle=\int_{I}\psi\bm{u^{\prime}}(t)\odif x$ (Property 1 of
Lemma 5.53)
$\displaystyle=\int_{I}\psi\partial_{x}\partial_{x}\bm{u}(t){\downarrow}\odif
x$ (Propositions 5.18 and 5.21)
$\displaystyle=-\int_{I}(\partial_{x}\psi)(\partial_{x}\bm{u}(t){\downarrow})\odif
x$ (Integration by parts, Lemma 5.21)
$\displaystyle=\int_{I}(\partial_{x}\psi)\left(-\frac{\partial_{x}\bm{u}(t){\downarrow}}{\bm{u}(t)}\right)\bm{u}(t)\odif
x=\int_{\overline{I}}(\partial_{x}\psi)v_{t}\odif\varrho_{t}\,.$

######  Corollary 6.9 ($P_{t}$ is mass conserving).

Let $u\in L^{2}(I)$. Then for all $t>0$, $\int_{I}P_{t}u\odif x=\int_{I}u\odif
x$.

###### Proof.

If $u$ satisfies 6.7, then this follows from Proposition 6.8 by taking any
constant test function $\psi$ in Definition 6.3.

If $u\in H^{1}(I)$, then it is bounded, so let $\alpha>0$,
$\beta\in\mathbb{R}$ be such that $v\vcentcolon=\alpha u+\beta\in H^{1}(I)$
satisfies 6.7. Then $u=\frac{1}{\alpha}v-\frac{\beta}{\alpha}$ and, by
Proposition 5.63 and the above,

$\int_{I}P_{t}u\odif
x=\int_{I}\left(\frac{1}{\alpha}P_{t}v-\frac{\beta}{\alpha}\right)\odif
x=-\frac{\beta}{\alpha}+\frac{1}{\alpha}\int_{I}P_{t}v\odif
x=-\frac{\beta}{\alpha}+\frac{1}{\alpha}\int_{I}v\odif
x=\int_{I}\left(\frac{1}{\alpha}v-\frac{\beta}{\alpha}\right)\odif
x=\int_{I}u\odif x\,.$

Finally, let $u\in\mathcal{U}$ be arbitrary. Since $H^{1}(I)$ is dense in
$L^{2}(I)$, let $(u_{n})_{n}$ be a sequence in $H^{1}(I)$ such that $u_{n}\to
u$ in $L^{2}(I)$. By the continuity of $P_{t}$ from $L^{2}(I)$ to $L^{2}(I)$,
and using the above,

$\int_{I}P_{t}u\odif x=\lim_{n\to\infty}\int_{I}P_{t}u_{n}\odif
x=\lim_{n\to\infty}\int_{I}u_{n}\odif x=\int_{I}u\odif x\,.\qed$

By passing to the limit $P_{t}u\to P_{\infty}u$ in $L^{2}(I)$, we also
conclude

######  Corollary 6.10 ($P_{\infty}$ is mass conserving).

Let $u\in L^{2}(I)$. Then $\int_{I}P_{\infty}u\odif x=\int_{I}u\odif x$.

The definition of the velocity field $v_{t}$ in Proposition 6.8, along with
the fact that $\partial_{x}\bm{u}(t){\downarrow}\leq 0$, gives $v_{t}\geq 0$,
which intuitively says that “particles only move to the right”. This
suggests that the solution $\bm{u}(t)$ can only lose mass in any prefix of the
interval $(0,1)$ over time. The following items make this observation
rigorous.

######  Lemma 6.11.

Let $\delta\in I$ and let $J\vcentcolon=(0,\delta)$. Let $u\in L^{2}(I)$. Then
for all $t>0$,

$\int_{J}(P_{t}u)\odif x\leq\int_{J}u\odif x\,.$

###### Proof.

Suppose $u$ satisfies 6.7; the general case will then follow by the same
arguments as in the proof of Corollary 6.9. Let
$J^{\prime}\vcentcolon=(-\infty,\delta)$, let
$\chi_{J^{\prime}}:\mathbb{R}\to[0,1]$ be the characteristic function for the
set $J^{\prime}$, and for each $\epsilon>0$, let
$\chi_{J^{\prime}}^{\epsilon}$ be its mollification by the standard mollifier.
Note that each $\chi_{J^{\prime}}^{\epsilon}$ is nonincreasing. Also note that
$\chi_{J^{\prime}}=\chi_{J}$ pointwise in $I$. Now, fix $T>0$. Since the
family $(\varrho_{t},v_{t})_{t\in[0,T]}$ given by Proposition 6.8 solves the
continuity equation, applying (23) with test function
$\chi_{J^{\prime}}^{\epsilon}\in C^{\infty}(\overline{I})$ gives that
$t\mapsto\int_{I}\chi_{J^{\prime}}^{\epsilon}\bm{u}(t)\odif x$ is AC and, for
a.e. $t\in[0,T]$,

$\partial_{t}\int_{I}\chi_{J^{\prime}}^{\epsilon}\bm{u}(t)\odif
x=\int_{I}\underbrace{(\partial_{x}\chi_{J^{\prime}}^{\epsilon})}_{\leq
0}\underbrace{v_{t}}_{\geq 0}\underbrace{\bm{u}(t)}_{\geq 0}\odif x\leq 0\,.$

Hence $t\mapsto\int_{I}\chi_{J^{\prime}}^{\epsilon}\bm{u}(t)\odif x$ is
nonincreasing, and obtain

$\left\langle\chi_{J^{\prime}}^{\epsilon},u\right\rangle_{L^{2}(I)}=\int_{I}\chi_{J^{\prime}}^{\epsilon}\bm{u}(0)\odif
x\geq\int_{I}\chi_{J^{\prime}}^{\epsilon}\bm{u}(T)\odif
x=\left\langle\chi_{J^{\prime}}^{\epsilon},P_{T}u\right\rangle_{L^{2}(I)}\,.$

Now, since $\chi_{J^{\prime}}\in L^{2}_{\mathrm{loc}}(\mathbb{R})$, we have
that $\chi_{J^{\prime}}^{\epsilon}\to\chi_{J^{\prime}}$ in
$L^{2}_{\mathrm{loc}}(\mathbb{R})$, so in particular
$\chi_{J^{\prime}}^{\epsilon}\to\chi_{J^{\prime}}=\chi_{J}$ in $L^{2}(I)$.
Thus we have

$\int_{J}(P_{T}u)\odif
x=\left\langle\chi_{J},P_{T}u\right\rangle_{L^{2}(I)}=\lim_{\epsilon\to
0}\left\langle\chi_{J^{\prime}}^{\epsilon},P_{T}u\right\rangle_{L^{2}(I)}\leq\lim_{\epsilon\to
0}\left\langle\chi_{J^{\prime}}^{\epsilon},u\right\rangle_{L^{2}(I)}=\left\langle\chi_{J},u\right\rangle_{L^{2}(I)}=\int_{J}u\odif
x\,.\qed$

######  Corollary 6.12.

Let $\delta\in I$ and let $J\vcentcolon=(0,\delta)$. Let $u\in L^{2}(I)$. Then

$\int_{J}(P_{\infty}u)\odif x\leq\int_{J}u\odif x\,.$

###### Proof.

This follows from Lemma 6.11 along with the convergence $P_{t}u\to
P_{\infty}u$ in $L^{2}(I)$. ∎

######  Definition 6.13.

Let $\mu$ and $\nu$ be probability measures on $\mathbb{R}$. We say $\mu$
_dominates_ $\nu$, and write $\mu\succeq\nu$, if for every $x\in\mathbb{R}$ we
have $\mu(-\infty,x)\geq\nu(-\infty,x)$.

######  Corollary 6.14.

Let $u\in\mathcal{U}$ satisfy 6.7. Define the measures
$\odif\mu\vcentcolon=u\odif x$ and
$\odif\mu_{\infty}\vcentcolon=(P_{\infty}u)\odif x$. Then
$\mu\succeq\mu_{\infty}$.

###### Proof.

We may view $\mu$ and $\mu_{\infty}$ as absolutely continuous measures on all
of $\mathbb{R}$, taking zero outside $[0,1]$. Now, for each $x\in[0,1]$,
Corollary 6.12 gives

$\mu(-\infty,x)=\int_{(0,x)}u\odif y\geq\int_{(0,x)}(P_{\infty}u)\odif
y=\mu_{\infty}(-\infty,x)\,.$

On the other hand, for $x<0$ both sides are zero, and for $x>1$, both sides
are $1$ by Corollary 6.10. ∎

We can now upper bound the Wasserstein distance between $u\in\mathcal{U}$ and
its monotone equilibrium by combining the Benamou-Brenier formula with the
exponential decay of the directed Dirichlet energy, at least as long as $u$ is
positive and bounded away from zero.

###### Proof.

Let $a\vcentcolon=\inf u$. By Corollaries 5.56 and 6.9, $P_{t}u$ satisfies 6.7
with $P_{t}u\geq a$ a.e. for all $t\geq 0$. For each $t\geq 0$, define the
measure $\odif\varrho_{t}\vcentcolon=(P_{t}u)\odif x$. Let
$\ell\in\mathbb{Z}_{\geq 0}$. Then Proposition 6.8 applies to time interval
$[\ell,\ell+1]$ (we refrain from introducing excessive notation for such
relabeling of intervals) with initial/final states $\varrho_{\ell}$ and
$\varrho_{\ell+1}$, respectively, so that Proposition 6.6 gives

$\displaystyle W_{2}^{2}(\varrho_{\ell},\varrho_{\ell+1})$
$\displaystyle\leq\int_{\ell}^{\ell+1}\int_{I}\left(-\frac{\partial_{x}\bm{u}(t){\downarrow}}{\bm{u}(t)}\right)^{2}\bm{u}(t)\odif
x\odif
t=\int_{\ell}^{\ell+1}\int_{I}\frac{(\partial_{x}\bm{u}(t){\downarrow})^{2}}{\bm{u}(t)}\odif
x\odif t$
$\displaystyle\leq\frac{1}{a}\int_{\ell}^{\ell+1}\int_{I}(\partial_{x}\bm{u}(t){\downarrow})^{2}\odif
x\odif t=\frac{2}{a}\int_{\ell}^{\ell+1}\mathcal{E}^{-}(P_{t}u)\odif t$
$\displaystyle\leq\frac{2}{a}\mathcal{E}^{-}(u)\int_{\ell}^{\ell+1}e^{-Kt}\odif
t<\frac{2}{a}\mathcal{E}^{-}(u)\int_{\ell}^{\infty}e^{-Kt}\odif
t=\frac{2e^{-K\ell}}{aK}\mathcal{E}^{-}(u)\,,$

where the penultimate inequality and the constant $K$ come from Proposition
5.43. Now, by the triangle inequality (on the distance $W_{2}$, not the
squared distance $W_{2}^{2}$), for every $\ell\in\mathbb{Z}_{\geq 1}$ we have

$\displaystyle W_{2}(\varrho_{0},\varrho_{\ell})$
$\displaystyle\leq\sum_{j=0}^{\ell-1}W_{2}(\varrho_{j},\varrho_{j+1})\leq\left(\frac{2}{aK}\mathcal{E}^{-}(u)\right)^{1/2}\sum_{j=0}^{\infty}e^{-Kj/2}=\left(\frac{2}{aK}\mathcal{E}^{-}(u)\right)^{1/2}\frac{1}{1-e^{-K/2}}\,.$

Therefore, extracting the appropriate constant $C$ from the terms involving
$K$, we obtain

$W_{2}(\varrho_{0},\varrho_{\ell})\leq\frac{C}{a^{1/2}}(\mathcal{E}^{-}(u))^{1/2}\,.$

Recall that $P_{t}u\rightharpoonup P_{\infty}u$ weakly in $L^{2}(I)$ by
Proposition 5.65. Hence, defining the absolutely continuous measure
$\odif\varrho_{\infty}\vcentcolon=(P_{\infty}u)\odif x$, we conclude that
$\varrho_{t}\rightharpoonup\varrho_{\infty}$ as $t\to\infty$. Then by
Corollary 6.2,

$W_{2}(\varrho_{0},\varrho_{\infty})=\lim_{\ell\to\infty}W_{2}(\varrho_{0},\varrho_{\ell})\leq\frac{C}{a^{1/2}}(\mathcal{E}^{-}(u))^{1/2}\,.\qed$

###  6.3 Directed optimal transport

We now introduce directed versions of basic optimal transport concepts and
theory, with the goal of combining a directed version of weak Kantorovich
duality with a multidimensional directed version of the transport-energy
inequality from Section 1.2.3 to obtain our desired directed Poincaré
inequality via a perturbation argument. In the interest of space, we refrain
starting a systematic study of directed optimal transport, but rather limit
ourselves to the results we require.

######  Definition 6.15 (Quasimetric).

Let $X$ be a set. An extended real-valued function $d:X\times
X\to[0,+\infty]$, not necessarily symmetric, is called a _quasimetric_ on $X$
if it satisfies the following:

  1. 1.

For all $x,y\in X$, $d(x,y)\geq 0$ with $d(x,y)=0$ if and only if $x=y$.

  2. 2.

For all $x,y,z\in X$, $d(x,z)\leq d(x,y)+d(y,z)$.

######  Definition 6.16.

Let $\Omega\subset\mathbb{R}^{d}$ be a Borel set, and let $\mu$ and $\nu$ be
probability measures on $\Omega$. We define the set $\Pi(\mu\to\nu)$ of
_directed couplings_ , or _directed transport plans_ , from $\mu$ to $\nu$ as

$\Pi(\mu\to\nu)\vcentcolon=\left\\{\gamma\in\Pi(\mu,\nu):\int_{\Omega\times\Omega}\chi_{\\{x\not\preceq
y\\}}\odif\gamma(x,y)=0\right\\}\,,$

where the integral is well-defined because the set
$\\{(x,y)\in\mathbb{R}^{d}\times\mathbb{R}^{d}:x\not\preceq y\\}$ is open and
hence Borel measurable. Note that the condition could also be written as
$\gamma\left(\\{x\not\preceq y\\}\right)=0$, or $\gamma\left(\\{x\preceq
y\\}\right)=1$.

######  Definition 6.17 (Directed Wasserstein distance).

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded Borel set and let
$p\in[1,\infty)$. Given two probability distributions $\mu,\nu$ over $\Omega$,
we define the _directed $p$-Wasserstein distance from $\mu$ to $\nu$_ by

$W_{p}(\mu\to\nu)\vcentcolon=\inf_{\gamma\in\Pi(\mu\to\nu)}C_{p}(\gamma)\,,$

and we write $W_{p}^{p}(\mu\to\nu)\vcentcolon=W_{p}(\mu\to\nu)^{p}$. Note that
we may have $W_{p}(\mu\to\nu)=+\infty$, and that $W_{p}(\cdot\to\cdot)$ is not
symmetric in general.

######  Observation 6.18.

For all measures $\mu$ and $\nu$, we have $\Pi(\mu\to\nu)\subset\Pi(\mu,\nu)$
and hence $W_{p}(\mu,\nu)\leq W_{p}(\mu\to\nu)$.

######  Lemma 6.19 (Gluing lemma; see e.g. [San15, Lemma 5.5]).

Let $\Omega\subset\mathbb{R}^{d}$ be a Borel set. Let $\mu,\varrho,\nu$ be
probability measures on $\Omega$, and let $\gamma^{+}\in\Pi(\mu,\varrho)$ and
$\gamma^{-}\in\Pi(\varrho,\nu)$. Then there exists a probability measure
$\sigma$ on $\Omega\times\Omega\times\Omega$ such that
${(\pi_{x,y})}_{\\#}{\sigma}=\gamma^{+}$ and
${(\pi_{y,z})}_{\\#}{\sigma}=\gamma^{-}$.

The proof of that $W_{p}(\cdot\to\cdot)$ is a quasimetric follows the
presentation of [San15, Lemma 5.4], with a simple additional argument to
handle _directed_ couplings.

######  Lemma 6.20 (Composition of directed transport plans).

In Lemma 6.19, if $\gamma^{+}\in\Pi(\mu\to\varrho)$ and
$\gamma^{-}\in\Pi(\varrho\to\nu)$, then
${(\pi_{x,z})}_{\\#}{\sigma}\in\Pi(\mu\to\nu)$.

###### Proof.

Let $\gamma\vcentcolon={(\pi_{x,z})}_{\\#}{\sigma}$. First, since
${(\pi_{x})}_{\\#}{\gamma}={(\pi_{x}\circ\pi_{x,z})}_{\\#}{\sigma}={(\pi_{x}\circ\pi_{x,y})}_{\\#}{\sigma}={(\pi_{x})}_{\\#}{\gamma^{+}}=\mu$,
and similarly ${(\pi_{z})}_{\\#}{\gamma}=\nu$, we have
$\gamma\in\Pi(\mu,\nu)$. Moreover, by definition of pushforward measure we
have

$\displaystyle\int_{\Omega\times\Omega}\chi_{\\{x\not\preceq
z\\}}\odif\gamma(x,z)$
$\displaystyle=\int_{\Omega\times\Omega\times\Omega}\chi_{\\{x\not\preceq
z\\}}\odif\sigma(x,y,z)\leq\int_{\Omega\times\Omega\times\Omega}\left(\chi_{\\{x\not\preceq
y\\}}+\chi_{\\{y\not\preceq z\\}}\right)\odif\sigma(x,y,z)$
$\displaystyle=\int_{\Omega\times\Omega\times\Omega}\chi_{\\{x\not\preceq
y\\}}\odif\sigma(x,y,z)+\int_{\Omega\times\Omega\times\Omega}\chi_{\\{y\not\preceq
z\\}}\odif\sigma(x,y,z)$
$\displaystyle=\int_{\Omega\times\Omega}\chi_{\\{x\not\preceq
y\\}}\odif\gamma^{+}(x,y)+\int_{\Omega\times\Omega}\chi_{\\{y\not\preceq
z\\}}\odif\gamma^{-}(y,z)=0\,,$

the last equality since $\gamma^{+}$ and $\gamma^{-}$ are directed couplings.
Hence $\gamma\in\Pi(\mu\to\nu)$ as claimed. ∎

######  Proposition 6.21.

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded Borel set and $p\in[1,\infty)$.
Then $W_{p}(\cdot\to\cdot)$ is a quasimetric on $P(\Omega)$.

###### Proof.

It is clear that $W_{p}(\mu\to\nu)\geq 0$ always. If $\mu=\nu$ then the
identity coupling shows that $W_{p}(\mu\to\nu)=0$, and conversely if
$W_{p}(\mu\to\nu)=0$ then, by 6.18, $W_{p}(\mu,\nu)=0$ and hence $\mu=\nu$. It
remains to show that $W_{p}(\cdot\to\cdot)$ satisfies the triangle inequality.

Let $\mu,\varrho,\nu$ be probability measures on $\Omega$. Let
$\gamma^{+}\in\Pi(\mu\to\varrho)$ and $\gamma^{-}\in\Pi(\varrho\to\nu)$. Then
since we also have $\gamma^{+}\in\Pi(\mu,\varrho)$ and
$\gamma^{-}\in\Pi(\varrho,\nu)$, apply Lemma 6.19 to obtain a probability
measure $\sigma$ on $\Omega\times\Omega\times\Omega$ such that
${(\pi_{x,y})}_{\\#}{\sigma}=\gamma^{+}$ and
${(\pi_{y,z})}_{\\#}{\sigma}=\gamma^{-}$. Let
$\gamma\vcentcolon={(\pi_{x,z})}_{\\#}{\sigma}$, so that
$\gamma\in\Pi(\mu\to\nu)$ by Lemma 6.20. Then

$\displaystyle W_{p}(\mu\to\nu)$
$\displaystyle\leq\left(\int_{\Omega\times\Omega}|x-z|^{p}\odif\gamma(x,z)\right)^{1/p}=\left(\int_{\Omega\times\Omega\times\Omega}|x-z|^{p}\odif\sigma(x,y,z)\right)^{1/p}$
$\displaystyle=\||x-z|\|_{L^{p}(\sigma)}\leq\||x-y|+|y-z|\|_{L^{p}(\sigma)}\leq\||x-y|\|_{L^{p}(\sigma)}+\||y-z|\|_{L^{p}(\sigma)}$
$\displaystyle=\left(\int_{\Omega\times\Omega\times\Omega}|x-y|^{p}\odif\sigma(x,y,z)\right)^{1/p}+\left(\int_{\Omega\times\Omega\times\Omega}|y-z|^{p}\odif\sigma(x,y,z)\right)^{1/p}$
$\displaystyle=\left(\int_{\Omega\times\Omega}|x-y|^{p}\odif\gamma^{+}(x,y)\right)^{1/p}+\left(\int_{\Omega\times\Omega}|y-z|^{p}\odif\gamma^{-}(y,z)\right)^{1/p}\,.$

Since $\gamma^{+}\in\Pi(\mu\to\varrho),\gamma^{-}\in\Pi(\varrho\to\nu)$ were
arbitrary, $W_{p}(\mu\to\nu)\leq W_{p}(\mu\to\varrho)+W_{p}(\varrho\to\nu)$. ∎

The directed Wasserstein distance arises naturally in the one-dimensional case
when one probability measure dominates the other in the sense of Definition
6.13, as we now show.

######  Proposition 6.22 (Specialization of [San15, Theorem 2.9]).

Let $p\in(1,+\infty)$. Let $\mu$ and $\nu$ be two absolutely continuous
probability measures on $\overline{I}$ with strictly positive densities. Then
there exists a unique $\gamma\in\Pi(\mu,\nu)$ attaining
$C_{p}(\gamma)=W_{p}(\mu,\nu)$, and in fact
$\gamma={(\mathsf{id},T_{\mathrm{mon}})}_{\\#}{\mu}$ where
$T_{\mathrm{mon}}:\overline{I}\to\overline{I}$ is a nondecreasing function
that is a.e. uniquely determined, and $\mathsf{id}$ denotes the identity map.

######  Corollary 6.23.

If in Proposition 6.22 we have $\mu\succeq\nu$, then in fact
$\gamma\in\Pi(\mu\to\nu)$. As a consequence,
$W_{p}(\mu\to\nu)=W_{p}(\mu,\nu)$.

###### Proof.

We claim that $T_{\mathrm{mon}}(x)\geq x$ for every $x\in I$. Indeed, suppose
$T_{\mathrm{mon}}(x)<x$ for some $x\in I$. Then

$\nu(0,T_{\mathrm{mon}}(x))=({T_{\mathrm{mon}}}_{\\#}{\mu})(0,T_{\mathrm{mon}}(x))=\mu(T_{\mathrm{mon}}^{-1}(0,T_{\mathrm{mon}}(x)))\geq\mu(0,x)\,,$

the inequality because certainly $(0,x)\subseteq
T_{\mathrm{mon}}^{-1}(0,T_{\mathrm{mon}}(x))$, but we do not rule out at this
point that $T$ remains constant for a while after $x$. But since
$T_{\mathrm{mon}}(x)<x$ and $\nu$ has strictly positive density by assumption,
we conclude that

$\nu(0,x)>\nu(0,T_{\mathrm{mon}}(x))\geq\mu(0,x)\,,$

contradicting the assumption that $\mu\succeq\nu$. Hence
$T_{\mathrm{mon}}(x)\geq x$ for every $x\in I$ as claimed. We now show that
$\gamma\in\Pi(\mu\to\nu)$. Let
$S\vcentcolon=\\{(x,y)\in\overline{I}\times\overline{I}:x\leq y\\}$. Since
$\gamma={(\mathsf{id},T_{\mathrm{mon}})}_{\\#}{\mu}$, we have

$\gamma(S)=\mu((\mathsf{id},T_{\mathrm{mon}})^{-1}(S))=\mu(\\{x\in\overline{I}:x\leq
T_{\mathrm{mon}}(x)\\})=\mu(\overline{I})=1\,,$

so $\gamma\in\Pi(\mu\to\nu)$ as claimed. ∎

We now introduce the formal language for two related operations: constructing
a probability measure by its marginal and conditional distributions, and the
inverse process of extracting marginal and conditionals from a probability
measure, which is also called _disintegration_. We refer the reader to [AGS05,
Section 5.3] for an overview of these ideas from a measure-theoretic
perspective.

We use the following definition from [AGS05]. For $X$ and $Y$ separable metric
spaces and $x\in X\mapsto\mu_{x}\in P(Y)$ a measure-valued map, we say
$\mu_{x}$ is a Borel map if $x\mapsto\mu_{x}(B)$ is a Borel map for any Borel
set $B\subset Y$, or equivalently if this holds for any open set $A\subset Y$.
In this case we also have that

$x\mapsto\int_{Y}f(x,y)\odif{\mu_{x}(y)}$

is Borel for any bounded (or nonnegative) Borel function $f:X\times
Y\to\mathbb{R}$.

######  Definition 6.24 (Construction and disintegration).

Let $S$ be a finite set and let $I\subset S$. Let $x_{S\setminus
I}\in\mathbb{R}^{S\setminus I}\mapsto{\mu}_{|x_{S\setminus I}}\in
P(\mathbb{R}^{I})$ be a Borel map (the _conditionals_). Then for any bounded
(or nonnegative) Borel function $f:\mathbb{R}^{S}\to\mathbb{R}$, the
$\mathbb{R}^{S\setminus I}\to\mathbb{R}$ map

$x_{S\setminus I}\mapsto\int_{\mathbb{R}^{I}}f(x_{S\setminus
I},x_{I})\odif{{\mu}_{|x_{S\setminus I}}(x_{I})}$ (24)

is Borel. Therefore, for any ${\mu}_{S\setminus I}\in P(\mathbb{R}^{S\setminus
I})$ (the _marginal_), we define $\mu\in P(\mathbb{R}^{S})$ implicitly by

$\int_{\mathbb{R}^{S}}f\odif\mu=\int_{\mathbb{R}^{S\setminus
I}}\odif{{\mu}_{S\setminus I}(x_{S\setminus
I})}\int_{\mathbb{R}^{I}}f(x_{S\setminus I},x_{I})\odif{{\mu}_{|x_{S\setminus
I}}(x_{I})}\,,$ (25)

and we formally write $\mu=\int_{\mathbb{R}^{S\setminus
I}}{\mu}_{|x_{S\setminus I}}\odif{{\mu}_{S\setminus I}}$.

Conversely, given $\mu\in P(\mathbb{R}^{S})$, we let ${\mu}_{S\setminus
I}\vcentcolon={(\pi_{S\setminus I})}_{\\#}{\mu}\in P(\mathbb{R}^{S\setminus
I})$ and write $({\mu}_{|x_{S\setminus I}})_{x_{S\setminus
I}\in\mathbb{R}^{S\setminus I}}\subset P(\mathbb{R}^{I})$ for the
${\mu}_{S\setminus I}$-a.e. uniquely determined Borel family of probability
measures such that $\mu=\int_{\mathbb{R}^{S\setminus I}}{\mu}_{|x_{S\setminus
I}}\odif{{\mu}_{S\setminus I}}$, and we call this decomposition a
_disintegration_ of $\mu$.

##### Examples and simplified notational conventions.

Since the index notation in the definition above can be somewhat laborious to
parse, let us briefly give two concrete settings we will use below, and
introduce simplified notational conventions for those settings. The
simplifications are supposed to be mnemonic for what we have already defined
for projections.

  1. 1.

For a probability measure $\mu\in P(\mathbb{R}^{d})$ and index $i\in[d]$, we
may disintegrate $\mu$ into a marginal on index set $[d]\setminus\\{i\\}$ and
conditionals supported along $i$-th coordinate, and write
$\mu=\int_{\mathbb{R}^{[d]\setminus\\{i\\}}}{\mu}_{|x_{[d]\setminus\\{i\\}}}\odif{{\mu}_{[d]\setminus\\{i\\}}}$.
We simplify this notation by writing
$\mu=\int_{\mathbb{R}^{d}_{-i}}{\mu}_{|x_{-i}}\odif{{\mu}_{-i}}$.

  2. 2.

For a transport plan $\gamma\in P(\mathbb{R}^{d}\times\mathbb{R}^{d})$, we
write the index set as $[2d]=S=S_{x}\cup S_{y}$ for $S_{x}\vcentcolon=[d]$ and
$S_{y}\vcentcolon=[2d]\setminus[d]$, so we write each element in the support
of $\gamma$ as a tuple $(x,y)$ with $x\in\mathbb{R}^{S_{x}}$ and
$y\in\mathbb{R}^{S_{y}}$. Then, given index set $I\subset[d]$, we may
disintegrate $\gamma$ into a marginal on index set
$S_{M}\vcentcolon=(S_{x}\setminus I_{x})\cup(S_{y}\setminus I_{y})$ for
$I_{x}\vcentcolon=I$ and $I_{y}\vcentcolon=\\{d+i:i\in I\\}$, so that we think
of the marginal as determining all but the $I$-indexed coordinates of the
points $x$ and $y$; and we view the conditionals as supported along the
$I_{x}\cup I_{y}$ directions. This disintegration is
$\gamma=\int_{\mathbb{R}^{S_{M}}}{\gamma}_{|x_{S_{M}}}\odif{{\gamma}_{S_{M}}}$.
To be clear, here we have ${\gamma}_{S_{M}}\in P(\mathbb{R}^{S_{M}})$ and
${\gamma}_{|x_{S_{M}}}\in P(\mathbb{R}^{I_{x}\cup I_{y}})$ for each
$x_{S_{M}}\in\mathbb{R}^{S_{M}}$. We make the notation somewhat more mnemonic
by writing
$\gamma=\int_{\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I}}{\gamma}_{|x_{-I},y_{-I}}\odif{{\gamma}_{-I_{x}-I_{y}}}$.
When we are considering a singleton $I=\\{i\\}$, we further simplify the
notation by directly writing $i$ in the place of $I$ in this formula.

For absolutely continuous probability measures on $[0,1]^{d}$, the following
more familiar characterization of disintegrations will be useful:

######  Proposition 6.25.

Let $\mu$ be an absolutely continuous probability measure whose density is
Borel measurable and supported in $[0,1]^{d}$. Namely, write $\odif\mu=u\odif
x$ where $u:[0,1]^{d}\to[0,+\infty)$ is a Borel map. Let $i\in[d]$. Then the
disintegration
$\mu=\int_{\mathbb{R}^{d}_{-i}}{\mu}_{|x_{-i}}\odif{{\mu}_{-i}}$ is as
follows: ${\mu}_{-i}$ and each ${\mu}_{|x_{-i}}$ are absolutely continuous
probability measures with Borel densities supported in $[0,1]^{d}_{-i}$ and
$[0,1]$ respectively; and writing $\odif{{\mu}_{-i}}={u}_{-i}\odif x_{-i}$,
and $\odif{{\mu}_{|x_{-i}}}={u}_{|x_{-i}}\odif x_{i}$ for each $x_{-i}$, where
the (Borel) density functions are ${u}_{-i}:[0,1]^{d}_{-i}\to[0,+\infty)$ and
${u}_{|x_{-i}}:[0,1]\to[0,+\infty)$ for each $x_{-i}\in[0,1]^{d}_{-i}$, we
have (marginal density)

${u}_{-i}(x_{-i})=\int_{I}u(x_{-i},x_{i})\odif x_{i}$

for each $x_{-i}\in[0,1]^{d}_{-i}$, and (conditional density)

${u}_{|x_{-i}}(x_{i})=\frac{u(x)}{{u}_{-i}(x_{-i})}$

for each $x\in[0,1]^{d}$ where the denominator is nonzero, or $1$ by
convention otherwise.

###### Proof.

First, note that it is standard that ${u}_{-i}$ is a Borel map since it is the
integral of the section $u(x_{-i},\cdot)$ at each point $x_{-i}$. It is hence
immediate that each ${u}_{|x_{-i}}$ is also Borel. Thus ${\mu}_{-i}$ and all
${\mu}_{|x_{-i}}$ are indeed absolutely continuous probability measures.

We now check the conditions of Definition 6.24. We claim that the map
$x_{-i}\in\mathbb{R}^{d}_{-i}\mapsto{\mu}_{|x_{-i}}\in P(\mathbb{R})$ is
Borel. Let $B\subset\mathbb{R}$ be a Borel set; we may assume that
$B\subset[0,1]$ without loss of generality, since all measures assign zero
outside this interval. Then the map
$x_{-i}\in\mathbb{R}^{d}_{-i}\mapsto{\mu}_{|x_{-i}}(B)$ takes value zero
outside $[0,1]^{d}_{-i}$, and for $x_{-i}\in[0,1]^{d}_{-i}$, it is

$x_{-i}\mapsto\frac{1}{{u}_{-i}(x_{-i})}\int_{B}u(x_{-i},x_{i})\odif x_{i}\,.$

This is again, at each point, the integral of a section over the subspace $B$,
so it is standard that this map is Borel. Hence the claim holds.

It remains to verify (25). Let $f:\mathbb{R}^{d}\to[0,+\infty)$ be a
nonnegative Borel function (and the result for bounded $f$ will also follow).
By the definition of the measures and Tonelli’s theorem, we have

$\displaystyle\int_{\mathbb{R}^{-i}}\odif{{\mu}_{-i}(x_{-i})}\int_{\mathbb{R}}f(x_{-i},x_{i})\odif{{u}_{|x_{-i}}(x_{i})}=\int_{[0,1]^{d}_{-i}}{u}_{-i}(x_{-i})\odif
x_{-i}\int_{I}f(x_{-i},x_{i}){u}_{|x_{-i}}(x_{i})\odif x_{i}$
$\displaystyle\qquad=\int_{[0,1]^{d}}f(x){u}_{-i}(x_{-i}){u}_{|x_{-i}}(x_{i})\odif
x=\int_{[0,1]^{d}}f(x)u(x)\odif x\,,$

where the last equality holds because
$u(x)={u}_{-i}(x_{-i}){u}_{|x_{-i}}(x_{i})$ whenever ${u}_{-i}(x_{-i})>0$, and
if ${u}_{-i}(x_{-i})=0$ then $u(x_{-i},\cdot)=0$ almost everywhere, so
$\left\\{x\in[0,1]^{d}:u(x)>0\text{ and }{u}_{-i}(x_{-i})=0\right\\}$ is Borel
and has measure zero (by another application of Tonelli’s theorem). ∎

######  Definition 6.26 (Aligned transport plans).

Let $I\subsetneq[d]$ be nonempty and let $\mu,\nu\in P(\mathbb{R}^{d})$. Let
$\gamma\in\Pi(\mu,\nu)$ and write its disintegration
$\gamma=\int_{\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I}}{\gamma}_{|x_{-I},y_{-I}}\odif{{\gamma}_{-I_{x}-I_{y}}}$.
We say $\gamma$ is _$I$ -aligned_ if its marginal ${\gamma}_{-I_{x}-I_{y}}\in
P(\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I})$ satisfies the following:

${\gamma}_{-I_{x}-I_{y}}\left(\left\\{(x_{-I},y_{-I})\in\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I}:x_{-I}\neq
y_{-I}\right\\}\right)=0\,.$

Note that this condition is well-defined because the set being measured is
open and hence Borel. By convention, we say that every $\gamma\in\Pi(\mu,\nu)$
is $[d]$-aligned. For any nonempty $I\subseteq[d]$, we denote the set of
$I$-aligned transport plans by $\Pi_{I}(\mu,\nu)$. When we have a singleton
$I=\\{i\\}$, we write $i$ directly in the place of $I$ in this definition.

######  Definition 6.27.

For nonempty $I\subseteq[d]$ and $\mu,\nu\in P(\mathbb{R}^{d})$, write
$\Pi_{I}(\mu\to\nu)\vcentcolon=\Pi(\mu\to\nu)\cap\Pi_{I}(\mu,\nu)$.

######  Lemma 6.28 (Alternative characterization of $I$-aligned plans).

Let $I\subsetneq[d]$ be nonempty and let $\mu,\nu\in P(\mathbb{R}^{d})$. Then
$\gamma\in\Pi(\mu,\nu)$ is $I$-aligned if and only if

$\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif\gamma(x,y)=0\,.$

###### Proof.

Writing the disintegration
$\gamma=\int_{\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I}}{\gamma}_{|x_{-I},y_{-I}}\odif{{\gamma}_{-I_{x}-I_{y}}}$,
we have

$\displaystyle{\gamma}_{-I_{x}-I_{y}}(\\{x_{-I}\neq y_{-I}\\})$
$\displaystyle=\int_{\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif{{\gamma}_{-I_{x}-I_{y}}}\underbrace{\int_{\mathbb{R}^{I}\times\mathbb{R}^{I}}\odif{{{\gamma}_{|x_{-I},y_{-I}}(x_{I},y_{I})}}}_{=1}$
$\displaystyle=\int_{\mathbb{R}^{d}_{-I}\times\mathbb{R}^{d}_{-I}}\odif{{\gamma}_{-I_{x}-I_{y}}}\int_{\mathbb{R}^{I}\times\mathbb{R}^{I}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif{{{\gamma}_{|x_{-I},y_{-I}}(x_{I},y_{I})}}$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif\gamma\,.\qed$

While we defined $I$-aligned transport plans for probability measures on all
of $\mathbb{R}^{d}$, we extend the definition to measures on subsets
$\Omega\subset\mathbb{R}^{d}$ (e.g. on the cube) by canonically extending any
such measure to all of $\mathbb{R}^{d}$ by assigning zero outside of $\Omega$.

######  Lemma 6.29 (Composition of aligned transport plans).

Let $I,J\subseteq[d]$ be nonempty. Then in Lemma 6.19, if
$\gamma^{+}\in\Pi_{I}(\mu,\varrho)$ and $\gamma^{-}\in\Pi_{J}(\varrho,\nu)$,
then ${(\pi_{x,z})}_{\\#}{\sigma}\in\Pi_{I\cup J}(\mu,\nu)$.

###### Proof.

Let $\gamma\vcentcolon={(\pi_{x,z})}_{\\#}{\sigma}$ and $K\vcentcolon=I\cup
J$. As in the proof of Lemma 6.20, we do have $\gamma\in\Pi(\mu,\nu)$, so it
remains to show that $\gamma$ is $K$-aligned (and we may assume that
$K\subsetneq[n]$, otherwise there is nothing to prove). We have

$\displaystyle\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-K}\neq
z_{-K}\\}}\odif\gamma$
$\displaystyle\qquad=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-K}\neq
z_{-K}\\}}(x,z)\odif\sigma(x,y,z)$ (Pushforward)
$\displaystyle\qquad\leq\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\left(\chi_{\\{x_{-K}\neq
y_{-K}\\}}(x,y)+\chi_{\\{y_{-K}\neq z_{-K}\\}}(y,z)\right)\odif\sigma(x,y,z)$
$\displaystyle\qquad=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-K}\neq
y_{-K}\\}}\odif\sigma+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{y_{-K}\neq
z_{-K}\\}}\odif\sigma$
$\displaystyle\qquad=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-K}\neq
y_{-K}\\}}\odif\gamma^{+}+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{y_{-K}\neq
z_{-K}\\}}\odif\gamma^{-}$ (Pushforward)
$\displaystyle\qquad\leq\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif\gamma^{+}+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{y_{-J}\neq
z_{-J}\\}}\odif\gamma^{-}$ ($I,J\subseteq K$) $\displaystyle\qquad=0\,,$

the last step since $\gamma^{+}$ is $I$-aligned and $\gamma^{-}$ is
$J$-aligned and by Lemma 6.28; and again by the latter, $\gamma$ is
$K$-aligned. ∎

The following results let us find aligned transport plans by transporting mass
only within $i$-aligned lines.

######  Lemma 6.30 (Measurable selection; specialization of [Vil09, Corollary
5.22]).

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded set. Let $p\in[1,\infty)$. Let
$A$ be a measurable space and let $a\mapsto(\mu_{a},\nu_{a})$ be a measurable
function $A\to P(\mathbb{R})\times P(\mathbb{R})$ with each $\mu_{a}$ and
$\nu_{a}$ having bounded support. Then there is a measurable choice
$a\mapsto\gamma_{a}$ such that, for each $a\in A$,
$\gamma_{a}\in\Pi(\mu_{a},\nu_{a})$ and
$C_{p}(\gamma_{a})=W_{p}(\mu_{a},\nu_{a})$.

######  Lemma 6.31.

Let $p\in[1,\infty)$, $i\in[d]$, and let $\mu,\nu\in P(\mathbb{R}^{d})$ be
supported inside some bounded set. Suppose ${\mu}_{-i}={\nu}_{-i}$. Then there
exists $\gamma\in\Pi_{i}(\mu,\nu)$ satisfying

$C_{p}(\gamma)^{p}=\int_{\mathbb{R}^{d}_{-i}}W_{p}^{p}({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})\odif\mu_{-i}(x_{-i})\,.$

###### Proof.

Let $A\vcentcolon=\mathbb{R}^{d}_{-i}$ be equipped with the Borel
$\sigma$-algebra on $\mathbb{R}^{[d]\setminus\\{i\\}}$. Then the $A\to
P(\mathbb{R})\times P(\mathbb{R})$ function
$x_{-i}\mapsto({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})$ is Borel measurable because,
by disintegration, both $x_{-i}\mapsto{\mu}_{|x_{-i}}$ and
$x_{-i}\mapsto{\nu}_{|x_{-i}}$ are Borel measurable. By Lemma 6.30, we obtain
a Borel measurable map $x_{-i}\mapsto\gamma_{x_{-i}}$ such that, for each
$x_{-i}\in\mathbb{R}^{d}_{-i}$,
$\gamma_{x_{-i}}\in\Pi({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})$ and
$C_{p}(\gamma_{x_{-i}})^{p}=W_{p}^{p}({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})$.

Define $\gamma\in P(\mathbb{R}^{d}\times\mathbb{R}^{d})$ by its disintegration
$\gamma=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}{\gamma}_{|x_{-i},y_{-i}}\odif{{\gamma}_{-i_{x}-i_{y}}}$
as follows. First, for the marginal ${\gamma}_{-i_{x}-i_{y}}\in
P(\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i})$, for each Borel
$Z\subset\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}$ we set

${\gamma}_{-i_{x}-i_{y}}(Z)\vcentcolon=\int_{\mathbb{R}^{d}_{-i}}\chi_{Z}(x_{-i},x_{-i})\odif{{\mu}_{-i}(x_{-i})}\,,$

where the integral is well-defined because the set
$\\{x_{-i}\in\mathbb{R}^{d}_{-i}:(x_{-i},x_{-i})\in Z\\}$ is Borel by standard
arguments. We remark that ${\gamma}_{-i_{x}-i_{y}}$ is indeed a probability
measure because it is nonnegative, it assigns zero to the empty set, it is
additive over countable disjoint unions, and

${\gamma}_{-i_{x}-i_{y}}(\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i})=\int_{\mathbb{R}^{d}_{-i}}\odif{{\mu}_{-i}}=1\,.$

It follows that for every bounded (or nonnegative) Borel function
$f:\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}\to\mathbb{R}$, we have

$\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}f(x_{-i},y_{-i})\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}=\int_{\mathbb{R}^{d}_{-i}}f(x_{-i},x_{-i})\odif{{\mu}_{-i}(x_{-i})}\,.$
(26)

Second, for the conditionals, we simply set
${\gamma}_{|x_{-i},y_{-i}}\vcentcolon=\gamma_{x_{-i}}$ for each
$(x_{-i},y_{-i})\in\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}$.

We claim that $\gamma\in\Pi_{i}(\mu,\nu)$. The $i$-aligned condition holds by
construction, so it remains to show that $\gamma\in\Pi(\mu,\nu)$. We first
verify that ${(\pi_{1})}_{\\#}{\gamma}=\mu$. It suffices to verify that these
measures agree on each rectangle $X\vcentcolon=X_{-i}\times X_{i}$ of Borel
sets $X_{-i}\subset\mathbb{R}^{d}_{-i}$, $X_{i}\subset\mathbb{R}$. We have

$\displaystyle({(\pi_{1})}_{\\#}{\gamma})(X)$
$\displaystyle\quad=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{X}(x)\odif\gamma(x,y)$
(Pushforward)
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}\chi_{X}(x_{-i},x_{i})\odif{{{\gamma}_{|x_{-i},y_{-i}}(x_{i},y_{i})}}$
(Disintegration)
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\chi_{X_{-i}}(x_{-i})\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}\chi_{X_{i}}(x_{i})\odif{{\gamma_{x_{-i}}(x_{i},y_{i})}}$
(Definition of $X,{\gamma}_{|x_{-i},y_{-i}}$)
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\chi_{X_{-i}}(x_{-i})({(\pi_{1})}_{\\#}{\gamma_{x_{-i}}})(X_{i})\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}$
(Pushforward)
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\chi_{X_{-i}}(x_{-i}){\mu}_{|x_{-i}}(X_{i})\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}$
($\gamma_{x_{-i}}\in\Pi({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})$)
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}}\chi_{X_{-i}}(x_{-i}){\mu}_{|x_{-i}}(X_{i})\odif{{\mu}_{-i}(x_{-i})}$
(Application of (26))
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}}\chi_{X_{-i}}(x_{-i})\odif{{\mu}_{-i}(x_{-i})}\int_{\mathbb{R}}\chi_{X_{i}}(x_{i})\odif{{{\mu}_{|x_{-i}}(x_{i})}}$
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}}\odif{{\mu}_{-i}(x_{-i})}\int_{\mathbb{R}}\chi_{X}(x_{-i},x_{i})\odif{{{\mu}_{|x_{-i}}(x_{i})}}$
(Definition of $X$)
$\displaystyle\quad=\int_{\mathbb{R}^{d}}\chi_{X}\odif\mu=\mu(X)$
$\displaystyle\text{(Disintegration)}\,,$

as desired. Thanks to the hypothesis that ${\mu}_{-i}={\nu}_{-i}$, an
analogous calculation on the second variable yields that
${(\pi_{2})}_{\\#}{\gamma}=\nu$ and hence $\gamma\in\Pi(\mu,\nu)$. Thus
$\gamma\in\Pi_{i}(\mu,\nu)$ as claimed.

Finally, we compute the cost of the plan $\gamma$. Again using its
disintegration and (26), we have

$\displaystyle C_{p}(\gamma)^{p}$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-y|^{p}\odif\gamma(x,y)$
$\displaystyle=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}|(x_{-i},x_{i})-(y_{-i},y_{i})|^{p}\odif{{{\gamma}_{|x_{-i},y_{-i}}(x_{i},y_{i})}}$
$\displaystyle=\int_{\mathbb{R}^{d}_{-i}}\odif{{{\mu}_{-i}(x_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}|(x_{-i},x_{i})-(x_{-i},y_{i})|^{p}\odif{{{\gamma}_{|x_{-i},x_{-i}}(x_{i},y_{i})}}$
$\displaystyle=\int_{\mathbb{R}^{d}_{-i}}\odif{{{\mu}_{-i}(x_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}|x_{i}-y_{i}|^{p}\odif{{\gamma_{x_{-i}}(x_{i},y_{i})}}$
$\displaystyle=\int_{\mathbb{R}^{d}_{-i}}\odif{{{\mu}_{-i}(x_{-i})}}C_{p}(\gamma_{x_{-i}})^{p}=\int_{\mathbb{R}^{d}_{-i}}W_{p}^{p}({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})\odif{{{\mu}_{-i}(x_{-i})}}\,.\qed$

We may combine the above to find _directed_ transport plans by transporting
mass within axis-aligned lines, as long as the first probability distribution
dominates the second in each such line.

######  Lemma 6.32.

Let $p\in(1,+\infty)$, and let $\mu,\nu\in P([0,1]^{d})$ be absolutely
continuous probability measures with strictly positive densities. Let
$i\in[d]$ and suppose that 1) ${\mu}_{-i}={\nu}_{-i}$ and 2)
${\mu}_{|x_{-i}}\succeq{\nu}_{|x_{-i}}$ for each $x_{-i}\in[0,1]^{d}_{-i}$.
Then there exists $\gamma\in\Pi_{i}(\mu\to\nu)$ satisfying

$C_{p}(\gamma)^{p}=\int_{[0,1]^{d}_{-i}}W_{p}^{p}({\mu}_{|x_{-i}},{\nu}_{|x_{-i}})\odif{{\mu}_{-i}(x_{-i})}\,.$
(27)

###### Proof.

Let $\gamma\in\Pi_{i}(\mu,\nu)$ be the plan obtained from Lemma 6.31, which
satisfies (27). After recalling the definition of $\gamma$ in Lemma 6.31,
Corollary 6.23 implies that
${\gamma}_{|x_{-i},y_{-i}}\in\Pi({\mu}_{|x_{-i}}\to{\nu}_{|x_{-i}})$ for each
$(x_{-i},y_{-i})\in[0,1]^{d}_{-i}\times[0,1]^{d}_{-i}$. It remains to confirm
that $\gamma\in\Pi(\mu\to\nu)$. Indeed,

$\displaystyle\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x\not\preceq
y\\}}\odif\gamma(x,y)$
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}\chi_{\\{x\not\preceq
y\\}}((x_{-i},x_{i}),(y_{-i},y_{i}))\odif{{{\gamma}_{|x_{-i},y_{-i}}(x_{i},y_{i})}}$
(Disintegration)
$\displaystyle\quad\leq\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}\int_{\mathbb{R}\times\mathbb{R}}\chi_{\\{x_{-i}\not\preceq
y_{-i}\\}}(x_{-i},y_{-i})\odif{{{\gamma}_{|x_{-i},y_{-i}}(x_{i},y_{i})}}$
$\displaystyle\quad\qquad+\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\odif{{{\gamma}_{-i_{x}-i_{y}}(x_{-i},y_{-i})}}\underbrace{\int_{\mathbb{R}\times\mathbb{R}}\chi_{\\{x_{i}>y_{i}\\}}(x_{i},y_{i})\odif{{{\gamma}_{|x_{-i},y_{-i}}(x_{i},y_{i})}}}_{\text{$=0$
since ${\gamma}_{|x_{-i},y_{-i}}$ is a directed plan}}$
$\displaystyle\quad=\int_{\mathbb{R}^{d}_{-i}\times\mathbb{R}^{d}_{-i}}\chi_{\\{x_{-i}\not\preceq
y_{-i}\\}}\odif{{{\gamma}_{-i_{x}-i_{y}}}}\underbrace{\int_{\mathbb{R}\times\mathbb{R}}\odif{{{\gamma}_{|x_{-i},y_{-i}}}}}_{=1}=\int_{\mathbb{R}^{d}_{-i}}\underbrace{\chi_{\\{x_{-i}\not\preceq
x_{-i}\\}}}_{=0}\odif{{\mu}_{-i}(x_{-i})}$ (By (26)) $\displaystyle\quad=0\,.$

Now, we show that the directed, aligned $2$-Wasserstein distance enjoys a nice
“Pythagorean” composition property: composing $I$-aligned and $J$-aligned
directed transport plans, for $I$ and $J$ disjoint, yields an $I\cup
J$-aligned directed transport plan whose cost is obtained from the costs of
the two other plans via the Pythagorean theorem. Arguments of this nature are
well-known in the undirected case, e.g. a similar statement appears in [Vil09,
p. 572].

###### Proof.

Using Lemma 6.19, we obtain $\sigma\in
P(\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d})$ such that
${(\pi_{1,2})}_{\\#}{\sigma}=\gamma^{+}$ and
${(\pi_{2,3})}_{\\#}{\sigma}=\gamma^{-}$, and by Lemma 6.20 and Lemma 6.29, we
have $\gamma\vcentcolon={(\pi_{1,3})}_{\\#}{\sigma}\in\Pi_{I\cup
J}(\mu\to\nu)$. To compute the cost of $\gamma$, we first claim that $\sigma$
assigns zero measure to points $x,y,z$ such that $x_{-I}\neq y_{-I}$ or
$y_{-J}\neq z_{-J}$. Indeed,

$\displaystyle 0$
$\displaystyle\leq\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}(1-\chi_{\\{x_{-I}=y_{-I}\text{
and }y_{-J}=z_{-J}\\}})\odif\sigma(x,y,z)$
$\displaystyle\leq\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif\sigma(x,y,z)+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{y_{-J}\neq
z_{-J}\\}}\odif\sigma(x,y,z)$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}\neq
y_{-I}\\}}\odif\gamma^{+}(x,y)+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{y_{-J}\neq
z_{-J}\\}}\odif\gamma^{-}(y,z)=0\,,$

the last step by Lemma 6.28 since $\gamma^{+}$ and $\gamma^{-}$ are $I$\- and
$J$-aligned, respectively. Hence $\int f\odif\sigma=\int
f\chi_{\\{x_{-I}=y_{-I}\text{ and }y_{-J}=z_{-J}\\}}\odif\sigma$ for any
bounded or nonnegative Borel function $f$. Therefore, using the assumption
that $I$ and $J$ are disjoint to apply the Pythagorean theorem, we obtain

$\displaystyle C_{2}(\gamma)^{2}$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-z|^{2}\odif\gamma(x,z)=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-z|^{2}\odif\sigma(x,y,z)$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}=y_{-I}\text{
and }y_{-J}=z_{-J}\\}}|x-z|^{2}\odif\sigma(x,y,z)$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\chi_{\\{x_{-I}=y_{-I}\text{
and }y_{-J}=z_{-J}\\}}\left(|x-y|^{2}+|y-z|^{2}\right)\odif\sigma(x,y,z)$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}\left(|x-y|^{2}+|y-z|^{2}\right)\odif\sigma(x,y,z)$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-y|^{2}\odif\sigma(x,y,z)+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}\times\mathbb{R}^{d}}|y-z|^{2}\odif\sigma(x,y,z)$
$\displaystyle=\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|x-y|^{2}\odif\gamma^{+}(x,y)+\int_{\mathbb{R}^{d}\times\mathbb{R}^{d}}|y-z|^{2}\odif\gamma^{-}(y,z)$
$\displaystyle=C_{2}(\gamma^{+})^{2}+C_{2}(\gamma^{-})^{2}\,.\qed$

######  Lemma 6.33 (Induction over coordinates).

Let $(\mu^{(i)})_{i\in[d+1]}$ be a family of absolutely continuous probability
measures on $[0,1]^{d}$ with strictly positive densities. Suppose that, for
each $i\in[d]$, we have 1) ${\mu^{(i)}}_{-i}={\mu^{(i+1)}}_{-i}$ and 2)
${\mu^{(i)}}_{|x_{-i}}\succeq{\mu^{(i+1)}}_{|x_{-i}}$ for each
$x_{-i}\in[0,1]^{d}_{-i}$. Then

$W_{2}^{2}(\mu^{(1)}\to\mu^{(d+1)})\leq\sum_{i=1}^{d}\int_{[0,1]^{d}_{-i}}W_{2}^{2}({\mu^{(i)}}_{|x_{-i}},{\mu^{(i+1)}}_{|x_{-i}})\odif{{\mu^{(i)}}_{-i}(x_{-i})}\,.$

###### Proof.

For each $k\in[d+1]$, let $A(k)$ be the following proposition: there exists
$\gamma\in\Pi_{[k]}(\mu^{(1)}\to\mu^{(k+1)})$ satisfying

$C_{2}(\gamma)^{2}=\sum_{i=1}^{k}\int_{[0,1]^{d}_{-i}}W_{2}^{2}({\mu^{(i)}}_{|x_{-i}},{\mu^{(i+1)}}_{|x_{-i}})\odif{{\mu^{(i)}}_{-i}(x_{-i})}\,.$

Note that $A(d)$ implies the result we want to prove, while $A(1)$ holds by
Lemma 6.32. Now let $2\leq k\leq d$. Suppose $A(k-1)$ holds, and thus let
$\gamma^{+}\in\Pi_{[k-1]}(\mu^{(1)}\to\mu^{(k)})$ satisfy

$C_{2}(\gamma^{+})^{2}=\sum_{i=1}^{k-1}\int_{[0,1]^{d}_{-i}}W_{2}^{2}({\mu^{(i)}}_{|x_{-i}},{\mu^{(i+1)}}_{|x_{-i}})\odif{{\mu^{(i)}}_{-i}(x_{-i})}\,.$

Apply Lemma 6.32 with $i=k$ to obtain a plan
$\gamma^{-}\in\Pi_{k}(\mu^{(k)}\to\mu^{(k+1)})$ satisfying

$C_{2}(\gamma^{-})^{2}=\int_{[0,1]^{d}_{-k}}W_{2}^{2}({\mu^{(k)}}_{|x_{-k}},{\mu^{(k+1)}}_{|x_{-k}})\odif{{\mu^{(k)}}_{-k}(x_{-k})}\,.$

Then, use Section 1.2.4 with $I=[k-1]$ and $J=\\{k\\}$ to obtain
$\gamma\in\Pi_{[k]}(\mu^{(1)}\to\mu^{(k+1)})$ satisfying

$C_{2}(\gamma)^{2}=C_{2}(\gamma^{+})^{2}+C_{2}(\gamma^{-})^{2}=\sum_{i=1}^{k}\int_{[0,1]^{d}_{-i}}W_{2}^{2}({\mu^{(i)}}_{|x_{-i}},{\mu^{(i+1)}}_{|x_{-i}})\odif{{\mu^{(i)}}_{-i}(x_{-i})}\,,$

which implies that $A(k)$ holds. Thus $A(d)$ follows by induction. ∎

###  6.4 From one-dimensional PDE to optimal transport in the cube

We now tie most of the foregoing theory together by showing how to inductively
transform a function on the unit cube into a monotone function, one coordinate
at a time via the directed heat semigroup machinery, keeping track of the cost
using the directed optimal transport machinery.

For this part of the proof, it is convenient to restrict our attention to
Lipschitz functions. As we will see, the Lipschitz property of
$f:[0,1]^{d}\to\mathbb{R}$ is preserved under taking the monotone equilibrium
along each axis-aligned direction, which enables an inductive argument over
the coordinates.

Denote the set of Lipschitz functions $f:[0,1]^{d}\to\mathbb{R}$ by
$\operatorname{\mathsf{Lip}}$ (the dimension $d$ will always be clear from
context).

######  Observation 6.34.

A function $f:[0,1]^{d}\to\mathbb{R}$ is Lipschitz if and only if its
restriction to every axis-aligned line is uniformly Lipschitz, i.e. there
exists some $M>0$ such that, for every $i\in[d]$ and
$x_{-i}\in[0,1]^{d}_{-i}$, $f(x_{-i},\cdot):[0,1]\to\mathbb{R}$ is
$M$-Lipschitz.

######  Definition 6.35 ($i$\- and $I$-monotonicity).

Let $i\in[d]$. We say $f:[0,1]^{d}\to\mathbb{R}$ is _$i$ -monotone_ if its
every restriction along direction $i$ is nondecreasing, i.e. if for every
$x_{-i}\in[0,1]^{d}_{-i}$, $f(x_{-i},\cdot)$ is nondecreasing. For
$I\subseteq[d]$, we say $f$ is _$I$ -monotone_ if it is $i$-monotone for every
$i\in I$. We write $\operatorname{\mathsf{Mon}}_{I}$ for the set of
$I$-monotone functions.

######  Observation 6.36.

$\operatorname{\mathsf{Mon}}_{[d]}$ is simply the set of monotone functions on
$[0,1]^{d}$.

The following useful characterization of $i$-monotone functions in terms of
$k$-aligned lines follows by definition:

######  Observation 6.37.

Let $i,k\in[d]$ be distinct. Then for all $f:[0,1]^{d}\to\mathbb{R}$, $f$ is
$i$-monotone if and only if the following holds: for every
$z_{-i-k}\in[0,1]^{d}_{-i-k}$, and for all $x_{i}\leq y_{i}$ in $[0,1]$,
$f(z_{-i-k},x_{i},\cdot)\leq f(z_{-i-k},y_{i},\cdot)$.

######  Definition 6.38 (Monotone equilibrium operator).

Let $k\in[d]$. We define the operator
$\mathcal{M}_{k}:\operatorname{\mathsf{Lip}}\to([0,1]^{d}\to\mathbb{R})$ as
follows: for each function $f\in\operatorname{\mathsf{Lip}}$,
$x_{-k}\in[0,1]^{d}_{-k}$ and $x_{k}\in[0,1]$, the function
$\mathcal{M}_{k}f:[0,1]^{d}\to\mathbb{R}$ satisfies

$(\mathcal{M}_{k}f)(x_{-k},x_{k})\vcentcolon=(P_{\infty}f(x_{-k},\cdot))(x_{k})\,,$

where the application of $P_{\infty}$ is well-defined because each
$f(x_{-k},\cdot)$ is Lipschitz and hence in $L^{2}(I)$.

The following lemma shows that $\mathcal{M}_{k}$ preserves Lipschitzness,
confirming that $\mathcal{M}_{k}f$ is well-defined pointwise as a real-valued
function, as opposed to only defined up to sets of measure zero.

######  Lemma 6.39.

Let $k\in[d]$. Let $f\in\operatorname{\mathsf{Lip}}$. Then
$\mathcal{M}_{k}f\in\operatorname{\mathsf{Lip}}$.

###### Proof.

Let $M>0$ be such that every axis-aligned line restriction of $f$ is
$M$-Lipschitz. We claim that every axis-aligned line restriction of
$\mathcal{M}_{k}f$ is $M$-Lipschitz. We first check the $k$-aligned lines. For
any $x_{-k}\in[0,1]^{d}_{-k}$ Proposition 5.72 implies that

$\psi((\mathcal{M}_{k}f)(x_{-k},\cdot))=\psi(P_{\infty}f(x_{-k},\cdot))\leq\psi(f(x_{-k},\cdot))\leq
M\,,$

the last inequality and the conclusion that $(\mathcal{M}_{k}f)(x_{-k},\cdot)$
is $M$-Lipschitz by Fact 5.37.

It remains to check the other line restrictions. Let $i\in[d]\setminus\\{k\\}$
and let $z_{-i-k}\in[0,1]^{d}_{-i-k}$. Let $x_{i}\neq y_{i}\in[0,1]$. Since
the axis-aligned line restrictions of $f$ (in particular, in direction $i$)
are $M$-Lipschitz, we have

$\sup_{w_{k}\in[0,1]}\left\lvert
f(z_{-i-k},x_{i},w_{k})-f(z_{-i-k},y_{i},w_{k})\right\rvert\leq
M|x_{i}-y_{i}|\,.$

Proposition 5.74 implies that

$\left\lvert
P_{\infty}f(z_{-i-k},x_{i},\cdot)-P_{\infty}f(z_{-i-k},y_{i},\cdot)\right\rvert\leq\operatorname*{ess\,sup}_{w_{k}\in[0,1]}\left\lvert
f(z_{-i-k},x_{i},w_{k})-f(z_{-i-k},y_{i},w_{k})\right\rvert\leq
M|x_{i}-y_{i}|$ (28)

a.e. in $I$, and since $P_{\infty}f(z_{-i-k},x_{i},\cdot)$ and
$P_{\infty}f(z_{-i-k},y_{i},\cdot)$ are (Lipschitz) continuous as observed
above, we conclude that (28) holds pointwise. Since (28) holds for every
$i\in[d]\setminus\\{k\\}$ and $z_{-i-k}\in[0,1]^{d}_{-i-k}$, and since

$\left\lvert(\mathcal{M}_{k}f)(z_{-i-k},x_{i},\cdot)-(\mathcal{M}_{k}f)(z_{-i-k},y_{i},\cdot)\right\rvert=\left\lvert
P_{\infty}f(z_{-i-k},x_{i},\cdot)-P_{\infty}f(z_{-i-k},y_{i},\cdot)\right\rvert$

pointwise, we get that every axis-aligned line restriction of
$\mathcal{M}_{k}f$ is $M$-Lipschitz, and
$\mathcal{M}_{k}f\in\operatorname{\mathsf{Lip}}$. ∎

######  Lemma 6.40 ($\mathcal{M}_{k}$ application makes progress).

Let $k\in[d]$, and let $I\subseteq[d]\setminus\\{k\\}$. Let
$f\in\operatorname{\mathsf{Lip}}\cap\operatorname{\mathsf{Mon}}_{I}$. Then
$\mathcal{M}_{k}f\in\operatorname{\mathsf{Lip}}\cap\operatorname{\mathsf{Mon}}_{I\cup\\{k\\}}$.

###### Proof.

The fact that $\mathcal{M}_{k}f\in\operatorname{\mathsf{Lip}}$ is given by
Lemma 6.39, and the fact that $\mathcal{M}_{k}f$ is $k$-monotone follows from
the definition of $\mathcal{M}_{k}$ and the fact that the monotone equilibrium
$P_{\infty}f(x_{-k},\cdot)$ is nondecreasing. Now, let $i\in I$, so that $f$
is $i$-monotone. By 6.37, we have that for every
$z_{-i-k}\in[0,1]^{d}_{-i-k}$, and for all $x_{i}\leq y_{i}$ in $[0,1]$,
$f(z_{-i-k},x_{i},\cdot)\leq f(z_{-i-k},y_{i},\cdot)$. By Corollary 5.68, we
obtain that $P_{\infty}f(z_{-i-k},x_{i},\cdot)\leq
P_{\infty}f(z_{-i-k},y_{i},\cdot)$ a.e. in $I$, and since these two functions
are (Lipschitz) continuous, this inequality holds pointwise. Thus
$(\mathcal{M}_{k}f)(z_{-i-k},x_{i},\cdot)\leq(\mathcal{M}_{k}f)(z_{-i-k},y_{i},\cdot)$,
and again by 6.37, $\mathcal{M}_{k}f$ is $i$-monotone. This concludes the
proof. ∎

######  Lemma 6.41 ($\mathcal{M}_{k}$ preserves bounds).

Let $f\in\operatorname{\mathsf{Lip}}$. Let $a\leq b$ be real numbers and
suppose $a\leq f\leq b$. Then $a\leq\mathcal{M}_{k}f\leq b$.

###### Proof.

This follows by the definition of $\mathcal{M}_{k}$ together with Corollaries
5.68 and 5.70. ∎

######  Lemma 6.42 ($\mathcal{M}_{k}$ is nonexpansive in $L^{2}$).

Let $f,g\in\operatorname{\mathsf{Lip}}$. Then

$\int_{[0,1]^{d}}(\mathcal{M}_{k}f-\mathcal{M}_{k}g)^{2}\odif
x\leq\int_{[0,1]^{d}}(f-g)^{2}\odif x\,.$

###### Proof.

This follows from Proposition 5.69 and Tonelli’s theorem, as follows:

$\displaystyle\int_{[0,1]^{d}}(\mathcal{M}_{k}f-\mathcal{M}_{k}g)^{2}\odif x$
$\displaystyle=\int_{[0,1]^{d}_{-k}}\odif
x_{-k}\int_{I}\left[(\mathcal{M}_{k}f)(x_{-k},x_{k})-(\mathcal{M}_{k}g)(x_{-k},x_{k})\right]^{2}\odif
x_{k}$ $\displaystyle=\int_{[0,1]^{d}_{-k}}\odif
x_{-k}\int_{I}\left[(P_{\infty}f(x_{-k},\cdot))(x_{k})-(P_{\infty}g(x_{-k},\cdot))(x_{k})\right]^{2}\odif
x_{k}$ $\displaystyle\leq\int_{[0,1]^{d}_{-k}}\odif
x_{-k}\int_{I}\left[f(x_{-k},x_{k})-g(x_{-k},x_{k})\right]^{2}\odif
x_{k}=\int_{[0,1]^{d}}(f-g)^{2}\odif x\,.$

Similarly, combining Corollary 6.10 with Fubini’s theorem yields

######  Lemma 6.43.

Let $f\in\operatorname{\mathsf{Lip}}$. Then

$\int_{[0,1]^{d}}(\mathcal{M}_{k}f)\odif x=\int_{[0,1]^{d}}f\odif x\,.$

The following result is essentially an $L^{2}$ version of [Fer23, Proposition
3.14].

######  Lemma 6.44 (Effect of $\mathcal{M}_{k}$ on the directed Dirichlet
energy).

Let $i,k\in[d]$ be distinct and let $f\in\operatorname{\mathsf{Lip}}$. Then

$\int_{[0,1]^{d}_{-i}}\mathcal{E}^{-}((\mathcal{M}_{k}f)(x_{-i},\cdot))\odif
x_{-i}\leq\int_{[0,1]^{d}_{-i}}\mathcal{E}^{-}(f(x_{-i},\cdot))\odif
x_{-i}\,.$

###### Proof.

For any $g\in\operatorname{\mathsf{Lip}}$, $i\in[d]$ and
$h\in\mathbb{R}\setminus\\{0\\}$, define the
$Q_{g,i,h}:[0,1]^{d}\to\mathbb{R}$ by

$Q_{g,i,h}(x)\vcentcolon=\begin{cases}\frac{g(x_{-i},x_{i}+h)-g(x_{-i},x_{i})}{h}&\text{if
$x_{i}+h\in(0,1)$}\\\ 0&\text{otherwise.}\end{cases}$ (29)

By Rademacher’s theorem on the open domain $(0,1)^{d}$, $g$ is differentiable
almost everywhere. In particular, the function
$D_{g,i}:[0,1]^{d}\to\mathbb{R}$ given by

$D_{g,i}(x)\vcentcolon=\begin{cases}\lim_{h\to 0}Q_{g,i,h}(x)&\text{if the
limit exists}\\\ 0&\text{otherwise}\end{cases}$

gives the partial derivative of $g$ in direction $i$ for a.e. $x\in[0,1]^{d}$,
and is measurable as the limit of the measurable functions $Q_{g,i,h}$.
Moreover, letting $M>0$ be such that the axis-aligned line restrictions of $g$
are $M$-Lipschitz, we have $\left\lvert Q_{g,i,h}\right\rvert\leq M$
pointwise. Finally, fixing any $x_{-i}\in[0,1]^{d}_{-i}$, $g(x_{-i},\cdot)$ is
(Lipschitz and hence) in $H^{1}(I)$, and the partial derivative
$D_{g,i}(x_{-i},\cdot)$ agrees with any weak derivative
$\partial_{x_{i}}g(x_{-i},\cdot)$ a.e. in $[0,1]$. In particular, all of these
considerations apply to the functions $f$ and $\mathcal{M}_{k}f$ by Lemma
6.39. We conclude that, for $g=f$ or $g=\mathcal{M}_{k}f$ and each
$x_{-i}\in[0,1]^{d}_{-i}$,

$\displaystyle\mathcal{E}^{-}(g(x_{-i},\cdot))$
$\displaystyle=\frac{1}{2}\int_{I}\left[\partial_{x_{i}}^{-}g(x_{-i},x_{i})\right]^{2}\odif
x_{i}=\frac{1}{2}\int_{I}\left[D_{g,i}(x_{-i},x_{i})^{-}\right]^{2}\odif
x_{i}$ $\displaystyle=\frac{1}{2}\int_{I}\lim_{h\to
0}\left[Q_{g,i,h}(x_{-i},x_{i})^{-}\right]^{2}\odif x_{i}\,,$

where for simplicity we write the limit in the last expression with the
understanding that it is only defined almost everywhere in $[0,1]$. Note that
the measurability of the last integrand over $[0,1]^{d}$ justifies writing the
integrals in the statement of the lemma via Tonelli’s theorem (as the
computation below shows). Define for each $h\in\mathbb{R}$ the set

$I_{h}\vcentcolon=\begin{cases}(0,1-h)&\text{if }h\geq 0\\\
(-h,1)&\text{otherwise,}\end{cases}$

so that $Q_{g,i,h}(x)$ is defined by the first case of (29) if and only if
$x_{i}\in I_{h}$. Then, by repeated applications of Tonelli’s theorem and the
dominated convergence theorem, we obtain

$\displaystyle\int_{[0,1]^{d}_{-i}}\mathcal{E}^{-}((\mathcal{M}_{k}f)(x_{-i},\cdot))\odif
x_{-i}=\frac{1}{2}\int_{[0,1]^{d}_{-i}}\odif x_{-i}\int_{I}\lim_{h\to
0}\left[Q_{\mathcal{M}_{k}f,i,h}(x_{-i},x_{i})^{-}\right]^{2}\odif x_{i}$
$\displaystyle\,=\frac{1}{2}\int_{[0,1]^{d}}\lim_{h\to
0}\underbrace{\left[Q_{\mathcal{M}_{k}f,i,h}(x)^{-}\right]^{2}}_{\leq
M^{2}}\odif x=\frac{1}{2}\lim_{h\to
0}\int_{[0,1]^{d}}\left[Q_{\mathcal{M}_{k}f,i,h}(x)^{-}\right]^{2}\odif x$
$\displaystyle\,=\frac{1}{2}\lim_{h\to 0}\int_{[0,1]^{d}_{-i-k}}\odif
x_{-i-k}\int_{I}\odif
x_{i}\int_{I}\left[Q_{\mathcal{M}_{k}f,i,h}(x_{-i-k},x_{i},x_{k})^{-}\right]^{2}\odif
x_{k}$ $\displaystyle\,=\frac{1}{2}\lim_{h\to 0}\int_{[0,1]^{d}_{-i-k}}\odif
x_{-i-k}\int_{I_{h}}\odif
x_{i}\int_{I}\left[\left(\frac{(\mathcal{M}_{k}f)(x_{-i-k},x_{i}+h,x_{k})-(\mathcal{M}_{k}f)(x_{-i-k},x_{i},x_{k})}{h}\right)^{-}\right]^{2}\odif
x_{k}$ $\displaystyle\,=\frac{1}{2}\lim_{h\to 0}\int_{[0,1]^{d}_{-i-k}}\odif
x_{-i-k}\int_{I_{h}}\odif
x_{i}\int_{I}\left[\left(\frac{(P_{\infty}f(x_{-i-k},x_{i}+h,\cdot))(x_{k})-(P_{\infty}f(x_{-i-k},x_{i},\cdot))(x_{k})}{h}\right)^{-}\right]^{2}\odif
x_{k}$ $\displaystyle\,\leq\frac{1}{2}\lim_{h\to
0}\int_{[0,1]^{d}_{-i-k}}\odif x_{-i-k}\int_{I_{h}}\odif
x_{i}\int_{I}\left[\left(\frac{f(x_{-i-k},x_{i}+h,x_{k})-f(x_{-i-k},x_{i},x_{k})}{h}\right)^{-}\right]^{2}\odif
x_{k}$ $\displaystyle\,=\frac{1}{2}\lim_{h\to 0}\int_{[0,1]^{d}_{-i-k}}\odif
x_{-i-k}\int_{I}\odif
x_{i}\int_{I}\left[Q_{f,i,h}(x_{-i-k},x_{i},x_{k})^{-}\right]^{2}\odif x_{k}$
$\displaystyle\,=\frac{1}{2}\lim_{h\to
0}\int_{[0,1]^{d}}\underbrace{\left[Q_{f,i,h}(x)^{-}\right]^{2}}_{\leq
M^{2}}\odif x\,=\frac{1}{2}\int_{[0,1]^{d}}\lim_{h\to
0}\left[Q_{f,i,h}(x)^{-}\right]^{2}\odif x$
$\displaystyle\,=\frac{1}{2}\int_{[0,1]^{d}_{-i}}\odif
x_{-i}\int_{I}\lim_{h\to 0}\left[Q_{f,i,h}(x_{-i},x_{i})^{-}\right]^{2}\odif
x_{i}=\int_{[0,1]^{d}_{-i}}\mathcal{E}^{-}(f(x_{-i},\cdot))\odif x_{-i}\,,$

the inequality by Proposition 5.67 via the identity
$\left[(a-b)^{+}\right]^{2}=\left[(b-a)^{-}\right]^{2}$. ∎

######  Definition 6.45 (Coordinate-wise monotone equilibrium).

For each $f\in\operatorname{\mathsf{Lip}}$, define
$f^{*}\in\operatorname{\mathsf{Lip}}$, the _coordinate-wise monotone
equilibrium_ of $f$, by

$f^{*}\vcentcolon=\mathcal{M}_{d}\mathcal{M}_{d-1}\dotsm\mathcal{M}_{1}f\,.$

Note that indeed $f^{*}$ is indeed Lipschitz by Lemma 6.39, and it is
monotone:

######  Proposition 6.46.

Let $f\in\operatorname{\mathsf{Lip}}$. Then $f^{*}$ is monotone.

###### Proof.

This follows by repeatedly applying Lemma 6.40, and recalling that
$\operatorname{\mathsf{Mon}}_{[d]}$ is the set of monotone functions on
$[0,1]^{d}$. ∎

By Proposition 5.73 and the definition of $\mathcal{M}_{k}$, we also observe
that the coordinate-wise monotone equilibrium behaves nicely with respect to
certain affine transformations:

######  Observation 6.47.

Let $f\in\operatorname{\mathsf{Lip}}$. Let $\alpha>0$ and
$\beta\in\mathbb{R}$. Then $(\alpha f+\beta)^{*}=\alpha f^{*}+\beta$.

Similarly, Lemma 6.42 implies that taking the coordinate-wise monotone
equilibrium is a nonexpansive operation:

######  Observation 6.48.

Let $f,g\in\operatorname{\mathsf{Lip}}$. Then
$\|f^{*}-g^{*}\|_{L^{2}((0,1)^{d})}\leq\|f-g\|_{L^{2}((0,1)^{d})}$. Since
$0^{*}=0$ by Lemma 5.70 (where we write $0$ for the constant zero function),
we conclude in particular that
$\|f^{*}\|_{L^{2}((0,1)^{d})}\leq\|f\|_{L^{2}((0,1)^{d})}$.

We are now ready to combine the ingredients from the previous sections and
establish the transport-energy inequality. Note that this inequality is most
effective when the function $f$ is pointwise bounded close to $1$.

###### Proof.

Define the family $(f^{(i)})_{i\in[d+1]}\subset\operatorname{\mathsf{Lip}}$ by
$f^{(1)}\vcentcolon=f$ and $f^{(i+1)}\vcentcolon=\mathcal{M}_{i}f^{(i)}$ for
each $i\in[d]$, so that $f^{(d+1)}=f^{*}$. Note that we have $1-a\leq
f^{(i)}\leq 1+a$ and $\int_{[0,1]^{d}}f^{(i)}\odif x=1$ for each $i\in[d+1]$
by Lemmas 6.41 and 6.43. Then, define the family $(\mu^{(i)})_{i\in[d+1]}$ of
probability measures on $[0,1]^{d}$ by $\odif\mu^{(i)}\vcentcolon=f^{(i)}\odif
x$, so that $\mu^{(1)}=\mu$ and $\mu^{(d+1)}=\mu^{*}$.

We claim $(\mu^{(i)})_{i\in[d+1]}$ satisfies the conditions of Lemma 6.33.
First, they have strictly positive densities because $1-a\leq f^{(i)}\leq 1+a$
for each $i\in[d+1]$ as observed above. They are also absolutely continuous
with Borel measurable density because each $f^{(i)}$ is continuous and hence
Borel.

Recall that, by Proposition 6.25, for each $i,j\in[d+1]$ we may characterize
the disintegration
$\mu^{(j)}=\int_{\mathbb{R}^{d}_{-i}}{\mu^{(j)}}_{|x_{-i}}\odif{{\mu^{(j)}}_{-i}}$
as follows: the marginal and all conditional measures are both absolutely
continuous supported in $[0,1]^{d}_{-i}$ and $[0,1]$ respectively, and we have
$\odif{{\mu^{(j)}}_{-i}}={f^{(j)}}_{-i}\odif x_{-i}$ where

${f^{(j)}}_{-i}(x_{-i})=\int_{I}f^{(j)}(x_{-i},x_{i})\odif x_{i}$ (30)