For Wasserstein space $W_p := (\mathcal P_p(X), W_p)$, the standard considerations from fluid mechanics tell us that the density $\mu_t$ of a family of particles may be interpreted as the continuity
$$
\partial_t \mu_t + \nabla \cdot (\mu_t v_t) = 0
$$
with $L^p$ vector $v_t$. Moreover, we want to connect the $L^p$ norm of $v_t$ with the metric derivative $|\mu’|(t)$.

Theorem 1. Let $(\mu_t)_{t\in[0,1]}$ be an absolutely continuous curve in $W_p(\Omega)$ (for $p>1$, and $\Omega\subset \mathbb R^d$ an open domain). Then for a.e. $t\in[0,1]$, there exists a vector field $v_t\in L^p(\mu_t;\mathbb R^d)$ such that

• the continuity equation $\partial_t \mu_t + \nabla\cdot(\mu_t v_t)=0$ is satisfied in the sense of distributions,
• for a.e. $t$, we have
  $$
  \Vert v_t\Vert_{L^p(\mu_t)} \le |\mu’|(t),
  $$
  where $|\mu’|(t)$ denotes the metric derivative at the time $t$ of the curve $t\mapsto \mu_t$ w.r.t. the distance $W_p$.

Conversely, if $(\mu _t)$ is a family of measures in $\mathcal P_p (\Omega)$ and for each $t$, we have a vector field $v_t\in L^p(\mu_t;\mathbb R^d)$ with $\int_0^1 \Vert v_t\Vert _{L^p(\mu_t)}\ dt<+\infty$ solving

$$
\partial_t \mu_t + \nabla\cdot(\mu_t v_t)=0,
$$

then $(\mu _t)$ is absolutely continuous in $W _p(\Omega)$ and for a.e. $t$, we have

$$
|\mu’|(t)\le \Vert v_t\Vert_{L^p(\mu_t)}.
$$

Note that as a consequence of the second part of the statement, the vector field $v_t$ introduced in the first part must satisfy
$$
\Vert v_t\Vert_{L^p(\mu_t)} = |\mu’|(t).
$$

McCann’s displacement interpolation

Theorem 2. If $\Omega\subset \mathbb R^d$ is convex, then all the spaces $W_p(\Omega)$ are length spaces and if $\mu$ and $\nu$ belong to $W_p(\Omega)$ and $\gamma$ is the optimal transport plan from $\mu$ to $\nu$ for the cost $c_p(x,y)=|x-y|^p$, then the curve
$$
\mu^\gamma(t):=(\pi_t)_\sharp \gamma
$$
where $\pi_t:\Omega\times\Omega\to\Omega$ is given by
$$
\pi_t(x,y)=(1-t)x+ty
$$
is a constant-speed geodesic from $\mu$ to $\nu$. In the case $p>1$, all constant-speed geodesics are of this form, and, if $\mu$ is absolutely continuous, then there is only one geodesic and it has the form
$$
\mu_t=(T_t)_\sharp \mu,\qquad T_t=(1-t)\operatorname{id}+tT,
$$
where $T$ is the optimal transport map from $\mu$ to $\nu$. In this case, the velocity field $V_t$ of geodesic $\mu_t$ is given by
$$
v_t=(T-\operatorname{id})\circ (T_t)^{-1}.
$$

In particular, for $t=0$, we have $v_0=-\nabla\varphi$ and for $t=1$, we have $v_1=-\nabla\psi$, where $\varphi$ is the Kantorovich potential in the transport from $\mu$ to $\nu$ and $\psi=\varphi^c$.

Using the characterization of constant speed geodesics as minimizers of a strictly convex kinetic energy, we have

• Looking for an optimal transport for the cost $c(x,y)=|x-y|^p$ is equivalent to looking for constant-speed geodesics in $W_p$.
• Constant-speed geodesics may be found by minimizing $\int_0^1 |\mu’|(t)^p\ dt$.
• In the case of $W_p$, we have $|\mu’|(t)^p=\int_\Omega |v_t|^p\ d\mu_t$, where $v$ is a velocity field solving the continuity equation together with $\mu$.

As a consequence of these considerations, for $p>1$, solving the kinetic energy minimization problem
$$
\min \left\{ \int_0^1 \int_\Omega |v_t|^p\ d\rho_t\ dt \ ; \ \partial_t \rho_t+\nabla\cdot(\rho_t v_t)=0,\ \rho_0=\mu,\ \rho_1=\nu \right\}
$$
selects constant-speed geodesics connecting $\mu$ to $\nu$ and hence allows to find the optimal transport between $\mu$ and $\nu$. This is what is usually called Benamou–Brenier formula.

Minimizing Movement Scheme in the Wasserstein Space and Evolution PDEs

From now on, we consider $W_2(\Omega)$, consider the MMS,
$$
\rho_{k+1}^\tau \in \operatorname{argmin}_\rho \left\{ F(\rho)+\frac{W_2^2(\rho,\rho_k^\tau)}{2\tau} \right\}. \tag{1}
$$

and denote
$$
\mathcal T_c(\rho,\nu):=\min\left\{ \int c(x,y)\ d\gamma,\ \gamma\in\Pi(\rho,\nu) \right\},
$$
for $\nu=\rho_k^\tau$, $c(x,y)=|x-y|^2$.

Given a functional $G:\mathcal P(\Omega)\to\mathbb R$, we call $\dfrac{\delta G}{\delta \rho}(\rho)$, if it exists, the unique (up to additive constants) function such that
$$
\left.\frac{d}{d\varepsilon}G(\rho+\varepsilon\chi)\right|_{\varepsilon=0}
=
\int \frac{\delta G}{\delta \rho}(\rho)\ d\chi,
$$
for every perturbation $\chi$ such that at least for $\varepsilon\in[0,\bar\varepsilon]$, $\rho+\delta\chi\in\mathcal P(\Omega)$. The function $\dfrac{\delta G}{\delta \rho}(\rho)$ is called the first variation of functional $G$ at $\rho$.

Examples: Let $f:\mathbb R\to\mathbb R$ be a convex superlinear function, $V:\Omega\to\mathbb R$, $W:\mathbb R^d\to\mathbb R$ be regular enough, and $W$ is taken symmetric i.e.
$$
W(z)=W(-z).
$$

We have three functionals
$$
\mathcal F(\rho)=
\begin{cases}
\displaystyle \int f(\rho(x))\ dx & \text{if }\rho\ll \operatorname{leb},\
+\infty & \text{otherwise},
\end{cases}
$$
$$
\mathcal V(\rho)=\int V\ d\rho,\qquad
\mathcal W(\rho)=\frac12\iint W(x-y)\ d\rho(x)\ d\rho(y).
$$

Then we have
$$
\frac{\delta\mathcal F}{\delta \rho}(\rho)=f’(\rho),\qquad
\frac{\delta\mathcal V}{\delta \rho}(\rho)=V,\qquad
\frac{\delta\mathcal W}{\delta \rho}(\rho)=W\star \rho.
$$

Proof : We have

$$
\frac{d}{d\varepsilon} \mathcal F(\rho+\varepsilon\chi)\Big| _{\varepsilon=0}
=
\lim _{\varepsilon\to 0} \frac{\mathcal F(\rho+\varepsilon\chi)-\mathcal F(\rho)}{\varepsilon}
=
\lim _{\varepsilon\to 0}\int \frac{f(\rho(x)+\varepsilon\chi(x))-f(\rho(x))}{\varepsilon}\ dx
=
\int f^\prime(\rho(x))\chi(x)\ dx
=
\int f^\prime(\rho(x))\ d\chi(x).
$$

Hence

$$
\frac{\delta\mathcal F}{\delta \rho}(\rho)=f’(\rho).
$$

Moreover,
$$
\frac{d}{d\varepsilon}\mathcal V(\rho+\varepsilon\chi)\Big| _{\varepsilon=0}
=
\lim _{\varepsilon\to 0}\frac{\mathcal V(\rho+\varepsilon\chi)-\mathcal V(\rho)}{\varepsilon}
=
\lim _{\varepsilon\to 0}\int V\ \frac{d(\rho+\varepsilon\chi)-d\rho}{\varepsilon}
=
\int V\ d\chi.
$$
Hence
$$
\frac{\delta\mathcal V}{\delta \rho}(\rho)=V.
$$

Finally,

$$
\begin{aligned}
\frac{d}{d\varepsilon}\mathcal W(\rho+\varepsilon\chi)\Big| _{\varepsilon=0}
&=
\lim _{\varepsilon\to 0}\frac{\mathcal W(\rho+\varepsilon\chi)-\mathcal W(\rho)}{\varepsilon}
=
\lim _{\varepsilon\to 0}\frac{1}{2\varepsilon}\iint W(x-y)\Big[d((\rho+\varepsilon\chi)(x))\ d((\rho+\varepsilon\chi)(y))-d\rho(x)\ d\rho(y)\Big]\\\
&=
\frac12\iint W(x-y)\ d\chi(x)\ d\rho(y)+\frac12\iint W(x-y)\ d\rho(x)\ d\chi(y)
=
\iint W(x-y)\ d\rho(y)\ d\chi(x).
\end{aligned}
$$

Hence
$$
\frac{\delta\mathcal W}{\delta \rho}(\rho)
=
\int W(x-y)\rho(y)\ dy
=
W\star \rho.
$$

Proposition 1. Let $c:\Omega\times\Omega\to\mathbb R$ be a continuous cost function. Then the functional
$$
\rho\mapsto \mathcal T_c(\rho,\nu)
$$
is convex and its subdifferential at $\rho_0$ coincides with the set of Kantorovich potentials
$$
\left\{ \varphi\in C^0(\Omega):\ \int \varphi\ d\rho_0+\int \varphi^c\ d\nu = \mathcal T_c(\rho,\nu) \right\}.
$$

Moreover, if there is a unique $c$-concave Kantorovich potential $\varphi$ from $\rho_0$ to $\nu$, up to additive constants, then we also have
$$
\frac{\delta \mathcal T_c(\rho,\nu)}{\delta \rho}(\rho_0)=\varphi.
$$

For (1) the objection function is just
$$
F(\rho)+\frac{1}{\tau} \mathcal T_{c/2}(\rho,\rho_k^\tau).
$$

The optimal condition is
$$
\frac{\delta F}{\delta \rho}(\rho_{k+1}^\tau)+\frac{\varphi}{\tau}=\text{const},
$$
where $\varphi$ is the Kantorovich potential for the cost $\dfrac{c}{2}=\dfrac12|x-y|^2$.

Combining the fact that the optimal transport map $T(x)=x-\nabla\varphi(x)$, we get

$$
-v(x) := \frac{T(x)-x}{\tau}=-\frac{\nabla\varphi(x)}{\tau} = \nabla\left( \frac{\delta F}{\delta \rho} (\rho) \right)(x).
$$

This suggest that at the limit $\tau\to 0$, we will find a solution of
$$
\partial_t \rho_t-\nabla\cdot\left(\rho \nabla\left[\frac{\delta F}{\delta \rho}(\rho)\right]\right)=0.
$$

Examples:

• For $\mathcal F(\rho)=\int f(\rho(x))\ dx$ with $f(u)=u\log u$, we have

$$
\frac{\delta\mathcal F}{\delta \rho}(\rho)=f’(\rho)=\log\rho+1,\qquad
\nabla\frac{\delta\mathcal F}{\delta \rho}(\rho)=\frac{\nabla\rho}{\rho}
$$
we have
$$
\partial_t \rho_t=\Delta \rho_t
$$
which is just the Heat equation.

• For $F(\rho)=\int f(\rho(x))\ dx+\int V(x)\ d\rho(x)$, we get
  $$
  \frac{\delta F}{\delta \rho}(\rho)=\log\rho+1+V
  \Longrightarrow
  \nabla\frac{\delta F}{\delta \rho}=\frac{\nabla\rho}{\rho}+\nabla V.
  $$

  We get the Fokker–Planck equation
  $$
  \partial_t \rho_t-\Delta \rho-\nabla\cdot(\rho\nabla V)=0.
  $$

Reference

Santambrogio, F. {Euclidean, metric, and Wasserstein} gradient flows: an overview. Bull. Math. Sci. 7, 87–154 (2017).

The cover image in this article was taken on North Stradbroke Island, Brisbane, Australia.

Preliminaries

• Metric derivative. Given a curve $x:[0,T]\to X$ valued in a metric space, we can define the speed

$$
|x’|(t):=\lim_{h\to0}\frac{d(x(t),x(t+h))}{|h|}
$$
provided the limit exists.

• Slope and modulus of the gradient.

  Upper gradient. If $g:X\to\mathbb{R}$ such that for every Lipschitz curve $x$,

$$
|F(x(t_1))-F(x(t_0))|\le \int_0^1 g(x(t))|x’|(t)\ dt.
$$

If $F$ is Lipschitz continuous, a possible choice is the local Lipschitz constant
$$
|\nabla F|(x):=\limsup_{y\to x}\frac{|F(y)-F(x)|}{d(x,y)}.
$$

• Descending slope (slope in short)

$$
|\nabla^-F|(x):=\limsup_{y\to x}\frac{[F(x)-F(y)]_+}{d(x,y)}.
$$

• Geodesic convexity. On a geodesic metric space, we say a function $F$ is geodesically convex if for every pair $(x(0),x(1))$ there exists a geodesic $x$ with constant speed connecting these two points and such that

$$
F(x(t))\le (1-t)F(x(0))+tF(x(1)).
$$

We can also define $\lambda$-convex by
$$
F(x(t))\le (1-t)F(x(0))+tF(x(1))-\lambda\frac{t(1-t)}{2}d^2(x(0),x(1)).
$$

Existence of a gradient flow

Let us suppose that the space $X$ and the function $F$ are such that every sub-level set $\{F\le c\}$ is compact in $X$, either for the topology induced by the distance $d$, or for a weaker topology such that $d$ is lower semi-continuous. $F$ is required to be l.s.c. in the same topology. This is the minimal framework to guarantee existence of the minimizers at each step.

Even if estimate (11) is enough to provide compactness and thus the existence of GMM, it will be never enough to characterize the limit curve (indeed, it is satisfied by any discrete evolution where $x_{k+1}^\tau$ gives a better value than $x_k^\tau$, without any need for optimality).

Variational interpolation (by De Giorgi). Once we fix $x_k^\tau$, for every $\theta\in(0,1]$, consider
$$
\min_x\ F(x)+\frac{d^2(x,x_k^\tau)}{2\theta\tau}
$$
and call $x(\theta)$ any minimizer of this problem and $\varphi(\theta)$ the minimal value.

Then we have

(1) for $\theta\to0^+$, we have $x(\theta)\to x_k^\tau$ and $\varphi(\theta)\to F(x_k^\tau)$.

Proof. Since $x(\theta)$ is the minimizer, we have
$$
F(x(\theta))\le F(x(\theta))+\frac{d^2(x(\theta),x_k^\tau)}{2\theta\tau}\le F(x_k^\tau).
$$
That is
$$
\frac{d^2(x(\theta),x_k^\tau)}{2\theta\tau}\le F(x_k^\tau)-F(x(\theta)).
$$
Since $x(\theta)\in\{F\le F(x_k^\tau)\}$ which is compact, the RHS above is bounded. Therefore $d(x(\theta),x_k^\tau)\to0$.

For $\varphi(\theta)$, we have $\varphi(\theta)\le F(x_k^\tau)$.

On the other hand
$$
\varphi(\theta)=F(x(\theta))+\frac{d^2(x(\theta),x_k^\tau)}{2\theta\tau}\ge F(x(\theta)).
$$
By the l.s.c. of $F$, we have
$$
F(x_k^\tau)\le \liminf_{\theta\to0^+}F(x(\theta))
\le \liminf_{\theta\to0^+}\varphi(\theta)
\le \limsup_{\theta\to0^+}\varphi(\theta)\le F(x_k^\tau).
$$
Hence
$$
\lim_{\theta\to0^+}\varphi(\theta)=F(x_k^\tau).\quad \square
$$

(2) for $\theta=1$, we get back to the original problem with minimizer $x_{k+1}^\tau$.

(3) the function $\varphi$ is non-increasing and hence a.e. differentiable. Moreover
$$
\varphi’(\theta)=-\frac{d^2(x(\theta),x_k^\tau)}{2\theta^2\tau}.
$$
which also means $d(x(\theta),x_k^\tau)$ does not depend on the minimizer $x(\theta)$ for all $\theta$ such that $\varphi’(\theta)$ exists.

Proof. Let $G_\theta(x):=F(x)+\dfrac{d^2(x,x_k^\tau)}{2\theta\tau}$, then
$$
\varphi(\theta)=\min_x G_\theta(x)=G_\theta(x(\theta)).
$$

For sufficiently small $h>0$, we have
$$
\varphi(\theta+h)\le G_{\theta+h}(x(\theta))
=F(x(\theta))+\frac{d^2(x(\theta),x_k^\tau)}{2(\theta+h)\tau}
$$
and
$$
\varphi(\theta)=F(x(\theta))+\frac{d^2(x(\theta),x_k^\tau)}{2\theta\tau}.
$$
Hence
$$
\frac{\varphi(\theta+h)-\varphi(\theta)}{h}
\le \frac{d^2(x(\theta),x_k^\tau)}{2\tau}\frac{\frac{1}{\theta+h}-\frac{1}{\theta}}{h}
= -\frac{d^2(x(\theta),x_k^\tau)}{2\theta(\theta+h)\tau}.
$$
Then
$$
\varphi’(\theta)=\lim_{h\to0}\frac{\varphi(\theta+h)-\varphi(\theta)}{h}
\le -\frac{d^2(x(\theta),x_k^\tau)}{2\theta^2\tau}.
\tag{$\star$}
$$

On the other hand
$$
\varphi(\theta)\le G_\theta(x(\theta+h))
=F(x(\theta+h))+\frac{d^2(x(\theta+h),x_k^\tau)}{2\theta\tau}.
$$
Meanwhile,
$$
\varphi(\theta+h)=F(x(\theta+h))+\frac{d^2(x(\theta+h),x_k^\tau)}{2(\theta+h)\tau}.
$$
We have
$$
\frac{\varphi(\theta+h)-\varphi(\theta)}{h}
\ge \frac{d^2(x(\theta+h),x_k^\tau)}{2\tau}\frac{\frac{1}{\theta+h}-\frac{1}{\theta}}{h}
= -\frac{d^2(x(\theta+h),x_k^\tau)}{2\theta(\theta+h)\tau}.
$$
Let $h\to0$, we have
$$
\varphi’(\theta)\ge -\limsup_{h\to0}\frac{d^2(x(\theta+h),x_k^\tau)}{2\theta(\theta+h)\tau}.
\tag{$\star\star$}
$$

Since all $x(\theta+h)\in\{F\le F(x_k^\tau)\}$ which is compact, there exists $h_j\to0$ such that
$$
x(\theta+h_j)\to \bar x.
$$

Since $F$ and $d$ are l.s.c., then $G_\theta$ is also l.s.c. Hence
$$
\begin{aligned}
G_\theta(\bar x)\le \liminf_{j\to\infty}G_\theta(x(\theta+h_j))&=\liminf_{j\to\infty}\left[G_{\theta+h_j}(x(\theta+h_j))
+\frac{d^2(x(\theta+h_j),x_k^\tau)}{2\tau}
\left(\frac{1}{\theta}-\frac{1}{\theta+h_j}\right)\right]\\\
&=\lim_{j\to\infty}\varphi(\theta+h_j)=\varphi(\theta)\le G_\theta(\bar x),
\end{aligned}
$$

where, we can similarly show that $d^2(x(\theta+h_j),x_k^\tau)$ is bounded and since $\varphi$ is differentiable at $\theta$, and then also continuous at $\theta$, and that $x(\theta)$ is the minimizer of $G_\theta$. Therefore
$$
G_\theta(\bar x)=\varphi(\theta).
$$
$\bar x$ is the minimizer at $\theta$. On the other hand
$$
\varphi(\theta+h_j)=F(x(\theta+h_j))
+\frac{d^2(x(\theta+h_j),x_k^\tau)}{2(\theta+h_j)\tau},
$$
then by the l.s.c. of $F$ and $d$,
$$
\varphi(\theta)=\liminf_{j\to\infty}\varphi(\theta+h_j)
\ge F(\bar x)+\frac{d^2(\bar x,x_k^\tau)}{2\theta\tau}
=G_\theta(\bar x)=\varphi(\theta).
$$
Hence all the inequality by l.s.c. becomes equality, especially
$$
\lim_{j\to\infty}\frac{d^2(x(\theta+h_j),x_k^\tau)}{2(\theta+h_j)\tau}
=\frac{d^2(\bar x,x_k^\tau)}{2\theta\tau}.
$$
Therefore,
$$
d(x(\theta+h_j),x_k^\tau)\to d(\bar x,x_k^\tau),
$$
and then by ($\star\star$),
$$
\varphi’(\theta)\ge -\lim_{j\to\infty}\frac{d^2(x(\theta+h_j),x_k^\tau)}{2(\theta+h_j)\tau}
= -\frac{d^2(\bar x,x_k^\tau)}{2\theta^2\tau}.
$$

Combine ($\star$) and ($\star\star$), we get (note that ($\star$) is true for all minimizer $x(\theta)$ and $\bar x$ is also a minimizer)
$$
\varphi’(\theta)= -\frac{d^2(x(\theta),x_k^\tau)}{2\theta^2\tau}.\quad \square
$$

(4) we have
$$
|\nabla^-F|(x(\theta))\le \frac{d(x(\theta),x_k^\tau)}{\theta\tau}.
$$

Proof. Since $x(\theta)$ is optimal, for all $y$
$$
F(y)+\frac{d^2(y,x_k^\tau)}{2\theta\tau}
\ge F(x(\theta))+\frac{d^2(x(\theta),x_k^\tau)}{2\theta\tau}.
$$
We have
$$
F(x(\theta))-F(y)\le \frac{1}{2\theta\tau}
\big(d^2(y,x_k^\tau)-d^2(x(\theta),x_k^\tau)\big)=\frac{1}{2\theta\tau}\big(d(x(\theta),x_k^\tau)+d(y,x_k^\tau)\big)d(x(\theta),y).
$$

Therefore by the l.s.c. of $d$
$$
\limsup_{y\to x(\theta)}
\frac{[F(x(\theta))-F(y)]_+}{d(x(\theta),y)}
\le \frac{1}{2\theta\tau}\cdot 2d(x(\theta),x_k^\tau)
=\frac{d(x(\theta),x_k^\tau)}{\theta\tau}.\quad \square
$$

(5) due to the possible singular part of the derivative for monotone functions, we have
$$
\varphi(0)-\varphi(1)\ge -\int_0^1 \varphi’(\theta)\ d\theta.
$$
Together with the inequality
$$
-\varphi’(\theta)=\frac{d^2(x(\theta),x_k^\tau)}{2\theta^2\tau}
\ge \frac{\tau}{2}|\nabla^-F|^2(x(\theta)),
$$
we get
$$
F(x_k^\tau)-\left(F(x_{k+1}^\tau)+\frac{d^2(x_{k+1}^\tau,x_k^\tau)}{2\tau}\right)
\ge \frac{\tau}{2}\int_0^1 |\nabla^-F|^2(x(\theta))\ d\theta.
$$

If we sum up for $k=0,1,\cdots$ and take the limit $\tau\to0$, under some suitable assumptions, we can prove for every GMM $x$ we have
$$
F(x(t))+\frac12\int_0^t |x’|^2(r)\ dr
+\frac12\int_0^t |\nabla^-F|^2(x(r))\ dr
\le F(x(0)).
$$

But it is not exactly the EDE

• it is an inequality
• just compare instants $t$ and $0$ instead of $t$ and $s$

If we want equality for every pair $(t,s)$ we need to require the slope to be an upper gradient. Indeed, in this case,
$$
F(x(0))-F(x(t))
\le \int_0^t |\nabla^-F|(x(r))|x’|(r)\ dr
$$
and then we get the equality and also allows to subtract the equalities for $t$ and $s$ and get for $s<t$,
$$
F(x(t))+\frac12\int_s^t |x’|^2(r)\ dr
+\frac12\int_s^t |\nabla^-F|^2(x(r))\ dr
=F(x(s)).
$$

Magically, as it happens that the assumption that $F$ is $\lambda$-geodesically convex make all the assumptions hold true.

Uniqueness and contractivity

Between EDE and EVI, by Savaré, we have

• All curves which are gradient flows in EVI sense also satisfy the EDE condition
• The EDE condition is not in general enough to guarantee uniqueness of the gradient flow. A simple example is $X=\mathbb{R}^2$, with $l^\infty$ distance
  $$
  d((x_1,y_1),(x_2,y_2))=|x_1-y_1|\vee |x_2-y_2|
  $$
  and take $F(x_1,x_2)=x_1$, then any curve $(x_1(t),x_2(t))$ with $x_1’(t)=-1$ and $|x_2’(t)|\le1$ satisfies EDE.

Proof. It is easy to see that $|\nabla^-F|(x)=1$ for all $x\in X$, and
$$
|x’|(t)=\lim_{h\to0}\frac{d(x(t+h),x(t))}{|h|}
=\lim_{h\to0}\frac{\max\{|x_1(t+h)-x_1(t)|,\ |x_2(t+h)-x_2(t)|\}}{|h|}
=\max\{|x_1’(t)|,\ |x_2’(t)|\}.
$$
If $x_1’(t)=1$, $|x_2’(t)|\le1$, $|x’|(t)=1$. Now, we need to check EDE: for $s<t$
$$
F(x(s))-F(x(t))
=\int_s^t \frac12|x’|^2(r)+\frac12|\nabla^-F|^2(x(r))\ dr.
$$
LHS $=x_1(s)-x_1(t)$, by $x_1’(t)=1$, we have $x_1(t)=x_1(0)-t$ for all $t$, hence
$$
\text{LHS}=(x_1(0)-s)-(x_1(0)-t)=t-s. \quad \text{RHS}=\int_s^t \frac12+\frac12\ dr=t-s.
$$
Hence,the EDE holds. $\square$

• Existence of gradient flow of EDE sense is easy to get
• The EVI condition is general too strong to get existence, but always guarantees uniqueness and stability.

Proposition. If two curves $x,y:[0,T]\to X$ satisfy the $\mathrm{EVI}_\lambda$ condition, then we have
$$
\frac{d}{dt}d(x(t),y(t))^2\le -2\lambda\ d(x(t),y(t))^2
$$
and
$$
d(x(t),y(t))\le e^{-\lambda t}d(x(0),y(0)).
$$

Proof. By $\mathrm{EVI}_\lambda$ condition, for all $y\in X$
$$
\frac{d}{dt}\frac12 d(x(t),y)^2\le F(y)-F(x(t))-\frac{\lambda}{2}d(x(t),y)^2.
$$

Take $y=y(t_0)$.
$$
\left.\frac{d}{dt}\frac12 d(x(t),y(t_0))^2\right|_{t=t_0}
\le F(y(t_0))-F(x(t_0))-\frac{\lambda}{2}d(x(t_0),y(t_0))^2.
$$

Similarly,
$$
\left.\frac{d}{ds}\frac12 d(x(t_0),y(s))^2\right|_{s=t_0}
\le F(x(t_0))-F(y(t_0))-\frac{\lambda}{2}d(x(t_0),y(t_0))^2.
$$

Add them up, we get
$$
\frac{d}{dt}\frac12 d(x(t),y(t_0))^2\Big| _{t=t_0}
=\frac{d}{dt}\frac12 d(x(t),y(t_0))^2\Big| _{t=t_0}
+\frac{d}{ds}\frac12 d(x(t_0),y(s))^2\Big| _{s=t_0}\le -\lambda\ d(x(t_0),y(t_0))^2.
$$

Hence
$$
\frac{d}{dt}d(x(t),y(t))^2\le -2\lambda\ d(x(t),y(t))^2.
$$
By Gronwall inequality
$$
d(x(t),y(t))\le e^{-\lambda t}d(x(0),y(0)).\quad \square
$$

If we want a satisfying theory for gradient flows which includes uniqueness, we just need to prove the existence of curves which satisfy the EVI condition, accepting that this will probably require additional assumptions.

This assumption, that we will call $C^2G^2$ (Compatible Convexity along Generalized Geodesics), is the following: suppose that for every pair $(x_0,x_1)$, and every $y\in X$, there is a curve $x(t)$ connecting $x(0)=x_0$ to $x(1)=x_1$, such that

• ($F$ is $\lambda$-convex)

$$
F(x(t))\le (1-t)F(x_0)+tF(x_1)-\lambda\frac{t(1-t)}{2}d^2(x_0,x_1)
$$

• ($x\mapsto d^2(x,y)$ is $2$-convex)

$$
d^2(x(t),y)\le (1-t)d^2(x_0,y)+t\ d^2(x_1,y)-t(1-t)d^2(x_0,x_1).
$$

Reference

Santambrogio, F. {Euclidean, metric, and Wasserstein} gradient flows: an overview. Bull. Math. Sci. 7, 87–154 (2017).

The cover image in this article was taken on Tinian Island, a U.S. territory in the Northern Mariana Islands.

If one has a metric space $(X,d)$ and a l.s.c. function $F:X\to \mathbb{R}\cup\{+\infty\}$ (under suitable compactness assumptions to guarantee existence of the minimum), one can define
$$
x_{k+1}^{\tau}\in \arg\min_{x\in X}\left\{F(x)+\frac{d(x,x_k^{\tau})^2}{2\tau}\right\} \tag{9}
$$
and study the limit as $\tau\to 0$. Then we use the piecewise constant interpolation
$$
x^{\tau}(t):=x_k^{\tau}\qquad \text{for every } t\in ((k-1)\tau,k\tau].\tag{10}
$$
and study the limit of $x^{\tau}$ as $\tau\to 0$.

Definition 1. A curve $x:[0,T]\to X$ is called Generalized Minimizing Movements (GMM) if there exists a sequence of time steps $\tau_j\to 0$ such that the sequence of curves $x^{\tau_j}$, defined in (10) using the iterated solutions of (9), uniformly converges to $x$ in $[0,T]$.

We start from
$$
F(x_{k+1}^{\tau})+\frac{d(x_{k+1}^{\tau},x_k^{\tau})^2}{2\tau}\le F(x_k^{\tau})
$$
and
$$
\sum_{k=0}^{\ell}\frac{d(x_{k+1}^{\tau},x_k^{\tau})^2}{2\tau}\le F(x_0^{\tau})-F(x_{\ell+1}^{\tau})\le C\tau.
$$

The Cauchy–Schwarz inequality gives for $t<s$, $t\in [k\tau,(k+1)\tau]$ and $s\in [k’\tau,(k’+1)\tau]$ (which implies $|k’-k|\le \frac{|t-s|}{\tau}+1$),
$$
d(x^{\tau}(t),x^{\tau}(s))
\le d(x_k^{\tau},x_{k’}^{\tau})
\le \sum_{j=k}^{k’-1} d(x_{j+1}^{\tau},x_j^{\tau})
\le \left(\sum_{j=k}^{k’-1} d(x_{j+1}^{\tau},x_j^{\tau})^2\right)^{1/2}(k’-k)^{1/2}.
$$

For $t<s$, $t\in [i\tau,(i+1)\tau]$, $s\in [j\tau,(j+1)\tau]$ (which implies $j-i\le \frac{|t-s|}{\tau}+1$),

$$
d(x^{\tau}(t),x^{\tau}(s))
\le d(x_{i+1}^{\tau},x_{j+1}^{\tau})
\le \sum_{k=i+1}^{j} d(x_{k+1}^{\tau},x_k^{\tau})\le \left(\sum_{k=i+1}^{j} d(x_{k+1}^{\tau},x_k^{\tau})^2\right)^{1/2}(j-i)^{1/2}
\le \left(\sum_{k=0}^{\ell} d(x_{k+1}^{\tau},x_k^{\tau})^2\right)^{1/2}\left(\frac{|t-s|}{\tau}\right)^{1/2}
\le C|t-s|^{1/2}.
$$

This shows the curves $x^{\tau}$ are morally equi-Hölder with exponent $\frac12$. Since they start from the same $x^{\tau}(0)=x_0$, they are also equibounded. By applying Ascoli–Arzelà theorem, we can extract a converging subsequence.

Curves and geodesics in metric spaces

Definition 2. If $w:[0,1]\to X$ is a curve valued in the metric space $(X,d)$, we define the metric derivative of $w$ at the time $t$, denoted by $|w’|(t)$ through
$$
|w’|(t):=\lim_{h\to 0}\frac{d(w(t+h),w(t))}{|h|}
$$
provided this limit exists.

In the spirit of Rademacher theorem, it is possible to prove that if $w:[0,1]\to X$ is Lipschitz continuous, then the metric derivative $|w’|(t)$ exists for a.e. $t$. Moreover, we have for $t_0<t_1$
$$
d(w(t_0),w(t_1))\le \int_{t_0}^{t_1}|w’|(s)\ ds.
$$

Definition 3. A curve $w:[0,1]\to X$ is said to be absolutely continuous whenever there exists $g\in L^1[0,1]$ such that
$$
d(w(t_0),w(t_1))\le \int_{t_0}^{t_1} g(s)\ ds
\qquad \text{for every } t_0<t_1.
$$

The set of absolutely continuous curves defined on $[0,1]$ and valued in $X$ is denoted by $AC(X)$.

It is well-known that every absolutely continuous curve can be reparameterized in time (through a monotone increasing reparametrization) and become Lipschitz continuous, and the existence of the metric derivative is also true for $w\in AC(X)$ with i.e.t.

Definition 4. For a curve $w:[0,1]\to X$, let us define
$$
\operatorname{Length}(w):=\sup\left\{\sum_{k=0}^{n-1} d\bigl(w(t_k),w(t_{k+1})\bigr): n\ge 1,\ 0=t_0<t_1<\cdots<t_n=1\right\}.
$$

It is easy to see that all curves $w\in AC(X)$ satisfy
$$
\operatorname{Length}(w)\le \int_0^1 g(t)\ dt<+\infty.
$$

Also we can prove that for any curve $w\in AC(X)$, we have
$$
\operatorname{Length}(w)=\int_0^1 |w’|(t)\ dt.
$$

Definition 5. A curve $w:[0,1]\to X$ is said to be a geodesic between $x_0$ and $x_1\in X$ if $w(0)=x_0$, $w(1)=x_1$, and
$$
\operatorname{Length}(w)=\min\left\{\operatorname{Length}(\widetilde w): \widetilde w(0)=x_0,\ \widetilde w(1)=x_1\right\}.
$$

A space $(X,d)$ is said to be a length space if for every $x$ and $y$, we have
$$
d(x,y)=\inf\left\{\operatorname{Length}(w): w\in AC(X),\ w(0)=x,\ w(1)=y\right\}.
$$

A space $(X,d)$ is said to be a geodesic space if for every $x$ and $y$, we have
$$
d(x,y)=\min\left\{\operatorname{Length}(w): w\in AC(X),\ w(0)=x,\ w(1)=y\right\}.
$$
i.e. if it is a length space and there exist geodesics between arbitrary points.

In a length space, a curve $w:[t_0,t_1]\to X$ is said to be a constant-speed geodesic between $w(t_0)$ and $w(t_1)$ if it satisfies
$$
d(w(t),w(s))=\frac{|t-s|}{t_1-t_0}d(w(t_0),w(t_1))
\qquad \text{for all } t,s\in [t_0,t_1].
$$

The following three facts are equivalent:

1. $w$ is a constant-speed geodesic defined on $[t_0,t_1]$ and joining $x_0$ and $x_1$.
2. $w\in AC(X)$ and
   $$
   |w’|(t)=\frac{d(w(t_0),w(t_1))}{|t_1-t_0|}
   \qquad \text{a.e.}
   $$
3. $w$ solves
   $$
   \min\left\{\int_{t_0}^{t_1}|w’(t)|^p\ dt:\ w(t_0)=x_0,\ w(t_1)=x_1\right\}
   \qquad \text{for all } p>1.
   $$

Come back to the interpolation of the points obtained through the Minimizing Movement Scheme, if $(X,d)$ is a geodesic space, then the piecewise affine interpolation that we used in the Euclidean space may be helpfully replaced via a piecewise geodesic interpolation. This means defining a curve
$$
x^{\tau}:[0,T]\to X
$$
such that
$$
x^{\tau}(k\tau)=x_k^{\tau}
$$
and such that $x^{\tau}$ restricted to any interval $[k\tau,(k+1)\tau]$ is a constant-speed geodesic with speed equal to $\frac{d(x_k^{\tau},x_{k+1}^{\tau})}{\tau}$. Then the same equicontinuity will hold.

The next question is how to characterize the limit curve obtained when $\tau\to 0$. In a general metric space
$$
x’(t)=-\nabla F(x(t))
$$
has no meaning!

• EDE (Energy Dissipation Equality) viewpoint.

For $s<t$
$$
F(x(s))-F(x(t))
=\int_s^t -\nabla F(x(r))\cdot x’(r)\ dr
\le \int_s^t |\nabla F(x(r))|\ |x’(r)|\ dr
\le \int_s^t \left(\frac12|x’(r)|^2+\frac12|\nabla F(x(r))|^2\right)\ dr.
$$

Here the first inequality becomes equality if and only if $x’(r)$ and $\nabla F(x(r))$ are vectors with opposite direction for a.e. $r$, and the second inequality becomes equality if and only if their norms are the same. Therefore
$$
F(x(s))-F(x(t))
=\int_s^t \left(\frac12|x’(r)|^2+\frac12|\nabla F(x(r))|^2\right)\ dr
\qquad \text{for all } s<t
$$
if and only if
$$
x’(t)=-\nabla F(x(t))
\qquad \text{a.e. } t.
$$

• EVI (Evolution Variational Inequality) viewpoint.

If $F$ is $\lambda$-convex, the inequality that characterize the gradient is
$$
F(y)\ge F(x)+\frac{\lambda}{2}|y-x|^2+p\cdot (y-x)
\qquad \text{for all } y\in \mathbb{R}^n.
$$

Therefore
$$
\frac{d}{dt}\frac12|x(t)-y|^2=(y-x(t))\cdot \bigl(-x’(t)\bigr)
\le F(y)-F(x(t))-\frac{\lambda}{2}|x(t)-y|^2
\qquad \text{for all } y.
$$
which will be equivalent to
$$
-x’(t)\in \partial F(x(t)).
$$

By $\mathrm{EVI}_{\lambda}$, take two curves $x(t)$ and $y(s)$, we have
$$
\frac{d}{dt}\frac12 d(x(t),y(s))^2
\le F(y(s))-F(x(t))-\frac{\lambda}{2}d(x(t),y(s))^2
\tag{12}
$$
and
$$
\frac{d}{ds}\frac12 d(x(t),y(s))^2
\le F(x(t))-F(y(s))-\frac{\lambda}{2}d(x(t),y(s))^2.
\tag{13}
$$

Define
$$
E(t)=\frac12 d(x(t),y(t))^2,
\qquad
G(t,s)=\frac12 d(x(t),y(s))^2.
$$
Then
$$
\frac{d}{dt}E(t)=\frac{d}{dt}G(t,t)=\partial_t G(t,t)+\partial_s G(t,t)
\le -\lambda d(x(t),y(t))^2=-2\lambda E(t).
$$

By Gronwall inequality, this provides uniqueness and stability.

Reference

Santambrogio, F. {Euclidean, metric, and Wasserstein} gradient flows: an overview. Bull. Math. Sci. 7, 87–154 (2017).

The cover image in this article was taken at Château de Chillon in Switzerland.

Given a function $F:\mathbb{R}^n \to \mathbb{R}$, smooth enough and a point $x_0 \in \mathbb{R}^n$. A gradient flow is just defined as a curve $X(t)$, with starting point at $t=0$ given by $x_0$, which moves by choosing at each instant of time the direction which makes the function $F$ decrease as much as possible. More precisely, we consider the solution of the Cauchy problem
$$
\begin{cases}
x’(t) = -\nabla F(x(t)) & \text{for } t>0, \\\
x(0)=x_0
\end{cases}
\tag{1}
$$

This is a standard Cauchy problem which has a unique solution if $\nabla F$ is Lipschitz continuous.

If $F$ is convex, we can replace the gradient with subdifferential, looking for absolutely continuous curve $x:[0,T]\to \mathbb{R}^n$
$$
\begin{cases}
x’(t)\in -\partial F(x(t)) & \text{for a.e. } t>0, \\\
x(0)=x_0
\end{cases}
\tag{2}
$$
where
$$
\partial F(x):=\left\{ p\in \mathbb{R}^n \mid F(y)\ge F(x)+p\cdot (y-x)\ \text{for all } y\in \mathbb{R}^n \right\}.
$$

Properties of subdifferential

1. $F$ is differentiable at $x$ iff $\partial F(x)=\{\nabla F(x)\}$.
2. $\partial F(x)$ is a convex set.
3. $\partial F(x)$ is non-empty for $x\in \operatorname{dom} F$.
4. For every $x_1,x_2$ and $p_1\in \partial F(x_1)$, $p_2\in \partial F(x_2)$,
   $$
   (x_1-x_2)\cdot (p_1-p_2)\ge 0.
   $$

We denote by $\partial^\circ F(x)$ its element of minimal norm.

Proposition 1. Suppose that $F$ is convex and let $x_1,x_2$ be two solutions of (2). Then we have
$$
|x_1(t)-x_2(t)|\le |x_1(0)-x_2(0)|
$$
for every $t\ge 0$. In particular, this gives uniqueness of solution of the Cauchy problem.

Proof. Let us consider
$$
g(t)=\frac12 |x_1(t)-x_2(t)|^2.
$$
Then we have
$$
g’(t)=(x_1(t)-x_2(t))\cdot (x_1’(t)-x_2’(t)).
$$
Since $x_1’(t)\in -\partial F(x_1(t))$, $x_2’(t)\in -\partial F(x_2(t))$, we have
$$
g’(t)\le 0.
$$
Therefore
$$
g(t)\le g(0).\quad \square
$$

Remark. We recall that $F$ is semi-convex means that there exists $\lambda\in \mathbb{R}$ such that
$$
x\mapsto F(x)-\frac{\lambda}{2}|x|^2
$$
is convex.

For $\lambda$-convex function $F$, we can define the subdifferential by
$$
\partial F(x)=\left\{ p\in \mathbb{R}^n \ \middle| \ F(y)\ge F(x)+p\cdot (y-x)+\frac{\lambda}{2}|y-x|^2 \ \text{for all } y\in \mathbb{R}^n \right\}.
$$

Again, we define $\partial^\circ F$ the element of minimal norm of $\partial F$.

If $F$ is $\lambda$-convex, then for $x_1,x_2,p_1,p_2$ with $p_i\in \partial F(x_i)$, $i=1,2$, one has
$$
(x_1-x_2)\cdot (p_1-p_2)\ge \lambda |x_1-x_2|^2.
$$

This implies
$$
g’(t)\le -2\lambda g(t).
$$
By Gronwall’s inequality
$$
g(t)\le g(0)e^{-2\lambda t}.
$$

If $\lambda>0$, then $F$ is strictly convex and admits a unique minimizer $\bar{x}$.

Take $x_2(t)\equiv \bar{x}$, which is also a solution since $0\in \partial F(\bar{x})$. Then we get
$$
|x_1(t)-\bar{x}|\le e^{-\lambda t}|x_1(0)-\bar{x}|.
$$

Proposition 2. Suppose that $F$ is $\lambda$-convex and let $x$ be a solution of (2). Then, for all times $t_0$ such that both $t\mapsto x(t)$ and $t\mapsto F(x(t))$ are differentiable at $t=t_0$, the subdifferential $\partial F(x(t_0))$ is contained in a hyperplane orthogonal to $x’(t_0)$. In particular, we have
$$
x’(t)=-\partial^\circ F(x(t)) \qquad \text{for a.e. } t.
$$

Proof. Let $t_0$ be as in the statement and $p\in \partial F(x(t_0))$. By the definition of subdifferential
$$
F(x(t))\ge F(x(t_0))+p\cdot (x(t)-x(t_0))+\frac{\lambda}{2}|x(t)-x(t_0)|^2
\qquad \text{for all } t,
$$
and it is equality when $t=t_0$. Hence
$$
t\mapsto F(x(t))-F(x(t_0))-p\cdot (x(t)-x(t_0))-\frac{\lambda}{2}|x(t)-x(t_0)|^2
$$
is minimal for $t=t_0$, and then
$$
\frac{d}{dt}F(x(t))\Big|_{t=t_0}=p\cdot x’(t_0).
$$
Hence
$$
\partial F(x(t_0))\subseteq \left\{ p\in \mathbb{R}^n \mid p\cdot x’(t_0)=\mathrm{const} \right\},
$$
which is a hyperplane orthogonal to $x’(t_0)$.

Since for a.e. $t_0$,
$$
-x’(t_0)\in \partial F(x(t_0)),
$$
we have
$$
\mathrm{const}=-x’(t_0)\cdot x’(t_0)=-|x’(t_0)|^2.
$$
The hyperplane is
$$
\left\{ p\in \mathbb{R}^n \mid p\cdot x’(t_0)=-|x’(t_0)|^2 \right\}.
$$
Hence the orthogonal projection of $0$ onto the hyperplane which contains $\partial F(x(t_0))$ is
$$
\frac{x’(t_0)}{|x’(t_0)|^2}\cdot \big(-|x’(t_0)|^2\big)=-x’(t_0).
$$
This provides
$$
x’(t_0)=-\partial^\circ F(x(t_0))
$$
for a.e. $t_0$. $\square$

Discretization in time

Fix a small step time parameter $\tau>0$, define

$$
x_{k+1}^\tau \in \operatorname*{argmin} _x \left\{ F(x)+\frac{|x-x_k^\tau|^2}{2\tau} \right\}.
\tag{3}
$$

and we say $(x_k^\tau) _k$ the Minimizing Movement Scheme.

If $F$ is smooth, then

$$
x _{k+1}^\tau \in \operatorname*{argmin} _x \left\{ F(x)+\frac{| x-x_k^\tau |^2}{2\tau} \right\}
\implies
\nabla F(x _{k+1}^\tau)+\frac{x _{k+1}^\tau-x _k^\tau}{\tau}=0.
$$

Hence,
$$
\frac{x_{k+1}^\tau-x_k^\tau}{\tau}=-\nabla F(x_{k+1}^\tau),
$$
which is exactly the discrete-time implicit Euler scheme.

Given ODE
$$
\begin{cases}
x’(t)=V(x(t)),\\\
x(0)=x_0,
\end{cases}
$$

• Explicit scheme
  $$
  x_{k+1}^\tau=x_k^\tau+\tau V(x_k^\tau), \qquad x_0=x_0.
  $$

• Implicit scheme
  $$
  x_{k+1}^\tau=x_k^\tau+\tau V(x_{k+1}^\tau), \qquad x_0=x_0.
  $$

Now, we want to show that for $\tau\to 0$, the sequence $(x_k^\tau)$ we found, suitably interpolated, converges to the solution of (2).

First, define
$$
v_{k+1}^\tau:=\frac{x_{k+1}^\tau-x_k^\tau}{\tau}.
$$
Then we set
$$
x^\tau(t):=x_{k+1}^\tau, \qquad \text{for } t\in (k\tau,(k+1)\tau],
$$
$$
\tilde x^\tau(t):=x_k^\tau+(t-k\tau)v_{k+1}^\tau, \qquad \text{for } t\in (k\tau,(k+1)\tau],
$$
$$
v^\tau(t):=v_{k+1}^\tau, \qquad \text{for } t\in (k\tau,(k+1)\tau].
$$

It is easy to see that $\tilde x^\tau$ is continuous and pairwise affine, satisfying
$$
(\tilde x^\tau)’=v^\tau.
$$

What’s more, $x^\tau$ is not continuous, but by definition
$$
v^\tau(t)\in -\partial F(x^\tau(t)).
$$

By the definition of $x_{k+1}^\tau$, we have
$$
F(x_{k+1}^\tau)+\frac{|x_{k+1}^\tau-x_k^\tau|^2}{2\tau}\le F(x_k^\tau).
\tag{4}
$$

If $F(x_0)<+\infty$ and $\inf F>-\infty$, summing over $k$, we get
$$
\sum_{k=0}^{\ell}\frac{|x_{k+1}^\tau-x_k^\tau|^2}{2\tau}
\le
\big(F(x_0^\tau)-F(x_{\ell+1}^\tau)\big)\le C.
\tag{5}
$$

Note that
$$
\frac{|x_{k+1}^\tau-x_k^\tau|^2}{2\tau}
=
\frac{\tau}{2}\left|\frac{x_{k+1}^\tau-x_k^\tau}{\tau}\right|^2
=
\frac{\tau}{2}|v_{k+1}^\tau|^2
=
\frac12\int_{k\tau}^{(k+1)\tau}|(\tilde x^\tau)’(t)|^2\ dt.
$$

This means that we have
$$
\int_0^T \frac12 |(\tilde x^\tau)’(t)|^2\ dt\le C.
\tag{6}
$$

Hence $\tilde x^\tau$ is bounded in $H^1$ and $v^\tau$ in $L^2$. By Hölder inequality, for $t\ge s$,
$$
|\tilde x^\tau(t)-\tilde x^\tau(s)|
=
\left|\int_s^t (\tilde x^\tau)’(u)\ du\right|
\le
\int_s^t |(\tilde x^\tau)’(u)|\ du
\le
\left(\int_s^t |(\tilde x^\tau)’(u)|^2\ du\right)^{1/2}|t-s|^{1/2}
\le C|t-s|^{1/2}.
\tag{7}
$$

Therefore $\tilde x^\tau$ is equicontinuous for all $\tau>0$. Moreover,
$$
|\tilde x^\tau(t)-x_0|
=
|\tilde x^\tau(t)-\tilde x^\tau(0)|
\le Ct^{1/2}\le CT^{1/2}.
$$
Hence $\tilde x^\tau$ is also equibounded. Furthermore, since $x^\tau(t)=\tilde x^\tau(s)$ for a certain $s=k\tau$ with $|s-t|\le \tau$, we have
$$
|x^\tau(t)-\tilde x^\tau(t)|\le C\tau^{1/2}.
\tag{8}
$$

Proposition 3. Let $\tilde x^\tau$, $x^\tau$ and $v^\tau$ be constructed as above using the minimizing movement scheme. Suppose $F(x_0)<+\infty$ and $\inf F>-\infty$, and lower-semicontinuous. Then, up to a subsequence $\tau_j\to 0$ (still denoted by $\tau$), both $\tilde x^\tau$ and $x^\tau$ converge uniformly to a same curve $x\in H^1$, and $v^\tau$ weakly converges in $L^2$ to a vector function $v$ such that $x’=v$ and

1. if $F$ is $\lambda$-convex, we have
   $$
   v(t)\in -\partial F(x(t)) \qquad \text{for a.e. } t,
   $$
   i.e. $x$ is a solution of (2);

2. if $F$ is $C^1$, we have
   $$
   v(t)=-\nabla F(x(t)) \qquad \text{for all } t,
   $$
   i.e. $x$ is a solution of (1).

Proof. By the above analysis, $\tilde x^\tau$ is continuous, equicontinuous and equibounded. By applying Arzelà–Ascoli theorem, we get a uniformly converging subsequence (still denoted by $\tau$)
$$
\tilde x^{\tau}\to x
\qquad \text{as } \tau\to 0
$$

By (8), $x^\tau$ also uniformly converges to the same limit $x$. Moreover by (6), we know that $v^\tau$ is equibounded in $L^2$ which is a reflexive Banach space. Thus, up to an extra subsequence extraction (still denoted by $\tau$),
$$
v^\tau \rightharpoonup v \qquad \text{in } L^2 \quad \text{as } \tau\to 0.
$$

Therefore, as a consequence of distributional convergence,
$$
x’=v.
$$

To prove (1), fix a point $y\in \mathbb{R}^n$. By $v^\tau(t)\in -\partial F(x^\tau(t))$, we have
$$
F(y)\ge F(x^\tau(t)) - v^\tau(t)\cdot (y-x^\tau(t)) + \frac{\lambda}{2}|y-x^\tau(t)|^2.
$$

For all positive measurable function $a:[0,T]\to \mathbb{R}_+$,
$$
\int_0^T a(t)\left( F(y)-F(x^\tau(t))+v^\tau(t)\cdot (y-x^\tau(t))-\frac{\lambda}{2}|y-x^\tau(t)|^2 \right)\ dt \ge 0.
$$

Since $x^\tau\to x$ uniformly and hence $L^2$ strong, $v^\tau\rightharpoonup v$, and $F$ is lower-semicontinuous, by letting $\tau\to 0$, we get
$$
\int_0^T a(t)\left( F(y)-F(x(t))+v(t)\cdot (y-x(t))-\frac{\lambda}{2}|y-x(t)|^2 \right)\ dt \ge 0.
$$

From the arbitrariness of $a$, we get
$$
F(y)\ge F(x(t)) - v(t)\cdot (y-x(t)) + \frac{\lambda}{2}|y-x(t)|^2
\qquad \text{for a.e. } t.
$$

i.e. there exists a negligible set $N_y$ such that
$$
G_y(t):=F(y)-F(x(t))+v(t)\cdot (y-x(t))-\frac{\lambda}{2}|y-x(t)|^2 \ge 0
\qquad \text{for } t\notin N_y.
$$

Let $D$ be a countable dense set in the interior of $\operatorname{dom}F$, where $F$ is continuous, and
$$
N=\bigcup_{y\in D}N_y
$$
which is also negligible. For all $t\notin N$, and all $y\in \operatorname{dom}F$, there exists $y_m\in D$ such that $y_m\to y$ as $m\to\infty$ and then
$$
G_y(t)=\lim_{m\to\infty} G_{y_m}(t)\ge 0
\qquad \text{for all } t\notin N.
$$
Hence,
$$
v(t)\in -\partial F(x(t)).
$$

To prove (2), we have
$$
-\nabla F(x^\tau(t))=v^\tau(t)=(\tilde x^\tau)’(t).
$$
Since $x^\tau\to x$ uniformly and $F\in C^1$, we get, $v^\tau\rightharpoonup v$, we get
$$
v(t)=-\nabla F(x(t)) \qquad \text{a.e. } t.
$$
Since $t\mapsto -\nabla F(x(t))$ is uniformly continuous, the above equality holds for all $t$. $\square$

Remark. If the sequence $x_k^\tau$ is defined by

$$
x_{k+1}^\tau \in \operatorname*{argmin} _x \left\{ 2F\left(\frac{x+x _k^\tau}{2}\right)+\frac{|x-x _k^\tau |^2}{2\tau} \right\},
$$

then we have

$$
\frac{x_{k+1}^\tau-x_k^\tau}{\tau}
=
-\nabla F\left(\frac{x_{k+1}^\tau+x_k^\tau}{2}\right),
$$

and the convergence is of order $\tau^2$.

Reference

Santambrogio, F. {Euclidean, metric, and Wasserstein} gradient flows: an overview. Bull. Math. Sci. 7, 87–154 (2017).

The cover image in this article was taken in Hobbiton, New Zealand.

Definition 2.1 Given a nondecreasing and right continuous function $F:\mathbb{R}\to[0,1]$, its pseudo-inverse is the function $F^{[-1]}:[0,1]\to\overline{\mathbb{R}}$ given by
$$
F^{[-1]}(x)=\inf\{t\in\mathbb{R}\mid F(t)\ge x\}.
$$
Thanks to right continuity of $F$, the infimum is minimum as soon as the set is nonempty.

Lemma 2.1 We have

1. $F^{[-1]}(x)\le a \iff F(a)\ge x$,
2. $F^{[-1]}(x)>a \iff F(a)<x$.

Proof : We only prove (1). If $F^{[-1]}(x)\le a$, then the set
$$
\{t\in\mathbb{R}\mid F(t)\ge x\}\neq\varnothing
$$
and then the infimum is minimum. Suppose $F(a)<x$, then for each $b\le a$, $F(b)\le F(a)<x$, hence
$$
b\notin\{t\in\mathbb{R}\mid F(t)\ge x\}\Rightarrow (-\infty,a]\subset\{t\in\mathbb{R}\mid F(t)<x\}.
$$
Hence by definition, $F^{[-1]}(x)>a$, which gives a contradiction. Conversely, if $F(a)\ge x$, it is obvious that $F^{[-1]}(x)\le a$. $\square$

Proposition 2.1 If $\mu\in\mathcal{P}(\mathbb{R})$ and $F_\mu^{[-1]}$ is the pseudo-inverse of its cumulative distribution function $F_\mu$, then

$$
(F_\mu^{[-1]}) _{\sharp}\big(\mathcal{L}^1\llcorner[0,1]\big) =\mu.
$$

Moreover, given $\mu,\nu\in\mathcal{P}(\mathbb{R})$, if we set
$$
\eta:=\big(F_\mu^{[-1]},F_\nu^{[-1]}\big)_{\sharp}\big(\mathcal{L}^1\llcorner[0,1]\big),
$$
then $\eta\in\Pi(\mu,\nu)$ and
$$
\eta\big((-\infty,a]\times(-\infty,b]\big)=F_\mu(a)\wedge F_\nu(b).
$$

Proof :
$$
\mathcal{L}^1\big(\{x\in[0,1]\ ;\ F_\mu^{[-1]}(x)\le a\}\big)
=\mathcal{L}^1\big(\{x\in[0,1]\ ;\ x\le F_\mu(a)\}\big)=F_\mu(a).
$$
Hence
$$
\{(-\infty,a)\ ;\ a\in\mathbb{R}\}\subset \mathcal{G}:=\{A\in\mathcal{B}(\mathbb{R})\ ;\ (\mathcal{L}^1\llcorner[0,1])((F_\mu^{[-1]})^{-1}(A))=\mu(A)\}.
$$
By monotone class theorem, we get $\mathcal{G}=\mathcal{B}(\mathbb{R})$, which gives the desired result. Moreover $\eta\in\Pi(\mu,\nu)$ is just a consequence of the above statement.

Further,
$$
\eta((-\infty,a]\times(-\infty,b])
=\mathcal{L}^1\big(\{x\in[0,1]\ ;\ F_\mu^{[-1]}(x)\le a,\ F_\nu^{[-1]}(x)\le b\}\big)
$$
and by Lemma 2.1 this equals
$$
\mathcal{L}^1\big(\{x\in[0,1]\ ;\ x\le F_\mu(a),\ x\le F_\nu(b)\}\big)
=F_\mu(a)\wedge F_\nu(b),
$$
which is the desired equality. $\square$

Definition 2.2 We will call the transport plan

$$
\eta:=\big(F_\mu^{[-1]},F_\nu^{[-1]}\big) _{\sharp}\big(\mathcal{L}^1\llcorner[0,1]\big)
$$

the co-monotone transport plan between $\mu$ and $\nu$ and denote it by
$$
\gamma_{\mathrm{mon}}:=\eta.
$$

Lemma 2.2 If $\mu\in\mathcal{P}(\mathbb{R})$ is atomless, then $(F_\mu)_{\sharp}\mu=\mathcal{L}^1\llcorner[0,1]$. As a consequence, for every $\ell\in[0,1]$, the set $\{x:F_\mu(x)=\ell\}$ is $\mu$-negligible.

Proof : First, $F_\mu$ is continuous since $\mu$ is atomless. Hence for $a\in(0,1)$, the set
$$
\{x\ ;\ F_\mu(x)\le a\}=(-\infty,x_a],
$$
where $F_\mu(x_a)=a$. Hence
$$
\mu\big(F_\mu^{-1}((-\infty,a])\big)=\mu((-\infty,x_a])=F_\mu(x_a)=a.
$$
Therefore

$$
(F_\mu) _{\sharp}\mu=\mathcal{L}^1\llcorner[0,1].
$$

As a consequence, for $\ell\in[0,1]$, the set
$$
\{x\ ;\ F_\mu(x)=\ell\}
$$
is $\mu$-negligible. Otherwise, if
$$
\mu(\{x\ ;\ F_\mu(x)=\ell\})>0,
$$
then
$$
0=\mathcal{L}^1(\{\ell\})=(F_\mu)_{\sharp}\mu(\{\ell\})>0,
$$
which is a contradiction. $\square$

Theorem 2.1 Given $\mu,\nu\in\mathcal{P}(\mathbb{R})$, suppose that $\mu$ is atomless. Then there exists a unique nondecreasing map $T_{\mathrm{mon}}:\mathbb{R}\to\mathbb{R}$ such that
$$
(T_{\mathrm{mon}})_{\sharp}\mu=\nu.
$$

Proof : Define
$$
T_{\mathrm{mon}}(x):=F_\nu^{[-1]}(F_\mu(x)).
$$
By Lemma 2.2, it is easy to see that $T_{\mathrm{mon}}$ is well-defined. The fact that $T_{\mathrm{mon}}$ is monotone nondecreasing is obvious. Now, we only need to show $(T_{\mathrm{mon}})_{\sharp}\mu=\nu$.

In fact,

$$
(T_{\mathrm{mon}}) _{\sharp}\mu
=(F_\nu^{[-1]}\circ F_\mu) _{\sharp}\mu
=(F_\nu^{[-1]}) _{\sharp}\big((F_\mu) _{\sharp}\mu\big)
=(F_\nu^{[-1]}) _{\sharp}\big(\mathcal{L}^1\llcorner[0,1]\big)=\nu.
$$

Finally, we need to show the uniqueness. From monotonicity,
$$
T^{-1}((-\infty,T(x)])\supset (-\infty,x].
$$
We deduce
$$
F_\mu(x)=\mu((-\infty,x])\le \mu\big(T^{-1}((-\infty,T(x)])\big)
=\nu((-\infty,T(x)])=F_\nu(T(x)),
$$
which means
$$
T(x)\ge F_\nu^{[-1]}(F_\mu(x)).
$$
Suppose $T(x)>F_\nu^{[-1]}(F_\mu(x))$. This means there exists $\varepsilon_0>0$ such that
$$
F_\nu(T(x)-\varepsilon)\ge F_\mu(x)\qquad\text{for all }\varepsilon\in(0,\varepsilon_0).
$$
On the other hand, from
$$
T^{-1}((-\infty,T(x)-\varepsilon])\subset (-\infty,x],
$$
we get
$$
F_\nu(T(x)-\varepsilon)=\nu((-\infty,T(x)-\varepsilon])
=\mu\big(T^{-1}((-\infty,T(x)-\varepsilon])\big)\le \mu((-\infty,x])=F_\mu(x).
$$
Therefore
$$
F_\nu(T(x)-\varepsilon)=F_\mu(x)\qquad\text{for every }\varepsilon\in(0,\varepsilon_0).
$$
Thus $F_\nu(T(x)-\varepsilon)$ is the value that $F_\nu$ takes on an interval where it is constant. These intervals are countable, hence the values of $F_\nu$ on those intervals are also countable. We can call $\ell_i$ these values. As a consequence, the points $x$ where
$$
T(x)>F_\nu^{[-1]}(F_\mu(x))
$$
are contained in
$$
\bigcup_i \{x\ ;\ F_\mu(x)=\ell_i\},
$$
which is $\mu$-negligible. Thus, $T(x)=F_\nu^{[-1]}(F_\mu(x))$ $\mu$-a.e. $\square$

Lemma 2.3 Let $\gamma\in\Pi(\mu,\nu)$ be a transport plan between two measures $\mu,\nu\in\mathcal{P}(\mathbb{R})$. Suppose that it satisfies the property:
$$
(x,y),(x’,y’)\in\operatorname{Supp}(\gamma),\ x<x’ \Rightarrow y\le y’. \tag{1}
$$
Then we have $\gamma=\gamma_{\mathrm{mon}}$. In particular, there is a unique $\gamma$ satisfying (1). Moreover, if $\mu$ is atomless, then $\gamma=T_{\mathrm{mon}}$.

Proof : First, we need to prove
$$
\gamma((-\infty,a]\times(-\infty,b])=F_\mu(a)\wedge F_\nu(b).
$$
Consider
$$
A:=(-\infty,a]\times(b,+\infty),\qquad B:=(a,+\infty)\times(-\infty,b].
$$
Claim: It is not possible to have both $\gamma(A)>0$ and $\gamma(B)>0$.

Indeed, if not, $\gamma(A)>0$ and $\gamma(B)>0$, then for some $(x,y)\in A$ and $(x’,y’)\in B$ we have
$$
x<x’,\qquad y>y’,
$$
with $(x,y),(x’,y’)\in\operatorname{Supp}(\gamma)$, which contradicts (1). Therefore,
$$
\gamma((-\infty,a]\times(-\infty,b])
=\gamma\big(((-\infty,a]\times\mathbb{R})\setminus A\big)
\wedge
\gamma\big((\mathbb{R}\times(-\infty,b])\setminus B\big)
$$
and hence
$$
\gamma((-\infty,a]\times(-\infty,b])=F_\mu(a)\wedge F_\nu(b).
$$
Hence $\gamma=\gamma_{\mathrm{mon}}$.

Now, assume $\gamma$ is atomless, then we can define $I_x$ as the minimal interval such that
$$
\operatorname{Supp}(\gamma)\cap(\{x\}\times\mathbb{R})\subset \{x\}\times I_x.
$$
The assumption (1) shows that the interior of all these intervals are disjoint. There can be at most countable points such that $I_x$ is not singleton. Since $\mu$ is atomless, we can define a map $T$ such that $\gamma$ is concentrated on graph of $T$, $\mu$-a.e. By Theorem 2.1, $T=T_{\mathrm{mon}}$. $\square$

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken in Sydney, Australia.

Theorem 1.12 Let $\Omega \subset \mathbb{R}^d$ be compact and $c$ be $C^1$ cost function satisfying the twist condition on $\Omega \times \Omega$. Suppose that $\mu \in \mathcal{P}(\Omega)$ and $\varphi \in c\text{-cone}(\Omega)$ are given, that $\varphi$ is differentiable $\mu$-a.e. and that $\mu(\partial \Omega)=0$. Suppose that the map $T$ satisfies
$$
\nabla_x c(x,T(x))=\nabla \varphi(x).
$$
Then $T$ is optimal for the transport cost $c$ between the measures $\mu$ and
$$
\nu:=T_{\sharp}\mu.
$$

Proof : Since $\varphi \in c\text{-cone}(\Omega)$, $\varphi^{cc}=\varphi$, take $\psi=\varphi^c$ we have $\varphi=\psi^c$.

Fix $x_0\in \Omega$ such that $\nabla \varphi(x_0)$ exists and $x_0\notin \partial \Omega$. By compactness and continuity $\exists\ y_0\in \Omega$ such that
$$
\varphi(x_0)=\inf_y [c(x_0,y)-\psi(y)]=c(x_0,y_0)-\psi(y_0).
$$
This gives,
$$
\varphi(x)\leq c(x,y_0)-\psi(y_0)
$$
and
$$
\varphi(x_0)=c(x_0,y_0)-\psi(y_0).
$$
and hence
$$
x\mapsto c(x,y_0)-\varphi(x)
$$
is minimal at $x=x_0$. As a consequence, we get
$$
\nabla \varphi(x_0)=\nabla_x c(x_0,y_0).
$$
By assumption, $y\mapsto \nabla_x c(x,y)$ is injective, and $\nabla_x c(x_0,T(x_0))=\nabla \varphi(x_0)$. We have $y_0=T(x_0)$. This proves
$$
\varphi(x_0)+\psi(T(x_0))=c(x_0,T(x_0)),
$$
and this same equality is true for $\mu$-a.e. $x_0$.

Integrate with respect to $\mu$, we have
$$
\int_{\Omega} c(x,T(x))\ d\mu(x)
=
\int_{\Omega} \varphi(x)+\psi(T(x))\ d\mu(x)
=
\int_{\Omega} \varphi\ d\mu+\int_{\Omega} \psi\ d\nu.
$$
Hence
$$
\max(DP)\geq \int_{\Omega}\varphi\ d\mu+\int_{\Omega}\psi\ d\nu
=
\int_{\Omega} c(x,T(x))\ d\mu(x)
\geq \min(KP)\geq \max(DP).
$$
We have $T$ is optimal. $\square$

Theorem 1.13 Suppose $\mu\in \mathcal{P}(\mathbb{R}^d)$ is such that
$$
\int_{\mathbb{R}^d} |x|^2\ d\mu(x)<+\infty,
$$
and that $u:\mathbb{R}^d\to \mathbb{R}\cup\{+\infty\}$ is convex and differentiable $\mu$-a.e.

Set
$$
T=\nabla u
$$
and suppose
$$
\int_{\mathbb{R}^d} |T(x)|^2\ d\mu(x)<+\infty.
$$
Then $T$ is optimal for the transport cost
$$
c(x,y):=\frac{1}{2}|x-y|^2
$$
between the measures $\mu$ and
$$
\nu:=T_{\sharp}\mu.
$$

Proof : Note that for a convex function $u$
$$
u(x)+u^{\star}(y)\geq x\cdot y
$$
for all $x,y\in \mathbb{R}^d$ and
$$
u(x)+u^{\star}(y)=x\cdot y
$$
if $y=\nabla u(x)$.

Now consider $\gamma\in \Pi(\mu,\nu)$
$$
\int_{\mathbb{R}^d\times \mathbb{R}^d} x\cdot y\ d\gamma(x,y)
\leq
\int_{\mathbb{R}^d\times \mathbb{R}^d} (u(x)+u^{\star}(y))\ d\gamma(x,y)
=
\int_{\mathbb{R}^d} u(x)\ d\mu(x)+\int_{\mathbb{R}^d} u^{\star}(T(x))\ d\mu(x)=
\int_{\mathbb{R}^d} x\cdot T(x)\ d\mu(x)
=
\int_{\mathbb{R}^d\times \mathbb{R}^d} x\cdot y\ d\gamma_T.
$$

Therefore,
$$
\int_{\mathbb{R}^d\times \mathbb{R}^d} \frac{1}{2}|x-y|^2\ d\gamma(x,y)
=
\int_{\mathbb{R}^d\times \mathbb{R}^d} \frac{1}{2}(|x|^2+|y|^2)\ d\gamma(x,y)
-
\int_{\mathbb{R}^d\times \mathbb{R}^d} x\cdot y\ d\gamma(x,y)=
\int_{\mathbb{R}^d\times \mathbb{R}^d} \frac{1}{2}(|x|^2+|y|^2)\ d\gamma_T
-
\int_{\mathbb{R}^d\times \mathbb{R}^d} x\cdot y\ d\gamma_T=
\int_{\mathbb{R}^d\times \mathbb{R}^d} \frac{1}{2}|x-y|^2\ d\gamma_T,
$$

which shows $T$ is optimal. $\square$

Theorem 1.14 Suppose that $\gamma\in \mathcal{P}(X\times Y)$ is given, that $X$ and $Y$ are Polish spaces, that $c:X\times Y\to \mathbb{R}$ is uniformly continuous and bounded, and that $\operatorname{supp}(\gamma)$ is $c$-CM.

Then $\gamma$ is an optimal transport plan between its marginals
$$
\mu=(\pi_x) _{\sharp}\gamma
\quad \text{and} \quad
\nu=(\pi_y) _{\sharp}\gamma
$$
for the cost $c$.

Proof : By Theorem 1.6, there exists a $c$-concave function $\varphi$ such that
$$
\operatorname{supp}(\gamma)\subset \{(x,y):\varphi(x)+\varphi^c(y)=c(x,y)\}
$$
and both $\varphi$ and $\varphi^c$ are continuous and bounded.

By Theorem 1.8, we know the duality result, hence
$$
\min(KP)\leq \int_{X\times Y} c(x,y)\ d\gamma
=
\int_{X\times Y} \varphi(x)+\varphi^c(y)\ d\gamma
=
\int_X \varphi\ d\mu+\int_Y \varphi^c\ d\nu\leq \max(DP)\leq \min(KP),
$$

which shows that $\gamma$ is optimal. $\square$

Definition 1.9 (Hausdorff distance) For a compact metric space $X$, we define the Hausdorff distance on pair of compact subsets of $X$ by
$$
d_H(A,B):=
\max\left\{
\max\{d(x,A):x\in B\},
\max\{d(x,B):x\in A\}
\right\}.
$$

Remark : We have the following equivalent definition.

1. $d_H(A,B)=\max\{|d(x,A)-d(x,B)|:x\in X\}$.

2. $d_H(A,B)=\inf\{\varepsilon>0:A\subset B_{\varepsilon},\ A_{\varepsilon}\supset B\}$, where $A_{\varepsilon},B_{\varepsilon}$ stands for the $\varepsilon$-neighborhood of $A$ and $B$ respectively.

Theorem 1.15 (Blaschke) $d_H$ is a distance.

Proposition 1.5 If $d_H(A_n,A)\to 0$ and $\mu_n$ is a sequence of positive measures such that
$$
\operatorname{supp}(\mu_n)\subset A_n
$$
with $\mu_n\rightharpoonup \mu$, then
$$
\operatorname{supp}(\mu)\subset A.
$$

Theorem 1.16 Suppose that $X$ and $Y$ are compact metric spaces and that $c:X\times Y\to \mathbb{R}$ is continuous. Suppose that $\gamma_n\in \mathcal{P}(X\times Y)$ is a sequence of transport plan which are optimal for cost $c$ between their own marginals

$$
\mu_n=(\pi_x) _{\sharp}\gamma_n
\quad \text{and} \quad
\nu_n=(\pi_y) _{\sharp}\gamma_n
$$
and suppose $\gamma_n\rightharpoonup \gamma$. Then

$$
\mu_n \to \mu_{\infty}:=(\pi_x) _{\sharp} \gamma,
\quad
\nu_n \to \nu:= (\pi_y) _{\sharp} \gamma,
$$

and $\gamma$ is optimal in the transport between $\mu$ and $\nu$.

Proof : Set $\Gamma_n:=\operatorname{supp}(\gamma_n)$, up to subsequence, we can assume $\Gamma_n\to \Gamma$ in Hausdorff distance. Each $\Gamma_n$ is a $c$-CM set, now we claim $\Gamma$ is also a $c$-CM set.

Fix $(x_1,y_1),\dots,(x_k,y_k)\in \Gamma$, there are points $(x_1^n,y_1^n),\dots,(x_k^n,y_k^n)\in \Gamma_n$ such that
$$
(x_i^n,y_i^n)\to (x_i,y_i),\qquad i=1,2,\dots,k.
$$
The cyclical monotonicity of $\Gamma_n$ gives
$$
\sum_{i=1}^k c(x_i^n,y_i^n)\leq \sum_{i=1}^k c(x_i^n,y_{\sigma(i)}^n).
$$
Take $n\to\infty$
$$
\sum_{i=1}^k c(x_i,y_i)\leq \sum_{i=1}^k c(x_i,y_{\sigma(i)}).
$$
Hence $\Gamma$ is a $c$-CM set. Since $\gamma_n\rightharpoonup \gamma$ and $\Gamma_n\to \Gamma$ in Hausdorff distance, we have
$$
\operatorname{supp}(\gamma)\subset \Gamma,
$$
which is also a $c$-CM set and implies the optimality of $\gamma$. $\square$

Notation : For a given cost $c:X\times Y\to \mathbb{R}$ and $\mu\in \mathcal{P}(X)$, $\nu\in \mathcal{P}(Y)$, define
$$
J_c(\mu,\nu):=
\min\left\{
\int_{X\times Y} c(x,y)\ d\gamma;\ \gamma\in \Pi(\mu,\nu)
\right\}.
$$

Theorem 1.17 Suppose that $X$ and $Y$ are compact metric spaces and that $c:X\times Y\to \mathbb{R}$ is continuous. Suppose that $\mu_n\in \mathcal{P}(X)$, $\nu_n\in \mathcal{P}(Y)$ with
$$
\mu_n\to \mu
\qquad \text{and} \qquad
\nu_n\to \nu.
$$
Then we have
$$
J_c(\mu_n,\nu_n)\to J_c(\mu,\nu).
$$

Proof : Let $\gamma_n$ be an optimal transport plan from $\mu_n$ to $\nu_n$ for the cost $c$. Up to subsequences, we can assume
$$
\gamma_n\rightharpoonup \gamma.
$$
By Theorem 1.16, $\gamma$ is optimal for $\mu$ to $\nu$. This means
$$
J_c(\mu_n,\nu_n)=\int_{X\times Y} c(x,y)\ d\gamma_n
\to
\int_{X\times Y} c\ d\gamma
=
J_c(\mu,\nu),
$$
which proves the claim.

Theorem 1.18 Suppose that $X$ and $Y$ are compact metric spaces, and that $c:X\times Y\to \mathbb{R}$ is continuous. Suppose that $\mu_n\in \mathcal{P}(X)$ and $\nu_n\in \mathcal{P}(Y)$, with
$$
\mu_n\to \mu
\qquad \text{and} \qquad
\nu_n\to \nu.
$$
Let $(\varphi_n,\psi_n)$ be a pair of $c$-concave Kantorovich potentials for the cost $c$ in the transport from $\mu_n$ to $\nu_n$. Then up to a subsequences, we have
$$
\varphi_n\to \varphi
\qquad \text{and} \qquad
\psi_n\to \psi
$$
where convergence is uniform and $(\varphi,\psi)$ is a pair of Kantorovich potentials for $\mu$ and $\nu$.

Proof : It is easy to see that $\{\varphi_n\}$ and $\{\psi_n\}$ are equibounded and equicontinuity.

By Arzelà–Ascoli theorem, we have
$$
\varphi_n\to \varphi,\qquad \psi_n\to \psi
$$
up to a subsequence, and the convergence is uniform.

Since
$$
\varphi_n(x)+\psi_n(y)\leq c(x,y),
$$
we have
$$
\varphi(x)+\psi(y)\leq c(x,y).
$$
Moreover
$$
J_c(\mu_n,\nu_n)
=
\int_X \varphi_n\ d\mu_n+\int_Y \psi_n\ d\nu_n
\to
\int_X \varphi\ d\mu+\int_Y \psi\ d\nu.
$$
By Theorem 1.17, we also have
$$
J_c(\mu_n,\nu_n)\to J_c(\mu,\nu).
$$
Hence
$$
J_c(\mu,\nu)=\int_X \varphi\ d\mu+\int_Y \psi\ d\nu,
$$
which implies they are Kantorovich potentials. $\square$

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken in Sydney, Australia.

Theorem 1.8 Suppose that $X$ and $Y$ are Polish spaces, and that $c:X \times Y \to \mathbb{R}$ is uniformly continuous and bounded. Then the problem $(\mathrm{DP})$ admits a solution $(\varphi,\varphi^c)$ and we have
$$
\max (\mathrm{DP}) = \min (\mathrm{KP}).
$$

Proof. Consider the minimization problem $(\mathrm{KP})$. Since $c$ is continuous, it admits a solution $\gamma$. By Theorem 1.7, the support $\Gamma=\operatorname{supp}(\gamma)$ is $c$-cyclically monotone. Since $c$ is real-valued, we can apply Theorem 1.6 and obtain the existence of a $c$-concave function $\varphi$ such that
$$
\Gamma \subset \{(x,y)\in X\times Y:\ \varphi(x)+\varphi^c(y)=c(x,y)\}.
$$

Since $c$ is continuous, we have $\varphi$ and $\varphi^c$ are continuous. Moreover, from the global upper bound on $\varphi^c$, which turns into a lower bound on $\varphi$,

$$
\varphi^c(y)=\inf_{x\in X}[c(x,y)-\varphi(x)].
$$

Symmetrically, we can also obtain upper bounds on $\varphi$ and lower bounds on $\varphi^c$, which proves that $\varphi$ and $\varphi^c$ are both continuous and bounded.

Hence, $(\varphi,\varphi^c)$ can be used as an admissible pair in $(\mathrm{DP})$. Consider now

$$
\int_X \varphi\ d\mu + \int_Y \varphi^c\ d\nu
= \int_{X\times Y} (\varphi(x)+\varphi^c(y))\ d\gamma
= \int_{\operatorname{supp}(\gamma)} (\varphi(x)+\varphi^c(y))\ d\gamma= \int_{\operatorname{supp}(\gamma)} c(x,y)\ d\gamma
= \int_{X\times Y} c(x,y)\ d\gamma.
$$

Hence

$$
(\mathrm{DP})\ge \int_X \varphi\ d\mu+\int_Y \varphi^c\ d\nu
= \int_{X\times Y} c(x,y)\ d\gamma
= (\mathrm{KP}).
$$

Since we already saw $(\mathrm{DP})\le (\mathrm{KP})$, we get $(\mathrm{DP})=(\mathrm{KP})$ and $(\varphi,\varphi^c)$ is optimal for $(\mathrm{DP})$. $\square$

Theorem 1.9 Let $\mu$ and $\nu$ be probability measures over $\mathbb{R}^d$, and $c(x,y):=\frac12|x-y|^2$. Suppose
$$
\int_X |x|^2\ \mu(dx)<+\infty
\qquad\text{and}\qquad
\int_Y |y|^2\ \nu(dy)<+\infty.
$$

Consider the following variant of $(\mathrm{DP})$:

$$
(\mathrm{DP-var})\qquad
\sup\left\{
\int_{\mathbb{R}^d}\varphi\ d\mu+\int_{\mathbb{R}^d}\psi\ d\nu
:\ \varphi\in L^1(\mu),\ \psi\in L^1(\nu),\ \varphi\oplus\psi\le c
\right\}.
$$

Then $(\mathrm{DP-var})$ admits a solution $(\varphi,\psi)$, and the functions

$$
x\mapsto \frac12|x|^2-\varphi(x)
\qquad\text{and}\qquad
y\mapsto \frac12|y|^2-\psi(y)
$$

are convex and conjugate to each other for the Legendre transform. Moreover, we have

$$
\max (\mathrm{DP-var})=\min (\mathrm{KP}).
$$

Proof. Since $c$ is continuous, by Theorem 1.7 the support $\Gamma=\operatorname{supp}(\gamma)$ is $c$-cyclically monotone and then by Theorem 1.6, we get the existence of a pair $(\varphi,\psi)$ with $\psi=\varphi^c$ such that

$$
\varphi(x)+\psi(y)=c(x,y)
\qquad\text{for all }(x,y)\in \Gamma.
$$

From Proposition 1.2, we know

$$
x\mapsto \frac12|x|^2-\varphi(x),\qquad
y\mapsto \frac12|y|^2-\psi(y)
$$

are convex and conjugate to each other, hence bounded from below by a linear function; hence $\varphi$ and $\psi$ are bounded by a second order polynomial. Since

$$
\int_X |x|^2\ d\mu<+\infty
\qquad\text{and}\qquad
\int_Y |y|^2\ d\nu<+\infty,
$$

we know $\varphi_+\in L^1(\mu)$ and $\psi_+\in L^1(\nu)$.

Since

$$
\int_{\mathbb{R}^d}\varphi\ d\mu+\int_{\mathbb{R}^d}\psi\ d\nu
=
\int_{\mathbb{R}^d\times\mathbb{R}^d} \varphi\oplus\psi\ d\gamma
=
\int_{\mathbb{R}^d\times\mathbb{R}^d} c(x,y)\ d\gamma
\ge 0,
$$

this proves

$$
\int_{\mathbb{R}^d}\varphi\ d\mu,\ \int_{\mathbb{R}^d}\psi\ d\nu >-\infty.
$$

Hence $\varphi\in L^1(\mu)$, $\psi\in L^1(\nu)$.

Therefore

$$
\sup(\mathrm{DP-var})\ge
\int_{\mathbb{R}^d}\varphi\ d\mu+\int_{\mathbb{R}^d}\psi\ d\nu
=
\int_{\mathbb{R}^d\times\mathbb{R}^d} c(x,y)\ d\gamma
=
\min(\mathrm{KP}).
$$

However, by $\varphi\oplus\psi\le c$, we know $\sup(\mathrm{DP-var})\le \min(\mathrm{KP})$. We get

$$
\max(\mathrm{DP-var})=\min(\mathrm{KP})
$$

and $(\varphi,\psi)$ is the solution of $(\mathrm{DP-var})$. $\square$

Lemma 1.8 Suppose that $c_k$ and $c$ are lower semi-continuous, and bounded from below and that $c_k$ converges increasingly to $c$. Then
$$
\lim_{k\to\infty}\min\left\{\int_{X\times Y} c_k\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}
=
\min\left\{\int_{X\times Y} c\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}.
$$

Proof. Since $c_k\le c$, it is obvious that $LHS\le RHS$.

Now consider a sequence $\gamma_k\in\Pi(\mu,\nu)$ which is the minimizer for each cost $c_k$.

Up to subsequences, due to the tightness of $\Pi(\mu,\nu)$, we can suppose

$$
\gamma_k \rightharpoonup \overline{\gamma}.
$$

Fix now an index $j$. Since for $k\ge j$ we have $c_k\ge c_j$, and then

$$
\lim_k \min\left\{\int_{X\times Y} c_k\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}
=
\lim_k \int_{X\times Y} c_k\ d\gamma_k
\ge
\liminf_k \int_{X\times Y} c_j\ d\gamma_k.
$$

Since $c_j$ is lower semi-continuous, and $\gamma_k\rightharpoonup \overline{\gamma}$,

$$
\liminf_k \int_{X\times Y} c_j\ d\gamma_k
\ge
\int_{X\times Y} c_j\ d\overline{\gamma}.
$$

Hence, we obtain

$$
\lim_k \min\left\{\int_{X\times Y} c_k\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}
\ge
\int_{X\times Y} c_j\ d\overline{\gamma}.
$$

Since $j$ is arbitrary and

$$
\lim_{j\to\infty}\int_{X\times Y} c_j\ d\overline{\gamma}
=
\int_{X\times Y} c\ d\overline{\gamma}
$$

by monotone convergence theorem, we also have

$$
\lim_k \min\left\{\int_{X\times Y} c_k\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}
\ge
\int_{X\times Y} c\ d\overline{\gamma}
\ge
\min\left\{\int_{X\times Y} c\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}.
$$

This completes the proof and $\overline{\gamma}$ is optimal for cost $c$. $\square$

Theorem 1.10 If $X,Y$ are Polish spaces and $c:X\times Y\to \mathbb{R}\cup\{+\infty\}$ is lower semi-continuous and bounded from below, then the duality formula
$$
\min(\mathrm{KP})=\sup(\mathrm{DP})
$$

holds.

Proof. Consider a sequence $c_k$ of bounded $k$-Lipschitz functions approaching $c$ increasingly. Then the dual result holds for $c_k$ and then

$$
\min\left\{\int_{X\times Y} c_k\ d\gamma:\ \gamma\in\Pi(\mu,\nu)\right\}
=
\max\left\{
\int_X \varphi\ d\mu+\int_Y \psi\ d\nu:\ \varphi\in C_b(X),\ \psi\in C_b(Y),\ \varphi\oplus\psi\le c_k
\right\}\le
\sup\left\{
\int_X \varphi\ d\mu+\int_Y \psi\ d\nu:\ \varphi\oplus\psi\le c
\right\}.
$$

Let $k\to\infty$, by Lemma 1.8, we get the desired result. $\square$

Remark: For the cost $c$, we can not guarantee the existence of a maximizing pair $(\varphi,\psi)$.

Theorem 1.11 If $c$ is lower semi-continuous and $\gamma$ is an optimal transport plan, then $\gamma$ is concentrated on a $c$-CM set $\Gamma$.

Proof. By Theorem 1.10, we can take a sequence of maximizing pairs $(\varphi_n,\psi_n)$ in the dual problem, we have

$$
\int_{X\times Y} (\varphi_n(x)+\psi_n(y))\ d\gamma
=
\int_X \varphi_n(x)\ d\mu+\int_Y \psi_n(y)\ d\nu
\to
\int_{X\times Y} c\ d\gamma,
$$

where $\gamma$ is optimal for $c$.

Moreover, since

$$
f_n(x,y):=c(x,y)-\varphi_n(x)-\psi_n(y)\ge 0,
$$

we have

$$
\int_{X\times Y} |f_n(x,y)|\ d\gamma
=
\int_{X\times Y} f_n(x,y)\ d\gamma
\to 0
\qquad\text{as }n\to\infty,
$$

that is $f_n\to 0$ in $L^1(X\times Y,\gamma)$. Therefore, up to a subsequence, they also converge pointwisely $\gamma$-a.e. to $0$. Let $\Gamma\subset X\times Y$ be the set with $\gamma(\Gamma)=1$, where the convergence holds.

Take any $k$, $\sigma$ and $(x_1,y_1),\cdots,(x_k,y_k)\in \Gamma$, we have

$$
\sum_{i=1}^k c(x_i,y_i)
=
\lim_{n\to\infty}\sum_{i=1}^k(\varphi_n(x_i)+\psi_n(y_i))
=
\lim_{n\to\infty}\sum_{i=1}^k(\varphi_n(x_i)+\psi_n(y_{\sigma(i)}))
\le
\sum_{i=1}^k c(x_i,y_{\sigma(i)}),
$$

which proves that $\Gamma$ is $c$-CM. $\square$

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken from the Luzern–Interlaken Express while passing Lake Lungern in Switzerland.

Proposition 1.3 Let $c:X \times Y \to \mathbb{R}$. For $\varphi:X \to \mathbb{R} \cup \{-\infty\}$, we have
$$
\varphi^{c\bar c} \ge \varphi.
$$

Moreover, the equality holds if and only if $\varphi$ is $c$-concave. In general, $\varphi^{c\bar c}$ is the smallest $c$-concave function that is larger than $\varphi$.

Proof : First, we show $\varphi^{c\bar c} \ge \varphi$. For any $x \in X$, $y \in Y$,
$$
\varphi^c(y)=\inf_{x\in X}[c(x,y)-\varphi(x)] \le c(x,y)-\varphi(x).
$$

Therefore,
$$
\varphi^{c\bar c}(x)=\inf_{y\in Y}[c(x,y)-\varphi^c(y)]
\ge \inf_{y\in Y}[c(x,y)-(c(x,y)-\varphi(x))]=\varphi(x).
$$

Then, let us prove $\varphi^{c\bar c}=\varphi$ if and only if $\varphi$ is $c$-concave. It is obvious that if $\varphi^{c\bar c}=\varphi$ then $\varphi=(\varphi^c)^{\bar c}$, which shows that $\varphi$ is $c$-concave. If $\varphi$ is $c$-concave, then there exists a function $\zeta:Y\to\mathbb{R}\cup\{-\infty\}$ such that
$$
\varphi=\zeta^{\bar c}.
$$

Hence
$$
\varphi^c=\zeta^{\bar c c}\ge \zeta,
$$
and then
$$
\varphi^{c\bar c}\le \zeta^{\bar c}=\varphi,
$$
which implies $\varphi^{c\bar c}=\varphi$.

Finally, if $\psi\ge \varphi$ and $\psi$ is $c$-concave, we need to show $\psi\ge \varphi^{c\bar c}$. Suppose $\psi=\eta^{\bar c}$, then
$$
\psi=\eta^{\bar c}\ge \varphi
\Rightarrow \eta^{\bar c c}\le \varphi^c
\Rightarrow \eta\le \varphi^c
\Rightarrow \psi=\eta^{\bar c}\ge \varphi^{c\bar c}.\quad \square
$$

Recall the sub-differential for convex functions

Definition 1.5 For every convex function $f:\mathbb{R}^d\to\mathbb{R}\cup\{+\infty\}$, we define its subdifferential at $x$ as the set
$$
\partial f(x):=\{p\in\mathbb{R}^d:\ f(y)\ge f(x)+p\cdot (y-x),\ \forall y\in\mathbb{R}^d\}.
$$

Proposition 1.4 For every convex function $f:\mathbb{R}^d\to\mathbb{R}$, we have

(1) $\partial f(x)\neq \emptyset$ if $x\in \operatorname{int}(\operatorname{dom} f)$.

(2) If $f$ is differentiable at $x$, then $\partial f(x)=\{\nabla f(x)\}$.

(3) $p\in \partial f(x)$ if and only if $x\in \partial f^\star(p)$ if and only if
$$
f(x)+f^\star(p)=x\cdot p.
$$

(4) If $p_1\in \partial f(x_1)$, $p_2\in \partial f(x_2)$, then
$$
(p_1-p_2)\cdot (x_1-x_2)\ge 0.
$$

Definition 1.6 Let us define the graph of the subdifferential of a convex function as
$$
\operatorname{Graph}(\partial f):=\{(x,p):\ p\in \partial f(x)\}
=\{(x,p):\ f(x)+f^\star(p)=x\cdot p\}.
$$

Definition 1.7 A set $A\subset \mathbb{R}^d\times \mathbb{R}^d$ is said to be cyclically monotone if for every $k\in \mathbb{N}$, every permutation $\sigma$ and every finite family of points $(x_1,p_1),\ldots,(x_k,p_k)\in A$, we have
$$
\sum_{i=1}^k x_i\cdot p_i \ge \sum_{i=1}^k x_i\cdot p_{\sigma(i)}.
$$

Note that if we take $k=2$, we get the usual definition of monotonicity
$$
x_1\cdot p_1+x_2\cdot p_2\ge x_1\cdot p_2+x_2\cdot p_1
\Leftrightarrow
(p_1-p_2)\cdot (x_1-x_2)\ge 0.
$$

By Rockafellar theorem, every cyclically monotone set is contained in the graph of the subdifferential of a convex function.

c-Cyclical Monotonicity

Definition 1.8 Once a function $c:X\times Y\to \mathbb{R}\cup\{+\infty\}$ is given, we say that a set $\Gamma\subset X\times Y$ is $c$-cyclically monotone (briefly $c$-CM) if for every $k\in\mathbb{N}$, every permutation $\sigma$ and every finite family of points $(x_1,y_1),\ldots,(x_k,y_k)\in \Gamma$, we have
$$
\sum_{i=1}^k c(x_i,y_i)\le \sum_{i=1}^k c(x_i,y_{\sigma(i)}).
$$

Remark : Take $c(x,y)=\frac12|x-y|^2$, we get $\Gamma$ is cyclically monotone.

Theorem 1.6 If $\Gamma\neq \emptyset$ is a $c$-CM set in $X\times Y$ and $c:X\times Y\to\mathbb{R}$ (note that $c$ is required not to take the value $+\infty$), then there exists a $c$-concave function $\varphi:X\to \mathbb{R}\cup\{-\infty\}$ (different from the constant $-\infty$ function) such that
$$
\Gamma\subset \{(x,y)\in X\times Y:\ \varphi(x)+\varphi^c(y)=c(x,y)\}.
$$

Proof : Let us fix a point $(x_0,y_0)\in \Gamma$. For $x\in X$ set
$$
\varphi(x)=\inf \Big\{
c(x,y_n)-c(x_n,y_n)+c(x_n,y_{n-1})-c(x_{n-1},y_{n-1})
+\cdots +c(x_1,y_0)-c(x_0,y_0): n\in \mathbb{N},\ (x_i,y_i)\in \Gamma \text{ for all } i=1,2,\ldots,n
\Big\}.
$$

Since $c$ is real valued and $\Gamma$ is nonempty, $\varphi$ never takes the value $+\infty$.

If we set for $y\in Y$
$$
-\psi(y)=\inf \Big\{
-c(x_n,y)+c(x_n,y_n)-c(x_{n-1},y_n)+c(x_{n-1},y_{n-1})-\cdots +c(x_1,y_0)-c(x_0,y_0): n\in \mathbb{N},\ (x_i,y_i)\in \Gamma \text{ for all } i=1,2,\ldots,n,\ \text{and } y_n=y
\Big\}.
$$

By definition
$$
\psi(y)>-\infty
\Leftrightarrow
y\in (\pi_Y)(\Gamma)=\{y\in Y:\ \exists x\in X \text{ s.t. } (x,y)\in \Gamma\}.
$$

Next, we prove that $\varphi=\psi^c$, which implies $\varphi$ is $c$-concave. Indeed, if $y\notin (\pi_Y)(\Gamma)$, $\psi(y)=-\infty$, then $c(x,y)-\psi(y)=+\infty$. Hence, we only need to consider $y\in (\pi_Y)(\Gamma)$. For $y\in (\pi_Y)(\Gamma)$,

$$
\begin{aligned}
c(x,y)-\psi(y)
&=
c(x,y)+\inf \Big\{
-c(x_n,y)+c(x_n,y_n)-c(x_{n-1},y_n)+\cdots +c(x_1,y_0)-c(x_0,y_0):(x_i,y_i)\in \Gamma \text{ for all } i\ge 1,\ldots,n,\ \text{and } y_n=y
\Big\}\\\
&=
\inf \Big\{
c(x,y_n)-c(x_n,y_n)+c(x_n,y_{n-1})-c(x_{n-1},y_{n-1})+\cdots +c(x_1,y_0)-c(x_0,y_0):(x_i,y_i)\in \Gamma \text{ for all } i=1,2,\ldots,n,\ \text{and } y_n=y
\Big\}.
\end{aligned}
$$

Take the infimum of $y=y_n$, we get
$$
\inf_y[c(x,y)-\psi(y)]=\varphi(x),
$$
that is
$$
\varphi=\psi^c.
$$

Then, we claim: $\varphi(x_0)\ge 0$ and hence $\varphi \not\equiv -\infty$. In fact, for all $n\in \mathbb{N}$ and $(x_i,y_i)\in \Gamma$, $i=1,2,\ldots,n$,
$$
c(x_0,y_n)-c(x_n,y_n)+c(x_n,y_{n-1})-c(x_{n-1},y_{n-1})+\cdots +c(x_1,y_0)-c(x_0,y_0)=
\sum_{i=0}^n c(x_{i+1},y_i)-\sum_{i=0}^n c(x_i,y_i)\ge 0.
$$

Hence $\varphi(x_0)\ge 0$.

Now, we need to show $\varphi(x)+\varphi^c(y)\ge c(x,y)$ for $(x,y)\in \Gamma$. Since $\varphi=\psi^c$ and we have $\varphi^c=\psi^{c\bar c}\ge \psi$, we only need to show
$$
\varphi(x)+\psi(y)\ge c(x,y)
\qquad \text{for } (x,y)\in \Gamma.
$$

Fix $(x,y)\in \Gamma$. Since
$$
\varphi(x)=\psi^c(x)=\inf_{y\in (\pi_Y)(\Gamma)}[c(x,y)-\psi(y)],
$$
for all $\varepsilon>0$, there exists $\bar y\in (\pi_Y)(\Gamma)$ such that
$$
c(x,\bar y)-\psi(\bar y)<\varphi(x)+\varepsilon. \tag{1}
$$

Claim :
$$
-\psi(y)\le -c(x,y)+c(x,\bar y)-\psi(\bar y).
$$

In fact, since $\bar y\in (\pi_Y)(\Gamma)$, there exists a chain
$$
(x_1,y_1),\ldots,(x_n,y_n)\in \Gamma,
\qquad \text{with } y_n=\bar y.
$$

Since $(x,y)\in \Gamma$, we can add it to the chain above, and hence
$$
-\psi(y)\le -c(x,y)+c(x,\bar y)-c(x_n,\bar y)+c(x_n,y_n)-c(x_{n-1},y_n)+c(x_{n-1},y_{n-1})+\cdots +c(x_1,y_0)-c(x_0,y_0).
$$

Take the infimum, we get
$$
-\psi(y)\le -c(x,y)+c(x,\bar y)-\psi(\bar y). \tag{2}
$$

Therefore, combined by (1) and (2), we have
$$
-\psi(y)<-c(x,y)+\varphi(x)+\varepsilon,
$$
by the arbitrarity of $\varepsilon>0$, we get
$$
\varphi(x)+\psi(y)\ge c(x,y).
$$

Since $\varphi^c\ge \psi$, which completes the prove. $\square$

Theorem 1.7 If $\gamma$ is an optimal transport plan for the cost $c$ and $c$ is continuous, then $\operatorname{Supp}(\gamma)$ is a $c$-CM set.

Proof : Suppose by contradiction that there exists $k$, $\sigma$ and $(x_1,y_1),\ldots,(x_k,y_k)\in \operatorname{Supp}(\gamma)$ such that
$$
\sum_{i=1}^k c(x_i,y_i)>\sum_{i=1}^k c(x_i,y_{\sigma(i)}).
$$

Take
$$
0<\varepsilon<\frac{1}{2k}\left(\sum_{i=1}^k c(x_i,y_i)-\sum_{i=1}^k c(x_i,y_{\sigma(i)})\right).
$$

By continuity of $c$, there exists $r>0$ such that for all $i=1,2,\ldots,k$, we have
$$
c(x,y)>c(x_i,y_i)-\varepsilon
\qquad \text{for all } (x,y)\in B(x_i,r)\times B(y_i,r),
$$
and
$$
c(x,y)<c(x_i,y_{\sigma(i)})+\varepsilon
\qquad \text{for all } (x,y)\in B(x_i,r)\times B(y_{\sigma(i)},r).
$$

Now, consider
$$
V_i:=B(x_i,r)\times B(y_i,r).
$$

Note that since $(x_i,y_i)\in \operatorname{supp}(\gamma)$, we have
$$
\gamma(V_i)>0
\qquad \text{for all } i=1,\ldots,k.
$$

Define measures

$$
\gamma_i:=\gamma \llcorner V_i/\gamma(V_i)
\quad \text{and} \quad
\mu_i=(\pi_x)_\sharp \gamma_i,\ \nu_i=(\pi_y)_\sharp \gamma_i.
$$

Take
$$
\varepsilon_0<\frac{1}{k}\min_i \gamma(V_i).
$$

For every $i$, build a measure $\widetilde\gamma_i\in \Pi(\mu_i,\nu_{\sigma(i)})$ and define
$$
\widetilde\gamma:=\gamma-\varepsilon_0\sum_{i=1}^k \gamma_i+\varepsilon_0\sum_{i=1}^k \widetilde\gamma_i.
$$

Claim : $\widetilde\gamma\in \Pi(\mu,\nu)$ and
$$
\int_{X\times Y} c\ d\widetilde\gamma<\int_{X\times Y} c\ d\gamma.
$$

First, we check $\widetilde\gamma$ is a positive measure. It is sufficient to prove that $\gamma-\varepsilon_0\sum_{i=1}^k \gamma_i$ is positive, and for that, the condition
$$
\varepsilon_0\gamma_i<\frac1k\gamma
$$
is enough.

Since
$$
\varepsilon_0\gamma_i=\frac{\varepsilon_0}{\gamma(V_i)}\ \gamma\llcorner V_i,
$$
and
$$
\frac{\varepsilon_0}{\gamma(V_i)}<\frac1k,
$$
we get the desired result.

Now, let us check the marginals of $\widetilde\gamma$. We have

$$
(\pi_x)_\sharp \widetilde \gamma = \mu-\varepsilon_0 \sum _{i=1} ^k \mu_i+\varepsilon_0\sum _{i=1} ^k \mu_i = \mu
$$

and

$$
(\pi_y)_\sharp \widetilde\gamma = \nu-\varepsilon _0 \sum _{i=1} ^k \nu_i + \varepsilon_0 \sum _{i=1} ^k \nu _{\sigma(i)} = \nu.
$$

Finally, let us estimate
$$
\int_{X\times Y} c\ d\gamma-\int_{X\times Y} c\ d\widetilde\gamma.
$$
We have
$$
\begin{aligned}
\int_{X\times Y} c\ d\gamma-\int_{X\times Y} c\ d\widetilde\gamma
&=
\varepsilon_0\sum_{i=1}^k \int_{X\times Y} c\ d\gamma_i
-\varepsilon_0\sum_{i=1}^k \int_{X\times Y} c\ d\widetilde\gamma_i\\\
&\ge
\varepsilon_0\sum_{i=1}^k (c(x_i,y_i)-\varepsilon)
-\varepsilon_0\sum_{i=1}^k (c(x_i,y_{\sigma(i)})+\varepsilon)\\\
&=
\varepsilon_0\left(
\sum_{i=1}^k c(x_i,y_i)-\sum_{i=1}^k c(x_i,y_{\sigma(i)})-2k\varepsilon
\right)>0,
\end{aligned}
$$

where we use the fact that $\gamma_i$ is concentrated on $B(x_i,r)\times B(y_i,r)$ and $\widetilde\gamma_i$ is concentrated on $B(x_i,r)\times B(y_{\sigma(i)},r)$. $\square$

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken in Cromwell, New Zealand.

From now on, we assume the duality result
$$
\min (\mathrm{KP})=\max (\mathrm{DP})
$$
holds true, which means for the optimal transport plan $\gamma$ and a Kantorovich potential $\varphi$, we have
$$
\int_{\Omega \times \Omega} c \ d\gamma
=
\int_{\Omega} \varphi \ d\mu + \int_{\Omega} \varphi^c \ d\nu.
$$

Since we have $\varphi(x)+\varphi^c(y)\le c(x,y)$ for every $x,y\in \Omega$, we have
$$
\varphi(x)+\varphi^c(y)=c(x,y)\quad \gamma-a.s.
$$

Furthermore, since $\varphi^c$ and $c$ are continuous, we have $\varphi(x)+\varphi^c(y)=c(x,y)$ on $\operatorname{supp}(\gamma)$.

Fix $(x_0,y_0)\in \operatorname{supp}(\gamma)$. By the definition of $\varphi^c$,
$$
\varphi^c(y_0)=\inf_{x\in \Omega}[c(x,y_0)-\varphi(x)]\le c(x,y_0)-\varphi(x)
\qquad \forall x\in \Omega.
$$

On the other hand, since $(x_0,y_0)\in \operatorname{supp}(\gamma)$,
$$
\varphi(x_0)+\varphi^c(y_0)=c(x_0,y_0),
$$
which implies
$$
c(x_0,y_0)-\varphi(x_0)=\varphi^c(y_0)\le c(x,y_0)-\varphi(x)
\qquad \forall x\in \Omega.
$$

Therefore
$$
x\longmapsto c(x,y_0)-\varphi(x)
$$
is minimal at $x=x_0$.

If $\varphi$ and $c(\cdot,y_0)$ are differentiable at $x_0$ and $x_0\notin \partial \Omega$, then the first order condition tells us
$$
\nabla_x c(x_0,y_0)-\nabla \varphi(x_0)=0,
$$
namely
$$
\nabla \varphi(x_0)=\nabla_x c(x_0,y_0).
$$

Proposition 1.1 If $c$ is $C^1$, $\varphi$ is a Kantorovich potential for the cost $c$ in the transport from $\mu$ to $\nu$, and $(x_0,y_0)$ belongs to the support of an optimal transport plan $\gamma$. Then
$$
\nabla \varphi(x_0)=\nabla_x c(x_0,y_0),
$$
provided $\varphi$ is differentiable at $x_0$.

In particular, the gradients of two different Kantorovich potentials coincide on every point $x_0\in \operatorname{supp}(\mu)$ where both the potentials are differentiable.

Proof : The proof is contained in the above considerations.

Definition 1.4 (Twist condition) For $\Omega \subset \mathbb{R}^d$, we say that $c:\Omega \times \Omega \to \mathbb{R}$ satisfies the twist condition whenever $c$ is differentiable w.r.t. $x$ at every point and the map
$$
y\longmapsto \nabla_x c(x,y)
$$
is injective for every $x_0\in \Omega$.

Remark By Proposition 1.1, we know for $(x_0,y_0)\in \operatorname{supp}(\gamma)$,
$$
\nabla_x c(x_0,y_0)=\nabla \varphi(x_0).
$$

If $c$ satisfies the twist condition, for given $x_0$, there is a unique $y_0$ such that
$$
(x_0,y_0)\in \operatorname{supp}(\gamma).
$$

This shows that $\gamma$ is concentrated on a graph.

For $c(x,y)=h(x-y)$ with $h$ strictly convex, if $\varphi$ and $h$ are differentiable at $x_0$ and $x_0-y_0$, respectively, and $x_0\notin \partial \Omega$. Then
$$
\nabla_x c(x,y)=\nabla h(x-y).
$$

If $\nabla_x c(x,y_1)=\nabla_x c(x,y_2)$, then
$$
\nabla h(x-y_1)=\nabla h(x-y_2).
$$
Since $h$ is strictly convex,
$$
x-y_1=x-y_2,
$$
which implies $y_1=y_2$ and therefore $h$ satisfies twist condition.

Since
$$
\nabla \varphi(x_0)=\nabla_x c(x_0,y_0)=\nabla h(x_0-y_0),
$$
then
$$
x_0-y_0=(\nabla h)^{-1}(\nabla \varphi(x_0)).
$$

Hence
$$
y_0=x_0-(\nabla h)^{-1}(\nabla \varphi(x_0)).
$$

By Rademacher theorem, $\varphi$ is differentiable $\mathcal{L}^d$-a.e. If $\mu\ll \mathcal{L}^d$, then $\nabla \varphi$ exists $\mu$-a.s.

Theorem 1.5 Given $\mu$ and $\nu$ probability measures on a compact domain $\Omega\subset \mathbb{R}^d$, there exists an optimal transport plan $\gamma$ for the cost $c(x,y)=h(x-y)$ with $h$ strictly convex. It is unique and of the form $(id \times T)_\sharp \mu$ provided $\mu\ll \mathcal{L}^d$ and $\partial \Omega$ is negligible. Moreover, there exists a Kantorovich potential $\varphi$ and $T$ and the potentials $\varphi$ are linked by
$$
T(x)=x-(\nabla h)^{-1}(\nabla \varphi(x)).
$$

Proof : The proof is contained in the previous considerations and the uniqueness is by Proposition 1.1, which implies $\nabla \varphi$ is unique, then the optimal $T$. $\square$

Remark. Every time we know that any optimal $\gamma$ must be induced by a map $T$, then we have uniqueness.

Proof : If we have two different plans
$$
\gamma_1=\gamma_{T_1}, \qquad \gamma_2=\gamma_{T_2}
$$
are optimal. Then
$$
\bar{\gamma}:=\frac12\gamma_1+\frac12\gamma_2
$$
is also optimal, but it cannot be induced by a map unless $T_1=T_2$, $\mu$-a.e., which gives a contradiction.

Indeed, let
$$
A:=\{x\in X:\ T_1(x)\ne T_2(x)\}.
$$
Suppose $\mu(A)>0$ and there exists a map $T:X\to Y$ such that
$$
\bar{\gamma}=\gamma_T=(id\times T)_\sharp\mu.
$$

For $B\in \mathcal{B}(X)$, consider
$$
\operatorname{Graph}(T|_B):=\{(x,T(x)):\ x\in B\}.
$$
We have
$$
\begin{aligned}
\bar{\gamma}(\operatorname{Graph}(T|_B)) &= (id\times T)_\sharp\mu\big(\{(x,T(x)):\ x\in B\}\big)\\\
&=\mu\big((id\times T)^{-1}\{(x,T(x)):\ x\in B\}\big)\\\
&=\mu(B).
\end{aligned}
$$

Define
$$
B_1=\{x\in A:\ T(x)=T_1(x)\}, \qquad B_2=\{x\in A:\ T(x)=T_2(x)\}.
$$
By the definition of $A$, we have $B_1\cap B_2=\varnothing$.

First, we claim $\mu\big(A\setminus (B_1\cup B_2)\big)=0.$

If not, $\mu(C)>0$, where $C=A\setminus (B_1\cup B_2)$. Consider
$$
E:=\operatorname{Graph}(T|_C)=\{(x,T(x)):\ x\in C\}.
$$
We have
$$
\bar{\gamma}(E)=\mu(C)>0.
$$

However,
$$
\gamma_1(E)
=
\mu\big(\{x\in C:\ (x,T_1(x))\in E\}\big)
=
\mu\big(\{x\in C:\ T_1(x)=T(x)\}\big)=0.
$$
Similarly $\gamma_2(E)=0$. Therefore,
$$
\bar{\gamma}(E)=\frac12\gamma_1(E)+\frac12\gamma_2(E)=0,
$$
which gives a contradiction. Therefore
$$
\mu\big(A\setminus (B_1\cup B_2)\big)=0.
$$

Next, since
$$
A=B_1\cup B_2\cup A\setminus (B_1\cup B_2)
$$
and $B_1\cap B_2=\varnothing$,
$$
\mu(A)=\mu(B_1)+\mu(B_2)>0.
$$
Therefore, either $\mu(B_1)>0$ or $\mu(B_2)>0$. Without loss of generality, we assume $\mu(B_1)>0$.

Finally, consider
$$
F:=\operatorname{Graph}(T_2|_{B_1})=\{(x,T_2(x)):\ x\in B_1\}.
$$
We have
$$
\gamma_2(F)=\mu\big(\{x\in B_1:\ (x,T_2(x))\in F\}\big)=\mu(B_1),
$$
therefore
$$
\bar{\gamma}(F)\ge \frac12\gamma_2(F)=\frac12\mu(B_1)>0.
$$

However,
$$
\bar{\gamma}(F)
=
\mu\big(\{x\in X:\ (x,T(x))\in F\}\big)
=
\mu\big(\{x\in B_1:\ T(x)=T_2(x)\}\big)
=
\mu(\varnothing)=0,
$$
which gives a contradiction. $\square$

The quadratic case in $\mathbb{R}^d$: $c(x,y)=\frac12|x-y|^2$

Proposition 1.2 Given a function $\chi:\mathbb{R}^d\to \mathbb{R}\cup\{+\infty\}$, let us define
$$
u_\chi:\mathbb{R}^d\to \mathbb{R}\cup\{+\infty\}
$$
$$
u_\chi(x)=\frac12|x|^2-\chi^c(x).
$$

Then we have
$$
u_{\chi^c}=(u_\chi)^\star
$$
where $f^\star$ denotes the Legendre-Fenchel transform of $f$.

In particular, a function $\varphi$ is $c$-concave if and only if
$$
x\longmapsto \frac12|x|^2-\varphi(x)
$$
is convex and lower semi-continuous.

Proof :
$$
\begin{aligned}
u_{\chi^c}(x)& =\frac12|x|^2-\chi^c(x)\\\
&=\sup_y\left[\frac12|x|^2-\frac12|x-y|^2+\chi(y)\right]\\\
&=
\sup_y\left[x\cdot y-\left(\frac12|y|^2-\chi(y)\right)\right]\\\
&=(u_\chi)^\star.
\end{aligned}
$$

If $\varphi$ is $c$-concave, which means there exists a function $\chi$ s.t.
$$
\varphi=\chi^c,
$$
then
$$
u_\varphi(x)=\frac12|x|^2-\varphi(x)=(u_\chi)^\star
$$
which is convex and lower semi-continuous.

Conversely, if $u_\varphi=\frac12|x|^2-\varphi(x)$ is convex and lower-semi-continuous, then there exists a function $\chi$ such that
$$
u_\varphi=\chi^\star.
$$

Then
$$
\begin{aligned}
\varphi(x)&=\frac12|x|^2-\chi^\star=
\frac12|x|^2-\sup_y[x\cdot y-\chi(y)]\\\
&=
\inf_y\left[\frac12|x-y|^2-\left(\frac12|y|^2-\chi(y)\right)\right]=
\left(\frac12|y|^2-\chi(y)\right)^c,
\end{aligned}
$$

which shows that $\varphi$ is $c$-concave. $\square$

By Theorem 1.5, there exists a optimal transport map
$$
T(x)=x-\nabla \varphi(x)=\nabla\left(\frac12|x|^2-\varphi(x)\right)=\nabla u_\varphi(x).
$$

By Proposition 1.2, $u_\varphi$ is convex.

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken upon arriving in Kagoshima, Japan, aboard a Royal Caribbean cruise.

Let us express the constraint $\gamma \in \Pi(\mu,\nu)$ in the following way: if $\gamma \in M_+(X\times Y)$ then we have
$$
\sup_{\varphi,\psi}
\left[
\int_X \varphi \ d\mu
+\int_Y \psi \ d\nu
-\int_{X\times Y} (\varphi(x)+\psi(y)) \ d\gamma
\right]
=
\begin{cases}
0, & \text{if } \gamma \in \Pi(\mu,\nu),\\\
+\infty, & \text{otherwise.}
\end{cases}
$$
where $\varphi,\psi$ are bounded continuous.

Proof : If $\gamma \in \Pi(\mu,\nu)$, it is trivial. If $\gamma \notin \Pi(\mu,\nu)$, then there exists a pair $(\varphi_0,\psi_0)$ such that
$$
I(\varphi_0,\psi_0)
:=
\int_X \varphi_0 \ d\mu
+\int_Y \psi_0 \ d\nu
-\int_{X\times Y} (\varphi_0(x)+\psi_0(y)) \ d\gamma
\neq 0.
$$
W.L.O.G. we can assume $I>0$, then if we choose $(a\varphi_0,a\psi_0)$ for $a>0$,
$$
I(a\varphi_0,a\psi_0)=a I(\varphi_0,\psi_0).
$$
Hence
$$
\mathrm{LHS}=\sup_{\varphi,\psi} I(\varphi,\psi)=+\infty.
$$

Hence the Kantorovich problem is equivalent to the following problem
$$
\min_{\gamma \in M_+(X\times Y)}
\left\{
\int_{X\times Y} c \ d\gamma
+\sup_{\varphi,\psi}
\left[
\int_X \varphi \ d\mu
+\int_Y \psi \ d\nu
-\int_{X\times Y} (\varphi(x)+\psi(y)) \ d\gamma
\right]
\right\}.
$$

Consider interchanging $\sup$ and $\inf$
$$
\sup_{\varphi,\psi}
\left\{
\int_X \varphi \ d\mu
+\int_Y \psi \ d\nu
+\inf_{\gamma \in M_+(X\times Y)}
\left[
\int_{X\times Y} \bigl(c(x,y)-(\varphi(x)+\psi(y))\bigr)\ d\gamma
\right]
\right\}.
$$

Now we denote $(\varphi \oplus \psi)(x,y):=\varphi(x)+\psi(y)$, then the infimum in $\gamma$ can be written as
$$
\inf_{\gamma \in M_+(X\times Y)} \int_{X\times Y} (c-\varphi\oplus\psi)\ d\gamma
=
\begin{cases}
0, & \text{if } \varphi\oplus\psi \le c \text{ on } X\times Y,\\\
-\infty, & \text{otherwise.}
\end{cases}
$$

Proof : Let
$$
\widetilde{J}(\gamma):=\int_{X\times Y} (c-\varphi\oplus\psi)\ d\gamma.
$$
If $\varphi\oplus\psi \le c$ on $X\times Y$, then for every $\gamma \in M_+(X\times Y)$,
$$
\widetilde{J}(\gamma)\ge 0.
$$
Since for all $a>0$,
$$
\widetilde{J}(a\gamma)=a\widetilde{J}(\gamma),
$$
we have
$$
\inf_{\gamma \in M_+(X\times Y)} \widetilde{J}(\gamma)=0.
$$

If $\varphi+\psi \le c$ is not true, then for every $\gamma \in M_+(X\times Y)$, $\widetilde{J}(\gamma)<0$. Since for all $a>0$ we have $\widetilde{J}(a\gamma)=a\widetilde{J}(\gamma)$, we have
$$
\inf_{\gamma \in M_+(X\times Y)} \widetilde{J}(\gamma)=-\infty.
$$

Therefore, we get the following dual optimization problem:

Problem 1.3. Given $\mu \in \mathcal{P}(X)$, $\nu \in \mathcal{P}(Y)$ and the cost function $c:X\times Y \to [0,+\infty]$, we consider the problem
$$
(\mathrm{DP})\qquad
\sup
\left\{
\int_X \varphi \ d\mu + \int_Y \psi \ d\nu
\ \middle|\
\varphi \in C_b(X),\ \psi \in C_b(Y),\ \varphi\oplus\psi \le c
\right\}.
$$

Lemma 1.3. $\sup(\mathrm{DP}) \le \min(\mathrm{KP})$.

Proof : For $\gamma \in \Pi(\mu,\nu)$, then for every $\varphi \in C_b(X)$, $\psi \in C_b(Y)$ with $\varphi\oplus\psi \le c$, we have
$$
\int_X \varphi \ d\mu + \int_Y \psi \ d\nu
=
\int_{X\times Y} \varphi\oplus\psi \ d\gamma
\le
\int_{X\times Y} c \ d\gamma.
$$

Definition 1.1. Given a function $\chi:X\to \overline{\mathbb R}$, we define its $c$-transform (also called $c$-conjugate function)
$$
\chi^c:Y\to \overline{\mathbb R}
$$
by
$$
\chi^c(y):=\inf_{x\in X}\ [c(x,y)-\chi(x)].
$$

We also define the $\overline{c}$-transform of $\xi:Y\to \overline{\mathbb R}$ by
$$
\xi^{\overline{c}}:X\to \overline{\mathbb R}
$$
$$
\xi^{\overline{c}}(x):=\inf_{y\in Y}\ [c(x,y)-\xi(y)].
$$

Moreover, we say that a function $\psi$ defined on $Y$ is $\overline{c}$-concave if there exists $\chi:X\to \overline{\mathbb R}$ such that $\psi=\chi^c$, and a function $\varphi$ on $X$ is said to be $c$-concave if there exists $\xi:Y\to \overline{\mathbb R}$ such that $\varphi=\xi^{\overline{c}}$.

We denote by $c\text{-conc}(X)$ and $\overline{c}\text{-conc}(Y)$ the sets of $c$-concave and $\overline{c}$-concave functions, respectively.

(When $X=Y$, and $c$ is symmetric, the distinction between $c$ and $\overline{c}$ will play no more any rule and will be dropped as soon as possible.)

Definition 1.2. A function $f:X\to \mathbb R$ is said to have modulus of continuity $\omega:\mathbb R_+\to \mathbb R_+$ if for all $x,y\in X$

$$
|f(x)-f(y)|\le \omega(d(x,y)).
$$

Lemma 1.4. If $c:X\times Y\to \overline{\mathbb R}$ is continuous and finite on a compact set, then there exists an increasing continuous function $\omega:\mathbb R_+\to \mathbb R_+$ with $\omega(0)=0$ such that
$$
|c(x,y)-c(x’,y’)|\le \omega\bigl(d(x,x’)+d(y,y’)\bigr).
$$

Proof : Let $\omega:\mathbb R_+\to \mathbb R_+$ be defined by
$$
\omega(t):=
\sup\left\{
|c(x,y)-c(x’,y’)|
\ :\
d(x,x’)+d(y,y’)\le t
\right\}.
$$
It is easy to see that $\omega(0)=0$ and $\omega(t)$ is increasing.

Let $r=d(x,x’)+d(y,y’)$, then
$$
|c(x,y)-c(x’,y’)|\le \omega(r)=\omega\bigl(d(x,x’)+d(y,y’)\bigr).
$$
Since $c$ is uniformly continuous on its domain, $\omega$ is continuous.

Lemma 1.5. Let $(f_\alpha)$ be a family of functions, all satisfying the same modulus of continuity $\omega:\mathbb R_+\to \mathbb R_+$,

$$
|f_\alpha(x)-f_\alpha(x’)|\le \omega(d(x,x’)),\qquad \forall \alpha.
$$

Then
$$
f:=\inf_\alpha f_\alpha
$$
also satisfies the same modulus of continuity.

Proof : For each $\alpha$,
$$
f_\alpha(x)\le f_\alpha(x’)+\omega(d(x,x’)),
$$
which implies
$$
f(x)\le f_\alpha(x’)+\omega(d(x,x’))
$$
since $f\le f_\alpha$. Taking the infimum at the RHS, we get
$$
f(x)\le f(x’)+\omega(d(x,x’)).
$$
Interchanging $x$ and $x’$, we obtain
$$
|f(x)-f(x’)|\le \omega(d(x,x’)).
$$

Lemma 1.6. If $c:X\times Y\to \overline{\mathbb R}$ has the modulus of continuity $\omega:\mathbb R_+\to \mathbb R_+$, then $\chi^c$ shares the same modulus of continuity.

Proof : Since
$$
\chi^c(y)=\inf_{x\in X}\bigl(c(x,y)-\chi(x)\bigr)=: \inf_{x\in X} g_x(y),
$$
and for $x\in X$, $y,y’\in Y$,
$$
|g_x(y)-g_x(y’)|=|c(x,y)-c(x,y’)|\le \omega(d(y,y’)),
$$
by Lemma 1.5 we get the desired result.

Lemma 1.7. Let
$$
DP(\varphi,\psi):=\int_X \varphi \ d\mu+\int_Y \psi \ d\nu.
$$
For $\varphi\in C_b(X)$, $\psi\in C_b(Y)$ with $\varphi\oplus\psi\le c$. Then
$$
DP(\varphi,\varphi^c)\ge DP(\varphi,\psi).
$$

Proof : Since $\varphi\oplus\psi\le c$, we have
$$
\psi(y)\le c(x,y)-\varphi(x).
$$
Furthermore
$$
\psi(y)\le \inf_x [c(x,y)-\varphi(x)]=\varphi^c(y).
$$
Hence
$$
DP(\varphi,\psi)\le \int_X \varphi \ d\mu+\int_Y \varphi^c \ d\nu = DP(\varphi,\varphi^c).
$$

Remark. Similarly, we will have
$$
DP(\varphi,\psi)\le DP(\varphi,\varphi^c)\le DP(\varphi^{c\overline{c}},\varphi^c)\le DP(\psi^{\overline{c}},\psi^{\overline{c}c})\le DP(\varphi^{c\overline{c}c},\varphi^{c\overline{c}})\le \cdots
$$

Theorem 1.4. Suppose that $X$ and $Y$ are compact and $c$ is continuous. Then there exists a solution $(\varphi,\psi)$ to problem $(\mathrm{DP})$ and it has the form $\varphi\in c\text{-conc}(X)$, $\psi\in \overline{c}\text{-conc}(X)$ and $\psi=\varphi^c$. In particular,
$$
\max(\mathrm{DP})
=
\max_{\varphi\in c\text{-conc}(X)}
\left[
\int_X \varphi \ d\mu+\int_Y \varphi^c \ d\nu
\right].
\tag{$\star$}
$$

Proof : From the considerations above, we can take a maximizing sequence $(\varphi_n,\psi_n)$ and improve it by means of $c$- and $\overline{c}$-transforms, and we can assume they have the same modulus of continuity as $c$. Instead of renaming the sequence, we will still call $(\varphi_n,\psi_n)$ the new sequence obtained after these transforms.

By Ascoli-Arzelà’s theorem, we only need to check the equiboundedness. For every constant $a\in \mathbb R$,
$$
DP(\varphi,\psi)=DP(\varphi+a,\psi-a)
$$
and
$$
(\varphi+a)\oplus(\psi-a)\le c.
$$
Since $\varphi_n$ is continuous on a compact set and then bounded, W.L.O.G. we can assume
$$
\min \varphi_n=0
$$
and then we get
$$
\max \varphi_n\le \omega(\operatorname{diam} X).
$$
If we have chosen $\psi_n=\varphi_n^c$, we also have
$$
\psi_n(y)=\inf_{x\in X}[c(x,y)-\varphi_n(x)]
\in
\bigl[\min c-\omega(\operatorname{diam} X),\ \max c\bigr].
$$

Hence equibounded. By Ascoli-Arzelà’s theorem, there exists a subsequence
$$
\varphi_{n_k}\to \varphi,\qquad \psi_{n_k}\to \psi\quad \text{(uniform convergence).}
$$

Therefore
$$
\int_X \varphi_{n_k}\ d\mu+\int_Y \psi_{n_k}\ d\nu
\to
\int_X \varphi\ d\mu+\int_X \psi\ d\nu.
$$
and
$$
\varphi_{n_k}(x)+\psi_{n_k}(y)\le c(x,y)
\quad\Rightarrow\quad
\varphi(x)+\psi(y)\le c(x,y).
$$
This shows that $(\varphi,\psi)$ is an admissible pair for $(\mathrm{DP})$ and then optimal.

And if $\psi\neq \varphi^c$, we can improve the objective functional in $(\mathrm{DP})$, which contradicts to $(\varphi,\psi)$ is optimal. Similarly $\varphi=\psi^{\overline{c}}$.

Hence $\psi=\varphi^c\in \overline{c}\text{-conc}(X)$ and similarly $\varphi=\psi^{\overline{c}}\in c\text{-conc}(X)$.

Remark. If $\min(\mathrm{KP})=\max(\mathrm{DP})$, we also have
$$
\min(\mathrm{KP})
=
\max_{\varphi\in c\text{-conc}(X)}
\left[
\int_X \varphi \ d\mu+\int_Y \varphi^c \ d\nu
\right]
$$
which also shows that the minimum value of $(\mathrm{KP})$ is a convex function of $(\mu,\nu)$ as it is a supremum of linear functionals.

Definition 1.3. The functions $\varphi$ realizing the maximum in $(\star)$ are called Kantorovich potentials for the transport from $\mu$ and $\nu$.

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken in Auckland, New Zealand.

Monge Problem

Problem 1.1 (Monge Problem). Given $\mu \in \mathcal{P}(X)$, $\nu \in \mathcal{P}(Y)$ and a Borel measurable cost function
$$
c:X \times Y \to [0,+\infty],
$$
solve
$$
(\mathrm{MP}) \qquad \inf \left\{ M(T):=\int_X c(x,T(x))\ d\mu(x) \ \middle|\ T_\sharp\mu=\nu \right\},
$$
where
$$
T_\sharp\mu(A)=\mu\bigl(T^{-1}(A)\bigr)
$$
for all $A \in \mathcal{B}(Y)$. Find also the optimal transport map $T$.

Remark. Difficulty of Monge Problem is that the constraint
$$
\{T:X \to Y \mid T_\sharp\mu=\nu\}
$$
is often not closed under weak convergence.

Example. $X=[0,2\pi]$, $\mu=\dfrac{1}{2\pi}\ dx$, $T_n(x)=\sin(nx)$, $Y=\mathbb R$.

Claim. For all $n \in \mathbb{N}$,

$$
(T_n)_\sharp\mu=\nu,
$$

where

$$
\nu(dy)=\frac{1}{\pi\sqrt{1-y^2}}\ 1_{(-1,1)}(y)\ dy.
$$

Proof : For all $\varphi \in C_b(\mathbb R)$,

$$
\begin{aligned}
\int_Y \varphi(y)\ ((T_n)_\sharp\mu)(dy)
&=
\int_X \varphi(\sin nx)\ \mu(dx)
=
\frac{1}{2\pi}\int_0^{2\pi}\varphi(\sin nx)\ dx \\\
&=\frac{1}{2\pi n} \int_0^{2\pi n}\varphi(\sin u)\ du
=\frac{1}{2\pi}\int_0^{2\pi}\varphi(\sin u)\ du
=\int_Y \varphi(y)\ \nu(dy).
\end{aligned}
$$

Hence $(T_n)_\sharp\mu=\nu$ for all $n \in \mathbb{N}$.

However, by Riemann-Lebesgue Lemma, in $L^p[0,2\pi]$, $1 \leq p < +\infty$, we have for all $g \in L^q[0,2\pi]$
$$
\int_0^{2\pi} g(x)\sin(nx)\ dx \to 0.
$$
Hence
$$
T_n \rightharpoonup 0.
$$
But for $T=0$, $T_\sharp\mu=\delta_0 \neq \nu$, which shows that
$$
\left\{ T \in L^p([0,2\pi]) \ ; \ T_\sharp\mu=\nu \right\}
$$
is not closed under weak topology of $L^p[0,2\pi]$.

Kantorovich Problem

Problem 1.2 (Kantorovich Problem). Given $\mu \in \mathcal{P}(X)$, $\nu \in \mathcal{P}(Y)$, and a Borel measurable cost function
$$
c:X \times Y \to [0,\infty],
$$
find
$$
(\mathrm{KP}) \qquad \inf \left\{ K(\gamma):=\int_{X \times Y} c\ d\gamma \ \middle|\ \gamma \in \Pi(\mu,\nu) \right\},
$$
where $\Pi(\mu,\nu)$ is the set of so-called transport plans, i.e.
$$
\Pi(\mu,\nu)
=
\left\{
\gamma \in \mathcal{P}(X \times Y)
\ \middle|\
(\pi_x)_\sharp\gamma=\mu,\ (\pi_y)_\sharp\gamma=\nu
\right\},
$$
where $\pi_x$ and $\pi_y$ are two projections of $X \times Y$ onto $X$ and $Y$, respectively.

Remark. If $T_\sharp\mu=\nu$ then the plan
$$
\pi=(\mathrm{id} \times T)_\sharp\mu \in \Pi(\mu,\nu).
$$
Conversely, if $\pi=(\mathrm{id} \times T)_\sharp\mu \in \Pi(\mu,\nu)$, then $T_\sharp\mu=\nu$.

Proof : First, if $T_\sharp\mu=\nu$, then for all $A \in \mathcal{B}(X)$, $B \in \mathcal{B}(Y)$,
$$
\pi(A \times Y)
=
((\mathrm{id} \times T)_\sharp\mu)(A \times Y)
=
\mu\bigl((\mathrm{id} \times T)^{-1}(A \times Y)\bigr)
=
\mu(A),
$$
and
$$
\pi(X \times B)
=
((\mathrm{id} \times T)_\sharp\mu)(X \times B)
=
\mu\bigl((\mathrm{id} \times T)^{-1}(X \times B)\bigr)
=
\mu(T^{-1}(B))
=
\nu(B).
$$
Hence $\pi \in \Pi(\mu,\nu)$.

Second, if $\pi \in \Pi(\mu,\nu)$, then for all $B \in \mathcal{B}(Y)$,
$$
\nu(B)
=
\pi(X \times B)
=
((\mathrm{id} \times T)_\sharp\mu)(X \times B)
=
\mu\bigl((\mathrm{id} \times T)^{-1}(X \times B)\bigr)
=
\mu(T^{-1}(B)).
$$
Hence $T_\sharp\mu=\nu$. $\square$

Remark. In the problem of Monge, it may happen that there is no map $T$ such that $T_\sharp\mu=\nu$. Indeed, if $\mu=\delta_{x_0}$, $x_0 \in X$ and $\nu$ is any measure that is different from an atomic measure on one atom, then there is no such map, as
$$
T_\sharp\mu=\delta_{T(x_0)}.
$$
But the set $\Pi(\mu,\nu)$ is always non-empty, as
$$
\mu \otimes \nu \in \Pi(\mu,\nu).
$$

Weak-$\star$ Convergence and Weak Convergence

Let $\mathcal{M}(X)$ be the set of finite signed measures on $X$. By Riesz Representation theorem, if $X$ is separable and locally compact metric space, let $\mathcal{X}=C_0(X)$ be the set of continuous functions on $X$ vanishing at infinity, i.e. $f \in C_0(X)$ if and only if $f \in C_b(X)$ and for every $\varepsilon>0$, there exists a compact subset $K \subset X$ such that
$$
|f(x)|<\varepsilon
$$
on $X \setminus K$.

And endow this space with sup norm. Since $C_0(X) \subset C_b(X)$, then $C_0(X)$ is a closed subset of $C_b(X)$ also a Banach space. By Riesz Representation Theorem for all $\xi \in \mathcal{X}’$, there exists a unique $\lambda \in \mathcal{M}(X)$ such that
$$
\langle \xi,\varphi\rangle=\int_X \varphi\ d\lambda
$$
for every $\varphi \in \mathcal{X}=C_0(X)$.

Moreover $\mathcal{X}^\prime \simeq \mathcal{M}(X)$ endowed with the norm $\Vert\lambda\Vert:=|\lambda|(X).$

For signed measures $\mathcal{M}(X)$:

• weak-$\star$ convergence: the convergence in the duality with $C_0(X)$, that is $\mu_n \overset{\mathrm{weak}-\star}{\longrightarrow} \mu$ if and only if for all $\varphi \in C_0(X)$

$$
\int_X \varphi\ d\mu_n \to \int_X \varphi\ d\mu
\qquad \text{as } n \to \infty.
$$

• weak convergence (also called narrow convergence): $\mu_n \rightharpoonup \mu$ if and only if for all $\varphi \in C_b(X)$

$$
\int_X \varphi\ d\mu_n \to \int_X \varphi\ d\mu
\qquad \text{as } n \to \infty.
$$

If $X$ is compact, then $C_0(X)=C_b(X)=C(X)$, weak-$\star$ convergence and weak convergence are the same.

Existence of Solutions

Theorem 1.1. Suppose that $X,Y$ are compact and $c:X \times Y \to \mathbb R$ is continuous. Then the Kantorovich Problem admits a minimizer.

Proof : We just need to show that the set $\Pi(\mu,\nu)$ is compact and that
$$
\gamma \longmapsto K(\gamma)=\int_{X \times Y} c\ d\gamma
$$
is continuous. Since $X \times Y$ compact, $C_0(X \times Y)=C_b(X \times Y)=C(X \times Y)$, $\gamma \mapsto K(\gamma)$ is continuous.

For the compactness, take $(\gamma_n) \subset \Pi(\mu,\nu)$, since they are probability measures, they are bounded in $\mathcal{M}(X \times Y) \simeq (C(X \times Y))’$. By Banach-Alaoglu theorem, there exists a subsequence $\gamma_{n_k}$,
$$
\gamma_{n_k} \overset{\mathrm{weak}-\star}{\longrightarrow} \gamma \in \mathcal{M}(X \times Y).
$$
Since $X \times Y$ is compact,
$$
\gamma(X \times Y)
=
\int_{X \times Y} 1\ d\gamma
=
\lim_{k \to \infty}\int_{X \times Y} 1\ d\gamma_{n_k}
=
1.
$$
Hence $\gamma \in \mathcal{P}(X \times Y)$. Moreover, for all $\varphi \in C(X)$, and since $\gamma_{n_k} \in \Pi(\mu,\nu)$,
$$
\int_{X \times Y} \varphi\ d\gamma_{n_k}
=
\int_X \varphi\ d\mu.
$$
Let $k \to \infty$, we have
$$
\int_{X \times Y} \varphi\ d\gamma
=
\int_X \varphi\ d\mu.
$$
Hence $(\pi_x)_\sharp\gamma=\mu$ and similarly, $(\pi_y)_\sharp\gamma=\nu$. Hence $\gamma \in \Pi(\mu,\nu)$. Therefore, $\Pi(\mu,\nu)$ is compact. $\square$

Lemma 1.1. Let $f:X \to \mathbb R \cup \{+\infty\}$ be a function bounded from below. Then $f$ is lower semi-continuous if and only if there exists a sequence $f_k$ of $k$-Lipschitz functions such that for every $x \in X$, $f_k(x)$ converges increasingly to $f(x)$. Furthermore, $f_k$ can also be made bounded.

Proof : On the one hand, if
$$
f(x)=\lim_{k \to \infty} f_k(x)
$$
for all $x \in X$ with $f_k$ $k$-Lipschitz and increasing, then
$$
f=\sup_k f_k
$$
is also lower semi-continuous.

On the other hand, if $f$ is lower semi-continuous and bounded from below, define
$$
f_k(x)=\inf_y \{ f(y)+k\ d(x,y) \}.
$$
It is easy to show that $f_k$ is $k$-Lipschitz. For fixed $x \in X$, $f_k(x)$ is increasing and we have
$$
\inf f \leq f_k(x) \leq f(x).
$$
We just need to show that
$$
\ell:=\lim_{k \to \infty} f_k(x)=\sup_k f_k(x)=f(x).
$$
Suppose by contradiction $\ell<f(x)$, which implies $\ell<+\infty$. For every $k$, choose $y_k$ such that
$$
f(y_k)\leq f(y_k)+k\ d(x,y_k)<f_k(x)+\frac{1}{k}\leq \ell+\frac{1}{k}. \tag{1}
$$
We get
$$
d(y_k,x)\leq \frac{\ell+\frac{1}{k}-f(y_k)}{k}\leq \frac{C}{k}
$$
(since $\ell<\infty$, $f$ bounded from below). Hence $y_k \to x$. Let $k \to \infty$ in $(1)$. Since $f$ is lower semi-continuous,
$$
f(x)\leq \liminf_{k \to \infty} f(y_k)\leq \lim_{k \to \infty} f_k(x)=\ell,
$$
which gives a contradiction.

Finally, $f_k$ can be made bounded by taking $f_k \wedge k$. $\square$

Lemma 1.2. If $f:X \to \mathbb R \cup \{+\infty\}$ is a lower semi-continuous function, bounded from below on a metric space $X$, then the functional
$$
J:\mathcal{M}_+(X)\to \mathbb R\cup\{+\infty\}
$$
defined on positive measures on $X$ through
$$
J(\mu):=\int_X f\ d\mu
$$
is lower semi-continuous for the weak convergence of measures.

Proof : By Lemma 1.1, there exists a sequence $f_k$ of continuous and bounded functions converging increasingly to $f$. Then write
$$
J(\mu)=\sup_k J_k(\mu):=\int_X f_k\ d\mu
$$
(Actually $J_k \leq J$ and $J_k(\mu)\to J(\mu)$ for every $\mu$ by monotone convergence.)

Since every $J_k$ is continuous for the weak convergence, hence $J$ is a lower semi-continuous functional. $\square$

Theorem 1.2. Let $X$ and $Y$ be compact metric spaces, $\mu \in \mathcal{P}(X)$, $\nu \in \mathcal{P}(Y)$ and
$$
c:X \times Y \to \mathbb R\cup\{+\infty\}
$$
be lower semi-continuous and bounded from below. Then the Kantorovich Problem $(\mathrm{KP})$ admits a solution.

Proof : By Lemma 1.2
$$
\gamma \mapsto K(\gamma)=\int_{X \times Y} c\ d\gamma
$$
is lower semi-continuous. By the proof of Theorem 1.1, we know $\Pi(\mu,\nu)$ is compact. $\square$

Theorem 1.3. Let $X$ and $Y$ be Polish spaces, $\mu \in \mathcal{P}(X)$, $\nu \in \mathcal{P}(Y)$ and
$$
c:X \times Y \to \mathbb R\cup\{+\infty\}
$$
is lower semi-continuous. Then $(\mathrm{KP})$ admits a solution.

Proof : We only need to prove the compactness of $\Pi(\mu,\nu)$.

Since $\mu$ and $\nu$ are finite Borel measures on Polish spaces, by Ulam theorem, $\mu$ and $\nu$ are tight, there exist $K_X \subset X$ and $K_Y \subset Y$ such that
$$
\mu(X \setminus K_X)<\frac{\varepsilon}{2},
\qquad
\nu(Y \setminus K_Y)<\frac{\varepsilon}{2}.
$$
Then the set $K_X \times K_Y$ is compact in $X \times Y$ and for any $(\gamma_n)\subset \Pi(\mu,\nu)$
$$
\gamma_n\bigl((X \times Y)\setminus (K_X \times K_Y)\bigr)
\leq
\gamma_n\bigl((X \setminus K_X)\times Y\bigr)+\gamma_n\bigl(X \times (Y \setminus K_Y)\bigr)=
\mu(X \setminus K_X)+\nu(Y \setminus K_Y)
<
\frac{\varepsilon}{2}+\frac{\varepsilon}{2}
=
\varepsilon.
$$
Hence $(\gamma_n)$ is tight. Hence by Prokhorov theorem, there exists $\gamma \in \mathcal{P}(X \times Y)$ and a subsequence $(\gamma_{n_k})$ such that
$$
\gamma_{n_k}\rightharpoonup \gamma.
$$

Now, we need to show that $\gamma \in \Pi(\mu,\nu)$. For all $\varphi \in C_b(X)$, since $\gamma_{n_k}\in \Pi(\mu,\nu)$,
$$
\int_{X \times Y} \varphi(x)\ d\gamma_{n_k}(x,y)
=
\int_X \varphi(x)\ d\mu(x).
$$
Let $k \to \infty$, we have
$$
\int_{X \times Y} \varphi(x)\ d\gamma(x,y)
=
\int_X \varphi(x)\ d\mu(x).
$$
Hence $(\pi_x)_\sharp\gamma=\mu$ and similarly $(\pi_y)_\sharp\gamma=\nu$. Therefore, $\gamma \in \Pi(\mu,\nu)$. $\square$

Reference

Santambrogio, Filippo. Optimal transport for applied mathematicians. Birkäuser, NY 55.58-63 (2015): 94.

The cover image of this article was taken in Saipan, Commonwealth of the Northern Mariana Islands (U.S.).

Notation:
$$
L^1=L,\qquad L^{n+1}=[L^n,L]\qquad (n\geq 1)
$$

Lemma 1. If $I,J$ are ideals of $L$, so is $[I,J]$.

Proof: Let $x\in I,\ y\in J,\ z\in L$, then
$$
[[x,y],z]=[x,[y,z]]-[y,[x,z]]\in [I,J].\quad \square
$$

Proposition 1. (i) $L^n$ is an ideal of $L$,

(ii) Also
$$
L=L^1\supseteq L^2\supseteq L^3\supseteq \cdots
$$

Proof: (i) Follows from the above lemma.

(ii)
$$
L^{n+1}=[L^n,L]\subseteq L^n.\quad \square
$$

Definition 2. A Lie algebra $L$ is nilpotent if $L^n=0$ for some $n\geq 1$,

Remark: Every subalgebra and every quotient algebra of a nilpotent Lie algebra are nilpotent.

Notation:
$$
L^{(0)}=L,\qquad L^{(n+1)}=[L^{(n)},L^{(n)}]\qquad (n\geq 0).
$$

Proposition 2. (i) $L^{(n)}$ is an ideal of $L$.

(ii)
$$
L=L^{(0)}\supseteq L^{(1)}\supseteq L^{(2)}\supseteq \cdots
$$

Proof: Similar to above. $\square$

Definition 3. A Lie algebra $L$ is soluble if $L^{(n)}=0$ for some $n\geq 0$.

Proposition 3. (i)
$$
[L^m,L^n]\subseteq L^{m+n}\qquad \text{for }m,n\geq 1,
$$

(ii)
$$
L^{(n)}\subseteq L^{2^n}\qquad \text{for }n\geq 0,
$$

(iii) nilpotent implies soluble.

Proof: (i) We prove by induction on $n$. Base case $n=1$ is clear. Inductive step:

Suppose (i) holds for $n\leq r$,
$$
[L^m,L^{r+1}]
=[L^m,[L^r,L]]
=[ [L^m,L],L^r ]+[ [L^m,L^r],L ]
$$
by Jacobi identity. Hence
$$
[L^m,L^{r+1}]
\subseteq [L^{m+1},L^r]+[L^{m+r},L]
\subseteq L^{m+r+1}.
$$
(ii) We prove by induction on $n$. Base case $n=0$ is clear. Inductive step:

Suppose (ii) holds for $n\leq r$,
$$
L^{(r+1)}=[L^{(r)},L^{(r)}]
\subseteq [L^{2^r},L^{2^r}]
\subseteq L^{2^r+2^r}=L^{2^{r+1}}.
$$

(iii) Suppose $L$ is nilpotent. Then $L^n=0$ for some $n$. We can pick $k$ s.t. $2^k\geq n$.

$$
L^{(k)}\subseteq L^{2^k}\subseteq L^n=0.
$$
Hence $L^{(k)}=0$. So $L$ is soluble. $\square$

Proposition 4. Let $I$ be an ideal of $L$. If $I$ and $L/I$ are soluble, so is $L$.

Proof: We have $(L/I)^{(n)}=0$ for some $n$. Then
$$
L^{(n)}\subseteq I.
$$

Also, $I^{(m)}=0$, for some $m$. Hence
$$
L^{(n+m)}=(L^{(n)})^{(m)}\subseteq I^{(m)}=0.
$$
So $L$ is soluble. $\square$

Proposition 5. (i) Every finite dimensional Lie algebra $L$ contains a unique maximal soluble ideal $R$.

(ii) Moreover, $L/R$ contains no non-zero soluble ideal.

Proof: (i) Let $I,J$ be soluble ideals of $L$. Then $I+J$ is an ideal of $L$.

Also, $I$ is a soluble ideal of $I+J$. And

$$
(I+J)/I \cong J/(I\cap J),
$$
which is soluble. Therefore, by proposition 4, $I+J$ is soluble.

We have thus shown that the sum of two soluble ideals is soluble, it follows that $L$ has a unique maximal soluble ideal $R$. $\square$

(ii) Let $I/R$ be a soluble ideal of $L/R$. Then $I$ is a soluble ideal of $L$. By part (i), we have

$$
I\subseteq R.
$$
Therefore,
$$
I=R,
$$
and hence
$$
I/R=0.\quad \square
$$
Definition 4. $R$ is called the soluble radical of $L$.

Definition 5. A Lie algebra $L$ is semisimple if $R=0$. In other words, $L$ is semisimple if and only if $L$ has no non-zero soluble ideal.

Definition 6. $L$ is simple if $L$ has no proper ideal.

Example 1. Suppose $L$ has $\dim L=1$. Then $L$ has a basis $\{e\}$. Since $[e,e]=0$, we have
$$
L^2=0.
$$
So $L$ is abelian. Also, $L$ is simple. $L$ is called the trivial simple Lie algebra.

Proposition 6. Every non-trivial simple Lie algebra is semisimple.

Proof: Suppose $L$ is simple but not semisimple. Then
$$
R\neq 0.
$$

Since $L$ is simple and $R$ is an ideal of $L$, we must have
$$
R=L.
$$

Because $R$ is soluble, there exists some $n\geq 0$ such that
$$
L^{(n)}=0.
$$

Then
$$
L^{(1)}\neq L,
$$
for otherwise we would have
$$
L^{(n)}=L\neq 0
$$
for all $n$, which is a contradiction.

Since $L^{(1)}$ is an ideal of $L$ and $L$ is simple, it follows that
$$
L^{(1)}=0.
$$

That is,
$$
[L,L]=0.
$$

Therefore, every subspace of $L$ is an ideal of $L$.

But $L$ is simple, so necessarily
$$
\dim L=1.
$$

This contradicts the assumption that $L$ is non-trivial. Hence $L$ must be semisimple. $\square$

Reference

This series of articles on Lie algebras is based on Carter R., Lie Algebras of Finite and Affine Type (Cambridge University Press, 2005), as well as the class notes for Introduction to Lie Algebras by Chenwei Ruan at BIMSA.

The cover image of this article was taken in Wellington, New Zealand.