(You may see the rest of the lecture notes here).

See my handwritten notes.

The following is a rough transcription of my handwritten notes, processed with an LLM.

Today:

• Heuristics for Brenier’s theorem
• The $c$-exponential map
• Rademacher’s theorem

Heuristics for Brenier’s Theorem

(or: how to characterize optimal $(\phi,\psi)$ ??)

Consider a pair $(\phi,\psi)$ solving the Dual Problem, with $\phi = \psi^c$, $\psi = \phi^c$.

Fix $\eta \in C^0(Y)$, and set

\[\psi_s(y) = \psi(y) + s\eta(y)\]

Then $((\psi_s)^c, \psi_s)$ is an admissible pair for every $s$… What is the value of

\[\frac{d}{ds}\Big|_{s=0} \left( \int_X (\psi_s)^c(x)\; d\mu(x) + \int_Y \psi_s(y)\; d\nu(y) \right)\]

(Note: the point is whatever this is, it equals zero!)

Clearly,

\[\frac{d}{ds}\int_Y \psi_s(y)\; d\nu(y) = \int_Y \eta(y)\; d\nu(y)\]

The $\psi_s^c$ is more delicate. What are the possible values of

\[\frac{d}{ds}\Big|_{s=0} (\psi_s)^c(x) \;?\]

Working assumption: $(\psi^c)(x)$ is such that for every $x$ there exists a unique $y(x)$ such that

\[\mathop{\rm argmin}\limits_y \{ c(x,y) - \psi(y) \} = \{ y(x) \}\]

In fact, let’s assume* an even stronger form of this assumption: $\exists$ $\forall\, s \in (-\varepsilon,\varepsilon)$ $(\varepsilon > 0)$ and every $x$ there is a unique $y_s(x)$ such that

(* this is not needed, as we shall see)

\[\mathop{\rm argmin}\limits_y \{ c(x,y) - \psi_s(y) \} = \{ y_s(x) \}\]

In particular, $\forall\, s \in (-\varepsilon,\varepsilon)$, $\forall\, x \in X$ we have

\[(\psi_s)^c(x) = c(x, y_s(x)) - \psi(y_s(x)) - s\eta(y_s(x))\]

Key fact: As $s \to 0$, for any fixed $x$,

\[\text{“}\ \frac{d}{ds}\Big|_{s=0} \big( c(x, y_s(x)) - \psi(y_s(x)) \big) \geq 0\ \text{”}\]

Indeed, for a given $x$, the map

\[y \mapsto c(x,y) - \psi(y)\]

is achieving its minimum at $y(x)$. Thus

\[c(x, y_s(x)) - \psi(y_s(x)) \geq c(x,y) - \psi(y)\]

i.e. $s \mapsto c(x, y_s(x)) - \psi(y_s(x))$ achieves its minimum at $0$. It follows that

\[\varliminf_{s\to 0} \frac{1}{s}\left( (\psi_s)^c(x) - \psi(x) \right) \geq -\eta(y(x))\]

This used implicitly that as $s \to 0$ we have $y_s(x) \to y(x)$ (this follows from the working assumption)

Exercise: Prove this last claim.

(Hint: Show that

\[\lim_{s\to 0} (\psi_s)^c(x) = c(x, y(x)) - \psi(y(x))\]

and use the fact that there is a unique minimizer of $y \mapsto c(x,y) - \psi(y)$ to conclude that $y_s(x) \to y(x)$ as $s \to 0$.)

If all of the above holds, then we expect that

\[\begin{aligned} 0 = \frac{d}{ds}\Big|_{s=0} & \left( \int_X (\psi_s)^c(x)\; d\mu + \int_Y \psi_s(y)\; d\nu \right) \\ & \geq -\int_X \eta(y(x))\; d\mu + \int_Y \eta(y)\; d\nu \end{aligned}\]

or,

\[\int_X \eta(y(x))\; d\mu \geq \int_Y \eta(y)\; d\nu\]

Since $\eta(y) \in C^0(Y)$ was arbitrary, it follows that

\[\int_X \eta(y(x))\; d\mu(x) = \int_Y \eta(y)\; d\nu(y).\]

The significance of what we just discussed is that if we knew that for each $x$ we had

\[(\partial^c \phi)(x) = \{ y(x) \}\]

then $y(x)$ has to be a Brenier solution of the Monge-problem.

This raises some questions:

(1) When is $y(x)$ even well-defined?

(2) How regular is it?

Well, these questions can be answered at least $\mu$-a.e. provided there is some differentiability (ie $X, Y$ are manifolds, $c(x,y)$ is smooth) and some regularity of the target measure. Then, Rademacher’s theorem shall come to the rescue!