# Strict convexity of the cost function is enough to ensure the existence and uniqueness of the optimal transport map

Let $$X=Y = \mathbb R^d$$ and $$c:X \times Y \to [0, \infty)$$ be Borel measurable. Let $$\mu, \nu$$ be Borel probability measures on $$X,Y$$ respectively. Let $$\mathcal T$$ be the set of all Borel measurable maps $$T:X \to Y$$ such that $$T_\sharp \mu = \nu$$. Let $$\mathbb M (T) := \int_X c(x, T(x)) \mathrm d \mu(x) \quad \forall T \in \mathcal T.$$

Then we are interested in the Monge's transportation problem $$\mathrm{MP} : \quad\inf_{T \in \mathcal T} \mathbb M (T).$$

The existence and uniqueness of the solution of $$\mathrm{MP}$$ is guaranteed if $$c$$ is strictly convex and the supports of $$\mu, \nu$$ are compact [1]. We can remove the assumption of compact supports by stronger conditions on $$c$$ [2].

Let $$\Pi (\mu, \nu)$$ be the set of Borel probability measures on $$X \times Y$$ with marginals $$\mu$$ on $$X$$ and $$\nu$$ on $$Y$$. Recently, I have seen a strong theorem from this lecture note, i.e.,

Theorem 3.14. Assume

1. $$h:\mathbb R^d \to [0, \infty)$$ is strictly convex and $$c(x, y) := h(x-y)$$ for all $$(x, y) \in X \times Y$$.
2. $$\int_{X \times Y} c \mathrm{d} \gamma <\infty$$ for some $$\gamma \in \Pi(\mu, \nu)$$,
3. $$\mu\left(\left\{x \in X: \int_Y c(x, y) \mathrm{d} \nu(y)<\infty\right\}\right)>0$$,
4. $$\nu\left(\left\{y \in Y: \int_X c(x, y) \mathrm{d} \mu(x)<\infty\right\}\right)>0$$,
5. $$\mu$$ is absolutely continuous with respect to Lebesgue measure.

Then $$\mathrm{MP}$$ has a unique (up to $$\mu$$-a.e. solution).

The conditions 2, 3, 4 are very mild and just to ensure the dual of the dual of the corresponding Kantorovich's problem has a solution in a form of a pair of $$c$$-conjugates. Theorem 3.14. is striking because it does not require the supports of $$\mu, \nu$$ to be compact nor any condition on $$c$$ besides strict convexity.

Could you elaborate if there are some references of Theorem 3.14.?

[1] Caffarelli, Luis A. "Allocation maps with general cost functions." Partial differential equations and applications. Routledge, 2017. 29-35.

[2] Gangbo, Wilfrid, and Robert J. McCann. "The geometry of optimal transportation." Acta Mathematica 177.2 (1996): 113-161.

First a comment: you write (before stating your Theorem 3.14) that "The existence and uniqueness of the solution of the Monge Problem is guaranteed if $$c$$ is strictly convex and the supports of $$\mu,\nu$$ are compact", but this is absolutely not true: you really need some conditions on the starting point, i-e that $$\mu$$ does not charge "small sets" in some sense (this is exactly assumption 5 in your theorem 3.14, but this can be relaxed). If $$\mu=\delta_x$$ is a Dirac delta there exists no transport map $$T$$ from $$\mu$$ to $$\nu$$, unless $$\nu=\delta_{T(x)}$$, so clearly the Monge problem is ill-posed in general.
Next, my real answer: this precise statement is often called the Brenier-McCann theorem. You can find an extremely general version in [1], Theorem 10.38. Note in particular that this is stated without any compactness assumptions or behaviour at infinity, and that the strict convexity $$c(x,y)=h(|x-y|)$$ is not needed (only the so-called and weaker twist condition). This result is usually credited to Brenier, Rachev and Rüschendorf for the quadratic cost in Euclidean spaces, and then R. McCann extended to Riemannian manifolds.
• It seems the assumption (Super) ($c$ is superdifferentiable everywhere) may not be salified for the cost function $c(x, y) := h(x-y)$ with $h$ being strictly convex... Jan 9 at 17:25
• Agreed, but it is in practice satisfied for all the reasonably smooth cases, e.g. as soon as $h$ is actually differentiable. Are you interested in a borderline case where this fails? Jan 9 at 17:43
• I just would like to verify Theorem 3.14. in the lecture note. Of course, convex function on $\mathbb R^n$ is locally Lipschitz and thus differentiable almost everywhere w.r.t. Lebesgue measure $\lambda$. I wonder if the condition (Super) can be weakened to "$c$ is superdifferentiable $\lambda$-a.e." If not, then I think Theorem 3.14. is not true as stated. Jan 9 at 17:54