Traversing the Schrödinger Bridge Strait: Robert Fortet’s Marvelous Proof Redux

Abstract

In the early 1930s, Erwin Schrödinger, motivated by his quest for a more classical formulation of quantum mechanics, posed a large deviation problem for a cloud of independent Brownian particles. He showed that the solution to the problem could be obtained through a system of two linear equations with nonlinear coupling at the boundary (Schrödinger system). Existence and uniqueness for such a system, which represents a sort of bottleneck for the problem, was first established by Fortet in 1938/1940 under rather general assumptions by proving convergence of an ingenious but complex approximation method. It is the first proof of what are nowadays called Sinkhorn-type algorithms in the much more challenging continuous case. Schrödinger bridges are also an early example of the maximum entropy approach and have been more recently recognized as a regularization of the important optimal mass transport problem. Unfortunately, Fortet’s contribution is by and large ignored in contemporary literature. This is likely due to the complexity of his approach coupled with an idiosyncratic exposition style and due to missing details and steps in the proofs. Nevertheless, Fortet’s approach maintains its importance to this day as it provides the only existing algorithmic proof, in the continuous setting, under rather mild assumptions. It can be adapted, in principle, to other relevant optimal transport problems. It is the purpose of this paper to remedy this situation by rewriting the bulk of his paper with all the missing passages and in a transparent fashion so as to make it fully available to the scientific community. We consider the problem in ${\mathbb {R}}^d$ rather than in ${\mathbb {R}}$ and use as much as possible his notation to facilitate comparison.

Appendix

1.1 Proof of (33) from Theorem I

Let $Z \subset {\mathfrak {I}}^2$ be the set of $\{ y \in {\mathfrak {I}}^2{:}\,g(x_0,y) = 0 \}$.

Define $Z_k = \{ y \in {\mathfrak {I}}^2{:}\,g(x_0,y) < \frac{1}{k} \}$ for $k \in {\mathbb {N}}^*$. We have $Z_{k+1} \subset Z_k$, and $Z_k \downarrow Z$ as $k \rightarrow +\infty$.

By assumption (H.vi) we know that Z has Lebesgue measure 0. From the continuity of g (H.iv), we also know that Z is closed.

Hence, $m(Z_k) \rightarrow 0$ as $k \rightarrow +\infty$.

Denote ${\mathfrak {I}}^2_k = {\mathfrak {I}}^2 \backslash Z_k$. Then we have ${\mathfrak {I}}^2_k \subset {\mathfrak {I}}^2_{k+1}$ and ${\mathfrak {I}}^2_k \uparrow {\mathfrak {I}}^2 \backslash Z$ as $k \rightarrow +\infty$.

Since

$$\begin{aligned} H_n'(x_0) = \int _{{\mathfrak {I}}^2} g(x_0,y) \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y \rightarrow 0, \quad \text { as } n \rightarrow +\infty \end{aligned}$$

$\forall \epsilon >0$, we have for n large enough:

$$\begin{aligned} \int _{{\mathfrak {I}}^2} g(x_0,y) \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y < \epsilon \end{aligned}$$

Fix $\epsilon >0, k \in {\mathbb {N}}^*$. We then have for n large enough:

$$\begin{aligned} 0 \le \int _{{\mathfrak {I}}^2_{k}} g(x_0,y) \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y + \int _{{\mathfrak {I}}^2\backslash {\mathfrak {I}}^2_{k}} g(x_0,y) \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y < \epsilon \end{aligned}$$

and in particular, by nonnegativity, the first integral yields:

$$\begin{aligned} 0 \le \int _{{\mathfrak {I}}^2_{k}} \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y < k \epsilon \end{aligned}$$

This implies that the measure $\frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y$ converges weakly to 0 on ${\mathfrak {I}}^2_{k}$. Indeed, it is the case when evaluated on any step function with support included in ${\mathfrak {I}}^2_{k}$, and step functions are dense in the family of bounded continuous functions.

We would like the measure $\frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y$ to converge to 0 for any step function whose support I is included in ${\mathfrak {I}}^2$, and not merely on ${\mathfrak {I}}^2_k$.

Pick a subset $I \subset {{\mathfrak {I}}^2}$, and consider:

$$\begin{aligned} \int _{{\mathfrak {I}}^2} \mathbb 1_{I}(y) \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y = \int _{{\mathfrak {I}}^2_{k} \cap I} \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y + \int _{({\mathfrak {I}}^2_{k} \cap I)^C} \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y \end{aligned}$$

The first integral converges to 0 as $n \rightarrow +\infty$, since the measure $\frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y$ converges weakly to 0 on ${\mathfrak {I}}^2_{k}$.

As for the second integral, we have that $H_n \le H_1$, so $\frac{\omega _2(y)}{G(H_n,y)} \le \frac{\omega _2(y)}{G(H_1,y)}$ which implies:

$$\begin{aligned} \int _{({\mathfrak {I}}^2_{k} \cap I)^C} \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y \le \int _{({\mathfrak {I}}^2_{k} \cap I)^C} \frac{\omega _2(y)}{G(H_1,y)} {\mathrm{d}}y \le \int _{Z_{k}} \frac{\omega _2(y)}{G(H_1,y)} {\mathrm{d}}y \end{aligned}$$

where the last inequality comes from $({\mathfrak {I}}^2_{k} \cap I)^C \subset Z_{k}$.

Condition ($\star$) states that $\int _{{\mathfrak {I}}^2} \frac{\omega _2(y)}{G(H_1,y)} {\mathrm{d}}y < +\infty$; thus, we know that the measure $\frac{\omega _2(y)}{G(H_1,y)} {\mathrm{d}}y$ is absolutely continuous with respect to the Lebesgue measure m on ${\mathfrak {I}}^2$. This implies that the second integral converges to 0, as $k \rightarrow +\infty$ since $m(Z_{k}) \rightarrow 0$.

Hence, for any measurable $I \subset {\mathfrak {I}}^2$, $\int _{{\mathfrak {I}}^2} \mathbb 1_{I}(y) \frac{\omega _2(y)}{G(H_n,y)} {\mathrm{d}}y \rightarrow 0$ as $n \rightarrow +\infty$.