Small Noise Limit and Convexity for Generalized Incompressible Flows, Schrödinger Problems, and Optimal Transport

Abstract

This paper is concerned with six variational problems and their mutual connections: the quadratic Monge–Kantorovich optimal transport, the Schrödinger problem, Brenier’s relaxed model for incompressible fluids, the so-called Brödinger problem recently introduced by Arnaudon et al. (An entropic interpolation problem for incompressible viscid fluids, arXiv preprint arXiv:1704.02126, 2017) the multiphase Brenier model, and the multiphase Brödinger problem. All of them involve the minimization of a kinetic action and/or a relative entropy of some path measures with respect to the reversible Brownian motion. As the viscosity parameter $\nu \rightarrow 0$ we establish Gamma-convergence relations between the corresponding problems, and prove the convergence of the associated pressures arising from the incompressibility constraints. We also present new results on the time-convexity of the entropy for some of the dynamical interpolations. Along the way we extend previous results by Lavenant (Calc Var Partial Differ Equ 56(6):170, 2017) and Benamou et al. (Generalized incompressible flows, multi-marginal transport and Sinkhorn algorithm, arXiv preprint arXiv:1710.08234, 2017).

Appendices

Appendices

1.1 A. Existence and Uniqueness for ${\mathsf {MBr}\ddot{\mathsf {o}}}$

Here we establish

Theorem A.1

Take $\varGamma$ bistochastic in average satisfying the entropy condition (14), and let $\nu >0$. Then ${\mathsf {MBr}\ddot{\mathsf {o}}}_{\nu }(\varGamma )$ admits a unique solution.

This is perhaps not completely standard in this form due to our choice of exposition in terms of traffic plans, and we include the details for the sake of completeness.

Proof

For the existence it suffices to show that there exists at least one admissible traffic plan ($\varvec{{\mathcal {H}}}_{\nu }$ being proper and lower semicontinuous, the direct method in the calculus of variations applies). In order to find such a traffic plan, one can either adapt the proof of [3, Cor. 5.2], or also observe that, given an admissible traffic plan ${\varvec{P}}$ for ${\mathsf {MREu}}(\varGamma )$, the traffic plan ${\varvec{P}}^{\nu }=\varPhi ^\nu _{\#}{\varvec{P}}$ constructed in the proof of Theorem 2.4 is admissible for ${\mathsf {MBr}\ddot{\mathsf {o}}}_{\nu }(\varGamma )$—in particular (49) ensures that $\varvec{{\mathcal {H}}}_\nu ({\varvec{P}}^\nu )< +\infty$.

For the uniqueness part, we first show that if ${\varvec{P}}$ is a solution to ${\mathsf {MBr}\ddot{\mathsf {o}}}_{\nu }(\varGamma )$, then the conditional law ${\varvec{P}}^{\rho _0, \rho _1} :={\varvec{P}}(\,\bullet \, | {\varvec{X}}_0 = \rho _0 , {\varvec{X}}_1 = \rho _1)$ is a Dirac mass for $\varGamma$-almost all $(\rho _0, \rho _1)$. In other words, ${\varvec{P}}$ is supported on the graph of a measurable map, which assigns to any $(\rho _0, \rho _1)$ a unique curve $m=m[\rho _0,\rho _1] \in C([0,1]; {\mathcal {P}}({\mathbb {T}}^d))$ joining $\rho _0$ to $\rho _1$. Indeed, let us define the average as

$$\begin{aligned} m[\rho _0, \rho _1] :=\int \rho {{\,\mathrm{d}\,}}{\varvec{P}}^{\rho _0, \rho _1}(\rho ) \quad \in C([0,1];{\mathcal {P}}({\mathbb {T}}^d)). \end{aligned}$$

This curve is well defined for $\varGamma$-almost all $(\rho _0, \rho _1)$, and we claim that ${\varvec{P}}^{\rho _0, \rho _1} = {\varvec{\delta }}_{m[\rho _0, \rho _1]}$ for $\varGamma$-almost all $(\rho _0, \rho _1)$. To check this, let us define

$$\begin{aligned} {\widetilde{{\varvec{P}}}} :=\int {\varvec{\delta }}_{m[\rho _0, \rho _1]} {{\,\mathrm{d}\,}}\varGamma (\rho _0, \rho _1). \end{aligned}$$

Because $m[\rho _0,\rho _1]$ has endpoints $\rho _0,\rho _1$ one can check that ${\widetilde{{\varvec{P}}}}_{0,1} = \varGamma$, and in the same spirit it is easy to see that ${\widetilde{{\varvec{P}}}}$ is incompressible in average (because ${\varvec{P}}$ is). By strict convexity of ${\mathcal {H}}_{\nu }$ and Jensen’s inequality, we have for $\varGamma$-almost all $(\rho _0, \rho _1)$ that

$$\begin{aligned} {\mathcal {H}}_{\nu }(m[\rho _0, \rho _1]) ={\mathcal {H}}_\nu \left( \int \rho {{\,\mathrm{d}\,}}{\varvec{P}}^{\rho _0, \rho _1}(\rho )\right) \leqq \int {\mathcal {H}}_{\nu }(\rho ) {{\,\mathrm{d}\,}}{\varvec{P}}^{\rho _0, \rho _1}(\rho ), \end{aligned}$$

(59)

with equality if and only if ${\varvec{P}}^{\rho _0, \rho _1}$ is a Dirac mass. To verify that equality holds as desired, let us integrate (59) with respect to $\varGamma$: by definition of ${\widetilde{{\varvec{P}}}}$ on the left-hand side, and using the disintegration formula (17) with respect to ${\varvec{P}}$ on the right-hand side, we get

$$\begin{aligned} \varvec{{\mathcal {H}}}_\nu ({{\widetilde{{\varvec{P}}}}})&= \int {\mathcal {H}}_\nu (\rho ){{\,\mathrm{d}\,}}{{\widetilde{{\varvec{P}}}}}(\rho )=\iint {\mathcal {H}}_\nu (\rho ){{\,\mathrm{d}\,}}{\varvec{\delta }}_{m[\rho _0,\rho _1]}(\rho ){{\,\mathrm{d}\,}}\varGamma (\rho _0,\rho _1)\\&=\int {\mathcal {H}}_\nu (m[\rho _0,\rho _1]) {{\,\mathrm{d}\,}}\varGamma (\rho _0,\rho _1) \\&\overset{(59)}{\leqq } \int \left( \int {\mathcal {H}}_{\nu }(\rho ) {{\,\mathrm{d}\,}}{\varvec{P}}^{\rho _0, \rho _1}(\rho )\right) {{\,\mathrm{d}\,}}\varGamma (\rho _0,\rho _1)\\&= \int {\mathcal {H}}_\nu (\rho ){{\,\mathrm{d}\,}}{\varvec{P}}(\rho )=\varvec{{\mathcal {H}}}_\nu ({\varvec{P}}). \end{aligned}$$

Since ${\varvec{P}}$ is a minimizer and ${{\widetilde{{\varvec{P}}}}}$ is admissible the reverse inequality $\varvec{{\mathcal {H}}}_\nu ({\varvec{P}})\leqq \varvec{{\mathcal {H}}}_\nu ({{\widetilde{{\varvec{P}}}}})$ holds as well, thus we must have equality in (59) for $\varGamma$-a.e. $(\rho _0,\rho _1)$ and therefore ${\varvec{P}}^{\rho _0, \rho _1} = {\varvec{\delta }}_{m[\rho _0, \rho _1]}$ as claimed.

Finally, if ${\varvec{P}}_1$ and ${\varvec{P}}_2$ are two solutions to ${\mathsf {MBr}\ddot{\mathsf {o}}}(\varGamma )$ then, because $\varvec{{\mathcal {F}}}_{\nu }$ is affine, ${\varvec{P}}_3 :=({\varvec{P}}_1 + {\varvec{P}}_2)/2$ is a solution as well and must be supported on a graph. But, ${\varvec{P}}_1,{\varvec{P}}_2$ being themselves supported on a graph, ${\varvec{P}}_1+{\varvec{P}}_2$ can be supported on a graph if and only if ${\varvec{P}}_1$ and ${\varvec{P}}_2$ coincide. Hence, uniqueness is proved. $\quad \square$

Remark A.2

From this proof it is clear that we established a slightly stronger statement, namely that any minimizer for ${\mathsf {MBr}\ddot{\mathsf {o}}}$ must be supported on a graph $(\rho _0,\rho _1)\mapsto \delta _{m[\rho _0,\rho _1]}$. This shows that the framework of traffic plans is not much more general than multiphase flows in the sense of Brenier’s parametric setting, i.e. when the phases $\rho =(\rho ^a)_{a}$ are labeled using the initial and final positions $a=(x,y)\in {\mathbb {T}}^d\times {\mathbb {T}}^d$ and the incompressibility reads $\int _{{\mathbb {T}}^d\times {\mathbb {T}}^d}\rho ^a_t {{\,\mathrm{d}\,}}a={{\,\mathrm{Leb}\,}}$ for all t. Somehow we just proved that one can allow labeling on couples $(\rho _0,\rho _1)\in {\mathcal {P}}({\mathbb {T}}^d)\times {\mathcal {P}}({\mathbb {T}}^d)$ instead of $(x,y)\in {\mathbb {T}}^d\times {\mathbb {T}}^d$, but no better.

Remark A.3

In addition to being a sufficient condition as stated above in Theorem A.1, the entropy condition (14) is in fact also necessary for ${\mathsf {MBr}\ddot{\mathsf {o}}}(\varGamma )$ to admit a (unique) solution. Indeed, by the classical Logarithmic Sobolev Inequality, the Fisher Information controls the entropy $H(\rho )\leqq C_d F(\rho )$. Since $\varvec{{\mathcal {F}}}({\varvec{P}})=\int _0^1\int F(\rho _t){{\,\mathrm{d}\,}}{\varvec{P}}(\rho ) {{\,\mathrm{d}\,}}t <\infty$ there exists at least a time $t_0\in [0,1]$ such that the average entropy $\int H(\rho _{t_0}){{\,\mathrm{d}\,}}{\varvec{P}}(\rho )\leqq C\int F(\rho _{t_0}){{\,\mathrm{d}\,}}{\varvec{P}}(\rho )<\infty$. Moreover, for an $AC^2$ curve the time derivative of the entropy can be computed by the chain rule $\frac{d}{dt}H(\rho _t)=\int \nabla \log \rho _t\cdot c_t\,{{\,\mathrm{d}\,}}\rho _t$, where $c_t(x)$ corresponds to the metric speed ${{\dot{\rho }}}_t$ in Theorem 2.1. By the definition of $\varvec{{\mathcal {H}}}_\nu$, any plan with finite entropy $\varvec{{\mathcal {H}}}_\nu ({\varvec{P}})<\infty$ has both its metric speed ${{\dot{\rho }}}_t$ and Fisher information $F(\rho _t)$ controlled in the $L^2$ sense, hence $\frac{d}{dt} H(\rho _t)$ is controlled in $L^1$. This $L^1$ bound allows one to propagate $\int H(\rho _{t_0}){{\,\mathrm{d}\,}}{\varvec{P}}(\rho )<\infty$ to $\int H(\rho _{t}){{\,\mathrm{d}\,}}{\varvec{P}}(\rho )<\infty$ the whole interval $t\in [0,1]$, in particular to $t=0$ and $t=1$.

1.2 B. Properties of the Brownian Motion on ${\mathbb {T}}^d$

Here we give detailed proofs of some technical results used in Section 3 for the Brownian motion and bridges on the torus, mainly Lemma 3.3 and Lemma 3.5. Throughout this “Appendix”, barred quantities will live in ${\mathbb {R}}^d$, while unbarred quantities will live in the torus. Typically, we shall write ${\overline{\omega }}\in C([0,1];{\mathbb {R}}^d)$ and $\omega \in C([0,1];{\mathbb {T}}^d)$, and the canonical processes will read ${\overline{X}}_t:{\bar{\omega }}\mapsto {\bar{\omega }}_t$ and $X_t:\omega \mapsto \omega _t$. The pinned Brownian motions read accordingly ${\overline{R}}{}^\nu _{{\bar{x}}}$ on the whole space and $R^\nu _x$ on the torus.

1.2.1 Brownian Bridges on the Torus

By definition, the pinned Brownian motion on ${\mathbb {T}}^d$ is nothing but the projection of a pinned Brownian motion on ${\mathbb {R}}^d$, i.e. $R^\nu _x=\varPi _{\#}{\overline{R}}{}^\nu _{{\bar{x}}}$ as soon as the starting points are chosen consistently, $x=\pi ({\bar{x}})$. However, the bridges on ${\mathbb {T}}^d$,

$$\begin{aligned} R^{\nu ,x,y} :=R^\nu _x(\,\bullet \,|\,X_1=y) = R^{\nu }(\,\bullet \,|\,X_0=x,X_1=y), \end{aligned}$$

are not the projection of the bridges on ${\mathbb {R}}^d$,

$$\begin{aligned} {\overline{R}}{}^{\nu ,{\bar{x}},{\bar{y}}} :={\overline{R}}{}^{\nu }_{{\bar{x}}}(\,\bullet \,|\,{\overline{X}}_1={\bar{y}}). \phantom {= R^{\nu }(\,\bullet \,|\,X_0=x,X_1=y)} \end{aligned}$$

The former can however be expressed as mixtures of the latter, as follows:

Lemma B.1

Take x and y in ${\mathbb {T}}^d$, and choose any lifts ${{\bar{x}}}$ and ${{\bar{y}}}$ in ${\mathbb {R}}^d$ such that $\pi ({{\bar{x}}})=x$ and $\pi ({{\bar{y}}})=y$. Then

$$\begin{aligned} R^{\nu , x, y} = \frac{1}{Z^{\nu ,x,y}}\sum _{{\bar{l}} \in {\mathbb {Z}}^d} \exp \left( - \frac{|{{\bar{y}}}-{{\bar{x}}} + {\bar{l}}|^2}{2 \nu } \right) \varPi _{\#}{\overline{R}}{}^{\nu , {{\bar{x}}}, {{\bar{y}}} + {\bar{l}}}, \end{aligned}$$

where $Z^{\nu , x,y}:=\sum \limits _{{\bar{l}} \in {\mathbb {Z}}^d} \exp \left( - \frac{|{{\bar{y}}}-{{\bar{x}}} + {\bar{l}}|^2}{2 \nu } \right) >0$ is a normalization constant.

Remark that because of (20),

$$\begin{aligned} Z^{\nu ,x,y} =(2 \pi \nu )^{d/2} \tau _{\nu }(y-x). \end{aligned}$$

(60)

Proof

We only give the proof for $x=0$, the general case is identical by shift invariance. We first observe that the right-hand side in our statement obviously does not depend on the particular choice of the lifts, hence for $x=0$ it suffices to establish equality with $\overline{x} = 0$. Let $\overline{R}{}_0^\nu$ and $R_0^\nu = \varPi _{\#}\overline{R}{}^\nu _0$ be the Brownian motions started from the origin on ${\mathbb {R}}^d$ and ${\mathbb {T}}^d$, respectively, and consider $i: {\mathbb {T}}^d \rightarrow {\mathbb {R}}^d$ a measurable right inverse of the canonical projection $\pi : {\mathbb {R}}^d \rightarrow {\mathbb {T}}^d$. Applying (18) to disintegrate $p = \overline{R}{}_0^\nu$ with respect to $\varPhi = {\overline{X}}_1$, we have

$$\begin{aligned} \overline{R}{}_0^\nu = \int _{{\mathbb {R}}^d} \overline{R}{}^{\nu ,0,\overline{y}}\frac{1}{\sqrt{2\pi \nu }^d} \exp \left( - \frac{|\overline{y}|^2}{2 \nu } \right) {{\,\mathrm{d}\,}}\overline{y}. \end{aligned}$$

Partitioning ${\mathbb {R}}^d$ in cubes $\left\{ \overline{l} + i({\mathbb {T}}^d)\right\} _{\overline{l} \in {\mathbb {Z}}^d}$ leads to

$$\begin{aligned} \overline{R}{}_0^\nu = \int _{{\mathbb {T}}^d} \frac{1}{\sqrt{2\pi \nu }^d}\sum _{\overline{l} \in {\mathbb {Z}}^d}\exp \left( - \frac{|i(y) + \overline{l}|^2}{2 \nu } \right) \overline{R}{}^{\nu ,0,i(y) + \overline{l}}{{\,\mathrm{d}\,}}y, \end{aligned}$$

hence by the linearity of the pushforward operation $\varPi _{\#}$, we get

$$\begin{aligned} R_0^\nu = \varPi _{\#}\overline{R}{}^\nu _0= \int _{{\mathbb {T}}^d} \Bigg \{ \frac{1}{\sqrt{2\pi \nu }^d} \sum _{\overline{l} \in {\mathbb {Z}}^d}\exp \left( - \frac{|i(y) + \overline{l}|^2}{2 \nu } \right) \varPi _{\#}\overline{R}{}^{\nu ,0,i(y) + \overline{l}} \Bigg \}{{\,\mathrm{d}\,}}y. \end{aligned}$$

Remark next that, for all y and by definition of the inverse $\pi \circ i(y)=y$, the integrand in this right-hand side is a measure supported on the fiber $\{X_1 = y\}$. By the uniqueness in the disintegration theorem [2, Thm. 5.3.1] one simply reads off of the last equality the conditioning

$$\begin{aligned} R^{\nu ,0,y}=R^\nu _0(\bullet \,|\,X_1=y) =\frac{1}{\sqrt{2\pi \nu }^d} \sum _{\overline{l} \in {\mathbb {Z}}^d}\exp \left( - \frac{|i(y) + \overline{l}|^2}{2 \nu } \right) \varPi _{\#}\overline{R}{}^{\nu ,0,i(y) + \overline{l}}, \end{aligned}$$

and the result follows. $\quad \square$

1.2.2 Proof of Lemma 3.3

Let us start with estimate (25). Since the increments of the Brownian motion are independent and stationary we have first that

$$\begin{aligned} \int \exp \left( \alpha \frac{A_N(\omega )}{\nu }\right) {{\,\mathrm{d}\,}}R^\nu (\omega )= & {} \int \exp \left( \alpha \sum \limits _{n=0}^{N-1}\frac{{\mathsf {d}}^2(\omega _{t_n},\omega _{t_{n+1}})}{2\nu \tau }\right) {{\,\mathrm{d}\,}}R^{\nu }(\omega ) \nonumber \\= & {} \left[ \int \exp \left( \alpha \frac{{\mathsf {d}}^2(\omega _{0},\omega _{\tau })}{2\nu \tau }\right) {{\,\mathrm{d}\,}}R^{\nu }(\omega )\right] ^N, \end{aligned}$$

(61)

with $\tau =1/N$. Whence, by definition of the reversible Brownian motion $R^{\nu }$,

$$\begin{aligned}&\int \exp \left( \alpha \frac{{\mathsf {d}}^2(\omega _{0},\omega _{\tau })}{2\nu \tau }\right) {{\,\mathrm{d}\,}}R^{\nu }(\omega ) \\&\quad = \frac{1}{(2\pi \nu \tau )^{d/2}} \int _{{\mathbb {R}}^d} \exp \left( \alpha \frac{{\mathsf {d}}^2(0, \pi ({\bar{y}}))}{2\nu \tau }\right) \exp \left( -\frac{|{\bar{y}}|^2}{2\nu \tau } \right) {{\,\mathrm{d}\,}}{\bar{y}} \\&\quad \leqq \frac{1}{(2\pi \nu \tau )^{d/2}} \int _{{\mathbb {R}}^d} \exp \left( (\alpha -1) \frac{|{\bar{y}}|^2}{2\nu \tau }\right) {{\,\mathrm{d}\,}}{\bar{y}}, \end{aligned}$$

where we used ${\mathsf {d}}(0,\pi ({\bar{y}}))\leqq |{\bar{y}}-0|$ in the last line. Because we were cautious enough to choose $\alpha < 1$ this quantity is finite, and changing variables ${\bar{z}}=\sqrt{\frac{1-\alpha }{\nu \tau }} {\bar{y}}$ in the integral yields

$$\begin{aligned} \int \exp \left( \alpha \frac{{\mathsf {d}}^2(\omega _{0},\omega _{\tau })}{2\nu \tau }\right) {{\,\mathrm{d}\,}}R^{\nu }(\omega )&\leqq \frac{1}{(2\pi (1-\alpha ) )^{d/2}} \int _{{\mathbb {R}}^d} \exp \left( -\frac{|{\bar{z}}|^2}{2} \right) {{\,\mathrm{d}\,}}{\bar{z}} \nonumber \\&= \frac{1}{(1-\alpha )^{d/2}}. \end{aligned}$$

(62)

Gathering (61) and (62) gives, exactly, (25).

For the conditioned version (26), choose arbitrary lifts ${{\bar{x}}},{\bar{y}}\in {\mathbb {R}}^d$ of $x,y\in {\mathbb {T}}^d$. From Lemma B.1 we deduce that

$$\begin{aligned}&\int \exp \left( \frac{\alpha }{\nu } A_N(\omega )\right) {{\,\mathrm{d}\,}}R^{\nu ,x,y}(\omega )\nonumber \\&\quad = \frac{1}{Z^{\nu ,x,y}} \sum _{{\bar{l}} \in {\mathbb {Z}}^d} \exp \left( - \frac{|{{\bar{y}}}-{{\bar{x}}} + {\bar{l}}|^2}{2\nu } \right) \int \exp \left( \frac{\alpha }{\nu } A_N\circ \varPi ({\overline{\omega }})\right) {{\,\mathrm{d}\,}}{\overline{R}}{}{}^{\nu ,{{\bar{x}}},{{\bar{y}}}+{\bar{l}}}({\overline{\omega }}).\nonumber \\ \end{aligned}$$

(63)

For arbitrary points ${{\bar{p}}},{{\bar{q}}}\in {\mathbb {R}}^d$ we first estimate the terms

$$\begin{aligned} \varLambda ({{\bar{p}}},{{\bar{q}}})&:=\int \exp \left( \frac{\alpha }{\nu } A_N\circ \varPi ({\overline{\omega }}))\right) {{\,\mathrm{d}\,}}\overline{R}{}^{\nu ,{{\bar{p}}},{{\bar{q}}}}({\overline{\omega }})\\&= \int \exp \left( \frac{\alpha }{\nu } \sum \limits _{n=0}^{N-1}\frac{{\mathsf {d}}^2(\pi ({\overline{\omega }}_{t_n}),\pi ({\overline{\omega }}_{t_{n+1}}))}{2\tau }\right) {{\,\mathrm{d}\,}}\overline{R}{}^{\nu ,{{\bar{p}}},{{\bar{q}}}}({\overline{\omega }}), \end{aligned}$$

appearing in the summand of (63). Since ${\mathsf {d}}(\pi ({\overline{u}}), \pi ({\overline{v}})) \leqq |{\overline{v}}-{\overline{u}}|$, we have

$$\begin{aligned} \varLambda ({\bar{p}},{\bar{q}})&\leqq \int \exp \left( \frac{\alpha }{2 \nu \tau } \sum _{n=0}^{N-1} |{\overline{\omega }}_{t_{n+1}} - {\overline{\omega }}_{t_n}|^2\right) {{\,\mathrm{d}\,}}\overline{R}{}^{\nu , {\bar{p}}, {\bar{q}}}({\overline{\omega }})\\&={\mathbb {E}}_{\overline{R}{}^{\nu , {\bar{p}},{\bar{q}}}}\left[ \exp \left( \frac{\alpha }{2 \nu \tau } \sum _{n=0}^{N-1} |{\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}|^2\right) \right] . \end{aligned}$$

Moreover, the law of the canonical process ${\overline{X}}_t$ under the bridge $\overline{R}{}^{\nu , {\bar{p}},{\bar{q}}}$ is the same as the law of ${\overline{Y}}_t:={\overline{X}}_t + (1-t)({\bar{p}}-{\overline{X}}_0) + t({\bar{q}}-{\overline{X}}_1)$ under the law of the standard Brownian motion ${\overline{R}}{}^{\nu }_0$ started from the origin. Thus

$$\begin{aligned} \varLambda ({\bar{p}},{\bar{q}})&\leqq {\mathbb {E}}_{\overline{R}{}^{\nu }_0}\left[ \exp \left( \frac{\alpha }{2 \nu \tau } \sum _{n=0}^{N-1} |{\overline{Y}}_{t_{n+1}} - {\overline{Y}}_{t_n}|^2\right) \right] \\&= {\mathbb {E}}_{\overline{R}{}^{\nu }_0}\left[ \exp \left( \frac{\alpha }{2 \nu \tau } \sum _{n=0}^{N-1} \Big |({\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}) + \tau \Big \{({\bar{q}} - {\bar{p}}) - ({\overline{X}}_1 - {\overline{X}}_0)\Big \}\Big |^2\right) \right] . \end{aligned}$$

Expanding $(a+b-c)^2$ in the sum and recalling that $N \tau = 1$, it is easy to get

$$\begin{aligned}&\sum _{n=0}^{N-1} \left| ({\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}) + \tau ({\bar{q}} - {\bar{p}}) - \tau ({\overline{X}}_1 - {\overline{X}}_0)\right| ^2 \\&\quad = \sum _{n=0}^{N-1} \left| {\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}\right| ^2 + \sum _{n=0}^{N-1} \tau ^2 \left| {\bar{q}} - {\bar{p}}\right| ^2 + \sum _{n=0}^{N-1} \tau ^2 \left| {\overline{X}}_1 - {\overline{X}}_0\right| ^2 \\&\qquad +2 \sum _{n=0}^{N-1}({\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n})\cdot \tau ({\bar{q}} - {\bar{p}}) -2 \sum _{n=0}^{N-1}\tau ({\bar{q}}-{\bar{p}}) \cdot \tau ({\overline{X}}_1 - {\overline{X}}_0)\\&\qquad - 2 \sum _{n=0}^{N-1}({\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n})\cdot \tau ({\overline{X}}_1 - {\overline{X}}_0) \\&\quad = \sum _{n=0}^{N-1} |{\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}|^2 + \tau | {\bar{q}} - {\bar{p}}|^2 + \tau |{\overline{X}}_1 - {\overline{X}}_0|^2\\&\qquad + 2\tau ({\overline{X}}_1-{\overline{X}}_0)({\bar{q}}-{\bar{p}}) -2\tau ({\bar{q}}-{\bar{p}})({\overline{X}}_1-{\overline{X}}_0) - 2\tau |{\overline{X}}_1-{\overline{X}}_0|^2 \\&\quad = \sum _{n=0}^{N-1} |{\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}|^2 + \tau | {\bar{q}} - {\bar{p}}|^2 - \tau |{\overline{X}}_1 - {\overline{X}}_0|^2. \end{aligned}$$

As a consequence, and by the independence of the Brownian increments,

$$\begin{aligned} \varLambda&({\bar{p}},{\bar{q}}) \\&\leqq \exp \left( \frac{\alpha }{2 \nu } |{\bar{q}} - {\bar{p}}|^2 \right) {\mathbb {E}}_{\overline{R}{}^{\nu }_0}\left[ \exp \left( \frac{\alpha }{ 2\nu \tau } \left\{ \sum _{n=0}^{N-1} |{\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}|^2 - \tau |{\overline{X}}_1 - {\overline{X}}_0|^2 \right\} \right) \right] \\&\leqq \exp \left( \frac{\alpha }{2 \nu } |{\bar{q}} - {\bar{p}}|^2 \right) {\mathbb {E}}_{\overline{R}^{\nu }_0}\left[ \exp \left( \frac{\alpha }{ 2\nu \tau } \sum _{n=0}^{N-1} |{\overline{X}}_{t_{n+1}} - {\overline{X}}_{t_n}|^2 \right) \right] \\&= \exp \left( \frac{\alpha }{2 \nu } |{\bar{q}} - {\bar{p}}|^2 \right) \times \left( {\mathbb {E}}_{\overline{R}^{\nu }_0}\left[ \exp \left( \frac{\alpha }{ 2\nu \tau }|{\overline{X}}_{\tau } - {\overline{X}}_{0}|^2 \right) \right] \right) ^N \\&= \displaystyle {\exp \left( \frac{\alpha }{2 \nu } |{\bar{q}} - {\bar{p}}|^2 \right) }\times \frac{1}{(1 - \alpha )^{Nd/2}}, \end{aligned}$$

where the last equality follows from the same explicit computation as in (62) with ${\overline{X}}_0=0$ for ${\overline{R}}^\nu _0$-almost all $\omega$.

Finally, setting ${\bar{p}}={\bar{x}}$ and ${\bar{q}}={\bar{y}}+{\bar{l}}$ as in (63), using formulas (20) and (60), and the dimensional bounds (19) on the heat kernel, we get, when $\nu \leqq 1$, that

$$\begin{aligned}&\int \exp \Big (\frac{\alpha }{\nu } A_N(\omega )\Big ) {{\,\mathrm{d}\,}}R^{\nu ,x,y}(\omega ) \\&\quad \leqq \frac{1}{(1 - \alpha )^{dN/2}} \times \frac{1}{Z^{\nu ,x,y}} \sum _{{\bar{l}} \in {\mathbb {Z}}^d} \exp \left( - (1 - \alpha ) \frac{|{\bar{y}}-{\bar{x}} + {\bar{l}}|^2}{2 \nu } \right) \\&\quad = \frac{1}{(1 - \alpha )^{dN/2}} \times \frac{\tau _{\frac{\nu }{1-\alpha }}(y-x)}{\tau _{\nu }(y-x)} \\&\quad \leqq \frac{1}{(1 - \alpha )^{dN/2}} \times \frac{K_d\exp \left( -\frac{(1-\alpha )}{2\nu } {\mathsf {d}}^2(x,y) \right) }{k_d\exp \left( -\frac{1}{2\nu } {\mathsf {d}}^2(x,y) \right) }\\&\quad \leqq \frac{C_d}{(1 - \alpha )^{dN/2}} \exp \left( \frac{\alpha }{2\nu } {\mathsf {d}}^2(x,y) \right) , \end{aligned}$$

and the proof is achieved. $\quad \square$

1.2.3 Proof of Lemma 3.5

We first establish the corresponding result in the whole space. As in the torus in (27), we define the translation operator in ${\mathbb {R}}^d$ as

$$\begin{aligned} {\overline{T}}_{\overline{\omega }}:\quad \overline{\alpha } \in C([0,1]; {\mathbb {R}}^d) \quad \mapsto \quad \overline{\omega } + \overline{\alpha } \in C([0,1]; {\mathbb {R}}^d). \end{aligned}$$

We recall that ${\overline{B}}{}^{\nu } = {\overline{R}}{}^{\nu , 0,0}$ is the Brownian bridge of diffusivity $\nu$ on ${\mathbb {R}}^d$ joining 0 to 0. We have

Lemma B.2

Let $\overline{\omega } \in AC^2([0,1]; {\mathbb {R}}^d)$ and $\nu >0$. Then

$$\begin{aligned} \nu H\left( {\overline{T}}_{\overline{\omega }} {}_{\#}{\overline{B}}{}^{\nu } \,|\, {\overline{R}}{}^{\nu , \overline{\omega }_0, \overline{\omega }_1}\right) = \frac{1}{2} \int _0^1 |\dot{\overline{\omega }}_t|^2 {{\,\mathrm{d}\,}}t - \frac{|\overline{\omega }_1 - \overline{\omega }_0|^2}{2}. \end{aligned}$$

Proof

We will rather establish the following equivalent formula: if ${{\overline{\alpha }}} \in AC^2([0,1]; {\mathbb {R}}^d)$ satisfies ${{\overline{\alpha }}}_0 = {{\overline{\alpha }}}_1 = 0$, then, for all $\nu >0$ and ${\bar{x}},{\bar{y}} \in {\mathbb {R}}^d$,

$$\begin{aligned} \nu H\left( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{R}}{}^{\nu , {\bar{x}}, {\bar{y}}} \, \Big | \, {\overline{R}}{}^{\nu , {\bar{x}}, {\bar{y}}}\right) = \frac{1}{2}\int _0^1 |\dot{{{\overline{\alpha }}}}_t|^2 {{\,\mathrm{d}\,}}t. \end{aligned}$$

(64)

If ${\overline{\xi }}_t:=(1-t) \overline{\omega }_0 + t \overline{\omega }_1$, it will then suffice to apply this formula with ${{\overline{\alpha }}}_ t:=\overline{\omega }_t - {\overline{\xi }}_t$ and to use the identities ${\overline{T}}_{\overline{\omega }} = {\overline{T}}_{{{\overline{\alpha }}}} \circ {\overline{T}}_{{\overline{\xi }}}$ and ${\overline{T}}_{\xi } {}_{\#}{\overline{B}}{}^{\nu } = {\overline{R}}{}^{\nu , \overline{\omega }_0, \overline{\omega }_1}$.

Let us prove (64). We fix ${{\overline{\alpha }}} \in AC^2([0,1]; {\mathbb {R}}^d)$ with ${{\overline{\alpha }}}_0 = {{\overline{\alpha }}}_1 = 0$ and $\nu >0$. First, by the standard Cameron–Martin formula, if $\overline{R}{}^{\nu }$ is any $\nu$ Brownian motion on ${\mathbb {R}}^d$, then

$$\begin{aligned} \nu H({\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}\overline{R}{}^{\nu } \, | \, \overline{R}{}^{\nu }) = \frac{1}{2} \int _0^1 |\dot{{{\overline{\alpha }}}}_t|^2 {{\,\mathrm{d}\,}}t. \end{aligned}$$

Noticing that the marginals $(\overline{R}{}^{\nu })_{0,1}$ and $(T_{{{\overline{\alpha }}}} {}_{\#}\overline{R}{}^{\nu })_{0,1}$ coincide (because ${{\overline{\alpha }}}_0={{\overline{\alpha }}}_1=0$), we can apply Proposition 2.10 in order to condition on the endpoints $(X_0, X_1)$ and get

$$\begin{aligned} H({\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}\overline{R}{}^{\nu } \, | \, \overline{R}{}^{\nu }) = 0 + \int H\Big ( ({\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}\overline{R}{}^{\nu })^{{\bar{x}}, {\bar{y}}} \, \Big | \, {\overline{R}}{}^{\nu ,{\bar{x}}, {\bar{y}}} \Big ) {{\,\mathrm{d}\,}}\overline{R}{}^{\nu }_{0,1}({\bar{x}}, {\bar{y}}). \end{aligned}$$

Gathering these two formulas and observing that $({\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}\overline{R}{}^{\nu })^{{\bar{x}}, {\bar{y}}} = {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}({\overline{R}}{}^{\nu ,{\bar{x}}, {\bar{y}}})$, we get

$$\begin{aligned} \nu \int H( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{R}}{}^{\nu ,{\bar{x}},{\bar{y}}} \, | \, {\overline{R}}{}^{\nu , {\bar{x}},{\bar{y}}} ) {{\,\mathrm{d}\,}}\overline{R}{}^{\nu }_{0,1}(x,y) = \frac{1}{2} \int _0^1 |\dot{{{\overline{\alpha }}}}_t|^2 {{\,\mathrm{d}\,}}t. \end{aligned}$$

(65)

Finally, take ${\bar{x}},{\bar{y}}\in {\mathbb {R}}^d$ and consider the geodesic ${\overline{\xi }}_t:=(1-t) {\bar{x}} + t {\bar{y}}$. Then ${\overline{R}}{}^{\nu , {\bar{x}},{\bar{y}}} = {\overline{T}}_{{\overline{\xi }}} {}_{\#}{\overline{B}}{}^{\nu }$, the translations ${\overline{T}}_{{\overline{\xi }}}$ and ${\overline{T}}_{{{\overline{\alpha }}}}$ commute, and ${\overline{T}}_{{\overline{\xi }}}$ is invertible, whence, by Proposition 2.11,

$$\begin{aligned} H( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{R}}{}^{\nu ,{\bar{x}}, {\bar{y}}} \, | \, {\overline{R}}{}^{\nu , {\bar{x}},{\bar{y}}} )&= H( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{T}}_{{\overline{\xi }}} {}_{\#}{\overline{B}}{}^{\nu } \, | \, {\overline{T}}_{{\overline{\xi }}} {_{\#}} {\overline{B}}{}^{\nu } ) \\&=H( {\overline{T}}_{\xi } {}_{\#}{\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{B}}{}^{\nu } \, | \, {\overline{T}}_{{\overline{\xi }}} {_{\#}} {\overline{B}}{}^{\nu } ) =H( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{B}}{}^{\nu } \, | \, {\overline{B}}{}^{\nu } ). \end{aligned}$$

Finally, exploiting (65), we get, for all ${\bar{x}}$ and ${\bar{y}}$ in ${\mathbb {R}}^d$,

$$\begin{aligned} \nu H( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{R}}{}^{\nu ,{\bar{x}},{\bar{y}}} | {\overline{R}}{}^{\nu , {\bar{x}},{\bar{y}}} ) = \nu H( {\overline{T}}_{{{\overline{\alpha }}}} {}_{\#}{\overline{B}}{}^{\nu } | {\overline{B}}{}^{\nu } ) = \frac{1}{2} \int _0^1 |\dot{{{\overline{\alpha }}}}_t|^2 {{\,\mathrm{d}\,}}t, \end{aligned}$$

and the proof is complete. $\quad \square$

In order to deduce Lemma 3.5 from Lemma B.2 we need a canonical construction of a process on ${\mathbb {R}}^d$ out of a process on ${\mathbb {T}}^d$. To this end, we choose $i : {\mathbb {T}}^d \rightarrow {\mathbb {R}}^d$ a measurable right inverse of the projection $\pi$ with bounded image. For $\omega \in C([0,1]; {\mathbb {T}}^d)$ we denote by $I(\omega )$ the unique lift of $\omega$ starting from $i(\omega _0)$, and I is of course a measurable right inverse of $\varPi$. The entropy is invariant under the canonical projection in the following sense:

Lemma B.3

Take $\overline{P}$ a probability measure and $\overline{R}$ a finite positive Radon measure on $C([0,1]; {\mathbb {R}}^d)$. Suppose that $\overline{P} \ll \overline{R}$ and that, $\overline{R}$-almost surely, ${\overline{X}}_0 = i(\pi ({\overline{X}}_0))$. Then

$$\begin{aligned} H(\varPi _{\#}\overline{P}\,|\, \varPi _{\#}\overline{R}) = H(\overline{P}\,|\,\overline{R}). \end{aligned}$$

Proof

On the set $\{ {\overline{X}}_0 = i(\pi ({\overline{X}}_0)) \}$ we have $\overline{R}$-almost surely $I \circ \varPi = {{\,\mathrm{Id}\,}}$, and our statement is a direct consequence of Proposition 2.11. $\quad \square$

With the above definition of the lift $I(\omega )$, observe that for all $\omega \in C([0,1]; {\mathbb {T}}^d)$ the shifted bridge $B^{\nu }_{\omega }$ from Definition 3.4 satisfies that

$$\begin{aligned} B^{\nu }_{\omega } :=T_{\omega } {}_{\#}\varPi _{\#}{\overline{B}}{}^{\nu } = \varPi _{\#}{\overline{T}}_{I(\omega )} {}_{\#}{\overline{B}}{}^{\nu }. \end{aligned}$$

We are finally in position to establish Lemma 3.5.

Proof of Lemma 3.5

For notational convenience, we denote the lift of $\omega$ by

$$\begin{aligned} {\overline{\omega }}:=I(\omega )\in C([0,1];{\mathbb {R}}^d). \end{aligned}$$

By Lemma B.1, we have

$$\begin{aligned} R^{\nu , \omega _0, \omega _1}&= \frac{1}{Z^{\nu ,\omega _0,\omega _1}} \sum _{{\bar{l}} \in {\mathbb {Z}}^d} \exp \left( - \frac{|{\overline{\omega }}_1-{\overline{\omega }}_0 + {\bar{l}}|^2}{2 \nu } \right) \varPi _{\#}{\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1 + {\bar{l}}}\\&= \varPi _{\#}\Bigg ( \underbrace{ \frac{1}{Z^{\nu ,\omega _0,\omega _1}} \sum _{{\bar{l}} \in {\mathbb {Z}}^d} \exp \left( - \frac{|{\overline{\omega }}_1-{\overline{\omega }}_0 + {\bar{l}}|^2}{2 \nu } \right) {\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1 + {\bar{l}}}}_{ :=\overline{B} {}^{\nu , \omega _0, \omega _1}} \Bigg ). \end{aligned}$$

(The superscripts in $\overline{B} {}^{\nu , \omega _0, \omega _1}$ do not stand for the conditioning of some ${\overline{B}}^\nu$, but rather emphasizes the dependence of the measure $\overline{B} {}^{\nu , \omega _0, \omega _1}$ on the fixed endpoints $\omega _0,\omega _1\in {\mathbb {R}}^d$.) Observe that

• all the measures involved in the definition of $\overline{B} {}^{\nu , \omega _0, \omega _1}$ are mutually singular (because ${\overline{R}}{}^{\nu ,{\overline{a}},{\overline{b}}}\perp {\overline{R}}{}^{\nu ,{\overline{a}},{\overline{b}}'}$ as soon as ${\overline{b}}\ne {\overline{b}}'$),

• ${\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu } \ll {\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1}$ by Lemma B.2,

• ${\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1} \ll \overline{B} {}^{\nu , \omega _0, \omega _1}$ (because ${\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1}$ appears in the sum defining the measure $\overline{B} {}^{\nu , \omega _0, \omega _1}$ for ${\bar{l}}=0$),

As a consequence, ${\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu } \ll \overline{B} {}^{\nu , \omega _0, \omega _1}$, and the corresponding Radon–Nikodym derivative only involves the ${\bar{l}}=0$ contribution in ${\overline{B}} {}^{\nu ,\omega _0,\omega _1}$.

Moreover by definition 3.4 of $B^\nu _\omega =T_\omega {}_{\#}\varPi {}_{\#}{\overline{B}} ^\nu$ and because $T_\omega \circ \varPi =\varPi \circ {\overline{T}}_{{\overline{\omega }}}$ one can write $B^\nu =\varPi {}_{\#}{\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu }$. Since $\overline{B} {}^{\nu , \omega _0, \omega _1}$ and, almost surely, ${\overline{X}}_0 = i(\pi ({\overline{X}}_0))$, we can apply Lemma B.3 as follows:

$$\begin{aligned} H(B^{\nu }_{\omega }\, | \, R^{\nu , \omega _0, \omega _1})&= H( \varPi {_{\#}}{\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu }\, | \,\varPi {}_{\#}\overline{B} {}^{\nu , \omega _0, \omega _1}) \\&= H( {\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu }\, | \,\overline{B} {}^{\nu , \omega _0, \omega _1}) \\&=H\left( {\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu } \,\Bigg | \, \frac{1}{Z^{\nu ,\omega _0,\omega _1}} \exp \left( - \frac{| {\overline{\omega }}_1-{\overline{\omega }}_0 |^2}{2\nu }\right) {\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1} \right) \\&= H\left( {\overline{T}}_{{\overline{\omega }}} {}_{\#}{\overline{B}}{}^{\nu } \, | \, {\overline{R}}{}^{\nu , {\overline{\omega }}_0, {\overline{\omega }}_1}\right) + \frac{| {\overline{\omega }}_1-{\overline{\omega }}_0 | ^2}{2\nu } + \log Z^{\nu ,\omega _0,\omega _1}, \end{aligned}$$

where we used Proposition 2.9 in the last equality. We can compute the first entropy term in the right hand side using Lemma B.2 (remark that the kinetic action of $\omega$ on the torus coincides with that of its lift ${\overline{\omega }}$) and we can estimate the last term using (60), (19), which leads to

$$\begin{aligned} \nu H(B^{\nu }_{\omega }\,|\, R^{\nu , \omega _0, \omega _1})&\leqq \left( \frac{1}{2}\int _0^1|{{\dot{{\overline{\omega }}}}}_t|^2\,{{\,\mathrm{d}\,}}t - \frac{1}{2}|{\overline{\omega }}_1-{\overline{\omega }}_0|^2\right) \\&\quad + \nu \left( \frac{|{\overline{\omega }}_1-{\overline{\omega }}_0|^2}{2\nu } + \log K_d - \frac{{\mathsf {d}}^2(\omega _0, \omega _1)}{2 \nu }\right) \\&=\frac{1}{2} \int _0^1 |{\dot{\omega }}_t|^2 {{\,\mathrm{d}\,}}t - \frac{{\mathsf {d}}^2(\omega _0, \omega _1)}{2 \nu } + \nu \log K_d, \end{aligned}$$

and concludes the proof with $C :=\log K_d$. $\quad \square$