Long-time Behaviour of Entropic Interpolations

Abstract

In this article we investigate entropic interpolations. These measure valued curves describe the optimal solutions of the Schrödinger problem Schrödinger (Sitzungsberichte Preuss Akad. Wiss. Berlin. Phys. Math. 144:144–153 1931), which is the problem of finding the most likely evolution of a system of independent Brownian particles conditionally to observations. It is well known that in the short time limit entropic interpolations converge to the McCann-geodesics of optimal transport. Here we focus on the long-time behaviour, proving in particular asymptotic results for the entropic cost and establishing the convergence of entropic interpolations towards the heat equation, which is the gradient flow of the entropy according to the Otto calculus interpretation. Explicit rates are also given assuming the Bakry-Émery curvature-dimension condition. In this respect, one of the main novelties of our work is that we are able to control the long time behavior of entropic interpolations assuming the CD(0,n) condition only.

Appendix : A: Details About the Examples

1.1 A.1 A (0,n)-convex Function

To understand what happened in the (0,n)-convex case, let’s begin by an example on the real line. The prototypal (0,1)-convex function is $F(x)=-\log x$, x > 0. This is a (0,1)-convex function since $F^{\prime \prime }=(F^{\prime })^2.$ Let x,T > 0, for simplicity we just treat the case where x = y. The gradient flow from x > 0, denoted by (S_t(x))_t≥ 0 is the solution of the ODE $\dot X_t={1}/{X_t}$ starting from x, hence for all t > 0, $S_t(x)=\sqrt {2t+x^2}$.

The Newton system associated is

$$ \left\{ \begin{array}{l} \ddot X_{t}=-\frac{1}{{X_{t}^{3}}}, \\ X_{0}=X_{T}=x. \end{array} \right. $$

Now $(X_t^T)_{t \in [0,T]}$ denote the entropic interpolation between x and x. The conserved quantity is given by $E_{T}(x,x)=\dot {X_t^T}^2-\frac {1}{{X_t^T}^2}.$ Thus $\left |\dot X_t^T\right |=\sqrt {E_{T}(x,x)+F^{\prime }(X_t^T)^2}$ and we can deduce that

$$ \left\{\begin{array}{cc} \dot {X_{t}^{T}}=\sqrt{E_{T}(x,x)+F^{\prime}({X_{t}^{T}})^{2}},& t \in [0,T/2]; \\ \dot {X_{t}^{T}} = - \sqrt{E_{T}(x,x)+F^{\prime}({X_{t}^{T}})^{2}},& t \in (T/2,T]. \end{array} \right. $$

In this example we have enough information to compute explicitly the conserved quantity.

Proposition A.1

For T > 0 and x ∈R, $E_{T}(x,x)= 2\frac {-x^2- \sqrt {x^4+T^2}}{T^2}.$

Proof

By the continuity in T/2 of $\left (X_t^T \right )_{t \in [0,T ]}$ we can deduce that $E_{T}(x,x)=-F^{\prime }(X_{T/2}^T)^2$. Notice that for all t ∈ [0,T/2]

$$ \frac{\dot {X_{t}^{T}}}{F^{\prime}({X_{t}^{T}})}= \sqrt{1+ \frac{E_{T}(x,x)}{F^{\prime}({X_{t}^{T}})^{2}}} $$

and

$$ \frac{d}{dt} \frac{\dot {X_{t}^{T}}}{F^{\prime}({X_{t}^{T}})}=-\frac{F^{\prime\prime}({X_{t}^{T}})}{F^{\prime}({X_{t}^{T}})^{2}}\mathcal{E}_{T}(x,x)=-\mathcal{E}_{T}(x,x). $$

By integration of this inequality we see that for every t ∈ [0,T/2)

$$ \sqrt{1+\frac{E_{T}(x,x)}{F^{\prime}({X_{t}^{T}})^{2}}}-\sqrt{1+\frac{E_{T}(x,x)}{F^{\prime}(x)^{2}}}=\frac{TE_{T}(x,x)}{2}. $$

(1)

When t = T/2 we get $\frac {T^2}{4}E_{T}(x,x)^2- \frac {E_{T}(x,x)}{F^{\prime }(x)^2}-1=0$ and since E_T(x,x) ≤ 0 we deduce that

$$ E_{T}(x,x)=\frac{- \frac{1}{F^{\prime}(x)^{2}}-\sqrt{\frac{1}{F^{\prime}(x)^{4}}+T^{2}}}{T^{2}/2}=\frac{-x^{2}-\sqrt{x^{4}+T^{2}}}{T^{2}/2}. $$

□

Hence E_T(x,x) is of order 1/T in this case. From Eq. 1, we can deduce an explicit formula for $X_t^T$.

Proposition A.2

For x ∈R and T > 0, the entropic interpolation from x to x is given by

$$ {X_{t}^{T}}= \sqrt{x^{2}+t^{2} E_{T}(x,x)+2t\sqrt{1+E_{T}(x,x)x^{2}}}, 0 \leq t \leq T. $$

Furthermore $X_t^T \rightarrow S_t(x)$ when $T \rightarrow \infty$, more precisely

$$ {X_{t}^{T}}-S_{t}(x) \underset{T \rightarrow \infty}{\sim} E_{T}(x,x)\frac{t+t^{2}x^{2}}{2\sqrt{x^{2}+2t}}, $$

hence there exists a constant C > 0 such that $|X_t^T-S_t(x)| \underset {T \rightarrow \infty }{\sim } \frac {C}{T}.$

In this particular case we can compute the cost in an explicit way.

Proposition A.3

For every x ∈R, $C_{T}(x,x) \underset {T \rightarrow \infty }{\sim } 2 \log (T).$

Proof

By the very defnition of the cost,

$$ \begin{array}{ll} C_{T}(x,x)&={{\int}_{0}^{T}} \left( \dot {{X_{t}^{T}}}^{2}+\frac{1}{{{X_{t}^{T}}}^{2}} \right)dt=2{\int}_{0}^{T/2} \left( \dot {{X_{t}^{T}}}^{2}+\frac{1}{{{X_{t}^{T}}}^{2}} \right)dt \\ &=4 {\int}_{0}^{T/2} \dot {{X_{t}^{T}}}^{2}dt+2 {\int}_{0}^{T/2} \left( \frac{1}{{X_{t}^{T}}}- \dot {{X_{t}^{T}}}^{2} \right)dt \\ &=4 {\int}_{0}^{T/2} \dot {X_{t}^{T}} \frac{\sqrt{1+{{X_{t}^{T}}}^{2}E_{T}(x,x)}}{{X_{t}^{T}}}-T E_{T}(x,x) \\ &={\int}_{\sqrt{-E_{T}(x,x)}x}^{1} \frac{\sqrt{1-v^{2}}}{v}dv -T E_{T}(x,x). \end{array} $$

Hence, $C_{T}(x,x) \underset {T \rightarrow \infty }{\sim } 2 \log (T).$□

1.2 A.2 The Example of two Gaussians on R

This example take place on R. This is a flat space of dimension one, that mean it verify the CD(0, 1) condition. Recall that for m ∈R and σ > 0 the normal law of expected value m and variance σ² is the probability measure on R with density against the Lebesgue measure,

$$ \mathcal{N}(m,\sigma^{2})(x)=\frac{1}{\sqrt{2\pi \sigma^{2}}}\exp \left( - \frac{(x-m)^{2}}{2 \sigma^{2}} \right). $$

In this case we know an expression for the heat semigroup, for f ∈ L^∞(R), we have

$$ \forall t>0, \ P_{t}f=\mathcal{N}(0,2t)*f. $$

Furthermore, for all m ∈R and σ,t > 0 we know that from elementary probability theory

$$ \mathcal{N}(0,2t)*\mathcal{N}(m,\sigma^{2})=\mathcal{N}(m,\sigma^{2}+2t). $$

Thanks to all of these considerations we can make explicit calculus in this case. For simplicity here we are gonna consider the case of two centered gaussian measure, that is σ² = 1. Let x₀,x₁ ∈R, T > 0, $\mu =\mathcal {N}(x_{0},1)$ and $\nu = \mathcal {N}(x_1,1)$. We can solve explicitely the Schrödinger system

$$ \left\{ \begin{array}{cc} \mu=fP_{T}g , \\ \nu=gP_{T}f, \end{array} \right. $$

by searching solutions of the form $x \mapsto a \exp \left (- \frac {(x-b)^2}{2c^2} \right )$ with a,b,c ∈R. We can make explicit computations to find two solutions given by for all x ∈R

$$ \left\{ \begin{array}{cc} f(x)=\frac{1}{\sqrt{2 \pi \mathcal{D}_{T}^{2}}} \exp \left( - \frac{\left( x - \frac{\mathcal{D}_{T}^{2}(\mathcal{D}_{T}^{2}+2T)^{2}}{(\mathcal{D}_{T}^{2}+2T)^{2}-\mathcal{D}_{T}^{2}} \left( x_{0}-\frac{\mathcal{D}_{T}^{2}}{\mathcal{D}_{T}^{2}+2T}x_{1} \right) \right)^{2}}{2\mathcal{D}_{T}^{2}}\right), \\ g(x)= \sqrt{\mathcal{D}_{T}^{2}+2T} \exp \left( - \frac{\left( x - \frac{\mathcal{D}_{T}^{2}(\mathcal{D}_{T}^{2}+2T)^{2}}{(\mathcal{D}_{T}^{2}+2T)^{2}-\mathcal{D}_{T}^{2}} \left( x_{1}-\frac{\mathcal{D}_{T}^{2}}{\mathcal{D}_{T}^{2}+2T}x_{0} \right)\right)^{2}}{2\mathcal{D}_{T}^{2}}\right), \end{array} \right. $$

where the parameter $\mathcal {D}_{T}$ is given by $\mathcal {D}_{T}^2=\sqrt {(T-1)^2+2T}-(T-1).$ Observe that f is the density of the normal law $\mathcal {N}\left (\frac {\mathcal {D}_{T}^2(\mathcal {D}_{T}^2+2T)^2}{(\mathcal {D}_{T}^2+2T)^2-\mathcal {D}_{T}^2} \left (x_{0}-\frac {\mathcal {D}_{T}^2}{\mathcal {D}_{T}^2+2T}x_1 \right ),\mathcal {D}_{T}^2 \right )$. This is an arbitrary choice, because there is only unicity up to the trivial transform (f,g)↦(cf,g/c) for some c ∈R. From those expressions we can easily deduce a formula for the entropic interpolation $(\mu _t^T)_{t \in [0,T]}$ between μ and ν, actually it’s a normal law $\mathcal {N}(x_t^T,\sigma _t^T)$ where the parameter are given by

$$ \left\{\begin{array}{cc} {x_{t}^{T}}=\frac{T-t}{T}x_{0}+\frac{t}{T}x_{1}, \\ {\sigma_{t}^{T}}=1+2\frac{t(T-t)}{\mathcal{D}_{T}^{2}+T}. \end{array} \right. $$

We want to quantify the convergence of μ^T toward the gradient flow $\left (P_t^{*} (\mu ) \right )_{t \in [0,T]}$. The gradient flow is given by $P_t^{*}(\mu )=P_t\left (\frac {d\mu }{dm} \right )dm=\mathcal {N}(x_{0},\mathcal {D}_{T}^2+2t)$. Actually the Wasserstein distance between two gaussian measures can be explicitely computed. Indeed let $\mu = \mathcal {N}(m_{0},\sigma _{0}^2)$ and $\nu =\mathcal {N}(m_1,\sigma _1^2)$, the map $T:x \mapsto \frac {\sigma _1}{\sigma _{0}}(x-m_{0})+m_1$ verify T#μ = ν, hence by Brenier theorem $W_2^2(\mu ,\nu )={\int \limits } |x-T(x)|^2 d \mu (x)$. From this expression and some easy computations we find

$$ {W_{2}^{2}}(\mu,\nu)=|\sigma_{0}-\sigma_{1}|^{2}+|m_{0}-m_{1}|^{2}. $$

For the detail and the extension to Gaussian vectors we refer to [31, Remark 2.31]. Hence we can compute explicitly the Wasserstein distance between the entropic interpolation and the gradient flow.

Proposition A.4

In the notations of this subsection

$$ {W_{2}^{2}}\left( {\mu_{t}^{T}},P_{t}^{*} \mu \right)=\frac{t^{2}}{T^{2}}(x_{0}-x_{1})^{2}+\left|\sqrt{{\sigma_{t}^{T}}}-\sqrt{2t+1} \right|^{2}, $$

and there exists a constant C > 0 such that $W_2^2(\mu _t^T,P_t^{*} \mu ) \underset {T \rightarrow \infty }{\sim } \frac {C}{T^2}.$

The velocity of $\left (\mu _t^T \right )_{t \in [0,T]}$ is given by

$$ \forall t\in[0,T], \ \forall x \in {\mathbf{R}}, \ \dot {\mu_{t}^{T}}(x)= \frac{\dot {\sigma_{t}^{T}}}{2{\sigma_{t}^{T}}}\left( x-x_{t/T}\right)+ \frac{1}{T} \left( x_{1} - x_{0} \right). $$

Now we have all the element we need to compute the conserved quantity, and the following proposition follow from basic integration.

Proposition A.5

In the notations of this subsection we have the following equality for every T > 0,

$$ \mathcal{E}_{T}(\mu,\nu):={\left| \dot {\mu_{t}^{T}} \right|_{{\mu_{t}}^{T}}^{2}}- {\left|\nabla \log \left( {\mu_{t}^{T}} \right) \right|_{{\mu_{t}}^{T}}^{2}}=\frac{\dot {{\sigma_{t}^{T}}}^{2}}{4 {\sigma_{t}^{T}}}+\frac{1}{T^{2}}(x_{1}-x_{0})^{2} - \frac{1}{{\sigma_{t}^{T}}}, 0 \leq t \leq T. $$

In particular we can take t = T/2 to find, $\mathcal {E}_{T}(\mu ,\nu ) \underset {T \rightarrow \infty }{\sim }\frac {(x_1-x_{0})^2}{T^2}-\frac {2}{T+2}$ and finally we get

$$ \mathcal{C}_{T}(\mu,\nu) \underset{T \rightarrow \infty}{\sim} 2 \log (T). $$