Consider a strongly convex potential $U: \mathbb{R}^d \to \mathbb{R}$ and the Langevin diffusion $$dX = -\nabla U(X) dt + dW \qquad (*)$$ where $W$ is a standard Brownian motion. If $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are two solutions driven by the same Brownian motion, one can check that $t \mapsto \mathbb{E} \| X_t-Y_t \|^2$ is decreasing since $d \| X_t-Y_t \|^2 = -2 \langle X_t-Y_t, \nabla U(X_t) - \nabla U(Y_t)\rangle \leq -2 \lambda \|X_t - Y_t\|^2$ where $\textrm{Hessian}(x) \geq \lambda I_d$ on $\mathbb{R}^d$. In other words, given two starting positions $x_0, y_0 \in \mathbb{R}^d$, one can construct a coupling $(X_t,Y_t)$ of the diffusion $(*)$, one starting from $x_0$ and the other one from $y_0$, such that $\mathbb{E} \|X_t-Y_t\| \leq e^{- \lambda t} \|x_0-y_0\|$. Indeed, if the function $U$ is not strongly convex, one can only say that $\mathbb{E} \|X_t-Y_t\|$ can be made non-increasing. This can also be expressed in terms of Wasserstein distance between $P_t(x,dx)$ and $P_t(y,dy)$ where $P$ is the transition operator of the diffusion $(*)$.
Question: given $x_-,x^+,y_-,y^+ \in \mathbb{R}^d$, can we find a coupling $(X_t,Y_t)$ of the conditioned diffusion $(*)$ with $$X_0=x_-, \quad X_T=x^+, \quad Y_0=y_-, \quad Y_T=y^+,$$ and such that there is still a contraction property of the type $$\mathbb{E} \|X_{T/2}-Y_{T/2}\| \leq \beta \big( \|x_- - y_-\| + \|x_+ - y_+\|\big)$$ where $0<\beta<\frac12$ is some constant that does not depend on $(x_-,x^+,y_-,y^+)$. Is this known or have already been studied? Any reference welcome.
PS: one can check that this is true for a quadratic potential $U(x) = \frac12 \|x\|^2$ since the Ornstein-Uhlenbeck process is easy to study. The case of a brownian bridge (i.e. vanishing potential) is also straightforward and corresponds to the case $\beta = \frac12$.
PPS: it is important to exploit the structure of the SDE. Indeed, one can find counter examples of Markov processes $(X_t)_{t \geq 0}$ that have the contraction property $\mathbb{E} \|X_t-Y_t\| \leq e^{-\lambda t} \|x_0 - y_0\|$ but that do not verify the 'conditioned' contraction property.