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I’m trying to understand if, when I move the marginals of a Wasserstein geodesic along a contractive flow, the geodesic between the new probability measures is “near” to the geodesic connecting the original measures.

Consider, in the Wasserstein space $(\mathcal{P}_2(\mathbb{R}^n),\mathbb{R}^n)$, two probability measures $\mu_0, \mu_1$ (absolutely continuous w.r.t. the Lebesgue measure) and the constant speed geodesic $\mu(\cdot)$ connecting them. Consider now the heat semigroup $(\mathcal{S}_t)_{t>0}$, fix $\sigma>0$ (small) and denote with $\nu_i:=\mathcal{S}_\sigma(\mu_i)$ for $i=0,1$. Finally, if $\nu(\cdot)$ is the constant speed geodesic connecting $\nu_0$ and $\nu_1$, is it true that $$ W_2(\mu(t),\nu(t))\leq \sigma\quad \forall t\in[0,1]\,? $$

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As currently asked the answer is NO, because your desired upper bound already fails for $t=0$ (or equivalently, $t=1$). Indeed, it is well understood that the small-time deviation along the heat flow, as measured in the quadratic Wasserstein distance, is related to the Fisher information $$ \mathcal I(\rho):=\int_{\mathbb R^n}|\nabla\log\rho|^2d \rho. $$ More precisely, if $\rho_0$ is any reasonably smooth measure (with finite Fisher information) and $\rho_\sigma=K_\sigma\ast \rho_0$ denotes the time-$\sigma$ heat flow, one has $$ W_2(\rho_0,\rho_\sigma)\approx \sigma\sqrt{\mathcal I(\rho_0)} \qquad\text{as }\sigma\to 0^+. $$ Again, I don't want to go into the technical details, so let me justtify this claim very informally: the heat flow is of course the Wasserstein gradient-flow $$ \dot\rho_s=-\operatorname{grad}_W \mathcal H(\rho_s) $$ of the free entropy $\mathcal H(\rho)=\int\rho\log\rho$. With Otto's calculus one has, at least informally along reasonably smooth solutions, that $$ \|\dot\rho_s\|^2_{\rho_s}= \|\operatorname{grad}_W \mathcal H(\rho_s)\|^2_{\rho_s}= \mathcal I(\rho_s). $$ (This can be made rigorous if for example $\rho_0$ is a Gaussian, in which case everything can be computed explicitly.) Whence for small $\sigma$ $$ W(\rho_0,\rho_\sigma)\approx \|\dot\rho_0\|_{\rho_0}\sigma =\sqrt{\mathcal I(\rho_0)}\sigma. $$ Since one can choose the Fisher information arbitrarily large, you see that clearly your desired inequality must fail at $t=0$.

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