Let $\nu = (\nu_t)_{t \in [0,T]} \in C( [0,T], \mathcal{P}_2(\mathbb{R}) ) ,$ where $\mathcal{P}_2(\mathbb{R})$ denotes the space of probability measures with finite moment equipped with the Wasserstein distance. Is it possible to define a function \begin{align} \Phi : C( [0,T], \mathcal{P}_2(\mathbb{R})) &\rightarrow C( [0,T], \mathcal{P}_2(\mathbb{R}^2)) \\ \mu \mapsto \pi, \end{align} where $\pi$ satisfies for all $t \in [0,T]$ that $\pi_t \circ pr_2^{-1} = \nu_t $ and $pr_2$ denotes the projection on the second component. Edit: It should also hold that $\pi_t \circ pr_1^{-1} = \mu_t$ for all $t \in [0,T].$
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$\begingroup$ Are you sure you're stating the question correctly? The input $\mu$ to your function has no bearing on the condition you require of the output. The way you have stated it, the answer is trivially yes. For example, take $\Phi(\mu)_t = \delta_0 \times \nu_t$. $\endgroup$– DanCommented Dec 13, 2018 at 16:07
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$\begingroup$ thank you, the first marginal should match $\mu.$ $\endgroup$– WhiteCommented Dec 13, 2018 at 16:16
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$\begingroup$ Then just take the product measure $\Phi(\mu)_t = \mu_t \times \nu_t$... $\endgroup$– DanCommented Dec 14, 2018 at 18:50
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$\begingroup$ In this case is $ t \mapsto \mu_t \times \nu_t$ still continuous ? $\endgroup$– WhiteCommented Dec 15, 2018 at 20:27
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$\begingroup$ Yes, because the "product measure" operation is continuous. That is, $\mathcal{P}_2(\mathbb{R})^2 \ni (\mu,\nu) \mapsto \mu \times \nu \in \mathcal{P}_2(\mathbb{R}^2)$ is continuous. I don't know of a reference for this off the top of my head, but its proof is pretty straightforward. $\endgroup$– DanCommented Dec 16, 2018 at 13:45
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