What is the relationship between the spaces $X\_1\triangleq \mathscr{P}(C([0,1],\mathbb{R}))$ and $X\_2\triangleq C([0,1],\mathscr{P}(\mathbb{R}))$; where $\mathscr{P}(\cdot)$ denotes the Borel probability measures on a space and it is equipped with the total-variation topology. Specifically, I wonder, is $X\_1$ continuously embeded in $X\_2$ or the converse?



Embeddings of spaces of probability measures
This is an extended version of my comment above. It is not an answer, or at least: not a positive answer. (Perhaps it is sort of a negative answer?)




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There is a (sort of natural) candidate for an embedding $\Phi$ from $X\_2$ into $X\_1$, defined as follows. Take a family of measures $\mu = (\mu\_t, t \in [0,1])$ in $X\_2$, and denote by $F\_t(x)$ the corresponding distribution functions. Let $F\_t^{-1}$ be the generalised inverse function of $F\_t$, so that if $U$ is a random variable uniformly distributed on $[0, 1]$, then
$$ F\_t^{-1}(U) $$
is a random variable with distribution $\mu\_t$. Finally, define a stochastic process
$$ x\_t = F\_t^{-1}(U) ,$$
and set $\Phi(\mu)$ to be the law of the process $x\_t$ (again with $U$ distributed uniformly over $[0,1]$).


The above construction is natural in the following sense: $x\_t$ is the unique process with one-dimensional distributions (a.k.a. marginals) $\mu\_t$ with the following monotonicity property: if $x\_t(\omega\_1) \leqslant x\_t(\omega\_2)$ for some $t$, then $x\_t(\omega\_1) \leqslant x\_t(\omega\_2)$ for all $t$.




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The above 'embedding' looks nice, but it remains to prove that $\Phi(\mu)$ is indeed in $X\_1$. In other words: $x\_t$ has almost surely continuous paths. Equivalently: $t \mapsto F\_t^{-1}(u)$ is continuous for almost every $u \in [0, 1]$. This need not be true, though!


Set $\mu\_t = (1 - t) \delta\_0 + t \delta\_1$. Then $x\_t = 0$ for $t < U$ and $x\_t = 1$ for $t > U$, which means that the path of $x\_t$ is almost surely discontinuous!




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Clearly, $\mu$ constructed above is not a family of one-dimensional distributions of a continuous stochastic process, so in a sense there is no hope to fix the above construction. The only thing one can easily say is that if $t \mapsto \mu\_t$ is a function of bounded variation (with respect to the total variation distance), then $x\_t$ is almost surely a function of bounded variation:
$$ \mathbb E \sum\_i |x\_{t\_i}-x\_{t\_{i-1}}| = \sum\_i \|\mu\_{t\_i} - \mu\_{t\_{i-1}}\|\_{TV} \leqslant \operatorname{TV}(t \mapsto \mu\_t) , $$
and hence
$$ \mathbb E \operatorname{TV}(t \mapsto x\_t) \leqslant \operatorname{TV}(t \mapsto \mu\_t) . $$


