This is the infinite-dimensional sequel to my question, Conditional probabilities are measurable functions - when are they continuous?.
Let $\Omega = \Omega_1 \times \Omega_2$ be a probability space which is Banach, $\mathcal F$ the Borel $\sigma$-algebra on $\Omega$, and $\mathbb P$ a probability measure which need not be the product measure. e.g., $\Omega = C(U_1) \times C(U_2)$ for disjoint, compact $U_1, ~U_2 \subseteq {\mathbb R}$. Let $\mathcal F_2 = \sigma(\Omega_2^*)$, where $\Omega_2^*$ is the dual space to $\Omega_2$. In the case of continuous functions, this is the $\sigma$-algebra generated by the evaulation maps $\pi_x$ for $x \in U_2$. Note that these evaluations $\pi_x$ are random variables on $\Omega$.
Goal: I would like an explicit expression for conditional expectations with respect to $\mathcal F_2$. Namely, I want a "reasonable" linear operator $P : \Omega^* \to \Omega^*$ such that $$(*) \qquad \mathbb E(\pi_x | \mathcal F_2) = P\pi_x.$$Ideally, "reasonable" will mean continuous.
I can do this in the case that $\Omega$ is a Gaussian Hilbert space. Decompose the covariance operator of $\mathbb P$ as $$K = \binom{K_{11} ~ K_{12}}{K_{21} ~ K_{22}}.$$(This should be matrix formatted but that doesn't seem to work here). Let $$P = \binom{~~~0 ~~~~~~~~ 0}{K_{22}^{-1} K_{21} ~~ I_2},$$ where $I_2$ is the identity operator on $\Omega_2^*$. Then using the Gaussian structure, I can show that $(*)$ holds; using the technology in Anderson & Trapp's Shorted Operators, II I can show that $P$ is continuous.
This is overkill! I think I can adapt my Gaussian calculation without too much trouble to the generic case (since it only deals with covariance operators). On the other hand, I don't know how to show that such an operator $P$ is continuous without relying on heavy-duty functional analysis. Surely this has been studied before, but I can't seem to find a good reference.
Note: Brownian motion is a very special case of this, where $\Omega_2 = C(${$0$}$) = \mathbb R$, the starting value $B_0$. There, one skirts the issue of this conditioning by the Markov property: the "future" $\Omega_1 = C((0,\infty))$ is independent of the "present" $\Omega_2$ up to the starting value $B_0$.
Note: The variance of $\pi_x$ for $x \in U_1$ is the Schur complement $\pi_x^* (K_{11} - K_{12}^* K_{22}^{-1} K_{21})\pi_x$.
