pr.probability - Conditional probabilities in Banach spaces

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 $$(*) qquad mathbb E(pi_x | mathcal F_2) = Ppi_x.$$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 $pi_x^* (K_{11} - K_{12}^* K_{22}^{-1} K_{21})pi_x$.