Disclaimer: This is not a homework problem, so full answers/explanations are appreciated.
Setup:
Let $S$ be a second-countable compact Hausdorff space, and let $B(S)$ denote its Borel $\sigma$-algebra. Let $M(S)$ be the Banach space of Radon measures on $(S,B(S))$ equipped with the total variation norm $\|\cdot\|_\text{TV}$, and let $P(S) \subseteq M(S)$ be the set of probability measures on $(S,B(S))$. Let $M(S)^*$ denote the space of bounded linear functionals on $M(S)$ equipped with the operator norm induced by $\|\cdot\|_\text{TV}$.
Let $U\subseteq M(S)$ be open in the strong topology (i.e., the topology generated by $\|\cdot\|_\text{TV}$, and let $F \colon U \to \mathbb{R}$ be Frechet differentiable in the sense that, for all $\mu\in U$ there exists a bounded linear functional $DF(\mu) \colon M(S) \to \mathbb{R}$ such that
\begin{equation} \lim_{\epsilon \to 0} \frac{|F(\mu+\epsilon) - F(\mu) - DF(\mu)\epsilon|}{\|\epsilon\|_\text{TV}} = 0, \end{equation}
where the limit is taken with respect to the strong topology. Furthermore, let $G \colon P(S) \to M(S)$ be continuous with respect to the weak$*$ topology on $M(S)$.
(Broad) Question:
Under what conditions is the map $\mu \mapsto DF(\mu) G(\mu)$ weak$*$-continuous?
My attempt:
In the case $S = \{1,2,\dots,n\}$, we can identify $M(S)$ with $\mathbb{R}^n$ and $P(S)$ with the simplex $\Delta^n = \{\mu \in \mathbb{R}^n : \sum_{i=1}^n \mu_i = 1, ~ \mu_i \ge 0 ~ \text{for all $i$}\}$. In this case, $DF(\mu)$ is the linear operator defined by dot product with the gradient, i.e., $DF(\mu) \nu = \nabla F(\mu) \cdot \nu$. Since the strong and weak$*$ topologies coincide in this case, and the dot product is continuous, the map $\mu \mapsto DF(\mu) G(\mu) = \nabla F(\mu) \cdot G(\mu)$ is continuous whenever $\nabla F \colon \mu \mapsto \nabla F(\mu)$ is continuous.
In the case where $S$ is infinite, more care must be taken since the strong and weak$*$ topologies no longer coincide. In particular, I do not think it is necessarily true that $DF$ is weak$*$-continuous when $F$ is continuously Frechet differentiable, since even when $DF$ is strongly continuous, there may be an open set $U\subseteq M(S)^*$ such that the preimage $DF^{-1}(U)$ is strongly open but not weak$*$-open (as the weak$*$ topology is a subset of the strong topology).
To make things easier, let us assume that $DF$ is weak$*$-continuous (in the sense that the preimage of every open set $U\subseteq M(S)^*$ under $DF$ is weak$*$-open). In this case, I'm not even sure that the evaluation of the generalized "dot product" $M(S)^*\times M(S) \to \mathbb{R}$ given by $(L,\mu)\mapsto L\mu$ for $L\in M(S)^*$ and $\mu\in M(S)$ is weak$*$-continuous, since the coupling between $M(S)$ and its dual $M(S)^*$ is defined in terms of the total variation norm.
According to the related paper [1, Section 5.1], the dual of $M(S)$ is the space of bounded continuous functions $C_b(S)$ (which equals the entire space of continuous functions $C(S)$ since $S$ is compact), which may be helpful since then every $L\in M(S)^*$ can be represented by a continuous function $f\in C(S)$ such that
\begin{equation} L\mu = \int_S f d\mu. \end{equation}
However, I don't think that this claim is necessarily true, since $C(S)$ is reflexive only when $S$ is finite [2]. On the other hand, the topological dual of $M(S)$ endowed with the weak$*$ topology appears to indeed be $C(S)$ [3], but I'm not quite sure how to use this in my problem.
(Specific) Question:
If $DF$ and $G$ are both weak$*$-continuous, is the map $\mu \mapsto DF(\mu) G(\mu)$ also weak$*$-continuous?
Any and all help is very appreciated!
References:
[1] "Pairwise comparison dynamics for games with continuous strategy space" by Man-Wah Cheung, Journal of Economic Theory, 2014. URL: https://doi.org/10.1016/j.jet.2014.07.001