Question: Let $F$ be a real Banach space, $\mu$ be a Borel probability measure on a compact Hausdorff space $X$, and let $ f : X \rightarrow F$ be a continuous mapping. Then, using a compactness argument show the existence of a point $ y \in \overline{co}(f(X))$ such that $$ \Psi(y) = \int _X \Psi \circ f d\mu \quad \textrm{for every}\; \Psi \in F'.$$
How can I show this property? can you give me hint?
We can use the following Theorem
Theorem: Let $F$ be a real Banach space, $\mu$ be a Borel probability measure on a compact Hausdorff space $X$, and let $ f : X \rightarrow F$ be a continuous mapping. Given $ \Psi_1 , \ldots \Psi_n \in F'$ let $$ \nu _j = \int _X \Psi \circ f d \mu \quad \textrm{for}\quad j = 1, \ldots ,n,$$ and let $T : F \rightarrow \mathbb{R}^n$ linear function defined by $$ T(y) = (\Psi_1(y),\ldots, \Psi_n(y)) \quad \textrm{for every} \quad y \in F.$$ then, $$ (\nu_1,\ldots,\nu_n) \in co(T\circ f(X)) = T(co(f(X))),$$ where $co(B)$ denotes the convex hull of the set B.
I used this theorem like that
For any $ \Psi \in F'$, we take $$ \nu = \int _X \Psi \circ f d \mu$$ and let $ T: F \rightarrow \mathbb{R}$ be defined by $$ T(y) = \Psi(y). $$ Hence, applying above theorem we have $$ \Psi(\int_X f d\mu) = \int_X \Psi \circ f d \mu \in co(\Psi (f(X))) = \Psi (co(f(X)))$$ I stack here can you give me any hints ?
I write $C = \overline{co}(f(X))$ and for any finite subset $L$ of $F'$ I write $F_L$ for the set of $v \in F$ such that for each $\Psi \in L$, $\Psi(v) = \int \Psi \circ f d\mu$. Finally, I write $C_L = F_L \cap C$.
With this notation in hand, we want to show that $$\bigcap_{\substack{L \subseteq F' \\ |L| < \infty}} C_L \neq \emptyset.$$
Now $C$ is compact and for each $L$, $C_L$ is a closed subset of $C$. Hence, by the finite intersection property for compact sets (see here), it suffices to see that given $L_1, \dots, L_n \subseteq F'$ with $|L_i|< \infty$ we have that $\bigcap_{i=1}^n C_{L_i} \neq \emptyset$. This follows immediately from the observation that $C_{L_1} \cap C_{L_2} = C_{L_1 \cup L_2}$ and the theorem that you are allowed to assume since, in my notation, that theorem says that for each finite $L \subseteq F'$, $C_L \neq \emptyset$.