How can I show by using Hahn Banach separation theorem? The convex hull of a set of points S in n dimensions is the intersection of all convex sets containing $S$. For $N$ points $p_1, ..., p_N$, the convex hull $C$ is then given by the expression
$$ C = \left\{ \sum_{j=1}^ N \lambda _j p_j : \lambda _j \geq 0 \quad \text{for all} \, j \, \text{and} \, \sum_{j=1}^ N \lambda _j =1  \right\}. $$
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.
Let linear continuous functionals $ \phi _1 , \ldots ,\phi_n \in F'$ and
$$ \nu_j = \int _{X}\phi_j \circ f d\mu \quad \textrm{for} \; j = 1,\ldots,n.$$
Let $ T \in L({F; \mathbb{R}^n})$ be defined by
$$ Ty := (\phi_1(y),\ldots\phi_n(y)) \quad \textrm{for every} \; y \in F. $$
How can I  show that  by using the Hahn-Banach separation theorem,
$$ (\nu_1,\ldots \nu_n) \in co(T\circ f(X)) = T(co(f(X))), $$
where $co(B)$ denotes the convex hull of the set B ?
 A: If $$ \nu=(\nu_1,\ldots \nu_n) \notin co(T\circ f(X))  , $$
then there must be a linear functional $\psi:\mathbb R^n \to \mathbb R$
that satisfies $$\psi(\nu)>\max_{x \in X} \psi \circ T\circ f(x) \,. \tag{*}$$
Now $\psi$ has the form $\psi(y)=\sum_{j=1}^n a_j y_j$ for some real coefficients $a_j$, so we can rewrite $(*)$ as
$$\sum_{j=1}^n a_j \nu_j>\max_{x \in X} \, \sum_{j=1}^n a_j \phi_j(f(x))  \,. \tag{**}$$
This yields the desired contradiction, since the integral of a function cannot exceed its maximum.
Addendum  Given a  linear functional $\psi \in F'$, define
$$K_\psi:=\{y \in \overline{co}(f(X)) : \, \psi(y) =\int \psi \circ f \, d\mu \} \,.$$ Then $K_\psi$ is a  closed subset  of the compact set $\overline{co}(f(X))$, see Thm 5.35 in [1] or Thm 3.20 in Rudin, "Functional Analysis" for the compactness. Moreover, the previous discussion shows that these sets have the finite intersection property
$$\forall  \phi_1,\dots,\phi_n \in F',  \quad \cap_{j=1}^n K_{\phi_j} \ne \emptyset \,.$$
Indeed, recall the notation
$$ \nu_j = \int _{X}\phi_j \circ f d\mu \quad \textrm{for} \; j = 1,\ldots,n.$$  The relation proved above,
$$ (\nu_1,\ldots \nu_n) \in  T(co(f(X)))  $$
means that  there exists $y \in co(f(X))$ such that
$$(\phi_1(y),\dots,\phi_n(y))=Ty=(\nu_1,\dots,\nu_n) \,,$$
and this $y$ is in $\cap_{j=1}^n K_{\phi_j}$.
Compactness yields that the intersection  $\displaystyle \bigcap_{\psi \in F'} K_\psi$ is nonempty.
[1] http://books.google.com/books?id=4hIq6ExH7NoC&pg=PA185
