I have been studying the applications of geometric Hahn-Banach Theorems. I came across this book Functional Analysis by S. Kesavan where the geometric Hahn-Banach Theorem is applied on Vector-Valued Integration(here, vector-valued integration does not mean the integration of vector-valued function. It means that the integral of a function must be a vector in a Banach space.). But I have a very hard time understanding exactly how it is applied.
Theorem Let $\varphi:[0,1]\rightarrow V$ be a continuous mapping into a real Banach space $X$. Then, the integral of $\varphi$ over $[0,1]$ exists.
Proof:
Since $[0,1]$ is compact, the set $\overline{P}$ which is the closure (in $X$) of the set $P$ which is the convex hull of $\varphi([0,1])$,(i.e., the smallest convex set containing $\varphi([0,1])$)), is compact, by completeness of $X$.
Let $L$ be an arbitrary finite collection of continuous linear functionals on $V$. Define $$E_L=\{y\in \overline{P}:f(y)=\int_{0}^{1}f(\varphi(t))\mathrm{d}t\;\mathrm{for\;all}\; f\in L\}.$$ It is immediate to see that $_L$ is a closed set.
Step 1: For any such finite collection $L$ of continuous linear functionals, $E_L\ne \phi$. To see this, let $L=\{f_1,...,f_k\}$. Define $\mathcal{A}:X\rightarrow \mathbb{R}^{k}$ by $$\mathcal{A}(x)=(f_1(x),...,f_k(x)).$$ Then, $\mathcal{A}$ is a continuous linear transformation and so $K=\mathcal{A}(\overline{P})$ is a compact and convex set. If $(t_1,...,t_k)\notin K$, then, by the Hahn-Banach Theorem, we can find constants $c_1,...,c_k$ such that $$\sum_{i=1}^{k}c_iu_i<\sum_{i=1}^{k}c_it_i$$ for all $(u_1,...,u_k)\in K$. In particular, for all $t\in [0,1]$, we have $$\sum_{i=1}^{k}c_if_i(\varphi(t))< \sum_{i=1}^{k}c_it_i.$$ Integrating this inequality over $[0,1]$, we get $$\sum_{i=1}^{k}c_im_i < \sum_{i=1}^{k}c_it_i$$ where $$m_i=\int_{0}^{1}f_i(\varphi(t))\mathrm{d}t.$$ In other words, if $(t_1,...,t_k)\notin K$, then $(t_1,...,t_k)\ne (m_1,...,m_k)\in K$. Thus, there exists $y\in \overline{P}$ such that, for $1\le i\le k$, we have $$m_i=f_i(y).$$ This means that $y\in E_L$, i.e., $E_L$ is non-empty.
Step 2: Let $I$ be a finite indexing set and $L_i$ be finite collections of elements of $X^*$ for each $i\in I$. Then, $L=\underset{i \in I}{\bigcup} L_i$ is still finite and further, it is easy to see that, $$\underset{i\in I}{\bigcap} E_{L_i}= E_L.$$
It now follows from the previous step that the class of closed sets, $$\{E_L: L \;\mathrm{is\; a\; finite\; subset\; of}\; X^*\},$$ has finite intersection property. Since $\overline{P}$ is compact, it now follows that $$\underset{L,\;\mathrm{finite\; subset\; of}\; X^*}{\bigcap} E_L \ne \phi.$$
In particular, there exists $y$ such that $y\in E_{(f)}$ for every $f\in X^*$, i.e., $y$ satisfies $$f(y)=\int_{0}^{1}f(\varphi(t))\mathrm{d}t,$$ for every $f\in V^*$. Thus, $y=\int_{0}^{1}\varphi(t)\mathrm{d}t$. This completes the proof.
According to the article, the Hahn-Banach Theorem applied here is:
Let $A$ and $B$ be non-empty and disjoint convex sets in a real normed linear space $X$. Assume that $A$ is closed and that $B$ is compact. Then, $A$ and $B$ can be separated strictly by a closed hyperplane, i.e. there exists $f\in X^*,\; a\in \mathbb{R}$ and $\epsilon >0$ such that $f(x)\le \alpha -\epsilon$ and $f(y)\ge \alpha -\epsilon $ for all $x\in A$ and $y\in B$.
The problem I have is this:
We have, $K=\mathcal{A}(\varphi(Q))$ as a compact and convex set.
But for the theorem, mentioned above, to be applied, we need another closed convex set and that must be disjoint from $K$ and constant $\alpha$. The article does not specify the other set. Why? Can somebody elaborate on it, please?
The other set is the singleton $\{(t_1, \dots, t_k)\}$. There is then a linear functional $g \in (\mathbb{R}^k)^*$ such that $g((t_1, \dots, t_k)) > g(v)$ for all $v \in K$. But a linear functional $g$ on $\mathbb{R}^k$ is necessarily of the form $g((v_1, \dots, v_k)) = \sum_{i=1}^k c_i v_i$ for some scalars $c_1, \dots, c_k$, so we get the result.