# Applying Geometric Hahn-Banach Theorem on Vector-Valued Integration

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.