# Using Hahn-Banach theorem show that $\sum\limits_{i = 1}^{n} \alpha_i \beta_i\gt \lambda \geq \left \|\sum\limits_{i = 1}^{n} \beta_i f_i \right \|.$ [closed]

Let $$V$$ be a real Banach space and $$B$$ be the closed unit ball in $$V.$$ Let $$f_1, \cdots, f_n \in V^{\ast}$$ be norm continuous linear functionals on $$V$$ and $$\alpha_1, \cdots, \alpha_n \in \mathbb R$$ be scalars. Let $$\overline {\alpha} = (\alpha_1, \cdots, \alpha_n).$$ Define a linear operator $$\mathcal A : V \longrightarrow \mathbb R^n$$ by $$\mathcal A (x) = \left (f_1 (x), \cdots, f_n (x) \right ),$$ $$x \in V.$$ If $$\overline {\alpha} \notin \overline {\mathcal A (B)}$$ then show that there exist scalars $$\beta_1, \cdots, \beta_n, \lambda \in \mathbb R$$ such that $$\sum\limits_{i = 1}^{n} \alpha_i \beta_i \gt \lambda \geq \left \|\sum\limits_{i = 1}^{n} \beta_i f_i \right \|.$$

This is used in one of the proofs in my book. Anybody have any idea on it?

Choose $$A = \overline {\mathcal A (B)}$$ and $$B = \{\overline {\alpha}\}.$$ Then $$A$$ and $$B$$ are respectively closed convex and compact convex subsets of $$\mathbb R^n$$ and $$A \cap B = \varnothing.$$ Apply Hahn-Banach theorem to get hold of a linear functional $$f \in \left (\mathbb R^n \right )^{\ast}$$ and $$\lambda \in \mathbb R$$ such that for all $$\overline {x} \in A$$ we have $$f(\overline {x}) \lt \lambda \lt f(\overline {\alpha}).$$
In particular, for all $$\overline {x} \in \mathcal A (B)$$ we have $$f(\overline {x}) \lt \lambda \lt f(\overline {\alpha})$$ i.e. for all $$x \in B$$ we have $$f(f_1(x), \cdots, f_n (x)) \lt \lambda \lt f (\alpha_1, \cdots, \alpha_n).$$
Let $$\beta_i = f (e_i),$$ where $$e_i$$ is the $$i$$-th coordinate vector of $$\mathbb R^n.$$
Then we have for all $$x \in B$$ $$\sum\limits_{i = 1}^{n} \beta_i f_i (x) \lt \lambda \lt \sum\limits_{i = 1}^{n} \alpha_i \beta_i.$$
Finally take the supremum over all $$x \in B$$ and conclude.