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?


1 Answer 1


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.


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