I'm trying to solve below exercise, i.e.,
Let $(X, |\cdot|)$ be a normed vector space and $M>0$. Fix $n$ linear functionals $f_1, \ldots, f_n$ and $n$ real numbers $\alpha_1, \ldots, \alpha_n$. Then the following statements are equivalent.
- (a) For all $\varepsilon>0$, there is $x_\varepsilon \in E$ such that $|x_\varepsilon| \le M+ \varepsilon$ and $\langle f_i, x_\varepsilon \rangle = \alpha_i$ for all $i=1, \ldots,n$.
- (b) For any $(\beta_1, \ldots, \beta_n) \in \mathbb R^n$ we have $|\sum_{i=1}^n \beta_i \alpha_i| \le M \| \sum_{i=1}^n \beta_i f_i\|$.
Could you please have a check on my attempt?
My attempt: The direction (a) $\implies$ (b) is trivial. Let's prove the converse. WLOG, we assume $\{f_1, \ldots, f_m\}$ is a maximal linearly independent subset of $\{f_1, \ldots, f_n\}$ for some $m \le n$. Consider $$ F:X \to \mathbb R^m, x \mapsto (f_1(x), \ldots, f_m(x)). $$
Let $B$ be the closed unit ball of $X$. Assume the contrary that $(\alpha_1 \ldots, \alpha_m) \notin H :=F((M+\varepsilon)B) \subset \mathbb R^m$. Notice that $H$ is a convex subset of a finite-dimensional normed vector space, so we apply Hahn-Banach theorem to separate $H$ and $\{(\alpha_1, \ldots, \alpha_m) \}$. There is $\beta :=(\beta_1, \ldots, \beta_m) \in \mathbb R^m \setminus \{0\}$ such that $$ G(x):=\sum_{i=1}^m \beta_i f_i(x) \le \sum_{i=1}^m \beta_i \alpha_i \quad \forall x \in X. $$
We claim that $G(x) = 0$ for all $x \in X$. if not, there is $a \in X$ such that $G(a) \neq 0$. Then $G$ can be arbitrarily large or arbitrarily small by picking $x = k a$ with $k \in \mathbb Z$. It follows that $\sum_{i=1}^m \beta_i f_i = 0$ and thus $\beta_1 = \cdots = \beta_m=0$, which is a contradiction. Hence there is $a\in X$ such that $F(a) = (\alpha_1 \ldots, \alpha_m)$.
Let $p \in \{m+1, \ldots, n\}$. We assume $f_p = \lambda_1 f_1 +\cdots +\lambda_m f_m$ for some $(\lambda_1, \ldots, \lambda_m) \in \mathbb R^m \setminus \{0\}$. Then $f_p - \lambda_1 f_1 - \cdots -\lambda_m f_m=0$. By (b), we get $\alpha_p - \lambda_1 \alpha_1 - \cdots -\lambda_m \alpha_m=0$. On the other hand, $f_p (a) = \lambda_1 f_1 (a)+\cdots +\lambda_m f_m (a) = \lambda_1 \alpha_1 + \cdots + \lambda_m \alpha_m$. It follows that $f_p(a) = \alpha_p$. This completes the proof.