I'm doing Ex 3.14 in Brezis's book of Functional Analysis.
Let $E$ be a reflexive Banach space and $I$ a set of indices. Consider a collection $(f_{i})_{i \in I}$ in the dual space $E'$ and a collection $(\alpha_{i})_{i \in I}$ in $\mathbb{R}$. Let $M>0$. Show that the following properties are equivalent:
- There exists some $a \in E$ with $|a| \le M$ such that $\langle f_{i}, a\rangle=\alpha_{i}$ for every $i \in I$.
- One has $|\sum_{i \in J} \beta_{i} \alpha_{i}| \leq M\|\sum_{i \in J} \beta_{i} f_{i}\|$ for every collection $(\beta_{i})_{i \in J}$ in $\mathbb{R}$ with $J \subset I, J$ finite.
My below solution uses net characterization of compactness, which is probably not the author's intention.
Could you have a check on my attempt and suggest another solution without using net?
My attempt: WLOG, we can assume $M:=1$. The direction $1 \implies 2$ is trivial. Let's prove the converse. Let $B_E$ be the closed unit ball of $E$. Let $\mathcal J$ be the collection of all finite subsets of $I$. By Helly's theorem (Lemma 3.3 in the book), for each $(J,n) \in \mathcal J \times \mathbb N$, there is $x_{J,n} \in B_E$ such that $$ |\langle f_i, x_{J,n} \rangle - \alpha_i| < 1/n \quad \forall i \in J. $$
We endow $\mathcal J \times \mathbb N$ with a partial order $\le$ defined by $(J, n) \le (J', n') \iff J \subseteq J' \text{ and } n \le n'$. Then $(\mathcal J \times \mathbb N, \le)$ is a directed set and thus $(x_{(J,n)})$ a net in $B_E$. Because $E$ is reflexive, $B_E$ is compact in weak topology $\sigma(E', E)$. Then there are $a \in B_E$ and a subnet $(x_{\varphi(d)})_{d\in D}$ such that $x_{\varphi(d)} \rightharpoonup a$. Next we prove that $a$ satisfies the requirement of 1.
Assume the contrary that there is $i_0 \in I$ such that $\langle f_{i_0}, a \rangle \neq \alpha_{i_0}$. There is $n_0$ such that $|\langle f_{i_0}, a \rangle - \alpha_{i_0}| > 1/n_0$. Let $J_0 := \{i_0\}$ and $$ U := \{x\in E \mid |\langle f_{i_0}, x \rangle - \alpha_{i_0}| > 1/n_0\}. $$
Then $U$ is an open neighborhood of $a$ in $\sigma(E', E)$. There is $d_0$ such that $x_{\varphi(d)} \in U$ for all $d \ge d_0$. There is $d_1 \ge d_0$ such that $\varphi(d_1) \ge (J_0, n_0+1)$. It follows that $$ 1/(n_0+1) > |\langle f_{i_0}, x_{\varphi(d_1)} \rangle - \alpha_{i_0}| > 1/n_0. $$
This is a contradiction. This completes the proof.