Let $\varphi:E^\star \to \mathbb R$ be linear and continuous in $\sigma(E^\star, E)$. Then there is $e \in E$ such that $\varphi = Je$

Question

I'm trying to prove Prop 3.14 in Brezis' book of Functional Analysis. I posted my proof as an answer below. Could you have a check on my attempt?

Let $(E, | \cdot|)$ be a normed linear space and $E^\star$ its topological dual. Let $\sigma(E^\star, E)$ be the weak$^\star$ topology on $E^\star$. Consider the canonical injection $J:E \to E^{\star\star}, x \mapsto (f \mapsto \langle f, x \rangle)$. Let $\varphi:E^\star \to \mathbb R$ be linear and continuous in $\sigma(E^\star, E)$. Then there is $e \in E$ such that $\varphi = Je$.

Answer 1

Because $\varphi$ is continuous at $0$ in $\sigma(E^\star, E)$, there is a neighborhood $U$ of $0$ in $\sigma(E^\star, E)$ such that $$| \langle \varphi, f \rangle | < 1, \quad \forall f\in U.$$

By construction of $\sigma(E^\star, E)$, there are $x_1, \ldots, x_n \in E$ such that $$U \supseteq V := \{f \in E^\star \mid \forall k = 1, \ldots, n: |\langle Jx_k, f \rangle| < 1\}.$$

Let $V_k := V/k := \{f/k \mid f \in V\}$ and $U_k := U/k$. It's easy to show that $$\bigcap_{k=1}^n \ker (Jx_k) = \bigcap_{k \in \mathbb N^\star} V_k \subseteq \bigcap_{k \in \mathbb N^\star} U_k = \ker \varphi.$$

It follows that $\varphi$ is linearly dependent of $Jx_1, \ldots, Jx_n$. This means $\varphi = \lambda_1 Jx_1 + \cdots + \lambda_n Jx_n = J (\lambda_1 x_1 + \cdots + \lambda_n x_n)$. Hence $e:= \lambda_1 x_1 + \cdots + \lambda_n x_n$ satisfies the requirement.