[proofplan]
The forward direction "$f \in F \Rightarrow f$ is $\sigma(E,F)$-continuous" is the defining property of the weak topology $\sigma(E, F)$: it is the coarsest topology on $E$ making every element of $F$ continuous. For the converse, suppose $f$ is $\sigma(E, F)$-continuous. The unit-modulus neighbourhood $V = \{x : |f(x)| < 1\}$ of $0$ is then $\sigma(E, F)$-open, so by the structure of basic open neighbourhoods in the weak topology there exist $g_1, \ldots, g_n \in F$ and $\varepsilon > 0$ with $\{x : |g_i(x)| < \varepsilon\} \subseteq V$. A scaling argument upgrades this neighbourhood inclusion to the kernel inclusion $\bigcap_i \ker(g_i) \subseteq \ker(f)$. The [Kernel Lemma](/theorems/2648) then gives $f \in \operatorname{span}\{g_1, \ldots, g_n\} \subseteq F$.
[/proofplan]
[step:Recall the basic open neighbourhoods of $0$ in $\sigma(E, F)$]
By definition, $\sigma(E, F)$ is the coarsest topology on $E$ making every $g \in F$ continuous as a map $g: E \to \mathbb{F}$ (where $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$). It is generated by the sub-basis $\{g^{-1}(U) : g \in F, U \subseteq \mathbb{F} \text{ open}\}$. A basis of neighbourhoods of $0$ in $\sigma(E, F)$ is given by sets of the form
\begin{align*}
N(g_1, \ldots, g_n; \varepsilon) := \{x \in E : |g_i(x)| < \varepsilon \text{ for } i = 1, \ldots, n\},
\end{align*}
for $g_1, \ldots, g_n \in F$ and $\varepsilon > 0$. (Each such set is a finite intersection of preimages under $g_i$ of the open ball $\{|t| < \varepsilon\} \subseteq \mathbb{F}$.)
In particular, every $g \in F$ is $\sigma(E, F)$-continuous, by definition of $\sigma(E, F)$.
[/step]
[step:Forward direction: if $f \in F$ then $f$ is $\sigma(E, F)$-continuous]
Suppose $f \in F$. By Step 1, every element of $F$ is $\sigma(E, F)$-continuous, so $f$ is $\sigma(E, F)$-continuous. This direction is exactly the universal property of the weak topology.
[/step]
[step:Converse setup: extract a basic neighbourhood of $0$ contained in $\{x : |f(x)| < 1\}$]
Suppose $f : E \to \mathbb{F}$ is linear and $\sigma(E, F)$-continuous. The set
\begin{align*}
V := \{x \in E : |f(x)| < 1\} = f^{-1}(\{t \in \mathbb{F} : |t| < 1\})
\end{align*}
is the preimage under $f$ of the open unit ball in $\mathbb{F}$, so $V$ is $\sigma(E, F)$-open. Since $f(0) = 0 \in \{|t| < 1\}$, $V$ is an open neighbourhood of $0$ in $\sigma(E, F)$.
By Step 1, there exist $g_1, \ldots, g_n \in F$ and $\varepsilon > 0$ with
\begin{align*}
N(g_1, \ldots, g_n; \varepsilon) = \{x \in E : |g_i(x)| < \varepsilon,\ i = 1, \ldots, n\} \subseteq V.
\end{align*}
[/step]
[step:Promote the neighbourhood inclusion to the kernel inclusion $\bigcap_i \ker(g_i) \subseteq \ker(f)$]
Fix $x \in \bigcap_{i=1}^n \ker(g_i)$, so $g_i(x) = 0$ for all $i$. We show $f(x) = 0$.
For any scalar $\lambda \in \mathbb{F}$, linearity of each $g_i$ gives $g_i(\lambda x) = \lambda g_i(x) = 0$, in particular $|g_i(\lambda x)| = 0 < \varepsilon$. So $\lambda x \in N(g_1, \ldots, g_n; \varepsilon) \subseteq V$, meaning $|f(\lambda x)| < 1$. By linearity of $f$, $|f(\lambda x)| = |\lambda||f(x)|$, so
\begin{align*}
|\lambda||f(x)| < 1 \quad \text{for all } \lambda \in \mathbb{F}.
\end{align*}
If $f(x) \neq 0$, choosing $\lambda \in \mathbb{F}$ with $|\lambda| > 1/|f(x)|$ contradicts this inequality. Therefore $f(x) = 0$, i.e. $x \in \ker(f)$.
This shows $\bigcap_{i=1}^n \ker(g_i) \subseteq \ker(f)$.
[/step]
[step:Apply the Kernel Lemma to conclude $f \in F$]
By the [Kernel Lemma](/theorems/2648), the inclusion $\ker(f) \supseteq \bigcap_{i=1}^n \ker(g_i)$ established in Step 4 implies
\begin{align*}
f \in \operatorname{span}\{g_1, \ldots, g_n\}.
\end{align*}
We verify the hypothesis of the Kernel Lemma: $E$ is a vector space (given), and $f, g_1, \ldots, g_n$ are linear functionals on $E$ ($f$ by hypothesis, the $g_i$ as elements of $F$, where $F$ is by setup a vector space of linear functionals on $E$).
Since $F$ is a vector subspace of the algebraic dual $E^\#$ and contains $g_1, \ldots, g_n$, it contains $\operatorname{span}\{g_1, \ldots, g_n\}$. Therefore $f \in F$.
Combining the forward direction (Step 2) and the converse (Steps 3-5):
\begin{align*}
\{f : E \to \mathbb{F} \text{ linear and } \sigma(E,F)\text{-continuous}\} = F,
\end{align*}
which is precisely the statement $(E, \sigma(E, F))^* = F$.
[/step]
