[proofplan] The [weak* topology](/page/Weak*%20Topology) $\sigma(X^*, X)$ is the initial [topology](/page/Topology) generated by the evaluation maps $\{\hat{x}\}_{x \in X}$, where $\hat{x}(g) = g(x)$. The structure is parallel to the [Sequential Characterisation of Weak Convergence](/theorems/255): the forward direction follows from [continuity](/page/Continuity) of each evaluation map, and the reverse direction reduces an arbitrary weak* neighbourhood to finitely many evaluation conditions and finds a uniform index. [/proofplan] [step:Define the evaluation maps and the weak* topology] For each $x \in X$, define the evaluation map \begin{align*} \hat{x}: X^* &\to \mathbb{R} \\ g &\mapsto g(x). \end{align*} The [weak* topology](/page/Weak*%20Topology) $\sigma(X^*, X)$ is the coarsest [topology](/page/Topology) on $X^*$ making every evaluation map $\hat{x}$ [continuous](/page/Continuity). [/step] [step:Deduce pointwise convergence from weak* convergence] Suppose $f_n \to f$ in $\sigma(X^*, X)$. Let $x \in X$ be arbitrary. By definition of $\sigma(X^*, X)$, the evaluation map $\hat{x}: X^* \to \mathbb{R}$ is continuous. Since $f_n \to f$ in $\sigma(X^*, X)$ and $\hat{x}$ is continuous, $f_n(x) = \hat{x}(f_n) \to \hat{x}(f) = f(x)$ in $\mathbb{R}$. [/step] [step:Deduce weak* convergence from pointwise convergence on all of $X$] Suppose $f_n(x) \to f(x)$ for every $x \in X$. Let $W$ be an arbitrary $\sigma(X^*, X)$-neighbourhood of $f$. By the definition of the initial [topology](/page/Topology) generated by $\{\hat{x}\}_{x \in X}$, $W$ contains a basic neighbourhood of $f$: there exist finitely many vectors $x_1, \ldots, x_m \in X$ and a tolerance $\varepsilon > 0$ such that \begin{align*} U &:= \{ g \in X^* : |g(x_i) - f(x_i)| < \varepsilon \text{ for all } i = 1, \ldots, m \} \subseteq W. \end{align*} By hypothesis, for each $i \in \{1, \ldots, m\}$, the numerical [sequence](/page/Sequence) $(f_n(x_i))_{n=1}^\infty$ converges to $f(x_i)$, so there exists $N_i \in \mathbb{N}$ such that $|f_n(x_i) - f(x_i)| < \varepsilon$ for all $n \geq N_i$. Define $N := \max\{N_1, \ldots, N_m\}$. For every $n \geq N$ and every $i \in \{1, \ldots, m\}$, the bound $|f_n(x_i) - f(x_i)| < \varepsilon$ holds, so $f_n \in U \subseteq W$. Since $W$ was arbitrary, $f_n \to f$ in $\sigma(X^*, X)$. [guided] We prove the harder direction: pointwise convergence $f_n(x) \to f(x)$ for all $x \in X$ implies $f_n \to f$ in the [weak* topology](/page/Weak*%20Topology). The argument mirrors the weak convergence case ([Theorem 255](/theorems/255)), with evaluation maps $\hat{x}$ playing the role of the functionals $f \in X^*$. Let $W$ be any $\sigma(X^*, X)$-neighbourhood of $f$. The initial topology structure guarantees that $W$ contains a basic neighbourhood determined by finitely many vectors $x_1, \ldots, x_m \in X$ and $\varepsilon > 0$: \begin{align*} U &:= \{ g \in X^* : |g(x_i) - f(x_i)| < \varepsilon \text{ for all } i = 1, \ldots, m \} \subseteq W. \end{align*} The finiteness of the collection $\{x_1, \ldots, x_m\}$ is essential. For each $i$, the hypothesis $f_n(x_i) \to f(x_i)$ provides an index $N_i$ beyond which $|f_n(x_i) - f(x_i)| < \varepsilon$. Taking $N = \max\{N_1, \ldots, N_m\}$ ensures all $m$ conditions hold simultaneously for $n \geq N$: \begin{align*} |f_n(x_i) - f(x_i)| &< \varepsilon \quad \text{for all } i = 1, \ldots, m \text{ and all } n \geq N. \end{align*} Therefore $f_n \in U \subseteq W$ for $n \geq N$. Since $W$ was arbitrary, $f_n \to f$ in $\sigma(X^*, X)$. [/guided] [/step] [step:Combine both directions to conclude the equivalence] By the previous steps, $f_n \to f$ in $\sigma(X^*, X)$ if and only if $f_n(x) \to f(x)$ for every $x \in X$. [/step]