[proofplan] The weak [topology](/page/Topology) $\sigma(X, X^*)$ on a [Banach space](/page/Banach%20Space) $X$ is the initial topology generated by $X^*$. The forward direction is immediate: each $f \in X^*$ is continuous in the [weak topology](/page/Weak%20Topology) by definition, so it preserves convergent sequences. The reverse direction takes an arbitrary weak neighbourhood of $x$, extracts the finitely many functionals and tolerance defining it, and uses the pointwise convergence hypothesis to find a uniform index beyond which every $x_n$ belongs to the neighbourhood. [/proofplan] [step:Deduce pointwise convergence of functionals from weak convergence] Suppose $x_n \to x$ in $\sigma(X, X^*)$. Let $f \in X^*$ be arbitrary. By definition of the weak [topology](/page/Topology) $\sigma(X, X^*)$, the map $f: X \to \mathbb{R}$ is [continuous](/page/Continuity) with respect to $\sigma(X, X^*)$ and the standard topology on $\mathbb{R}$. Since $x_n \to x$ in $\sigma(X, X^*)$ and $f$ is continuous, $f(x_n) \to f(x)$ in $\mathbb{R}$. [/step] [step:Deduce weak convergence from pointwise convergence of all functionals] Suppose $f(x_n) \to f(x)$ for every $f \in X^*$. Let $W$ be an arbitrary $\sigma(X, X^*)$-neighbourhood of $x$. By the definition of the initial [topology](/page/Topology), $W$ contains a basic neighbourhood of $x$: there exist finitely many functionals $f_1, \ldots, f_m \in X^*$ and a tolerance $\varepsilon > 0$ such that \begin{align*} U &:= \{ y \in X : |f_i(y) - f_i(x)| < \varepsilon \text{ for all } i = 1, \ldots, m \} \subseteq W. \end{align*} By hypothesis, for each $i \in \{1, \ldots, m\}$, the [sequence](/page/Sequence) $(f_i(x_n))_{n=1}^\infty$ converges to $f_i(x)$ in $\mathbb{R}$, so there exists $N_i \in \mathbb{N}$ such that $|f_i(x_n) - f_i(x)| < \varepsilon$ 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 $|f_i(x_n) - f_i(x)| < \varepsilon$ holds, so $x_n \in U \subseteq W$. Since $W$ was arbitrary, $x_n \to x$ in $\sigma(X, X^*)$. [guided] We want to show that if $f(x_n) \to f(x)$ for every $f \in X^*$, then $x_n \to x$ in the weak topology. The key structure to exploit is that every weak neighbourhood is determined by finitely many functionals. Let $W$ be an arbitrary $\sigma(X, X^*)$-neighbourhood of $x$. By the definition of the initial topology generated by $X^*$, $W$ contains a basic neighbourhood of $x$ defined by finitely many functionals $f_1, \ldots, f_m \in X^*$ and a tolerance $\varepsilon > 0$: \begin{align*} U &:= \{ y \in X : |f_i(y) - f_i(x)| < \varepsilon \text{ for all } i = 1, \ldots, m \} \subseteq W. \end{align*} Why does finiteness matter? Because we need to find a single index $N$ that works for all conditions simultaneously. For each $i$, the hypothesis $f_i(x_n) \to f_i(x)$ provides $N_i$ with $|f_i(x_n) - f_i(x)| < \varepsilon$ for $n \geq N_i$. Taking $N = \max\{N_1, \ldots, N_m\}$ handles all $m$ conditions at once: for $n \geq N$ and all $i$, \begin{align*} |f_i(x_n) - f_i(x)| &< \varepsilon, \end{align*} so $x_n \in U \subseteq W$. Since $W$ was an arbitrary weak neighbourhood of $x$, this establishes $x_n \to x$ in $\sigma(X, X^*)$. Note that this argument would fail if the neighbourhood base involved infinitely many functionals -- the maximum of infinitely many indices need not be finite. This is precisely why the initial topology is defined using finite intersections of sub-basic sets. [/guided] [/step] [step:Combine both directions to conclude the equivalence] By the previous two steps, $x_n \to x$ in $\sigma(X, X^*)$ if and only if $f(x_n) \to f(x)$ for every $f \in X^*$. [/step]