[proofplan] We must show two things: (1) the function $d$ is a well-defined metric on $X$, and (2) the topology induced by $d$ agrees with $\sigma(X, X^*)$ on every norm-bounded subset. For (2), we fix a bounded set $B \subset X$ and show that the identity map from $(B, \sigma(X, X^*))$ to $(B, d)$ is a homeomorphism by verifying [continuity](/page/Continuity) in both directions. The weak-to-$d$ direction follows from dominated convergence applied to the defining series. The $d$-to-weak direction uses the density of $\{f_n\}$ in the unit ball of $X^*$ together with a uniform approximation argument that crucially requires boundedness of $B$. [/proofplan] [step:Verify that $d$ is a metric on $X$] Let $\{f_n\}_{n=1}^\infty$ be a [dense subset](/page/Dense%20Subset) of the closed unit ball $\overline{B}_{X^*} := \{f \in X^* : \|f\|_{X^*} \le 1\}$. Define \begin{align*} d: X \times X &\to [0, \infty) \\ (x, y) &\mapsto \sum_{n=1}^\infty \frac{1}{2^n} |f_n(x - y)|. \end{align*} The series converges absolutely for every $(x, y) \in X \times X$, since $\|f_n\|_{X^*} \le 1$ gives $|f_n(x - y)| \le \|x - y\|_X$, so \begin{align*} \sum_{n=1}^\infty \frac{1}{2^n} |f_n(x - y)| \le \|x - y\|_X \sum_{n=1}^\infty \frac{1}{2^n} = \|x - y\|_X. \end{align*} Symmetry $d(x, y) = d(y, x)$ follows from $|f_n(x - y)| = |f_n(y - x)|$. The triangle inequality follows from $|f_n(x - z)| \le |f_n(x - y)| + |f_n(y - z)|$ and summing term by term. For definiteness, suppose $d(x, y) = 0$. Then $f_n(x - y) = 0$ for all $n \ge 1$. Since $\{f_n\}$ is dense in $\overline{B}_{X^*}$ and the map $f \mapsto f(x - y)$ is continuous on $(X^*, \|\cdot\|_{X^*})$, we get $f(x - y) = 0$ for all $f \in \overline{B}_{X^*}$, hence for all $f \in X^*$ by scaling. By a corollary of the [Hahn-Banach Theorem](/theorems/879) (if $x - y \neq 0$, there exists $f \in X^*$ with $f(x - y) = \|x - y\|_X > 0$), we conclude $x = y$. [guided] We verify each metric axiom individually. **Well-definedness and finiteness.** Each $f_n$ lies in the unit ball of $X^*$, so $|f_n(x - y)| \le \|f_n\|_{X^*} \|x - y\|_X \le \|x - y\|_X$. Therefore \begin{align*} \sum_{n=1}^\infty \frac{1}{2^n} |f_n(x - y)| \le \|x - y\|_X \sum_{n=1}^\infty \frac{1}{2^n} = \|x - y\|_X < \infty. \end{align*} **Symmetry.** Since $f_n$ is linear, $|f_n(x - y)| = |{-f_n(y - x)}| = |f_n(y - x)|$, giving $d(x, y) = d(y, x)$. **Triangle inequality.** For any $x, y, z \in X$ and each $n$, the triangle inequality for absolute values and linearity of $f_n$ give \begin{align*} |f_n(x - z)| = |f_n(x - y) + f_n(y - z)| \le |f_n(x - y)| + |f_n(y - z)|. \end{align*} Multiplying by $2^{-n}$ and summing over $n$ yields $d(x, z) \le d(x, y) + d(y, z)$. **Definiteness.** Suppose $d(x, y) = 0$. Since each term $2^{-n} |f_n(x - y)|$ is nonneg and the sum vanishes, $f_n(x - y) = 0$ for every $n \ge 1$. Now let $g \in \overline{B}_{X^*}$ be arbitrary. Since $\{f_n\}$ is dense in $\overline{B}_{X^*}$, there exists a subsequence $f_{n_k}$ with $\|f_{n_k} - g\|_{X^*} \to 0$. Then \begin{align*} |g(x - y)| = |g(x - y) - f_{n_k}(x - y)| \le \|g - f_{n_k}\|_{X^*} \|x - y\|_X \to 0, \end{align*} so $g(x - y) = 0$. Since $g \in \overline{B}_{X^*}$ was arbitrary, $f(x - y) = 0$ for every $f \in X^*$ (any $f \in X^* \setminus \{0\}$ can be rescaled into $\overline{B}_{X^*}$). The [Hahn-Banach Theorem](/theorems/879) guarantees that for any nonzero $v \in X$, there exists $f \in X^*$ with $f(v) \neq 0$. Taking $v = x - y$, the condition $f(x - y) = 0$ for all $f \in X^*$ forces $x - y = 0$, i.e. $x = y$. [/guided] [/step] [step:Show that [weakly convergent](/page/Weak%20Topology) nets in a bounded set are $d$-convergent] Let $B \subset X$ be norm-bounded, say $\|x\|_X \le M$ for all $x \in B$. We show that if a net $(x_\alpha)$ in $B$ converges to $x \in B$ in the weak topology $\sigma(X, X^*)$, then $d(x_\alpha, x) \to 0$. Fix $\varepsilon > 0$. Choose $N \in \mathbb{N}$ such that $\sum_{n=N+1}^\infty 2^{-n} \le \varepsilon / (4M)$. Since $\|f_n\|_{X^*} \le 1$ and $\|x_\alpha - x\|_X \le 2M$ for all $\alpha$, the tail satisfies \begin{align*} \sum_{n=N+1}^\infty \frac{1}{2^n} |f_n(x_\alpha - x)| \le 2M \sum_{n=N+1}^\infty \frac{1}{2^n} \le \frac{\varepsilon}{2}. \end{align*} Since $x_\alpha \rightharpoonup x$ weakly, we have $f_n(x_\alpha) \to f_n(x)$ for each $n = 1, \ldots, N$. Choose $\alpha_0$ such that for all $\alpha \ge \alpha_0$, \begin{align*} \sum_{n=1}^{N} \frac{1}{2^n} |f_n(x_\alpha - x)| < \frac{\varepsilon}{2}. \end{align*} Combining both estimates, $d(x_\alpha, x) < \varepsilon$ for all $\alpha \ge \alpha_0$. [guided] This direction says: [weak convergence](/page/Weak%20Convergence) in $B$ implies $d$-convergence. The argument splits the defining series into a finite head and a tail, and uses boundedness to control the tail uniformly. Fix $\varepsilon > 0$. We split the series defining $d(x_\alpha, x)$ at a cutoff $N$: \begin{align*} d(x_\alpha, x) = \underbrace{\sum_{n=1}^{N} \frac{1}{2^n} |f_n(x_\alpha - x)|}_{\text{finite head}} + \underbrace{\sum_{n=N+1}^\infty \frac{1}{2^n} |f_n(x_\alpha - x)|}_{\text{tail}}. \end{align*} **Controlling the tail.** Since $x_\alpha, x \in B$ and $B$ is bounded with $\|x\|_X \le M$ for all $x \in B$, we have $\|x_\alpha - x\|_X \le 2M$. Each $f_n$ has $\|f_n\|_{X^*} \le 1$, so $|f_n(x_\alpha - x)| \le 2M$. Therefore \begin{align*} \sum_{n=N+1}^\infty \frac{1}{2^n} |f_n(x_\alpha - x)| \le 2M \sum_{n=N+1}^\infty \frac{1}{2^n}. \end{align*} Choose $N$ large enough that $\sum_{n=N+1}^\infty 2^{-n} < \varepsilon/(4M)$, giving a tail bound of $\varepsilon/2$. This bound is **uniform** in $\alpha$ — this is where boundedness of $B$ is consumed. **Controlling the finite head.** Since $x_\alpha \rightharpoonup x$ in $\sigma(X, X^*)$, we have $f_n(x_\alpha) \to f_n(x)$ for each fixed $f_n \in X^*$. In particular, for each of the finitely many $n = 1, \ldots, N$, $|f_n(x_\alpha - x)| \to 0$. A finite sum of terms each tending to zero tends to zero, so there exists $\alpha_0$ such that for $\alpha \ge \alpha_0$, \begin{align*} \sum_{n=1}^{N} \frac{1}{2^n} |f_n(x_\alpha - x)| < \frac{\varepsilon}{2}. \end{align*} Combining: for $\alpha \ge \alpha_0$, $d(x_\alpha, x) < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Why does boundedness matter here? Without it, the tail bound $2M \sum_{n > N} 2^{-n}$ is replaced by $\|x_\alpha - x\|_X \sum_{n > N} 2^{-n}$, which could grow without bound as $\alpha$ varies. Weak convergence does not imply norm-boundedness in general nets (though it does for sequences, by the Uniform Boundedness Principle). On unbounded sets, the $d$-topology can be strictly finer than the weak topology. [/guided] [/step] [step:Show that $d$-convergent nets in a bounded set are weakly convergent] Conversely, suppose $(x_\alpha)$ is a net in $B$ with $d(x_\alpha, x) \to 0$ for some $x \in B$. We must show $f(x_\alpha) \to f(x)$ for every $f \in X^*$. **Step for the dense set.** For each fixed $n$, since $2^{-n} |f_n(x_\alpha - x)| \le d(x_\alpha, x) \to 0$, we obtain $f_n(x_\alpha) \to f_n(x)$. **Extension to all of $X^*$.** Let $f \in X^*$ be arbitrary with $f \neq 0$. Set $g := f / \|f\|_{X^*}$, so $g \in \overline{B}_{X^*}$. Fix $\varepsilon > 0$. Since $\{f_n\}$ is dense in $\overline{B}_{X^*}$, there exists $f_{n_0}$ with $\|g - f_{n_0}\|_{X^*} < \varepsilon / (4M)$. Then for every $x_\alpha \in B$, \begin{align*} |g(x_\alpha - x)| &\le |g(x_\alpha - x) - f_{n_0}(x_\alpha - x)| + |f_{n_0}(x_\alpha - x)| \\ &\le \|g - f_{n_0}\|_{X^*} \|x_\alpha - x\|_X + |f_{n_0}(x_\alpha - x)| \\ &\le \frac{\varepsilon}{4M} \cdot 2M + |f_{n_0}(x_\alpha - x)| \\ &= \frac{\varepsilon}{2} + |f_{n_0}(x_\alpha - x)|. \end{align*} Since $f_{n_0}(x_\alpha) \to f_{n_0}(x)$, there exists $\alpha_0$ such that $|f_{n_0}(x_\alpha - x)| < \varepsilon / 2$ for $\alpha \ge \alpha_0$. Thus $|g(x_\alpha - x)| < \varepsilon$ for $\alpha \ge \alpha_0$, giving $g(x_\alpha) \to g(x)$. By linearity, $f(x_\alpha) = \|f\|_{X^*} \, g(x_\alpha) \to \|f\|_{X^*} \, g(x) = f(x)$. [guided] This is the key direction: $d$-convergence implies weak convergence on bounded sets. The argument is a standard "$\varepsilon/3$" density argument, and it is here that both density of $\{f_n\}$ and boundedness of $B$ are consumed simultaneously. **Convergence on the dense set.** From $d(x_\alpha, x) \to 0$, each individual term in the series tends to zero: \begin{align*} 0 \le \frac{1}{2^n} |f_n(x_\alpha - x)| \le d(x_\alpha, x) \to 0, \end{align*} so $f_n(x_\alpha - x) \to 0$ for every $n \ge 1$. **Extension to all of $X^*$.** Let $f \in X^*$ with $f \neq 0$. We normalise: set $g := f / \|f\|_{X^*} \in \overline{B}_{X^*}$. Fix $\varepsilon > 0$. Since $\{f_n\}$ is dense in $\overline{B}_{X^*}$, there exists $f_{n_0}$ with $\|g - f_{n_0}\|_{X^*} < \varepsilon / (4M)$. We decompose \begin{align*} |g(x_\alpha - x)| &= |(g - f_{n_0})(x_\alpha - x) + f_{n_0}(x_\alpha - x)| \\ &\le \|g - f_{n_0}\|_{X^*} \, \|x_\alpha - x\|_X + |f_{n_0}(x_\alpha - x)|. \end{align*} The first term is bounded by $(\varepsilon / (4M)) \cdot 2M = \varepsilon / 2$, using $\|x_\alpha - x\|_X \le 2M$ since both $x_\alpha$ and $x$ lie in $B$. The second term $|f_{n_0}(x_\alpha - x)| \to 0$ as established above, so it falls below $\varepsilon/2$ for all $\alpha$ sufficiently large. Therefore $|g(x_\alpha - x)| < \varepsilon$ eventually, and since $\varepsilon > 0$ was arbitrary, $g(x_\alpha) \to g(x)$. Rescaling back: $f(x_\alpha) = \|f\|_{X^*} \, g(x_\alpha) \to \|f\|_{X^*} \, g(x) = f(x)$. Why does boundedness enter this direction? Because we need the approximation error $\|g - f_{n_0}\|_{X^*} \, \|x_\alpha - x\|_X$ to be small. The density of $\{f_n\}$ lets us make $\|g - f_{n_0}\|_{X^*}$ arbitrarily small, but the factor $\|x_\alpha - x\|_X$ must remain bounded for the product to be small. Without a bound on the norms of $x_\alpha$, this product could diverge even as $\|g - f_{n_0}\|_{X^*} \to 0$. [/guided] [/step] [step:Combine both directions to obtain the [metrizability](/page/Metrizable%20Space) conclusion] Let $B \subset X$ be any norm-bounded set. The previous two steps show that for any net $(x_\alpha)$ in $B$ and any $x \in B$: \begin{align*} x_\alpha \xrightarrow{\sigma(X, X^*)} x \quad \Longleftrightarrow \quad d(x_\alpha, x) \to 0. \end{align*} A subset $V \subset B$ is open in the restriction $\sigma(X, X^*)|_B$ if and only if for every $x \in V$, every net in $B$ converging weakly to $x$ is eventually in $V$; and the analogous characterisation holds for the $d$-topology. Since the two notions of net convergence coincide on $B$, the two topologies have the same [open sets](/page/Open%20Set) in $B$. Therefore $\sigma(X, X^*)|_B$ is metrizable, with $d$ serving as a compatible metric. [/step]