[proofplan] We show $X' = X^*$ (continuous linear functionals equal bounded linear functionals on a normed space) by proving both inclusions. Bounded implies continuous via an $\varepsilon$-$\delta$ argument at the origin. Continuous implies bounded by exhibiting a ball on which $|f| < 1$ and using linearity to bound the operator norm. The topological conclusion (2) is then immediate since both weak* topologies are defined by the same construction on the same set. [/proofplan] [step:Prove bounded linear functionals are continuous] Let $f: X \to \mathbb{R}$ be linear with $\|f\|_{X^*} < \infty$. It suffices to show continuity at $0$. Let $\varepsilon > 0$ and set $\delta := \varepsilon / (\|f\|_{X^*} + 1)$. For any $x \in X$ with $\|x\|_X < \delta$: \begin{align*} |f(x)| &\leq \|f\|_{X^*} \cdot \|x\|_X < \|f\|_{X^*} \cdot \delta < \varepsilon. \end{align*} Thus $f^{-1}((-\varepsilon, \varepsilon)) \supseteq \{x : \|x\|_X < \delta\}$, an open neighbourhood of $0$. [/step] [step:Prove continuous linear functionals are bounded] Continuity at $0$ gives an open neighbourhood $U$ of $0$ with $|f(x)| < 1$ for all $x \in U$. There exists $r > 0$ with $B(0, r) \subseteq U$. For any $x$ with $\|x\|_X \leq 1$, the vector $(r/2)x$ satisfies $\|(r/2)x\|_X < r$, so $|f((r/2)x)| < 1$, giving $|f(x)| < 2/r$. Taking the supremum: $\|f\|_{X^*} \leq 2/r < \infty$. [/step] [step:Conclude the topologies coincide] By the previous two steps, $X' = X^*$ as sets. Both $\sigma(X', X)$ and $\sigma(X^*, X)$ are the coarsest topology making $\operatorname{ev}_x: f \mapsto f(x)$ continuous for every $x \in X$. Since they are the same construction on the same set, they produce the same topology. [/step]