[proofplan] For $(1) \Rightarrow (2)$: WOT convergence gives pointwise boundedness of the evaluation maps $x \mapsto f(T_k x)$, and the [Uniform Boundedness Principle](/theorems/549) upgrades this to $\sup_k \|T_k\| < \infty$; convergence on dense subsets is immediate. For $(2) \Rightarrow (1)$: we use the uniform bound $M = \sup_k \|T_k\|$ together with an $\varepsilon/3$-argument — approximate any $x \in X$ by $d \in D$ and any $f \in Y^*$ by $g \in E$, controlling the error in three terms, each bounded by $\varepsilon/3$. [/proofplan] [step:Prove $(1) \Rightarrow (2)$: WOT convergence implies uniform boundedness and convergence on dense subsets] Assume $T_k \to T$ in the WOT: for every $x \in X$ and every $f \in Y^*$, $f(T_k x) \to f(Tx)$. **Uniform boundedness.** Fix $x \in X$. For each $k$, define the bounded linear functional \begin{align*} \varphi_k: Y^* &\to \mathbb{R} \\ f &\mapsto f(T_k x). \end{align*} Each $\varphi_k$ lies in $Y^{**}$ via the canonical embedding, and WOT convergence gives $\sup_{k} |\varphi_k(f)| = \sup_k |f(T_k x)| < \infty$ for every $f \in Y^*$. Since $Y^*$ is a Banach space (the dual of a normed space is always complete), the [Uniform Boundedness Principle](/theorems/549) applied to the family $(\varphi_k)_{k \in \mathbb{N}} \subset \mathcal{L}(Y^*, \mathbb{R})$ yields \begin{align*} \sup_{k \in \mathbb{N}} \|\varphi_k\|_{Y^{**}} < \infty. \end{align*} By the definition of the canonical embedding, $\|\varphi_k\|_{Y^{**}} = \|T_k x\|_Y$ (this follows from the [Hahn-Banach Theorem](/theorems/879), which guarantees the existence of a norming functional). Therefore $\sup_k \|T_k x\|_Y < \infty$ for each $x \in X$. Since $X$ is a Banach space, applying the [Uniform Boundedness Principle](/theorems/549) a second time to the family $(T_k)_{k \in \mathbb{N}} \subset \mathcal{L}(X, Y)$ gives \begin{align*} \sup_{k \in \mathbb{N}} \|T_k\|_{\mathcal{L}(X,Y)} < \infty. \end{align*} **Convergence on dense subsets.** Since WOT convergence requires $f(T_k x) \to f(Tx)$ for *all* $x \in X$ and *all* $f \in Y^*$, the convergence holds a fortiori for $x$ in any dense subset $D \subset X$ and $f$ in any dense subset $E \subset Y^*$. [guided] Assume $T_k \to T$ in the WOT, so $f(T_k x) \to f(Tx)$ for every $x \in X$ and $f \in Y^*$. We must establish two properties. **Uniform boundedness.** The argument requires two applications of the [Uniform Boundedness Principle](/theorems/549), each upgrading a different layer of pointwise boundedness. *First application (at the level of $Y^*$).* Fix $x \in X$. For each $k \in \mathbb{N}$, define \begin{align*} \varphi_k: Y^* &\to \mathbb{R} \\ f &\mapsto f(T_k x). \end{align*} Each $\varphi_k$ is a bounded linear functional on $Y^*$ (it is the image of $T_k x \in Y$ under the canonical embedding $Y \hookrightarrow Y^{**}$), so $\varphi_k \in \mathcal{L}(Y^*, \mathbb{R})$. The WOT convergence hypothesis gives $\sup_k |\varphi_k(f)| = \sup_k |f(T_k x)| < \infty$ for each $f \in Y^*$. Since $Y^*$ is a Banach space (the dual of any normed space is complete), the hypotheses of the [Uniform Boundedness Principle](/theorems/549) are satisfied for the family $(\varphi_k) \subset \mathcal{L}(Y^*, \mathbb{R})$. The principle gives $\sup_k \|\varphi_k\|_{Y^{**}} < \infty$. Now, $\|\varphi_k\|_{Y^{**}} = \sup_{\|f\|_{Y^*} \le 1} |f(T_k x)| = \|T_k x\|_Y$, where the last equality uses the fact that the canonical embedding $Y \hookrightarrow Y^{**}$ is isometric (a consequence of the [Hahn-Banach Theorem](/theorems/879)). Therefore $\sup_k \|T_k x\|_Y < \infty$ for each fixed $x \in X$. *Second application (at the level of $X$).* We have just shown that the family $(T_k) \subset \mathcal{L}(X, Y)$ is pointwise bounded. Since $X$ is a Banach space, the [Uniform Boundedness Principle](/theorems/549) applies again: \begin{align*} \sup_{k \in \mathbb{N}} \|T_k\|_{\mathcal{L}(X,Y)} < \infty. \end{align*} This double application is the reason the theorem needs $X$ to be a Banach space and not merely a normed space. **Convergence on dense subsets.** The WOT convergence $f(T_k x) \to f(Tx)$ for all $x \in X$, $f \in Y^*$ immediately restricts to any subsets $D \subset X$ and $E \subset Y^*$. [/guided] [/step] [step:Prove $(2) \Rightarrow (1)$: uniform boundedness plus convergence on dense subsets implies WOT convergence] Let $M := \sup_{k \in \mathbb{N}} \|T_k\|_{\mathcal{L}(X,Y)} < \infty$. By hypothesis (2), we also have $\|T\|_{\mathcal{L}(X,Y)} \le \liminf_{k \to \infty} \|T_k\|_{\mathcal{L}(X,Y)} \le M$ (by [SOT Convergence Implies Uniform Boundedness](/theorems/1239), since WOT convergence on a dense set together with the uniform bound implies SOT convergence on $D$... but we do not need this). In fact, we verify directly: for any $x \in X$ and $f \in Y^*$ with $\|x\|_X \le 1$ and $\|f\|_{Y^*} \le 1$, taking $d_m \to x$ in $D$ gives $|f(Tx)| = \lim_m |f(Td_m)| = \lim_m \lim_k |f(T_k d_m)| \le M$, so $\|T\| \le M$. Fix $x \in X$, $f \in Y^*$, and $\varepsilon > 0$. Choose $d \in D$ with $\|x - d\|_X < \frac{\varepsilon}{3(M + \|T\|)\|f\|_{Y^*} + 1}$ and $g \in E$ with $\|f - g\|_{Y^*} < \frac{\varepsilon}{3(M + \|T\|)\|d\|_X + 1}$ (using density of $D$ in $X$ and $E$ in $Y^*$). Write \begin{align*} f(T_k x) - f(Tx) &= f(T_k x - T_k d) + (f - g)(T_k d - Td) + g(T_k d - Td) + (g - f)(Td) + f(Td - Tx). \end{align*} We regroup more cleanly. Write $f(T_k x) - f(Tx) = A_k + B_k + C_k$ where \begin{align*} A_k &:= f(T_k(x - d)) - f(T(x - d)) = f((T_k - T)(x - d)), \\ B_k &:= (f - g)(T_k d - Td), \\ C_k &:= g(T_k d - Td). \end{align*} **Bound on $|A_k|$:** $|A_k| \le \|f\|_{Y^*}(\|T_k\| + \|T\|)\|x - d\|_X \le \|f\|_{Y^*}(M + M)\|x - d\|_X < \varepsilon/3$ by the choice of $d$. **Bound on $|B_k|$:** $|B_k| \le \|f - g\|_{Y^*}(\|T_k\| + \|T\|)\|d\|_X \le \|f - g\|_{Y^*} \cdot 2M \cdot \|d\|_X < \varepsilon/3$ by the choice of $g$. **Bound on $|C_k|$:** Since $d \in D$ and $g \in E$, the hypothesis gives $g(T_k d) \to g(Td)$, so $|C_k| < \varepsilon/3$ for all $k$ sufficiently large. Therefore $|f(T_k x) - f(Tx)| \le |A_k| + |B_k| + |C_k| < \varepsilon$ for all $k$ sufficiently large. [guided] We now prove the harder direction: the uniform bound $M := \sup_k \|T_k\| < \infty$ combined with convergence $g(T_k d) \to g(Td)$ for $d \in D$, $g \in E$, implies $f(T_k x) \to f(Tx)$ for all $x \in X$, $f \in Y^*$. First, we need $\|T\| \le M$. For $x \in X$ with $\|x\| \le 1$ and $f \in Y^*$ with $\|f\| \le 1$: pick $d_m \in D$ with $d_m \to x$. Then $|f(Tx)| = |f(T(x - d_m))| + |f(Td_m)|$. For the second term, $f(Td_m) = \lim_k f(T_k d_m)$, so $|f(Td_m)| \le \liminf_k \|f\| \cdot \|T_k\| \cdot \|d_m\| \le M\|d_m\|$. Taking $m \to \infty$ gives $|f(Tx)| \le M\|x\| \le M$. The idea for the main argument is the classical *$\varepsilon/3$-trick*: decompose the error $f(T_k x) - f(Tx)$ into three pieces — two that are small by the approximation quality of $d \approx x$ and $g \approx f$, and one that vanishes by the hypothesis. Fix $x \in X$, $f \in Y^*$, and $\varepsilon > 0$. Choose $d \in D$ and $g \in E$ to be specified. We decompose: \begin{align*} f(T_k x) - f(Tx) &= \underbrace{f((T_k - T)(x - d))}_{A_k} + \underbrace{(f - g)((T_k - T)d)}_{B_k} + \underbrace{g((T_k - T)d)}_{C_k}. \end{align*} To verify this decomposition: $A_k + B_k + C_k = f(T_k(x-d)) - f(T(x-d)) + f(T_k d) - f(Td) - g(T_k d) + g(Td) + g(T_k d) - g(Td)$. Simplifying: $f(T_k(x-d)) - f(T(x-d)) + f(T_k d) - f(Td) = f(T_k x) - f(Tx)$, confirming the identity. **Controlling $A_k$.** We estimate: \begin{align*} |A_k| = |f((T_k - T)(x - d))| \le \|f\|_{Y^*} \|T_k - T\|_{\mathcal{L}(X,Y)} \|x - d\|_X \le 2M \|f\|_{Y^*} \|x - d\|_X. \end{align*} Choose $d \in D$ with $\|x - d\|_X < \frac{\varepsilon}{6M\|f\|_{Y^*} + 1}$. This gives $|A_k| < \varepsilon/3$ for all $k$. **Controlling $B_k$.** We estimate: \begin{align*} |B_k| = |(f - g)((T_k - T)d)| \le \|f - g\|_{Y^*} \|(T_k - T)d\|_Y \le 2M \|f - g\|_{Y^*} \|d\|_X. \end{align*} Choose $g \in E$ with $\|f - g\|_{Y^*} < \frac{\varepsilon}{6M\|d\|_X + 1}$. This gives $|B_k| < \varepsilon/3$ for all $k$. **Controlling $C_k$.** Since $d \in D$ and $g \in E$, the hypothesis gives $g(T_k d) \to g(Td)$ as $k \to \infty$. Therefore there exists $K$ such that $|C_k| = |g(T_k d - Td)| < \varepsilon/3$ for all $k \ge K$. Combining: $|f(T_k x) - f(Tx)| \le |A_k| + |B_k| + |C_k| < \varepsilon$ for all $k \ge K$. [/guided] [/step]