[proofplan] We identify $\overline{B}_{\mathcal{L}(X,Y)}$ with a subset of the product space $\prod_{x \in X} \overline{B}_Y(0, \|x\|)$ via the evaluation embedding $T \mapsto (Tx)_{x \in X}$, and show this image is closed in the product topology. When $Y$ is reflexive, each factor $\overline{B}_Y(0, \|x\|)$ is weakly compact (by the [Characterisation of Reflexivity](/theorems/977)), and the product topology on the product of weak-topology spaces coincides with the WOT on the image, so [Tychonoff's Theorem](/theorems/953) gives compactness. Conversely, when $Y$ is not reflexive, we show the ball $\overline{B}_Y$ embeds into $\overline{B}_{\mathcal{L}(X,Y)}$ as a WOT-closed subset, and if the latter were WOT-compact, $\overline{B}_Y$ would be weakly compact, contradicting non-reflexivity. [/proofplan] [step:Identify the WOT on $\overline{B}_{\mathcal{L}(X,Y)}$ with a product topology via the evaluation embedding] Define the evaluation embedding \begin{align*} \Phi: \mathcal{L}(X, Y) &\to \prod_{x \in X} Y \\ T &\mapsto (Tx)_{x \in X}, \end{align*} where the product $\prod_{x \in X} Y$ carries the product topology induced by the weak topology $\sigma(Y, Y^*)$ on each factor $Y$. By definition, a net $(T_\alpha)$ converges to $T$ in the WOT if and only if $f(T_\alpha x) \to f(Tx)$ for every $x \in X$ and $f \in Y^*$. This is precisely the statement that $T_\alpha x \to Tx$ in the weak topology $\sigma(Y, Y^*)$ for every $x \in X$, which is convergence of $\Phi(T_\alpha)$ to $\Phi(T)$ in the product topology. Therefore $\Phi$ is a homeomorphism from $(\mathcal{L}(X, Y), \text{WOT})$ onto its image in $\prod_{x \in X} (Y, \sigma(Y, Y^*))$. For $T \in \overline{B}_{\mathcal{L}(X,Y)}$ and each $x \in X$, $\|Tx\|_Y \le \|T\| \cdot \|x\|_X \le \|x\|_X$. Therefore \begin{align*} \Phi(\overline{B}_{\mathcal{L}(X,Y)}) \subset \prod_{x \in X} \overline{B}_Y(0, \|x\|_X), \end{align*} where $\overline{B}_Y(0, r) = \{y \in Y : \|y\|_Y \le r\}$. [/step] [step:Show the image $\Phi(\overline{B}_{\mathcal{L}(X,Y)})$ is closed in the product topology] We show that $\Phi(\overline{B}_{\mathcal{L}(X,Y)})$ is closed in $\prod_{x \in X} (Y, \sigma(Y, Y^*))$. Let $(T_\alpha)$ be a net in $\overline{B}_{\mathcal{L}(X,Y)}$ with $\Phi(T_\alpha) \to \varphi \in \prod_{x \in X} Y$ in the product topology. This means that for each $x \in X$, $T_\alpha x \to \varphi(x)$ weakly in $Y$. We must show $\varphi = \Phi(T)$ for some $T \in \overline{B}_{\mathcal{L}(X,Y)}$. **Linearity.** For $x, y \in X$ and $\lambda \in \mathbb{R}$, fix $f \in Y^*$. Then $f(T_\alpha(\lambda x + y)) = \lambda f(T_\alpha x) + f(T_\alpha y)$ by linearity of $T_\alpha$. Taking limits: $f(\varphi(\lambda x + y)) = \lambda f(\varphi(x)) + f(\varphi(y))$. Since $f \in Y^*$ was arbitrary and $Y^*$ separates points of $Y$ (by the [Hahn-Banach Theorem](/theorems/879)), $\varphi(\lambda x + y) = \lambda \varphi(x) + \varphi(y)$. **Boundedness.** For each $x \in X$ and $f \in Y^*$, $|f(\varphi(x))| = \lim_\alpha |f(T_\alpha x)| \le \|f\|_{Y^*} \liminf_\alpha \|T_\alpha x\|_Y \le \|f\|_{Y^*} \|x\|_X$ (using $\|T_\alpha\| \le 1$). Taking the supremum over $f$ with $\|f\| \le 1$: $\|\varphi(x)\|_Y \le \|x\|_X$. Therefore the linear map $T := \varphi$ satisfies $\|T\| \le 1$, so $T \in \overline{B}_{\mathcal{L}(X,Y)}$. [guided] We need the image of $\Phi$ restricted to the unit ball to be a closed subset of the product. This is where we use the linearity and norm constraints that define $\mathcal{L}(X, Y)$. Let $(T_\alpha)$ be a net in $\overline{B}_{\mathcal{L}(X,Y)}$ with $T_\alpha x \to \varphi(x)$ weakly in $Y$ for every $x \in X$. We must verify two things about the limit function $\varphi: X \to Y$. **Linearity.** For $x, y \in X$, $\lambda \in \mathbb{R}$, and any $f \in Y^*$: \begin{align*} f(\varphi(\lambda x + y)) &= \lim_\alpha f(T_\alpha(\lambda x + y)) = \lim_\alpha [\lambda f(T_\alpha x) + f(T_\alpha y)] = \lambda f(\varphi(x)) + f(\varphi(y)) \\ &= f(\lambda \varphi(x) + \varphi(y)). \end{align*} Since this holds for every $f \in Y^*$, and $Y^*$ separates points (by the [Hahn-Banach Theorem](/theorems/879)), $\varphi(\lambda x + y) = \lambda \varphi(x) + \varphi(y)$. **Operator norm bound.** For each $x$ with $\|x\|_X \le 1$, the net $(T_\alpha x)$ lies in $\overline{B}_Y(0, 1)$ (since $\|T_\alpha x\| \le \|T_\alpha\| \|x\| \le 1$). The weak limit of a net in a closed convex set remains in that set (closed convex sets are weakly closed, by the Hahn-Banach separation theorem). Therefore $\varphi(x) \in \overline{B}_Y(0, 1)$, i.e., $\|\varphi(x)\|_Y \le 1$. Taking the supremum: $\|\varphi\|_{\mathcal{L}(X,Y)} \le 1$. Thus $T := \varphi \in \overline{B}_{\mathcal{L}(X,Y)}$ and $\Phi(T) = \varphi$, confirming that the image is closed. [/guided] [/step] [step:Apply Tychonoff's theorem when $Y$ is reflexive to obtain WOT compactness] Assume $Y$ is reflexive. By the [Characterisation of Reflexivity](/theorems/977), the closed unit ball $\overline{B}_Y$ is weakly compact. For each $x \in X$, the ball $\overline{B}_Y(0, \|x\|_X) = \|x\|_X \cdot \overline{B}_Y$ is weakly compact (as the continuous image of $\overline{B}_Y$ under the scaling map $y \mapsto \|x\|_X \, y$, which is weak-to-weak continuous). By [Tychonoff's Theorem](/theorems/953), the product \begin{align*} \prod_{x \in X} \overline{B}_Y(0, \|x\|_X) \end{align*} is compact in the product topology (where each factor carries $\sigma(Y, Y^*)$). The image $\Phi(\overline{B}_{\mathcal{L}(X,Y)})$ is a closed subset of this compact product (by the previous step), hence it is compact. Since $\Phi$ is a homeomorphism onto its image, $\overline{B}_{\mathcal{L}(X,Y)}$ is WOT-compact. The "more generally" statement follows: if $S \subset \mathcal{L}(X, Y)$ is WOT-closed and norm-bounded by $M$, then $S \subset M \cdot \overline{B}_{\mathcal{L}(X,Y)}$. Since $M \cdot \overline{B}_{\mathcal{L}(X,Y)}$ is WOT-compact (by the same argument applied with $\|x\|_X$ replaced by $M\|x\|_X$), and $S$ is a WOT-closed subset of a WOT-compact set, $S$ is WOT-compact. [/step] [step:Prove the converse: WOT compactness of the unit ball implies reflexivity of $Y$] Assume $\overline{B}_{\mathcal{L}(X,Y)}$ is WOT-compact. Fix any $x_0 \in X$ with $\|x_0\|_X = 1$ and any $f_0 \in X^*$ with $f_0(x_0) = 1$ and $\|f_0\|_{X^*} = 1$ (which exists by the [Hahn-Banach Theorem](/theorems/879)). Define the embedding \begin{align*} \Theta: Y &\to \mathcal{L}(X, Y) \\ y &\mapsto f_0(\cdot) \, y. \end{align*} This maps $\overline{B}_Y$ into $\overline{B}_{\mathcal{L}(X,Y)}$, since $\|\Theta(y)\| = \|f_0\| \cdot \|y\| = \|y\|$. We claim $\Theta$ is continuous from $(Y, \sigma(Y, Y^*))$ to $(\mathcal{L}(X, Y), \text{WOT})$, and that $\Theta(\overline{B}_Y)$ is WOT-closed. **Continuity.** Let $y_\alpha \to y$ weakly in $Y$. For any $x \in X$ and $g \in Y^*$, $g(\Theta(y_\alpha) x) = g(f_0(x) y_\alpha) = f_0(x) g(y_\alpha) \to f_0(x) g(y) = g(\Theta(y) x)$. So $\Theta(y_\alpha) \to \Theta(y)$ in the WOT. **Closedness.** Let $(T_\alpha)$ be a net in $\Theta(\overline{B}_Y)$ with $T_\alpha \to T$ in the WOT. Write $T_\alpha = \Theta(y_\alpha)$ with $y_\alpha \in \overline{B}_Y$. Evaluating at $x_0$: $T_\alpha x_0 = f_0(x_0) y_\alpha = y_\alpha$. Since $T_\alpha \to T$ in the WOT, $y_\alpha = T_\alpha x_0 \to T x_0$ weakly in $Y$. Set $y := Tx_0$. For any $x \in X$ and $g \in Y^*$: $g(Tx) = \lim_\alpha g(T_\alpha x) = \lim_\alpha f_0(x) g(y_\alpha) = f_0(x) g(y) = g(\Theta(y) x)$. Since $Y^*$ separates points, $T = \Theta(y)$. Since $\|y\|_Y = \|Tx_0\|_Y \le \|T\| \cdot \|x_0\| \le 1$, we have $y \in \overline{B}_Y$ and $T \in \Theta(\overline{B}_Y)$. Therefore $\Theta(\overline{B}_Y)$ is a WOT-closed subset of the WOT-compact set $\overline{B}_{\mathcal{L}(X,Y)}$, hence WOT-compact. Since $\Theta$ is a weak-to-WOT homeomorphism onto its image (the evaluation map $T \mapsto Tx_0$ provides the continuous inverse), $\overline{B}_Y$ is weakly compact. By the [Characterisation of Reflexivity](/theorems/977), $Y$ is reflexive. [guided] The converse requires extracting weak compactness of $\overline{B}_Y$ from WOT compactness of $\overline{B}_{\mathcal{L}(X,Y)}$. The idea is to embed $\overline{B}_Y$ as a WOT-closed subset of $\overline{B}_{\mathcal{L}(X,Y)}$ via rank-one operators. We pick any $x_0 \in X$ with $\|x_0\| = 1$ and a norming functional $f_0 \in X^*$ with $f_0(x_0) = 1$, $\|f_0\| = 1$ (guaranteed by the [Hahn-Banach Theorem](/theorems/879)). The map $\Theta(y) = f_0(\cdot) y$ sends $y \in \overline{B}_Y$ to the rank-one operator $x \mapsto f_0(x) y$, which has norm $\|f_0\| \cdot \|y\| = \|y\| \le 1$. The crucial point is that evaluation at $x_0$ recovers $y$: $\Theta(y)(x_0) = f_0(x_0) y = y$. This means the embedding $\Theta$ has a left inverse — the evaluation map $\text{ev}_{x_0}: T \mapsto Tx_0$ — which is continuous from the WOT to the weak topology on $Y$ (since WOT convergence at $x_0$ is precisely weak convergence of $Tx_0$). Therefore: $\Theta(\overline{B}_Y)$ is WOT-closed in $\overline{B}_{\mathcal{L}(X,Y)}$ (shown above), hence WOT-compact (closed subset of compact). The map $\text{ev}_{x_0} \circ \Theta^{-1}$ is continuous from the WOT to the weak topology and equals the identity on $\overline{B}_Y$. So $\overline{B}_Y$ is the continuous image of the WOT-compact set $\Theta(\overline{B}_Y)$ under the weak-continuous map $\text{ev}_{x_0}$, hence weakly compact. By the [Characterisation of Reflexivity](/theorems/977), $Y$ is reflexive. [/guided] [/step]