[proofplan] We verify each inclusion by unwinding the definitions. For norm convergence implying SOT convergence, we bound $\|T_\alpha x - Tx\|_Y \le \|T_\alpha - T\|_{\mathcal{L}(X,Y)} \|x\|_X$ for each $x \in X$. For SOT convergence implying WOT convergence, we bound $|f(T_\alpha x - Tx)| \le \|f\|_{Y^*} \|T_\alpha x - Tx\|_Y$ for each $f \in Y^*$. We then show that neither inclusion reverses in infinite dimensions by exhibiting explicit counterexamples. [/proofplan]

[step:Show that norm convergence implies SOT convergence]Let $(T_\alpha)$ be a net in $\mathcal{L}(X, Y)$ with $\|T_\alpha - T\|_{\mathcal{L}(X,Y)} \to 0$. Fix $x \in X$. By the definition of the operator norm, \begin{align*} \|T_\alpha x - Tx\|_Y = \|(T_\alpha - T)x\|_Y \le \|T_\alpha - T\|_{\mathcal{L}(X,Y)} \|x\|_X. \end{align*} Since $\|x\|_X$ is a fixed constant and $\|T_\alpha - T\|_{\mathcal{L}(X,Y)} \to 0$ by hypothesis, the right-hand side tends to zero. Therefore $T_\alpha x \to Tx$ in $Y$ for every $x \in X$, which is precisely SOT convergence.[/step]

[guided]We must show that every norm-convergent net is SOT-convergent. The SOT requires convergence at each individual vector, so we fix an arbitrary $x \in X$ and estimate the error $\|T_\alpha x - Tx\|_Y$. The key tool is the defining property of the operator norm: for any $S \in \mathcal{L}(X, Y)$ and any $x \in X$, $\|Sx\|_Y \le \|S\|_{\mathcal{L}(X,Y)} \|x\|_X$. Applying this with $S = T_\alpha - T$: \begin{align*} \|T_\alpha x - Tx\|_Y = \|(T_\alpha - T)x\|_Y \le \|T_\alpha - T\|_{\mathcal{L}(X,Y)} \|x\|_X. \end{align*} The factor $\|x\|_X$ is independent of $\alpha$. The factor $\|T_\alpha - T\|_{\mathcal{L}(X,Y)}$ tends to zero by hypothesis. Therefore $\|T_\alpha x - Tx\|_Y \to 0$, and since $x$ was arbitrary, $T_\alpha \to T$ in the SOT. This argument reveals why the norm topology is strictly finer: norm convergence forces a *uniform* rate of convergence $\|T_\alpha x - Tx\|_Y \le \varepsilon \|x\|_X$ across all $x$ simultaneously, whereas SOT convergence only requires convergence for each fixed $x$ — the rate may depend on $x$.[/guided]

[step:Show that SOT convergence implies WOT convergence]Let $(T_\alpha)$ be a net in $\mathcal{L}(X, Y)$ with $T_\alpha \to T$ in the SOT. Fix $x \in X$ and $f \in Y^*$. Since $T_\alpha x \to Tx$ in the norm of $Y$ and $f \in Y^*$ is a bounded (hence continuous) linear functional on $Y$, \begin{align*} |f(T_\alpha x) - f(Tx)| = |f(T_\alpha x - Tx)| \le \|f\|_{Y^*} \|T_\alpha x - Tx\|_Y \to 0. \end{align*} Since this holds for every $x \in X$ and every $f \in Y^*$, we have $T_\alpha \to T$ in the WOT.[/step]

[guided]We must show that convergence at each vector (SOT) implies convergence of all linear functionals applied to each vector (WOT). Fix $x \in X$ and $f \in Y^*$. The WOT requires $f(T_\alpha x) \to f(Tx)$. By hypothesis, $T_\alpha x \to Tx$ in the norm topology of $Y$. Since $f: Y \to \mathbb{R}$ is a bounded linear functional, it is continuous with respect to the norm topology on $Y$. Therefore $f$ preserves convergence of nets: \begin{align*} |f(T_\alpha x) - f(Tx)| = |f(T_\alpha x - Tx)| \le \|f\|_{Y^*} \|T_\alpha x - Tx\|_Y. \end{align*} The right-hand side tends to zero because $\|T_\alpha x - Tx\|_Y \to 0$ (SOT convergence at $x$) and $\|f\|_{Y^*}$ is a fixed constant. Since $x \in X$ and $f \in Y^*$ were arbitrary, $T_\alpha \to T$ in the WOT. The distinction between SOT and WOT is now visible: the SOT tests convergence against all vectors in $X$, producing convergent sequences in $Y$. The WOT additionally composes with all functionals in $Y^*$, but since continuous functionals preserve convergence, this is a weaker requirement — not a stronger one.[/guided]

[step:Show that the inclusions are strict in infinite dimensions] It remains to show that neither implication reverses when $\dim X = \infty$. **SOT does not imply norm convergence.** Let $H$ be a separable infinite-dimensional Hilbert space with orthonormal basis $(e_k)_{k \in \mathbb{N}}$. Define the orthogonal projection \begin{align*} P_n: H &\to H \\ x &\mapsto \sum_{k=1}^{n} (x, e_k)_H \, e_k. \end{align*} For each fixed $x \in H$, the Parseval identity gives $\|P_n x - x\|_H^2 = \sum_{k=n+1}^{\infty} |(x, e_k)_H|^2 \to 0$ as $n \to \infty$, so $P_n \to I$ in the SOT. However, $\|P_n - I\|_{\mathcal{L}(H,H)} = 1$ for every $n$ since $(I - P_n)e_{n+1} = e_{n+1}$ has norm $1$. Therefore $P_n \not\to I$ in the norm topology. **WOT does not imply SOT convergence.** Using the same Hilbert space $H$ and basis, define the sequence of operators $T_k := (\cdot, e_k)_H \, e_1$, that is, \begin{align*} T_k: H &\to H \\ x &\mapsto (x, e_k)_H \, e_1. \end{align*} For any $x \in H$ and $f \in H^*$, the [Riesz Representation Theorem](/theorems/218) identifies $f$ with some $y \in H$ via $f(\cdot) = (\cdot, y)_H$. Then $f(T_k x) = (T_k x, y)_H = (x, e_k)_H (e_1, y)_H$. By Bessel's inequality, $(x, e_k)_H \to 0$ as $k \to \infty$, so $f(T_k x) \to 0$ for every $x$ and $f$. Thus $T_k \to 0$ in the WOT. However, $\|T_k e_k\|_H = \|e_1\|_H = 1$ for every $k$, so $T_k e_k \not\to 0$ in $H$, and $T_k \not\to 0$ in the SOT. [/step]