> [!definition] > > Let $\cx, \cy$ be [[Banach Space|Banach spaces]], and $L(\cx, \cy)$ be the [[Space of Bounded Linear Maps|space of bounded linear maps]]. The [[Weak Topology|weak topology]] generated by the evaluation maps $T \mapsto Tx$ ($x \in \cx$) is the **strong operator topology**, where $T_\alpha \to T$ strongly if and only if $T_\alpha x \to Tx$ for all $x \in \cx$ strongly. > > The weak topology generated by the [[Bounded Linear Functional|bounded linear functionals]] $T \mapsto f(Tx)$ ($x \in \cx$, $f \in \cy^*$) is the **weak operator topology**, where $T_\alpha \to T$ weakly if and only if $T_\alpha x \to Tx$ for all $x \in \cx$ weakly. > [!theorem] > > Let $\seq{T_n} \subset L(\cx, \cy)$ be a [[Sequence|sequence]] of bounded linear maps such that $\sup_{n \in \nat}\norm{T_n} < \infty$, and $T \in L(\cx, \cy)$. If $\norm{T_nx - Tx} \to 0$ for all $x$ in a [[Dense|dense]] subset $D$ of $\cx$, then $T_n \to T$ strongly. > > *Proof*. Let $M = \max(\sup_{n \in \nat}\norm{T_n}, \norm{T})$, $\varepsilon > 0$, $x \in \cx$, $x' \in D$ such that $\norm{x' - x} < \varepsilon/3M$, and $N \in \nat$ such that $\norm{(T_n - T)x'} < \varepsilon/3$ for all $n \ge N$, then > $ > \begin{align*} > \norm{T_nx - Tx} &\le \norm{T_n(x' - x)} + \norm{T(x - x')} + \norm{(T_n - T)x'} \\ > &< \norm{T_n}\norm{x' - x} + \norm{T}\norm{x' - x} + \varepsilon/3 \\ > &< \varepsilon/3 + \varepsilon/3 + \varepsilon /3 = \varepsilon > \end{align*} > $ > for all $n \ge N$. Therefore $T_nx \to Tx$ for all $x \in \cx$, and $T_n \to T$ strongly.