Strong and weak operator topologies

Let $X,Y$ be normed spaces over the same field $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$. On $\mathcal{B}(X,Y)$, the bounded scalar-linear operators of The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, define:

• The strong operator topology (SOT) is the initial topology of all maps $T\mapsto Tx$ to normed $Y$, for $x\in X$.
• The weak operator topology (WOT) is the initial topology of all scalar maps $T\mapsto f(Tx)$, for $x\in X$ and $f\in Y^{\ast }$.

These topologies exist by The initial topology of a family of maps into spaces and the final topology of a family of maps out of spaces, and the subspace topology as the model initial topology. Equivalently WOT is initial for $T\mapsto Tx$ with the weak topology on $Y$ from Weak topology on a normed space. At $T$, basic neighborhoods impose finitely many inequalities $\mathrm{\parallel }(S-T)x_j\mathrm{\parallel }<\epsilon$ for SOT, or $\mathrm{\mid }{f}_j((S-T)x_j)\mathrm{\mid }<\epsilon$ for WOT, with $\epsilon >0$. Empty lists give the whole operator space.

For a net of bounded operators with specified limit $T\in \mathcal{B}(X,Y)$, SOT convergence means $\mathrm{\parallel }(T_i-T)x\mathrm{\parallel }\to 0$ for every fixed $x$, and WOT convergence means $f(T_{i}x)\to f(Tx)$ for every fixed $x,f$. Both equivalences follow by testing one coordinate and then using a common upper bound for the finitely many eventual indices in a basic neighborhood. No uniformity in $x$ is part of either definition. These are choice-free constructions, also when one space is zero and the operator space is a singleton.