Is the weak operator topology equal to $\sigma(L(X,Y), X\times Y^*)$?

Question

I'm asking myself if the weak operator topology is equal to the weak topology $\sigma(L(X,Y), X \times Y^{*},b)$ with $$ \begin{align} b:L(X,Y)\times (X \times Y^{*})\to& [0, \infty)\\ (T,(x,y') \mapsto& \langle y', Tx\rangle \end{align} $$ I should only check that $(L(X,Y), X \times Y^{*},b)$ is a dual pair, right? Because then I could conclude using the fact that $\sigma(L(X,Y), X \times Y^{*},b)$ is also generated by the same family of seminorms which defines the weak operator topology.

Answer 1

You need to be a little careful as your mapping $b$ here isn't bilinear : you can check that when $T\in L(X,Y)$ is nonzero, the mapping $(x,y')\mapsto b(T,(x,y'))$ from $X\times Y^\ast$ to $\mathbb R$ is bilinear, not linear.

Instead of $X\times Y^\ast$, you should consider the algebraic tensor product $X\otimes Y^\ast$. There is a unique bilinear mapping $B$ from $L(X,Y^\ast)\times(X\otimes Y^\ast)$ to $\mathbb R$ such that $B(T,x\otimes y')=b(T,(x,y'))$ for all $T\in L(X,Y)$, $x\in X$ and $y'\in Y^\ast$.

And then, you can check that the weak operator topology on $L(X,Y)$ is indeed $\sigma(L(X,Y),X\otimes Y^\ast,B)$.