Continuity of a bilinear form with respect to weak$^*$ topology

Question

Let $X$ be a normed linear space and let $X^*$ be its topological dual. The bilinear form $\psi:X\times X^*\to F$ is defined by $$\psi(x,x^*)=x^*(x).$$ Is $\psi$ continuous with respect to the topology $\tau$ on $X\times X^*$, where members of $\tau$ is of the form $$\{U\times V:~U\text{ is open in norm topology in }X, V\text{ is open in weak${}^*$ topology in }X^*\}.$$ A detailed answer will be of very much help! Thanks in advance!

Answer 1

Your map is not continuous. The basic problem is that while a weak* convergent sequence of operators is bounded, a weak* convergent net may be unbounded.

Work over real Banach spaces for simplicity. Consider $\phi^{-1}(-1, 1).$ This set is not open in the topology you describe, which we prove as follows.

Assume that it was open. Note that this set definitely includes the pair $(0, 0)$. So, it should contain some open set $U\times V$ where $U$ is a norm open set containing the 0 vector, and $V$ is a weak* open set containing the 0 functional.

The problem: any weak* open set $V$ containing the 0 functional has to contain a line, i.e. a one-dimensional space of functionals. (In fact, it contains many such lines. The idea is to look at the basic open sets of the weak* topology.)

Take some $f\neq 0$ so that the line generated by $f$ is contained in $V.$ Then pick $u \in U$ so that $u \not\in\ker f.$ This is possible since in the norm topology, an open ball at 0 contains vectors in every direction.

We must have that $\lambda f(u) \in (-1, 1)$ for every possible $\lambda,$ since $\lambda f \in V$ and $u\in U$ for every choice of $\lambda.$ But by scaling, since $f(u)\neq 0,$ this is clearly impossible!