Let $X$ be a Banach space with topological dual $X^*$. Then the dual product $$ (x,x^*)\in X\times X^*\to x^*(x)\in\mathbb{R} $$ is strongly$\times$strongly continuous on $X\times X^*$, mainly because every strongly convergent sequence is bounded. That does not hold if the topology considered on $X^*$ is the weak-star topology. So my question is:
Q. Is the dual product strongly$\times$weakly-star continuous on $X\times X^*$?
The problem does not originate from the weak-* topology, it also is present within the weak topology on Hilbert spaces.
On $\ell^2$, consider the set \begin{equation*} B := \{ \sqrt{n} \, e_n : n \in \mathbb N\}, \end{equation*} where $e_n$ are the canonical unit sequences.
Now, it is possible to show that $0$ belongs to the weak closure of $B$, but no sequence in $B$ convergences weakly to $0$.
By the characterization of the weak closure, there is a net $\{x_i\}_{i \in I}$ which converges weakly to $0$ and we can write \begin{equation*} x_i = \sqrt{n_i} \, e_{n_i}. \end{equation*} Now, we define the net \begin{equation*} y_i = \frac{1}{\sqrt{n_i}} \, e_{n_i} \end{equation*} and it is not hard to show that $y_i \to 0$ in $\ell^2$. However, \begin{equation*} ( x_i, y_i ) = 1 \not\to 0 = (0,0). \end{equation*}
However (just for the record -- I think this is already clear for you), this situation cannot appear if we merely consider sequences. That is, for sequences $\{x_k\}$, $\{x_k^*\}$ with $x_k \to x$ and $x_k^* \stackrel*\rightharpoonup x^*$ we always (here, the completeness of $X$ is crucial) have \begin{equation*} \langle x_k^* , x_k \rangle \to \langle x^*, x\rangle. \end{equation*} This relies on the boundedness of $\{x_k^*\}$.