Suppose I have a Banach space $X$ with continuous dual. In that space, I have a sequence $(x_n)_{n = 1}^\infty$ in $X$ converging weakly to $y$, and a sequence $(\phi_n)_{n = 1}^\infty$ in $X^*$ satisfying $\phi_n \to \psi$ in the weak$^*$ topology. Suppose further that $\| x_n \| \leq 1, \| \phi_n \| \leq 1$ in their respective norms. Is it necessarily true that $\phi_n x_n \to \psi y$?
I have the inequality $$ |\psi y - \phi_n x_n| $$ that I want to make small, but whenever I try to expand it and apply a triangular inequality to several terms, but I keep ending up with a term like $|\phi_n (y - x_n)|$ or $|(\psi - \phi_n) x_n|$, a term which I can't make small by attending to only of the modes of convergence. Because both "parts" of the term are moving at once, and I don't have norm-convergence for either, I don't know how to make the term small because both the functionals and the vectors they act on are $n$-dependent.
So is this result true? Am I missing something? Or does this kind of thing just not work in general?
This result is false. Take $X=c_0$, $x_n$ the usual basis in $c_0$, and $\phi_n$ the usual basis in $X^*=\ell_1$. Then $x_n$ tends weakly to zero in $c_0$, and $\phi_n$ tends to zero in the weak-star topology in $\ell_1$, but $\phi_n(x_n)=1$ for all $n$.