In a Banach space, $x_n \to x$ and $f_n \to f$ weak* implies $f_n(x_n) \to f(x)$

Question

Let $X$ be a Banach space. Assume that $ x_n \to x$ and $f_n \in X^*$ such that $f_n \to f$ in the $w^*$ topology on $X^*$. Show that $ f_n(x_n) \to f(x)$.

I know that $f_n \to f$ in $w^*$ topology iff $f_n(x) \to f(x)$ for all $x \in X$. But I can't go further. Need some help.

Answer 1

I will elaborate a little bit on the hint by zhw.

You can use the uniform boundedness principle to infer the boundedness of $\{f_n\}$ in $X^*$ (here, it is important that $X$ is complete).

Then, you can show the desired convergence by using $0 = f_n(x) - f_n(x)$ and the triangle inequality in $\mathbb{R}$.