Q:

A sequence of functionals, of norm at most $1$, on a Banach space converging point-wise to a functional of norm $1$

Let $X$ be a Banach space, let $\{f_n\}$ be a sequence in $X^*$ and $f \in X^*$ such that $\{f_n\}$ converges to $f$ pointwise i.e. $f_n(x) \to f(x), \forall  x\in X $ and $\|f_n\| \le 1=\|f\|, \forall n\in \mathbb N$. 
Then is it true that $\|f_n\| \to \|f\|$  ? 

A:

The norm is weak-* lower semicontinuous, meaning if $f_n \xrightarrow{w^*} f$ then $\|f\| \le \liminf_{n\to\infty} \|f_n\|$.
Indeed, for any $x \in X$ we have $$|f(x)| = \lim_{n\to\infty} |f_n(x)| \le  \liminf_{n\to\infty} \|x\|\|f_n\| = \|x\| \liminf_{n\to\infty} \|f_n\| $$
so we conclude $\|f\| \le \liminf_{n\to\infty} \|f_n\|$.
Now we have
$$1 = \|f\| \le \liminf_{n\to\infty} \|f_n\| \le \limsup_{n\to\infty} \|f_n\| \le 1$$
so $\liminf_{n\to\infty} \|f_n\| = \limsup_{n\to\infty} \|f_n\| = 1$ which implies $\|f_n\| \xrightarrow{n\to\infty} \|f\|$. 

