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

Question

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 $\|f_n\| \to \|f\|$ ?

Answer 1

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 $\|f_n\| \xrightarrow{n\to\infty} \|f\|$.

Answer 2

For any $\epsilon>0$ there is $x_0\in X$, with $\|x_0\|=1$ and $$\|f\|-\epsilon\leq |f(x_0)|$$

Then $\|f\|-\epsilon\leq |f(x_0)|\leq |f(x_0)-f_n(x_0)|+|f_n(x_0)|$.

There is $N$ such that for $n>N$ we have $|f(x_0)-f_n(x_0)|<\epsilon$.

Therefore, $$\|f\|-2\epsilon\leq |f_n(x_0)|\leq \|f_n\|\leq 1=\|f\|$$

or $$0\leq \|f\|-\|f_n\|\leq 2\epsilon$$

So, for every $\epsilon>0$ there is $N$ such that for all $n>N$

$$0\leq |\|f\|-\|f_n\||\leq 2\epsilon$$