Question about dual of a Banach space

Question

Let $X $ be a Banach space and $X^*$ its dual space. Let $\{f_n\}_{n\in \mathbb N} $ be a sequence in $X^*$ and let $f \in X^*$ such that $f_n (x) \to f (x) $ for all $x\in X $ (as $n $ goes to infinity). I must show that $||f_n|| \le M, \ \forall n\in \mathbb N,$ with $M $ real, and that $||f||\le $ lim inf $||f_n||$, as $n\to \infty $. Now, the first point is direct consequence of Banach-Steinhaus theorem, however I don't have ideas for the second point; to be honest I would say that necessarily $||f||=$ lim inf $||f_n||$ for $n\to \infty$. I would like a clarify, or even a example of a situation where the equality doesn't hold; thank you.

Answer 1

To show that $||f||\leq \liminf_n||f_n||$, observe that if $||x||=1$ then $$ |f(x)|=\lim_{n\to\infty}|f_n(x)|\leq \liminf_{n\to\infty}||f_n||$$ since $|f_n(x)|\leq ||f_n||$ for each $n$, and then taking the supremum over all $x$ with $||x||=1$ we get $$||f||\leq \liminf_{n\to\infty}||f_n||$$

Now consider the case when $X=\ell^2(\mathbb{N})$ and $f_n(x)=\langle x, e_n\rangle$, where $e_n$ has a $1$ in the $n$th position and $0$s elsewhere. Then we have $f_n(x)\to 0$ for all $x$, but $||f_n||=1$ for each $n$, and so $$ ||f||=0<1=\liminf_{n\to\infty}||f_n||$$