If $\|f_n\| \leq c$ and $f_n \rightarrow f$ in weak-star topology, then $\|f\| \leq c$?

Question

Let $E=(E,\|\cdot \|_E)$ be a Banach space over $\mathbb{C}$. Consider $E'$ the dual of $E$, with norm $\|\cdot\|_{E'}$. Let $(f_n)_{n \in \mathbb{N}} \subset E'$ such that $f_n \rightarrow f$ in the weak-star topology of the $E'$, for some $f \in E'$.

Question. If there exists $c>0$ such that $$\|f_n\|_{E'} \leq c, \; \forall \; n \in \mathbb{N} \tag{1}$$ then $$\|f\|_{E'} \leq c? \tag{2}$$

I think if $f_n \rightarrow f$ in the strongly topology of the $E'$, then $(2)$ holds. But I wonder if the same is true in the weak-star topology. Or $(2)$ also holds, since $\|\cdot\|_{E'}$ is defined via $\sup$ and $f_n \rightarrow f$ in the weak-star topology implies $$f_n(x) \rightarrow f(x), \; \forall \; x \in E, $$ (strongly) in $\mathbb{C}$?

Answer 1

You do not need heavy machinery like Banach-Alaoglu.

Let $x \in E$ be arbitrary. From (1) you get $$|f_n(x)| \le c \, \|x\|_E.$$ Due to weak-star convergence, you can pass to the limit and get $$|f(x)| \le c \, \|x\|_E \qquad\forall x \in E.$$ By definition of the dual norm $$\| f \|_{E'} = \sup\{ |f(x)| \;\mid\; \|x\|_E \le 1\} \le c.$$

Answer 2

Yep! The closed ball $B_{E'}[0; c]$ is weak$^*$ compact (Banach-Alaoglu), and hence weak$^*$ closed, and thus $E' \setminus B[0; c]$ is weak$^*$ open. If $f \in E' \setminus B[0; c]$, then by definition of sequence convergence, the sequence $f_n$ must eventually lie in $E' \setminus B[0; c]$, i.e. $\|f_n\| > c$ for sufficiently large $n$, contradicting our assumption.