I'm trying to prove this result. Could you have a check on my proof?
Let $(E, | \cdot|)$ be a normed linear space and $E^\star$ its topological dual. Let $\sigma(E^\star, E)$ be the weak$^\star$ topology on $E^\star$. Let $f\in E^\star$ and $(f_n)$ be a sequence in $E^\star$ such that $f_n \overset{\star}{\rightharpoonup} f$ in $\sigma(E^\star, E)$. Then $\|f\| \le \liminf \|f_n\|$.
My attempt: Let $B := \{x \in X \mid |x|=1\}$ be the unit sphere. We have $\lim_n \langle f_n, x \rangle = \langle f, x \rangle$ for all $x\in X$. So $$\sup_{x\in B} \lim_n \langle f_n, x \rangle = \sup_{x\in B} \langle f, x \rangle = \|f\|.$$
Hence it suffices to show that $$\sup_{x\in B} \lim_n \langle f_n, x \rangle \le \liminf_n \|f_n\|.$$
In fact, we have $\langle f_n, x \rangle \le \sup_{x\in B} \langle f_n, x \rangle = \|f_n\|$ and thus $$\lim_n \langle f_n, x \rangle = \liminf_n \langle f_n, x \rangle \le \liminf_n \|f_n\|.$$ The claim then follows by taking the supremum on both sides of above inequality.