is operatory norm weak* - lower semicontinuous?

Question

Let $X$ be a normed space with norm $\|.\|$. We already know that the norm $\| . \|: X \to R$ is weakly lower semicontinuous in the sense, if $x_n \overset{w}{\longrightarrow} x_0$ as $n \to \infty$, then $\| x_0 \| \leq \liminf_{n \to \infty} \| x_n \|.$

Now my question: Is this property holds in the dual of $X$ i.e $X^*$ with usual operatory norm and weak* topology ?

If not, does it hold when we additionally assume $X$ is reflexive?

Answer 1

True. Let $x_n^{*} \to x^{*}$ in weak* topology and $\|x\| \leq 1$. Then $|x_n^{*}(x)| \leq \|x_n^{*}\|$ for each $n$. Take liminf on both sides. You get $|x^{*}(x)|\leq \lim \inf \|x_n^{*}\|$. Take sup over all $x$ with $\|x\| \leq 1$ we get $\|x^{*}\| \leq\lim \inf \|x_n^{*}\|$.