For any Banach space $F,$ let $B_F$ be the closed unit ball of $F,$ that is, $B_F = \{x\in F: \|x\| \leq 1\}.$ Also, let $ext B_F$ be the set of extreme points of $B_F$ (Recall that $x$ is an extreme point of $B_F$ if $x = \frac{1}{2}(x_1+x_2)$ for some $x_1,x_2\in B_F$ implies that $x= x_1=x_2.)$
Let $F^*$ be the continuous dual space of $F.$
Question: Let $F$ be a Banach space and $x\in F\setminus\{0\}.$ Is it true that there exists $x^*\in ext B_{F^*}$ such that $x^*(x) = \|x\|?$
I think it is true. Assume that $\|x\| = 1.$ Consider $$S = \{x^*\in B_{F^*}: x^*(x) = 1\}.$$ By Hahn-Banach Theorem, $S$ is nonempty. Clearly $S$ is convex.
I would like to show that $S$ is weak-star closed in $B_{F^*}$ so that it is weak-star compact (Banach-Alaoglu states that $B_{F^*}$ is weak-star compact).
Then by Krein-Milman Theorem, $S$ has an extreme point, say $z^*.$ It is easy to see that $z^*$ is also an extreme point of $B_{F^*}.$ Such $z^*$ is our desired bounded linear functional.
However, I am not sure how to show that $S$ is weak-star closed. Any hint is appreciated.
For a fixed $x\in F$, the map $$ \phi_x : F^*\to\mathbb K,\; x^*\mapsto x^*(x) $$ is continuous from the weak-star topology of $F^*$ to the ordinary topology of $\mathbb K$, where $\mathbb K\in\{\mathbb R,\mathbb C\}$. This is because the weak-star topology of $F^*$ is the coarsest topology such that all functionals $\phi_y:F^*\to\mathbb K$ for $y\in F$ are continuous (this is sometimes even used as the definition of the weak-star topology).
Therefore, the preimage $\phi_x^{-1}(\{1\})$ of the closed set $\{1\}$ is weak-star closed. It follows that the set $S= B_{F^*}\cap \phi_x^{-1}(\{1\})$ is weak-star closed.