Let $E$ be a Banach space.
Question: For any $\varepsilon>0, u\in E$ and $y^*\in\text{ext}\left( B_{E^*} \right)$ such that $|y^*(u) - \|u\|| < \varepsilon,$ does there exist $x^*\in E^*$ such that
$(1)$ $x^*\in\text{ext}\left( B_{E^*} \right),$
$(2)$ $x^*(u) = \|u\|,$
$(3)$ $\|y^*-x^*\|_{op}<\varepsilon?$
By defining the set $$\{x^*\in E^*: \|x^*\|\leq 1, x^*(u)=\|u\|\},$$ one can show existence of $x^*$ that satisfies the $(1)$ and $(2),$ but may not satisfy $(3)$ (a proof involves Hahn-Banach to ensure the set is nonempty, Banach-Alaoglu to ensures that the set is compact in weak$^*$ topology and lastly, Krein-Milman to obtain extreme point). However, if we define $$\{x^*\in E^*: \|x^*\|\leq 1, x^*(u)=\|u\|, \|y^*-x^*\|_{op}<\varepsilon\},$$ I am not able to prove that the set is nonempty.
Notations: Let $E$ be a Banach space. Denote $E^*$ to be its dual space, that is, a space containing bounded linear functionals on $E.$ Let $B_{E^*}$ be the closed unit ball of $E^*.$ We say that $x^*$ is an extreme point of $B_{E^*}$ if it cannot be expressed as midpoint of two elements from $B_{E^*}.$
For any $x^*\in E^*,$ its operator norm is defined by $$\|x^*\|_{op} = \sup_{\|x\|\leq 1}|x^*(x)|.$$