# Application of Bishop-Phelps Theorem

Consider a real Banach space $$X$$ and its continuous dual $$X^*$$. The Bishop-Phelps Theorem states that the set $$A^*=\{x^*\in X^* \mid x^* \text{ attains its supremum on } \text{ the unit ball } \mathbb{B} \}$$ is norm-dense in $$X^*$$. Clearly, $$\overline{A^*\cap \mathbb{S}^*} \subset \mathbb{S}^*$$, with $$\mathbb{S}^*$$ be the dual unit sphere in $$X^*$$. Here $$\overline{A}$$ denotes the norm-closure of $$A$$. When $$X$$ is reflexive, $$A^*=X^*$$ (by James's Theorem) and the equality $$$$\overline{A^*\cap \mathbb{S}^*} = \mathbb{S}^* \label{1}\tag{1}$$$$ holds.

1. Does the equality \eqref{1} holds in arbitrary non-reflexive spaces $$X$$?
2. If the answer is 1. is false, in which non-reflexive space $$X$$ that the equality \eqref{1} holds?

• I'm confused: let $f\in \mathbb{S}^*$ and by Bishop-Phelps you have $f_k\in A^*$ such that $f_k \to f$ in $X^*$. WLOG $f_k \neq 0$. Then isn't it the case that $f_k / \|f_k\|_*$ (here $\|\cdot\|_*$ denotes the dual norm) belongs to $A^*\cap \mathbb{S}^*$ and converges to $f$ in $X^*$? (Here we use that $A^* = \lambda A^*$.) Am I missing something in this sketch? Commented Aug 1, 2022 at 17:00