2
$\begingroup$

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 \begin{equation} \overline{A^*\cap \mathbb{S}^*} = \mathbb{S}^* \label{1}\tag{1} \end{equation} 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?

Thank you for your solutions.

Improve this question
$\endgroup$
1
  • 6
    $\begingroup$ 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? $\endgroup$
    – Willie Wong
    Commented Aug 1, 2022 at 17:00

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.