Does Bounded Closed Subset of Banach Space Attain Suprermum?

Question

Let $B$ be a Banach space and $A$ be a bounded closed convex subset of $B$. I wonder if every $x^*\in B^*$ attains its supremum in $A$. In detail, I wonder if there is a point $x\in X$ such that $$ x^*(x)=\sup_{y\in A} x^*(y). $$

By Uniform Bounded Principle, I know that $\sup_{y\in A}x^*(y)=\alpha < \infty$. Hence I can find a sequence $\{x_n\}$ with the following property: $$ \alpha-\frac{1}{n} < x^*(x_n) \leq \alpha. $$

Problem is that $x_n$ may not converge. Maybe I can regard $\{x_n\}$ as a sequence in $B/\textrm{Ker} x^*$, but even if it converges, I am not sure that the limit is in $A$.

By Alaogu Theorem, for each $x\in A$, I also have $x^*$ with $\|x^*\|=1$ and $$ \|x\| = \sup_{\|y^*\|=1}|y^*(x)|= |x^*(x)| $$but I am not sure this will help in any way.

Is there any counter example?

Thank you in advance.

Answer 1

Assume that $A$ is the closed unit ball. James showed that each Banach space such that each continuous linear functional is norm attaining has to be reflexive. Therefore, a non-reflexive Banach space is a counter-example.

Answer 2

I found a partial result, and this was what I exactly wanted. Bishop-Phelps Theorem states that if $A$ is a bounded closed convex subset of a real Banach space $X$, then a set of functionals attaining maximum in $A$ is dense in $X^*$.