text: # Strongly exposed points/Exposed points
  
  I was studying and I got the next doubt:
  
  We suppose that $(X,\|\cdot\|)$ is a Banach space and $C$ it is a convex closed subset of X.
  
  We say that $x\in C$ it is an exposed point of $C$ if $\exists x^*\in X^*$ such that $x$ it is the only one point of $C$ that $x^*(x)=\displaystyle\sup_C x^*$.
  
  We say that $x\in C$ it is an strongly exposed point of $C$ if $\exists x^*\in X^*$ such that for every sequence $(x_n)\subset C$ that verifies $x^*(x_n)\to \displaystyle\sup_C x^*$ then $\|x_n-x\|\to 0$.
  
  It's true that every strongly exposed point it is an exposed point. So I try to view if every exposed point it is an strongly exposed point (it would be false!).
  
  Suppose that $x\in C$ it is an exposed point, then there is $x^*\in X^*$ such that $x^*(x)=\displaystyle\sup_C x^*$. Let $(x_n)\in C$ such that $x^*(x_n)\to \displaystyle\sup_C x^*$. $x_n$ must converge ($x^*$ is continuous), so we suppose that $x_n\to y\in C$ ($C$ is closed). Then, by continuity $x^*(x_n)\to x^*(y)=\displaystyle\sup_C x^*$.
  
  So by the uniqueness, $x=y$ and then $\|x_n-x\|=\|x_n-y\|\to 0$. Then $x$ it is an strongly exposed point.
  
  The question is: what is wrong in my reasoning? There is a counterexample of Lindenstrauss for example (7.73 in the a book of Fabian: Functional Analysis and Infinite-Dimensional Geometry).
  
  Many thanks beforehand.
  
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  You probably mean example 8.73, at least if it is the same edition of the book - see here. – Martin Sleziak May 4 '12 at 12:57
  Ah, I refer to 7.73 of another book of Fabian: Banach Space (the basis for...). It's similar. There are counterexamples, so my reasoning is wrong, but I don't know where and it has me baffled. – Shrek May 4 '12 at 13:44
  Your problem is with the line "$x_n$ must converge". You have that $x^*(x_n) \to x^*(x)$ for one fixed $x^*$, not for all $x^*$. Even if you had $x^*(x_n) \to x^*(x)$ for all $x^* \in X^*$ (which you do not have), this would be weak convergence and not convergence in the norm topology. – Tom Cooney May 4 '12 at 15:02
  Amm, thanks. That was the fail. – Shrek May 4 '12 at 22:35
  
  You have to prove $x_n$ converges to $x$ in norm. You have assumed that $x_n$ converges to $y$. This assumption is wrong. First of all, from the hypothesis (i.e., $x^*(x_n)$ converges to $x^*(x)$) we could only conclude $x_n$ converges to $x$ w. r. to the weak topology on $X$, provided $x_n$ is a bounded (norm) sequence.
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