Let $(E, \lVert \cdot \rVert_{E}) $ be a normed vector space: I'd like to find an example where $B_{E'}$ is weak$^{\star}$ sequentially compact but not weak sequentially compact.
Theory: If $E$ is separable, then $B_{E'}$ is weak$^{\star}$ sequentially compact.
(Weak sequential compactness) If $E$ is reflexive, then $B_E$ is weak sequentially compact.
Summing these results together, I find $L^{\infty}((0,1))$ might be a good candidate: being the dual space of $L^1((0,1))$, which is separable, we have $B_{L^{\infty}}$ is weak$^{\star}$ sequentially compact, and since $L^{\infty}((0,1))$ is not reflexive, I have the feeling it is neither weak sequentially compact.
Question: How do I prove $B_{L^{\infty}}$ is not weak sequentially compact?
I have somme problems as in the course i attended we never saw a characterization of weak convergence in $L^{\infty}((0,1))$, so I'd like to understand if my example is correct, and also how to prove the weak sequential compactness of the ball in $L^{\infty}((0,1))$.
Any hint, help or reference would be much appreciate, thanks in advance.
Best Answer
If you just want any example, $\ell^1$ is probably easier than $L^\infty$ (because we have a nice description of its dual).
Denote by $e_n$ the sequence whose $n$-th term is $1$ and every other term is $0$. If $x\in c_0$, then $\langle e_n,x\rangle=x(n)\to 0$. Hence any weak limit of a subsequence would have to be $0$. On the other hand, $\langle e_n,1\rangle= 1$ for every $n$. Thus $(e_n)$ does not have any weakly convergent subsequence.
In $L^\infty(0,1)$ you can look at the functions $e_n\in L^\infty(0,1)$ given by $e_n(x)=e^{-2\pi i n x}$. By the Riemann-Lebesgue Lemma, $\int e_n(x) f(x)\,dx=\hat f(n)\to 0$ for every $f\in L^1(0,1)$. To see that $e_n$ does not have a weakly convergent subsequence we need Hahn-Banach (or at least I don't see an easier way).
Let $\phi\colon C([0,1])\to \mathbb{R},\,f\mapsto f(0)$. This is a bounded linear functional. By Hahn-Banach it can be extended to a bounded linear functional on $L^\infty(0,1)$. Since $\phi(e_n)=1$ for every $n$, the sequence $(e_n)$ cannot have a weakly convergent subsequence.