Prove that Helly Theorem is not true in $L^{\infty}[0,1]$ Prove that Helly Theorem is not true in $X=L^{\infty}[0,1]$
Helly's Theorem: Let $X$ be a separable normed linear space and $\{T_n \}$ a sequence in its dual space $X^*$ that is bounded, that is, there is an $M > 0$ for which $|T_n(f)|\leq M \cdot ||f||$ for all $f$ in $X$ and all $n$. Then there is a subsequence $\{T_{n_k}\}$ of $\{T_n\}$ and $T$ in $X^*$ for which $\lim\limits_{k \to \infty} T_{n_k} =T(f)$ for all $f$ in $X$
My thoughts: Since one of the condition of the theorem is $X$ is separable normed linear space, I thought that will be enough to prove that $L^{\infty} [0,1]$ is not separable (and we can do this by contradiction) but I think it is not enough and maybe a counterexample (that I can't see) will solve this problem easily, any clues or solutions? Thanks 
 A: A little bit late but might be helpful for future visitors:
Define $$ f_n(x)=\begin{cases} 2^n & x\in \left[\frac{1}{2^n},\frac{1}{2^{n-1}}\right] \\ 0 & \text{else} \end{cases} $$ Then $f_n \in L^1[0,1]$ so it induces a linear functional on $L^\infty[0,1]$ by $T_n(g)=\int_{[0,1]}f_n\, g $. 
Assume $T_n$ has a weak-* convergent subsequence, $ T_{n_k} \rightharpoonup T $. Define $ f(x) $ by $$ f(x)= \begin{cases} (-1)^k & x\in \left[\frac{1}{2^{n_k}},\frac{1}{2^{n_k-1}} \right) \\ 0 & \text{else} \end{cases} $$ Clearly, $ f\in L^\infty[0,1] $ but $T_{n_k}(f)=(-1)^k $ which clearly doesn't converge. 
A: For the space of bounded sequences  $\ell^\infty$ this is easy: The projections
$\pi_n((x_k)_{k\in\mathbb N})=x_n$ do not have a $\sigma((\ell^\infty)^*,\ell^\infty))$ convergent subsequence because this would be a single subsequence $(n_j)_j$ such that $\pi_{n_j}(x)=x_{n_j}$ converges for all bounded sequences.
For $L^\infty$ you can find an isometric embedding $\ell^\infty\to L^\infty$ 
and extend the $\pi_n$ by Hahn-Banach to continuous linear functionals on $L^\infty$.
