How do I show $L \not\in \widehat{(l^1)} $?

Question

Let $l^\infty$ be the space of all real bounded sequences equipped with supremum norm. Let $S$ be the shift operator defined on $l^\infty$ by $(Sx)_n=x_{n+1}$, $n \in \mathbb{N}$ for all $x \in l^{\infty}$.

I proved that by using Hahn Banach theorem that there exists $L \in (l^\infty)^*$ such that

i) $\liminf_{n \rightarrow \infty} x_n \leq L(x) \leq \limsup_{n \rightarrow \infty} x_n$

ii) $L(Sx)=L(x)$

I also proved that $l^\infty \cong (l^1)^*$, how do I show $ L \not\in \widehat{(l^1)} $?

Can anyone give some hint for the last part?

Answer 1

As I said in the comments, you need to show that no $(y_n)_n \in \ell^1$ exists so that $$L(x_n) = \sum_{n=1}^\infty x_n y_n$$ holds for all $(x_n)_n \in \ell^\infty$. Suppose this were the case, and consider $e^m = (e^m_n)_n \in \ell^1$, defined by $e^m_n = \delta_{mn}$ (i.e. the standard basis). Note that $e^m_n \to \infty$ as $n \to \infty$, so $L(e^m) = 0$. But then, $$0 = L(e^m) = \sum_{n=1}^\infty e^m_n y_n = \sum_{n=1}^\infty y_n\delta_{mn} = y_m,$$ hence $(y_n)_n$ must be the $0$ sequence. But, $L$ doesn't send every sequence to $0$ (e.g. the constantly $1$ sequence), so no such $(y_n)_n \in \ell^1$ exists.

Answer 2

If $ L \in \widehat{(l^1)} $ then there exists $(a_n) \in l^{1}$ such that $\lim \inf x_n \leq L(x)=\sum a_nx_n$ for all $x=(x_n) \in l^{\infty}$. Taking $x=e_k$ we see that $1 \leq a_k$ for all $k$. But then $(a_n) \notin l^{1}$.