Let $S \subset l^{\infty}$ be the set of complex convergent sequences. Define $L:S \to \mathbb{C}$ as $L(x) = \lim\limits_{n\to\infty} x\lbrack n\rbrack$. I have shown that $S$ is a closed subspace of $l^\infty$ and that $L \in S^{'}$ (dual of $S$). But I'm struggling with showing the following:
Problem
Show that $\nexists y \in l^1$ such that $\forall x \in S: L(x) = \sum\limits_{n\ge 1}(x y)\lbrack n \rbrack$.
Attempt
I assume that such $y$ exists and try to find a contradiction. Let $x_c := (c, c, ....)$. Then $L(x_c) = c = c \sum y[n] \implies \sum y[n] = 1$. From here, I tried to find a sequence $x$ that shows that $\sum y[n] > 1$, hence contradiction and $y$ does not exist. But I do not manage that.
Any tips?
Take $x=(0,0,...,0,1,0,0...)$ which has $1$ in the $k-$th place and $0$ elsewhere. We get $0=Lx=\sum x[n]y[n]=y[k]$. Thus $y[k]=0$ and this is true for all $k$. But then $Lx=0$ for all $x$ which is contradiction since $L(1,1,1...) \neq 0$.