I have a question concerning the identification\begin{equation}l_1(\mathbb{N})=c_0(\mathbb{N})' \end{equation}where $c_0(\mathbb{N}):=\{(x_j)_{j\in \mathbb{N}} : x_j \to 0, j \to \infty\}\subset l_\infty(\mathbb{N})$ and $E'$ denotes the dual space of some normed vector space $E$.
My problem is that I don't see why in this proof $c_0$ can't be replaced with $l_\infty$.
The proof which was presented to us was as follows: First consider the map \begin{equation}I:l_1\to c_0', y=(y_j)\mapsto I(y)=\left( x=(x_j)\mapsto \sum x_j y_j\right).\end{equation} This is well defined and injective. Now consider \begin{equation}I':c_0'\to l_1, f\mapsto (y_j),\end{equation} where $y_j:=f( (\delta_{ij})_i).$ Now for each $j$ we can choose some $\epsilon_j$ in the corresponding field ($\mathbb{R}$ or $\mathbb{C}$) with \begin{equation}\vert \epsilon_j\vert=1 \text{ and }\vert y_j\vert =\epsilon_jy_j.\end{equation} Now for all $N\in \mathbb{N}$ estimate \begin{equation}\sum_{j=1}^N \vert y_j\vert=\sum \epsilon_j y_j=f((\epsilon_1,...,\epsilon_N,0,0,...))\leq \Vert f \Vert_{op} \Vert (\epsilon_1,...,\epsilon_N,0,0,...)\Vert_\infty= \Vert f \Vert_{op} , \end{equation}which proves that $I'$ is well defined. Because $I'$ is injective, we are done.
So that was the proof. I'd be glad for any hints why this proof is wrong when replacing $c_0$ by $l_\infty.$ Thanks in advance.
This is the part that does not carry over to $\ell^\infty$. How do you know that $I'$ is injective? Because the linear span of basis vectors $e_j = (\delta_{ij})_i$ is dense in $c_0$, that is why. On $\ell_\infty$ this is no longer true. Consequently, one can have a linear functional that vanishes on all $e_j$ (so on all $c_0$) but is not zero on $\ell_\infty$. Such functionals are not presented explicitly, but they exist by the Hahn-Banach theorem.