In Conway's functional analysis text, he claims that there is a linear functional $L:l^{\infty}\rightarrow \mathbb{R}$ such that if $x\in c$, $L(x)=\lim x_{n}$. Here $c$ will denote the space of sequences that converge to a point and $c_{0}$ the space of sequences that converge to $0$.
To prove this, he first defines a shift operator, so that if $x = (x_{1},x_{2},\dots)$ then \begin{align*} Sx &= (x_{2},x_{3}, \dots)\\ S^{2}x &= (x_{3},x_{4},\dots)\\ &\vdots\\ S^{n}x &= (x_{n+1},x_{n+2},\dots). \end{align*}
He observes that for $x\in c_{0}$,
$$S^{n+1}x - x = [S^{n+1}x - S^{n}x] + \dots + [Sx-x]\in M$$
where $M = \{Sx - x : x\in l^{\infty}\}$ and it is proven earlier that $L(M) =0$. This is similar to a telescoping sum and concludes that $L$ is a shift-invariant operator. Now, as $x\in c_{0}$, we fix an $\epsilon >0$ and $n$ such that $\lvert x_{m}\rvert < \epsilon$ for $m>n$ and write
$$\lvert L(x)\rvert = \lvert L(S^{n}x)\rvert \leq \lVert S^{n}x\rVert_{\infty} = \sup\{\lvert x_{m}\rvert: m> n\} < \epsilon.$$
Conway now concludes that since $x\in \operatorname{Ker}L$, $c_{0}\subseteq \operatorname{Ker}L$ and hence if $x\in c$, $L(x) = \lim x_{n}$ with no details whatsoever and it isn't clear to me.
Why does $c_{0}\subseteq \operatorname{Ker} L$ imply that if $x\in c$, $L(x) = \lim x_{n}$?
