TITLE: Dual space of $c_0$
QUESTION [6 upvotes]: I want to prove that the dual of $c_0=\{(x_n)_{n\in\mathbb N}\subset\mathbb R : \lim x_n=0 \text{  and }\|x\|_\infty=\sup\limits_n|x_n|\}$ is $l^1$ . 
So I defined the following map,
$$T:l^1\rightarrow(c_0)^*$$
$$ x_n\mapsto f_x$$
where $f_x(a)=\sum\limits_{n=0}^\infty x_na_n$.
In a next step I showed that tis map is well defined.
Now I want to show that T is an isometry and thats where I am not sure how to do it.
Could someone help me? Thanks

REPLY [5 votes]: The map $T$ is linear, hence we have to show that $\lVert f_x\rVert_{(c_0)^*}=\lVert x\rVert_1$.
Notice that 
$$|f_x(a)|\leqslant\sum_{j=1}^{+\infty}|a_j||x_j|\leqslant \sup_{j\geqslant 1}|a_j|\cdot \lVert x\rVert_1=\lVert a\rVert_\infty \cdot \lVert x\rVert_1,$$
hence $\lVert f_x\rVert\leqslant \lVert x\rVert_1$.
In order to show the converse inequality, consider for a fixed integer $N$ the sequence $a:=(\epsilon(1),\dots,\epsilon(N),0,\dots)$, an element of $c_0$, where $\epsilon(j)=1$ if $x_j\geqslant 0$ and $-1$ otherwise. The norm of $a$ is $1$ and we have $$f_x(a)=\sum_{j=1}^N|x_j|.$$