Dual space of $l^1$ I m taking a course in functional analysis. The book state that the dual space of $l^1$, the set of real valued absolutely summable sequence, is $l^\infty$. Can anyone explain why the dual space of $l^1$ is $l^\infty$. I read a proof online http://math.uga.edu/~clayton/courses/608/608_5.pdf (Wayback Machine, new link). I don't understand the correspondence between $l^1$ and $l^\infty$ they mentioned. Can some one explain more about this.
Thanks
 A: Hint.
Step I. $\ell^\infty \subset (\ell^1)^*$. This is clear as every bounded sequence (i.e., member of $\ell^\infty$) defines a bounded linear functional on $\ell^1$.
Step II. If $\varphi\in(\ell^1)^*$, and $e_n=(0,0,\ldots,1,\ldots)\in\ell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set
$$
u_n=\varphi(e_n).
$$
Then
$$
\lvert u_n\rvert=\lvert\varphi(e_n)\rvert \le \|\varphi\|_{(\ell^1)^*} \|e_n\|_{\ell^1}=\|\varphi\|_{(\ell^1)^*},
$$
and hence $\{u_n\}$ is a bounded sequence, i.e., $\{u_n\}\in\ell^\infty$.
Step III. It remains to show that $\varphi(x)=\sum_{n=1}^\infty u_nx_n$, for all $x=\{x_n\}\in\ell^1$.
Setting $x^n=(x_1,x_2,\ldots,x_n,0,0,\ldots,0,\ldots)$, we have $\|x^n-x\|\to 0$ and
$$
\varphi(x^n)=\sum_{k=1}^nx_k\varphi(e_k)=\sum_{k=1}
^nx_ku_k.
$$
Hence
$$
\Big|\varphi(x)-\sum_{n=1}^\infty u_nx_n\Big|\le 
|\varphi(x)-\varphi(x_n)|+
\Big|\varphi(x^n)-\sum_{n=1}^\infty u_nx_n\Big| \\ 
\le \|\varphi\|_{(\ell^1)^*}\|x^n-x\|_{\ell^1}+\Big|\sum_{k=n+1}^\infty u_kx_k\Big|\to 0.
$$
