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
Best Answer
Hint.
Step I. $\ell^1 \subset \ell^\infty$. 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 $x=\{x_n\}\in\ell^1$.