# 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). I don't understand the correspondence between $$l^1$$ and $$l^\infty$$ they mentioned. Can some one explain more about this.

Thanks

• If $y=\{ y_{j}\}_{j=1}^{\infty} \in l^{\infty}$, then there is a correspondence $y \mapsto L_{y}\in (l^{1})^{\star}$ given by $L_{y}(x)=\sum_{j=1}^{\infty}x_{j}y_{j}$ for all $x \in l^{1}$. You can show that $\|L_{y}\|_{(l^{1})^{\star}}=\|y\|_{l^{\infty}}$. So the correspondence $y\mapsto L_{y}$ is isometric. And this correspondence is a surjective linear map. – DisintegratingByParts Mar 4 '14 at 19:04

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 $$x=\{x_n\}\in\ell^1$$.
• I think $\ell^1\subset \ell^\infty$. Since $(1,1,\cdots)\in \ell^\infty$ but $(1,1,\cdots)\not\in \ell^1$ – user62498 Sep 12 '14 at 14:50
• @user62498. Of course. $l^{\infty}$ contains all bounded sequences. Absolutely summable sequences (members of $l^1$) are necessarily bounded sequences. – DanielWainfleet Jan 3 at 9:28