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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

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    $\begingroup$ 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. $\endgroup$ – DisintegratingByParts Mar 4 '14 at 19:04
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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 $x=\{x_n\}\in\ell^1$.

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

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