Dual space of $l^1$












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



Thanks






functional-analysis







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  • 2




    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 $ymapsto L_{y}$ is isometric. And this correspondence is a surjective linear map.
    – DisintegratingByParts
    Mar 4 '14 at 19:04


















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



Thanks






functional-analysis







share|cite|improve this question




















  • 2




    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 $ymapsto L_{y}$ is isometric. And this correspondence is a surjective linear map.
    – DisintegratingByParts
    Mar 4 '14 at 19:04
















1












1








1


3





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



Thanks






functional-analysis







share|cite|improve this question















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



Thanks






functional-analysis




functional-analysis






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share|cite|improve this question













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edited Apr 11 '15 at 16:07









Pete L. Clark

80.1k9161311




80.1k9161311










asked Mar 2 '14 at 22:26









nerdnerd

708411




708411








  • 2




    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 $ymapsto L_{y}$ is isometric. And this correspondence is a surjective linear map.
    – DisintegratingByParts
    Mar 4 '14 at 19:04
















  • 2




    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 $ymapsto L_{y}$ is isometric. And this correspondence is a surjective linear map.
    – DisintegratingByParts
    Mar 4 '14 at 19:04










2




2




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 $ymapsto L_{y}$ is isometric. And this correspondence is a surjective linear map.
– DisintegratingByParts
Mar 4 '14 at 19:04






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 $ymapsto L_{y}$ is isometric. And this correspondence is a surjective linear map.
– DisintegratingByParts
Mar 4 '14 at 19:04












1 Answer
1






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oldest

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0














Hint.



Step I. $ell^inftysubset ell^1$. This is clear as every bounded sequence (i.e., member of $ell^infty$) defines a bounded linear functional on $ell^1$.



Step II. If $varphiin(ell^1)^*$, and $e_n=(0,0,ldots,1,ldots)inell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set
$$
u_n=varphi(e_n).
$$
Then
$$
lvert u_nrvert=lvertvarphi(e_n)rvert le |varphi|_{(ell^1)^*} |e_n|_{ell^1}=|varphi|_{(ell^1)^*},
$$
and hence ${u_n}$ is a bounded sequence, i.e., ${u_n}inell^infty$.



Step III. It remains to show that $varphi(x)=sum_{n=1}^infty u_nx_n$, for all $x={x_n}inell^1$.






share|cite|improve this answer

















  • 2




    I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
    – 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











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1 Answer
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1 Answer
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active

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0














Hint.



Step I. $ell^inftysubset ell^1$. This is clear as every bounded sequence (i.e., member of $ell^infty$) defines a bounded linear functional on $ell^1$.



Step II. If $varphiin(ell^1)^*$, and $e_n=(0,0,ldots,1,ldots)inell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set
$$
u_n=varphi(e_n).
$$
Then
$$
lvert u_nrvert=lvertvarphi(e_n)rvert le |varphi|_{(ell^1)^*} |e_n|_{ell^1}=|varphi|_{(ell^1)^*},
$$
and hence ${u_n}$ is a bounded sequence, i.e., ${u_n}inell^infty$.



Step III. It remains to show that $varphi(x)=sum_{n=1}^infty u_nx_n$, for all $x={x_n}inell^1$.






share|cite|improve this answer

















  • 2




    I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
    – 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
















0














Hint.



Step I. $ell^inftysubset ell^1$. This is clear as every bounded sequence (i.e., member of $ell^infty$) defines a bounded linear functional on $ell^1$.



Step II. If $varphiin(ell^1)^*$, and $e_n=(0,0,ldots,1,ldots)inell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set
$$
u_n=varphi(e_n).
$$
Then
$$
lvert u_nrvert=lvertvarphi(e_n)rvert le |varphi|_{(ell^1)^*} |e_n|_{ell^1}=|varphi|_{(ell^1)^*},
$$
and hence ${u_n}$ is a bounded sequence, i.e., ${u_n}inell^infty$.



Step III. It remains to show that $varphi(x)=sum_{n=1}^infty u_nx_n$, for all $x={x_n}inell^1$.






share|cite|improve this answer

















  • 2




    I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
    – 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














0












0








0






Hint.



Step I. $ell^inftysubset ell^1$. This is clear as every bounded sequence (i.e., member of $ell^infty$) defines a bounded linear functional on $ell^1$.



Step II. If $varphiin(ell^1)^*$, and $e_n=(0,0,ldots,1,ldots)inell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set
$$
u_n=varphi(e_n).
$$
Then
$$
lvert u_nrvert=lvertvarphi(e_n)rvert le |varphi|_{(ell^1)^*} |e_n|_{ell^1}=|varphi|_{(ell^1)^*},
$$
and hence ${u_n}$ is a bounded sequence, i.e., ${u_n}inell^infty$.



Step III. It remains to show that $varphi(x)=sum_{n=1}^infty u_nx_n$, for all $x={x_n}inell^1$.






share|cite|improve this answer












Hint.



Step I. $ell^inftysubset ell^1$. This is clear as every bounded sequence (i.e., member of $ell^infty$) defines a bounded linear functional on $ell^1$.



Step II. If $varphiin(ell^1)^*$, and $e_n=(0,0,ldots,1,ldots)inell^1$, the sequence with zeros everywhere except on the $n-$position where there is an $1$, set
$$
u_n=varphi(e_n).
$$
Then
$$
lvert u_nrvert=lvertvarphi(e_n)rvert le |varphi|_{(ell^1)^*} |e_n|_{ell^1}=|varphi|_{(ell^1)^*},
$$
and hence ${u_n}$ is a bounded sequence, i.e., ${u_n}inell^infty$.



Step III. It remains to show that $varphi(x)=sum_{n=1}^infty u_nx_n$, for all $x={x_n}inell^1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 2 '14 at 22:40









Yiorgos S. SmyrlisYiorgos S. Smyrlis

62.8k1383163




62.8k1383163








  • 2




    I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
    – 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














  • 2




    I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
    – 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








2




2




I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
– user62498
Sep 12 '14 at 14:50






I think $ell^1subset ell^infty$. Since $(1,1,cdots)in ell^infty$ but $(1,1,cdots)notin ell^1$
– 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




@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


















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