Dual of a subspace












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In question Can a subspace have a larger dual?, they discussed the following type of situtation:



Suppose that $Ysubset X$ is a proper embedding of Banach spaces (i.e. the inclusion of a proper closed subspace). By Hahn-Banach, we may view $Y^*subset X^*$, but this seems to me to be only at the level of topology. What I mean by that is that this doesnt seem to be an an embedding of Banach spaces anymore. For example, it may happen that the extensions you choose for $x,yin Y^*$ may have the property that their sum is not the extension you chose for $x+y$. Can one realize $Y^*subset X^*$ as a linear subspace? Or better still, as a closed linear subspace?



There are examples of this happening that I can think of; for example, if you have a surjection $phi:Xto Y$, the pullback map gives an embedding $ell^1(Y)subset ell^1(X)$ and we do have that $(ell^1(Y))^*subset (ell^1(X))^*$ (as this is just the pullback embedding of $ell^infty(Y)subset ell^infty(X)$). I dont see how to make this work at the level of choosing extensions using Hahn-Banach in a systematic way.










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  • In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.)
    – GEdgar
    Dec 30 '18 at 13:51


















3














In question Can a subspace have a larger dual?, they discussed the following type of situtation:



Suppose that $Ysubset X$ is a proper embedding of Banach spaces (i.e. the inclusion of a proper closed subspace). By Hahn-Banach, we may view $Y^*subset X^*$, but this seems to me to be only at the level of topology. What I mean by that is that this doesnt seem to be an an embedding of Banach spaces anymore. For example, it may happen that the extensions you choose for $x,yin Y^*$ may have the property that their sum is not the extension you chose for $x+y$. Can one realize $Y^*subset X^*$ as a linear subspace? Or better still, as a closed linear subspace?



There are examples of this happening that I can think of; for example, if you have a surjection $phi:Xto Y$, the pullback map gives an embedding $ell^1(Y)subset ell^1(X)$ and we do have that $(ell^1(Y))^*subset (ell^1(X))^*$ (as this is just the pullback embedding of $ell^infty(Y)subset ell^infty(X)$). I dont see how to make this work at the level of choosing extensions using Hahn-Banach in a systematic way.










share|cite|improve this question






















  • In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.)
    – GEdgar
    Dec 30 '18 at 13:51
















3












3








3







In question Can a subspace have a larger dual?, they discussed the following type of situtation:



Suppose that $Ysubset X$ is a proper embedding of Banach spaces (i.e. the inclusion of a proper closed subspace). By Hahn-Banach, we may view $Y^*subset X^*$, but this seems to me to be only at the level of topology. What I mean by that is that this doesnt seem to be an an embedding of Banach spaces anymore. For example, it may happen that the extensions you choose for $x,yin Y^*$ may have the property that their sum is not the extension you chose for $x+y$. Can one realize $Y^*subset X^*$ as a linear subspace? Or better still, as a closed linear subspace?



There are examples of this happening that I can think of; for example, if you have a surjection $phi:Xto Y$, the pullback map gives an embedding $ell^1(Y)subset ell^1(X)$ and we do have that $(ell^1(Y))^*subset (ell^1(X))^*$ (as this is just the pullback embedding of $ell^infty(Y)subset ell^infty(X)$). I dont see how to make this work at the level of choosing extensions using Hahn-Banach in a systematic way.










share|cite|improve this question













In question Can a subspace have a larger dual?, they discussed the following type of situtation:



Suppose that $Ysubset X$ is a proper embedding of Banach spaces (i.e. the inclusion of a proper closed subspace). By Hahn-Banach, we may view $Y^*subset X^*$, but this seems to me to be only at the level of topology. What I mean by that is that this doesnt seem to be an an embedding of Banach spaces anymore. For example, it may happen that the extensions you choose for $x,yin Y^*$ may have the property that their sum is not the extension you chose for $x+y$. Can one realize $Y^*subset X^*$ as a linear subspace? Or better still, as a closed linear subspace?



There are examples of this happening that I can think of; for example, if you have a surjection $phi:Xto Y$, the pullback map gives an embedding $ell^1(Y)subset ell^1(X)$ and we do have that $(ell^1(Y))^*subset (ell^1(X))^*$ (as this is just the pullback embedding of $ell^infty(Y)subset ell^infty(X)$). I dont see how to make this work at the level of choosing extensions using Hahn-Banach in a systematic way.







real-analysis banach-spaces lp-spaces dual-spaces






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asked Dec 30 '18 at 9:17









Kyle AustinKyle Austin

837




837












  • In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.)
    – GEdgar
    Dec 30 '18 at 13:51




















  • In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.)
    – GEdgar
    Dec 30 '18 at 13:51


















In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.)
– GEdgar
Dec 30 '18 at 13:51






In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.)
– GEdgar
Dec 30 '18 at 13:51












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It is not quite correct to say that Hahn–Banach implies that $Y^*subset X^*$. It only says that $${f|_Ycolon fin X^*}=Y^*.$$



As for concrete examples where this may fail, let us take $X=C[0,1]$. By the Banach–Mazur theorem, $X$ contains an isometric copy of every separable Banach space so let $Y$ be a subspace of $X$ isometric to $ell_1$. The dual space of $X^*$ is weakly sequentially complete and this property passes to closed subspaces. On the other hand $Y^*$ is isometric to $ell_infty$, so it is not weakly sequentially complete.






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    1 Answer
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    1 Answer
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    active

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    1














    It is not quite correct to say that Hahn–Banach implies that $Y^*subset X^*$. It only says that $${f|_Ycolon fin X^*}=Y^*.$$



    As for concrete examples where this may fail, let us take $X=C[0,1]$. By the Banach–Mazur theorem, $X$ contains an isometric copy of every separable Banach space so let $Y$ be a subspace of $X$ isometric to $ell_1$. The dual space of $X^*$ is weakly sequentially complete and this property passes to closed subspaces. On the other hand $Y^*$ is isometric to $ell_infty$, so it is not weakly sequentially complete.






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      1














      It is not quite correct to say that Hahn–Banach implies that $Y^*subset X^*$. It only says that $${f|_Ycolon fin X^*}=Y^*.$$



      As for concrete examples where this may fail, let us take $X=C[0,1]$. By the Banach–Mazur theorem, $X$ contains an isometric copy of every separable Banach space so let $Y$ be a subspace of $X$ isometric to $ell_1$. The dual space of $X^*$ is weakly sequentially complete and this property passes to closed subspaces. On the other hand $Y^*$ is isometric to $ell_infty$, so it is not weakly sequentially complete.






      share|cite|improve this answer
























        1












        1








        1






        It is not quite correct to say that Hahn–Banach implies that $Y^*subset X^*$. It only says that $${f|_Ycolon fin X^*}=Y^*.$$



        As for concrete examples where this may fail, let us take $X=C[0,1]$. By the Banach–Mazur theorem, $X$ contains an isometric copy of every separable Banach space so let $Y$ be a subspace of $X$ isometric to $ell_1$. The dual space of $X^*$ is weakly sequentially complete and this property passes to closed subspaces. On the other hand $Y^*$ is isometric to $ell_infty$, so it is not weakly sequentially complete.






        share|cite|improve this answer












        It is not quite correct to say that Hahn–Banach implies that $Y^*subset X^*$. It only says that $${f|_Ycolon fin X^*}=Y^*.$$



        As for concrete examples where this may fail, let us take $X=C[0,1]$. By the Banach–Mazur theorem, $X$ contains an isometric copy of every separable Banach space so let $Y$ be a subspace of $X$ isometric to $ell_1$. The dual space of $X^*$ is weakly sequentially complete and this property passes to closed subspaces. On the other hand $Y^*$ is isometric to $ell_infty$, so it is not weakly sequentially complete.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 19:48









        Tomek KaniaTomek Kania

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        12.1k11943






























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