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.
real-analysis banach-spaces lp-spaces dual-spaces
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
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add a comment |
$begingroup$
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
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
$endgroup$
$begingroup$
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.)
$endgroup$
– GEdgar
Dec 30 '18 at 13:51
add a comment |
$begingroup$
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
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
$endgroup$
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
real-analysis banach-spaces lp-spaces dual-spaces
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
asked Dec 30 '18 at 9:17
Kyle AustinKyle Austin
857
857
$begingroup$
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.)
$endgroup$
– GEdgar
Dec 30 '18 at 13:51
add a comment |
$begingroup$
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.)
$endgroup$
– GEdgar
Dec 30 '18 at 13:51
$begingroup$
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.)
$endgroup$
– GEdgar
Dec 30 '18 at 13:51
$begingroup$
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.)
$endgroup$
– GEdgar
Dec 30 '18 at 13:51
add a comment |
1 Answer
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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.
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
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add a comment |
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$begingroup$
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.
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
$endgroup$
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.
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
answered Jan 4 at 19:48
Tomek KaniaTomek Kania
12.3k11945
12.3k11945
add a comment |
add a comment |
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$begingroup$
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.)
$endgroup$
– GEdgar
Dec 30 '18 at 13:51