Bidual Banach space
$begingroup$
I have a really hard time understanding what a bidual banach space is. Can someone give me an simple insight on this concept and an example maybe.
Thank you !
normed-spaces
asked Dec 16 '18 at 20:08
mimimimi
175
$endgroup$
add a comment |
$begingroup$
I have a really hard time understanding what a bidual banach space is. Can someone give me an simple insight on this concept and an example maybe.
Thank you !
normed-spaces
asked Dec 16 '18 at 20:08
mimimimi
175
$endgroup$
add a comment |
$begingroup$
I have a really hard time understanding what a bidual banach space is. Can someone give me an simple insight on this concept and an example maybe.
Thank you !
normed-spaces
asked Dec 16 '18 at 20:08
mimimimi
175
$endgroup$
I have a really hard time understanding what a bidual banach space is. Can someone give me an simple insight on this concept and an example maybe.
Thank you !
normed-spaces
normed-spaces
asked Dec 16 '18 at 20:08
mimimimi
175
asked Dec 16 '18 at 20:08
mimimimi
175
asked Dec 16 '18 at 20:08
mimimimi
175
asked Dec 16 '18 at 20:08
mimimimi
175
asked Dec 16 '18 at 20:08
mimimimi
175
175
add a comment |
add a comment |
1 Answer
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I shan't bore you with the definition, since you can look that up on wikipedia.
Before you understand the bidual, it's important that you understand the dual. The best examples to consider are the $ell_p$ spaces, i.e. the spaces of sequences $(a_n)_{n=1}^infty$ satisfying
$$|(a_n)_{n=1}^infty|_p:=left(sum_{n=1}^infty|a_n|^pright)^{1/p}<infty.$$
These are Banach spaces provided $1leq p<infty$. Define $p'$ as the conjugate to $p$, i.e. the number $1<p'leqinfty$ satisfying $frac{1}{p}+frac{1}{p'}=1$. The dual space to $ell_p$ is then $ell_{p'}$. (We haven't yet defined $ell_infty$ but you can look that up on wikipedia if you are curious.) Each functional $x^*inell_p^*$ can be identified with a sequence $(b_n)_{n=1}^inftyinell_{p'}$, under the action
$$x^*[(a_n)_{n=1}^infty]=sum_{n=1}^infty a_nb_n.$$
It turns out that if $p=2$ then $p'=2$ as well so that $ell_2^*=ell_2$. In other words, the dual of $ell_2$ is itself!
But $ell_2$ is a special case. It's the only separable Banach space which is its own dual. In fact, even in the nonseparable case, the only spaces which are their own duals are Hilbert spaces.
However, if you take the bidual, then $ell_p^{**}=ell_p$ for all $1<p<infty$. The class of Banach spaces for which this is true via a natural "canonical" identification are called reflexive spaces, although quasireflexive spaces are isomorphic to their biduals via noncanonical identifications. And the class of reflexive Banach spaces is very, very large.
Even when $X$ is a nonreflexive space, it still embeds isometrically into its bidual $X^{**}$. The most well-behaved nonreflexive spaces are $c_0$, $ell_1$, and $ell_infty$. In fact, we have $ell_1=c_0^*$, and $ell_infty=ell_1^*=c_0^{**}$. In particular, $ell_infty$ is the bidual of $c_0$, and $c_0$ is the subspace of $ell_infty$ spanned by the unit vectors.
So, I think the best way to envision biduals, aside from the obvious fact that they are duals of duals, are (possibly larger) spaces containing the original space.
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
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1 Answer
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1 Answer
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$begingroup$
I shan't bore you with the definition, since you can look that up on wikipedia.
Before you understand the bidual, it's important that you understand the dual. The best examples to consider are the $ell_p$ spaces, i.e. the spaces of sequences $(a_n)_{n=1}^infty$ satisfying
$$|(a_n)_{n=1}^infty|_p:=left(sum_{n=1}^infty|a_n|^pright)^{1/p}<infty.$$
These are Banach spaces provided $1leq p<infty$. Define $p'$ as the conjugate to $p$, i.e. the number $1<p'leqinfty$ satisfying $frac{1}{p}+frac{1}{p'}=1$. The dual space to $ell_p$ is then $ell_{p'}$. (We haven't yet defined $ell_infty$ but you can look that up on wikipedia if you are curious.) Each functional $x^*inell_p^*$ can be identified with a sequence $(b_n)_{n=1}^inftyinell_{p'}$, under the action
$$x^*[(a_n)_{n=1}^infty]=sum_{n=1}^infty a_nb_n.$$
It turns out that if $p=2$ then $p'=2$ as well so that $ell_2^*=ell_2$. In other words, the dual of $ell_2$ is itself!
But $ell_2$ is a special case. It's the only separable Banach space which is its own dual. In fact, even in the nonseparable case, the only spaces which are their own duals are Hilbert spaces.
However, if you take the bidual, then $ell_p^{**}=ell_p$ for all $1<p<infty$. The class of Banach spaces for which this is true via a natural "canonical" identification are called reflexive spaces, although quasireflexive spaces are isomorphic to their biduals via noncanonical identifications. And the class of reflexive Banach spaces is very, very large.
Even when $X$ is a nonreflexive space, it still embeds isometrically into its bidual $X^{**}$. The most well-behaved nonreflexive spaces are $c_0$, $ell_1$, and $ell_infty$. In fact, we have $ell_1=c_0^*$, and $ell_infty=ell_1^*=c_0^{**}$. In particular, $ell_infty$ is the bidual of $c_0$, and $c_0$ is the subspace of $ell_infty$ spanned by the unit vectors.
So, I think the best way to envision biduals, aside from the obvious fact that they are duals of duals, are (possibly larger) spaces containing the original space.
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
$endgroup$
add a comment |
$begingroup$
I shan't bore you with the definition, since you can look that up on wikipedia.
Before you understand the bidual, it's important that you understand the dual. The best examples to consider are the $ell_p$ spaces, i.e. the spaces of sequences $(a_n)_{n=1}^infty$ satisfying
$$|(a_n)_{n=1}^infty|_p:=left(sum_{n=1}^infty|a_n|^pright)^{1/p}<infty.$$
These are Banach spaces provided $1leq p<infty$. Define $p'$ as the conjugate to $p$, i.e. the number $1<p'leqinfty$ satisfying $frac{1}{p}+frac{1}{p'}=1$. The dual space to $ell_p$ is then $ell_{p'}$. (We haven't yet defined $ell_infty$ but you can look that up on wikipedia if you are curious.) Each functional $x^*inell_p^*$ can be identified with a sequence $(b_n)_{n=1}^inftyinell_{p'}$, under the action
$$x^*[(a_n)_{n=1}^infty]=sum_{n=1}^infty a_nb_n.$$
It turns out that if $p=2$ then $p'=2$ as well so that $ell_2^*=ell_2$. In other words, the dual of $ell_2$ is itself!
But $ell_2$ is a special case. It's the only separable Banach space which is its own dual. In fact, even in the nonseparable case, the only spaces which are their own duals are Hilbert spaces.
However, if you take the bidual, then $ell_p^{**}=ell_p$ for all $1<p<infty$. The class of Banach spaces for which this is true via a natural "canonical" identification are called reflexive spaces, although quasireflexive spaces are isomorphic to their biduals via noncanonical identifications. And the class of reflexive Banach spaces is very, very large.
Even when $X$ is a nonreflexive space, it still embeds isometrically into its bidual $X^{**}$. The most well-behaved nonreflexive spaces are $c_0$, $ell_1$, and $ell_infty$. In fact, we have $ell_1=c_0^*$, and $ell_infty=ell_1^*=c_0^{**}$. In particular, $ell_infty$ is the bidual of $c_0$, and $c_0$ is the subspace of $ell_infty$ spanned by the unit vectors.
So, I think the best way to envision biduals, aside from the obvious fact that they are duals of duals, are (possibly larger) spaces containing the original space.
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
$endgroup$
add a comment |
$begingroup$
I shan't bore you with the definition, since you can look that up on wikipedia.
Before you understand the bidual, it's important that you understand the dual. The best examples to consider are the $ell_p$ spaces, i.e. the spaces of sequences $(a_n)_{n=1}^infty$ satisfying
$$|(a_n)_{n=1}^infty|_p:=left(sum_{n=1}^infty|a_n|^pright)^{1/p}<infty.$$
These are Banach spaces provided $1leq p<infty$. Define $p'$ as the conjugate to $p$, i.e. the number $1<p'leqinfty$ satisfying $frac{1}{p}+frac{1}{p'}=1$. The dual space to $ell_p$ is then $ell_{p'}$. (We haven't yet defined $ell_infty$ but you can look that up on wikipedia if you are curious.) Each functional $x^*inell_p^*$ can be identified with a sequence $(b_n)_{n=1}^inftyinell_{p'}$, under the action
$$x^*[(a_n)_{n=1}^infty]=sum_{n=1}^infty a_nb_n.$$
It turns out that if $p=2$ then $p'=2$ as well so that $ell_2^*=ell_2$. In other words, the dual of $ell_2$ is itself!
But $ell_2$ is a special case. It's the only separable Banach space which is its own dual. In fact, even in the nonseparable case, the only spaces which are their own duals are Hilbert spaces.
However, if you take the bidual, then $ell_p^{**}=ell_p$ for all $1<p<infty$. The class of Banach spaces for which this is true via a natural "canonical" identification are called reflexive spaces, although quasireflexive spaces are isomorphic to their biduals via noncanonical identifications. And the class of reflexive Banach spaces is very, very large.
Even when $X$ is a nonreflexive space, it still embeds isometrically into its bidual $X^{**}$. The most well-behaved nonreflexive spaces are $c_0$, $ell_1$, and $ell_infty$. In fact, we have $ell_1=c_0^*$, and $ell_infty=ell_1^*=c_0^{**}$. In particular, $ell_infty$ is the bidual of $c_0$, and $c_0$ is the subspace of $ell_infty$ spanned by the unit vectors.
So, I think the best way to envision biduals, aside from the obvious fact that they are duals of duals, are (possibly larger) spaces containing the original space.
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
$endgroup$
I shan't bore you with the definition, since you can look that up on wikipedia.
Before you understand the bidual, it's important that you understand the dual. The best examples to consider are the $ell_p$ spaces, i.e. the spaces of sequences $(a_n)_{n=1}^infty$ satisfying
$$|(a_n)_{n=1}^infty|_p:=left(sum_{n=1}^infty|a_n|^pright)^{1/p}<infty.$$
These are Banach spaces provided $1leq p<infty$. Define $p'$ as the conjugate to $p$, i.e. the number $1<p'leqinfty$ satisfying $frac{1}{p}+frac{1}{p'}=1$. The dual space to $ell_p$ is then $ell_{p'}$. (We haven't yet defined $ell_infty$ but you can look that up on wikipedia if you are curious.) Each functional $x^*inell_p^*$ can be identified with a sequence $(b_n)_{n=1}^inftyinell_{p'}$, under the action
$$x^*[(a_n)_{n=1}^infty]=sum_{n=1}^infty a_nb_n.$$
It turns out that if $p=2$ then $p'=2$ as well so that $ell_2^*=ell_2$. In other words, the dual of $ell_2$ is itself!
But $ell_2$ is a special case. It's the only separable Banach space which is its own dual. In fact, even in the nonseparable case, the only spaces which are their own duals are Hilbert spaces.
However, if you take the bidual, then $ell_p^{**}=ell_p$ for all $1<p<infty$. The class of Banach spaces for which this is true via a natural "canonical" identification are called reflexive spaces, although quasireflexive spaces are isomorphic to their biduals via noncanonical identifications. And the class of reflexive Banach spaces is very, very large.
Even when $X$ is a nonreflexive space, it still embeds isometrically into its bidual $X^{**}$. The most well-behaved nonreflexive spaces are $c_0$, $ell_1$, and $ell_infty$. In fact, we have $ell_1=c_0^*$, and $ell_infty=ell_1^*=c_0^{**}$. In particular, $ell_infty$ is the bidual of $c_0$, and $c_0$ is the subspace of $ell_infty$ spanned by the unit vectors.
So, I think the best way to envision biduals, aside from the obvious fact that they are duals of duals, are (possibly larger) spaces containing the original space.
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
answered Dec 16 '18 at 20:38
Ben WBen W
2,283615
2,283615
add a comment |
add a comment |
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