infinite dimensional reflexive subspace of $C[0,1]$
$begingroup$
We know that $C[0,1]$, the space of functions continuous on interval $[0,1]$ equipped with maximum norm is not reflexive. Is there any infinite dimensional reflexive subspace of $C[0,1]$ or every infinite dimensional subspace is neccessarily non-reflexive ?
Thank you for your suggestions.
reflexive-space
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
$endgroup$
add a comment |
$begingroup$
We know that $C[0,1]$, the space of functions continuous on interval $[0,1]$ equipped with maximum norm is not reflexive. Is there any infinite dimensional reflexive subspace of $C[0,1]$ or every infinite dimensional subspace is neccessarily non-reflexive ?
Thank you for your suggestions.
reflexive-space
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
$endgroup$
add a comment |
$begingroup$
We know that $C[0,1]$, the space of functions continuous on interval $[0,1]$ equipped with maximum norm is not reflexive. Is there any infinite dimensional reflexive subspace of $C[0,1]$ or every infinite dimensional subspace is neccessarily non-reflexive ?
Thank you for your suggestions.
reflexive-space
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
$endgroup$
We know that $C[0,1]$, the space of functions continuous on interval $[0,1]$ equipped with maximum norm is not reflexive. Is there any infinite dimensional reflexive subspace of $C[0,1]$ or every infinite dimensional subspace is neccessarily non-reflexive ?
Thank you for your suggestions.
reflexive-space
reflexive-space
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
asked Dec 10 '18 at 8:36
ellipticelliptic
1008
1008
add a comment |
add a comment |
1 Answer
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$C[0,1]$ is a universal space for the collection of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. This result is called the Banach - Mazur theorem. [You can find a proof in 'Geometric Functional Analysis and its Applications' by Holmes]. Hence there are lots of reflexive infinite dimensional subspaces. [ Embed $ell ^{2}$ in $C[0,1]$, for example].
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
add a comment |
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1 Answer
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$begingroup$
$C[0,1]$ is a universal space for the collection of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. This result is called the Banach - Mazur theorem. [You can find a proof in 'Geometric Functional Analysis and its Applications' by Holmes]. Hence there are lots of reflexive infinite dimensional subspaces. [ Embed $ell ^{2}$ in $C[0,1]$, for example].
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
add a comment |
$begingroup$
$C[0,1]$ is a universal space for the collection of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. This result is called the Banach - Mazur theorem. [You can find a proof in 'Geometric Functional Analysis and its Applications' by Holmes]. Hence there are lots of reflexive infinite dimensional subspaces. [ Embed $ell ^{2}$ in $C[0,1]$, for example].
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
add a comment |
$begingroup$
$C[0,1]$ is a universal space for the collection of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. This result is called the Banach - Mazur theorem. [You can find a proof in 'Geometric Functional Analysis and its Applications' by Holmes]. Hence there are lots of reflexive infinite dimensional subspaces. [ Embed $ell ^{2}$ in $C[0,1]$, for example].
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
$C[0,1]$ is a universal space for the collection of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. This result is called the Banach - Mazur theorem. [You can find a proof in 'Geometric Functional Analysis and its Applications' by Holmes]. Hence there are lots of reflexive infinite dimensional subspaces. [ Embed $ell ^{2}$ in $C[0,1]$, for example].
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 10 '18 at 8:53
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
60.5k42161
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
add a comment |
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
$begingroup$
Thanks for the answer and for the book tip.
$endgroup$
– elliptic
Dec 11 '18 at 8:33
add a comment |
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