Banach space and Hilbert space topology












2












$begingroup$


Let $B$ be a Banach space. It is not necessarily true that
there exists a Hilbert space $H$ linearly isometric to $B$.



However, is it true that there exists a Hilbert space $H$
homeomorphic to $B$?






general-topology functional-analysis hilbert-spaces banach-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
    $endgroup$
    – user124910
    Apr 7 at 21:35






  • 1




    $begingroup$
    @user124910 We can extend this to non-separable as well. See my answer.
    $endgroup$
    – Henno Brandsma
    Apr 7 at 21:37
















2












$begingroup$


Let $B$ be a Banach space. It is not necessarily true that
there exists a Hilbert space $H$ linearly isometric to $B$.



However, is it true that there exists a Hilbert space $H$
homeomorphic to $B$?






general-topology functional-analysis hilbert-spaces banach-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
    $endgroup$
    – user124910
    Apr 7 at 21:35






  • 1




    $begingroup$
    @user124910 We can extend this to non-separable as well. See my answer.
    $endgroup$
    – Henno Brandsma
    Apr 7 at 21:37














2












2








2





$begingroup$


Let $B$ be a Banach space. It is not necessarily true that
there exists a Hilbert space $H$ linearly isometric to $B$.



However, is it true that there exists a Hilbert space $H$
homeomorphic to $B$?






general-topology functional-analysis hilbert-spaces banach-spaces







share|cite|improve this question











$endgroup$




Let $B$ be a Banach space. It is not necessarily true that
there exists a Hilbert space $H$ linearly isometric to $B$.



However, is it true that there exists a Hilbert space $H$
homeomorphic to $B$?






general-topology functional-analysis hilbert-spaces banach-spaces




general-topology functional-analysis hilbert-spaces banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 7 at 21:37









Henno Brandsma

116k349127




116k349127










asked Apr 7 at 21:30









user156213user156213

69238




69238








  • 1




    $begingroup$
    If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
    $endgroup$
    – user124910
    Apr 7 at 21:35






  • 1




    $begingroup$
    @user124910 We can extend this to non-separable as well. See my answer.
    $endgroup$
    – Henno Brandsma
    Apr 7 at 21:37














  • 1




    $begingroup$
    If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
    $endgroup$
    – user124910
    Apr 7 at 21:35






  • 1




    $begingroup$
    @user124910 We can extend this to non-separable as well. See my answer.
    $endgroup$
    – Henno Brandsma
    Apr 7 at 21:37








1




1




$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
Apr 7 at 21:35




$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
Apr 7 at 21:35




1




1




$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
Apr 7 at 21:37




$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
Apr 7 at 21:37










1 Answer
1






active

oldest

votes


















7












$begingroup$

Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $X$ and $Y$ are homeomorphic iff $d(X)=d(Y)$, where the density $d(X)$ is the minimal size of a dense subset of $X$.



So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $ell^2$ (and even to $mathbb{R}^omega$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $ell_2(kappa)$ as models. Finite dimensional we only have the $mathbb{R}^n$ up to homeomorphism, which are already Hilbert spaces.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    Apr 8 at 0:22






  • 2




    $begingroup$
    @user156213 This post has a reference.
    $endgroup$
    – David Mitra
    2 days ago












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









7












$begingroup$

Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $X$ and $Y$ are homeomorphic iff $d(X)=d(Y)$, where the density $d(X)$ is the minimal size of a dense subset of $X$.



So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $ell^2$ (and even to $mathbb{R}^omega$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $ell_2(kappa)$ as models. Finite dimensional we only have the $mathbb{R}^n$ up to homeomorphism, which are already Hilbert spaces.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    Apr 8 at 0:22






  • 2




    $begingroup$
    @user156213 This post has a reference.
    $endgroup$
    – David Mitra
    2 days ago
















7












$begingroup$

Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $X$ and $Y$ are homeomorphic iff $d(X)=d(Y)$, where the density $d(X)$ is the minimal size of a dense subset of $X$.



So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $ell^2$ (and even to $mathbb{R}^omega$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $ell_2(kappa)$ as models. Finite dimensional we only have the $mathbb{R}^n$ up to homeomorphism, which are already Hilbert spaces.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    Apr 8 at 0:22






  • 2




    $begingroup$
    @user156213 This post has a reference.
    $endgroup$
    – David Mitra
    2 days ago














7












7








7





$begingroup$

Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $X$ and $Y$ are homeomorphic iff $d(X)=d(Y)$, where the density $d(X)$ is the minimal size of a dense subset of $X$.



So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $ell^2$ (and even to $mathbb{R}^omega$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $ell_2(kappa)$ as models. Finite dimensional we only have the $mathbb{R}^n$ up to homeomorphism, which are already Hilbert spaces.






share|cite|improve this answer









$endgroup$



Yes, but this is quite a deep result. Two infinite-dimensional Banach spaces $X$ and $Y$ are homeomorphic iff $d(X)=d(Y)$, where the density $d(X)$ is the minimal size of a dense subset of $X$.



So any separable infinite-dimensional Banach space is homeomorphic to the Hilbert space $ell^2$ (and even to $mathbb{R}^omega$, because the result extends to locally convex completely metrisable TVS's as well). And for higher densities we have Hilbert spaces $ell_2(kappa)$ as models. Finite dimensional we only have the $mathbb{R}^n$ up to homeomorphism, which are already Hilbert spaces.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 7 at 21:36









Henno BrandsmaHenno Brandsma

116k349127




116k349127












  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    Apr 8 at 0:22






  • 2




    $begingroup$
    @user156213 This post has a reference.
    $endgroup$
    – David Mitra
    2 days ago


















  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    Apr 8 at 0:22






  • 2




    $begingroup$
    @user156213 This post has a reference.
    $endgroup$
    – David Mitra
    2 days ago
















$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
Apr 8 at 0:22




$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
Apr 8 at 0:22




2




2




$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
2 days ago




$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
2 days ago


















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