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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

edited Apr 7 at 21:37

Henno Brandsma

116k349127

asked Apr 7 at 21:30

user156213user156213

69238

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited Apr 7 at 21:37

Henno Brandsma

116k349127

asked Apr 7 at 21:30

user156213user156213

69238

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Apr 7 at 21:37

Henno Brandsma

116k349127

asked Apr 7 at 21:30

user156213user156213

69238

$endgroup$

share|cite|improve this question

edited Apr 7 at 21:37

Henno Brandsma

116k349127

asked Apr 7 at 21:30

user156213user156213

69238

share|cite|improve this question

edited Apr 7 at 21:37

Henno Brandsma

116k349127

asked Apr 7 at 21:30

user156213user156213

69238

edited Apr 7 at 21:37

Henno Brandsma

116k349127

edited Apr 7 at 21:37

Henno Brandsma

116k349127

asked Apr 7 at 21:30

user156213user156213

69238

asked Apr 7 at 21:30

user156213user156213

69238

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 $mathbbR^omega$, 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 $mathbbR^n$ up to homeomorphism, which are already Hilbert spaces.

share|cite|improve this answer

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 8 at 11:42
  

  
  
  
  

add a comment | 

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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 $mathbbR^omega$, 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 $mathbbR^n$ up to homeomorphism, which are already Hilbert spaces.

share|cite|improve this answer

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 8 at 11:42
  

  
  
  
  

add a comment | 

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 $mathbbR^omega$, 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 $mathbbR^n$ up to homeomorphism, which are already Hilbert spaces.

share|cite|improve this answer

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Apr 8 at 11:42
  

  
  
  
  

add a comment | 

$begingroup$

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 $mathbbR^omega$, 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 $mathbbR^n$ up to homeomorphism, which are already Hilbert spaces.

share|cite|improve this answer

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

share|cite|improve this answer

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127

answered Apr 7 at 21:36

Henno BrandsmaHenno Brandsma

116k349127