Arsthbt

Pronouncing Dictionary.com's W.O.D "vade mecum" in English

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Shell script can be run only with sh command

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Banach space and Hilbert space topology

Is any Banach space a dual space?A Banach space that is not a Hilbert spaceIs every Hilbert space a Banach algebra?Which Hilbert space is isometrically isomorphism with $B(E)$ for some Banach space $E$.Is every Banach space densely embedded in a Hilbert space?Existence of a $mathbb C$-Banach space isometric to a Hilbert Space but whose norm is not induced by an inner product?An example of a Banach space isomorphic but not isometric to a dual Banach spaceThe Hahn-Banach Theorem for Hilbert SpaceBanach spaces and Hilbert spaceBasis of infinite dimensional Banach space and separable hilbert space

1

$begingroup$

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 1 hour ago

Henno Brandsma

115k349125

asked 1 hour ago

user156213user156213

60338

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
  $endgroup$
  – user124910
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user124910 We can extend this to non-separable as well. See my answer.
  $endgroup$
  – Henno Brandsma
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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 1 hour ago

Henno Brandsma

115k349125

asked 1 hour ago

user156213user156213

60338

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
  $endgroup$
  – user124910
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user124910 We can extend this to non-separable as well. See my answer.
  $endgroup$
  – Henno Brandsma
  1 hour ago
  

  
  
  
  

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 1 hour ago

Henno Brandsma

115k349125

asked 1 hour ago

user156213user156213

60338

$endgroup$

share|cite|improve this question

edited 1 hour ago

Henno Brandsma

115k349125

asked 1 hour ago

user156213user156213

60338

share|cite|improve this question

edited 1 hour ago

Henno Brandsma

115k349125

asked 1 hour ago

user156213user156213

60338

edited 1 hour ago

Henno Brandsma

115k349125

edited 1 hour ago

Henno Brandsma

115k349125

asked 1 hour ago

user156213user156213

60338

asked 1 hour ago

user156213user156213

60338

1 Answer
1

active

oldest

votes

5

$begingroup$

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 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

add a comment | 

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5

$begingroup$

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 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

add a comment | 

5

$begingroup$

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 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

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 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125

answered 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125

answered 1 hour ago

Henno BrandsmaHenno Brandsma

115k349125