Etydhv

Japan - Plan around max visa duration

whey we use polarized capacitor?

Why are only specific transaction types accepted into the mempool?

What are these boxed doors outside store fronts in New York?

Continuity at a point in terms of closure

N.B. ligature in Latex

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

How to get the available space of $HOME as a variable in shell scripting?

What would the Romans have called "sorcery"?

Is there really no realistic way for a skeleton monster to move around without magic?

Is it possible to do 50 km distance without any previous training?

Compute hash value according to multiplication method

I probably found a bug with the sudo apt install function

Why Is Death Allowed In the Matrix?

Why is an old chain unsafe?

Question about Goedel's incompleteness Proof

Modification to Chariots for Heavy Cavalry Analogue for 4-armed race

New order #4: World

How can I hide my bitcoin transactions to protect anonymity from others?

Do Phineas and Ferb ever actually get busted in real time?

Set-theoretical foundations of Mathematics with only bounded quantifiers

XeLaTeX and pdfLaTeX ignore hyphenation

Example of a relative pronoun

Draw simple lines in Inkscape

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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3178808%2fbanach-space-and-hilbert-space-topology%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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