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Banach space and Hilbert space topology

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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











$endgroup$







  • 1




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






  • 1




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















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











$endgroup$







  • 1




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






  • 1




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













1












1








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











$endgroup$




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 3 hours ago









Henno Brandsma

115k349125




115k349125










asked 3 hours ago









user156213user156213

60338




60338







  • 1




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






  • 1




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












  • 1




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






  • 1




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







1




1




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




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




1




1




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




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










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









$endgroup$












  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    33 mins ago











Your Answer





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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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









$endgroup$












  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    33 mins ago















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









$endgroup$












  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    33 mins ago













5












5








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









$endgroup$



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 $mathbbR^omega$, 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












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









Henno BrandsmaHenno Brandsma

115k349125




115k349125











  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    33 mins ago
















  • $begingroup$
    Do you know of a reference with the proof of this?
    $endgroup$
    – user156213
    33 mins ago















$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
33 mins ago




$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
33 mins ago

















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