Banach space and Hilbert space topologyShowing that two Banach spaces are homeomorphic when their dimensions are equal.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
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Banach space and Hilbert space topology
Showing that two Banach spaces are homeomorphic when their dimensions are equal.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
$begingroup$
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
edited yesterday
Henno Brandsma
115k349125
asked yesterday
user156213user156213
69238
$endgroup$
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
edited yesterday
Henno Brandsma
115k349125
asked yesterday
user156213user156213
69238
$endgroup$
1
$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
yesterday
1
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
yesterday
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
edited yesterday
Henno Brandsma
115k349125
asked yesterday
user156213user156213
69238
$endgroup$
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
edited yesterday
Henno Brandsma
115k349125
asked yesterday
user156213user156213
69238
edited yesterday
Henno Brandsma
115k349125
asked yesterday
user156213user156213
69238
edited yesterday
Henno Brandsma
115k349125
edited yesterday
Henno Brandsma
115k349125
edited yesterday
Henno Brandsma
115k349125
115k349125
asked yesterday
user156213user156213
69238
asked yesterday
user156213user156213
69238
asked yesterday
user156213user156213
69238
69238
1
$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
yesterday
1
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
yesterday
add a comment |
1
$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
yesterday
1
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
yesterday
1
1
$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
yesterday
$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
yesterday
1
1
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
yesterday
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
add a comment |
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$begingroup$
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.
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
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.
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
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.
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
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.
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
answered yesterday
Henno BrandsmaHenno Brandsma
115k349125
115k349125
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
add a comment |
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
yesterday
2
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
15 hours ago
add a comment |
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$begingroup$
If $B$ is separable, then yes. All separable Banach Spaces are homeomorphic. So homeomorphic to $ell^2$
$endgroup$
– user124910
yesterday
1
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
yesterday