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Banach space and Hilbert space topology
The 2019 Stack Overflow Developer Survey Results Are InShowing 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 Apr 7 at 21:37
Henno Brandsma
116k349127
asked Apr 7 at 21:30
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 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
edited Apr 7 at 21:37
Henno Brandsma
116k349127
asked Apr 7 at 21:30
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 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
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
edited Apr 7 at 21:37
Henno Brandsma
116k349127
116k349127
asked Apr 7 at 21:30
user156213user156213
69238
asked Apr 7 at 21:30
user156213user156213
69238
asked Apr 7 at 21:30
user156213user156213
69238
69238
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 |
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
1
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
$begingroup$
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
1
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
Apr 7 at 21:37
$begingroup$
@user124910 We can extend this to non-separable as well. See my answer.
$endgroup$
– Henno Brandsma
Apr 7 at 21:37
add a comment |
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 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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1 Answer
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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 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.
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.
answered Apr 7 at 21:36
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
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 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
answered Apr 7 at 21:36
Henno BrandsmaHenno Brandsma
116k349127
116k349127
$begingroup$
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 |
$begingroup$
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
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
Apr 8 at 0:22
$begingroup$
Do you know of a reference with the proof of this?
$endgroup$
– user156213
Apr 8 at 0:22
2
2
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
Apr 8 at 11:42
$begingroup$
@user156213 This post has a reference.
$endgroup$
– David Mitra
Apr 8 at 11:42
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
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