An inner product space and its proper closed subspace with trivial orthogonal complement
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I am looking to do the following:
Construct an inner product space $X$ (with inner product $langle cdot, cdot rangle$ and a proper, closed subspace $Y$ of $X$ such that $Y^perp = {0}$, ie $langle x, y rangle = 0 forall y in Y iff x = 0$.
I see that we need $X$ to be infinite dimensional, hence isomorphic to $Bbb R^n$ with the standard dot product, and then clearly not possible, as a proper (closed) subspace has a smaller dimension.
If someone could give me a hint as to how to start this construction, then I'd be most grateful, as I'm fairly stuck beyond this! As always, please make sure that the hint is reasonably minor, ie doesn't give away too much - I still want to learn from this question, not just be told the answer!
The question is on a course in linear analysis.
functional-analysis inner-product-space orthogonality
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
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add a comment |
$begingroup$
I am looking to do the following:
Construct an inner product space $X$ (with inner product $langle cdot, cdot rangle$ and a proper, closed subspace $Y$ of $X$ such that $Y^perp = {0}$, ie $langle x, y rangle = 0 forall y in Y iff x = 0$.
I see that we need $X$ to be infinite dimensional, hence isomorphic to $Bbb R^n$ with the standard dot product, and then clearly not possible, as a proper (closed) subspace has a smaller dimension.
If someone could give me a hint as to how to start this construction, then I'd be most grateful, as I'm fairly stuck beyond this! As always, please make sure that the hint is reasonably minor, ie doesn't give away too much - I still want to learn from this question, not just be told the answer!
The question is on a course in linear analysis.
functional-analysis inner-product-space orthogonality
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
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– user642796
Nov 15 '14 at 21:06
add a comment |
$begingroup$
I am looking to do the following:
Construct an inner product space $X$ (with inner product $langle cdot, cdot rangle$ and a proper, closed subspace $Y$ of $X$ such that $Y^perp = {0}$, ie $langle x, y rangle = 0 forall y in Y iff x = 0$.
I see that we need $X$ to be infinite dimensional, hence isomorphic to $Bbb R^n$ with the standard dot product, and then clearly not possible, as a proper (closed) subspace has a smaller dimension.
If someone could give me a hint as to how to start this construction, then I'd be most grateful, as I'm fairly stuck beyond this! As always, please make sure that the hint is reasonably minor, ie doesn't give away too much - I still want to learn from this question, not just be told the answer!
The question is on a course in linear analysis.
functional-analysis inner-product-space orthogonality
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
$endgroup$
I am looking to do the following:
Construct an inner product space $X$ (with inner product $langle cdot, cdot rangle$ and a proper, closed subspace $Y$ of $X$ such that $Y^perp = {0}$, ie $langle x, y rangle = 0 forall y in Y iff x = 0$.
I see that we need $X$ to be infinite dimensional, hence isomorphic to $Bbb R^n$ with the standard dot product, and then clearly not possible, as a proper (closed) subspace has a smaller dimension.
If someone could give me a hint as to how to start this construction, then I'd be most grateful, as I'm fairly stuck beyond this! As always, please make sure that the hint is reasonably minor, ie doesn't give away too much - I still want to learn from this question, not just be told the answer!
The question is on a course in linear analysis.
functional-analysis inner-product-space orthogonality
functional-analysis inner-product-space orthogonality
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
edited Nov 7 '15 at 22:40
user147263
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
asked Nov 15 '14 at 16:40
Sam TSam T
3,9851031
3,9851031
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Comments are not for extended discussion; this conversation has been moved to chat.
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– user642796
Nov 15 '14 at 21:06
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– user642796
Nov 15 '14 at 21:06
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– user642796
Nov 15 '14 at 21:06
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– user642796
Nov 15 '14 at 21:06
add a comment |
1 Answer
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Following Daniel Fischer's hints in comments, here is an example:
- Begin with $ell^2$, its one-dimensional subspace $V$ spanned by vector $v = (1,1/2,1/3,dots)$, and its orthogonal complement $V^perp$. Then $(V^perp)^perp = V$.
- Intersect all of the above with the set $F$ of sequences that have only finitely many nonzero elements.
- Outcome: $X = ell^2cap F$ is an inner product space; $Y = V^perpcap F$ is closed in $X$, and $Y^perp = Vcap F = {0}$. Also, $Y$ is a proper subset of $X$ since it does not contain, say, $(1,0,0,dots)$.
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1 Answer
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1 Answer
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$begingroup$
Following Daniel Fischer's hints in comments, here is an example:
- Begin with $ell^2$, its one-dimensional subspace $V$ spanned by vector $v = (1,1/2,1/3,dots)$, and its orthogonal complement $V^perp$. Then $(V^perp)^perp = V$.
- Intersect all of the above with the set $F$ of sequences that have only finitely many nonzero elements.
- Outcome: $X = ell^2cap F$ is an inner product space; $Y = V^perpcap F$ is closed in $X$, and $Y^perp = Vcap F = {0}$. Also, $Y$ is a proper subset of $X$ since it does not contain, say, $(1,0,0,dots)$.
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add a comment |
$begingroup$
Following Daniel Fischer's hints in comments, here is an example:
- Begin with $ell^2$, its one-dimensional subspace $V$ spanned by vector $v = (1,1/2,1/3,dots)$, and its orthogonal complement $V^perp$. Then $(V^perp)^perp = V$.
- Intersect all of the above with the set $F$ of sequences that have only finitely many nonzero elements.
- Outcome: $X = ell^2cap F$ is an inner product space; $Y = V^perpcap F$ is closed in $X$, and $Y^perp = Vcap F = {0}$. Also, $Y$ is a proper subset of $X$ since it does not contain, say, $(1,0,0,dots)$.
$endgroup$
add a comment |
$begingroup$
Following Daniel Fischer's hints in comments, here is an example:
- Begin with $ell^2$, its one-dimensional subspace $V$ spanned by vector $v = (1,1/2,1/3,dots)$, and its orthogonal complement $V^perp$. Then $(V^perp)^perp = V$.
- Intersect all of the above with the set $F$ of sequences that have only finitely many nonzero elements.
- Outcome: $X = ell^2cap F$ is an inner product space; $Y = V^perpcap F$ is closed in $X$, and $Y^perp = Vcap F = {0}$. Also, $Y$ is a proper subset of $X$ since it does not contain, say, $(1,0,0,dots)$.
$endgroup$
Following Daniel Fischer's hints in comments, here is an example:
- Begin with $ell^2$, its one-dimensional subspace $V$ spanned by vector $v = (1,1/2,1/3,dots)$, and its orthogonal complement $V^perp$. Then $(V^perp)^perp = V$.
- Intersect all of the above with the set $F$ of sequences that have only finitely many nonzero elements.
- Outcome: $X = ell^2cap F$ is an inner product space; $Y = V^perpcap F$ is closed in $X$, and $Y^perp = Vcap F = {0}$. Also, $Y$ is a proper subset of $X$ since it does not contain, say, $(1,0,0,dots)$.
answered Nov 7 '15 at 22:39
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user147263
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– user642796
Nov 15 '14 at 21:06