$v subseteq H implies V^bot$ is a closed subspace of the Hilbert space $H$
$begingroup$
Exercise :
Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.
Attempt :
I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.
For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :
$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$
That's really straightforward and simple.
I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.
functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
$endgroup$
add a comment |
$begingroup$
Exercise :
Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.
Attempt :
I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.
For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :
$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$
That's really straightforward and simple.
I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.
functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35
$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39
$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41
$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25
add a comment |
$begingroup$
Exercise :
Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.
Attempt :
I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.
For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :
$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$
That's really straightforward and simple.
I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.
functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
$endgroup$
Exercise :
Show that if $H$ is a Hilbert space and $V subseteq H$, then $V^bot$ is a closed subspace of $H$.
Attempt :
I thought of two possible approaches. One would be a classic one, getting an arbitrary sequence $v_n in V^bot$ and showing that $lim v_n = v in V^bot$. The other one would be showing that $V^bot = (V^bot)^{botbot}$, which implies that $V^bot$ is a closed subspace of $H$.
For my first approach, let $v_n in V^bot$ and $v_n to h in H$ . Then, for all $v in V$, it is :
$$langle x,v rangle = langle lim v_n, vrangle = limlangle v_n, vrangle =0 implies x in V^bot $$
That's really straightforward and simple.
I am wondering how would one complete my thought for my second possible approach, meaningly showing that $V^bot = (V^bot)^{botbot}$. I seem a little rusty on that.
functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality
functional-analysis hilbert-spaces normed-spaces inner-product-space orthogonality
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
asked Dec 10 '18 at 19:31
RebellosRebellos
14.8k31248
14.8k31248
$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35
$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39
$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41
$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25
add a comment |
$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35
$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39
$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41
$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25
$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35
$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35
$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39
$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39
$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41
$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41
$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25
$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)
Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.
On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.
So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
$endgroup$
add a comment |
$begingroup$
Hint: An arbitrary intersection of closed subspaces is a closed subspace.
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)
Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.
On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.
So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
$endgroup$
add a comment |
$begingroup$
Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)
Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.
On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.
So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
$endgroup$
add a comment |
$begingroup$
Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)
Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.
On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.
So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
$endgroup$
Presumably you already know that (1) $Xsubseteq (X^bot)^bot$ for all sets $X$ and (2) if $Xsubseteq Y$ then $X^botsupseteq Y^bot$. (If not, just check that both facts follow immediately from the definition of ${ }^bot$.)
Apply (1) with $V$ as your $X$ to get $Vsubseteq(V^bot)^bot$ and then use (2) to get $V^botsupseteq((V^bot)^bot)^bot$.
On the other hand, if you apply (1) with $V^bot$ as your $X$, then you get $V^botsubseteq((V^bot)^bot)^bot$.
So you have inclusions in both directions, and therefore $V^bot=((V^bot)^bot)^bot$.
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
answered Dec 10 '18 at 22:06
Andreas BlassAndreas Blass
50k451108
50k451108
add a comment |
add a comment |
$begingroup$
Hint: An arbitrary intersection of closed subspaces is a closed subspace.
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
add a comment |
$begingroup$
Hint: An arbitrary intersection of closed subspaces is a closed subspace.
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
add a comment |
$begingroup$
Hint: An arbitrary intersection of closed subspaces is a closed subspace.
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
$endgroup$
Hint: An arbitrary intersection of closed subspaces is a closed subspace.
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
answered Dec 10 '18 at 19:34
Robert IsraelRobert Israel
324k23214468
324k23214468
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
add a comment |
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
$begingroup$
See my comment to the question. But actually, proving it for a generic $V$ as he does is not more complicated than proving ti for $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:37
add a comment |
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$begingroup$
$V^perp$, by definition, is the intersection of closed subspaces, once you show that it is true for a singleton $V={v}$
$endgroup$
– Federico
Dec 10 '18 at 19:35
$begingroup$
Also I believe that your second approach is a vicious circle, because how do you prove that $A=(A^perp)^perp$ implies that $A$ is closed? You invoke this result that you want to prove now
$endgroup$
– Federico
Dec 10 '18 at 19:39
$begingroup$
@Federico Thanks for your first input, but it's not needed as I've already proven what needs to be proven. My question is strictly regarding my second approach. It is already proven that $ A = A^{botbot}$ implies that $A$ is closed, thus no need for any elaboration.
$endgroup$
– Rebellos
Dec 10 '18 at 19:41
$begingroup$
Honestly, I really doubt that one can prove "$V=(V^perp)^perp implies V text{ is closed}$" without already knowing that $W^perp$ is closed for any $W$
$endgroup$
– Federico
Dec 11 '18 at 13:25