Is the sum of such two banach spaces also a banach space?
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Let $L^2(mu)$ and $L^2(nu)$ with respect to two different positive measures, then they are two Banach spaces. I'm considering whether the space $$L^2(mu)+L^2(nu)$$ is still a Banach space?
e.g. $mu$ be Lebesgue measure, $dnu=ln(1+|x|)dmu$, my idea is that since both $L^2(mu)$ and $L^2(nu)$ are continuous embedded to the measurable functions space $mathcal M$, it's done.
real-analysis functional-analysis measure-theory
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add a comment |
$begingroup$
Let $L^2(mu)$ and $L^2(nu)$ with respect to two different positive measures, then they are two Banach spaces. I'm considering whether the space $$L^2(mu)+L^2(nu)$$ is still a Banach space?
e.g. $mu$ be Lebesgue measure, $dnu=ln(1+|x|)dmu$, my idea is that since both $L^2(mu)$ and $L^2(nu)$ are continuous embedded to the measurable functions space $mathcal M$, it's done.
real-analysis functional-analysis measure-theory
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3
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What is your definition of $L^2(mu)+L^2(nu)$?
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– DisintegratingByParts
Dec 22 '18 at 20:30
$begingroup$
Are $mu, nu$ defined on the same sigma algebra?
$endgroup$
– Alex Vong
Dec 23 '18 at 20:43
add a comment |
$begingroup$
Let $L^2(mu)$ and $L^2(nu)$ with respect to two different positive measures, then they are two Banach spaces. I'm considering whether the space $$L^2(mu)+L^2(nu)$$ is still a Banach space?
e.g. $mu$ be Lebesgue measure, $dnu=ln(1+|x|)dmu$, my idea is that since both $L^2(mu)$ and $L^2(nu)$ are continuous embedded to the measurable functions space $mathcal M$, it's done.
real-analysis functional-analysis measure-theory
$endgroup$
Let $L^2(mu)$ and $L^2(nu)$ with respect to two different positive measures, then they are two Banach spaces. I'm considering whether the space $$L^2(mu)+L^2(nu)$$ is still a Banach space?
e.g. $mu$ be Lebesgue measure, $dnu=ln(1+|x|)dmu$, my idea is that since both $L^2(mu)$ and $L^2(nu)$ are continuous embedded to the measurable functions space $mathcal M$, it's done.
real-analysis functional-analysis measure-theory
real-analysis functional-analysis measure-theory
asked Dec 22 '18 at 17:57
Yixuan ZhangYixuan Zhang
4910
4910
3
$begingroup$
What is your definition of $L^2(mu)+L^2(nu)$?
$endgroup$
– DisintegratingByParts
Dec 22 '18 at 20:30
$begingroup$
Are $mu, nu$ defined on the same sigma algebra?
$endgroup$
– Alex Vong
Dec 23 '18 at 20:43
add a comment |
3
$begingroup$
What is your definition of $L^2(mu)+L^2(nu)$?
$endgroup$
– DisintegratingByParts
Dec 22 '18 at 20:30
$begingroup$
Are $mu, nu$ defined on the same sigma algebra?
$endgroup$
– Alex Vong
Dec 23 '18 at 20:43
3
3
$begingroup$
What is your definition of $L^2(mu)+L^2(nu)$?
$endgroup$
– DisintegratingByParts
Dec 22 '18 at 20:30
$begingroup$
What is your definition of $L^2(mu)+L^2(nu)$?
$endgroup$
– DisintegratingByParts
Dec 22 '18 at 20:30
$begingroup$
Are $mu, nu$ defined on the same sigma algebra?
$endgroup$
– Alex Vong
Dec 23 '18 at 20:43
$begingroup$
Are $mu, nu$ defined on the same sigma algebra?
$endgroup$
– Alex Vong
Dec 23 '18 at 20:43
add a comment |
1 Answer
1
active
oldest
votes
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If you have two Banach spaces $X,Y$ both continuously embedded in a Hausdorff topological vector space $Z$, you can endow the sum $X+Y$ with the norm $|z|=inf{|x|_X+|y|_Y: z=x+y}$. Then $X+Y$ is a quotient of the Banach space $Xtimes Y$ with respect to the subspace $Xcap Y$ and hence itself a Banach space.
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$begingroup$
Thank you! But for my case, what's the space $Z$?
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– Yixuan Zhang
Jan 3 at 8:31
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You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
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– Jochen
Jan 3 at 13:33
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
If you have two Banach spaces $X,Y$ both continuously embedded in a Hausdorff topological vector space $Z$, you can endow the sum $X+Y$ with the norm $|z|=inf{|x|_X+|y|_Y: z=x+y}$. Then $X+Y$ is a quotient of the Banach space $Xtimes Y$ with respect to the subspace $Xcap Y$ and hence itself a Banach space.
$endgroup$
$begingroup$
Thank you! But for my case, what's the space $Z$?
$endgroup$
– Yixuan Zhang
Jan 3 at 8:31
$begingroup$
You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
$endgroup$
– Jochen
Jan 3 at 13:33
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
add a comment |
$begingroup$
If you have two Banach spaces $X,Y$ both continuously embedded in a Hausdorff topological vector space $Z$, you can endow the sum $X+Y$ with the norm $|z|=inf{|x|_X+|y|_Y: z=x+y}$. Then $X+Y$ is a quotient of the Banach space $Xtimes Y$ with respect to the subspace $Xcap Y$ and hence itself a Banach space.
$endgroup$
$begingroup$
Thank you! But for my case, what's the space $Z$?
$endgroup$
– Yixuan Zhang
Jan 3 at 8:31
$begingroup$
You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
$endgroup$
– Jochen
Jan 3 at 13:33
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
add a comment |
$begingroup$
If you have two Banach spaces $X,Y$ both continuously embedded in a Hausdorff topological vector space $Z$, you can endow the sum $X+Y$ with the norm $|z|=inf{|x|_X+|y|_Y: z=x+y}$. Then $X+Y$ is a quotient of the Banach space $Xtimes Y$ with respect to the subspace $Xcap Y$ and hence itself a Banach space.
$endgroup$
If you have two Banach spaces $X,Y$ both continuously embedded in a Hausdorff topological vector space $Z$, you can endow the sum $X+Y$ with the norm $|z|=inf{|x|_X+|y|_Y: z=x+y}$. Then $X+Y$ is a quotient of the Banach space $Xtimes Y$ with respect to the subspace $Xcap Y$ and hence itself a Banach space.
answered Dec 23 '18 at 11:37
JochenJochen
7,0431023
7,0431023
$begingroup$
Thank you! But for my case, what's the space $Z$?
$endgroup$
– Yixuan Zhang
Jan 3 at 8:31
$begingroup$
You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
$endgroup$
– Jochen
Jan 3 at 13:33
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
add a comment |
$begingroup$
Thank you! But for my case, what's the space $Z$?
$endgroup$
– Yixuan Zhang
Jan 3 at 8:31
$begingroup$
You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
$endgroup$
– Jochen
Jan 3 at 13:33
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
$begingroup$
Thank you! But for my case, what's the space $Z$?
$endgroup$
– Yixuan Zhang
Jan 3 at 8:31
$begingroup$
Thank you! But for my case, what's the space $Z$?
$endgroup$
– Yixuan Zhang
Jan 3 at 8:31
$begingroup$
You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
$endgroup$
– Jochen
Jan 3 at 13:33
$begingroup$
You are right, there is a problem. If your measures are equivalent (i.e, have the same null sets) you can take $Z$ as the space of equivalence classes w.r.t. $mu$-a.e. equality endowed with convergence in measure. However, if the measures aren't equivalent I don't see a candidate for $Z$ (you need the Hausdorff vector space topology to deduce from the continuity of $(f,g)mapsto f-g$ that the kernel $Xcap Y$ is closed).
$endgroup$
– Jochen
Jan 3 at 13:33
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
$begingroup$
well, thank you very much!
$endgroup$
– Yixuan Zhang
Jan 5 at 13:18
add a comment |
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3
$begingroup$
What is your definition of $L^2(mu)+L^2(nu)$?
$endgroup$
– DisintegratingByParts
Dec 22 '18 at 20:30
$begingroup$
Are $mu, nu$ defined on the same sigma algebra?
$endgroup$
– Alex Vong
Dec 23 '18 at 20:43