Is the sum of such two banach spaces also a banach space?












3












$begingroup$


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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$endgroup$








  • 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












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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








3


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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