If the sections of a function are $mathscrL^2$, would the function be jointly $mathscrL^2$?
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Let $(varOmega,mathcalA,mu)$ and $(varXi,mathcalB,nu)$ be $sigma$-finite measure spaces, and $fcolonvarOmegatimesvarXitomathbbR$ be $big(mathcalAotimesmathcalBbig)$-$mathcalB(mathbbR)$-measurable. Suppose also that the sections of $f$ are $mathscrL^2$; i.e.,
$$f(cdot,xi)inmathscrL^2(varOmega,mu),~textfor $nu$-a.e.~xi,~textand~f(omega,cdot)inmathscrL^2(varXi,nu),~textfor $mu$-a.e.~omega.$$
Then, is it true that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$? If not, what restrictions do we need?
lebesgue-integral
asked Sep 10 at 19:02
sami.spricht.sprache
13610
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Let $(varOmega,mathcalA,mu)$ and $(varXi,mathcalB,nu)$ be $sigma$-finite measure spaces, and $fcolonvarOmegatimesvarXitomathbbR$ be $big(mathcalAotimesmathcalBbig)$-$mathcalB(mathbbR)$-measurable. Suppose also that the sections of $f$ are $mathscrL^2$; i.e.,
$$f(cdot,xi)inmathscrL^2(varOmega,mu),~textfor $nu$-a.e.~xi,~textand~f(omega,cdot)inmathscrL^2(varXi,nu),~textfor $mu$-a.e.~omega.$$
Then, is it true that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$? If not, what restrictions do we need?
lebesgue-integral
asked Sep 10 at 19:02
sami.spricht.sprache
13610
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(varOmega,mathcalA,mu)$ and $(varXi,mathcalB,nu)$ be $sigma$-finite measure spaces, and $fcolonvarOmegatimesvarXitomathbbR$ be $big(mathcalAotimesmathcalBbig)$-$mathcalB(mathbbR)$-measurable. Suppose also that the sections of $f$ are $mathscrL^2$; i.e.,
$$f(cdot,xi)inmathscrL^2(varOmega,mu),~textfor $nu$-a.e.~xi,~textand~f(omega,cdot)inmathscrL^2(varXi,nu),~textfor $mu$-a.e.~omega.$$
Then, is it true that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$? If not, what restrictions do we need?
lebesgue-integral
asked Sep 10 at 19:02
sami.spricht.sprache
13610
Let $(varOmega,mathcalA,mu)$ and $(varXi,mathcalB,nu)$ be $sigma$-finite measure spaces, and $fcolonvarOmegatimesvarXitomathbbR$ be $big(mathcalAotimesmathcalBbig)$-$mathcalB(mathbbR)$-measurable. Suppose also that the sections of $f$ are $mathscrL^2$; i.e.,
$$f(cdot,xi)inmathscrL^2(varOmega,mu),~textfor $nu$-a.e.~xi,~textand~f(omega,cdot)inmathscrL^2(varXi,nu),~textfor $mu$-a.e.~omega.$$
Then, is it true that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$? If not, what restrictions do we need?
lebesgue-integral
lebesgue-integral
asked Sep 10 at 19:02
sami.spricht.sprache
13610
asked Sep 10 at 19:02
sami.spricht.sprache
13610
asked Sep 10 at 19:02
sami.spricht.sprache
13610
asked Sep 10 at 19:02
sami.spricht.sprache
13610
asked Sep 10 at 19:02
sami.spricht.sprache
13610
13610
add a comment |Â
add a comment |Â
1 Answer
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The answer is no. You cannot conclude that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$, see this essentially the same question.
One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(omega, xi) = f_1(omega)f_2(xi)$. Since the $mathscrL^p$ play nice with tensor products:
$$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$
answered Sep 10 at 20:17
Joseph Adams
1167
1
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The answer is no. You cannot conclude that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$, see this essentially the same question.
One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(omega, xi) = f_1(omega)f_2(xi)$. Since the $mathscrL^p$ play nice with tensor products:
$$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$
answered Sep 10 at 20:17
Joseph Adams
1167
1
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
add a comment |Â
up vote
1
down vote
The answer is no. You cannot conclude that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$, see this essentially the same question.
One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(omega, xi) = f_1(omega)f_2(xi)$. Since the $mathscrL^p$ play nice with tensor products:
$$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$
answered Sep 10 at 20:17
Joseph Adams
1167
1
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The answer is no. You cannot conclude that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$, see this essentially the same question.
One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(omega, xi) = f_1(omega)f_2(xi)$. Since the $mathscrL^p$ play nice with tensor products:
$$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$
answered Sep 10 at 20:17
Joseph Adams
1167
The answer is no. You cannot conclude that $finmathscrL^2(varOmegatimesvarXi,muotimesnu)$, see this essentially the same question.
One sufficient condition for the claim to hold though is that $f$ be a tensor product: $f(omega, xi) = f_1(omega)f_2(xi)$. Since the $mathscrL^p$ play nice with tensor products:
$$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$
answered Sep 10 at 20:17
Joseph Adams
1167
edited Sep 10 at 20:24
answered Sep 10 at 20:17
Joseph Adams
1167
answered Sep 10 at 20:17
Joseph Adams
1167
answered Sep 10 at 20:17
Joseph Adams
1167
1167
1
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
add a comment |Â
1
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
1
1
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
A Tonelli-like answer to the question ‘what restrictions do we need?’ is that after, say, determining that $f(cdot,xi)inmathcalL^2(Omega,mu)$ for $nu$-almost every $xi$, let $g(xi)$ be $|f(cdot,xi)|_mathcalL^2(Omega,mu)$ (which has just been seen to be $nu$-almost everywhere finite) and demand that $ginmathcalL^2(Xi,nu)$. That $f(omega,cdot)inmathcalL^2(Xi,nu)$ for $mu$-almost every $omega$ then follows by Tonelli's Theorem, in addition to $finmathcalL^2(OmegatimesXi,muotimesnu)$ as desired.
– Toby Bartels
Sep 10 at 20:49
add a comment |Â
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