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?






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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?






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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







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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




      lebesgue-integral






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      asked Sep 10 at 19:02









      sami.spricht.sprache

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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:
          $$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$






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            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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          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:
          $$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$






          share|cite|improve this answer


















          • 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















          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:
          $$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$






          share|cite|improve this answer


















          • 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













          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:
          $$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$






          share|cite|improve this answer














          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:
          $$ |f_1 otimes f_2|_mathscrL^p(varOmegatimesvarXi,muotimesnu) = |f_1|_mathscrL^p(varOmega,mu) |f_2|_mathscrL^p(varXi,nu).$$







          share|cite|improve this answer














          share|cite|improve this answer



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          edited Sep 10 at 20:24

























          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













          • 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


















           

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