Expectation of $XY$ bounded for all bounded $Y$ implies $X$ is $L^p$












1












$begingroup$


I'm trying to prove:




Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces







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$endgroup$

















    1












    $begingroup$


    I'm trying to prove:




    Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




    My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






    real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I'm trying to prove:




      Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




      My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






      real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces







      share|cite|improve this question









      $endgroup$




      I'm trying to prove:




      Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.




      My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?






      real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces




      real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 19:54









      D FordD Ford

      675313




      675313






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
          $$
          |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
          $$           By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
          $$
          E[|X||Y|]le C|Y|_{L^q}
          $$           for all bounded $Y$. Then we can use monotone convergence theorem to conclude
          $$
          E[|X||Y|]le C|Y|_{L^q}
          $$           for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
            $endgroup$
            – D Ford
            Jan 11 at 21:30






          • 1




            $begingroup$
            In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
            $endgroup$
            – Song
            Jan 11 at 21:37



















          2












          $begingroup$

          Hint: for each fixed $n$, apply the assumption to the random variable
          $$
          Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
          $$
          where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
          This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



          Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






          share|cite|improve this answer











          $endgroup$














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            2 Answers
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            2 Answers
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            active

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            active

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            active

            oldest

            votes









            1












            $begingroup$

            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37
















            1












            $begingroup$

            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37














            1












            1








            1





            $begingroup$

            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.






            share|cite|improve this answer











            $endgroup$



            Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
            $$
            |X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
            $$             By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all bounded $Y$. Then we can use monotone convergence theorem to conclude
            $$
            E[|X||Y|]le C|Y|_{L^q}
            $$             for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 10 at 20:30

























            answered Jan 10 at 20:22









            SongSong

            18.6k21651




            18.6k21651












            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37


















            • $begingroup$
              Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
              $endgroup$
              – D Ford
              Jan 11 at 21:30






            • 1




              $begingroup$
              In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
              $endgroup$
              – Song
              Jan 11 at 21:37
















            $begingroup$
            Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
            $endgroup$
            – D Ford
            Jan 11 at 21:30




            $begingroup$
            Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
            $endgroup$
            – D Ford
            Jan 11 at 21:30




            1




            1




            $begingroup$
            In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
            $endgroup$
            – Song
            Jan 11 at 21:37




            $begingroup$
            In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
            $endgroup$
            – Song
            Jan 11 at 21:37











            2












            $begingroup$

            Hint: for each fixed $n$, apply the assumption to the random variable
            $$
            Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
            $$
            where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
            This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



            Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hint: for each fixed $n$, apply the assumption to the random variable
              $$
              Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
              $$
              where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
              This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



              Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hint: for each fixed $n$, apply the assumption to the random variable
                $$
                Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
                $$
                where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
                This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



                Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.






                share|cite|improve this answer











                $endgroup$



                Hint: for each fixed $n$, apply the assumption to the random variable
                $$
                Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
                $$
                where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
                This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.



                Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 12 at 10:34

























                answered Jan 10 at 20:25









                Davide GiraudoDavide Giraudo

                128k17156268




                128k17156268






























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